ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES
Chapter Outline
9.1 General introduction
9.2 Tetravalence of Carbon
9.3 Structural Representation of Organic Compounds
9.4 Classification of Organic Compounds
9.5 Nomenclature of Organic Compounds
9.6 Isomerism
9.7 Fundamental Concepts in Organic Reaction Mechanism
9.8 Methods of Purification of Organic Compounds
9.9 Qualitative Analysis of organic compounds
9.10 Quantitative Analysis of organic compounds
9.1 GENERAL INTRODUCTION
The element carbon has the unique property called catenation, due to which it forms covalent bonds with other carbon atoms. It also forms covalent bonds with atoms of other elements, like hydrogen, oxygen, nitrogen, sulphur, phosphorus, and halogens. The resulting compounds are studied under a separate branch of chemistry, called organic chemistry. However, according to modern definition, organic chemistry is the branch of chemistry dealing with the study of hydrocarbons and the compounds that
co uld be thought of as the derivatives of hydrocarbons.
In the early days, it was considered impossible to prepare organic compounds in the laboratory. Berzelius (1815) believed that organic compounds are from living organisms and these could only be produced by some mysterious force, called ‘vital force’, and they cannot be prepared artificially. According to him, inorganic compounds are from nonliving sources, like minerals.
However, this vital force theory was rejected in 1828 when Wohler prepared an organic compound, urea, by heating ammonium cyanate, an inorganic compound. Then, the synthesis of acetic acid by Kolbe (1845) and that of methane by Berthelot (1856) showed conclusively that organic compounds could be synthesised from inorganic sources in a laboratory.
NH4CNO heat NH2CONH2
Ammonium cyanate Urea
The reasons for a large number of carbon compounds include its high catenation capacity, i.e., the ability to form chains with its own atoms. It has the ability to form single and multiple bonds with its own atoms or other atoms. Carbon has the ability to form four single bonds, a double bond and two single bonds, a triple bond and a single bond, or two double bonds with its own atoms or other atoms, as per the requirements. All these bonds are strong. Carbon can also form
9: Organic Chemistry – Some Basic Principles and Techniques
ring compounds and fused ring compounds. Isomerism is one of the reasons of innumerable number of carbon compounds. The main natural sources of organic compounds are coal, petroleum, natural gas, animals, and plants.
9.2 TETRAVALENCE OF CARBON
Valency of carbon in almost all the compounds is four. This tetravalency of carbon is one of the reasons behind the existence of a large number of carbon compounds.
9.2.1 Shapes of Organic Compounds
The knowledge of fundamental concepts of molecular structure helps us in understanding and predicting the properties of organic compounds. Already, it was studied in chemical bonding that the tetravalence of carbon and the formation of covalent bonds by it are explained in terms of its electronic configuration and the hybridisation of s and p orbitals. The formation and shapes of molecules like methane (CH4), ethene (C2H4) and ethyne (C2H2) are explained in terms of the use of sp3, sp2, and sp hybrid orbitals by carbon atoms in the respective molecules ( Table 9.1).
Table 9.1 Hybridisaton of Carbon atoms
Hybridisation influences the bond length and bond enthalpy (strength) in organic compounds.
The sp hybrid orbital contains more s character and, hence, it is closer to its nucleus and forms shorter and stronger bonds than the sp3 hybrid orbital. The sp2 hybrid orbital is intermediate in s character between sp and sp 3. Hence, the length and enthalpy of the bonds it forms are also intermediate between them. The change in hybridisation affects the electronegativity of carbon. The
greater the s-character of the hybrid orbitals, the greater is the electronegativity. Thus, a carbon atom having an sp hybrid orbital with 50% s-character is more electronegative than that possessing sp2 or sp3 hybridised orbitals. This relative electronegativity is reflected in several physical and chemical properties of the molecules concerned.
9.2.2 Characteristic Features of π Bonds
In a π (pi) bond formation, parallel orientation of the two p orbitals on adjacent carbon atoms is necessary for a proper sideways overlapping. Thus, in CH2 = CH2 molecule, all the atoms must be in the same plane. The p orbitals are mutually parallel and both the p orbitals are perpendicular to the plane of the molecule. Rotation of one CH 2 fragment with respect to other interferes with maximum overlap of p orbitals and, therefore, such rotations about carbon-carbon double bond (C = C) is restricted. The electron charge cloud of the π bond is located above and below the plane of bonding atoms. This results in the electrons being easily available to the attacking reagents. In general, π bonds provide the most reactive centres in the molecules containing multiple bonds.
9.3 STRUCTURAL REPRESENTATION OF ORGANIC COMPOUNDS
In every molecule of carbon compounds, carbon is surrounded by four bonds, some of which may be pi-bonds. The following are representations of a few molecules.
The above structural formulas can be further abbreviated by omitting some or all of the dashes representing covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. This type is called a condensed structural formula.
Examples:
CH3CH3 H2C=CH2 HC ≡ CH
Ethane Ethylene Acetylene
CH3OH CH3CHBrCH3 (or) (CH3)2CHBr
Methyl alcohol Isopropyl bromide
9.3.1 Writing Systematic Lewis Structures
■ Find the molecular formula and the connectivity. Methyl nitrite has the formula CH3–O–N=O.
All hydrogens are connected to carbon atom and the order of atomic connections is H3CONO.
■ For a neutral molecule, valence electrons are equal to the sum of the valence electrons of the constituent atoms. Total number of valence electrons in CH 3 NO 2 is 24. (Each hydrogen has one, carbon has four, nitrogen has five, and each oxygen has six).
■ Connect bonded atoms by a shared electron pair bond (:), represented by a dash (–).
order of decreasing electronegativity. In CH 3–O–N=O, already 4 bonds are there around carbon atom with 8 electrons. Both oxygen atoms have 8 electrons, but nitrogen has only 6 electrons.
■ C o u nt the number of electrons in shared pairs and subtract this from the total number of electrons to get the number of electrons to be added to complete the structure. In the above structure, 6 bonds are equivalent to 12 electrons and the electrons to be further added are 12.
■ Add electrons in pairs to atoms to get octet around them. When the number of electrons is insufficient to provide an octet for all atoms, assign electrons to atoms in
■ For atoms that have fewer electrons than 8, use unshared pairs on adjacent atoms to form a double bond or triple bond to complete the octet. An electron pair on the terminal oxygen is shared with nitrogen to give double bond.
In this structure, hydrogen atoms have duet and all other atoms have octet, and are hence, most stable. There are two resonating structures for methyl nitrite.
In structure I, there are no formal charges and in structure II, there are formal charges. The structure II with formal charges is less stable than structure I with no formal charges. Formal charge,
NA = Number of electrons in valence shell of atom (Group number in periodic table)
NL.P. = Number of electrons in lone pairs
NB.P. = Number of electrons in bond pairs
C H H H O N O
C H H H O N O
In The Structure:
Formal charge on oxygen atom, a =
62 1 2 ()61
Formal charge on oxygen atom, b =
66 1 2 ()21
Formal charge on nitrogen atom
52 6 2 0
9.3.2 Bond-line Structural Formulas
In bond-line structural representation of organic compounds, carbon and hydrogen atoms are not shown and the lines representing carbon–carbon bonds are drawn in a zigzag fashion. The only atoms specifically written are oxygen, chlorine, nitrogen, etc. The terminals denote methyl (–CH3) groups, while the line junctions denote carbon atoms bonded to appropriate number of hydrogens required to satisfy the valency of the carbon ato ms.
Example:
Butane, CH3CH2CH2CH3 is written as 3-methyloctane
2–ethyl–4–methyl–1– pentanol, CH2 CH CH2 CH CH3 OH C2H5 CH3 is represented as
CH3 H3C is represented as
CH2 CH2 C
The structural formula of C C H H
is
Three-dimensional Representation of Organic Molecules: By using solid ( ) and dashed ( C (Bond away from observer)
Bond in the plane of paper (Bond towards observer) dashed wedge
CH3CH2CHCH2CH2CH2CH2CH3 CH3 is represented as Terminals represent methyl groups.
Cyclopropane, CH2 CH2 CH2 , is represented as
Cyclohexanone, CH2 C CH2 CH2 CH2 CH2 O , is represented as O
Bond in the plane of paper
Solid wedge ) wedge formula, the 3D image of an organic molecule from a twodimensional picture can be perceived. The solid wedge is used to indicate a bond projecting out of the plane of the paper, towards the observer. The dashed wedge is used to depict the bond projecting out of the plane of the paper and away from the observer. Wedges are shown in such a way that the broad end of the wedge is towards the observer. The bonds lying in plane of the paper are depicted by using a normal line (–). 3D representation of methane molecule is as shown: C H H H H
dashed wedge (Bond away from observer)
Solid wedge (Bond towards observer)
(i) 3-methyloctane can be represented in various forms:
Terminals represent methyl group
Various ways of representing 2-bromobutane are as follows:
■ CH3CHBrCH2CH3
■ H3C CH CH2 CH3 Br
■ Br
9.4 CLASSIFICATION OF ORGANIC COMPOUNDS
The organic compounds have been classified on the basis of carbon skeleton (structure),
or functional groups, or the concept of homology. A general classification is present in Fig.9.1
9.4.1 Classification Based on Carbon Chain
On the basis of chain of carbon atoms, compounds are named as acyclic compounds and cyclic compounds.
Ac yclic Compounds
Non-cyclic compounds are called aliphatic compounds and consist of single or branched chain compounds. They do not contain rings and are open chain compounds. They may be saturated, in which carbon atoms form only single bonds or they may be unsaturated, in which one or more carbon atoms form multiple bonds with other carbon atoms.
Example:
CH3CH3 CH3CH2CH3
Ethane Propane
CH2=CH2 CH3CH2OH
Ethylene Ethyl alcohol
In the above, only ethylene is unsaturated compound.
Organic compounds
Acyclic or open chain compounds
Saturated hydrocarbons (Alkanes)
Unsaturated hydrocarbons
Alkenes Alkynes
Fig. 9.1 Cl assification of organic compounds
Cyclic or closed chain or ring compounds
Homocyclic (Carbocyclic compounds)
Heterocyclic compounds
Alicyclic compounds Aromatic compounds Non-aromatic compounds Aromatic compounds
Benzenoid compounds Non-benzenoid compounds
Cyclic Compounds
These are the compounds in which carbon atoms are linked to each other or to the atoms of other elements in the form of a ring. These are closed chain compounds. The compounds with only one ring are monocyclic and those with more than one ring are polycyclic. Compounds having a ring or rings of only carbon atoms in the molecule are called homocyclic or carbocyclic compounds. They may again be of two types. Alicyclic and Aromatic compounds. Alicyclic compounds are cyclic but aliphatic in nature. The following are examples.
The above compounds are also known as benzenoid aromatics since they contain at least one benzene ring. However, there are aromatic compounds that have structural units different from benzenoid type and are known as non-benzenoid aromatics.
Examples: O
Cyclopentanol Cyclobutane
Cyclohexane Cyclohexene Cyclohexadiene
Aromatic compounds are the compounds with planar ring structures, with conjugate double bonds and possess (4n + 2) π electrons, which, according to Huckel’s rule, are responsible for aromaticity, where n = 0, 1, 2, 3,....... If the ring contains only carbon atoms, then they are carbocyclic or homocyclic aromatic compounds. If they are benzene related, they are called benzenoids. If the ring does not relate to benzene, they are called non-benzenoids (e.g., tropone, azulene). Aromatic compounds may also contain heterocyclic rings.
Examples:
cation
Azulene
Compou nds having ring or rings built up of more than one kind of atoms are called heterocyclic compounds. The most common hetero atoms are oxygen, nitrogen, and sulphur. Examples:
Cyclopropane
Benzene Napthalene
Benzoic acid Biphenyl
Tropone Cyclopentadienyl anion
Cycloheptatrienyl
The above heterocyclic compounds are aromatic but all heterocyclic compounds need not be aromatic.
Examples:
Pyrrolidine Tetrahydrofuran
The above heterocyclic compounds are not aromatic.
9.4.2 Classification Based on Functional Groups
The functional group may be defined as an atom or a group of atoms joined in a specific manner in its molecule that is responsible for the characteristic chemical properties of the compounds. Important functional groups are listed in Table 9.2
Table 9.2 Important functional groups in organic compounds
A group or a series of organic compounds, each containing a characteristic functional group, forms a homologous series and the members of the series are called homologues. The general characteristics of a homologous series are listed below:
i) They exhibit similar chemical properties because of the presence of same functional group.
ii) There is a regular gradation in their physical properties.
iii) They can be represented by a general formula.
iv) The successive members differ in molecular formula by a –CH2– unit and so, in molecular mass by 14.
(v) Their general methods of preparation are almost similar.
TEST YOURSELF
1. Ratio between σ C-C bond to σ C-H bond in molecule HC ≡ C–CH = CHCH3 is
(1) 4 : 3 (2) 2 : 3
(3) 3 : 2 (4) 3 : 4
2. The alicyclic compound is (1) cyclohexane
(2) hexene-2
(3) pyrrole
(4) hexane
3. Which of the following is an example of alicyclic compound?
(1) Hexane
(2) Benzene
(3) Furan
(4) Cyclohexane
4. Which of the following is an aromatic heterocyclic compound?
(1) Pyrrole (2) Pyrrolidine (3) Epoxyethane (4) Dioxane
5. Species in which all C–C bonds are equal is (1) (2) (3) (4)
Answer Key
(1) 2 (2) 1 (3) 4 (4) 1 (5) 2
9.5 NOMENCLATURE OF ORGANIC COMPOUNDS
I n the beg inning, the names of carbon compounds were given depending upon their source or certain characteristic properties. With the rapid growth of organic chemistry, it is necessary to have a systematic nomenclature based on structure.
9.5.1 Trivial System or Common System of Nomenclature
Before the IUPAC system of nomenclature, however, organic compounds assigned names based on their origin or certain properties. For instance, citric acid is named so because it is found in citrus fruits and the acid found in red ant is named formic acid, since the Latin word for ant is ‘formica’. Methane was called fire damp as it easily catches fire. Some common names are followed even today. For example, Buckminister fullerene is a common name given to the newly discovered C60 cluster (a form of carbon), noting its structural similarity to the geodesic domes popularised by the famous architect R. Buckminister Fuller.
9.5.2 The IUPAC System
A systematic name of an organic compound is generally derived by identifying the parent hydrocarbon and the functional group(s) attached to it.
Accordingly, the International Congress of Chemistry (ICC), held in Geneva in 1892, evolved a rational system of nomenclature, which was called Geneva system. A committee appointed by the International Union of Chemists (IUC) made amendments to the above system and this amended system was called the IUC system. The IUC system goes through a number of revisions, whenever needed, by a committee called the International Union of Pure and Applied Chemistry (IUPAC).
According to IUPAC nomenclature, a given compound can be assigned only one name, and a given name can have only one molecular structure. The name of the compound should be only one word, as far as possible. Each systematic name has two or three of the following parts: root word, primary suffix, secondary suffix, and prefix(es). Root word indicates the number of carbon atoms present in the longest continuous chain, as shown in Table 9.3. Primary suffixes are added to the root word to show saturation or unsaturation in a carbon chain, as shown in Table 9.4.
Table 9.3 Root words
Table 9.4 Primary suffixes
Chain
Saturated –ane Alkane
Unsaturated with one C=C double bond –ene Alkene
Unsaturated with one C ≡ C triple bond –yne Alkyne
Secondary suffix is added after the primary suffix to indicate the presence of a particular functional group in the carbon chain. Some secondary suffixes are given in Table 9.5.
Table 9.5 Some secondary suffixes
Hydrocarbon
Alcohol –OH –ol
Aldehyde –CHO –al
Ketone C = O –one
Carboxylic acid –COOH –oic acid
Amine –NH2 –amine
Sulphonic acid –SO3H –sulphonic acid
Cyanide –CN – nitrile
Isocyanide –NC – carbylamine
Ester –COOR –oate
Acyl halide –COX –oyl halide
Acid amide –CONH2 –amide
Anhydride –CO.O.OC– -oic anhydride
Prefix: Alkyl group forming branches of the parent chain are considered as side-chains. Atoms or group of atoms, such as halo (–X), nitro (–NO2), nitroso (–NO), and alkoxy (–OR), are referred to as substituents. Root words are prefixed with the name of the substituent or side chain. These are arranged as follows,
while writing the names: Prefix(es) + root word + primary suffix + secondary suffix.
Prefixes
Primary prefix: Primary prefix is meant only for alicyclic compounds and it is always ‘cyclo’. If the compound is not alicyclic, the primary prefix is absent.
Cyclobutane Cyclopentane
Secondary prefix: Secondary prefix tells us about the substituents or secondary grade functional groups. In polyfunctional compounds, after selecting the principal functional group, all other functional groups are taken as substituents. While writing them as substituents, there is a change in their designation, as in Table 9.6.
Table 9.6 Prefixes of some substituents Substituent
– X halo –NH2 amino > CO oxo, keto –NO2 nitro–
–R alkyl –NO nitroso
RO– alkoxy –CHO formyl
–CN cyano –OH hydroxy
–COOR alkoxycarbonyl –SO3H sulpho
–COX halocarbonyl –CONH2 carbamoyl
Types of Carbon and Hydrogen Atoms
There are four types of carbon atoms and three types of hydrogen atoms.
■ Primary (1°) carbon: A carbon atom attached to one or no other carbon atoms
■ Secondary (2°) carbon: A carbon atom attached to two other c arbon atoms
■ Tertiary (3°) carbon : A carbon atom attached to three other carbon atoms
■ Quaternary (4°) carbon: A carbon atom attached to four other carbon atoms
Hydrogens attached to primary, secondary, and tertiary carbon atoms are termed primary, se condary, and tertiary hydrogen atoms, respectively.
(30)
(40)
Total number of 1° hydrogen atoms – 15
Total number of 2° hydrogen atoms – 2
Total number of 3° hydrogen atoms – 1
Hydrocarbons
A branched chain hydrocarbon is named using the following general rules.
Longest Continuous Chain Rule
Select the longest continuous chain of carbon atoms as the parent chain. If some carboncarbon multiple bond is present, the parent chain must contain the carbon atoms involved in it. The number of carbon atoms in the parent chain determines the root word. The carbon atoms that are not involved in the parent chain are considered alkyl substituents and they determine the prefixes. A list of common alkyl, alkenyl, or alkynyl groups and their IUPAC names are listed in Table 9.7
Example:
In the above structure, parent chain is pentane and substituent is methyl group.
Example:
In the above structure, the parent chain is heptane and the substituents are ethyl and methyl groups.
If two equally long chains are possible, the chain with maximum number of side chains is selected as parent chain.
two substituents (case 1).
has one substituent (case 2).
The chain of (case 1) is chosen as the parent chain because it has two substituents.
The correct IUPAC name is 3-ethyl–2–methyl pentane.
three substituents (case 1).
has one substituent (case 2).
The chain of (case 1) is chosen as the parent chain because it has three substituents.
The correct IUPAC name is 3–ethyl–2, 2–dimethyl pentane.
Lowest Set of Locants
When two or more substituents are present, then end of the parent chain, which gives the lowest set of locants, is preferred for numbering. This rule is called lowest set of locants.
Table 9.7 Common and IUPAC names of alkyl groups
Methane CH4 CH3– Methyl Methyl–
Ethane CH3–CH3 CH3–CH2– Ethyl Ethyl–
Propane CH3–CH2–CH3 CH3–CH2–CH2–CH3–CH–CH3 | n- Propyl Isopropyl Propyl–Methylethyl–
Butane CH3–CH2–CH2–CH3 CH3–CH2–CH2–CH2–CH3–CH2–CH–CH3 | n-Butyl sec-Butyl Butyl–1-Methylpropyl–
Isobutane
This means that when two or more different sets of locants are possible, that set of locants which, when compared, term by term, with the other sets, each in order of increasing magnitude, as the lowest term is the first point of difference. This rule is used irrespective of the nature of the substituent.
locant 3, 4).
Set of locants 2, 3, 5 (correct)
Set of locants 2, 4, 5 (wrong)
The correct set of locants is 2, 3, 5 and not 2, 4, 5. The first set is lower than the second because, at the first, difference 3 is less than 4 (note that first locant is same in both sets and the first difference is with the second
Set of locants 2, 7, 8 (correct)
Set of locants 3, 4, 9 (wrong)
In the first preference 2 is less than 3
Example: Br Br
2, 3 - dibromo cyclohexene - 1 (wrong) 1, 6 - dibromo cyclohex - 1 - ene (correct)
Lowest locant rule is followed after giving priority to double bond (Table 9.8).
Compounds Containing Functional Groups
The longest chain of carbon atoms containing the functional group is numbered in such a way that the functional group containing carbon should get lowest number. In the case of polyfunctional compounds, one of the functional groups is chosen as the principal functional group and the compound is then named on that basis. The order of decreasing priority for some functional groups is listed in Table 9.9. The remaining functional groups are named as substituents. Less preferred functional groups are written in the beginning of the name, using the appropriate prefixes.
Names of chloro substituted hydrocarbons are listed in Table 9.10. Names of some important compounds containing nitrogen and oxygen atoms as functional groups and substituents are given in Table 9.11
The chain terminating groups, like –CHO, –CN, and –COOH, should always get the number 1 for their carbon atoms, irrespective of the above rules.
–CH2–CH2–CH2 –C – C –CH3 OH NO2 CH3 CH3
It we follow lowest sum rule, its name should be 3–hydroxy–2,3–dimethyl–2–nitro–heptan–7–oic acid. But its correct IUPAC name is 5–hydroxy–5,6– dimethyl–6–nitroheptanoic acid. Though this is against lowest sum rule, the chain terminating group, –COOH, should be given number 1.
In case the structure contains both C=C and C ≡ C, the C=C is given the lowest number, if both are at same position from either end.
CH ≡ C–CH=CH2
But –3–en–1–yne (wrong)
But –1–en–3–yne (correct)
It may be noted that double bond is preferred without violating least sum rule.
With More Than One Similar Functional Groups
Example:
4 3 2 1
CH3–CH–CH–CH2–OH
OH C2H5
2–Ethyl butane–1,3–diol
Example:
4 3 2 1 5
HOO C–CH 2–CH 2 –CH 2 –COO H
Pentane–1,5–dioic acid
Example:
4 3 2 1 5
OH C– CH 2 –C H = C H –C H O
Pent–2–ene–1,5–dial
If an unbranched chain is linked directly to more than two carboxyl groups, these groups are named as carboxylic acids indicating their locants.
COOH COOH
HOOC–CH2–CH–CH2–CH–CH2–COOH 4 3 2 1 5
Pentane–1,2,4,5– tetracarboxylic acid
The carboxyl groups that are not directly linked to the principal chain are expressed by ‘carboxyalkyl’ prefixes.
6
COOH COOH 5
4 3 2 1
CH3 –CH–CH2 –CH–CH–CH2–COOH C H2COOH
3–(Carboxymethyl) hexane–1,2,5–tricarboxylic acid
HOOC
Table
9.8 Names of some important acyclic hydrocarbons
Structural formula
name
1. CH3–CH2–CH3 Propane Propane
2. CH3–CH2–CH2–CH3 n-Butane Butane
name
3. CH3–CH(CH3)2 Iso-butane 2-Methylpropane
4. CH3(CH2)3CH3 n-Pentane Pentane
5. (CH3)2CH–CH2–CH3 Iso-pentane 2-Methylbutane
6. C(CH3)4 Neo-pentane Dimethylpropane
7. CH3(CH2)4CH3 n-Hexane Hexane
8. (CH3)2CH(CH2)2CH3 Iso-hexane 2-Methylpentane
9. (CH3)2CH–CH(CH3)2 Iso-hexane 2,3-Dimethylbutane
10. CH3–CH2–C(CH3)3 Neo-hexane 2,2-Dimethylbutane
11. CH3(CH2)6CH3 n-Octane Octane
12. (CH3)2CHCH2C(CH3)3 Neooctane 2,2,4-Trimethylpentane
13. (CH3)2CH–CH(C2H5)2 Isopropylpentane 3-Ethyl-2-methylpentane
14. (CH3CH2CH2)2CH–CH(CH3)2 Isopropylheptane 4-(Methylethyl)heptane
15. H2C=CH2 Ethylene Ethene
16. CH3–CH=CH2 Propylene Propene
17. CH2=C=CH2 Allene Propadiene
18. CH3–CH2–CH=CH2 Butylene 1-Butene
19. CH3–CH=CH–CH3 sym-Butylene 2-Butene
20. HC ≡ CH Acetylene Ethyne
21. CH3–C ≡ CH Methylacetylene Propyne
22. CH3–CH2–C ≡ CH Ethylacetylene 1-Butyne
23. CH3–C ≡ C–CH3 Dimethylacetylene 2-Butyne
24. CH3–CH2–C ≡ C–CH3 Ethylmethylacetylene 2-Pentyne
25. H2C=CH–C ≡ CH Vinylacetylene But-1-ene-3-yne
26. CH3–CH=CH–C ≡ CH Allylacetylene Pent-3-ene-1-yne
27. CH(C2H5)3 Triethylmethane 3-Ethylpentane
28. (CH3–CH2)2C=CH2 Ethylbutene
29. (CH3)2C=CH2 Isobutene or Isobutylene Methylpropene
30. CH CH CH CH CH CH CH 22 2 | 3-Ethenyl-1,4-pentadiene
Table 9.9 Order of decreasing priority for some functional groups
of compound
1. Carboxylic acid –COOH – carboxy – oic acid
2. Sulphonic acid – SO3H – sulpho – sulphonic acid
3. Ester – COOR alkoxycarbonyl – – oate
4. Acyl chloride –COCl chlorocarbonyl– – oyl chloride
5. Amide –CONH2 carbamoyl– – amide
6. Nitrile –CN cyano– – nitrile
7. Aldehyde –CHO formyl (or) oxo– – al
8. Ketone O C oxo– (keto) – one
9. Alcohol –OH hydroxy– – ol
10. Amine –NH2 amino– – amine
11. Alkene –CH = CH– alkene– – ene
12. Alkyne –C ≡ C– alkyne– –yne
13. Ether –O– alkoxy–
Table 9.10 Names of some important chlorohydrocarbons*
1. CH3-CH2Cl Ethyl chloride Chloroethane
2. ClCH2-CH2Cl Ethylene chloride 1,2-Dichloroethane
3. CH3-CHCl2 Ethylidene chloride 1,1-Dichloroethane
4. CF2Cl2 Freon Dichlorodifluoromethane
5. CHCl3 Chloroform Trichloromethane
6. CH3-CH2-CH2-Cl n-Propyl chloride 1-Chloropropane
7. CH3-CHCl-CH3 Iso-propyl chloride 2-Chloropropane
8. CH3-CH2-CHCl-CH3 sec-Butyl chloride 2-Chlorobutane
9. (CH3)2CH-CH2Cl Iso-butyl chloride 1-Chloro-2-methylpropane
10. Cl2CH-CHCl2 Westron 1,1,2,2-Tetrachloroethane
11. Cl3C-CCl2-CCl3 Perchloropropane Octachloropropane
12. H2C = CHCl Vinyl chloride Chloroethene
13. Cl2C = CHCl Westrol Trichloroethene
* If two halogen atoms are attached on the same carbon, then it is called geminal dihalide and if they are attached on the adjacent carbon atoms, it is called vicinal dihalide.
Table 9.11
Names of some important acyclic heteroatomic compounds
Structural formula Trivial name
1. CH3CH2NO2
name
Nitroethane Nitroethane
2. CH3CH2ONO Ethylnitrite Ethylnitrite
3. CCl3NO2 Chloropicrin Trichloronitromethane
4. CH3–CH2–NH2
5. H2N–CH2–CH2–NH2
Ethylamine Ethanamine
Ethylene diamine Ethane-1,2-diamine
6. CH3–CH2–CH2–NH2 Propylamine 1-Propanamine
7. CH3–CH(NH2)–CH3 Isopropylamine 2-Propanamine
8. CH3–CH2–NH–CH3 Ethylmethylamine N-Methylethanamine
9. (CH3)3N Trimethyl amine N,N-Dimethylmethanamine
10. CH3–CH(NO2)–CH2–CH2–NH2 Nitrobutylamine 3-Nitro-1- butanamine
11. CH3–CH2–OH Ethyl alcohol Ethanol
12. HO–CH2–CH2–OH Ethylene glycol 1,2-Ethanediol
13. HO–CH2–CHOH–CH2OH Glycerine 1,2,3-Propanetriol
14. HO–CH2–CH2–Cl Ethylene chlorohydrin 2-Chloroethanol
15. (CH3)2C(OH)–CCl3 Chloretone 1,1,1-Trichloro-2-methyl-2propanol
16. C2H5–O–C2H5 Diethyl ether Ethoxyethane
17. CH3–CHO Acetaldehyde Ethanal
18. CCl3–CHO Chloral Trichloroethanal
19. OHC–CHO Glyoxal Ethanedial
20. CH3–CO–CH3 Acetone Propanone
21. CH3COOH Acetic acid Ethanoic acid
22. HOOC–COOH Oxalic acid Ethanedioic acid
23. CH3CONH2 Acetamide Ethanamide
24. CH3COCl Acetyl chloride Ethanoyl chloride
25. CH3COOC2H5 Ethylacetate Ethylethanoate
26. CH3–CO–O–CO–CH3 Acetic anhydride Ethanoylethanoate or Ethanoic anhydride
27. CH3CH2–CN Ethylcyanide Propanenitrile
28. H2C = CH–CN Acrylonitrile Propenenitrile
29. CH3–NC Methylisocyanide Methylcarbylamine
30. H2N–CO–NH2 Urea Aminomethanamide
31. CH3–CHOH–COOH Lactic acid 2-Hydroxypropanoic acid
32. OHC–CHOH–CH2OH Glyceraldehyde 2,3-Dihydroxypropanal
33. CH3–CH2–SO3H Ethylsulphurous acid Ethanesulphonic acid
34. HOOC–CH2–NH2 Glycine 2-Aminoethanoic acid
35. HO–CH(NH2)–COOH Serine 2-Amino-2-hydroxyethanoic acid
Some Typical Examples
5 4 3 2 1
CH3 –CH –CO–CH–OCH 2CH3 O CH3 CH3
2–Ethoxy–4–methoxypentan–3–one CH3
C6H5 –O–C–CH–CH3 O 1 2 3
Phenyl–2–methylpropanoate
CH 3–CH–COOH CH2 CHO 2 1 3
3–Formyl–2–methylpropanoic acid
CH 3–CO–COOCH 2CH 3 3 2 1
Ethyl–(2–oxo) propanoate
CH3–CH–C OH C–CH–C HO CH3 1 2 3 4 6 5
2–Hydroxy–5–methylhex–3–ynal
8 7 6 5 4 3 2 1
CH3–CH–CH=CH–C–CH =CH –CH3 O CH3
7–Methylocta–2,5–dien–4–one
C H 3 –CH– C H 2 –CH– C H O C H O C H 3 5 4 3 2 1
2, 4–dimethyl pentane–1, 5–dial ≡ 1CH 2C 3C 4CH2 5CH=6CH2
CH3 CH3
3, 3 – Dimethylhex –5– ene –1–yne
CH ≡ C – CHO propynal
In alphabetical order, the prefixes Isoand Neo- are considered to be the part of the fundamental name of alkyl group. The prefixes sec- and tert- are not considered to be the part of the fundamental name.
If there happenes to be two chains of equal size, the chain to be selected should contain more number of side chains. After selection of the chain, numbering is to be done from the end closer to the substituent.
1CH3 2CH2 3CH2
CH(CH3)2 4CH 5CH 6CH2 7CH2 8CH2 9CH2 10CH3
CH3CHCH2CH3
4–(Methylethyl)–5–(1–methylpropyl) decane.
1CH3 2CH2 3CH2 4CH2
5CH 6CH2 7CH2 8CH2 9CH3
CH2
CH3 C CH3
CH3
5 – (2, 2–Dimethylpropyl) nonane 10CH3 9CH2 8CH2 7CH2 6CH2
CH3 3C CH3 CH2 2CH2 1CH3 CH2 CH3
5CH CH2 CH CH2 CH3
5–(2–Ethylbutyl)–3, 3–dimethyl decane
If more than one functional group of the same type are present, their number is indicated by adding di-, tri-, tetra-, etc., before the class suffix. Then, the full name of the parent alkane is written before the class suffix.
2 H N CH CH CHO
COOH COOH
2-amino3 -formylbutane-, 4-dioic acid
If more than one functional groups are present then consider highest priority as main functional group and remaining are considered as substituents.
Compounds with bond line formula
Example:
2,2-Dimethylbutane
The parent chain has four carbon atoms and two methyl groups are on second carbon atom.
Example:
2,3-Dimethylbutane
The parent chain has four carbon atoms and one methyl group is each on second and third carbon atoms.
Example: 1 2 3 4 5 6
Hexa-1,3,5-triene
The compound has six carbon atoms in the chain with three double bonds at first, third, and fifth carbon atoms.
Structural Representation of Organic Molecules
After observing the word root, write the continuous carbon chain. Number the carbon atoms in a suitable way and attach the functional groups, substituents, and multiple bonds at their respective carbon atoms.
Attach the required number of hydrogen atoms at each carbon atom to satisfy its tetravalency.
Example:
3–chloro–2–methyl hexan–2–ol
As per the word root, hex, we have C–C –C–C–C–C 1 2 3 4 5 6
As per the name,
C–C – C–C–C–C CH3 OH Cl
To satisfy the tetravalency of each carbon atom, hydrogen atoms are added as per requirement.
C – C – C – C – C – C CH3 H Cl OH H H H H H H H H H H
The complete structure of the compound is given as CH3 CH3 Cl OH C CH–CH2–CH2–CH3
Nomenclature of Some Cyclic Compounds
Cyclic compounds: A saturated monocyclic compound is named by prefixing ‘cyclo’ to the corresponding straight chain alkane. If side chains are present, then the rules given above are applied. Names of some cyclic compounds are given below.
Benzene Derivatives
Common names are written for benzene derivatives in brackets.
Alphabetical order of numbering
propylcyclohexane
branched carbon gets lower number
3-Ethyl-1, 1-dimethylcyclohexane (not 1-ethyl-3, 3-dimethylcyclohexane)
(toluene)
zene (anisole)
Substituent of the base compound is assigned number 1 and then the direction of numbering is chosen such that the next substituent gets the lowest number. The substituents appear in the name in alphabetical order.
2-Chloro-1-methyl-4nitrobenzene
1-Chloro-2,4dinitrobenzene
When a benzene ring is attached to an alkane with a functional group, it is considered as substituent, instead of a parent.
When no simple base name other than benzene is possible, the positions are numbered so as to give the lowest locant at the first point of difference. Thus metanitrochlorobenzene is called 1–chloro–3–nitroben zene.
Cl NO2 NO2 1 2 3 4 5 6
1–Chloro–2,4–dinitrobenzene CH2CH3 F 1 2 3 4 5 6 NO2
C2H5 C CH CH3 OH
3,3–Di phenyl pentan –2–ol
4–Ethyl–1–fluoro–2–nitrobenzene
COOC2H5 Br Ethyl–4–bromobenzene carboxylate
Phenylmethanol (benzyl alcohol)
1 - Phenylethanone (IUPAC) CHO
Benzenecarbaldehyde (Benzaldehyde)
Benzenecarboxylic acid (Benzoic acid) OH
Phenol CH=CH2
Phenylethene (Styrene) CH3 CH3
1,2–Dimethylbenzene (o-Xylene)
CH3 NO2 NO2 O2N 1 2 3 4 5 6
2- methyl–1, 3, 5 trinitrobenzene
CH3–CH2–CH–CH2OH
2–Phenylbutan–1–ol
The groups obtained from methyl benzene. H → 3 CH 2 CH
Benzyl H → H →
Benzal Benzo CH C
TEST YOURSELF
1. Incorrect combination of organic structure and its IUPAC name is
(1) HC ≡ C−CH2−CH = CH2 -Pent-1-en4-yne
(2) OHCCHCHCHCN 32 ….3-formyl butane nitrile
(3) cycloheptanoic acid
(4) CH OH CH CCHCHCH CH 3 32 3 3 || | …….4,4-dimethyl pentan-2-ol
2. The correct order of preference according to IUPAC nomenclature is
(1) −CHO > −COOH > −OH > −NH 2
(2) −COOH > −CHO > −NH2 > −OH
(3) −COOH > −OH > −NH2 > −CHO
(4)−COOH > −CHO > −OH > −NH 2
3. IUPAC name of H2NCHO is (1) amino formaldehyde
(2)methanamide
(3) aminomethanal
(4) formamide
4. Which of the following compounds has wrong IUPAC name?
(1) CH 3 CH 2 −CH 2 COO−CH 2 CH 3 Ethyl butanoate
(2) CH3CH(CH3)CH2CHO 3-methyl butanal
(3) CH3CH(OH)CH(CH3)2 2 methyl 3- butanol
(4) CH CH CCHCH CH O 32 3 3 ||| 2 methyl- 3- pentanone
5. Write the correct IUPAC name of the following.
(1)1–chloro – 2, 4–dinitrobenzene
(2) 6–chloro – 1, 3–dinitrobenzene
(3) 1–chloro – 4, 6–dinitrobenzene
(4) 2–chloro – 1, 5–dinitrobenzene
6. Which of the following compounds has wrong IUPAC name?
(1) CH2=CH-CH=CH2 buta– 1, 3 – diene
(2) -3, 4 dimethyl phenol
(3) -2-chloro-4-methyl anisole
(4) II 3222 CH-C-CH-CH-CH-COOH
5 – Carboxypentan-2-one
7. IUPAC name of OHC–Cl is (1) chloro methanal (2) methanoyl chloride (3) chloro methanone
(4) chloro formaldehyde
8. The maximum number of linear atoms in acetylene is (1) 3 (2) 4
(3) 2 (4) 6
9. Carbamoyl is the IUPAC prefix used for the group
(1) −COCl (2) −COOR
(3) −COOCO– (4) −CONH2
10. IUPAC name of allyl chloride is (1) 1 – chloro – 1 – propene (2) 1 – chloro – 2 – propene (3) 3 – chloro – 2 – propene (4) 3 – chloro – 1 – propene
11. IUPAC name of (CH3)2CH – CHBr2is (1) 1,1-dibromo–2–methyl propane (2) 2-methyl–3–bromo propane (3) iso propyl bromid (4) 3° butyl bromide
12. IUPAC name of HC ≡ C−CH = CH2 is (1) But –1–yn–3–ene (2) But –1–en–3– yne (3) But–1–yn –2–ene (4) None of the above
Answer Key
(1) 3 (2) 4 (3) 2 (4) 3
(5) 1 (6) 4 (7) 2 (8) 2
(9) 4 (10) 4 (11) 1 (12) 2
9.6 ISOMERISM
Th e term ‘is omer’ was first introduced by Berzelius. Organic compounds having same molecular formula but differing from each other in some physical or chemical properties or both are known as isomers (Greek: iso–same; meros– part) and the phenomenon is known as isomerism.
Isomerism is exhibited by compounds having directional bonds. There are two main types of isomerism: structural isomerism and stereo isomerism. They are further classified as shown in Fig.9.2.
9.6.1 Structural Isomerism
Structural isomerism is due to the difference in the way in which the constituent atoms or groups are linked to one another within the molecule, without any reference to space. Structural isomerism is of five types:
■ Chain isomerism
■ Position isomerism
■ Functional group isomerism
■ Metamerism
■ Tautomerism
Chain Isomerism
Chain isomerism arises due to difference in the arrangement of carbon atoms constituting the chain. It is also known as nuclear or skeletal isomerism. Chain isomerism is found
Isomerism
in compounds containing more than three carbon atoms.
Example:
Butane has two chain isomers. CH3–CH2–CH2–CH3,
n-Butane Isobutane (butane) (methylpropane)
In n-butane, 4 carbons are in continuous chain and in isobutane, only 3 carbons are in continuous chain.
Example:
Pentane has three chain isomers.
CH3CH2CH2CH2CH3 n-pentane (pentane)
Note: 2-methyl pentane and 3-methyl pentane are chain isomers.
The number of theoretical possible isomers of a given formula increases rapidly with increase in the number of carbon atoms, as shown in Table 9.12.
Structural isomerism (Constitutional isomerism)
1) Chain 2) Position 3) Functional 4) Metamerism 5) Tautomerism Stereo isomerism Configuration isomerism Geometrical isomerism
Fig.9.2 Isomerism
isomerism Conformational isomerism
Table 9.12 Number of chain isomers of alkanes
Formula Number of chain isomers
C2H6
C3H8
C4H10 2
C5H12 3
C6H14 5
C7H16 9
C8H18 18
C9H20 35
C10H22 75
Besides alkanes, this type of isomerism is possible in other homologous series also.
■ C4H8 has two chain isomers.
CH3–CH2–CH=CH2 CH CH CCH 3 32 |
1-Butene Methylpropene
■ C4H10O has two chain isomers.
CH3CH2CH2CH2OH CH CH CH OH CH 32 3 |
1–Butanol 2-Methyl-1-propanol
Position Isomerism
I somers which differ in the position of a functional group or multiple bond or substituent in the same carbon chain are called position isomers.
■ C3H7OH has two position isomers.
CH CH CH OH 32 2 | CH CH CH OH 32 |
n-propyl alcohol Isopropyl alcohol (1-propanol) (2-propanol)
■ Butene has two position isomers.
CH3CH2CH=CH2 CH3CH=CHCH3 1-Butene 2-Butene
■ Chlorobutane has two position isomers.
CH CH CH CH Cl 3 222 | CH CH CHCH Cl 32 3 |
n–Butyl chloride sec. Butylchloride (1-chlorobutane) (2-chlorobutane)
Functional Group Isomerism
Compounds having same molecular formula but different functional groups are called functional group isomers. Some examples are given below.
■ Alcohols and ethers
C2H6O has two functional isomers.
CH3CH2OH CH3–O–CH3
Ethyl alcohol Dimethylether (ethanol) (methoxymethane)
■ Aldehydes, ketones, and unsaturated alcohols
C 3 H 6 O has the following functional isomers.
CH3CH2CHO CH3COCH3
Propionaldehyde Acetone (propanal) (propanone)
CH2=CHCH2OH
Allyl alcohol (prop-2-en-1-ol)
■ Acids, esters, and hydroxy carbonyl compounds
C 3 H 6 O 2 has the following functional isomers.
CH3CH2COOH CH3COOCH3
Propionic acid Methyl acetate (propanoic acid) (methylethanoate)
CH CH CHO OH 3 | CH CCHOH O 32 ||
2-Hydroxypropanal Hydroxypropanone
■ Cyanides and isocyanides
C2H3N has two functional group isomers.
CH3CN CH3NC
Methyl cyanide Methylisocyanide
■ Nitro alkanes and alkyl nitrites
C2H5NO2 has two funcional group isomers.
C2H5 N O O C2H5–O–N=O
Nitroethane Ethyl nitrite
■ Amines (primary, secondary, and tertiary).
C3H9N has the following funcional group isomers.
CH3CH2CH2NH2
Propanamine (1°)
CH3CH2 H CH3 N
N–Methylethanamine (2°)
CH3 CH3 CH3 N
N, N–Dimethylmethanamine (3°)
■ Alkynes and alkadienes
C4H6 has the following functional isomers.
CH3–CH2–C ≡ CH 1–Butyne
CH2=CH–CH=CH2 1,3–Butadiene
■ O - Cresol CH3 OH and Benzyl alcohol
2 CH OH are functional isomers.
Metamerism
Metamerism is due to the presence of different alkyl groups attached to the same bivalent functional group.
Ethers, secondary amines, ketones, esters, etc. can exhibit metamerism.
■ C2H5OC2H5 C3H7OCH3
Diethylether Methyl propyl ether
■ C3H7NHC3H7 C2H5NHC4H9
Dipropyl amine Ethyl butyl amine
■ C2H5COC2H5 C3H7COCH3
Diethyl ketone Methyl propyl ketone
3 – Pentanone
2 – Pentanone
■ 2 – Pentanone and 3 - Pentanone are preferentially taken as metamers.
■ Methyl n – propyl ether and methyl isopropyl ether are positional isomers
2 2 3
one carbon three carbons One carbon Three carbons one carbon three carbons
One carbon Three carbons
In both these cases, number of carbons on each side are 1 and 3. Therefore, they cannot be called metamers. Since they differ in their position of methoxy group, they are called positional isomers.
Tautomerism
The structural isomers that are spontaneously interconvertible can exist in dynamic equilibrium and differ in the position of atoms (generally hydrogen) and bonds are called tautomers and this phenomenon is called tautomerism.
Tautomers of propanone:
Tautomerism can be classified into three categories:
■ Open system of tautomerism, or desmotrophism
■ Ring-chain tautomerism
■ Valence tautomerism
Keto-enol tautomerism: This type of tautomerism is more common and it is catalysed by both acids and bases.
Acid Catalysed Mechanism
Base Catalyst Mechanism
Carbonyl compounds with only one chirality centre at a carbon and at least one a hydrogen undergo racemisation in the presence of acid or base catalyst.
phenylbutan-1-one
(achiral) ( )-2-methyl-1-phenybutan-1-one (racemic form) +
The composition of the equilibrium mixture depends on the relative stabilities of the tautomers, the nature of solvent used, and temperature. The greater the stability of a tautomer, greater is its percentage in the equilibrium mixture.
Factors Affecting Stability of Enol Form
For simple aldehydes and ketones, keto form is stable because bond energy of C=O (~364 kJ/mol) is more than that of C=C bond energy (250 kJ/mol)
Since keto-enol tautomerism is a dynamic equilibrium, if a carbonyl compound is dissolved in heavy water in presence of acid or base catalyst, all the a hydrogens that can participate in tautomerism are exchanged with deuterium.
(> 99%)
Acetone
Cyclohexanone (98.8%) 1.2%
In enol form, if a multiple bond, such as phenyl, is in conjugation with C = C, then stability of enol form increases.
Table 9.13 Percentage of enol forms
Stabilty order is II > I > III
Another feature that stabilises the enol, with respect to the keto form, is the possibility of strong intramolecular hydrogen bond.
Example:
MeCOCH2COMe and (MeCOCH2CO2Et.)
Conditions for Tautomerism for Carbonyl Compounds
Carbonyl compounds having at least one active a – hydrogen shows tautomerism.
ix)
no a H ×
x) O H H H H a H attached sp2 carbon, but has g hydrogens ✓
xi) O no a H, bridged H-atoms cannot participate ×
xii) O no a H × xiii) O no a H, but has ghydrogen atoms ✓
xiv) O 1 a H ✓
xv) O O O 6 a H ✓
xvi) O Enol from becomes anti aromatic × xvii) O Enol form is aromatic ✓
xix) (CH3)3CNO2 no a H × xx) O C32 HCNH – –a -H atoms both on N & O ✓
xxi) O C32 HCNH – –O C32 HCNH – –O C32 HCNH – –Enol form is negligible ✓
xxii) O O Keto form is negligible
For nitro compounds: Nitro compounds having at least one active a -hydrogen shows tautomerism.
xviii) N
Enol form is aromatic ✓
Nitro form Acinitro form (Acidic form soluble in base)
■ H–C ≡ N and H–N C are tautomers (also functional isomers) while R–N C and R–C ≡ N are only functional isomers.
■ H N O O and H–O–N=O are tautomers.
9.6.2 Stereo Isomerism
Stereo isomerism is due to the difference in arrangement of atoms or groups in space and is also known as space isomerism. Stereo isomers have the same molecular formula, same structural formula, but different spatial arrangement of atoms or groups. It is mainly classified into two types:
■ Configurational isomerism
■ Conformational isomerism
Configurational Isomerism
Configurational isomers possess certain type of rigidity in the molecule and they are interconvertible only by bond breaking and reforming of covalent bonds. Conformational isomers are interconvertible by rotation about ‘ σ ’ bond. The configurational isomerism is further divided into two parts:
■ Geometrical isomerism
■ Optical isomerism
Geometrical Isomerism
Activation energy for the rotation about carbon–carbon double bond is very high (250 kJ mol–1). This is the contribution of p bond to the total C = C bond energy of 605 kJ mol–1 Because of high activation energy, rotation about C = C is restricted. As a result, two spatial arrangements of groups are possible at C = C. They are named as cis and trans isomers, or geometrical isomers.
The isomer which has similar groups on the same side of the double bond is called ‘cis’ isomer and the isomer which has similar groups on the opposite side of the double bond is known as ‘trans’ isomer. (The prefixes ‘cis’ and ‘trans’ are from the Latin, in which cis means ‘on this side’ and trans means ‘across’).
1.
dichloroethene
HOOC
= C COOH H Maleic acid (cis) Fumaric acid (trans)
3. CHO C C H5C6 H H H C C H5C6 H CHO
cis- cinnamaldehyde transcinnamaldehyde
4.
CH2(CH2)6COOH C C CH3(CH2)6CH2 H H
cis-oleic acid
5. 2 5 C H 3 CH 3 CH 2 5 C H C C
cis – 3, 4 - dimethylhex - 3 - ene 2 5 C H 3 CH 3 CH 2 5 C H C C
trans – 3, 4 - dimethylhex -3 - ene
Geometrical isomerism requires the two groups attached to the each double bond carbon to be different. Alkenes of the type abC = Cab, abC = Cax, and abC = Cxy show geometrical isomerism. Alkenes of the type a2C = Ca2, abC = Cx2, and abC = Ca2 do not show geometrical isomerism. With the molecular formula C4H8, four acyclic isomers are possible. Among them, two are geometrical isomers. CH3
CH3-CH2-CH = CH3 1-Butene CH3-C = CH2 isobutene ; µ = 0.5D µ = 0 33D
cis-2-butene (b.p. 277 K)
cis-2-butene
trans - 2 - butene
trans-2-butene (b.p. 274 K)
[ m = 0.33D] [ m = nearly zero] (more polar) (less polar)
Due to restricted free rotation about a ring, cis-trans isomerism is possible in cycloalkanes also. H HOOC H COOH
Cis–1,4–cyclohexane dicarboxylic acid
Trans–1,4–cyclohexane dicarboxylic acid H HOOC H
cis–isomer
isomer
Geometrical isomerism is not shown by
E and Z Configuration
Sequence rules (or) CIP rules:
■ Higher priority is assigned to the atoms of higher atomic number.
■ If isotopes of the same element are attached, the isotope of higher mass number is given higher order of precedence.
Geometrical isomerism is also possible in the compounds with C=N and –N=N–type bonds.
■ In the groups, the order of precedence is also decided on the basis of atomic number of first atom of the group. When the order of precedence of the groups cannot be settled on the first atom, the second atom or the substituent atoms in the groups are considered.
■ A double or triply bonded atom is considered equivalent to two or three such atoms.
(Higher priority)
C Br F Cl H (Higher priority)
Syn–acetaldoxime Anti–acetaldoxime
Syn–benzaldoxime
Anti–benzaldoxime
In this case, syn and anti are taken with respect to –H and –OH. If –H and –OH are present on same side, it is called syn, and if they are on opposite sides, it is called anti. Symmetrical ketoximes cannot show geometrical isomerism but unsymmetrical ketoximes can show it. Compounds having N = N can also show geometrical isomerism.
(L.P same side)
opposite side)
Z - configuration
If the groups of higher priorities are on the same side of the C = C double bond, Z-configuration is given. (Higher priority) C C (Higher priority) H5C2 CH3 H Br
E-configuration
If the groups of higher priorities are on the opposite sides of the C=C double bond, E-configuration is given C C (Higher priority) CH(CH3)2 CH2 –CH2 –OH Cl Br
E-configuration
Work outward from the point of attachment, comparing all the atoms attached to particular atom before proceeding further along the chain.
–CH(CH3)2 [–C(C,C,H)] outranks
–CH2–CH2OH [–C(C,H,H)]
(Higher priority)
CH(CH3)2 C(H,H,H) C C(H,H,H)
C C
Br CH2OH C(CH3)3
(Higher priority)
Cl
Z-configuration
While working outward from the point of attachment, evaluate substituent atoms one by one but not as a group.
–CH2–OH[–C(O,H,H)] outranks
–C(CH3)3[–C(C,C,C)]
(Higher priority) C O H
Br CH2OH
Cl
C C
(Higher priority)
E-configuration
An atom which is multiply bonded to another atom is considered to be replicated as a substituent on that atom.
( ) CO,[CO,O,H]outranks | H −=
CH2–OH[–C(O,H,H)]
Br (Higher priority)
Cl
CH CH2 C–C–H C C H outranks H
CH=CH2
C C
(Higher priority)
CH(CH3)2
Z-configuration H
Bond pair has more priority than lone pair. H3C C H
Isotope having higher atomic mass is senior and is given priority.
For the substituted phenyl groups, substituents having lower position are given higher priority.
Z - configuration
In case of cycloalkene, geometrical isomer starts from 8 carbon atom.
4, 6, 9)
Geometrical isomers differ in their physical properties such as melting points, boiling points, dip,ole moments, densities, solubilities, stabilities, etc. Their chemical properties are similar but not identical.
i) In general, cis isomer is more polar than the trans isomer because, in it, the individual dipoles do not cancel each other.
ii) In general, cis isomer has greater boiling point than the trans isomer due to greater polarities and, consequently, stronger attractive forces.
iii) Trans isomer has higher melting point than the cis isomer because its molecules are more closely packed in solid state due to symmetry.
iv) Trans isomer is more stable than the cis isomer due to less repulsive forces.
v) Cis isomer has higher solubility in polar solvents.
vi) Cis isomer has higher density, refractive index, dipole moment, and heat of combustion than its trans isomer.
General properteis of cis-trans isomer s
Molecule Polarity MP BP
Solubility in inert solvent
Cis Polar Lower Higher Higher trans Nonpolar Higher lower lower
Number of Geometrical Isomers
1. When compound has ‘n’ double bonds and ends of polyene are different, the number of Geometrical isomers = 2 n .
Example:
CH CH CH CH nI 32 5 121 2 (, )
2. When the number of double bonds is even, then the number of geometrical isomers
GI nn . / 2212 1
Example:
ClCH = CH – CH = CHCl () nG.I22 23 21 2 2 1
3. When the number of double bonds is odd, the number of geometrial isomers
GI n n . 22 1 1 2 1
Example:
ClCH = CH – CH = CH – CH = CHC l 31 1 31 2 n3,Gometricalisomers226 + = =+=
Optical Isomerism
The light ray which vibrates in a single plane is termed plane polarised or simply polarised light. Some substances have the ability to rotate the plane of polarised light. These substances are termed optically active and the property is called optical activity. The substance which rotates the plane of polarised light in the clockwise direction is called dextro rotatory and is denoted as d – (or) (+). If it rotates the plane of polarised light in the anti-clockwise direction, it is called leavo rotatory and is denoted as l – (or) ( – ). In Greek, ‘dextro’ means right and ‘leavo’ means left.
The necessary condition for a molecule to show optical activity is that the structure of the molecule should not be superimposable on its mirror image.
The most common feature that leads to chirality is carbon atom that is bonded to four different groups. Van’t Hoff named the carbon which is attached to four different atoms or group of atoms as asymmetric carbon, or chiral carbon, or stereogenic centre, or stereocenter. Asymmetric carbon is generally designated by placing an asterisk (*). Compounds having no asymmetric carbon are usually achiral. Compounds having just one asymmetric carbon are definitely chiral. This can be explained with an example bromochlorofluoro methane, which has asymmetric c arbon and
2 - bromopropane which has no asymmetric carbon.
Centre of symmetry: It is the point in a molecule through which, if a straight line is drawn from any part of molecule, the line encounters identical group at equal distance in opposite directions.
A molecule having any element of symmetry is always superimposable with its mirror image. In three dimensional approach, a molecule that has any one of elements of symmetry, like plane of symmetry, centre of symmetry, axis of symmetry, and alternating axis of symmetry, is said to be symmetrical and optically inactive. But a molecule that has no element of symmetry, or one that has only a simple axis of symmetry (and none of the other three elements of symmetry) is not superimposable with its mirror image and is optically active.
There are certain compounds which are not having chiral centre but are optically active. Plane of symmetry: It is the imaginary line in a molecule which bisects it in 2 equal halves in such a way that each half of the molecule is a mirror image of the other half.
Three-dimensional representations of molecules: Various methods have been developed for the two dimensional representation of three dimensional structures. They are ball and stick model, wedge and dash model, fisher projection formula, sawhorse formula, and Newman projection. In ball and stick model, atoms are considered as spheres and the bonds are represented as sticks.
In wedge and dash model, to represent the bonds, three types of lines are used. Continuous lines (––––, solid lines) represent the bonds in the plane of the paper; a solid wedge (.........., thick lines) represents the bond projecting above the plane of paper( i.e., towards observer); and a broken wedge (..........., dashed line) represents the bond below the plane of the paper (i.e., away from observer).
C
H away from observer towards the observer
in the plane of the paper
The Fischer projection looks like a cross, with the asymmetric carbon (carbon which is attached to four different groups) at the point where the lines cross. The vertical bonds at the asymmetric carbon are directed away from the viewer and the horizontal bonds point towards the observer. If the carbon chain is with more than one carbon atom, the carbon chain is to be written vertically.
Bromochlorofluoromethane is shown in different models below:
Ball–and–stick model Cl Br F H C Wedge–and–dash model Fischer projection F H
Wedge dash of compound having 2-chiral centres
Specific rotation [] cl where
[ a ] = specific rotation, q = angle of rotation, c = concentration of the solution in gram per millimeter, l = length of the tube in decimetres (1 dm = 10 cm)
The specific rotation also depends on the temperature and the wavelength of light that is employed.
2. Number of enantiomeric excess = molesofone enantiomer molesofother enantiomer totalnumberofbotth enantiomers 100
Enantiomers and Diasterecisomers
■ Molecules which are non-superimposable mirror images of each other are called enantiomers. These are optical isomers.
■ Enantiomers differ in their behaviour towards plane polarised light. One of them is dextrorotatory and the other one is laevorotatory (their specific rotation values are numerically the same).
■ They have identical physical properties (except optical rotation).
■ They cannot be separated from each other by normal physical methods.
■ Stereo isomers that are not mirror images of each other are called diastereoisomers or diastereomers.
■ Diastereomers differ in physical properties. They can be separated from each other by physical methods.
Specific rotation of polarised light: Rotation of plane polarising light is measured in degrees. Value of rotation of plane polarised light depends on the length of the polarimeter tube and the concentration of the solution.
■ The stereo isomer in which the like substituents of the two chiral carbons are on the same side of the main chain (R–C–C–R') of Fishcher projection formula is called ‘erythro’ form. The one in which the like substituents are on opposite sides
of the main chain of the Fischer projection formula is called the ‘threo form’.
Isomers of 2,3-dihydroxybutanoic Acid: 2,3–dihydroxy butanoic acid has two chiral carbons and no plane of symmetry. Thus, it has four optical isomers. All are optically active.
Erythro diastereo isomer contains the like substituents on the same side of the Fischer projection. If like substituents in Fischer projection are on opposite sides, the isomer is called threo diastereomer.
The stereo isomers of 2,3-dihydroxy butanoic acid are shown in below.
I and II as well as III and IV are enantiomers of each other. Either I or II are not the mirror images of either III or IV. The stereo isomer I is a diastereomer of III or IV. Similarly, II is a diasteromer of III or IV.
When two optically inactive compounds react, no optically active compound is formed. However, if the reaction is catalysed by an optically active reagent, even if the starting compounds are optically inactive, the product can be optically active.
Asymmetric induction is the use of a chiral reagent or catalyst to get a chiral product from an achiral reactant.
Racemic Mixture and Resolution
An equimolar mixture of enantiomers is called racemic mixture. A racemic mixture is optically inactive.
In this, the rotation caused by the molecules of one enantiomer is cancelled by the rotation caused by the molecules of other enantiomer. This type of compensation is called external compensation. A racemic mixture is represented by the prefix (d l)– or (±) before the name. The process of separation of constituent enantiomeric forms from the racemic mixture is known as resolution. Resolution can be carried out by mechanical separation, biological separation, or chemical separation.
Racemic mixture opticallyactive substance
Mixture of diastereoisomers separation
Separated diastereoisomers suitable reagent
Separated enantiomers →
(I) (II) (Mirror) Erythro Erythro H OH COOH CH3
H COOH CH3 H OH (III) (IV) (Mirror) Threo Threo
Stereo Isomers of 2,3-Dihydroxy Butanoic Acid
F or example, if the racemic mixture of a carboxylic acid ( ± ) A is treated with an amino acid base (+) B, enantiomer ( ± ) A reacts with (+)B giving (+) A (+) B, (–) A(+)B, two salts related to each other, as diastereoisomers. These diastereoisomers differ from each other in physical properties and also in rate of formation. They differ also in rate of crystallisation, solubility, etc. Using one of these differentiating properties, the diastereomers are separated.
(±)A + (+)B
Separation by any one of the physical method (+) A (+) B + (–) A (+) B
(+)A (+)B (–)A (+)B
(+)A +BHCl (–)A +BHCl HCl HCl diastereomers
Racemisation
Racemisation is the reverse of resolution, i.e., the conversion of (+) or (–) isomer into its racemic mixture ( ± ) is termed racemisation.
Racemisation involves the change of half of the active compound to the isomer of opposite rotation, resulting the formation of racemic mixture.
Example: When a solution of (+) or (–) tartaric acid in water is heated under pressure, it is transformed into a completely inactive mixture of racemic (or also meso) tartaric acid.
D, L - Notation
D,L - Notation was introduced by Fischer by taking glyceraldehyde as arbitrary standard. OH CHO H
CH2OH D-form (–OH group on right side)
CHO H HO
CH2OH L-form (–OH group on left side)
The other stereo isomers can be assigned D or L notation by comparison with the structural arrangements of D and L glyceraldehyde.
■ The most oxidised carbon attached to the chiral centre is placed on the top of vertical line.
■ The group with carbon atom forming a part of the chain is kept at the bottom of vertical line.
■ The remaining groups are assumed to be projected towards the viewer and are placed along the horizontal line.
The D, L system is commonly used in assigning stereochemistry to carbohydrates and amino acids. For a -amino acids; –NH2,–COOH, –R, and –H groups at C a atom are related to –OH, –CHO, –CH 2OH, and –H, respectively, of glyceral dehyde.
D–glyceraldehyde and D– a –amino acids are said to have the same configuration. D, L–configurations have no relation with d, l or (+), (–) notations.
CHO OH H OH OH H HO H H
CH2OH
D– (+)– Glucose
CHO H OH
CH2OH D– (+)– Glyceraldehyde D(+) Glucose D(+) Glyceraldehyde
CHO OH H H HO
CH2OH L– (– )– Glucose HO HO H H
CHO H OH
CH2OH L– (– )– Glyceraldehyde L-Glucose L -Glyceraldehyde
COOH NH2 H
CH3 D– Threonine OH H
CHO
CH2OH OH H D– (+)– Glyceraldehyde D - Threonine D-(+) Glyceraldehyde
Absolute (R, S) and Relative (d, l)
Configurations
The three dimensional arrangement of substituents at a stereogenic centre is called its absolute configuration. Neither the sign nor the magnitude of rotation by itself provides any information about the absolute configuration of a substance.
C H OH H3C
C2H5 and C H HO
CH3 C2H5 (a) (b)
Here, (a) and (b) are absolute configurations of 2 - butanol. But they do not represent sign and magnitude of their optical rotation. However, the configurations of thousands of compounds relative to one another (known as their relative configurations) were experimentally determined through chemical interconversion.
Example:
When (+)-3-buten-2-ol is hydrogenated, the product formed is (+)-2-butanol.
CH3 - CH- CH = CH2 + H2
OH
(+)-3-butene-2-ol
CH3 C H HO is (-) -2- butanol C2H5
Cahn–Ingold–Prelog’s R,S Notation
CH3 - CH - CH2- CH3
2- butanol Pd ®
Hydrogen of the C = C does not involve any of the bonds at the stereogenic centre. Therefore, the substituents must have been arranged in the same manner in both the reactants (+) – 3– buten – 2 – ol and (+2) – butanol. As they have same sign of rotation, their identical configuration is established by the hydrogenation reaction, which otherwise could not be predicted before of the experiment.
It is to be noted that some compounds with the same relative configuration could have opposite sign of optical rotation.
R,S notation was devised by Cahn, Ingold, and Prelog. The four different groups attached to the chiral carbon atom are assigned priorites, 1, 2, 3, or 4, based on sequence rules, as discussed in E and Z system of nomenclature in geometrical isomerism.
1 is the highest priority group and 4 is the lowest, i.e., 1 > 2 > 3 > 4.
Now, the molecule is visualised in such a way that atom or group with lowest priority is directed away from the eye. The remaining atoms or groups, now, are observed in decreasing order of priorities, i.e., from 1 to 2 to 3. In doing so, if the movement of eye is clockwise, the configuration is R. On the other hand, if eye moves in anti-clockwise direction, the configuration is S.
The priority order is as follows:
2- Methyl - 1- butanol
CH3 – CH2 – CH – CH2OH + HBr CH3 [ ]25 0 D 5.8 a = +
1- Bromo - 2- methylbutane
CH3 – CH2 – CH – CH2Br CH3 [ ]25 0 D 4.0 a = +
After determining the absolute configuration of a salt of (+) tartaric acid (1951), the absolute configuration of all compounds whose configuration had been related to (+) tartaric acid were established.
H5C2 C OH is (+)2 - butanol and CH3
I, Br, Cl, SO3H, F, OR, OH, NO2, NR2, NHR, NH2, COOR, COOH, CONH2, COCH3, CHO, CH2OH, CN, C6H5, –C ≡ CH, –CR3, –CH = CH2, –CHR2, –CH3, D and H
From the above results, it is understood that (+) – 2 – butanol has ‘S’ configuration and its mirror image, (–)–2 – butanol, has R configuration.
When the molecule contains more than one chiral centre, the same procedure is applied to each. The direction of rotation of plane polarised light has nothing to do with R–S notation. An R or S compound may be either dextrorotatory or laevorotatory.
Allowed notions of Fischer projections: A chiral carbon be viewed in different directions and we get different Fischer projections. We may get confused that all these are different, but, actually, all these are same. The Fischer projections of a compound, which are formed by 180� rotation, are same.
COOH H Rotate 1800
(I) (II)
Fischer projections I and II are same and they have same absolute configuration.
If a Fischer projection is rotated by 90° , we get Fischer projection with opposite configuration. Two Fischer projections that differ by a 90° rotation are enantiomers.
COOH H Rotate 900 CH3 OH H OH COOH H3C (III) (IV)
III and IV are enantiomers. Thus, rotation of Fischer projection by 180° is allowed but 90° is not allowed.
In Fischer projection formula, if any one of the groups at chiral carbon is placed steadily and the remaining three are rotated in clockwise or anti-clockwise direction, we get the Fischer projection with the same configuration.
I, II, and III have same configuration. Thus, it is also one of the allowed notions. This is also true for wedge dash model.
Keep the -COOH steadily Rotate remaining
COOH CH3 HO H (IV) three groups in clockwise direction C COOH HO H H3C (V)
Structures IV and V are same and they have same configuration. If the groups at chiral carbon are interconverted by even number of times, we get the Fischer projection with same configuration. But in case of odd number of interconversions, we get enantiomers. COOH OH H CH3 (I) 1st Inter change COOH
H (II)
(III) 2nd 3rd
COOH (V)
In the configuration of lactic acid (I to V), configurations I, III, and V have the same configuration—R. II and IV have the same configuration—S.
Steady I Rotate in clockwise direction Rotate in anticlockwise direction
Simple Method for Converting Wedge and Dash Model into Fischer Projection Formula
The group which is away from the observer (i.e., connected by, HO H3C H (1) 3 2 2R,3R configuration CH3 H OH (2) (3) (2) (1) (3) ) is kept at the bottom in the Fischer projection formula. The remaining three groups at asymmetric carbon are attached in the same sequence as they appear in the wedge and dash model. Some examples are given below :
1) C COOH CH3 H OH COOH OH CH3 H
2) C COOH OH COOH OH H Ph Ph H
3)
Chiral Molecules with Two Stereogenic Centres
Compounds like tartaric acid (HOOC. CHOH. CHOH.COOH),2,3–butanediol (CH3.CHOH. CHOH.CH 3 ), etc., though they have two stereogenic centres, only three (not four) isomers are possible due to the presence of plane of symmetry.
I and II are enantiomers of each other. They have equal but opposite optical rotations. III, however, is an achiral structure and it is superimposable on its (2S,3R) mirror image. Being achiral, III is optically inactive.
The achiral molecules that have stereogenic centres are called meso forms. Optical inactivity of III is due to the plane of symmetry.
Optical Isomerism of Tartaric Acid
Tartaric acid has two asymmetric carbon atoms linked to the same four different groups.
The spatial arrangement of various groups in tartaric acid can be represented in four ways, as shown in the figure below.
Structures I and II are mirror images to each other and are a pair of enantiomers. Structures III and IV also look like a pair of mirror images, but due to plane of symmetry, they are one and the same and optically inactive.
One of the asymmetric carbon atoms turns the plane polarised light to the right and the other to the left to the same extent, so that the rotation due to upper half is compensated by the lower half, i.e., internally compensated. Such compounds are meso compounds.
(III) (IV) (Mirror)
Stereo isomers of tartaric acid
Finding the Number of Optical Isomers
Case-I:
Constitutionally unsymetrical molecules. If ‘n’ is the number of asymmetric carbon atoms, number of optically active isomers (a) = 2 n number of racemic forms (r) = a/2 number of meso forms (m) = 0
∴ Total optical isomers = a + m
Example:
CH2OH–(CHOH)4–CHO (glucose)
a = 24 = 16, r = 8, m = 0
∴ Total = 16 + 0 = 16
Case-II:
Constitutionally symmetrical molecules. If ‘n’ is the number of asymmetric carbon atoms, where, n = even,
ar am n 22 2 1 ,/ , n 2 1
Example:
CH OH CHOH CH OH sorbital 24 2 () ()
ar m 28 42 2 41 21 ,,
∴ Total isomers = 8 + 2 = 10
Case III:
Constitutionally symmetrical molecules containing n number of chiral centres, where n is odd number.
Number of optically active isomers
=2n–1–2n/2–1/2
Number of optically inactive meso isomers
= 2n/2–1/2
Total number of stereo isomers = 2 n–1
TEST YOURSELF
1. N umber of open chain isomers with the formula C4H8 (including stereo isomers) is (1) 2 (2) 6
(3) 4 (4) 3
2. The number of possible isomers for the compound with molecular formula C2BrClFI is
(1) 3 (2) 4
(3) 5 (4) 6
3. 3,4-dimethyl hexane has (1) two pairs of enantiomers and a meso compound.
(2) a pair of enantiomers and a meso compound.
(3) a pair of enantiomers and two meso compounds.
(4) 2 pairs of enantiomers and two meso compounds.
4. Which of the following is likely to show geometrical isomerism?
(1)2-Butene (2) 1-Butene (3) 1-Pentene (4) Propene
5. Number of aldehydes possible with the formula C5H10O is (1) 2 (2) 3
(3)4 (4) 5
6. C3H9N cannot represent (1) 10 amine (2) 20 amine
(3) 30 amine (4) quaternary salt
7. Number of structurally isomeric nonterminal alkynes possible with the formula C5H8 is (1) 4 (2) 3
(3) 2 (4)1
8. Which of the following molecules have (2R, 3Z) configuration?
(1) (2) (3) (4)
9. The number of stereo isomers possible for a compound of the molecular formula CH 3 CH = CH−CH(OH)−Me is (1) 3 (2) 2 (3)4 (4) 6
10. The absolute configuration of the following is
(1) (2S, 3R)
(2) (2S, 3S)
(3) (2R, 3R)
(4) (2R, 3S)
11. How many cyclic and acyclic isomers are possible for the molecular formula C3H6O?
(1) 5 (2) 6 (3) 7 (4)8
12. Which of the following is not an isomer of CH3−CH2−CHO?
(1) CH2 = CH−CH2−OH
(2) (3) (4)
Answer Key
(1) 3 (2) 4 (3) 2 (4) 1 (5) 3 (6) 4 (7) 4 (8) 3 (9) 3 (10) 1 (11) 4 (12) 4
9.7 FUNDAMENTAL CONCEPTS IN ORGANIC REACTION MECHANISM
In an organic reaction, the organic molecule, refe rred to as substrate, reacts with an appropriate reagent and leads to the formation of one or more intermediate(s) or transition state and, finally, product(s).
Organic molecule Attacking reagent [Intermediate] or (Substrate) Transition state ↓
Product(s)
The det ailed step-by-step description of a chemical reaction is called reaction mechanism. To understand clearly the mechanism of various organic reactions, it is essential to have knowledge about the following:
■ Electronic displacements in covalent bonds
■ Cleavage of covalent bonds
■ Nature of attacking reagents
Electronic displacements: The electronic displacements in covalent bonds may occur either due to the presence of some atom or group in the molecule or under the influence of attacking reagent.
As a result of these electron displacements, centres of different electron densities are created and these centres are susceptible to attack by the reagents. Inductive effect and resonance effect are permanent polarisation effects on the covalent bond. Electromeric effect, or polarisability effect, is the temporary effect.
9.7.1 Inductive Effect
Consider the molecule CH3–CH2–CH2–CH2–Cl. In C– Cl bond, more electronegative chlorine attracts the σ electrons. As a result, C 1 gets partially positive charge.
The positive charge on C 1 attracts the electron pair shared between C1 and C2 towards itself. This will cause C 2 to acquire a small positive charge but this charge on C 2 will be smaller than on C1. Similarly, C3 will acquire positive charge that will be still smaller. This effect is still relayed further. The polarisation of σ bond caused by the polarisation of adjacent σ bond is called inductive effect.
CC CC Cl 43 21
Inductive effect tends to be insignificant beyond the third carbon atom. It is a permanent effect and sigma electrons are involved.
In the molecules having polar covalent bond, in addition to inductive effect, field effect is possible. Inductive effect operates through the ‘σ’ bond but field effect operates either through the space surrounding the molecule or through the solvent molecules,
in case of solutions. In many cases, it is not possible to distinguish inductive and field effects. Hence, by inductive effect, we mean the inclusion of the field effect also, if any.
The inductive effect is related to the ability of substituents to either withdraw or donate electron density to the attached carbon atom. Based on this ability, the substituents are classified as electron withdrawing and donating groups relative to hydrogen.
– I effect: An atom or group of atoms, which has greater tendency to attract electrons, when compared with hydrogen, is called electron withdrawing group and it is said to be exhibiting negative inductive (–I) effect. Halogens, –OH, –NH2, and some groups with multiple bonds, like –CN, –NO2, – CHO, and –COOH, are examples for electron withdrawing groups.
The order of electron withdrawing tendency or –I effect of some groups is as follows:
OR NR 23 NH 3 + > –NO 2 >–CN > –SO 3H > –CHO > –COOH > –COCl > – COOR > –CONH2 > –F > –Cl > – Br > –I > – OH > –OR > –NH2 >–C6H5 > –H
Electronegativity is the measure of tendency to attract shared electron pairs. In inductive effect, sigma bond electrons are involved. Thus, with decreasing the electronegativity, –I effect or electron withdrawing tendency decreases. Some –I effect orders are given below :
–F > –Cl > –Br > – I
–F > –OH > –NH2 > –CH3
–F > –OR > –NR2 > – CR3
–OR >–SR > –SeR
With a decrease in the percentage of ‘s’ character of hybrid orbital, electronegativity of hybrid orbital decreases, and hence, –I effect decreases.
–C ≡ CH > –CH = CH2 > –CH = CR ≡ CR > –CR = CR2 > –OR
+ I effect: An atom or group of atoms, which has less tendency to attract sigma electrons, when compared with hydrogen, is called electron releasing group and it is said to be exhibiting positive inductive (+I) effect.
Alkyl groups are examples of it. With increasing the electronegativity, +I effect or electron releasing tendency decreases. +I effect or electron releasing tendency of some groups is given below : O COO > – C(CH3)3 > – CH(CH3)2 > –
CH2CH3 > – CD3 > – CH3
3°-alkyl > 2°-alkyl > 1°-alkyl
–Se– > –S– > – O–; NR O
I nductive effect influences chemical reactivity and it affects physical properties. This effect is very helpful in organic chemistry in explaining a number of facts, like reactivity of alkyl halides, dipole moment, solubility, reaction rates, relative strength of carboxylic acids, and strength of bases.
Application of Inductive Effect
With the help of inductive effect, acidic strength of alcohols, phenols, and carboxylic acids and reactivity of alkyl halides are explained.
Reactivity of Alkyl Halides
The presence of halogen atom in the molecule of alkyl halide creates a centre of low electron density, which is readily attacked by the negatively charged reagents.
For example, CH3Cl is more reactive than CH 4 as inductive effect is present in CH 3Cl and no inductive effect is present in methane. The activity also increases from primary to seconda ry and from secondary to tertiary
halides, as the +I effect of methyl groups enhances –I effect of the halogen atom by repelling the electrons towards the tertiary carbon atoms. The reactivity order for alkyl halides is: 3° > 2° > 1° > methyl.
H3C X > H3C CH X >
CH2 X > CH3 X CH3
Relative Strength of Acids
The strength of an acid depends on the tendency to release proton when the acid is dissolved in water.
R–COOH RCOO– + H+
Stronger the acid, weaker is its conjugate base. More negative the charge on carboxylate ion, the more is its reactivity and more is the basic nature (less is acidic nature). Less More Most –ve charge –ve charge –ve charge (most stable) (less stable) (least stable)
Thus, the decreasing order of the acidic nature is: HCOOH > CH3COOH > (C2H5–COOH) > C3H7COOH > C4H9COOH >
Any group or atom showing +I effect decreases the acid strength as it increases the negative charge on the carboxylate ion, which holds the hydrogen firmly. Alkyl groups have +I effect.
Comparison of Acid Strength between Various Acids
FCH2COOH > ClCH2COOH > BrCH2COOH > ICH2COOH.
CCl3COOH > CHCl2COOH > CH2ClCOOH > CH3COOH.
Thus, the group or atom having –I effect increases the acidic strength as it decreases the negative charge on the carboxylate ion.
CH3CH2CH(Cl)COOH >
CH3CH(Cl)CH2COOH >
CH2(Cl)CH2CH2COOH >
CH3CH2CH2COOH
As compared to water, phenol is more acidic (–I effect), but ethyl alcohol is less acidic (+I effect).
OH H OH C2H5 OH > >
Phenol Water Ethyl alcohol
CH3OH > CH3CH2OH > (CH3)2CHOH > (CH3)3COH
C6H5OH < CH3COOH < HNO2 < H3O + < C6H5SO3H< HCl < HBr < H2SO4 < HI < H2SbF6 (Strongest acid, pKa = –12)
Except formic acid, all other aliphatic acids are weaker than aromatic acids.
HCOOH CH COOH CH COOH 65 3
CH NH HO HF 43 2 << < - Acidic order
F– is more stable anion corresponding acid. HF is most acidic.
CH CH CH33CCHNHCHNH 32 - Acidic Order
CH CCHC C
ve charge on more E.N atom and no +I effect IofC 3 H IofCH 3 3 NH CH NH 23 - Stability order
CH SH CH OH 3 3 - Acidic order
CH SCHO
ve ch eon biggersizeatom more stable anion 33 arg () - Stability order
Relative Stability of Carbocations
Stabilityofcarbocation Ieffect Ieffect
Electron releasing methyl groups (+ I effect) decreases the positive charge on carbon and increases the stability.
() () CH CCHC HCHC H 33 32 32
Dipole Moment
As – I effect increases across a bond, the dipole moment also increases.
CH 3 I D (. ) 164 CH Br D 3 (.179 ) CH Cl 3 (. )194D
9.7.2 Electromeric Effect
The complete transfer of the shared pair of pi electrons of a multiple bond to one of the atoms in the presence of the attacking agent is called electromeric effect. It is a temporary effect and observed only in organic compounds with multiple bonds in the presence of attacking reagent. It comes into play instantaneously at the demand of the attacking reagent and when it is removed, the original condition is restored. This effect is shown by a curved arrow. There are two types of electromeric effect:
+ E effect: When the transfer of electrons takes place towards the attacking reagent, the effect is called +E effect.
Example:
The addition of acids to alkenes C C H C C + + + → H
Propene 3 2 3 3 CH CH CH H CH C H CH
Since –CH 3 group is electron repelling, the electrons are transferred in the direction shown.
–E effect: When the transfer of electrons takes place away from the attacking reagent, the effect is called –E effect.
Example:
The addition of cyanide ion to carbonyl compounds
Example:
–Cl, –Br, –I, –NH2, –NHR, –NR2, –OH, –OR, –SH, –OCOR, –NHCOR
+M effect in vinyl chloride
When the atoms linked to a multiple bond are similar, the shift can occur to either direction. When the dissimilar atoms are linked on the two ends of the double bond, the shift is decided by the inductive effect. Electrons are always transferred towards the more electronegative atom.
In cases where inductive effect and electromeric effect simultaneously operate in opposite direction, usually, electromeric effect predominates. Since the electromeric effect takes place only at the time of attacking reagent, it always facilitates the reaction and never inhibits.
9.7.3 Mesomeric Effect
The complete displacement of electron pair caused by an atom or a group along a chain by a conjugative mechanism is called the mesomeric effect of that atom or group. It is possible when some multiple bonds are in conjugation (presence of single and double bonds alternatively) or when an atom with at least one lone pair is in conjugation with multiple bond. It is a permanent effect and it operates until conjugation ends. Lone pair and p-electrons are involved in it. It influences the physical properties, reaction rates, etc.
+M effect: When the transfer of electrons is away from the atom or group in conjugation with a p-bond, the mesomeric effect is said to be +M effect (+R effect). This increases the electron density of the rest of the molecule. Groups exhibiting +M effect possess an atom with at least one lone pair.
+M effect in aniline
Since the lone pair in the group is responsible for + M effect, with increasing the electron density +M effect increases. Down the group, electron density decreases due to increase in size, and +M effect decreases.
–F > –Cl > –Br > – I
–OH > –SH > –SeH
–OR > –SR > –SeR
With decrease in the basic strength and electron releasing tendency, +M effect decreases.
–NH2 > –OH > –F
–NR2 > –OR > –F
–O– > – OR
–M effect: When the transfer of electrons is towards the atom or group in conjugation with a p -bond, the mesomeric effect is called –M effect (–R effect). This decreases the electron density of the rest of the molecule. Groups exhibiting –M effect possess multiple bond.
Example:
–COOH, –CHO, –COR,
–NO2, –SO3H, O || COR , O || CNH 2
–M effect in propenal O H ↔ + H 2 H C=CH C= 2 H C CH=C O : ii
–M effect in nitrobenzene
+
–M groups withdraw the p electrons from the rest of the molecule. Thus, –M effect can be compared based on electronegativity. –M order of some groups is given below:
= O > = NR > = CR2, > ≡ N > NR 2 + > = NR.
> –NO 2 > –CN > –SO 3 H > –CHO > –COCH3 > – COOH.
Halogens and some groups, like –OH and –NH 2 , have –I effect because of high electronegativity, but with lone pairs, they show +M effect. The order of influence or strength of various effects is mesomeric
or resonance effect > hyperconjugation > inductive effect.
9.7.4 Resonance
Some compounds require more than one structure to explain all of its properties. For example, urea with the structure 2 2 H N C NH O has to exhibit the following properties. It should be diacidic base due to the two – NH2 groups, carbon–nitrogen bond length has to be equal to single bond length, and it should not have dipole moment. But urea is monoacidic base, carbon–nitrogen bond length is intermediate to single bond, and double bond and has dipole moment. To explain all the properties of urea, the following structures are suggested:
Different structures of same molecule used to explain all the properties of a the compound are called resonance or canonical structures of the compound. This phenomenon is called resonance. The real structure of the molecule is not possible to denote with one structure and it is only a hypothetical structure, known as resonance hybrid.
In nitromethane, bond length of two N–O bonds are equal. It can be explained by the following two resonance structures.
Some Important Features of Resonating
Structures
■ The positions of nuclei are same in all the structures but the dif ference is only in electronic arrangement.
■ The number of paired or unpaired electrons is same in all structures.
■ All atoms in the molecule are to be in the same plane.
■ Resonance structures are not interconvertible.
■ All resonance structures should have almost same energy and stability.
■ The structure which one has more number of covalent bonds, with less separation of opposite charges, more dispersal of charge, and a negative charge, on any more electronegative atom, a positive charge on any more electropositive atom is more stable than others.
■ More the delocalisation, more is the stability.
■ More stable resonance structure contributes more to the actual molecule.
■ The resonance structures are hypothetical and individually do not represent any real molecule. They contribute to the actual structure in proportion to their stability.
■ The bond distances of hybrid structures are intermediates of those of resonating forms.
The energy of actual structure of the molecule (the resonance hybrid) is lower than that of any of the canonical structures. Resonance energy is the energy difference between the real resonance hybrid structure and the most stable (the lowest energy) resonance structure. The more the number of important contributing structures, the more is the resonance energy and more is the stability. Resonance is particularly important when the contributing structures are equivalent in energy.
Application of Resonance
■ It explains the acidic nature of phenols and carboxylic acids as the ion formed after the release of proton (H+) is more resonance stabilised.
■ It explains the stability of conjugated dienes over non-conjugated dienes.
■ It explains why two O – O bond lengths in ozone molecule are same (1.28 A°), which are between O – O (1.48 A°) in H 2O2 and O = O (1.21 A°) in O2
■ It explains why C – C bond length in benzene is 1.39 A° [between C – C (1.54 A°) and C = C (1.34A°) ].
■ It explains why two C – O bond lengths in formate ion O C H O-are same (1.27 A°) which are between C – O (1.23 A°) and C = O (1.36A°).
■ It explains the stability of benzyl carbocation, free radicals, and carbanions.
■ Structure with more conjugation is more stable if nature of bonding is same.
■ >
■ If conjugation as well as nature of bonding is same, then aromatic compound is more stable than non-aromatic.
■ <
■ Structure with linear conjugation is more stable than cross conjugation (nature of bonding remains same).
>
Resonance effect is the polarity produced in a molecule by the interaction of two or more p bonds in the conjugation or between a p bond and a lone pair of electrons present on adjacent atom.
Although resonance effect and mesomeric effect appear both as the same, they are different. Mesomeric effect involves only p electrons and lone pair in the conjugative system. Resonance effect involves even electrons.
Applications of Resonance Effect
1. Lesser reactivity of aryl halides (C 6H5Cl) and vinyl halides (CH 2 = CH – Cl) than alkyl halides (R–CH 2–Cl) towards nucleophilic substitution can be explained in terms of +R effect of halogen.
2. A group with –R effect increases the acidity of phenol and carboxylic acid while the reverse is true for a group with +R effect.
5. Decreasing order of C - O bond length is CO
Similarities between Inductive and Mesomeric Effects
■ Both are permanent effects.
■ Both affect the physical properties of the molecule.
■ Both can either hinder or facilitate a particular reaction. Both affect the rates of reaction, the strength of acids or bases, the reactivity of halides, and the substitution of different aromatic species.
Main differences between inductive and mesomeric effects are summarised in Table 9.14.
Table 9.14 Differences between inductive effect and mesomeric effect
effect
+R effect
3. A group with –R effect decreases the basic nature of aromatic amines while a group with +R effect increases the same.
4. Higher the number of resonating structures, higher is the stability.
effect Mesomeric effect
It is operative in saturated compounds. It is operative in compounds especially having conjugated systems. It involves electrons of sigma bonds. It involves electrons of pi bonds or lone pair of electrons. The sigma electron pair is slightly displaced from its position and, thus, partial charges are developed. The electron pair is completely transferred and, thus, full positive and negative charges are developed. It is transmitted over quite a short distance. The effect becomes negligible after third carbon atom in the chain. It is transmitted from one end to other end of the chain, provided conjugation is present. It continues till conjugation is there.
9.7.5 Hyperconjugation
Hyp erconjugation is a general stabilisation interaction. Hyperconjugation is a special kind of resonance in which delocalisation of electrons takes place through overlap between σ bond orbital of a C– H bond and p bond orbital or empty p orbital. It is also known as σ – p conjugation or no-bond resonance. It arises due to partial overlap of a sp 3-s sigma bond orbital with the empty p orbital or p bond orbital of an adjacent carbon atom. Hyperconjugation is a permanent effect.
The delocalisation of the sigma bond electrons of the C–H bond of alpha-carbon with the pi-electrons of the double bond occurs when alkyl groups are attached to an unsaturated system, like a double bond or a benzene nucleus.
In propene molecule, one of the C–H bonds of methyl group can align in the plane of p bond orbital and then partial overlap can take place between sigma orbital and p bond orbital. This leads to delocalisation of p electrons and makes the molecule more stable. In general, greater the number of methyl groups attached to the double bonded carbon atoms, greater is the hyperconjugation and greater is the stability.
More substituted alkenes are more stable because the number of contributing structures increases with increase in number of alkyl substituents.
of substituted alkenes is as follows
B) Isobutene > 2 - butene
Hyperconjugative structures of toluene:
Hyperconjugation of propene
Hyperconjugation in propene is as follows
Hyperconjugation is possible in carbocations having at least one a C–H bond. In ethyl carbocation, the positively charged carbon has empty ‘p’ orbital. The electrons of one of the C–H bonds of methyl group orient in the plane of this empty orbital and delocalise into it. Hyperconjugation is due to a C–H bond possessing partial ionic character because of resonance.
Hyperconjugation in ethyl carbocation
Hyperconjugation in ethyl carbocation is as given below + 2 2 H | H H C CH H C CH | | H H H H | | ↔ = + ↔ 2 2 + HC=CH H C CH | H H ↔ =
With increase of the number of a C – H bonds, number of possible hyperconjugation structures increases and stability increases. Stability order of carbocations is:
() () CH CCHC HCHC HC H alkylalkyl a 33 32 32 3 00 3210 l lkyl CH CH CH CH 96 30
T he facts that can be explained by hyperconjugation are stability of alkenes, stability of carbonium ions, stability of alkyl free radicals, bond lengths, orientation effect of alkyl groups in aromatic ring, dipole moments, and difference in heats of hydrogenation. Hyperconjugation is not possible in alkyl carbanion.
Heat of Hydrogenation
R–CH=CH2 + H2→ R–CH2–CH3 ΔH= –ve (Heat of hydrogenation)
Heat evolved when any unsaturated hydrocarbon is hydrogenated is called heat of hydrogenation ( D H). More the heat of hydrogenation, lesser is the stability of alkene.
Heat of Heathydrogenation of drogenation ∝ 1 stabilityofalkene Resonance is more dominated than hyperconjugation. Hence, decreasing order of C = C bond length is
only resonance but aromatic > resonance + 4 H S 7 H.S. > > 4 H S
More the hyperconjugation, more is the single bond character in double bond.
Acidic Strength of Methyl Substituted Phenol
Phenol > meta > para > ortho
Phenol is more acidic than o, p, and m cresol. Amongst o, m, and p cresols, + I effect is observed in m-cresol, which is less dominated than hyperconjugation (observed in o and p cresol). Between o and para cresol, electronic effect is more dominated in o cresol, and hence, p cresol is more acidic than o cresol.
Applications of Hyperconjugation
1) It explains the relative stability of alkenes
2) It explains the stability order of carbocations and free radicals/alkenes.
Stability ∝ number of hyperconjugative structures ∝ number of alpha hydrogen atoms 3 ° > 2 ° > 1 ° > methyl (stability order)
3) Hyperconjugation explains the ortho and para directing nature of methyl group attached to benzene ring, e.g., toluene.
4) The electron donating power of alkyl groups, when attached to benzene ring, depends upon the number of hyperconjugative structures and number of a-hydrogens on a-carbon. The electron releasing power is in the order: CH 3 – > CH 3 – CH 2 – > (CH 3 ) 2 CH – > (CH3)3C–
5) It explains the preferential formation of non-terminal alkenes during dehydrohalogenation of alkyl halides or dehydration of alcohols (Saytzeff’s rule).
Reverse Hyperconjugation
The phenomenon of hyperconjugation is also observed in the system given below:
C C C where X = halogen
In such a system, the effect operates in the reverse direction. Hence, hyperconjugation in such a system is known as reverse hyperconjugation.
(on increasing size of ring from 3 - 6 stability increases)
Some other examples showing reverse hyperconjugation are:
Applications of Electron Migration Effect
Stability order of carbocations is as follows:
Stability order: IV > III > II > I
xi)
NO2 (I) CH2 NO2 (II) CH2 NO2 (III) CH2 (IV)
Stability order: IV > II > III > I xii)
OH I CH2 OH (II) CH2 OH (III) CH2 (IV)
Stability order: III > I > IV > II
Cyclopropylmethylium Ion C2 H +
i) These carbocations are very stable carbocations. They are more stable than benzyl carbocations.
ii) Stability of cyclopropyl methyl carbocations increases with every cyclopropyl group. Thus, additional cyclopropyl group has cumulative additive effect on the stability. Thus,
Relative Strength of Bases
Base strength is defined as the tendency to donate an electron pair for sharing. The +I effect increases the electron density while –I effect decreases. So, more the tendency to donate electron pair for coordination with proton, more is the basic nature, i.e., more the negative charge on nitrogen atom (due to +I effect of alkyl group), more is the basic nature.
Thus, the order of basic nature is:
(C2H5)2NH > C2H5NH2 > CH3NH2 > NH3
Aliphatic amines > Ammonia > Aromatic amines. R NH2 > H NH2 > NH2
On the other hand, if alkyl part of primary amine is tertiary butyl, then ammonia will be more basic than tert-butyl amine due to steric hindrance in protonation caused by bulky nature of tert-butyl group
iii) The special stability is a result of conjugation between the bent orbitals of the cyclopropyl ring and the vacant p orbital of the cationic carbon. This type of bonding is called banana bonding or bend bonding.
On the basis of +I effect, basic strength of amine should be as follows:
This order is true in vapour phase or nonpolar aprotic solvent.
Actual order of basic character is
> 1° > 3° (if R is – CH3)
> 3° > 1° (if R is – C2H5 )
Thus, experimental data of basic strength order will be combined effect of:
* Inductive effect
* Steric effect
* Solvent effect
Basic Strength of Amines
Basic strength of amine depends upon hybridised state of nitrogen sp3 (N) > sp2 (N–localised) > sp2 (N-delocalised)
amine of same nature due to more compact structure.
Amidines are more basic than 2° amine because its conjugate acid is resonance stabilised.
* Guanidines are further more basic than amidine because its conjugate acid has more resonating structure.
* Aniline is less basic than ammonia because lone pair of electrons present at N atom in aniline conjugate with π electrons of benzene ring and, as a result, electron density at N atom is decreased and basic strength is also decreased.
Basic Strength Order
Basic strength H accepting tendency ∝ lone pair donating tendency.
l.p is involved in No resonance of l.p resonance maximum basic lp is involved in No resonance of lp resonance maximum basic III 3 NH CH
Lone pair is involved in resonance and stabilised by + I of CH 3 group.
Basic order: II > III > I
Resonance stabilised Localised lp on more EN Localised lp on less EN
Basic order: III > II > I
More resonance of l.p Less resonance of less basic l.p more basic
More resonance of lp Less resonance of less basic lp more basic
Basic character: I < II
Basic character : I < II
l.p on more E.N, More - I of oxygen minimum basic lp on more EN More –I of oxygen minimum basic
(SIP – Steric inhibition of protonation)
Note:
SIR means steric inhibition of resonance
4) Steric Effect: Operative at ortho position.
* Larger the ortho substituents, greater will be ortho effect and more acidic will be compound
Acidic strength order is
Basic strength order: III > IV > II > I
In I and II lone pair localised due to steric inhibition of resonance. In III lone pair delocalised
Acidic Strength of Substituted Benzoic Acid
It depends upon
1) Inductive Effect: Operative at all positions
2) Mesomeric effect: Operative at ortho and para position
3) Hyperconjugation: Operative at ortho and para position
Ortho effect: Ortho effect occurs when a group present in the ortho position with respect to carboxylic group creates steric strain resulting in rotation of the carboxylic group and shifting it out of plane of the benzene ring. Thus the carboxylic group can no longer participate in ring resonance and thereby the acidity increases as delocalisation of negative charge on the conjugate base of the acid is improved. It is called Ortho effect.
COOH COOH COOH COOH COOH
OCH3 H CH3
Considering reverse hyperconjugation for above example,
Ortho effect III COOH 2 NO IV COOH - I and - M
Acidic order: I > III > II > IV
7) I COOH 2 NO M -I II COOH I Cl III COOH IV COOH 3 CH +I + H.C V COOH OH +M - I I COOH 2 NO M -I II COOH I Cl III COOH IV COOH 3 CH +I + H.C V COOH OH +M - I
Acidic order: I > II > III > IV > V
8) I COOH 3 OCH II COOH 3 OCH III OCH3 COOH IV COOH
Ortho effect III OCH3 COOH IV COOH - I and + M
Acidic order: I > II > IV > III 9) I COOH CH3 II COOH CH3
Ortho effect + Ieffect III COOH CH3 IV COOH + I and + R
Acidic order: I > IV > II > III I COOH X II COOH X IV COOH X III COOH
Acidic order: I > II > III > IV
Halo substituted phenol: Chlorine group attached to benzene ring as well as trichloromethyl group attached to benzene ring will show -I effect only.
Acidic strength:
Acidic strength of phenol having different substituents at para and meta phenol:
Stability of Carboanions
Stability order: II > I > III
Resonance stabilised No resonance
Acidic character
Stability order: II > I > III
Stability Orders of Ions
EN atom with (More EN atom with incomplete octet) incomplete octet)
(–ve charge on more (–ve charge EN oxygen atom) on less EN)
9.7.6 Fission of A Covalent Bond
The breaking of covalent bond can take place in two different ways. The bond breaking generally depends upon the nature of bond, nature of attacking reagent, and conditions of the reaction.
Homolytic Fission
During homolytic fission, each fragment formed gets one electron from the shared pair of electrons. In homolytic fission, single electron movement is shown by half headed curved arrow (fish hook). Homolytic fission is called symmetrical bond fission or bond cleavage. Such cleavage results in the formation of neutral species, which contains an unpaired electron. So, the species is called free radical. Free radicals are very reactive. Homolysis is favoured by conditions such as non-polar nature of the bond, high temperature, UV radiation, vapour phase, and presence of organic peroxides or catalysts.
stabilised by solvation. The energy required for heterolysis is more than the corresponding homolysis.
Reactions involving homolytic cleavage are called radical, or homopolar, or non-polar reactions. Reactions involving heterolytic cleavage are called ionic, or heteropolar, or polar reactions.
more electronegative than B
more electronegative than A
9.7.7 Reaction Intermediates
Short lived fragments called reaction intermediates result from homolytic and heterolytic bond fission. They are produced during the reaction but are lost before the reaction is completed.
Carbocations
Carbocation may be defined as a group of atoms that contains a carbon atom bearing positive charge and having only six electrons in its valence shell. These are formed by the heterolytic cleavage and are also known as carbonium ions.
Heterolytic Fission
During heterolytic fission, both the shared electrons of the covalent bond are taken away by one of the fragments. It results in the formation of charged species. Heterolysis, the unsymmetrical bond fission, is favoured by the polar nature of bond and polar solvents, like water or alcohol, and it is influenced by the presence of ions due to acid and base catalyst. Heterolysis is favoured by polar solvent because ions formed in the heterolysis are
The positively charged carbon atom in the carbocation is sp2 hybridised. This part of the carbocation is planar and the vacant p orbital is perpendicular to this plane. The shape of methyl carbocation is trigonal planar.
Order of stability of alkyl carbocations is:
30 > 20 > 10 > CH3.
This can be explained on the basis of dispersal of positive charge by hyperconjugation. The stability order of carbocations can also be explained by +I effect of alkyl groups.
Higher the number of alkyl groups, higher is the dispersal of the positive charge and, therefore, greater is the stability of carbocation.
■ Stability of benzyl, allyl, and tert alkyl carbocations are almost same in solutions. Their stability, in fact, cannot be compared.
■ t-butyl carbocation is slightly more stable than benzyl primary carbocation.
Rearrangement of Carbocation
A carbocation may involve rearrangement to form more stable carbocation. The rearrangement of carbocation may follow either by 1, 2-hydride ion shift or by 1, 2-alkyl ion shift. (a) 1, 2 - hydride ion shift
In methyl, it being an electron releasing group, carbocation is formed on secondary carbon.
In NO2, it being an electron withdrawing group, carbocation is formed on primary carbon.
Stability of Different Types of Carbocations in Decreasing Order
Carbanions
Carbanion may be defined as a group of atoms that contains a carbon atom bearing negative charge and having six bonding and two nonbonding electrons in its valence shell. These are formed by the heterolytic cleavage in which shared pair remains with carbon atom.
CC Y + Y +
In carbanions, the carbon carrying negative charge is sp3 hybridised. Its shape is pyramidal. If carbanion is part of conjugated system, then it is sp2 hybridised.
H H H
Methyl carbanion
The stability order is CH 3> 1° > 2° > 3°
Stability of alkyl carbanions can be explained by + I effect of alkyl groups. With increasing the number of electrons releasing alkyl groups at C–, negative charge is intensified and the stability of carbanion decreases.
The stability order is : 3 ° > 2° > 1° > CH3
Stability order of carbanion:
Carbon Free Radicals
A free radical may be defined as an atom or group of atoms which contains an unpaired electron. They are formed by homolysis. If the unpaired electron is on carbon atom, they are called alkyl free radicals. Free radicals are highly reactive due to their unpaired electrons.
Stability of alkyl free radicals can be explained by hyperconjugation. Alkyl free radicals are planar and the central carbon atom is in sp2 hybridisation. The unhybridised 2p orbital of the central carbon atom contains the unpaired electron.
Methyl Free Radicals
Carbenes
Carbenes are neutral electrophiles in which carbon atom has six electrons in the outer shell, out of which two are non-bonding and four are bonding. The carbene, :CH 2 , is reactive and short-lived. Carbon is in sp2 or sp hybridisation. Carbenes are generally produced in situ and can be generated by photolysis of diazomethane or ketene.
i. Carbenes are obtained from carbanions through a -elimination reaction.
ii. The simplest carbene is CH
, which is known as methylene.
iii. Substituted carbenes are simply named derivatives of carbenes. For example:
6 5 C H C H
R C R
2 C C l
Phenyl carbene
Dialkyl carbene
Dichloro carbene
iv. sp 2 hybrid carbene is of two types: sp 2 singlet carbene and sp 2 triplet carbene
sp2-singlet carbene sp2-triplet carbene
sp 2 -triplet carbene is more stable than sp2-singlet carbene
v. sp hybrid carbene is always a triplet carbene.
R – C – R
vi. CH •• 2 can exist in singlet as well as in triplet states, where triplet form is more stable.
vii. Singlet CX 2 •• is more stable than triplet CX2. The decreasing order of stability of different types of singlet carbenes are as follows (X is halogen):
2 2 2 2 CF CC CBr CI l (explained on the basis of back bonding)
viii. Reimer–Tiemann reaction and carbyl amine reaction involve the formation of carbenes as intermediate and proceed via carbene mechanism.
Calculation of spin multiplicity:
Spin multiplicity = 21 s
s = Sum of spin quantum number
1. sp2 triplet carbene – 22 1 2 13
2. sp2 singlet carbene –2 1 2 1 2 11 +- += +
Nitrenes
A neutral monovalent species on nitrogen with two pairs of non-bonding electrons on it is called a nitrene.
There is a possibility of two spin states for nitrene, depending on whether the two nonbonding electrons (the normal nitrogen lone
pair remains paired) have their spins paired or parallel R N . These two are lone pair of electrons.
In general, nitrenes obey Hund’s rule and the ground state triplet has two degenerate sp orbitals, containing a single electron each.
R N sp - Triplet nitrene
Alkyl nitrene Aryl nitrene Acyl nitrene
Nitrenes are the nitrogen analogues of carbenes. Like carbenes, nitrenes also exist in singlet and triplet states. R N = N = N + R N N N + hv or heat R N +N2
Benzyne
1. 1, 2-didehydrobenzene, C 6 H 4 , and its derivatives are called benzyne or arynes, and the simplest member is benzyne.
2. It is neutral reactive intermediate derived from benzene ring by removing two
substituents from ortho positions—one in the form of electrophile and the other in the form of nucleophile, leaving behind two electrons to be distributed between two orbitals.
3. Benzyne intermediate is aromatic in character.
4. When halobenzene is heated with sodamide, formation of benzyne takes place.
substrate molecule to create centres of high or low electron density is influenced by attacking species. The reagents can be classified into nucleophiles, electrophiles, and free radicals.
Electrophiles and Nucleophiles
Electrophiles: Electrophile means electronseeking species. Electrophiles are the reagents that attack a point of high electron density or negative site.
Electrophiles are electron deficient species. They are either positively charged or neutral, and attack regions of high electron density in substrate molecule. They act as Lewis acids.
Charged Neutral electrophiles electrophiles
H+, Cl+, Br+, I+, SO3 , BF3 , AlCl3
NO2+ , R3C+ FeCl3, ZnCl2, BeCl2
5. i) It behaves as dienophile and gives Diels–Alder reaction with diene.
ii) It reacts with strong nucleophiles like
Other examples:
9.7.8
Types of Reagents
Most of the attacking reagents carry either a positive or a negative charge. The fission of the
NO+, CH N 65 2 +
An electrophilic reagent attacks the electron-rich centre, tha t is, nucleophile.
Nucle ophiles: Nucleophile means nucleusseeking or electron-hating species. Nucleophiles are the reagents that attack a site of low electron density, or positive centres.
Nucleophiles are electron donating species. They are either negatively charged or neutral molecules with a free electron pair to donate. They act as Lewis bases .
Charged Neutral nucleophiles nucleophiles
Cl–, Br–, I– NH RN HR NH •• •• •• 32 2 ,, NH RNHR N 22 ,, RN 3 •• OR RCOO , HO 2 •• • • , RO H
RC CH COCH OH 33 2 ,, RO R
CN ,,NSHHSO , 33
HS RS HR SR 2
,,
Molecules having multiple bonds between similar atoms like C2H4, C2H2 and C6H6 also act as nucleophiles.
During a polar reaction, a nucleophile attacks at an electrophilic centre of the substrate. For example, in alkyl halides, due to the polarity of C–X bond, a partial positive charge is developed on the carbon atom and the nucleophile attacks it. 3 3 H C Br OH CH OH Br
The reactions involving the attack of nucleophiles are known as nucleophilic reactions.
Nucleophilicity
The reactivity of nucleophile is called its nucleophilicity. Nucleophilicity can be compared as given below:
■ A species with a negative charge is a stronger nucleophile than a similar species without a negative charge. Therefore, OH is a stronger nucleophile than HOH and SH is a stronger nucleophile than HSH.
■ Nucleophilicity decreases on going from left to right in the period of periodic table. Therefore,, 3 CH is more nucleophilic than
2 NH and 2 NH is more nucleophilic than F
■ Nucleophilicity increases on going down in the group of the periodic table. Therefore, Nucleophilicity order is
Similarly, SeH SH OH and 3 3 R P R N
■ Bulky group present on nucleophile centre decreases nucleophilicity. 3 CH
in decreasing order
■ In non-polar solvents, like CCl CS 42 , , nucleophilicity order is IBrClF (more reactive is I–)
■ In DMF, DMA, and DMSO (polar aprotic solvents), nucleophilicity order is FClBrI
■ In polar protic solvents (e.g., water), the nucleophilicity order is IBrClF
Differences between nucleophilicity and basicity:
Basicity represents affinity towards H + , and it is thermodynamic factor.
Nucleophilicity represents affinity for carbon atom, and it is kinetic factor.
Observed order of nucleophilicity:
RS HS CN ICHO OH 3 >>>>>
NBrPhO CH COO 3 3
NH Cl FNOH O 2 33 >>>> -
Fugacity or leaving group ability: Weaker the base, greater is the fugacity:
CF SO CF COOHCOO 33 3
PhCOOCHCOO PhO 3 IBrClF
9.7.9 Types of Reactions
The reactions of organic compounds can be classified into four main types:
i) Substitution
ii) Addition
iii) Elimination
iv) Rearrangement
Substitution Reactions
In substitution reactions, an atom or a group attached to a carbon atom in a substrate molecule is replaced by another atom or group. The product formed is called substitution product. Depending upon the nature of the attacking species, these may be further classified as:
i) Free-radical substitution
ii) Electrophilic substitution
iii) Nucleophilic substitution
Free Radical Substitution Reactions
Substitution reactions that are brought about by free radical are called free-radical substitution reactions.
Halogenation of Alkanes
When an alkane is treated with a halogen (Cl2 or Br2) in the presence of heat or light, a mixture of mono and polyhaloalkanes is formed by successive halogenation. For example,
CH Cl heat or 42 light
CH3Cl+ CH2Cl2+CHCl3 + CCl4 + HCl
If excess of alkanes is used in the reaction, mainly monohaloalkane is formed. For example,
CH excess Cl CH Cl HCl heat or hv 42 3 ()
This is a free-radical chain reaction. A chain reaction consists of three kinds of steps,
i) Initiation
ii) Propagation
iii) Termination
Electrophilic Substitution Reactions
Substitution reactions that are brought about by electrophiles are called electrohilic substitution reactions.
Nitration of Benzene
Nucleophilic Substitution Reactions
Substitution reactions affected by nucleophiles are called nucleophilic substitution reactions. + C L Nu C + : L
Nucleophile Substrate Nu : Substitution product Leaving group
CH3CH2Br + NaOH water CH3CH2OH + NaBr
Addition Reactions
In ad dition reactions, the attacking reagent adds on to the substrate molecule without elimination. In this process, a C ≡ C triple bond may be converted into C = C double bond or C – C single bonds and a C = C double bond is converted into C – C single bonds.
The following are different types of addition reactions
i) Electrophilic addition
ii) Free-radical addition
iii) Nucleophilic addition
Electrophilic Addition Reactions
An addition reaction that is initiated by an electrophile is called an electrophilic addition reaction. This type of reaction occurs with C=C and C ≡ C because these
are nucleophiles (electron donors). Addition of hydrogen halides and halogens to alkenes and alkynes are well k nown examples of electrophilic addition.
Addition of hydrogen halides to alkenes:
Alkenes undergo electrophilic addition with hydrogen halides to form alkyl halides. For example, + HBr + Br Br
Major product Minor product
Addition of halogens to alkenes:
Alkenes (nucleophiles) undergo addition with halogenes (electrophiles) to form vicinal dihalides.
CH3CH2CH = CHCH3 + Br2
Pent-2-ene
CH3CH2CHBrCHBrCH3
Free-radical Addition Reactions
An addition reaction that is initiated by a free radial is called a free-radical addition reaction.
Addition of HBr to alkene in presence of a peroxide proceeds by a free-radical mechanism and gives an alkyl bromide with anti-Markovnikov orientation.
CH CH CH peroxide 32 +HBr dibenzoyl
CH3CHBrCH3 + CH3CH2CH2Br 4 % 96 %
Anti-Markovnikov product
Nucleophilic Addition Reactions
An addition reaction that is initiated by a nucleophile is called a nucleophilic addition reaction. An isolated C=C bond acts as a nucleophile and, hence, does not undergo nucleophilic addition. But a C=O bond of an aldehyde or a ketone acts as an electrophile, since its carbon and oxygen atoms bear a small positive and negative charge,
respectively, due to resonance in carbonyl group C O C O +. The carbon atom of the carbonyl group is electrophilic. For this reason, nuclophiles attack the carbon of aldehydic and ketonic carbonyl groups and, eventually, lead to nucleophilic additions.
Elimination Reactions
The reverse of addition reaction is elimination reaction. Depending upon the relative positions of the atoms or groups eliminated, these are classified as a- or b- or g-elimination reactions.
a Elimination:
In an a -elimination (1,1-elimination), both groups are lost from the same atom of the substrate to give a carbene. For example,
b Elimination:
In a b -elimination (1,2-elimination), two atoms or groups are lost from adjacent atoms of the substrate so that a bond is formed between them. For example,
g Elimination:
In g -elimination (1, 3-elimination), the two groups are lost from the atoms at relative positions 1, 3 and a three-membered ring is formed. For example,
2,3-Dibromopentane
Na HO H O CH2 CH2 Cl 2- chloroethanol →
E 1 reactions (Elimination unimolecular reaction):
H2O Na O CH2 Cl CH2 O CH2 CH2 NaCl Oxirane → +
Na HO H O CH2 CH2 Cl 2- chloroethanol →
H2O Na O CH2 Cl CH2 O CH2 CH2 NaCl Oxirane →
3 2 2 CH CH CH
2 2 CH C H
3 2 2 3 CH CH CH CH → 2 2 C CH CH l
3 2 CH CH CH CH
3 2 2 2 CH CH CH CH CH 2 CH ( elimination ) γ Cl 2 CH 2 2 CH CH ( elimination ) δ
Mo st common elimination reactions are those in which the atoms or groups from two adjacent carbon atoms in the substrate molecule are removed and multiple bond is formed. In this process, two σ bonds are lost and a new p bond is formed, i.e., state of hybridisation of carbon atom changes from sp3 to sp2 or sp2 to sp.
CH3 – CH2X + KOH alcohol
CH2 = CH2 + KX + H2O
CH3CH2CH2OH conc HSO C 24 1700
CH3 – CH = CH2 + H2O
CH2Br – CH2Br + Zn heat alcohol
CH2 = CH2 + ZnBr2
■ As name indicates, it is first order reaction, i.e., rate of reaction depends upon concentration of only one of the substrates.
■ It proceeds via carbonium ion intermediate. Hence, rearrangement takes place.
■ Reaction proceeds via carbonium ion. Hence, decreasing order of reactivity of substrate will be 3�>2�>1�.
■ Less polar protic will be more favourable substrate.
■ The H+ ion to be eliminated is present at adjacent carbon atom of carbonium ion does not need any specific orientation but will follow Saytzeff rule, according to which H + will be eliminated from that adjacent carbon, which has less number of hydrogen atoms, as more and more alkylated carbon=carbon double bond alkene will be more stable.
In dehydration of alcohols, removal of H2O occurs.
Reagent used is - conc.
HSOorH PO 24 34 // DD
In this, elimination of proton occurs according to Saytzeff rule -β-elimination.
favourable substrate will be tertiary alkyl halide. Thus, decreasing order of reactivity of substrate is 3� > 2� > 1�.
Alc. KOH used as reagent. In E2 mechanims, no rearrangment, occurs.
E 2 reactions (Elimination bimolecular reactions):
■ It is 2nd order reaction, i.e., rate of reaction depends upon concentration of substrate as well as base.
■ Eliminated proton needs specific orientation with respect to leaving group, i.e., it must be anti-parallel to leaving group as transition state formed in this condition will be more stable to favour E
(around Repulsion)
(T S= 5 atoms lie in same plane)
■ Since more and more alkylated carboncarbon double bond alkene is more stable,
this group shown, is so large that does not allow backside attack of base.
E1 the most preferred
Neopentylchloride
Some other examples of E 2 reaction:
TEST YOURSELF
1. +I effect is maximum in (1) –CH(CH3)2 (2) –CH2CH3
(3) –CH3
(4) –C(CH3)3
2. The correct decreasing order of –I effect is represented in
(1) −F > −Cl > −NO2 > −OH
(2) −NO2 > −F > −Cl > −OH
(3) −CN > −CHO > −F > −NH2
(4) −NO2 > −SO3H > −COOH > −CHO
3. Which of the following is the correct order of bond lengths?
(1) C − C > C ≡ C > C = C
(2) C = C > C = N > C = O
(3) C = N > C = O > C = C
(4) C − O > C – C > C - N
4. An example of an electron donating group relative to hydrogen is (1) – NO2 (2) – COOH
(3) – COOR (4) – CH3
Rearrangement Reactions
The reactions involving the migration of atoms or group of atoms from one position to another position within the substrate molecule under suitable conditions are called molecular rearrangement reactions. Generally, in these reactions, less stable organic compounds are rearranged to more stable organic compounds.
Example:
CH3CH2CH2CH3 anhydrous AlCl heat 3
CH3–CH–CH3 CH3
CH3CH2CH2Br AlBr heat 3 CH3CH(Br)CH3
NH CNOH NCONH
Ammonium cyanate 42 2 a
Ure
5. Which of the following substituents has +M (mesomeric) effect?
(1) −CN (2) −CHO (3)−NH2 (4) −NO2
6. Electromeric effect is not observed in the reactions involving (1) alkenes
(2) aldehydes
(3) nitriles
(4) alkanes
7. Which of the following group shows maximum negative inductive effect?
(1) –OCH3 (2) –F
(3)–CN (4) –OH
8. Hyperconjugation is not exhibited by (1) 2 – butene (2) 1 – butene (3) propene (4) ethene
9. Which of the following compounds does not exhibit resonance?
(1) CH3–NO2
(2) CH3CH2CH2CONH2
(3) CH3–CH2–CH2–COO–
(4) CH3–CH2–CH = CHCH2NH2
10. Hyperconjugation involves overlap of which of the following orbitals?
(1) σ−σ (2) σ−p
(3) p−p (4) π−π
11. Number of hyperconjugation structures in isopropyl radical is (1) 3 (2) 6
(3) 9 (4) 12
12. Maximum number of hyperconjugated H-atoms are present in
(1) ( ) 32 3 CHC CH • (2) ( ) 3 3 CHC •
(3) 3 CH • (4) 3 2 CH CH •
13. Which of the following is not an electrophile?
(1) BF 3 (2) AlCl3
(3) ZnCl2 (4) NH3
14. The correct stability order for the following carbon free radicals is
(I) • 3 CH
(II) • 3 3 CH– CH
(III) 3 • 2 H (H C C)
(IV) 3 • (CH)3 C
(1) I > II > III > IV
(2) IV > III > II > I
(3) II > I > III > IV
(4) I > II > IV > III
15. Which of the following orders is correct for the stability of these carbanions?
(I) 32 CH – CH
(II) 2 CHCH =
(III) HCC ≡
(1) I > II > III
(2) III > II > I
(3) II > I > III
(4) III > I > II
16. Out of the following, the one containing only nucleophiles is
(1) AlCl3, BF3, NH3
(2) NH3, CN–, CH3OH
(3) AlCl3, NH2–, H2O (4) RNH2, CX2, H–
17. Which of the following statements is incorrect?
(1) :CF2 is more stable than : CCl 2.
(2) :CCl2 is less stable than :CBr 2
(3) Singlet :CH2 is less stable than triplet :CH2
(4) Singlet :CH2 has planar geometry.
18. The decreasing order of nucleophilicity among the nucleophiles is (a) 3 CHCO || O
(b) CH3O–(c) CN–(d) (1) (a), (b), (c), (d) (2) (d), (c), (b), (a)
(3) (b), (c), (a), (d) (4) (c), (b), (a), (d)
19. Due to the presence of an unpaired electron, free radicals are (1) chemically reactive (2) chemically inactive (3) anions
(4) cations
20. Which of the following is an ambident nucleophile?
I) HSO3
II) CN–
III) NO2
(1) I and II only (2) II and III only
(3) I and III only (4) I, II, and III
21. The hybridisation of carbon in methyl carbocation is
(1) sp (2) sp2
(3) sp3 (4) sp3d
22. The correct combination is
(i) HBr 3233 CH-CH=CHCH-CHBr-CH→ electrophilic addition
(ii) 3 3 O CHCOCl 66 65 AlCl II CHCH-C-CH →
electrophilic substitution
(iii) 2 Cl 66 666 hv CHCHCl→ free radical addition
(iv) HBr 32 322 per oxide CH-CH=CH CHCHCHBr →
electrophilic addition
(1) i, ii, iii, and iv
(2) Only ii and iii
(3)Only i, ii, and iii
(4) Only ii and iv
23. Which of the following is not a rearrangement reaction?
(1) CH3–CH2–CH2–CH3 3 CH3 AndicAl 33 I CH-CH-CH → (2)
(3)
(4) 22222 || Br Br CH=CH+BrHC-CH →
24. Which of the following is nucleophilic substitution reaction among the following?
(1) OH I H+ 322 33 CH-CH=CH+HOCH-CH-CH →
(2) | 11 OH R-CHO+RgxR-C M H-R →
(3)CH 3–CH 2–CH 2–Br+NH 3 → CH 3–CH 2 –CH2–NH2
(4) | 33 OH CH-CHO+HCNCH-CH-CN →
25. The following reaction is an example of
(1) nucleophilic substitution
(2) electrophilic substitution
(3) electrophilic substitution
(4) nucleophilic addition
26. The most common type of reaction in aromatic compounds is
(1) elimination reaction
(2) addition reaction
(3) electrophilic substitution reaction
(4) rearrangement reaction
Answer Key
(1) 4 (2) 2 (3) 3 (4) 4
(5) 3 (6) 4 (7) 3 (8) 4
(9) 4 (10) 2 (11) 2 (12) 2 (13) 4 (14) 2 (15) 2 (16) 2
(17) 2 (18) 4 (19) 1 (20) 4
(21) 2 (22) 3 (23) 4 (24) 3
(25) 4 (26) 3
9.8 METHODS OF PURIFICATION OF ORGANIC COMPOUNDS
A carbon compound is either isolated from a natural source or synthesised in the laboratory by a suitable process. The compound thus obtained is generally impure and needs to be purified before its structural formula is determined.
Methods for purification of solids: Filtration, crystallisation, fractional crystallisation, sublimation, etc.
Methods for purification of liquids: Simple distillation, fractional distillation, distillation under reduced pressure, steam distillation, extraction with a solvent, etc.
Finally, the purity of a compound is ascertained by determining its melting or boiling point. Most of the pure compounds have sharp melting points and boiling points. New methods of checking the purity of an organic compound are based on different types of chromatographic and spectroscopic techniques.
9.8.1 Sublimation
Certain solids, on heating, change directly from the solid to vapour state and vapours, on cooling, change back into solid form. This process is used for the purification of sublimable solids.
If the compound to be purified has high vapour pressure below its melting point and sublimates readily on heating, and the impurities do not sublimate, the impure compound is taken in a beaker covered with a watch glass and heated on an electric plate or on a Bunsen burner. The compound sublimates and solidifies on the lower surface of the watch glass, as shown in Fig.9.3
If the sublimating substances have low vapour pressure or decompose on heating before sublimation, then sublimation is carried out under low pressure.
Iodine, benzoic acid, naphthalene, camphor, and other sublime substances can be purified.
9.8.2
Crystallisation
Crystallisation is the most common way of purifying organic solids. It is based on the difference in the solubilities of the compound and the impurities in a suitable solvent. The impure compound is dissolved in a solvent in which it is sparingly soluble at room temperature but appreciably soluble at higher temperature. The solution is concentrated to get a nearly saturated solution. On cooling the solution, pure compound crystallises out and is removed by filtration.
If the compound is highly soluble in one solvent and very less soluble in another solvent, crystallisation can be satisfactorily carried out in a mixture of these solvents. Coloured impurities are removed by adsorbing them over activated charcoal. The crystals are separated by filtering under reduced pressure, using Buchner funnel. The crystals are finally dried over sulphuric acid or calcium chloride in vacuum desiccator. Repeated crystallisation becomes necessary for the purification of compounds containing impurities of comparable solubilities.
9.8.3
Fractional Crystallisation
Fractional crystallisation is used for the separation of a mixture of two compounds which are soluble in the same solvent but their solubilities are different. The hot saturated solution of the mixture is allowed to cool when the less soluble component crystallises out earlier than the more soluble component.
The mother liquor is again concentrated and allowed to cool. The process is repeated several times as the crystals obtained in the
Fig. 9.3 Sublimation
beginning are contaminated with the other substance.
9.8.4 Distillation
Distillation is the most important method for purifying organic liquids. Several methods of distillation are used depending mainly on the nature of impurities present in the organic compound.
9.8.5 Simple Distillation
Simple distillation is applied only for liquids that boil without decomposition and are associated with non-volatile impurities and the liquids having sufficient difference in their boiling points. Liquids having different boiling points vapourise at different temperatures. The vapours are cooled and the liquids so formed are collected separately. The liquid mixture is taken in a round bottomed flask and heated carefully. On boiling, the vapours of lower boiling component are formed first. The vapours are condensed by using a condenser and the liquid is collected in a receiver. The vapour of higher boiling component forms later and the liquid can be collected separately, as shown in Fig.9.4 . This method is used to separate liquids, if only their boiling points differ by above 40°C.
Example:
Chloroform (b.p. 334 K) and aniline (b.p. 453 K) are easily separated by this technique.
9.8.6
Fractional Distillation
If the difference in boiling points of two liquids is not much, they can be separated by fractional distillation using a fractionating column, which is a long tube provided with obstructions to the passage of the vapour upwards and to the liquid downwards. In this technique, vapours of a liquid mixture are passed through a fractionating column before condensation, as shown i n Fig.9.5.
Fig.9.5 Experimental setup of fractional distillation
If A and B liquid mixture is subjected to this process (boiling point: A > B), vapours of the liquid with higher boiling point, A, condense before the vapours of the liquid with lower boiling point, B. Moreover, condensation being exothermic, the energy liberated is useful to keep the vapours of B still in vapour state. The vapours rising up in the fractionating column become richer in more volatile component. By the time the vapours reach to the top of the fractionating column, these are rich in the more volatile component, B. Fractional distillation is of great importance in refining of petroleum, distillation of coal tar, manufacture of alcohol, and many other industrial processes.
Fig.9.4 Simple distillation
Fractionating columns are available in various sizes and designs, as shown in Fig.9.6.
Examples:
In the sugar industry, the concentration of sugarcane juice, and in the soap industry, the separation of glycerol from the spent-lye are carried out by this technique.
9.8.8 Steam Distillation
Fig. 9.6. Various types of fractionating columns
9.8.7 Distillation under Reduced Pressure
This method is used for the purification of liquids that decompose at or below their boiling points and those whose boiling points are very high. In this technique ( Fig.9.7 ), the liquid is made to boil at a temperature much lower than its boiling point by reducing pressure to 0.1 mm of Hg or even less on its surface. It is also called vacuum distillation.
Liquids that are immiscible with water, volatile in steam, and possessing a fairly high vapour pressure are purified by steam distillation. This is based upon the fact that the vapour pressure of a mixture of two immiscible liquids is equal to the sum of the vapour pressures of the individual liquids, i.e., p=p1+p2, where p1 is the vapour pressure due to organic liquid and p2 is the vapour pressure due to water. Sum of vapour pressure of steam and liquid to be purified becomes equal to one atm. Since p1 is lower than p, the organic liquid vapourises at lower temperature than its boiling point. Thus, if one of the substances in the mixture is water and the other is a water insoluble substance, then the mixture will boil close to but below 373 K. A mixture of water and the substance is obtained, which can be separated by using a separating funnel. Aniline is separated by this technique from aniline-water mixture.
Glass beads
Bubble plate coloumn
Simple packed coloumn
Capillary
Fig. 9.7 Distillation under reduced pressure
If there are two liquids A and B,
where W A and W B are the masses of A and B, nA and nB are their number of moles; M A and M B are their molar masses, and P 0 A and P0B are their vapour pressures at a given temperature.
A mixture of aniline (b.p. 453 K, with decomposition) and water (b.p. 373 K), under normal atmospheric pressure, boils at 371 K. At this temperature, the vapour pressure of water is 717 mm and that of aniline is 43 mm and, therefore, the total vapour pressure is equal to 760 mm. Thus, in steam distillation, the liquid gets distilled at a temperature lower than its boiling point and any chances of decomposition are avoided. The proportion of water and liquid in the mixture that distils over is given by the following relation:
requirement of the solvent is that the compound should be more soluble in it than in water. The solvent should also be immiscible with water. The aqueous solution of the organic compound and a suitable organic solvent, like ether or chloroform, is taken in a separatory funnel, thoroughly shaken for sufficient time, and allowed to stand for some time. The organic solvent is later removed by distillation or by evaporation to get the compound back. The technique of continuous evaporation is employed in cases where the organic compound is less soluble in the organic solvent. The apparatus set up for differential extraction is shown in Fig.9.8.
Here, W1 and W2 stand for the masses of water and the organic liquid that distils over, p1 and p2 represent the vapour pressures of water and the liquid at the distillation temperature, and M is molecular mass of the liquid (molecular mass of water being 18). At 98.5°C; water and aniline have vapour pressures 717 torr and 43 torr. In steam distillation at 98.5°C, the relative masses obtained are
9.8.10
Chromatography
Tsw ett (1906), a botanist, separated chlorophyll, xanthophyll, and other compounds by percolating the vegetable extracts through a column of calcium carbonate. The calcium carbonate column acted as adsorbent.
Nitrobenzene–water mixture can also be separated by this method.
9.8.9 Differential Extraction
The process of isolating an organic compound from its aqueous solution by shaking with a suitable solvent is called differential extraction. It is also called solvent extraction. The basic
Different substances were adsorbed to different extents and, due to this, different coloured bands were obtained at different positions on the column. Tswett named the sy stem of coloured bands ‘chromatogram’ and the method, ‘chromatography’. Calcium carbonate column, being immovable, is called stationary phase.
The solution of vegetable extracts flows dow n the column and is called the mobile phase.
After extraction
Before extraction
organic compound in aqucous layer
organic compound in solvent layer aqucous layer solvent layer
Fig. 9.8 Differential extraction
Chromatography involves the following steps:
■ Adsorption and retention of a mixture of substances on the stationary phase and separation of adsorbed substances by the mobile phase to different distances on the stationary phase
■ Recovery of the substances separated by a continuous flow of the mobile phase and qualitative and quantitative analysis of the eluted substances
Chromatography processes are classified into several types, as given in Table 9.15.
Different types of chromatographic techniques are employed depending upon the nature of stationary and mobile phases. Important among them are adsorption chromatography and partition chromatography.
Column chromatography and thin layer chromatography are two types of adsorption chromatography that depend upon the principle of differential adsorption.
Column Chromatography
Column chromatography involves separation of a mixture over a column of adsorbent (stationary phase) packed in a glass tube. The column is fitted with a stopcock at its lower end. The mixture adsorbed on adsorbent is placed on the top of the adsorbent column packed in a glass tube. A suitable eluent, either a single solvent or a mixture of solvents, is allowed to flow down the column slowly.
Depending upon the degree to which the compounds are adsorbed, complete separation takes place. The most readily adsorbed substances are retained near the top and others come down, accordingly, to various distances in the column, as shown in Fig.9.9.
Solvent
Mixture of compounds (a + b + c) adsorbent (stationary phase)
wool
Thin Layer Chromatography
Thin layer chromatography (TLC) is another type of adsorption chromatography, which involves the separation of the substances of a mixture over a thin layer of an adsorbent coated on glass plate. A thin layer (about 0.2 mm thick) of an adsorbent (silica gel or alumina) is spread over a glass plate of suitable size. The plate is known as thin layer chromatography plate, or chromaplate. The solution of the mixture to be separated is applied as a small spot, about 2 cm above one end of the TLC plate. The glass plate is then placed in a closed jar containing the eluent ( Fig.9.10.).
6.
Glass
Fig. 9.9
Table 9.15 Cl assification of chromatography process
Fig.9.10 Thin layer chromatography, chromatogram being developed
As the eluent rises up the plate, the components of the mixture move up along with the eluent to different distances, depending on their degree of adsorption, and separation takes place (Fig.9.11.). The relative adsorption of each component of the mixture is expressed in terms of its retardation factor or retention factor, called Rf value.
Distancemoved by the substancefrombaseline(x)
R=
Distancemoved f b by the solventfrombaseline(y)
The spots of coloured compounds are visible on TLC plate due to their original colour. The spots of colourless compounds, which are invisible to the eye but fluoresce, can be detected by putting the plate under ultraviolet light. Another detection technique is to place the plate in a covered jar containing a few crystals of iodine. Spots of compounds, which absorb iodine, will show up as brown spots. Sometimes, an appropriate reagent may also be sprayed on the plate. For example, amino acids may be detected by spraying the plate with ninhydrin solution.
9.11
Partition Chromatography
Partition chromatography is based on continuous differential partitioning of components of a mixture between stationary and mobile phases. Paper chromatography is a type of partition chromatography. In paper chromatography, a special quality paper, known as chromatography paper, is used. Chromatography paper contains water trapped in it, which acts as the stationary phase. A strip of chromatography paper spotted at the base with the solution of the mixture is suspended in a suitable solvent or a mixture of solvents. This solvent acts as the mobile phase. The solvent rises up the paper by capillary action and flows over the spot. Paper chromatography is shown in Fig.9.12 with two different shapes of the paper commonly used.
The paper selectively retains different components, according to their differing partition in the two phases. The paper strip so developed is known as chromatogram. The spots of the separated coloured compounds are visible at different heights from the position of initial spot on the chromatogram. The spots of the separated colourless compounds may be observed either under ultraviolet light or by the use of an appropriate spray reagent, as discussed under thin layer chromatography.
Fig.
Thin layer chromatography, developed chromatogram
Fig.9.12 Paper chromatography
TEST YOURSELF
1. Steam volatile and water immiscible substances are separated by (1) fractional distillation
(2) distillation
(3) steam distillation
(4) distillation under reduced pressure
2. Select the correct statement among the following.
(1) Standard enthalpy of formation of H 2O (l) is zero.
(2) Naphthalene sublimes slower than solid CO2.
(3) Enthalpy of vaporisation of acetone is more than that of water.
(4) If water freezes, then the amount of heat given off to the surrounding will be equal to enthalpy of sublimation of water.
3. Retardation factor for four components A, B, C, and D in TLC is 0.8, 0.6, 0.7, and 0.4, respectively. Most adsorbed is
(1) A (2) C
(3) D (4) B
4. Glycerol is recovered from spent lye using (1) simple distillation
(2) distillation under reduced pressure
(3) steam distillation
(4) fractional distillation
5. The principle involved in the purification of solids by crystallisation is (1) difference in boiling points
(2) difference in solubilities
(3) difference in melting points
(4) difference in densities
6. A mixture of o-nitrophenol and p-nitrophenol can best be separated by (1) simple distillation
(2) steam distillation
(3) decantation
(4) fractional distillation
7. A mixture of benzene and toluene can be separated by
(1) crystallisation
(2) solubility
(3) separating funnel
(4) fractional distillation
8. Mixture of camphor and KCl can be separated by (1) evaporation
(2) sublimation
(3) filtration
(4) decantation
9. When an organic compound is present in an aqueous medium and is less soluble in any organic solvent, then it is separated by (1) solvent extraction
(2) distillation
(3) chromatography
(4) sublimation
Answer Key
(1) 3 (2) 2 (3) 3 (4) 2
(5) 2 (6) 2 (7) 4 (8) 2
(9) 1
9.9 QUALITATIVE ANALYSIS
The essential element of all organic compounds is carbon. Hydrogen is the element that commonly occurs in organic compounds. Oxygen, nitrogen, halogens, sulphur, and, in a few cases, phosphorus, boron, and metals, are also present in organic compounds.
9.9.1 Detection of Carbon and Hydrogen
Both the elements, carbon and hydrogen, can be tested together. These are detected by heating dry and powdered organic compound with freshly ignited cupric oxide. Carbon and hydrogen present in the organic compound are oxidised to carbon dioxide and water, respectively.
C + 2CuO → 2Cu + CO2↑ 2H + CuO → Cu + H2O
9: Organic Chemistry – Some Basic Principles and Techniques
Carbon dioxide is tested with lime water, which develops turbidity, and water is tested with anhydrous copper sulphate, which turns blue, as shown in Fig.9.13.
CO2 + Ca(OH)2 → CaCO3↓+ H2O
5H2O + CuSO4 → CuSO4.5H2O
White Blue
Dry CuO + organic substance
Water drops anhydrous CuSO4 turns blue
CO2 turns lime water milky
Fig. 9.13 Detection of carbon and hydrogen
If the compound to be analysed is a volatile liquid or a gas, then the vapour of the liquid or the gas itself is passed through heated copper oxide and the liberated gases are tested for carbon dioxide and water.
Detection
of Other Elements
Nitrogen, sulphur, and halogens present in an organic compound are detected by Lassaigne’s test. The elements present in the compound are converted from covalent form into the ionic form by fusing the compound with sodium metal. Cyanide, sulphide, and halide of sodium so formed on fusion with sodium are extracted from the fused mass by boiling it with distilled water. The extract obtained on filtration is known as sodium fusion extract.
Na + C + N NaCN
2Na + S Na2S
Na + C + N + S → NaSCN
Na + X NaX [X = Cl (or) Br (or) I]
9.9.2 Detection of Nitrogen
The sodium fusion extract, (Lassaigne’s extract) is boiled with ferrous sulphate and
then acidified with conc.H2SO4. On heating with conc. H2SO4, some Fe(II) ions are oxidised to Fe(III) ions, which react with sodium hexacyanoferrate(II) to produce iron (III) hexacyanoferrate (II), which is Prussian blue in colour.
The formulation of Prussian blue or green colour also confirms the presence of nitrogen. This test fails in case of diazo compounds.
6CN– + Fe2+→ [Fe(CN)6]4–
3[Fe(CN)6]4-+ 4Fe3+ xH O 2
Fe4[Fe(CN)6]3. xH2O
9.9.3 Detection of Sulphur
Lead acetate test: The sodium fusion extract is acidified with acetic acid and lead acetate solution is added. A black precipitate of lead sulphide indicates the presence of sulphur.
Na2S+(CH3COO)2Pb → PbS ↓ +2CH3COONa
Sodium nitroprusside test: The sodium fusion extract is treated with freshly prepared solution of sodium nitroprusside. Appearance of a purple (violet) colour indicates the presence of sulphur.
SFeCNNOFeCNNOS 2 5 2 5 4 () ()
9.9.4 Detection of Nitrogen and Sulphur
Detection of nitrogen and sulphur when present together: The sodium fusion extract is acidified with dil. HCl and then a few millilitres of ferric chloride solution is added. A bloodred colour confirms the presence of nitrogen and sulphur. This is due to the formation of ferric thiocyanamate complex.
Na + C + N + S → NaSCN Fe+3 + SCN– →[Fe(SCN)2+ (blood-red colour)
Precaution: If sodium fusion is carried out with excess of sodium, the thiocyanate decomposes to yield cyanide and sulphide. These ions give
their usual tests. Blood-red colour with KSCN is not obtained but individual tests of nitrogen and sulphur are positive.
NaSCN + 2Na → NaCN + Na 2S
9.9.5 Detection of Halogens
The sodium fusion extract is boiled with dilute nitric acid and cooled. The fusion extract is divided into three parts and silver nitrate solution is added to each part. Appearance of white precipitate, soluble in ammonium hydroxide, indicates the presence of chlorine. A pale yellow precipitate, sparingly soluble in ammonium hydroxide, indicates bromine, and yell ow precipitate, insoluble in ammonium hydroxide, indicates iodine in the organic compound, respectively.
X– + Ag+ → AgX (X = Cl, Br or I)
Significance of conc.HNO 3 : If sulphur or nitrogen is also present in the compound, the sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium formed during Lassaigne’s test. These ions would otherwise interfere with silver nitrate test for halogens.
NaCl + AgNO3 → AgC l ↓ + NaNO3
NaBr + AgNO3 → AgBr↓ + NaNO3
NaI + AgNO3 → AgI ↓ + NaNO3
9.9.6 Detection of Phosphorus
The organic compound is heated with an oxidising agent (sodium peroxide). The phosphorus present in the compound is oxidised to phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate (NH4)2MoO4. A yellow colouration or precipitate of ammonium phosphomolybdate indicates the presence of phosphorus.
Na3PO4 + 3HNO3→ H3PO4 + 3NaNO3 H3PO4 + 12 (NH4)2MoO4 + 21HNO3 → (NH4)3PO4.12MoO3+ 21NH4NO3+12H2O yellow
9.9.7
Detection of Oxygen
There is no satisfactory qualitative method for the detection of oxygen. However, its presence can be inferred indirectly. By knowing the functional groups, like alcohol, aldehyde, ketone, carboxylic acid, ester, amide, etc., which contain oxygen, the presence of oxygen is reported.
When organic compound is heated in pure nitrogen atmosphere, if water droplets are formed, it confirm s the presence of oxygen.
TEST YOURSELF
1. During Lassaigne’s test, N, S, and halogens in an organic compound are, respectively, converted into
(1) NaCN, NaS, Na2X
(2) Na2CN, Na2S, Na2X
(3) Na(CN)2, NaS, NaX2
(4) NaCN, Na2S, NaX
2. For which of the following compounds will the Lassaigne’s test of nitrogen fail?
(1) H2NCONH2 (2)H2NNH2.2HCl
(3) C6H5NH2 (4) CH3CONH2
3. The formula of the purple colour formed in Lassaigne’s test for sulphur using sodium nitroprusside is
(1) NaFe[Fe(CN)6]
(2) Na[Cr(NH3)2(NCS)4]
(3) Na2[Fe(CN)5 (NO)]
(4) Na4[Fe(CN)5 (NOS)]
4. The Lassaigne’s extract is boiled with dil. HNO3 before testing for halogens because
(1) AgCN is soluble in HNO3
(2) silver halide is soluble in HNO 3
(3) NaCN and Na 2S are decomposed by HNO3
(4) Ag2S is soluble in HNO3
5. Which of the following reagents is used to test for phosphate ion?
(1) Nessler’s reagent
(2) Ammonium molybdate
(3) Baeyer’s reagent
(4) Fenton’s reagent
6. Which of the following will give blood-red colour in Lassaigne’s test on adding FeCl3 solute?
(1) PhNH2
(2) PhNO2
(3)
(4) PhSO3H
7. In Lassaigne’s test, if the organic compound contains nitrogen, the Prussian blue colour is due to the formation of
(1) K4[Fe(CN)6]
(2) Fe4[Fe(CN)6]3
(3) Na3[Fe(CN)6]
(4) Cu2[Fe(CN)6]
8. Sodium fusion extract prepared from an organic compound is acidified with CH 3 COOH and then treated with (CH3COO)2Pb gives a black precipitate. This indicates the presence of_______in organic compound
(1) sulphur (2) chlorine
(3) phosphorous (4) bromine
9. Formula of ammonium phosphomolybdate is
(1) (NH4)3PO4.MoO4
(2) (NH4)2MoO3.PO4
(3) (NH4)3PO4.12MoO3
(4) (NH4)3PO4.6MoO4
10. In Lassaigne’s test, the reason behind the usage of sodium metal is
(1) sodium melts easily and reacts with elements easily
(2) sodium salts are ionic and water soluble
(3) sodium salts are slightly soluble in water
(4) sodium forms covalent compounds with elements of organic compounds
Answer Key
(1) 4 (2) 2 (3) 4 (4) 3
(5) 2 (6) 3 (7) 2 (8) 1
(9) 3 (10) 2
9.10 QUANTITATIVE ANALYSIS
After th e detection of various elements in the organic compound, the next step is the determination of their percentage composition. This is referred to as estimation of elements. The different methods employed for the estimation of various elements are based as follows.
9.10.1 Estimation of Carbon and Hydrogen
Carbon and hydrogen in the organic compound are estimated together in Liebig’s method. A known mass of the organic compound is burnt in the presence of excess of oxygen and dry copper (II) oxide. Oxygen should be free from moisture and carbon dioxide. Carbon and hydrogen in the compound are oxidised to carbon dioxide and water, respectively.
CH x y Ox CO y HO xy 42 22 2
The mass of water produced is determined by passing the mixture through a weighed U-tube containing anhydrous calcium chloride. Carbon dioxide is absorbed in another U-tube containing concentrated solution of potassium hydroxide. These tubes are connected in series, as shown in Fig.9.14.
Sample in platinumboat CuO pellets
Combustion tube
Pure dry oxygen
Furnace
Anhydrous CaCl2 Potash bulb Guard tube
Fig. 9.14 Apparatus for the estimation of carbon and hydrogen
Kn owing the masses of carbon dioxide and water vapours formed and the mass of the organic compound taken, the percentage of carbon and hydrogen can be calculated.
Let the mass of organic compound be ‘m’ g, and mass of water and carbon dioxide produced be ‘m1’ and ‘m2’ g, respectively.
Percentage of carbon:
44 g of CO2 contains 12 g of carbon.
↑ m2 g of CO2 contains 12 2 44 × m g of carbon.
12 2 44 × m g of carbon is present in ‘m’ g of organic compound.
Percentage of carbon in the organic compound:
= 12 100 44 2 ×× × m m
Percentage of hydrogen:
18 g of H 2 O contain 2 g of hydrogen m1 g of H2O contain 2 18 1 × m g of hydrogen 2 18 1 × m g of hydrogen is present in ‘m’ g of organic compound.
Percentage of hydrogen in the organic compound: = 2 100 18 1 ×× × m m
9.10.2 Estimation of Nitrogen
There are two methods for the estimation of nitrogen: (i) Dumas method and (ii) Kjeldahl’s method.
Dumas Method
A known mass of the organic compound is heated strongly with excess of copper oxide in an atmosphere of carbon dioxide, which yields
free nitrogen in addition to carbon dioxide and water. CH Nx y CuO xCO y HO z Nx y Cu xy z 2 2 22 2 2 22 2
If any oxide of nitrogen is produced, it is reduced to nitrogen gas by passing over hot reduced copper gauge. The mixture of gases so produced is collected over an aqueous solution of potassium hydroxide, which absorbs carbon dioxide. Nitrogen is collected in the upper part of the graduated tube. This equipment is shown in Fig.9.15. CuO
Fig. 9.15 Apparatus for estimation of nitrogen by Dumas method
Let the mass of organic compound be ‘m’ g, volume of nitrogen collected be V1 mL, and the room temperature be T 1 K.
Volume of nitrogen at STP = V = PV T 11 1 273 760 × × where P 1 and V 1 are the pressure and volume of dry nitrogen.
P1 = Atmospheric pressure –aqueous tension at that temperature
Calculations:
22400 mL N2 at STP weigh 28 g the weight of V mL N2 at STP = 28 22400 × V g.
Percentage of nitrogen = 28 100 22400 ×× × V m = V m 8 ×
Kjeldahl’s Method
Kjeldahl’s method is simpler and convenient and is largely used for the estimation of nitrogen in food, fertilisers, and drugs.
A known mass of the organic compound is heated with concentrated sulphuric acid. The nitrogen in the organic compound is quantitatively converted into ammonium sulphate. The resulting liquid is then distilled with excess of sodium hydroxide solution and the ammonia evolved is passed into a known excess volume of the standard acid (HCl or H 2 SO 4 ). The acid left unused is estimated by titration with some standard alkali. The amount of acid used against ammonia can thus be known and, from this, the percentage of nitrogen in the compound can be calculated. Apparatus used for the estimation of nitrogen by Kjeldahl’s method is shown in Fig.9.16.
Volume of the NaOH used for unused H2SO4 solution = V1 mL
Number of milli equivalents of H2SO4 used for NH 3 and also for the NaOH solution = NV = number of milli equivalents of NH 3 + number of milli equivalents of NaOH of NaOH, N1V1.
The number of milli equivalents of NH3 = (NV–N1V1) milli equivalents
Fig. 9.16 Apparatus for estimation of nitrogen by Kjeldahl’s method
Organic compound + H2SO4 → (NH4)2SO4 (NH4)2SO4+ 2NaOH → Na2SO4 + 2NH3 +2H2O 2NH3 + H2SO4 → (NH4)2SO4
Calculations:
Let the mass of the organic compound taken = W g
Normality of standard H 2 SO 4 solution taken = N
Volume of the H2SO4 used = V mL
Normality of the standard alkali, NaOH = N1
Weight of NH NVNV10 GEW of NH =−×
(or) ( ) 3 3 11 3
Weight of NH3 = 17(NV –N1V1) × 10–3 g
( GEW of NH3 = 17 3 )
17 g NH3 contains 14 g nitrogen.
17 (NV–N1V1) × 10–3 g NH3 contains = ( ) 3 11 1417NVNV10 17 ×−× g
Percentage of nitrogen in Wg compound = ( ) 3 11 14NVNV10 100 W −× ×
Limitation of Kjeldahl’s method: This method is not applicable to compounds containing nitrogen in nitro group and azo group compounds. It is also not applicable for nitrogen present in ring compounds, like pyridine, quinoline, etc. because these compounds do not produce ammonium sulphate under these compounds.
9.10.3 Halogens
Halogens are estimated by Carius method. A known mass of the organic compound containing halogen is heated with fuming nitric acid and a few crystals of silver nitrate in a sealed tube, called Carius tube, as shown in Fig.9.17 . Carbon and hydrogen present in the compound are oxidised to carbon dioxide and water.
Organic substance
Carius Tube Iron Tube
Furnace HNO3+AgNO3
Fig. 9.17 Carius method for estimation of halogens
The halogen present forms the corresponding silver halide, AgX. It is filtered, washed, dried, and weighed.
Let the mass of the organic substance = W g. Mass of silver halide (AgX) formed = xg
(108 + Atomic mass of X) parts by mass of AgX contain halogen = Atomic mass of X parts
x g by mass of AgX contains halogen
= Atomic mass of X 108 +AtomicmassofX xg ×
Hence, percentage of halogen
Atomic mass of X
Atomic mass of X x W 108 100
Percentage of chlorine = 35 5 143 5 100 ×× x W
Percentage of bromine = 80 188 100 ×× x W
Percentage of iodine = 127 235 100 ×× x W
Carius method does not give satisfactory results in case of estimation of iodine because AgI is slightly soluble in nitric acid. The results of Carius method have not been very accurate even in case of highly halogenated aromatic compounds.
9.10.4 Sulphur
A known mass of an organic compound is heated a in a Carius tube with an oxidising
from organic from
agent, like sodium peroxide or fuming nitric acid. The sulphur present in the compound is quantitatively oxidised to sulphuric acid. It is precipitated as barium sulphate by adding excess of barium chloride solution in water. The precipitate is filtered, washed, dried, and weighed. From the mass of barium sulphate formed, the percentage of sulphur can be calculated. compound 3 HNO
+ 2HCl
Calculations:
Let the mass of organic substance = W g
Mass of barium sulphate formed = x g
233 g of BaSO4 contain 32 g of sulphur x g of BaSO4 contains 32 233 × x g of sulphur
Percentage of sulphur = 32 100 233 ×× × x W
9.10.5 Phosphorus
A known mass of the organic compound is heated with fuming nitric acid. The phosphorus present is oxidised to phosphoric acid. It is precipitated as ammonium phosphomolybdate, (NH4)3PO4.12MoO3, by adding ammonia and ammonium molybdate.
Alternatively, phosphoric acid may be precipitated as magnesium ammonium phosphate by adding magnesia mixture, which on ignition yields magnesium pyrophosphate, Mg2P2O7.
Magnesia mixture is obtained by dissolving 100 g of MgCl2. 6H2O and 100 g of NH4Cl in water and then adding 50 mL of conc. NH4OH and diluting the solution to 1000 mL.
from organic from 3HNO 3 4 P 3H 4(O) H PO + + →
compound 3 HNO
H3PO4 + Magnesia mixture → MgNH4PO4
2MgNH4PO4 → Mg2P2O7 + 2NH3 + H2O
Let the mass of the organic compound = W g
Mass of Mg2 P2O7 obtained = x g
222 g of Mg 2 P 2 O 7 contains 62 g of phosphorus x g of Mg2P2O7 contains 62 222 × x g of phosphorus.
Percentage of phosphorus = 62 100 222 ×× × x W (or)
Mass of (NH4)3PO4.12MoO3 obtained = y g
1877 g of (NH4)3 PO4.12 MoO3 contain 31 g of phosphorus
7y g of (NH4)3 PO4. 12 MoO3 contain 31 1877 × y g of phosphorus
Percentage of phosphorus = 31 100 1877 ×× × y W
9.10.6 Estimation of Oxygen
The percentage of oxygen in an organic compound is usually found by difference between the total percentage composition (100) and the sum of the percentages of all other elements. The estimation of oxygen can also be carried out directly, as follows.
A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over red-hot coke, when all the oxygen is converted to carbon monoxide. This mixture is passed through warm iodine pentoxide, when carbon monoxide is oxidised to carbon dioxide producing iodine.
Organic compound heat
O2 + Other gaseous products
2C + O2 1373 K 2CO
I2O5 + 5CO → I2 + 5CO2
The percentage of oxygen can be derived from the amount of carbon dioxide or iodine produced. Let the mass of organic compound = Wg
Mass of carbon dioxide = x g
44 g of carbon dioxide contains 32 g of oxygen but half the oxygen is contributed from I2O5.
x g of carbo n dioxide contains 16 44 × x g of oxygen.
Percentage of oxygen = 16 100 44 ×× × x W
9.10.7 Determination of Molecular Mass of Carboxylic Acids (Silver Salt Method)
RCOOH →RCOONH4 → RCOOAg → Ag↓
Molecular mass of acid =
Basicity
Mass of silver salt
Mass of silver 108 107
9.10.8 Determination of Molecular Mass of Organic Bases
(Platinichloride method)
2RNH2 + H2PtCl6 → (RNH3)2PtCl6 (or) (RNH2)2H2PtCl6
Molecular mass of base = 1 2 × Acidity
Mass of salt
Mass of platinum 195 410
TEST YOURSELF
1. The reaction between ethylene and bromine is an example of (1) Electrophilic addition
(2) Electrophilic substitution
(3) Nucleophilic addition
(4) Nuclear substitution
2. 0.15 g of an organic compound, on combustion, gave 0.44 g of CO 2 and 0.27 g of H2O. The molecular formula of organic compound can be
(1) C2H4 (2) C2H6
(3) C2H2 (4) C4H8
3. Dumas method is used for the estimation of which one of the following elements?
(1) Halogen
(2) Nitrogen
(3) Sulphur
(4) Phosphorous
4. In sulphur estimation, 3.2 grams of an organic compound gives 2.33 grams of BaSO4. Then, the percentage of sulphur in the organic compound is
(1) 15% (2) 10%
(3) 20% (4) 25%
5. In the Dumas method for the estimation of nitrogen in an organic compound, nitrogen is determined in the form of (1) gaseous nitrogen
(2) gaseous ammonia
(3) ammonium sulphate
(4) sodium cyanide
6. In Liebig’s method for the estimation of C and H, if the compound also contains N, which of the following is kept near the exit of the combustion tube?
(1) Silver wire
(2) PbCrO4
(3) Both 1 and 2
(4) Cu gauge
7. The catalyst used in Kjeldahl’s method for the estimation of nitrogen is
(1) Copper
(2) Magnesium
(3) Mercury
(4) Sodium
8. 0.5 g of organic compound in Kjeldhal’s method liberated ammonia, which neutralised 60 ml of 0.1N H2SO4 solution. The percentage of nitrogen in the compound is
(1) 1.68
(2) 16.8
(3) 33.6
(4) 8.4
9. The substance used in the estimation of phosphorous is (1) conc. H2SO4
(2) fuming sulphuric acid and MgCl 2
(3) conc. HNO3; NaOH
(4) fuming nitric acid and magnesia mixture
10. In the Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is (at mass Ag = 108; Br = 80)
(1) 36
(2) 48
(3) 60
(4)24
11. 0.25 g of an organic compound gave 22.4 mL of N 2 at STP by Dumas method. The percentage of nitrogen in the compound is (1) 15%
(2) 28%
(3) 11.2%
(4) 14%
Answer Key
(1) 1 (2) 2 (3) 2 (4) 2
(5) 1 (6) 4 (7) 3 (8) 2
(9) 4 (10) 4 (11) 3
CHAPTER REVIEW
■ According to Berzelius, organic compounds are from living organisms and these could be produced by ‘vital force’.
■ By heating ammonium cyanate, urea, the first organic compound, was prepared by Wohler, which ruled out vital force theory.
■ Carbon has high catenation capacity and is usually tetravalent.
■ If all the four valencies of a carbon are satisfied by single bonds, it is called saturated carbon and carbon atom is in sp3 hybridisation.
■ If a carbon forms one pi bond, it is in sp 2 hybridisation and if a carbon forms two pi bonds, it is in ‘sp’ hybridisation.
■ Organic compounds are mainly classified as acyclic and cyclic compounds.
■ Cyclic compounds are classified as homocyclic and heterocyclic.
■ Homocyclic compounds may be alicyclic or aromatic.
■ Heterocyclic compounds may be aromatic or non-aromatic.
■ Acyclic compounds are classified as saturated and unsaturated.
■ Alkenes, alkynes, and their derivatives are examples of unsaturated hydrocarbons.
■ Furan, pyrrole, thiophene, pyridine, quinoline, etc. are heterocyclic. They are aromatic in nature.
■ Planar ring compounds with conjugate double bonds and obeying Huckel’s rule of (4n + 2) p electrons, are aromatic in nature.
■ An atom or a group of atoms in an organic substance, which is responsible for the characteristic chemical properties, is called a functional group.
■ Organic compounds belonging to the same homologous series exhibit similar chemical
properties and regular gradation in physical properties and the successive members differ by methylene group.
■ According to IUPAC, prefix(es) + root word + primary suffix + secondary suffix is the order in naming an organic compound.
■ Primary prefix is meant for alicyclic compounds.
■ Secondary prefix tells about the substituents or secondary grade functional groups.
■ Primary suffix tells about saturated or unsaturated compound.
■ Secondary suffix indicates the nature of the functional group.
■ A carbon atom attached to only one carbon atom directly is called primary (10) carbon, two other carbon atoms is secondary (2 0), three is tertiary (30), and four is quaternary (40) carbon.
■ Hydrogen atoms attached to primary, secondary, and tertiary carbon atoms are termed primary, secondary, and tertiary hydrogen atoms, respectively.
■ While naming a carbon compound according to IUPAC, longest continuous chain of carbon atoms is to be selected as the parent chain.
■ The parent chain must include carboncarbon multiple bonds present in the compound.
■ If two equally long chains are possible, the chain with maximum number of side chains is selected as parent chain.
■ The numbering of carbon atoms is done in such a way that the substituted carbon atoms have the lowest possible numbers.
■ The functional group bearing carbon should be given the lowest number or lowest sum rule is to be followed.
■ In case of unsaturated hydrocarbons, the carbon atoms involved in the multiple bond should get the lowest possible number.
■ If a number of groups are attached to the parent chain, the name is given to the compound following the alphabetical order.
■ The order of decreasing priority for functional groups is carboxylic acid > nitrile > aldehyde > ketone > alcohol > amine > double bond > triple bond.
■ The chain terminating groups like - COOH,–CHO,–CN, etc., should always get the number ‘1’ for the carbon atom.
■ When a benzene ring is attached to an alkane with a functional group, it is considered substituent, instead of a parent.
■ Carbon compounds with same molecular formula but different properties are called isomers and that phenomenon is called isomerism.
■ Structural isomerism is due to the difference in the linkage of atoms or groups without any reference to space.
■ Structural isomerism is divided into chain, position, functional group, metamerism, and tautomerism.
■ Stereo isomerism is due to the difference in arrangement of atoms or groups in space.
■ Stereo isomers have same molecular formula but differ in the spatial arrangement of atoms or groups.
■ Chain isomers differ in the nature of the carbon chain (skeleton).
■ Position isomerism is due to the difference in the position of a functional group, multiple bond, or substituent in the same carbon chain.
■ Compounds having same molecular formula but different functional groups are called functional isomers.
■ Alcohols–ethers, aldehydes–ketones, carboxylic acids–esters, etc. are examples of pairs of functional isomers.
■ Metamerism is due to the presence of different alkyl groups attached to the same bivalent functional group.
■ Ketones, secondary amines, ethers, etc. show metamerism.
■ Tautomerism is a special type of functional isomerism, where the isomers exist in equilibrium with each other.
■ Configurational isomerism is further divided into geometrical isomerism and optical isomerism.
■ Geometrical isomerism arises due to restricted rotation about carbon-carbon or carbon-nitrogen double bond.
■ In cis isomer, same or similar groups are present on the same side of double bond.
■ In trans isomer, similar groups are present on the opposite side of the double bond.
■ Unsymmetrical alkenes and their derivatives, cycloalkane dicarboxylic acids, oximes, azobenzene, azoxybenzene, etc., exhibit geometrical isomerism.
■ Generally, geometrical isomers possess similar chemical properties but differ in their physical properties.
■ Cis isomer is more polar than trans isomer; it possesses higher boiling point, lower melting point, greater solubility, higher density, higher refractive index, and high heat of combustion.
■ Cis isomer has lesser stability than the corresponding trans isomer.
■ The light ray which vibrates in a single plane is called plane polarised light.
■ The substances which have the ability of rotating plane polarised light are called optically active.
■ If the rotation is in clockwise direction, it is called dextro rotatory (d-) (or) (+) and in the anti-clockwise, laevo rotatory (l -) (or) (–).
■ A carbon atom that is bonded to four different atoms or groups is called an asymmetric carbon atom.
■ The electronic displacements in covalent bonds may occur either due to the presence of some atom or group in the molecule or under the influence of attacking reagent.
■ Inductive effect is the permanent displacement of electrons along a carbon chain when some atom or group of atoms with different electronegativity than carbon is attached to carbon chain.
■ The atom or group that has more power to attract (withdraw) electrons in comparison to hydrogen is said to have–I effect.
■ The order of –I effect is: NO2 > CN > SO3H > CHO > CO > COOH > COCl > COOR > CONH2 > F > Cl > Br > I > OR > C 6H5.
■ The atom or group that has less power to attract the electrons than hydrogen is said to have + I effect.
■ The order of +I effect is: (CH 3)C > (CH3)2 CH > CH3CH2 > CH3.
■ Inductive effect is a permanent effect and it tends to be insignificant beyond the third carbon atom.
■ Acidic nature order of various carboxylic acids and substituted carboxylic acids is HCOOH > CH3COOH > C2H5 COOH > C3H7COOH
■ Phenol is more acidic than water and methanol is more acidic than water.
■ Acidic nature order of some alcohols is CH3OH > CH3CH2OH > (CH3)2CHOH
■ Relative strength of bases can also be explained by inductive effect.
■ R–NH2 > H–NH2 > –NH2
■ Due to steric effect, tertiary amines may not be more basic than secondary amines.
■ Electromeric effect is the complete transfer of the shared pair of pi electrons of a multiple bond to one of the atoms in the presence of the attacking reagent.
■ Electromeric effect is a temporary effect and comes into play instantaneously at the demand of the attacking reagent.
■ When the transfer of electrons takes place towards the attacking reagent, it is +E effect, e.g., addition of acids to alkenes.
■ When the transfer of electrons takes place away from the attacking reagent, it is –E effect, e.g., addition of cyanide ion to carbonyl compounds.
■ When both inductive and electromeric effects simultaneously operate, usually, electromeric effect predominates.
■ Mesomeric or resonance effect is a permanent effect involving the transfer of electrons relayed through pi electrons of multiple bonds or a lone pair of electrons and multiple bonds in a chain of carbon atoms in a molecule.
■ Electromeric effect always facilitates the reaction and never inhibits.
■ Groups with +M effect increase the electron density of the rest of the molecule.
Examples: Cl,–Br,–I,–NH 2,–NHR,–NR 2, –OH,–OR,–OCOR,–NHCOR, etc.
■ Groups with –M effect decrease the electron density of the rest of the molecule.
Examples: –COOH,–COOR,–CHO,–COR,–CO–,–CN,–NO2,–SO3H, etc.
■ Resonance effect is called conjugative effect if it is transmitted through whole of the conjugated system.
■ The energy of actual structure of the molecule is lower than that of any of the resonance structures.
■ The difference between the real energy of the resonance hybrid structure and the most stable resonance structure is called resonance energy.
■ Resonance explains the stability of aromatic compounds, some unusual bond lengths in some molecules, behaviour of o- and p- directing and m-directing groups.
■ When alkyl groups are attached to an unsaturated system or a benzene nucleus, the sigma electrons present in –C–H bond of that alkyl group also involve in conjugation, known as hyperconjugation.
■ Hyperconjugation is also called σ – p* conjugation or no bond resonance.
■ Greater the number of methyl groups attached to the doubly bonded carbon atoms, greater is the hyperconjugation and greater is its stability.
■ Stability of alkenes, carbocations, alkyl free radicals, orienting effect of alkyl groups in aromatic ring, unexpected bond lengths in some molecules, etc. can be explained by hyperconjugation.
■ Homolytic fission of a covalent bond leads to the formation of neutral species, which contains an unpaired electron called free radicals.
■ Homolytic fission is favoured by conditions such as non-polar nature of the bond, high temperature, UV radiations, presence of peroxides, etc.
■ Heterolytic fission of a covalent bond leads to the formation of charged species.
■ Hetrolytic fission is favoured by polar nature of the bond, polar solvents, and presence of ions due to acid and base catalyst.
■ Carbocation is a group of atoms that contains a carbon atom bearing positive charge and has only six electrons in its valence shell.
■ The positively charged carbon atom in the carbocation is in sp2 hybridisation. It is trigonal planar in shape.
■ Order of stability of carbocations is: R3C+>R2CH+>RC+H2>C+H3
■ The above stability order can be explained by hyperconjugation and also by + I effect of alkyl groups.
■ Carbanion is a group of atoms that contains a carbon atom bearing negative charge.
■ The negatively charged carbon atom in the carbanion is in sp3 hybridisation. Its shape is pyramidal or tetrahedral with one lone pair.
■ The stability order of carbanions is: CH–3 > CH3CH–2 > (CH3)2CH– > (CH3)3C–
■ If an unpaired electron is present on a carbon atom in a group of atoms, which is formed due to homolysis of a covalent bond, it is called alkyl free radical.
■ Alkyl free radicals are planar and the central carbon atom is in sp2 hybridisation.
■ The unhybridised 2p orbital of the central carbon atom of alkyl free radical contains the unpaired electron.
■ The stability order of alkyl free radicals is: (CH3)3C > (CH3)2 CH > CH3 CH2 > CH3
■ The stability order of alkyl free radicals can be explained by hyperconjugation.
■ Carbenes ( CH2) are neutral species in which carbon atom has six electrons in the outer shell, out of which two constitute a lone pair and two are shared.
■ Carbon atom in carbene is in sp 2 hybridisation.
■ Based on the nature of the attacking site in the substrate, attacking reagents are classified as nucleophiles and electrophiles.
■ Nucleophiles donate a free electron pair to the electron deficient centre of the organic substrate.
■ Nucleophiles are either negatively charged or neutral.
■ Cl–, Br–, I–, NH–2, RNH–, R2N–, OR–, RCOO–, R3C–, CH3COCH2–, OH–, CN–, NH–3, SH–, HSO–3 etc. are charged nucleophiles.
■ Electrophiles are electron deficient and attack the substrate where the electron density is more.
■ Electrophiles are either positively charged or neutral.
■ H +. Cl +, Br +, I +, NO 2 + , R 3C +, NH 4+, NO + , C6H5N2+, etc. are charged electrophiles.
■ SO 3, BF 3, A l Cl 3, FeCl 3, ZnCl 2, BeCl 2, etc. are neutral electrophiles.
■ Nucleophiles act as Lewis bases and electrophiles act as Lewis acids.
■ Ambiphiles are those species that behave like electrophiles and nucleophiles.
Examples: H– O –H, R–O –H, carbon, etc.
■ In substitution reactions, an atom or a group attached to a carbon atom in a substrate molecule is replaced by another atom or group.
■ Substitution reactions are further classified as free radical, electrophilic, or nucleophilic substitution reactions.
■ Alkanes undergo free radical substitution, aromatic compounds undergo electrophilic substitution, and alkyl halides undergo nucleophilic substitution reactions.
■ If the attacking reagent adds on to the substrate molecule without elimination, it is called addition reaction.
■ In addition reactions, a triple bond is converted to double bond and a double bond to single bonds.
■ Addition reactions are also of three types: electrophilic, nucleophilic, and free radical addition reactions.
■ Unsaturated hydrocarbons undergo electrophilic addition and carbonyl compounds undergo nucleophilic addition.
■ In elimination reactions, generally, atoms or groups from adjacent carbon atoms in the substrate molecule are removed and a multiple bond is formed.
■ In elimination reactions, two sigma bonds are lost and a new pi bond is formed.
■ In the preparation of alkenes; dehydration of alcohols, dehydrohalogenation of alkyl halides, and dehalogenation of vicinal dihalides are called elimination reactions.
■ In a molecular rearrangement reaction, the product formed is different from that of the expected. Then new compound is actually the structural isomer of the expected product.
Methods of Purification of Organic Compounds
■ Methods of purification of organic solids are filtration, crystallisation, fractional crystallisation, sublimation, etc.
■ Methods of purification of organic liquids are distillation, solvent extraction, etc.
■ The solids which sublime readily on heating are purified by sublimation.
■ Crystallisation is the most common way of purifying organic solids.
■ If two compounds have different solubilities in the same solvent, they are purified by fractional crystallisation.
■ Distillation is the most important method of purifying organic liquids.
■ If the difference in the boiling points of two liquids is more than 400C, they can be purified by simple distillation, otherwise by fractional distillation.
■ Distillation under reduced pressure is the method used for purification of liquids which decompose at or below their boiling points.
■ Liquids which are immiscible with water, volatile in steam are purified by steam distillation.
■ The process of isolating an organic compound from its aqueous solution by shaking with a suitable solvent is called differential extraction.
■ Chromatography is used to separate mixtures into their components, purify compounds and to test the purity also.
■ Column chromatography and thin layer chromatography are two types of adsorption chromatography which depend upon the principle of differential adsorption.
■ Retardation factor is the ratio between the distance moved by the substance from base line and the distance moved by the solvent from base line.
■ Partition chromatography is based on continuous differential partitioning of components of a mixture between stationary and mobile phases.
Qualitative Analysis
■ By heating organic compound with cupric oxide, both carbon and hydrogen can be detected together.
■ Lassaigne’s test is used for detecting nitrogen, sulphur, and halogens. These elements are converted to sodium cyanide, sulphide, and halide, respectively.
■ Lassaigne’s extract containing organic nitrogen gives Prussian blue colour with ferrous sulphate and concentrated sulphuric acid.
■ Sulphur containing organic compounds give black precipitate with acetic acid and lead acetate solution.
■ Sodium fusion extract gives purple colour with sodium nitroprusside if the organic compound contains sulphur.
■ When sodium fusion extract is acidified with dilute nitric acid and treated with silver nitrate solution, precipitate is formed if halogen is present in the organic compound.
■ If phosphorus is present, yellow colouration or precipitate of ammonium phosphomolybdate is formed with ammonium molybdate.
■ Oxygen present in the organic compound is detected indirectly.
Quantitative Analysis
■ Nitrogen present in the organic compound is estimated by Dumas method or Kjeldahl’s method.
■ In Dumas method, nitrogen gas is liberated and its volume is measured.
■ In Kjeldahl’s method, nitrogen is converted to ammonia and is estimated by back titration
■ Halogens are estimated by Carius method as silver halides.
■ Estimation of sulphur is based on the principle of conversion of sulphur into barium sulphate precipitate.
■ Phosphorus present in the organic compound is precipitated as ammonium phosphomolybdate or magnesium pyrophosphate and then estimated.
■ Oxygen is estimated indirectly or by converting into carbon dioxide.
■ By using CHN, elemental analyser with few mg of sample carbon, hydrogen, and nitrogen present in the compound can be determined in a very short time.
Exercises
JEE MAIN LEVEL
Level - I
General introduction
Single Option Correct MCQs
1. Urea was prepared first time in the laboratory by heating
(1) Ammonium cyanate
(2) Ammonium cyanide
(3) Ammonium isocyanate
(4) Ammonium isocyanide
2. Who proposed the concept of ‘vital force’ for the formation of organic compounds?
(1) Wohler
(2) Kolbe
(3) Berzelius
(4) Berthelot
3. Which organic compound did F. Wohler synthesize to disprove the ‘vital force’ theory?
(1) Methane
(2) Urea
(3) Acetic acid
(4) Ammonium cyanate
4. What was synthesized by Kolbe in 1845?
(1) Methane
(2) Urea
(3) Acetic acid
(4) Ammonium cyanate
5. What did Berthelot synthesize in 1856, proving that organic compounds could be created from inorganic sources?
(1) Methane (2) Urea
(3) Acetic acid (4) Ethanol
Tetra Valence of Carbon: Shapes of Organic Compounds
Single Option Correct MCQs
6. Among the following compounds which does not have more than one type of hybridisation for carbon atoms is
(1) CH3-CH2-CH2-CH3
(2) CH3-CH=CH-CH3
(3) CH2=CH-C≡CH
(4) H-C≡C-CH3
7. Number of sigma bonds in ethane formed by the overlap ofsp3and s orbitals is
(1) 5 (2) 6
(3) 7 (4) 8
8. Ratio of π to σ bonds in benzene is
(1) 1 : 4 (2) 1 : 2
(3) 3 : 1 4) 1 : 6
9. Which of the following reactions does not involve change in hybridisation?
(1) Protonation at oxygen ofCH 2O
(2) Hydride ion addition at carbon ofCH2O
(3) Dissociation ofC-Clbond ofCH 3Cl
(4) Hydrogenation of ethene
10. In which of the following all carbon atoms are of same hybridisation?
(1) 1, 3 – Butadiene
(2) Hexane
(3) Acetylene
(4) All the above
11. In which of the following species is the underlined carbon having sp3 hybridisation?
(1) CH3COOH (2) CH3CH2OH
(3) CH3COCH3 (4) CH2=CH−CH3
12. Number ofσ− and π−bonds in acrylonitrile
CH2=CH-C ≡ N:is ____ .
(1) 3π and 6σ
(2) 7σ and 2π
(3) 7σ and 3π
(4) 2π and 6σ
13. The total number ofπ-bond electrons in the following structure is
(1) 12
(2) 16
(3) 4
(4) 8
Numerical Value Questions
14. Number of sp2 hybridized carbon atoms in benzene is
15. How many carbon atoms 1, 3-butadiene have tetrahedral orientation?
16. Find the number ofspcarbon atoms in benzyne?
Structural representation of Organic compounds
Single Option Correct MCQs
17. In molecular models, bonds are not shown
(1) Frame work model
(2) Ball–and–stick model
(3) Space filling model (4) Wedge and dash
18. The bond–line (________) structural representation of organic compounds, carbon–carbon bonds are drawn in
(1) Parallel (2) Horizontal
(3) Vertical (4) Zig–Zag
19. Number of primary, secondary, tertiary and neo carbons in the following compound respectively are
(1) 4,2,1,1 (2) 5.1.1.1
Numerical Value Questions
20. The maximum number of atoms that are present in the same plane for the following molecule for any conformation is:
HC≡C−C≡C−CH3
Classification of Organic Compounds
Single Option Correct MCQs
21. Heterocyclic compounds are ___________ in nature
(1) Aliphatic
(2) Aromatic
(3) cyclic or Aromatic
(4) Inorganic
22. The alicyclic compound is (1) Cyclohexane
(2) Hexene-2
(3) Pyrrole
(4) Hexane
23. Select the heterocyclic and nonaromatic compound?
(1) Furan
(2) Thiophene
(3) Pyridine
(4) Tetrahydro furan
24. Which among the following is not an example of alicyclic compound?
(1) Cyclohexane (2) Cyclohexene
(3) Acetic acid (4) Cyclobutane
25. Which of the following is acyclic
(1) Neopentane
(2) Azulene
(3) Cresol
(4) Naphthalene
26. Cyclohexene is _________ type of an organic compound
(1) Benzenoid aromatic
(2) Benzenoid nonaromatic
(3) Acyclic
(4) Alicyclic
27. Which of the following forms a homologous series?
(1) Ethane, ethylene, acetylene
(2) Ethane, propane, butanone
(3) Methanal, ethanol, propanoic acid
(4) Butane, 2 – methyl butane, 2, 3 – dimethyl butane
28. Which of the following statements is not true about a homologous series?
(1) Adjacent members of a series differ by a mass of 14
(2) Adjacent members of a series differ by one -CH2 group
(3) Members of a homologous series can be prepared by the same general methods
(4) Members of a homologous series have the same physical and chemical properties
29. Functional group present in sulphonic acids is
(1) SO3H
(2) SO2H
(3) SO4H
(4)
30. Which of the following statements is not true about a homologous series?
(1) Adjacent members of a group differ by a mass of 14.
(2) Adjacent members of a group differ by –CH2 group
(3) Members of a homologous series can be prepared by same general methods.
(4) Members of a homologous series have the same physical and chemical properties.
Numerical Value Questions
31. Find out number of aromatic compounds or ion from following.
32. The sum of all functional groups present in the given compound is
33. a, b, c, d, e are the successive members of homologous series in the increasing order of their Molecular weight. Atomic weight of b is 30 amu, what is the molecular weight of e in amu?
34. Number of functional groups in the below compound
35. How many types of functional groups are present in given compound.
Nomenclature of organic compounds
Single Option Correct MCQs
36. Structure of 4-Methylpent-2-enal is
(1)
(2)
(3)
(4)
37. IUPAC name of ethyl acetylene is (1) Methyl prop-1-yne
(2) But-2-yne
(3) Ethyl but - 1-yne (4) But-1-yne
38. The correct IUPAC name of (C 2H5)4C is (1) Tetraethyl methane
(2) 2-Ethylpentane
(3) 3, 3-Diethylpentane
(4) 2, 2-Diethylpentane
39. IUPAC name of is
(1) 1,1-dimethyl,3-Ethyl cyclohexane
(2) 3-Ethyl 1,1-dimethyl cyclohexane
(3) 1-Ethyl 3,3-dimethyl cyclohexane
(4) 1-Ethyl 5,5-dimethyl cyclohexane
40. The correct IUPAC name of following compound is ________.
(1) 2-methylhex-2-en-4-ol
(2) 2-methyl-4-hydroxyhex-2-ene
(3) 3-hydroxy-5-methylhex-4-ene
(4) 5-methylhex-4-en-3-o1
41. The correct IUPAC name of the following compound is
(1) Hex - 2 - en - 5 - yne
(2) Hex - 5 - ene - 2 - yne
(3) Hex - 1 - yn - 4 - ene
(4) Hex - 4 - en - 1 - yne
42. The IUPAC name of the following compound
(1) 3 - Ethyl - 4 - methyl hexene - 4
(2) 4, 4 - di ethyl - 3 - methyl Butene - 2
(3) 4 - Methyl - 3 - ethyl hexene - 4
(4) 4 - ethyl - 3 - methyl he xene - 2
43. IUPAC name of following hydrocarbon is
(1) 2 – Ethyl – 3, 6 - dimethyl heptane
(2) 2 – Ethyl –2, 6 - diethyl heptane
(3) 2, 5, 6 - Trimethyl octane
(4) 3, 4, 7 - Trimethyl octane
44. The IUPAC name of the following compound is:
(1) 3-Hydroxy-4-methylpentanoic acid
(2) 4-Methyl-3-hydroxypentanoic acid
(3) 4,4-Dimethyl-3-hydroxybutanoic acid
(4) 2-Methyl-3-hy droxypentan-5-oic acid
45. The IUPAC name of the compound is
(1) 1 – Bromo – 3 – methyl pentanone – 2
(2) 5 – Bromo – 3 – methyl pentanone – 4
(3) 4 – Bromo – 3 – ethyl butanone – 3
(4) 1 – Bromo – 3 – Butanone – 2
Numerical Value Questions
46. Number of carbons in the parent chain of
47. The number of carbons present in longest chain
48. The parent chain of given compound contains --------------- number of carbons
49. The position of –OH in the given molecule according to IUPAC numbering
50. In the following, keto group is at position
51. Total number of carbon atoms present in parent chain is:
Isomerism
Single Option Correct MCQs
52. Structures are
(1) Chain isomers
(2) Position isomers
(3) Both chain & position isomers
(4) Not isomers
53. Alcohols are functional isomers of (1) Aldehydes (2) Ethers
(3) Esters (4) Ketones
54. Which of the following pairs of compounds are positional isomers?
(1) n-butyl chloride and sec-butyl chloride
(2) n-butyl chloride and iso-butyl chloride
(3) 1-butene and iso-butene
(4) ethyl alcohol and dimethyl ether
55. Ortho, meta and para dichlorobenzenes are (1) Chain isomers
(2) Positional isomers
(3) Functional isomers
(4) Stereoisomers
56. The compound which shows metamerism is
(1) C4H10O
(2) C5H12
(3) C3H8O
(4) C3H6O
57. Aldehydes are functional isomers of (1) alcohols
(2) ethers
(3) esters
(4) ketones
58. In which compound, cis-trans nomenclature cannot be used?
(1) CH3-CH=CH-COOH
(2) C6H5-CH=CH-C6H5
(3) C6H5-CH=CHOH
(4) C6H5-CH=C(Cl)CH3
59. Dihedral angle of least stable conformer of ethane is
(1) 0°
(2) 120°
(3) 180°
(4) 60°
60. Meso isomer is not possible in (1) 2,3-dihydroxy butanoic acid
(2) Tartaric acid
(3) 2,3,4-trichloropentane (4) 2,3-butane diol
61. Number of optically active isomers possible for Butane-2,3-diol are (1) 4 (2) 3 (3) 2 (4) 6
62. Geometrical isomerism is shown by (1) CH2 = C(Br) I (2) CH3CH=C(Br) I (3)(CH3)2C = C(Br) I (4) CH3CH = CCl2
63. Which of the following is an E-isomer as well as also a cis isomer?
(1) (2) (3) (4)
64. Which of the following compound exhibits optical isomerism?
(1) 2-methylbutene-2 (2) 3-methylbutyne-1 (3) 2-methylpropanal
(4) 2-methylbutanoic acid
Numerical Value Questions
65. Amongst the following the total number of compounds will show geometrical isomerism2-Butene, Propene, 1-Phenylpropene, 2-Methyl-2-Butene
66.
Number of optical isomers possible for this compound is ________
67. The number of Aromatic isomers forC8H10?
68. Number of primary amines (structural) possible for the formulaC 4H11Nis _______
69. The number of different chain isomers for C7H16 is ______
70. The correct position number of C (* Labelled carbon) in the following molecule
71. The number of chiral carbons present in the molecule given below is
F undamental concepts in organic reaction mechanism
Single Option Correct MCQs
72. The –I effect is shown by?
(1) –COOH
(2) –CH3 (3) –CH3CH2 (4) –CHR2
73. Resonance is possible in (1) H3C – CO – CH3 (2) H2C = CH – CH = CH2 (3) H2C = CH – CH2 – CH3 (4) H2C = CH – CH2 – CH = CH2
74. – M effecting group in the following is (1) −OCOR (2) −NHCOR (3) – COOH (4) –OR
75. Decreasing –I effect of given groups is
a) CN
b) NO2
c) NH2
d) F
(1) c > b > a > d
(2) b > c > d > a
(3) c > b > d > a
(4) b > c > d > a
76. Hyper conjugation is not observed in (1) CH3 – CH = CH2
(2) ( ) 3 3 CHC +
(3) (CH3)3C−CH=CH2
(4) • 32 CHCH
77. Which of the following anion is most stabilized?
(1)
(2)
(3) (4) All have same stability
78. Correct order of stability for the following radicals is:
(1) II > III > I > IV
(2) III > II > I > IV
(3) III > II > I > IV
(4) III > II > I > IV
79. The paramagnetic species is (1) Carbonium ion
(2) Carbanion
(3) Free radical
(4) Both (1) and (2)
80. Nucleophilic reaction among the following is
(1) HBr 22 32 CHCHCHCHBr =→
(2) 2 Cl 43 h CHCHClHCl ν →+
(3) aq.NaOH 33 CHBrCHOHNaBr →+
(4) 3 FeCl 662 65 dark CHClCHClHCl +→+
81. Nitration of Benzene is (1) nucleophilic substitution (2) nucleophilic addition (3) electrophilic substitution (4) free radical substitution
Numerical Value Questions
82. How many carbocations are possible for molecular formula CH35+ ?
83. Total number of electrophiles present in the following are 23 43323 NO,SO,CN,F,CCl,NH,BCl,CO,AlCl +−−
84. The total number of contributing structures showing hyper conjugation for carbocation is
85. Find the number of electron-donating groups in resonance among the following:
a) –CONH2 b) –NO2
c) –OCOCH3 d) –COOCH3
e) –CHO f) –NHCOCH3
86. The number of no bond resonance structures in tertiary pentyl cation is _______
Methods of purification of organic compounds
Single Option Correct MCQs
87. In steam distillation, the vapour pressure of the volatile organic compound is
(1) Equal to atmospheric pressure
(2) Less than the atmospheric pressure
(3) More than the atmospheric pressure
(4) Just double the atmospheric pressure
88. Anthracene is purified by
(1) Filtration
(2) Crystallisation
(3) Distillation
(4) Sublimation
89. The best method to separate the mixture of ortho and para nitrophenol (1:1) is
(1) Steam distillation
(2) Crystallisation
(3) Vapourisation
(4) Colour spectrum
90. In column chromatography, the moving phase consists
A) A substance which is to be separated
B) Adsorbate
C) Adsorbent
(1) only A (2) only A, B
(3) only B (4) A, B or C
91. In paper chromatography
(1) Moving phase is liquid and stationary phase is solid
(2) Moving phase is liquid and stationary phase is liquid water
(3) Moving phase is solid and stationary phase is solid
(4) Moving phase is solid and stationary phase is liquid
92. Which of the following statements is not correct regarding purification of liquids by steam distillation?
(1) Impurities must be non-volatile
(2) The liquid must be completely immiscible with water
(3) The liquid must possess high boiling point
(4) The liquid must not be steaming volatile
93. Turpentine oil can be purified by this process
(1) Steam distillation
(2) Vacuum distillation
(3) Simple distillation
(4) Fractional distillation
94. Sugar containing an impurity of common salt can be purified by crystallization from
(1) Benzene
(2) Ether
(3) Alcohol
(4) Water
95. The stationary phase in TLC is
(1) Liquid
(2) Solid
(3) Solid or liquid
(4) Gas
Numerical Value Questions
96. The separation of two-coloured substance was done by paper chromatography. The distance travelled by solvent front, substance A and substance B from the base line are 4 cm,2cm and 1cm respectively. The ratio ofRfvaluesofAtoBis_____
97. Sublimation can be carried out for the purification of how many of the following compounds _____.
Dry ice, Naphthalene, Anthracene, Camphor, Iodine, AgCl and Benzaldehyde.
98. Calculate weight of aniline (in grams) distilled if weight of water distilled is 100g when
Porganic compound = 100mm and 2 HO P200mm =
Qualitative analysis of organic compounds
Single Option Correct MCQs
99. Lassaigen`s test gives a violet colouration with sodium nitroprusside, it indicates (1) N (2) S (3) O (4) Cl
100. Which of the following compounds will give blood red colour while doing the Lassaigne`s test for Nitrogen?
(1)(NH2)2C=O
(2)(NH2)2C=O
(3) H2N(C6H4)SO3H
(4) CHCl3
101. Which Lassaigne`s extract(Na2S)is acidified with acetic acid and then lead acetate solution is added to it, the colour of the precipitate is (1) Blue
(2) Black
(3) Red
(4) White
102. For detection of sulphur in an organic compound, sodium nitroprusside is added to the sodium extract. A violet colour is obtained which is due to the formation of (1) Fe4[FeCN6]3
(2) Fe(SCN)3
(3) Na4[Fe(CN)5NOS]
(4) Na2[Fe(CN)5NOS]
103.ClCH2COOH is heated with fuming HNO3 in the presence of AgNO3 is carius tube. After filtration and washing a white precipitate is obtained. The precipitate is of (1) Ag2SO4 (2) ClCH2COOAg (3) AgCl (4) AgCN
104.Lassaigne`s solution is boiled with conc. HNO3 before testing for halogens because (1) Silver halides are insoluble inHNO 3 (2) Na2SandNaCNare decomposed byHNO3 (3) Ag2Sis soluble inHNO3 (4) AgCNis soluble inHNO3
105.0.302 g of organic compound gave 0.268g of silver bromide. The % of bromine in the sample is (1) 20 (2) 50 (3) 37.75 (4) 75
106.The molecular formula of an organic compound is C4H9N.The volume of N2that will be given by 0.2g of the above compound at STP is __________ ml
(1) 31.5 (2) 50 (3) 63 (4) 93
107.If 0.75 g of an organic compound in Kjeldahl`s method neutralized 30ml of 0.25N H2SO4,the percentage of nitrogen in the compound is (1) 28 (2) 50 (3) 80 (4) 14
108.Carbon and hydrogen in an organic compound are detected as__________
(1) CaHCO3, CaCO3
(2) CaHCO3, CuSO4.5H2O
(3) CaCO3, CuSO4.5H2O
(4) CaCO3, Cu(OH)2
109.Kjeldahl`s method cannot be used for the estimation of nitrogen in (1) pyridine
(2) Nitro compounds
(3) Azo compounds
(4) All the three above
110.0.12g of an organic compound gave 0.22 Mg2P2O7by the usual analysis. The percentage of phosphorus in the compound is
(1) 15.23
(2) 38.75
(3) 51.20
(4) 60.92
111.If 0.02g of a volatile compound on heating displaces 11.2ml of dry air at STP, the molecular weight of the compound is (1) 20 (2) 30 (3) 40 (4) 50
112.In Duma’s method, the gas which is collected in nitro meter is (1) N2 (2) NO
(3) NH3 (4) H2
Numerical Value Questions
113. When sodium phosphate is treated with ammonium molybdate, a canary yellow precipitate is formed. The no. of atoms present in one molecule of the precipitate is 114. Amongst the following the total number of compounds which does not give Lassaigne’s test for nitrogen
i)
ii) H2N−CH2−COOH
iii) H2N−NH2
iv) H2N−OH
v) Urea
vi) PhNO2
vii) CH3CHO
viii) CH3CN
9: Organic Chemistry – Some Basic Principles and Techniques
115.The oxidation state of two different iron atoms in Prussian blue complex (Ferric Ferro cyanide) are x and y Then x+y=_______
116.How many of the following elements do not respond to sodium fusion extract? N, P, S, C, O, X
117.How many elements can be investigated by Lassaigne’s test or Beilstein’s test or both amongst the following? Cl, S, N, P, O, Br.
118.In the estimation of nitrogen by Duma’s method 1.18g of an organic compound gave 224ml of N2 at STP. The percentage of nitrogen in the compound is
119.In Sulphur estimation. 0.471 g of an organic compound gave 1.4439 g of barium sulphate. The percentage of Sulphur in the compound is ______
120.If 0.75gm of an organic compound in Kjeldhal’s method neutralized 30 ml of 0.25NH2SO4then the percentage of nitrogen in the compound is ….
121.On complete combustion 0.30g of an organic compound gave 0.20g of carbon dioxide and 0.10g of water.
The percentage of carbon in the given organic compound is_________
122.In the ‘S’ estimation 0.471 g of an organic compound gave 1.44 g of BaSO 4 the percentage of ‘S’(sulphur) in the compound is __________
Level - II
General Introduction
Single Option Correct MCQs
1. Number of primary, secondary, tertiary and neo carbons in the following compound respectively are
Tetravalence of Carbon
Single Option Correct MCQs
2. Identify the molecule with its correct shape?
(1) CH3Cl----linear
(2) CH2O-------trigonal planar
(3) CH2O----linear
(4) HCN-----trigonal planar
3. Incorrect combination is molecule σ and π bonds
(1) CH3COCH2CHO 10σ and 2π
(2) HCCCH2CHO 8σ and 3π
(3) HCCCH2CN 7σ and 4π
(4) CH3OCH2CHO 10σ and 1π
Structural Representation of Organic Compounds
Single Option Correct MCQs
4. Bond line representation of organic compound is given below CN CN OH
Identify correct statement about the compound. The compound contains
(1) 7 σ and 4 π bonds
(2) 7 σ and 2 π bonds
(3) 6 σ and 2 π bonds
(4) 6 σ and 4 π bonds
Classification of Organic Compounds
Single Option Correct MCQs
5. Which of the following statements is not true about a homologous series?
(1) Adjacent members of a group differ by a mass of 14.
(1) 4, 2, 1, 1 (2) 5, 1, 1, 1
(3) 3, 2, 2, 1 (4) 5, 0, 2, 1
(2) Adjacent members of a group differ by –CH2– group.
(3) Members of a homologous series have same general methods of preparation.
(4) Members of a homologous series have the same physical and chemical properties.
6. How many of the following is/are nonbenzenoid aromatic, benzene, aniline, naphthalene, tropolone, and furan (1) 3 (2) 2 (3) 1 (4) 4
Nomenclature of Organic Compounds
Single Option Correct MCQs
7. Correct IUPAC name of the compound is:
( ) ( ) 32323 CHCHCHCHCHCHOCHCH
(1) 2-Methyl-3-ethylpentanal
(2) 2,3-Diethylbutanal
(3) 2-Ethyl-3-methylpentanal
(4) 3-Methyl-2-ethylpentanal
8. The correct IUPAC name of the compound is:
( ) ( ) ( ) 323 CHCHClCHBrCHICHCH
(1) 4-Bromo-5-chloro-3-iodohexane
(2) 3-Bromo-2-chloro-4-iodohexane
(3) 3-Bromo-4-iodo-2-chlorohexane
(4) 2-Bromo-3-chloro-4-iodohexane
Isomerism
Single Option Correct MCQs
9. Which of the following compounds are structural isomers of C 5 H10O?
I) 2-Methylbutanal
II) Propylethanoate
III) Pentanal
(1) I and II (2) I and III
(3) II and III (4) I, II, and III
10. The maximum number of amide isomers (structural only) possible for C 3 H7ON is (1) 4 (2) 5 (3) 3 (4) 2
11. Which of the following is correct regarding stability of conformations of n-butane about C2–C3 bond?
I) anti II) gauche
III) partially eclipsed IV) fully eclipsed (1) I > II > III > 1V (2) I > II > IV > III
(3) IV > III > II > I (4) IV > II > I > III
12. The conformations of n-butane, commonly known as eclipsed, gauche and anticonformations can be interconverted by (1) rotation around C–H bond of methyl group
(2) rotation around C–H bond of methylene group
(3) rotation around C1−C2 linkage
(4) rotation around C2−C3 linkage
13. Isomerism shown in carbonyl compounds due to migration of a proton is known as (1) geometrical isomerism
(2) optical isomerism
(3) tautomerism
(4) position isomerism
14. According to CIP rules the correct order of priority of the given grousp is:
(1) OH > CHO > CH2OH > CH3
(2) OH > CHO > CH3 > CH2OH
(3) OH > CH3 > CHO > CH2OH
(4) CHO > OH > CH3 > CH2OH
15. Each question has two statements: statement I (S-I) and statement II (S-II).
S-I : In bond-line representation, the lines representing carbon–carbon bonds are drawn in a zig-zag fashion.
S-II : The line junctions in bondline representation denote carbon atoms bonded to appropriate number of hydrogens.
In light of the above statements, choose the correct answer from the options given below.
(1) Both S-I and S-II are true.
(2) Both S-I and S-II are false.
(3) S-I is true but S-II are false
(4) S-I is false but S-II is true.
16. Absolute configuration of the given molecule is
20. Which of the following would not rearrange to a more stable form?
(1) (2S, 3R) (2) (2S, 3S)
(3) (2R, 3R) (4) (2R, 3S)
17. The number of acyclic structural isomers (including geometrical isomers) for pentene are __________.
18. The total number of chiral compound/s from the following is ____________.
(1) + (2) + H (3) + (4) +
21. Arrange the following carbocations in decreasing order of stability. + + +
(1) A > C > B (2) B > A > C
(3) C > B > A (4) C > A > B
Fundamental Concept in Organic Reaction Mechanisms
Single Option Correct MCQs
19. Covalent bond fission in a molecule A–B is represented as follows:
A–B → A• + B• D H1
A–B → A+ + B– D H2
Identify correct relation between D H1 and D H2
(1) D H1 = D H2
(2) D H1 > D H2
(3) D H2 > D H1
(4) D H1 + D H2 = 0
22. Number of hyperconjugation structures in isopropyl radical is _____.
(1) 3 (2) 6 (3) 9 (4) 12
23. Stability order of carbene
(1) :CF2 > :CCl2 > :CBr2
(2) :CBr2 > :CCl2 > :CF2
(3) :CF2 > :CBr2 > :CCl2
(4) :CCl2 > :CF2 > :CBr2
24. Hybridisation of triple bonded carbons in benzyne
(1) sp, sp2 (2) sp2, sp2
(3) sp, sp3 (4) sp, sp
25. Order of increasing stability of the following carbanions is
(CH3)3C– (I), (CH3)2CH– (II), CH3CH2– (III), C6H5CH2– (IV),
(1) II < III < I < IV
(2) IV < III < II < I
(3) I < II < III < IV
(4) III < I < II < IV
26. The order of decreasing nucleophilicity of the following is
(1) CH3O– > OH– > CH3COO– > H2O
(2) H2O > OH– > CH3COO– > CH3O–
(3) CH3COO– > CH3O– > OH– > H2O
(4) OH– > CH3O– > CH3COO– > H2O
27. Which of the following is an example of α- elimination
(1) dehydration
(2) dehydrohalogenation of chloroform
(3) dehalogenation
(4) dehydrohalogenation
28. Presence of a nitro group in a benzene ring
(1) activates the ring towards electrophilic substitution.
(2) renders the ring basic.
(3) deactivates the ring towards nucleophilic substitution.
(4) deactivates the ring towards electrophilic substitution.
29. Arrange the following groups in order of decreasing –I effect
a) −NO2
b) –Cl
c) –CN
d) –CHO
(1) a > c > d > b
(2) a > b > c > d
(3) a > b > d > c
(4) d > a > b > c
30. Which of the following is least acidic? (1) COOH CD3 (2) COOH
(3)
(4)
31. Which among following statements are true with respect to electronic displacement in a covalent bond?
I) Inductive effect operates through π–bond.
II) Resonance effect operates through σ–bond.
III) Inductive effect operates through σ–bond.
IV) Resonance effect operate through π–bond.
(1) III and IV
(2) I and II
(3) II and IV
(4) I and III
32. Which of the following group shows the –M effect?
(1) −CMe3 (2) – O – S – O – R
(3) – NH – C – CH3
(4)
33. Which of the following has mesomeric and hyper conjugation effect
(1) CH3–CH=CH–CO–CH3
(2) CH3Cl
(3) CH2 = CH–CH = CH2
(4) CH3–CH = CH2
Numerical Value Questions
34. Consider the following carbocations
(1) (2) (3) (4) (5)
If most stable carbocation = x, least stable carbocation = y
Here, x and y are the numbers of above carbocations, then the value of x +y is _______.
35. How many α-H-atoms are present in the most stable alkene among the following? CH3 CH3 CH3
Methods of Purification of Organic Compounds
Single Option Correct MCQs
36. Fractional distillation is used to separate liquids that differ in their boiling point by (1) 5 °C
(2) 10 °C–20 °C
(3) 30 °C–80 °C
(4) 3 °C
37. Which of the following process is not used for the purification of solid impurities?
(1) Distillation
(2) Sublimation
(3) Crystallisation
(4) Fractional crystallisation
38. Simple distillation of liquids involves simultaneous.
(1) vaporisation and condensation
(2) heating and sublimation
(3) vaporisation and sublimation
(4) boiling and filtration
39. Which purification technique is used for high boiling organic liquid compound (decomposes near its boiling point)?
(1) Simple distillation
(2) Steam distillation
(3) Fractional distillation
(4) Reduced pressure distillation
40. The distillation technique most suited for separating glycerol from spent-lye in the soap industry is
(1) Fractional distillation
(2) Steam distillation
(3) Distillation under reduced pressure
(4) Simple distillation
41. A mixture of o -nitrophenol and p-nitrophenol can best be separated by
(1) Simple distillation
(2) Steam distillation
(3) Decantation
(4) Fractional distillation
42. If organic compound is quite less soluble in an organic solvent, then we require a very large quantity of solvent in instalments. The process is called
(1) solvent extraction
(2) continuous extraction
(3) repeated crystallisation
(4) slow extraction
Numerical Value Questions
43. Calculate the sum of retardation factor values for both compounds, A and B from the following data
Qualitative Analysis of Organic Compounds
Single Option Correct MCQs
44. An organic compound that produces a bluish-green coloured flame on heating in presence of copper is (1) chlorobenzene
(2) benzaldehyde
(3) aniline
(4) benzamide
45. In Liebig’s method for the estimation of C and H, if the compound also contains N, which of the following is kept near the exit of the combustion tube?
(1) Silver wire (2) PbCrO4
(3) Both a and b (4) Cu gauge
46. The blood-red colour in the combination test of nitrogen and sulphur in an organic compound is due to the formation of (1) ferric thiocyanate
(2) ferric acetate
(3) ferrous sulphocyanide
(4) ferric cyanide
47. The Beilstein test i s performed with an organic compound to detect the presence of (1) carbon (2) halogens
(3) nitrogen (4) sulphur
48. In Lassaigne’s test, thiourea is converted into
(1) NaCN (2) Na2S
(3) Na2SO4 (4) NaCNS
49. Which of the following elements present in an organic compound that cannot be directly detected?
(1) N (2) S (3) O (4) P
50. In organic quantitative analysis, CuO acts as
(1) reducing agent
(2) oxidising agent
(3) hydrolysing agent
(4) precipitating agent
Numerical Value Questions
51. During detection of sulphur in Lassaigne’s test, a violet compound with formula, Na x[Fe(CN)5NOS] is formed. What is x?
Quantitative Analysis
Single Option Correct MCQs
52. 0.36 g of an organic compound on combustion gave 1.1 g of CO 2 and 0.54 g of H 2 O. The percentages of carbon and hydrogen in the compound are (1) 75, 25 (2) 60, 40 (3) 83.33, 16.67 (4) 77.8, 22.2
53. If 0.02 g of a volatile compound on heating displaces 11.2 mL of dry air at STP, the molecular weight of the compound is (1) 20 (2) 30 (3) 40 (4) 50
54. In Carius method, halogen containing organic compound is heated with fuming nitric acid in the presence of:
(1) AgNO3 (2) HNO3
(3) BaSO4 (4) CuSO4
55. In sulphur estimation, 3.2 g of an organic compound gives 2.33 g of BaSO 4 . Then, the percentage of sulphur in the organic compound is
(1) 15% (2) 10%
(3) 20% (4) 25%
56. Phosphorous in organic compound is estimated as ___.
(1) Mg2P2O7
(2) (NH4)3PO4.12MoO3
(3) Mg3(PO4)2
(4) Both (1) and (2)
57. Substance used in the estimation of phosphorous is (1) conc. H2SO4 (2) fuming sulphuric acid and MgCl 2 (3) conc. HNO3; NaOH (4) Fuming nitric acid and magnesia mixture.
Numerical Value Questions
58. During estimation of nitrogen present in an organic compound by Kjeldhal method, the ammonia evolved from 0.5 g of the compound was exactly neutralised by 10 mL of 1 M H2SO4. Find out the percentage of nitrogen in the compound.
59. On complete combustion, 0.30 g of an organic compound gave 0.20 g of carbon dioxide and 0.10 g of water. The percentage of carbon in the given organic compound is _______. (nearest integer)
Multiple Concept Questions
Single Option Correct MCQs
60. The stablest conformer of CH2 – COO CH2 – NH3 ⊕
(1) Staggered (2) Skew (or) gauche (3) Partially eclipsed (4) Fully eclipsed
61. Which of the following is an ambident nucleophile?
62. Arrange the following compounds in increasing order of length of their C–N bond. (I) NH2
(1) I and II only (2) II and III only (3) I and III only (4) I, II, and III
(II) CH2 – NH2
(III) CH3–CH = NH
(IV) 32 CH-CO-NH
(1) IV < III < I < II (2) III < IV < I < II
(3) I < IV < III < II (4) IV < I < III < II 63.
I and II are geometrical isomers of each other because
(1) l1 = l2
(2) l1 > l2
(3) l2 > l1 (4) l1 and l2 cannot be compared
Numerical Value Questions
64. Degree of unsaturation for Q is ‘x’. Then the value of ‘x’ is OH [Q] i) (PhCO)2O. PhCO2Na,D ii) KOH, EtOH, reflux
65. How many of the following process are used for purification of organic compound?
[Sublimation, crystallisation, distillation, differential extraction, chromatography.]
66. Number of aromatic compounds in N ⊕
Level-III
1. IUPAC name of the following compound OH
(1) 7-Ethenyl-3-ethyl-2-methyldecan-2-ol
(2) 7-Vinyl-3-ethyl-2-methyldecan-2-ol
(3) 7-Ethenyl-2–ethyl-1,1-dimethyldecan -1-ol
(4) 3-Ethyl-2-methyldecan-2-ol
2.
If P, Q, and R compounds given then, sum of total number of stereo isomers of P+Q+R is
(1) 21 (2) 20 (3) 22 (4) 23
3. Which of the following molecules is expected to rotate the plane of plane-polarised light?
(1) CHO CH2OH HO H
4. Newman projections P, Q, R and S are shown below:
Which one of the following options represents identical molecules ?
(1) P and Q (2) Q and S
(3) Q and R (4) R and S
5. In the following three-dimensional structure of CH4, the bonds are labeled as W, X, Y, and Z. The bonds projecting out of the plane are C H H H H Y X Z W
(1) X, Y (2)W, Z (3) X, Z (4) W, Y
6. The number of chiral carbons present in the molecule given below is____.
7. The number of stereoisomers formed in a reaction of (±) Ph(C=O) C(OH)(CN)Ph with HCN is_____. O N N C H HO
8. Which of the following carbocations is most stable? (1) OCH3 ⊕ (2) H3CO ⊕ (3) H3CO ⊕ (4) OCH3 ⊕
9. The order of stability of the following carbanions
I) o-nitrobenzyl carbanion
II) m-nitrobenzyl carbanion
III) p-nitrobenzyl carbanion
IV) benzyl carbanion
(1) I > II > III > IV (2) IV > II > III > I
(3) I > III > II > IV (4) I > II > IV > III
10. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
Assertion (A): Carbenes act as electrophiles.
Reason (R): Only triplet carbenes act as biradicals (divalent free radicals) .
In light of the above statements, choose the correct answer from the options given below.
(1) Both (A) and (R) are true and (R) is the correct explanation of (A).
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(3) (A) is true but (R) is false.
(4) Both (A) and (R) are false.
11. Which of the following is a free radical addition reaction?
(1) CH3 → CHO + HCN → CH3(OH)CN
(2) CH3– CH = CH2 + HBr → CH3 – CH(Br) – CH3
(3)
12. Arrange the following in increasing order of reactivity towards nitration
A. p-xylene
B. bromobenzene
C. mesitylene
D. nitrobenzene
E. benzene
Choose the correct answer from the options given below:
(1) C < D < E < A < B
(2) D < B < E < A < C
(3) D < C < E < A < B
(4) C < D < E < B < A
13. Which of the following phenols has the highest pKa value?
(1) OH Cl (2) OH O2N (3) OH H3C (4) OH N ≡ C
14. The most stable canonical structure of th e following molecule is: O (1) O + –(2) O + –
15. The number of electrophilic centre in the given compound is ____.
19. From the figure of chromatography given below, identify incorrect statements. a b c
16. Select the correct statement among the following:
(1) Standard enthalpy of formation of H 2O (l) is zero.
(2) Naphthalene sublimes slower than solid CO2.
(3) Enthalpy of vapourisation of acetone is more than that of water.
(4) If water freezes then the amount of heat is given off to the surrounding will be equal to the enthalpy of sublimation of water.
17. A flask contains a mixture of isohexane (boiling point = 60°C) and 3-Methylpenate (boiling point = 63°C). What is the best way to separate the two liquids, and which one will be distilled out first?
(1) Fractional distillation, isohexane
(2) Simple distillation, siohexane
(3) Fractioonal distillationi, 3-Methylpentane
(4) Simple distillation, 3-Methylpentane.
18. Two volatile and miscible liquids can be separated by fractional distillation into pure components under the conditions when (1) they have low boiling points.
(2) the differences in their boiling points is large.
(3) the boiling points of the liquids are close to each other.
(4) they do not form azeotropic mixture.
A) Compound ‘c’ is more polar than ‘a’ and ‘b’.
B) Compound ‘a’ is least polar.
C) Compound ‘b’ comes out of the column before ‘c’ and after ‘a’.
D) Compound ‘a’ spends more time in the column.
Choose the correct answer from the options given below:
(1) A, B and D only
(2) A, B and C only
(3) B and D only
(4) B, C and D only
20. The separation of two coloured substances was done by paper chromatography. The distances travelled by solvent front, substance (A) and substance (B) from the base line are 3.25 cm, 2.08 cm, and 1.05 cm respectively. The ratio of Rf values of A to B is ______ (nearest integer value).
21. In Carius tube, an organic compound, ‘X’ is treated with sodium peroxide to form a mineral acid, ‘Y’. The solution of BaCl 2 is added to ‘Y’ to form a precipitate, ‘Z’. ‘Z’ is used for the quantitative estimation of an extra element. ‘X’ could be
(1) Methionine (2) Cytosine
(3) Chloroxylenol (4) A nucleotide
22. For detection of sulphur in an organic compound, sodium nitroprusside is added to the sodium extract. A violet colour is obtained which is due to the formation of
(1) Fe4[Fe(CN)6]3
(2) Fe(SCN)3
(3) Na4[Fe(CN)5NOS]
(4) Na2[Fe(CN)5NOS]
23. Lassaigne’s test is not used for detection of elements (X) and (Y). Identify (X) and (Y) respectively?
(1) X–C; Y–N (2) X–S; Y–Cl
(3) X–O; Y–N (4) X–C; Y–H
24. 120 kg of an organic compound that contains only carbon and hydrogen gives 330 g of CO2 and 270 g of water on complete combustion. The percentage of carbon and hydrogen, respectively are (1) 25 and 75 (2) 40 and 60 (3) 60 and 40 (4) 75 and 25
25. Kjeldhal method cannot be used for the estimation of nitrogen in I. pyridine II. nitro compounds III. azo compounds
(1) I and II only (2) II and III only (3) I and III only (4) I, II and III
26. In Dumas’ method for estimation of nitrogen, 0.3 g organic compound gave 50 mL of nitrogen which is collected at 300 K temperature and 715 mm pressure. Calculate the percentage composition of nitrogen in the compound (Aqueous tension at 300 K = 15 mm)
(1) 17.46% (2) 34.46%
(3) 64.46% (4) 8.46%
27. In a Carius tube, 1g of organic compound on heating with conc. HNO3 followed by the addition of AgNO3 solution gave 0.752 g of AgBr as a precipitate. Percentage of bromine in the organic compound is (Atomic weight of Br is 80 and Ag is 108)
(1) 16 (2) 32 (3) 64 (4) 28
28. An organic compound is heated with Na2O2 and then treated with conc. H2SO4 followed by addition of BaCl 2. 0.233 g of BaSO 4 is formed from 0.32 g of organic compound. The percentage of sulphur in the compound is
(1) 20 (2) 40 (3) 30 (4) 10
29. In the quantitative estimation of phosphorous (P) by using magnesia mixture, the formula used is: (w = mass of precipitate formed ; W = mass of organic compound taken)
(1) Percentage of 62W×100 P=× 222w
(2) Percentage of 31W×100 P=× 222w
(3) Percentage of 62w×100 P=× 222W
(4) Percentage of 31w×100 P=× 222W
30. Complete combustion of 1.80 g of an oxygen containing compound [Cx H yO2] gave 2.64 g of CO2 and 1.08 g of H2O. The percentage of oxygen in the organic compound is (1) 51.63 (2) 63.53 (3) 53.33 (4) 50.33
31. A sample of ‘a’ g of carbon is burnt with ‘b’ g of oxygen in a closed rigid container and no residue left. Select the correct option about relative amounts of oxygen and carbon:
(1) b a must be less than 1.33.
(2) a b must be less than 1.33.
(3) b a must be greater than 2.67.
(4) b a must be between 1.33 and 2.67.
32. If 0.2 g of an organic compound containing carbon, hydrogen and oxygen. On combustion, it yields 0.147 g of carbon dioxide and 0.12 g of water. What will be the content of oxygen in the substance?
(1) 7.34% (2) 73.4% (3) 37.4% (4) 3.74%
33. A 2.79 g of an organic compound when heated in Carius tube with conc. HNO 3 and H 3 PO 4 are converted into MgNH 4 PO4 precipitate. The precipitate on heating gave 1.332 g of Mg2P2O7. The percentage of phosphorous in the compound is (nearest integer)____.
34. Which of the following compounds would give the following set of qualitative analysis?
1. Fehling’s test : positive test
2. Sodium fusion extract upon treatment with sodium nitroprusside gives a blood red colour but not prussian blue.
(1) N N CHO O
(2) CHO O N
(3) N N CHO
(4) N S CHO
35. Which of the following orders is correct for the stability of these carbocations?
concentration = 0.13 g/mL, length of polarimeter = 1 dm, Wavelength = sodium D line, T = 25°C. The specific rotation of camphor is
(1) +44.3° (2) +26.7°
(3) –26.7° (4) –44.3°
37. The isomeric deuterated bromide with molecular formula C 4 H 8 DBr having two chiral carbon atoms is
(1) 2-Bromo-1-deuterobutane
(2) 2-Bromo-2-deuterobutane
(3) 2-Bromo-3-deuterobutane
(4) 2-Bromo-1-deuteron-2-methylpropane
38. The increasing order of basicity of the following compounds is:
(a) NH2 NH2 (b) NH
(c) NH NH2 (d) NHCH3
(1) (d) < (b) < (a) < (c)
(2) (a) < (b) < (c) < (d)
(3) (b) < (a) < (c) < (d)
(4) (b) < (a) < (d) < (c)
39. What is the correct order of acidity of the protons marked A–D in the given compounds? CH A B H D C COOH H
(1) I > II > III (2) III > II > I (3) I > III > II (4) II > III > I
36. The optical rotation of a solution of pure natural camphor is found to be +5.760 o Under the following conditions, the
(1) HD > HC > HB > HA
(2) HC > HA > HD > HB
(3) HC > HD > HA > HB
(4) HC > HD > HB > HA
40. Increasing order of stability of the resonance structures is:
(A) OHC NH2 + –
(B) NH2 OHC + –
(C) NH2 H O + –
(D) NH2 OHC + –
Choose the correct answer from the options given below:
(1) D, C, A, B
(2) C, D, B, A
(3) D, C, B, A
(4) C, D, A, B
41. The Kjeldhal method of nitrogen estimation fails for which of the following reaction products?
(a) NO2 Sn/HCl
(c) CN LiAlH4
(c) CH2CN (i) SnCl2 + HCl (ii) H2O
(d) NH2 NaNO2 HCl
(1) (a) and (d)
(2) (c) and (d)
(3) (b) and (c)
(4) (a), (c) and (d)
42. CH3 (x) CH3 (Number of plane of symmetry) (y)
(Number of meso isomer of 1, 2-dimethylcyclopentane)
Find the value of x + y in (1) and (2)
43. Three organic compounds A, B and C were allowed to run in thin layer chromatography using hexane and gave the following result (see figure). The Rf value of the most polar compound is ________ × 10 −2
44. 0.2 g of an organic compound was subject to estimation of nitrogen by Duma’s method in which volume of N2 evolved (at STP) was found to be 22.400 mL. The percentage of nitrogen in the compound is ____ (nearest integer). [Given: Molar mass of N2 is 28 g mol−1, molar volume of N2 at STP : 22.4 L]
45. When metal ‘M’ is treated with NaOH, a white gelatinous precipitate ‘X’ is obtained, which is soluble in excess of NaOH. Compound ‘X’ when heated strongly gives an oxide that is used in chromatography as an adsorbent. The atomic number of metal ‘M’ is ___.
46. The observed rotation of 10 mL of a solution containing 2 g of a compound when placed in 25 cm long polarimeter tube is +13.4°. The specific rotation of compound is
THEORY-BASED QUESTIONS
Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II). Mark the correct answer as
(1) if both statement I and statement II are correct.
(2) if both statement I and statement II are incorrect.
(3) if statement I is correct, but statement II is incorrect.
(4) if statement I is incorrect, but statement II is correct.
1. S-I : Some organic liquids can be purified by steam distillation.
S-II : Principle involved in steam distillation is that liquids immiscible in water possessing high boiling point are steam volatile.
2. S-I : Nitrogen present in nitrobenzene, azobenzene and pyridine cannot be estimated by Kjeldhal method.
S-II : Nitrogen of these compounds cannot be converted to ammonium sulphate under given conditions.
3. S-I : In chromatographic technique, the retardation factor (R f) does not depend on the polarity of solvent.
S-II : Thin layer chromatography is an example of absorption chromatography.
Assertion and Reason Questions
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as .
(1) if both (A) and (R) are true and (R) is the correct explanation of (A).
(2) if both (A) and (R) are true but (R) is
not the correct explanation of (A).
(3) if (A) is true but (R) is false.
(4) if both (A) and (R) are false.
4. (A) : In Duma’s method, when an organic compound is heated with cupric oxide, ‘N’ is converted to N 2 gas.
(R) : Cupric oxide oxidises carbon and hydrogen to CO2 and water vapour.
5. (A) : The boiling point of cis -1,2dichloro ethene is higher than that of corresponding trans-isomer.
(R) : cis-1,2-dichloro ethene has a higher dipole moment as compared to that of the trans-isomer.
6. (A) : Components of a mixture of red and blue inks can be separated by distributing the components between stationary and mobile phases in paper chromatography.
(R) : The coloured components of inks migrate at different rates because paper selectively retains different components according to the difference in their partition between the two phases.
7. (A) : CH3 – C ≡ O ⊕ is more stable than CH3 – C ⊕
(R) : Resonance structure in which all the atoms obey octet rule is more stable than resonance structure in which all atoms are not obeying octet rule.
8. (A) : CN H is called cyclo hexanecarbonitrile
(R) : It contains six carbon atoms in the ring and –CN as the functional group.
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1.
Product of this reaction is:
2. Which of the following is/are capable of showing geometrical as well as optical isomerism?
(1) (2)
3. According to IUPAC rules, select the correct statement(s) about the priority order of functional groups.
(1) –OR > –OH
(2) –CN > –CHO
(3) –OH > –NH2
(4) –COOH > –SO3 H
Correct statements are
(1) P and Q are functional isomers
(2) Q and S are metamers
(3) P and S are positional isomers
(4) Q is having higher dipole moment
5. Which of the following statements is/are wrong?
(1) Benzene diazonium chloride gives positive Lassaigne’s test.
(2) In Lassaigne’s test for nitrogen, prussian blue colour is due to the formation of ferroferric cyanide.
(3) When FeCl 3 solution is added to Lassaigne’s extract, a blue solution is obtained which indicates the presence of both nitrogen and sulphur.
(4) Aniline cannot give positive Lassaigne’s test.
6. Which of the following methods are used for recemic resolution?
(1) Biological methods by using special enzymes.
(2) By making their diastereomers.
(3) Chromatographic method using special adsorbents.
(4) Azeotropic distillation.
Correct statement (s) is/are
(1) Q is more stable.
(2) P is more stable.
(3) P is having higher dipole moment.
(4) Q is having higher dipole moment.
8. Structure of naturally occurring steroid cholesterol is given
(1) III > I (2) III > II
(3) II > I (4) I > II
11. Which of the following are functional isomers of methyl ethanoate?
(1) CH3–CH2–CHO
(2) CH3–CH(OH)–CHO
(3) CH3–O–CH2–CHO
(4) CH2(OH)–CO–CH3
12. True statement(s) about CH 3 CH = CHCH(OH)COOH is/are
(1) It has two chiral centres.
(2) It has four optically active forms.
Which of the following statements is/are correct
(1) There are nine chiral centres in this compound.
(2) It is a penta-cyclocompound.
(3) There are two 4° carbon atoms.
(4) There are seven 3° carbon atoms.
9. A solid mixture contains CaSO 4 and camphor. The components of the mixture may be separated by
(1) dissolution of the mixture in water followed by filtration, and crystallization of CaSO4 from the filtrate.
(2) sublimation.
(3) dissolution of the mixture in ether followed by filtration and recovering camphor from the filtrate by evaporating it to dryness.
(4) steam distillation.
10. The correct order of enol content.
(3) It shows both geometrical and optical isomerism.
(4) It shows conformational isomerism.
13. In the Newman's projection for 2,3Dimethylbutane, X, Y, and Z respectively are
CH3 Z Y X H H
(1) –H, –CH3 and –C2H5
(2) –H, –isopropyl and –H
(3) –C2H5 and –H and –CH3
(4) –CH3, –CH3 and –CH3
14. In the Newman's projection for 2, 2-Dimethylbutane, X and Y, respectively be
CH3 H3C Y X H H
(1) –H and –H
(2) –H and –C2H5
(3) –C2H5 and –H
(4) –CH3 and –CH3
15. In the following correct statements are:
(1) IUPAC name of pyruvic acid (CH3COCOOH) is 2-Oxopropanoic acid.
(2) CH3–CH(OH)–CH2–CH2–CH3 is also called iso–amylalcohol.
(3) IUPAC name of acetonitrile is ethanenitrile.
(4) Cinnamic acid (C6H5–CH=CH–COOH) is 3-Phenylprop-2-enoic acid.
Numerical Value Questions
16. 0.8 g of an organic compound was analysed by Kjeldhal method for the estimation of nitrogen. If the percentage of nitrogen in the compound was found to be 42%, then ______ mL of 1 M H2SO4 would have been neutralised by the ammonia evolved during the analysis.
17. 0.25 g of an organic compound containing chlorine gave 0.40 g of silver chloride in Carius estimation. The percentage of chlorine present in the compound is ____[in nearest integer] (given: molar mass of Ag is 108 g mol–1 and that of Cl is 35.5 g mol –1)
18. Aliphatic amines with nitrous acid give a quantitative yield of nitrogen gas and is the basis of Van–Slyke determination of amino nitrogen. The volume of nitrogen liberated at STP (in mL) when 0.01 mol of proline react with excess nitrous acid.
19. Optical activity of an enantiomeric mixture is +12.6° and the specific rotation of (+) isomer is +30°. The optical purity is ____%.
20. Using the provided information in the following paper chromatogram:
From the given chromatography for compounds A and B, the R f value of A is x × 10−1. The value of x is ____.
21. Identify the number of nucleophilic substitution reactions in the given reactions
(1) CH3–Cl + NaOH CH3–OH + NaCl
(2) + HBr Br
(3) + CH3 – Br CH3
(4) CH3–CH2–OH+PCl5 CH3–CH2–Cl+POCl3+HCl
(5) CH3–CH2–Cl+alc.KOH CH2=CH2+H2O+KCl
(6) H2N NH2 O C NH4CNO D
(7) H3C H3C CN OH H H HCN + O C C
(8) CH3–CH2–Cl+NH3 CH3–CH2 + H2O + KCl
22. How many of the following statements are correct?
i) Order of stability of carbocations ( ) ( ) 65655 223 6 CHCHCHCHCHC ⊕⊕⊕ <<
ii) Order of acidic strength: CH3COOH < C6H5COOH < HCOOH < (Cl)CH2COOH
iii) The hybridisation of carbocation and carbanion are sp2 and sp3
iv) The order of stability of free radicals is ( ) ( ) 3323332 CHCHCHCHCCHCH <−<<
23. For the estimation of nitrogen, 1.4 g of organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M 10 sulphuric acid. The unreacted acid required 20 mL of M 10 sodium hydroxide for complete neutralisation. The percentage of nitrogen in the compound is ______.
Integer Value Questions
24 The number of C-H bonds that can involve hyper conjugation in [P] is
Passage-based Questions
Q(30-31)
Electrophile is an electron deficient species, usually positively charged and neutral. Molecule or ion that donates a pair of electrons to form a new covalent bond is a nucleophile.
30. Among C H 3 CONH 2 , CH 3 Cl, HCHO, ( ) 33 CHC + and H + , the number of compounds that acts as both electrophile and nucleophile.
31. Among + 323223 FeCl,SnCl,CH:CCl,NO,CHCO,and , + 3232233 FeCl,SnCl,CH:CCl,NO,CHCO,and,BF , the number of neutral electrophile are __
Q(32-33)
When C−H electrons are in conjugation to pi–bond, then it is known as hyperconjugation. For any compound to show hyperconjugation.
25. The total number of isomers with molecular formula, C4H6 is x and the total number of optically active isomers of molecule, CH 3 –CH(Cl)–CH=CH–CH(Cl)–CH3 is y. Then, x y=?
26. The total number of cyclic isomers possible for molecule C5 H 10 is
27. 0.62 g of an organic compound gave 0.222 g of Mg2P2O7 (molecular weight = 222) by the usual analysis. The percentage of phosphorus in the compound is X, then 2 X is:
28. OH OH OH The sum of locants of secondary suffix(es) in the IUPAC nomenclature of the given compound is
29. How many different enols exist for 4-methyl3-hexanone?
i) Th compound should have altleast one sp2--hybridised carbon.
ii) α-carbon with respect to sp2 should be sp3
iii) α-carbon should contain at least one hydrogen atom.
(iv) The number of α-carbon ∞ the stability of cation and alkene.
32. Consider the following given compounds and identify how many of them does not exhibit hyperconjugation.
(a) Ethene
(b) 2,3-Dimethylbut-2-ene
(c) Propene
(d) 1-Phenylpropene
(e) 1-Chloro-1-butene
(f) Toluene
(g) 2-Methyl-2-phenylpropane
(h) 2-Methylpropane
(i) 2,2-Dimethylpropane
(j) Ethyne
(k) But-1-ene-3-yne
33. How many of the following are having more electron density than toluene?
(1) Ethylbenzene
(2) Iso-propylbenzene
(3) Tertiary-butylbenzene
(4) Nitrobenzene
(5) Chlorobenzene
(6) 1,3,5-Tri-nitrobenzene
Q(34-35)
Carbocations are species in which central carbon carries positive charge. If the charge on the carbocations gets concentrated or localised, the carbocation becomes unstable. The two factors that account the stability of carbocations are inductive effect and hyperconjugation (no bond resonance). Inductive effect is the minor factor, whereas hyperconjugation is the major factor.
34. Which is the most stable carbocation?
(1) + 3 CH (2) + 32CHCH
(3) ( ) + 32 CHCH (4) ( ) + 33 CHC
35. Hybridisation of carbon in carbocation is (1) sp (2) sp2 (3) sp3 (4) dsp2
Q(36-37)
Constitutional isomerism is also known as structural isomerism. The isomers that differ in the connectivity of their atoms is called constitutional isomers.
36. Which pair is correctly matched?
(1) n-butane and isobutane; metamers (2) 2-pentanol and 3-pentanol; position isomers
(3) Cyclohexene and cyclopentene; chain isomers.
(4) Methyl propylether and Di-ethylether; functional isomers
37. Which of the following compound will show tautomerism?
(1) acetic acid (2) cyclohexanol (3) cyclohexanone (4) hexanol
Q(38-39)
Substituted cyclohexanes can exhibit different isomeric phenomena like conformational, geometrical, and optical isomerism. To decide optical isomerism in cyclic compounds, one has to decide geometrical first, then has to apply symmetry elements to decide chirality. Every chiral molecule exhibit enantiomerism.
38. Which of the following molecules can exhibit enantiomerism? (1) COOH COOH (2) Br COOH (3)
Br (4) Br Br
39. How many geometrical isomers are possible for the following compound? = CH – CH = CH – CH = CH – CH = CH3 Br (1) 4 (2) 6 (3) 8 (4) 16
Q(40-41)
Molecules having same molecular formula but differ in structure (or) spatial orientation of atom are known isomers and the phenomenon known as isomerism. Molecules which differ in structural formula are known as structural isomers and phenomenon is known as structural isomerism. Structural isomerism can broadly classified as
i) Chain isomerism.
ii) Positional isomerism.
iii) Functional isomerism.
iv) Metamerism.
v) Tautomerism.
40. How many structural isomers possible for the given compound C4 H10O?
41. Number of positional isomers for Cl Br Br Br is ____.
Q(42-45)
Crixivan, a drug produced by Merck and Co. It is widely used in the fight against AIDS (Acquired Immune Deficiency Syndrome). The structure of crixivan is given below:
i) Delocalisation by p – p overlapping. ii) Delocalisation by p – overlapping. Delocalisation makes system stable. More is the number of resonating structures, more is the stability of the system.
46. How many resonating structures are possible for the squaric acid dianion (including given structure)?
42. How many 2° alcohol groups are present in the above compound?
43. How many amide groups are present in the compound?
44. How many 3° amine groups are present in the compound?
(1) Zero (2) 1
(3) 2 (4) 3
45. How many 2° amine groups are present in the compound?
(1) Zero (2) 1
(3) 2 (4) 3
Q(46-47)
Delocalisation of electrons take place in alternate single and multiple bonds involving carbon atoms. It may also occur in a conjugated system involving carbon and an atom other than carbon. There are also examples in which p –orbital and p–orbital (vacant or half-filled or filled) overlap. Thus delocalisations are following types:
47. The number of a-hydrogens for compounds A and B are X and Y respectively, then value of X × Y is:
Compound A Compound B
Q(48-49)
The delocalisation of σ electrons with p-orbital is known as hyperconjugation. It is also known as σ−π conjugation or no bond resonance. Presence of atleast one ‘α’ hydrogen at saturated carbon in an alkene is required. Carbocation and free radical shows hyperconjugation effect. The more is the number of hyperconjugation, the more is the stability. Bond lengths, dipole moments are also effected by hyperconjugation.
48. The number of α-hydrogens in (CH3)2C = CH(CH3) is
49. The number of hyperconjugative structures in (CH3)3C–CH = CH2 is
Q(50-51)
Restricted rotation is due to the steric crowding and it might be the reason for the development of chirality in the molecules. Examples are
CHAPTER 9: Organic Chemistry – Some Basic Principles and Techniques
ortho-substituted biphenyls (or) triphenyls and similar molecules. Based on this information, answer the following questions:
50. Which of the following molecules is chiral.
(1) CH3 CH3 NO2 COOH
(2)
(3) CH3 CH3 Cl CH3 CH3 Br C C COOH
(4)
51. Choose the correct statement regarding the following molecules.
OH OH
Matrix
Matching Questions
52. Consider the molecules in column-I and match them with their stereochemical properties from column-II
Column-I
Column-II
(A) 33 CHCHCHCHCHCHCH | OH −=−−=− (p) Have only 3 stereoisomers
(B) 33 CHCHCHCH || OHOH (q) Have 4 stereoisomers
(C) 323 CHCHCHCHCH || OHOH (r) Have only 2 optically active isomers
(D) 3 CHCHCHCHCl | Cl −−=− (s) Have more than 2 pairs of diastereomers
(A) (B) (C) (D)
(1) qrs pr qs qs
(2) p q r s
(3) pq qs r q
(4) pqr rs qr pq
53. Match list-I with list-II: List-I List-II
OH OH CH3 CH3 CH3
H3C Br CH3 x CH3 Br
(A) Neopentane (p) 1° and 3° carbons
(B) Isobutane (q) All 2° Carbons
OH Br Br OH CH3 CH3 CH3
H3C Br y CH3 CH3 Br
(1) 'x' is optically inactive; 'y' is optically active.
(2) 'x' is optically active; 'y' is optically inactive.
(3) Both are optically active.
(4) Both are optically inactive.
(C) Cyclohexane (r) 1° and 4° carbons
(D) Isopentane (s) Only 1° carbons
(t) 1°, 2° and 3° carbons
(A) (B) (C) (D)
(1) p t r q
(2) t r s p
(3) r p q t
(4) r p t r
54. Match column-I with column-II: Column-I Column-II
(A) aq.KOH 2525 CHBrCHOH →
aq.KOH 2525 CHBrCHOH → (p) Free radical substitution
(B) 2 Br/Fe 6665 dark CHCHBr →
2 Br/Fe 6665 dark CHCHBr → (q) Nucleophilic substitution
(C) 2Br 663652 773K CHCHCHCHBr →
2Br 663652 773K CHCHCHCHBr → (r) Nucleophilic substitution
(A) (B) (C)
(1) r p q
(2) q r p
(3) p q r
(4) p r q
55. Match list-I with list-II: List-I List-II
(A) Dimethyl ether, ethyl alcohol (p) Position isomers
(B) Acetaldehyde, Acetone (q) Metamers
(C) 2-Pentanone, 3-Pentanone (r) Functional isomers
(D) n-propyl alcohol, isopropyl alcohol (s) Chain isomers (t) None
(A) (B) (C) (D)
(1) t r q p
(2) r t q p
(3) q r p s
(4) r q p t
56. Match Column-I (functional group structure) with Column-II (IUPAC group prefix) Column-I Column-II
(A) –COOH (p) oic acid
(B) –COX (q) carbamoyl
(C) –CHO (r) halocarbonyl
(D) –CONH2 (s) oxo (t) carboxyl
(A) (B) (C) (D)
(1) p r s q
(2) t r s p
(3) t r s q
(4) p r s t
57. Match List-I with List-II: List-I Isomeric pairs List-II Type of isomers
(A) Propanamine and n-Methyl-ethanamine (p) Metamers
(B) Hexan-2-one and Hexan-3one (q) Positional isomers
(C) Ethanamide and Hydroxyethanimine (r) Functional isomers
(D) o-nitrophenol and p-nitrophenol (s) Tautomers
(A) (B) (C) (D)
(1) s r p q
(2) r p s q
(3) q r p s
(4) r s p q
58. Match Column-I with Column-II:
Column-I (Method)
Column-II (Test)
(A) + N ≡ N ] Br– (p) (p) Sodium fusion extract of the compound gives prussian blue colour with FeSO4
(B)
SO3H (q) Me (q) Sodium fusion extract of the compound gives blood red colour with FeSO4
(C)
SO3H (r) Br–[H3N ⊕ (r) Lassaigne’s extract (L.E) in CS2 and Cl2 water gives orange colour.
(D)
NH–NH3 ]] OOC ⊕ (s) Lassaigne’s extract with [Fe(CN)5 NO]2− gives violet colour.
(A) (B) (C) (D)
(1) r s q p
(2) p q s r
(3) r s p q
(4) q p r s
59. Match Column-I with Column-II:
Column-I
Column-II
(A) C7H16 (number of structural isomers) (p) 6
(B) C2FClBrI (number of structural isomers) (q) 3
(C) C6H4BrCl (number of Benzenoid aromatic isomers) (r) 5
(D) C5H10 (number of cyclic isomers) (s) 9
(A) (B) (C) (D)
(1) s p r q
(2) s p q r
(3) p r s q
(4) r s q p
60. Match List-I with List-II: List-I (Molecule) List-II (Orbitals involved)
(A) C2H6 (p) 12, 18
(B) C2H4 (q) 6, 4
(C) C2H2 (r) 6, 6
(D) C6H6 (s) 6, 8
(A) (B) (C) (D)
(1) s q r p
(2) q r s p
(3) r s q p
(4) s r q p
61. Match List-I with List-II: List-I (Mixture)
List-II (Purification Process)
(A) Chloroform and Aniline (p) Steam distillation
(B) Benzoic acid and Naphthalene (q) Sublimation
(C) Water and Aniline (r) Distillation
(D) Naphthalene and Sodium chloride (s) Crystallisation
(A) (B) (C) (D)
(1) s r p q
(2) r p s q
(3) r s q p
(4) r s p q
62. Match Column-I with Column-II.
Column-I (Colour) Column-II (Element)
(A) Prussian blue (p) [Fe(CN)5NOS]4–
(B) Violet (q) Fe4[Fe(CN)6]3
(C) Blood red (r) [Fe(SCN)(H2O)5]+2
(D) Colourless (s) AgCl
(t) Na4[Fe(CN)6]
(A) (B) (C) (D)
(1) r q q s
(2) s q r t
(3) q s r t
(4) q p r s
63. Match Column-I with Column-II.
Column-I (Compounds) Column-II (Numbers of isomers)
(A) Number of diastereomers of C2FClBrI (p) 4
(B) Number of stereoisomer of (q) 8
(C) Number of isomeric alcohols of C4H10O (r) 6
(D) Number of isomeric esters of C4H8O2 (s) 2
(A) (B) (C) (D)
(1) r s p p
(2) q s r s
(3) s p r p
(4) r s q p
64. Match the items given in Column-I with that in Column-II, and III.
Column-I Common name Column-II Characteristics (I)
Column-III Characteristics (II)
(A) Decalin (i) Aromatic (p) Contains one 3° C atoms
(B) Malic acid (ii) Bicyclo compound (q) Gives colour with FeCl3
(C) Maleic acid (iii) Contains4 C (r) Saturated
(D) Carbolic acid (iv) Dibasic acid (s) Decolorises Br2/CCl4 solution
(A) (B) (C) (D)
(1) i,r iii,iv,s iii,iv,r ii,q
(2) i,r iii,iv,s iii,iv,r ii,q
(3) iii,r ii,p i,s iv,r
(4) ii,p,r iii,iv,r iii,iv,s i,q
65. Match Column-I with Column-II.
Column-I Column-II
(A) Stability order: 3° > 2° > 1° (p) Toluene
(B) Stability order: 1° > 2° > 3° (q) Carbocations
(C) Hyper conjugation (r) Carbanion
(D) Two unshared pair of electrons (s) Carbene
(A) (B) (C) (D)
(1) q r p s
(2) s p r q
(3) r s p q
(4) p s q r
BRAIN TEASERS
1. The chemical reactions that occur in a single step via a cyclic transition state, bond making and bond breaking occurs simultaneously are known
(1) Polymerisation reaction
(2) Pericyclic reactions
(3) Isomerisation reactions
(4) Rearrangement reactions
2. Select correct statement out of the following:
(1) The eclipsed conformer of 1,2-dichloro ethane is more stable than its anticonformer.
(2) Gauche conformer of ethane-1,2-diol is more stable than its anti-conformer.
(3) The twisted boat form of cyclohexane is less stable than its boat form.
(4) The chair form of cyclohexane is optically active.
66. Match Column-I with Column-II.
Column-I Column-II
(A) carbocation (p) sextet and bivalent
(B) carbanion (q) sextet and trivalent
(C) methyl free radical (r) octet and trivalent
(D) Carbene (s) Heptet and trivalent
(A) (B) (C) (D)
(1) q s p r
(2) q r s p
(3) r s p q
(4) p s q r
(1) 75%(R), 25%(S)
(2) 25%(R), 75%(S)
(3) 50%(R), 50%(S)
(4) 67%(R), 33%(S)
4. A liquid (A) is treated with Na2CO3 solution and produces a mixture of two salts, (B) and (C) in the solution. The mixture is acidified with sulphuric acid followed by distillation, produces the liquid (A) again. Liquid (A), salts (B) and (C) are respectively
(1) Br2, NaBr, NaBrO3
(2) Cl2, NaCl, NaClO3
(3) I2, NaI, NaIO3
(4) HF, NaF, NaHF2
5. What is the percent composition of a mixture of ( ) ( ) 0 25 S-+-2-butanol ±=-13.52, D
3. What is the percent composition of a mixture of ( ) ( ) [ ] 250 S2butanol,13.52 D d −−±=+ , and ( ) ( ) [ ] 0 , 25 R2butanol,13.52 D −−±=− l with a specific rotation [ ] 250? 6.76 D ±=+
and
( ) ( ) 0 25 R–--2-Butanol, ±=+13.52, D with a specific rotation 25 ± =+6.76 D ?
(1) 75% (R) and 25% (S)
(2) 25% (R) and 75% (S)
(3) 50% (R) and 50% (S)
(4) 67% (R) and 33% (S)
6. Given below are two statements, statement 1 (S-I) and statement II (S-II)
S-I : The compound (A)
, is optically active.
S-II :
H is compound (B) and is the mirror image of compound (A).
In light of the above statements, choose the correct answer from the options given below.
(1) Both S-I and S-II are correct.
(2) Both S-I and S-II are incorrect.
(3) S-I is correct but S-II is incorrect.
(4) S-I is incorrect but S-II is correct.
7. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
(A) : Simple distillation can help in separating a mixture of propan–1–ol (boiling point 97°C) and propanone (boiling point 56°C).
(R) : Liquids with a difference of more than 20°C in their boiling points can be separated by simple distillation.
In light of the above statements, choose the correct answer from the options given below.
(1) Both (A) and (R) are true and (R) is the correct explanation of (A).
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(3) (A) is true but (R) is false.
(4) Both (A) and (R) are false.
8. A sample of 0.125 g of an organic compound when analysed by Duma's method yields 22.78 mL of nitrogen gas collected over KOH solution at 280 K and 759 mm Hg. The percentage of nitrogen in the given organic compound is _______ (nearest integer).
(a) The vapour pressure of water at 280 K is 14.2 mm Hg
(b) R = 0.082 L atm -1 mol-1 9.
Degrees of rotation
Configuration structure of which compound (s) correctly predicts the potential energy diagram.
10. In which of the following strength ∆G decreases, if there can be some intramolecular rearrangement?
11. Am ygdalin, a compound isolated from the pits of apricots, peaches and wild cherries. Although it has no known therapeutic value, it has been used as unsanctioned anticancer drug. The number of stereogenic centres present in amygdalin is (x) and the number of non -bonding electrons are ( y ). The degree of unsaturation is ( z ) . Find the value of
xyz ++
In the dyad tautomerism, hydrogen atom oscillates between two polyvalent atoms linked together. In triad tautomerism, hydrogen atoms travel from first to third atom in chain.
12. Among the isomeric nitroalkanes of formulaC4H9NO2, how many isomers exhibit tautomerism (excluding stereo)?
13. How many of the following molecules exhibit triad tautomerism?
14. Match List-I with List-II: List-I List-II
(A) Has more acid dissociation constant than conjugate acid of phenoxide ion. (p) O = S = O
(B) Releases carbon dioxide from the solution obtained by passing excess carbon dioxide in aq.NaHCO3 solution. (q)
(C) Has lower degree of dissociation than phenol at same concentration. (r) O
(D) Conjugate base has all equivalent energy resonance forms with same number of π-bonds. (s) OH CH3 (t) H3C O OH
The correct match is
(A) (B) (C) (D)
(1) pqr prt s pqrt
(2) prt prt s rt
(3) pqrt rt s prt
(4) pqrt prt s prt
15. Match List-I with List-II.
List-I
(A) N (I) (II) N O (III) (IV) N
Basic nature
(B) COOH NO2 NO2 COOH (I) (II) NO2 NO2 COOH COOH (III) (IV)
Acidic nature
(C) 3232 3 II I CHNH,(CH)NH,CH 23 IV III CONH,CHCN
Basic nature
(q) IV < III < II < I
(r) III > II > I > IV
List-II
(p) III < I < IV < II
(D) (I) (II) (III) (IV)
Basic nature (s) II > I > III > IV
The correct match is (A) (B) (C) (D)
(1) r r p p
(2) q r s p
(3) p q s p
(4) q r p p
16. Match List-I with List-II.
List-I (Compound) List-II (Enol content)
(A) ph (p) 89.2
(B) CH3 OCH3 O O (q) 1.1 10–6
(C) Ph CH3 O O (r) 80
(D) CH3 CH 3 O O (s) 8.4
The correct match is (A) (B) (C) (D)
(1) p q s r
(2) r s p q
(3) q s r p
(4) q s p r
FLASHBACK (Previous JEE Questions)
JEE Main
1. The technique used for purification of steam volatile water immiscible substance (2024)
(1) Fractional distillation
(2) Fractional distillation under reduced pressure
(3) Distillation
(4) Steam distillation
2. Bond line formula of HOCH(CN) 2 is (2024)
3. T he incorrect statement regarding conformations of ethane is (2024)
(1) Ethane has infinite number of conformations
(2) The dihedral angle in staggered conformation is 60°
(3) Eclipsed conformation is the most stable conformation.
(4) The conformations of ethane are interconvertible to one-another.
4. The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is (2024)
(1) electromeric energy
(2) resonance energy
(3) ionization energy
(4) hyperconjugation energy
5. Match List-I with List-II. List-I List-II
(A) (p) Spiro compound
(B) (q) Aromatic compound
(C) (r) Non-planar heterocyclic compound
(D) (s) Bicyclo compound
Choose the correct answer from the options given below: (2022) (A) (B) (C) (D)
(1) q p s r
(2) s r p q
(3) r s p q
(4) s r q p
6. A 0.166 g sample of an organic compound was digested with conc. H2SO4 and then distilled with NaOH. The ammonia gas evolved was passed through 50.0 mL of 0.5 N H2SO4. The acid used for ammonia if neutralised requires 30.0 mL of 0.25 N NaOH for complete neutralisation. The mass percentage of nitrogen in the organic compound is____. (2022)
7. In the given reaction H3C
(Where Et is –C2H5)
The number of chiral carbons in the molecule A is (2022)
8. A solution of m -Chloroaniline, m -Chlorophenol and m -Chlorobenzoic acid in ethyl acetate were extracted initially with a saturated solution of NaHCO3 to give fraction 'A'. The left-over organic phase was extracted with dilute NaOH solution to give fraction 'B'. The final organic layer was labelled as fraction 'C'. Fractions A, B, and C, respectively contains (2020)
(1) m-Chlorophenol, m-Chlorobenoic acid and m-Chloroaniline
(2) m-chlorobenzoic acid, m-chloroaniline and m-chlorophenol
(3) m-Chloroaniline, m-Chlorobenzoic acid and m-Chlorophenol
(4) m-Chlorobenzoic acid, m-Chlorophenol and m-Chloroaniline
9. A chemist has 4 samples of artificial sweetener A, B, C and D. To identify these samples, he performed certain experiments and noted the following observations:
i) Both A and D form blue-violet colour with ninhydrin..
ii) Lassaigne extract of C gives positive AgNO3 test but negative Fe4[Fe(CN)6]3 test.
iii) Lassaigne extract of B and D gives positive sodium nitroprusside test.
Based on these observations, which option is correct? (2020)
(1) A: Saccharin; B: Alitame; C: Sucralose; D: Aspartame
(2) A: Aspartame; B: Saccharin; C: Sucralose; D: Alitame
(3) A:Aspartame; B: Alitame; C:Sacharin; D: Sucralose
(4) A: Alitame; B: Saccharin; C:Aspartame; D: Sucralose
JEE Advanced
10. Consider the following reaction
On estimation of bromine in 1.00 g of R using Carius method, the amount of AgBr formed (in g) is ________.
[Given: Atomic mass of H = 1, C = 12, O = 16, P = 31, Br = 80, Ag = 108] (2022)
CHAPTER TEST - JEE MAIN
Section-A
1. The (C–H) bond length order in hydrocarbons in which the hybridisation of carbon is sp3 or sp2 or sp
(1) sp–s > sp2–s > sp3–s
(2) sp–s > sp2–s < sp3–s
(3) sp3–s > sp2–s > sp–s
(4) sp3–s > sp–s > sp2–s
2. The IUPAC name of neo-pentane is (1) 2-Methylbutane
(2) 2, 2-Di–methylpropane (3) 2-Methylpropane (4) 2, 2-Di-methylbutane
3. Which structure(s) represent(s) diastereomer(s) of I?
(1) V (2) III (3) IV (4) II
4. The decreasing order of acidity of the following carboxylic acids is (A) CH3COOH (B) COOH (C)
(D)
(1) A > B > C > D (2) B > C > A > D
(3) D > C > B > A (4) B > C > D > A
5. 0.5 g of organic compound in Kjeldhal method liberated ammonia, which neutralised 60 mL of 0.1 N H2SO4 solution. The percentage of nitrogen in the compound is
(1) 1.68 (2) 16.8 (3) 33.6 (4) 8.4
6. Which conformer of compound, 22 (2)(3) HOCHCHF is the most stable across C2−C3?
(1) Staggered (2) Eclipsed(partially) (3) Gauche (4) Fully eclipsed
7. Correct bond energy order is (1) C – C > C = C > C C
(2) C C > C = C > C = C
(3) C = C > C = C > C - C
(4) C – C > C C > C = C
8. What will be the total number of possible structural isomers with formula C 7 H 12Br 2 fulfilling the following conditions
(I) it should contain 6-membered ring
(II) none of the isomer should contain both bromine atom either on same or adjacent carbon atom.
(III) none of the isomers should contain bromine atom on tertiary carbon atom. (1) 5 (2) 8 (3) 3 (4) 7
9. Which of the following is correctly named?
(1) Cl Br = 1-Chloro-2-bromo benzene
(2) H2C = CH–CH2–C C–H = Pent-4en-1-yne
(3) 3-Ethyl-3, 3-dimethyl cyclohexane
(4) 3-Ethyl-3, 3-dimethyl cyclohexane
10. How many sigma and pi bonds are present in tetracyanoethylene?
(1) nine and nine (2) five and nine
(3) nine and seven (4) eight and eight
11. The correct order of nucleophilicity in polar protic solvent is
(1) I– > F– > Cl– > Br–
(2) I– > Br– > Cl– > F–
(3) F– > Cl– > Br– > I–
(4) Br– > F– > Cl– > I–
12. Which of the following cyclopentane derivative is optically inactive?
(1)
(2)
13. Correct order of basic strength is NH2 N N N H R S H Q P
(1) r > p > s > q (2) r > s > p > q
(3) q > s > p > r (4) p > q > r > s
14. Which of the following carbanion expected to be most stable?
(1) 2642pNOCHCH
(2) 2642 oNOCHCH
(3) 642 oCHOCHCH
(4) 642 pCHOCHCH
15. If 0.75 g of an organic compound in Kjeldhal’s method neutralised 30 mL of 0.25 N H2SO4, the percentage of nitrogen in the compound is
(1) 28 (2) 50 (3) 80 (4) 14
16. 1-Butanol and 2-Methylpropanol are a pair of which isomers?
(1) position (2) functional (3) metamers (4) chain
17. Which is more stable? (1)
18. Given below are two statements.One is labelled as Assertion(A) and the other is labelled as Reason(R).
(A) : Thin layer chromatography is an adsorption chromatography.
(R) : Thin layer of silica gel is spread over a glass plate of suitable size in thin layer chromatography which acts as an adsorbent.
In light of the above statements, choose the correct answer from the options given below.
(1) Both (A) and (R) are true and (R) is the correct explanation of (A).
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(3) (A) is true but (R) is false.
(4) Both (A) and (R) are false.
19. Which among the given molecules can exhibit tautomerism?
(1) III only (2) Both I and III
(3) Both I and II (4) Both II and III
20. The structural formula of 3-Ethyl-2-methyl hexane is
(1) CH3CH(CH3)CH(C2H5)CH2CH2CH3
(2) CH3CH2CH(C2H5)CH(CH3)CH2CH3
(3) CH3CH(C2H3)CH(CH3)CH2CH2CH3
(4) CH3CH(CH3)CH(CH3)CH2CH2CH3
Section-B
21. Identify the number of substituents that are deactivating but ortho and para directing in EAS reaction. 32
22. Butaclamol is potent antipsychotic that has been used clinically in the treatment of schizophrenia. Although patients are given a racemic mixture of the drug, only the (+)-enantiomer has pharmacological activity. How many chiral carbons does butaclamol have? OH
C(CH3)3
Butaclamol H H
23. How many of the following groups/atoms are considered as +I group (s)
–NH2, –O–, –D, –COOCH3, –Cl, –NO, –CH = CH2
24. How many of following undergo sublimation
1) Benzoic acid
2) Napthalene
3) Anthracene
4) Camphor
5) I2
6) NH4Br
7) Salicylic acid
8) Solid CO2
9) Nitrobenzene
10) Cl2
11) Br2
25. How many of the following act as electrophiles?
i) R+
ii) Br+
iii) BF 3
iv) OH(–)
v) NH2(–)
vi) SO3
vii) AlCl3
CHAPTER TEST - JEE ADVANCED
2023 P2 Model
Section-A
(Single Option Correct MCQs)
1. Consider the following reaction:
Which of the following statements about the above reaction are true?
(A) Formic acid is the strongest bronsted acid in the reaction.
(B) HF is the strongest Bronsted acid in the reaction.
(C) KF is the strongest Bronsted base in the reaction.
(D) KO2CH is the strongest Bronsted base in the reaction.
(E) The equilibrium favours the reactants.
(F) The equilibrium favours the products.
(G) Formic acid has a weaker conjugate base.
(H) HF has a weaker conjugate base.
(1) A, D, and F (2) B, D, and H
(3) A, C, and H (4) B, D, E, and H
2. In the given pair identify, most acidic compound in (A) and (B). Most basic in (C) and (D). (A) CO2H (I) (II) CO2H CH3 (B)
(1) A–I, B–II, C–I, D–II
(2) A–II, B–I, C–I, D–II
(3) A–II, B–II, C–II, D–II
(4) A–I, B–II, C–I, D–I
(II)
3. What is the relationship between I and II
(1) Enantiomers
(2) Diastereomers
(3) Constitutional isomers
(4) Identical molecules
4. The most stable carbocation for the following is:
(1) c (2) d (3) b (4) a
Section-B
(Multiple Option Correct MCQs)
5. Select the incorrect statements.
(1) Conjugate acid of pyrrole is antiaromatic.
(2) Me3C-CO-ND2 – tautomerism is not possible.
9: Organic Chemistry – Some Basic Principles and Techniques
(3) CH3–S– is less stable than CH 3–O–.
(4) 1,4-Dichloro-2-pentene does not show geometrical isomerism.
6. A solution of (R)-2-Iodobutane , [α] = 15.9° in acetone is treated with radioactive iodine.
1.0% of (S)-2-iodobutane is obtained whose specific rotation is −15.58°. Select the correct statement/s:
(1) It has 90% optical purity.
(2) Percent of R in solution is 99%.
(3) Racemic mixture is 2%.
(4) Final solution is dextrorotatory.
7. Pick the correct statements in the follwing.
(1) Mesomeric effect occurs in compounds having conjugate system of double bonds.
(2) Inductive effect is transmitted over to few carbon atoms.
(3) Due to mesomeric effect, electron pair is transferred completely.
(4) Inductive effect is due to the difference in electronegativity of atoms bonded together.
Section-C
(Integer Value Questions)
8. Some organic compounds given and each compound posses enantiomeric pairs which are indicated by P, Q, R, S
10. X is an open chain hydrocarbon that is opti cally active with the lowest possible molecular weight without isotopes. If
A) P is the maximum number of C-atoms present in straight line.
B) Q is the number of ‘H’ atoms in X.
C) R is the maximum number of carbon atoms lie in the same plane of X.
Then, value of P + Q − R is ___.
11. The concentration of cholesterol dissolved in chloroform is 6.15 g per 100 mL solution. A portion of this solution in a 5 cm polarimeter tube causes an observed rotation of −1.2°, if 10 mL of the solution is diluted to 20 mL and placed in same tube then observed rotation is −m × 10−1, find value of m
12. How many of the following statements are correct?
(i) Carbocations acts as electrophiles.
(ii) Free radicals are paramagnetic.
(iii) Carbenes has incomplete octet.
(iv) Allyl carbocation is more stable than propyl carbocation.
(v) Carbanions stability: 3° > 2°> 1° > methyl.
(vi) Benzyl carbocation is less stable than allyl carbocation.
(vii) Isopropyl free radical is less stable than 3°-butyl free radical.
(viii) Dichloro carbene has 2 lone pairs.
(ix) Carboanions are formed by heterolysis of a covalent bond.
13. Find the value of 3 2 x + where x = total number of compounds possible in tautomerism.
Then the sum of P + Q + R + S =?
9. The number of rings present in a compound of molecular formula C 8 H 12 that is hydrogenated to a compound of molecular formula C8H12 is X. Find the value of X?
(Passage-based Questions)
14. A mixture contains 3 g of (+)-2-Bromobutane and 2 g of (–)-2-Bromobutane. (+)-2-Bromobutane has a specific rotation of +23.1°
I. Calculate minor enantiomer.
II. What is the observed rotation of mixture?
ANSWER KEY
JEE Main Level
15. Study the following molecule to answer the given questions.
I. Total number of α-hydrogens to ketonic functions is
II. Degree of unsaturation = w
Total no of chiral centres = x
Total no. of sp2 carbons = y
Total number of enantiomers = z
Then, w + x + y + z =?
9: Organic Chemistry – Some Basic Principles and Techniques
(41) 2 (42) 2 (43) 1 (44) 1 (45) 4 (46) 1 (47) 2 (48) 4 (49) 3 (50) 2 (51) 4 (52) 3 (53) 3 (54) 1 (55) 2 (56) 4 (57) 4 (58) 56 (59) 18 (60) 2 (61) 4 (62) 2 (63) 3 (64) 11 (65) 5 (66) 5
Level -III (1) 4 (2) 1 (3) 1 (4) 3 (5) 2 (6) 5 (7) 3 (8) 3 (9) 3 (10) 2 (11) 4 (12) 2 (13) 3 (14) 2 (15) 3 (16) 2 (17) 1 (18) 4 (19) 2 (20) 2 (21) 1 (22) 3 (23) 4 (24) 4 (25) 4 (26) 1 (27) 2 (28) 4 (29) 3 (30) 3 (32) 4 (32) 2 (33) 13 (34) 2 (35) 1 (36) 1 (37) 3 (38) 4 (39) 4 (40) 4 (41) 2 (42) 3 (43) 25 (44) 14 (45) 13 (46) 26.8
Theory-based Questions (1) 1 (2) 1 (3) 2 (4) 2 (5) 1 (6) 1 (7) 1 (8) 1
Advanced Level
1,2,3 (10) 1,2,4 (11) 1,2,3,4(12) 2,3,4 (13) 2,4 (14) 2,4 (15) 2,3,4 (16) 12 (17) 40 (18) 0 (19) 42 (20) 4 (21) 3 (22) 3 (23) 10 (24) 7 (25) 5 (26) 7 (27) 5 (28) 9 (29) 9 (30) 3 (31) 5 (32) 6 (33) 0 (34) 4 (35) 2 (36) 2 (37) 3 (38) 3 (39) 3 (40) 7 (41) 5 (42) 2 (43) 2 (44) 3 (45) 1 (46) 4 (47) 108 (48) 9 (49) 0 (50) 1 (51) 1 (52) 1 (53) 3 (54) 2 (55) 2 (56) 3 (57) 2 (58) 1 (59) 2 (60) 4 (61) 4 (62) 4 (63) 1 (64) 4
Brain Teasers
Flashback
Chapter Test–JEE Main
Chapter Test–JEE Advanced
Chapter Outline
10.1 Alkanes
10.2 Alkenes
10.3 Alkynes
10.4 Aromatic Hydrocarbons
Organic compounds composed of only carbon and hydrogen are called hydrocarbons. Hydrocarbons can be broadly classified into two main classes, open chain (acyclic) hydrocarbons and closed chain (cyclic) hydrocarbons. Open chain hydrocarbons can again be classified as saturated and unsaturated hydrocarbons. Saturated hydrocarbons contain carbon-carbon and carbon-hydrogen single bonds. Unsaturated hydrocarbons contain carbon-carbon multiple bonds.
Hydrocarbons play an important role in our daily life. LPG is the abbreviated form of liquefied petroleum gas, CNG is compressed natural gas and LNG is liquefied natural gas. By the fractional distillation of petroleum found under the earth’s crust, petrol, diesel, and kerosene oil are obtained. Natural gas is found in upper strata during drilling of oil wells. LPG and kerosene oil are used as domestic fuels. LPG causes least pollution but kerosene oil causes some pollution. Higher hydrocarbons are used as solvents for paints and for the manufacture of many dyes and drugs.
10.1 ALKANES
Alkanes are aliphatic saturated hydrocarbons. Carbon-carbon single bonds are completely non-polar and carbon-hydrogen bonds are also
HYDROCARBONS
almost non-polar. These are all strong bonds. Alkanes are inert under normal conditions as they do not react with acids, bases, and other reagents. Hence, they are known as paraffins. The general formula of alkanes is C n H 2n + 2 Each carbon atom is sp3 hybridised; its four bonding orbitals are directed towards the four corners of a regular tetrahedron. Each carbon-carbon bond is formed by the overlap of sp3 orbitals, one from each carbon atom. All carbon-hydrogen bonds result in overlap of an sp3 hybrid orbital from carbon and an s-orbital from hydrogen. The bond lengths between carbon-carbon and carbon-hydrogen are 1.54� and 1.12�, respectively. The bond angles in alkanes are tetrahedral angles having a value of 109�5'. Carbon-carbon bond dissociation energy is 341.1 kJ mol–1 and carbon-hydrogen bond dissociation energy is 415.9 kJ mol –1
10.1.1 Nomenclature and Isomerism
In the common (trivial) system of nomenclature, the straight-chain paraffins are termed normal, those with a branched chain like
10: Hydrocarbons
(containing a tertiary carbon atom) are termed iso-, whereas others containing a quaternary carbon atom are called neo.
Higher alkanes containing four or more carbon atoms have different possible structures.
Isomers of butane CH CH CH CH 3 223 n - Butane (Butane) CH | isobutane (Methyl 3 CH CH CH 33 p propane) Isobutane (Methylpropane)
Isomers of pentane CH CH CH CH CH 3 2223 n-pentane (Pentane) CH3 – C – CH3
CH3 – CH – CH2 – CH3 CH3 | Isopentane (methylbutance)
CH3 | | Neopentane (Dimethylpropne)
(IUPAC names are given in parenthesis)
C6H14 has got five isomers. They are n-hexane, 2 - methyl pentane, 3-methylpentane, 2, 3-dimethylbutane, and 2, 2-dimethylbutane. C 7 H 16 has nine isomers, C 8 H 18 has eighteen isomers.
C9H20 thirty five isomers, C10H22 seventy five isomers, etc. With increase in the number of carbons, number of structural isomers also increases enormously. (Number of structural isomers with the formula C25H52 is 736 million). These isomers differ in the number and nature of carbon atoms and hydrogen atoms.
Some typical examples of IUPAC nomenclature are given below:
5-sec-butyl-4-isopropyldecane
(Isopropyl is taken as one word and sec is not considered while arranging alphabetically) (or) 4-(methylethyl)- 5 - (1 - methylpropyl) decane
3,3-diethyl-5-isopropyl-4-methyloctane (or) 3, 3-diethyl-5-(methylethyl)-4-methyloctane
Conformational Isomerism
The different arrangement of atoms in space that result from the free rotation of groups about C–C single bond are called conformations, or conformers, or conformational isomers, or rotational isomers, or rotamers. The phenomenon is known as conformational isomerism. Alkanes can thus have infinite number of conformations by rotation around C–C single bonds. The rotation around a C–C single bond is not completely free. The energy required is between 1–20 kJ mol –1 , which is very less. This type of repulsive interaction is called torsional strain.
Conformations of Ethane
In ethane molecule (CH3–CH3), the two carbon atoms are bonded by a single covalent bond and each of the carbon atoms is further linked to three hydrogen atoms. If one of the carbon
atoms is held still and the other carbon atom is allowed to rotate around the C–C bond, infinite number of different spatial arrangements of hydrogen atoms of one carbon atom with respect to the hydrogen atoms of the other carbon atom can be obtained. However, two conformations represent the extremes.
In staggered conformation, the hydrogen atoms of the two carbon atoms are oriented in such a way so that they lie far apart from one another. In eclipsed conformation, the hydrogen atoms of one carbon atom are lying directly behind the hydrogen atoms of the other. Any other intermediate conformation is called a skew conformation. In all these conformations, the bond angles and the bond lengths remain the same. Eclipsed and staggered conformations can be represented by Sawhorse or Newman or line–wedge projections.
In the staggered one, the distance between the hydrogen nuclei is 2.55 A� but in eclipsed, it is 2.29 A�. The energy difference between the two extreme forms is of the order of 12.5 kJ mol–1, which is very small. The energy vs dihedral angle plot of ethane is as shown in Fig.10.1
different conformational isomers of ethane. Sawhorse projections: In Sawhorse projections, the molecule is viewed along the molecular axis. The central carbon-carbon bond is represented by a straight line, slightly tilted to the left. The front carbon is shown at the lower end of the line, whereas the rear carbon is shown at the upper end. Each carbon has three lines attached to it corresponding to three hydrogen atoms inclined at an angle of 120° to each other. The extreme two conformations of ethane are depicted in Fig.10.2.
Fig. 10.1 Energy Vs dihedral angle
[E = Eclipsed and S = staggered form]
Even at ordinary temperatures, the ethane molecule gains thermal or kinetic energy sufficient enough to overcome this energy barrier of 12.5 kJ mol–1 through intermolecular collisions. Hence, rotation about carbon-carbon single bond in ethane is almost free for all practical purposes. It has not been possible to separate and isolate
Fig.10.2 Sawhorse projections of ethane
Newman projections: In Newman projections, the molecule is viewed from the front. The two carbon atoms forming the sigma bond are represented by two circles, one behind the other, so that only the front carbon is seen. The hydrogen atoms attached to the front carbon are represented by C–H bonds from the centre of the circle. The C–H bonds of the back carbon are drawn from the circumference of the circle. These can be represented for ethane as depicted in Fig.10.3
In staggered form of ethane, the electron clouds of carbon-hydrogen bonds are as far as possible. Thus, there are minimum repulsive forces, minimum energy, and maximum stability of the molecule. On the other hand, when the staggered form changes into the eclipsed form, the electron clouds of the carbon-hydrogen bonds come closer to each other, resulting in increase in electron cloud repulsions, possesses more energy, and thus, has lesser stability. The difference in the energy content of staggered and eclipsed conformations is 12.5 kJ mol –1
The repulsive interaction between the electron clouds, which affects the stability of a conformation, is called torsional strain (or) destabilisation associated with the eclipsing of bonds on adjacent atoms is called torsional strain and the angle of rotation about C–C bond is called dihedral angle or torsional angle. Line–wedge representation, also called wedge–and–dash structure of ethane, is shown as follows:
10.1.2 Methods of Preparation
Petroleum and natural gas are the main sources of alkanes. Petroleum is the main source of alkanes containing up to 40 carbon atoms. Natural gas contains mainly lower alkanes, about 80% methane, 10% ethane, and the remaining 10% being a mixture of higher members.
From Unsaturated Hydrocarbons
In the presence of finely divided nickel at 200–300�C, alkanes are obtained by hydrogenation of unsaturated hydrocarbons. This is known as Sabatier and Senderen’s reaction. Platinum and palladium catalyse the reaction at room temperature.
R–CH = CH2 + H2 RaneyNi C 200 300 R–CH2–CH3
R–C ≡ CH + 2H2 RaneyNi C 200 300 R–CH2–CH3
Ethylene or acetylene on catalytic (Ni or Pt or Pd) hydrogenation gives ethane.
CH2 = CH2 + H2 Pt Pd Ni // CH3–CH3
CH – CH + 2H2 Pt Pd Ni // CH3–CH3
CH3 – C ≡ C – H + 2H2 Pt Pd Ni // CH3–CH2 – CH3
Methane cannot be prepared by this method.
From Alkyl Halides
R eduction: A lkyl halides, except fluorides, undergo reduction with nascent hydrogen to alkane. The nascent hydrogen may be obtained from Zn–HCl (or) Zn–CH3COOH (or) Zn–Cu couple – C2H5OH (or) Zn – NaOH, etc.
R–X + 2[H] Zn H , R–H + H–X
When methyl bromide is reduced with zinc and hydrochloric acid, methane is formed.
CH3Br + 2[H] Zn H , CH4 + HBr
Ethyl chloride, on reduction with zinccopper couple and ethyl alcohol, forms ethane.
C2H5Cl + 2[H] Zn H , C2H6 + HCl
Reduction can also be done by any one of the reducing agents mentioned above.
CH3CH2CH2Cl + H2 Zn H , CH3CH2CH3
Wurtz Reaction
Alkyl halides, on treatment with sodium metal in dry ethereal solution, give higher alkanes. This reaction is known as Wurtz reaction and is a convenient method for the preparation of higher alkanes having even number of carbon atoms. Usually, bromides and iodides are preferred.
R–X + 2Na + X–R Dry ether R–R + 2NaX
Methyl bromide, on treating with sodium metal in dry ether, gives ethane.
CH3Br+2Na+BrCH3 Dry ether
CH3-CH3+2NaBr
Methane cannot be prepared by this method.
When a mixture of two different alkyl halides is used, a mixture of three different alkanes is obtained. This is called crossedWurtz reaction.
When a mixture of methyl bromide and ethyl bromide is used, the following reactions take place.
CH Br +2Na +BrC HCH- CH +2NaBr Propane 32 53 25 Dry ether
CH Br +2Na +BrCHCH- CH + 2NaBr Ethane 33 33
CH Br +2Na+BrC HC HC H+2NaBr 25 25 25 25 Butane
CH Br +2Na+BrC HC HC H+2NaBr 25 25 25 25 Butane
The separation of mixture of alkanes is not easy because their boiling points are very close. Hence, this method is not convenient to prepare alkanes having odd number of carbon atoms.
Frankland Reaction
CH3 Cl + Zn + Cl – CH3
CH3 – CH3 + ZnCl2 Dry ether
From Fatty Acids
Decarboxylation: Sodium salts of fatty acids, on heating with soda lime (NaOH + CaO), give alkanes containing one carbon atom less than the carboxylic acid. This is called decarboxylation.
RCOONa + NaOH CaO R–H + Na2CO3
Decarboxylation of sodium acetate with soda lime gives methane.
CH 3COONa + NaOH CaO CH 4 + Na2CO3
Decarboxylation of sodium propanoate with soda lime gives ethane.
C 2 H 5 COONa + NaOH CaO C 2 H 6 + Na2CO3
Kolbe’s electrolysis: When a concentrated aqueous solution of sodium or potassium salt of a carboxylic acid is electrolysed, alkane containing even number of carbon atoms along with carbon dioxide at anode and sodium or potassium hydroxide along with hydrogen are formed at cathode.
2RCOONa + 2H2O Electrolysis
R–R + 2CO2 + 2NaOH + H2
When a concentrated aqueous solution of sodium acetate is electrolysed, ethane is liberated at anode.
2CH3COONa + 2H2O Electrolysis
CH3–CH3 + 2CO2 + 2NaOH + H2
22 2 33 CH COONaCHCOO Na
At anode: 22 2 33 CH COOCHCOO e
22 2 33 2 CH CO OC HCO
CH CH CH CH 33 33
At cathode: 2H2O + 2e– → 2OH– + H2
Methane cannot be prepared by this method.
10.1.3
Physical Properties
Among many organic compounds, alkanes have weak intermolecular attractions. They are responsible for many of their physical properties.
Physical State
Alkanes are almost non-polar molecules. Weak van der Waals forces operate between their molecules. The strength of these forces increases with increase in their molecular weights. Due to this, C1 to C4 are gases, C5 to C 17 are liquids, and those containing more than 17 carbon atoms are solids at room temperature. Alkanes are colourless and odourless. Petrol is a mixture of hydrocarbons and is used as a fuel for automobiles. Grease, a mixture of higher hydrocarbons, is soluble in petrol. Hence, petrol is used to remove grease strains in dry cleaning.
Density
The density of gaseous alkanes increases very slowly with the rise in molecular mass until it becomes constant at about 0.8 g cm–3. All alkanes are lighter than water. All alkanes are heavier than air, except methane.
Solubility
Due to the non-polar nature of alkanes, they are insoluble in polar solvents, like water, but soluble in non-polar solvents, like ether, benzene, carbon tetrachloride, etc. It is generally observed that in relation to solubility of substances in solvents, ‘like dissolves like’.
Boiling Points
The boiling points of n-alkanes increase regularly with increasing number of carbon atoms. Except for the first few members, the boiling point difference between two successive members of the series is about 20–30 degrees. Among the isomeric alkanes, the normal isomer has a higher boiling point than the branched chain isomer. The greater the branching of the chain, the lower is the boiling point. This is due to the fact that the magnitude of the inter molecular forces depends upon the surface area. The branching of the chain makes the molecule more compact and brings it closer to a sphere. Thus, intermolecular forces are
weaker in branched chain isomers, thereby, they show, and so, lower boiling points.
(309K)
neopentane (283K)
Melting Points
The melting points of alkanes do not show a very smooth gradation with the increase in molecular mass. With the increase of molecular mass of normal alkanes, there is greater increase in melting point from alkane with odd number carbon atoms to the next higher homologue with even number carbon atoms. But from alkane, with even number carbon atoms to that with the next higher homologue with odd number of carbon atoms, there is no greater increase in melting point. Alkanes having odd number of carbon atoms are relatively more unsymmetrical (end carbon atoms on the same side) than with even number of carbon atoms (end carbon atoms on the opposite side). Alkanes with even number of carbon atoms pack closely in such a manner as to permit greater intermolecular attraction and, therefore, have slightly higher melting point. Variation of melting points (m.p.) and boiling points (b.p.) in alkanes is given Table 10.1.
Table 10.1 Variation of melting points and boiling points in alkanes
10.1.4 Chemical Properties
Alkanes are quite inert towards acids, bases, oxidising agents, and reducing agents. The characteristic reactions of alkanes are substitution reactions.
Substitution Reactions
One or more hydrogen atoms of alkanes can be replaced by halogens, nitro group, and sulphonic acid group.
Halogenation: When alkanes are treated with halogens (particularly chlorine or bromine) in the presence of sunlight or UV light or at elevated temperatures, the hydrogen atoms of alkanes are successively replaced by halogen atoms. When methane is chlorinated, we get a mixture of four different derivatives.
CH4 + Cl2 h
CH3Cl + Cl2 h
CH2Cl2 + Cl2 h
CHCl3 + Cl2 h
CH3Cl + HCl
CH2Cl2 + HCl
CHCl3 + HCl
CCl4 + HCl
If excess of methane is used, the main product formed is methyl chloride.
Ethane, on chlorination, gives mono-, di-, tri-, tetra-, pentachloroethane, and finally, hexachloroethane.
Ea se of r eplacement of hydrogens follows the order:
Tertiary > Secondary > Primary.
Reactivity of halogens towards alkanes: Fluorine > Chlorine > Bromine > Iodine.
Fluorination is violent and so, it is to be controlled. Iodination is very slow and also reversible. Iodination can be carried out in the presence of oxidising agent, like HNO3 or HIO3.
CH4 + I2 CH3I + HI
(HIO3 + 5 HI → 3 I2 + 3H2O)
2HNO3 + 2HI → I2 + 2NO2 + 2H2O
Halogenation of alkanes in light is a free radical substitution reaction. It involves three steps: Chain initiation: In presence of heat or light, chlorine molecule undergoes homolysis to form chlorine free radicals.
Cl Cl Cl Cl h
Chain propagation: Chlorine free radical attacks methane molecule breaking one of the C–H bonds and generating methyl free radical with the formation of HC l.
1. CH Cl CH HCl 43
The methyl free radical thus obtained attacks the second molecule of chlorine to form CH3Cl
with the liberation of another chlorine free radical.
2. CH3 + Cl – Cl CH3 + Cl – Cl
Steps (i) and (ii) are repeated several times and, thereby, set up a chain. Many other propagation steps are possible and may occur, thereby forming other chlorine substituted products.
Chain termination: This involves the combination of two free radicals, in which case, chain is terminated.
Cl + Cl Cl – Cl
CH3 + CH3 CH3 – CH3
CH3 + Cl CH3 – Cl
Reaction of chlorine with ethane, Chain initiation:
Cl Cl Cl h 2
(i) C2H6 + Cl • h C2H5 + HCl
(ii) C2H5 + Cl2 h C2H5 Cl + Cl
Chain termination: Cl + Cl Cl2
C2H5 + C2H5 C4H10
C2H5 + Cl C4H5Cl
Oxidation
Combustion: Alkanes are difficult to oxidise. On burning in the presence of excess air or oxygen, they are oxidised to carbon dioxide and water with the evolution of large amount of heat.
CH4 + 2O2→ CO2 + 2H2O; ΔH = – 890 kJ mol–1
2C2H6 + 7O2 → 4CO2 + 6H2O + Heat
C4H10 + 13 2 O2 → 4CO2 + 5H2O + 2875 kJ mol–1
The general combustion equation is given as
CH O n n n 22 2 31 2
nCO2 + (n+1)H2O + Heat
In complete oxidation: When alkanes are heated with insufficient amount of air or oxygen, carbon black is formed, or even carbon monoxide may be formed.
CH4 + O2 → C + 2H2O
2CH4 + 3O2 → 2CO + 4H2O
Controlled oxidation: Alkanes, on heating with a regulated supply of air or oxygen at high pressure and in the presence of suitable catalysts, give alcohols or aldehydes or ketones or carboxylic acids. These are also called catalytic oxidations.
At 250�C, in the presence of copper catalyst, methane is oxidised to methyl alcohol.
2CH4 + O2 Cu atm 100 2CH3OH
Methane, on oxidation with ozone or oxygen in presence of Mo2O3 catalyst, at 350 - 500�C, gives formaldehyde.
CH4 + O2 Mo O 23 HCHO + H2O
In the presence of manganese acetate catalyst, ethane is oxidised to acetic acid.
2CH3CH3 + 3O2 () CH32COOMn
2CH3COOH + 2H2O
Ordinarily, alkanes resist oxidation. Alkanes containing tertiary hydrogen atoms get converted into alcohols on oxidation with potassium permanganate. It is called chemical oxidation.
Isobutane, on oxidation with potassium permanganate, forms tertiary butyl alcohol.
Isomerisation
The process of conversion of one isomer into another isomer is called isomerisation. n-alkanes, when heated in presence of anhydrous A l C l 3 and HC l gas, isomerise to give one or more isomers. n-butane isomerises to isobutane. n-hexane isomerises to a mixture of 2-methyl pentane and 3-methylpentane.
Aromatisation
The conversion of aliphatic compounds into aromatic compounds is known as aromatisation. n-alkanes, having six or more carbon atoms, on heating to 500–600 0C at 10–20 atmospheric pressure in the presence of oxides of vanadium, molybdenum, or chromium, supported over alumina, get dehydrogenated and cyclised to benzene and its homologue. This can also be called reforming. n-hexane is aromatised to benzene and n-heptane to toluene.
Pyrolysis
The decomposition of a compound by heat is called pyrolysis or thermal decomposition. When pyrolysis occurs in alkanes, the process is termed as cracking. This is also known as fragmentation. When alkanes are heated to high temperatures in the absence of air, the higher alkane molecules break up into smaller molecules of lower alkanes, alkenes, and hydrogen. Pyrolysis or cracking is of great importance to the petroleum industry.
Heating methane in the absence of air at 1000�C gives carbon black. Carbon black is used in the manufacture of tyres, paints, black pigments, printing ink, filters, etc.
CH4
1000 C C + 2H2
When ethane is heated in absence of air at 450�C, it gives ethylene and hydrogen.
C2H6
C C2H4 + H2
C6H12 + H2
Hexene
C4H8 + C2H6
Butene Ethane
C3H6 + C2H4 + CH4
Propene Ethene Methane
Pyrolysis of alkanes is a free radical reaction. Preparation of oil gas or petrol gas from kerosene oil or petrol involves the principle of pyrolysis. For example, dodecane, a constituent of kerosene oil, on heating to 700�C in presence of platinum, palladium, or nickel, gives a mixture of heptane, pentene, and other products.
C12H26 → C7H16+C5H10 + Other products
Dodecane Heptane Pentene
At 10000C, methane reacts with steam in the presence of nickel to give hydrogen. This method is used for the industrial preparation of hydrogen.
CH4 + H2O Ni C 1000 CO + 3H2
1. Sodium salt of which fatty acid is needed to prepare butane? Give the equation.
Sol. Pentanoic acid has five carbon atoms. On decarboxylation, sodium salt of pentanoic acid gives butane.
C4H9COONa + NaOH CaO C 4 H 10 + Na2CO
2. Which alkane can be obtained by the electrolysis of the aqueous solution of sodium propanoate?
Sol. n-Butane is obtained
2CHCHCOO Na + 2HO 32 2 Kolbes electrolysis ’
CH CH CH CH + 2CO + 2NaOH + H 3 223 22 At anode At cathoode
Try yourself:
(3) 2525 CHClCHONa+→
(4) Agpowder 3 Ä CHCl →
2. The alkane which cannot be prepared by Wurtz reaction is (1) ethane
(2) 2, 3 – dimethyl butane
(3) n – hexane
(4) neo pentane
3. Sodium salt of X is heated with soda lime to give ethane. X is (1) ethanoic acid
(2) methanoic acid
(3) propanoic acid
(4) butanoic acid
4. A mixture of two organic compounds was treated with sodium metal in ether solution. Isobutane was obtained as one of chief products. The two chlorine compounds are
(1) methyl chloride and propyl chloride
(2) methyl chloride and ethyl chloride
(3) isopropyl chloride and methyl chloride
(4) isopropyl chloride and ethyl chloride
5. Which of the following liberates methane on treatment with water?
(1) Silicon carbide
(2) Calcium carbide
(3) Beryllium carbide
4MgX→CH3OH+CH5H2C + )X5H2Mg(OC
1. Predict the alkane formed when C 2H 5OH is treated with methyl magnesium halide. Write the equation. Ans:Methane is formed.
TEST YOURSELF
1. Wurtz reaction of methyliodide yields an organic compound, x. Which one of the following reactions also yields x?
(1) dryEther 25 CHClMg+→
(2) 25 4 CHClLiAlH+→
(4) Magnesium carbide
6. Which of the following is not prepared by Kolbe’s electrolytic process?
(1) CH4 (2) C2H6
(3) C4H10 (4) C6H14
7. Which of the following hydrocarbons is not formed when Wurtz reaction takes place between ethyl iodide and propyl iodide?
(1) Butane (2) Propane
(3) Pentane (4) Hexane
8. Which of the following does not give alkane with RMgX (R= alkyl group)?
(1) C6H5OH
(2) CH3–O–CH3
(3) CH3COOH
(4) HCl
9. Which of the following compounds has the highest boiling point?
(1) iso-butane (2) iso-pentane
(3) n-pentane (4) n-butane
10. The most volatile compound is (1) 2, 2 - dimethyl propane (2) isobutane
(3) n-pentane
(4) 2-methyl butane
11. The boiling points of three isomeric pentanes 1, 2, and 3 are 1) 9.5°C
2) 28°C 3) 36°C
They are, respectively,
(1) n-pentane, iso-pentane, neo-pentane
(2) iso-pentane, neo-pentane, n-pentane
(3) n-pentane, neo-pentane, iso-pentane
(4) neo-pentane, iso-pentane, n-pentane
12. Alkanes with how many carbons are solids at room temperature?
(1) above C17 (2) above C10
(3) above C8 (4) above C4
13. In the following reaction, ‘ X ’ is ( ) 3 Anhy.AlCl 323 4 HCl CHCHCH →
Major product (1) CH3CH(CH3)(CH2)2CH3 (2)
(3) Cl – CH2 – (CH2)4 CH2 – Cl
(4) CH3(CH2)4CH2Cl
14. Na 25 dryether CHI → A; total number of isomers
(structural and ster eo) obtained by monobromination of ‘A’ will be (1) 3 (2) 1 (3) 2 (4) 4
15. Catalyst and suitable conditions used in aromatisation of n-hexane is (1) HCl/anhydrous AlCl3
(2) Cr2O3/773 K, 10–20 atm (3) Cu / 523 K, 100 atm (4) Ni/1273 K
Answer Key (1) 2 (2) 4 (3) 3 (4) 3 (5) 3 (6) 1 (7) 2 (8) 2 (9) 3 (10) 1 (11) 4 (12) 1 (13) 1 (14) 1 (15) 2
10.2 ALKENES
Alkenes are unsaturated hydrocarbons containing carbon-carbon double bond. Their molecules have two hydrogen atoms less than the corresponding members of alkanes, with general formula, CnH2n
Alkenes are also known as olefins, since the first member, ethylene, was found to form an oily liquid on reaction with chlorine. Carboncarbon double b ond is made up of a sigma bond and a pi bond.
10.2.1 Nomenclature and Isomerism
Alkenes from C4H8 onwards exhibit isomersim.
CH2=CH–CH2–CH3 1-Butene (I)
CH3–CH=CH–CH3 2-Butene (II)
CH2=C–CH3 Methyl propene (III) | CH3
I and III and II and III are chain isomers. I and II are position isomers. For C 5 H 10, five structures are possible.
Alkenes, in addition to structural isomerism, mainly exhibit geometrical isomerism. Hydrogens present on sp2 carbon are vinylic hydrogens and hydrogens present on sp3 carbon attached to sp2 carbon are allylic hydrogens.
10.2.2 Structure of the Double Bond (> C = C <)
In C = C, there is one bond, a strong bond formed due to end–end overlap of sp2 hybrid orbital of one carbon with the sp2 hybrid orbital of other carbon atom, each containing one unpaired electron (bond enthalpy is 397 kJ mol–1). The other bond is a weak bond formed due to lateral overlap of p orbital of one carbon atom with the p orbital of the other carbon atom, each containing one unpaired electron (bond enthalpy is 284 kJ mol–1). The double bond is shorter in length (134 pm) than C –C single bond (154 pm). The bond contains loosely bonded electrons. These electrons attract electrophiles. Therefore, alkenes are more reactive than alkanes and undergo the characteristic electrophilic addition reactions.
70
De gree of unsaturation (DU) (or) Double bond equivalence (DBE): It is hydrogen deficiency index.
If DU = 1; 1 indicating the presence of double bond / 1 ring
If DU = 2; indicating the presence of 2 double bonds / 1 triple bond / 1 double bond + 1 ring / 2 rings
DBE = 22 2 12 nn
n1 = Number of carbons
n2 = Number of hydrogen atoms per molecule
Example: Cubane is C8 H 8
DBE = 28 28 2 5
10.2.3 Preparation
Alkenes can be prepared from alkynes, alcohols, haloalkanes, etc.
Dehydration of Alcohols
Alcohols on heating with concentrated sulphuric acid form alkenes with the elimination of one water molecule. Other dehydrating agents are alumina at 350�C or phosphoric acid at 200�C. This reaction is an example of b -elimination reaction since –OH group takes out one hydrogen atom from the b -carbon atom. Since a water molecule is eliminated from the alcohol molecule in the presence of an acid, this reaction is known as acidic dehydration. The ease of dehydration of alcohols is 3� > 2� > 1�.
R–CH2–CH2OH → R–CH = CH2+H2O
Ethyl alcohol, on heating with concentrated sulphuric acid at 170�C or with alumina at 350�C, gives ethylene.
CH3CH2OH → CH2 = CH2 + H2O
Saytzeff Rule
In the dehydration of secondary and tertiary alcohols, when there is a possibility of formation of two isomers, the hydrogen atom is preferentially eliminated from the adjacent carbon atom (adjacent to carbon having –OH group) with the fewer number of hydrogen atoms. This is called Saytzeff’s rule.
CH3-CH2-CH-CH3
CH3-CH=CH-CH3 + OH 2-Butene(80%)
CH3-CH2-CH=CH2 1-Butene (20%)
Dehydrohalogenation of Alkyl Halides
Alkyl halides, on heating with alcoholic potash (potassium hydroxide dissolved in alcohol), form alkenes with the elimination of one molecule of halogen acid. This is called dehydrohalogenation and is also an example of b -elimination reaction. Nature of halogen atom and the alkyl group determine rate of the reaction.
For halogens, the rate is: I – > Br– > Cl–
For alkyl groups , the rate is: 3� > 2� > 1�.
R–CH2–CH2X + KOH Alcohol
RCH = CH2 + KX + H2O
Ethyl chloride, on heating with alcoholic solution of potassium hydroxide, gives ethylene.
CH3CH2Cl + KOH Alcohol
CH2 = CH2 + KCl + H2O
According to Saytzeff’s rule, when two alkenes are likely to form, highly substituted alkene is the predominant product.
Dehalogenation of Vicinal Dihalides
Dihalides in which two halogen atoms are attached to two adjacent carbon atoms are called vicinal dihalides. When these are heated with zinc dust in alcohol solution, the two halogen atoms are removed and alkene is obtained. This reaction is called dehalogenation.
RCHCHR Zn
Alcohol ||
Br Br
RCH = CH R + ZnBr2
When ethylene bromide is heated with zinc dust in alcohol solution, ethylene is formed.
CH2Br–CH2Br + Zn Alcohol
CH2 = CH2 + ZnBr2
Propene is obtained from 1, 2-dichloropropane.
CH3CHBr – CH2Br + Zn →
CH3 – CH = CH2 + ZnBr2
Controlled Hydrogenation of Alkynes
Alkynes, on partial reduction with calculated amount of hydrogen in presence of Lindlar’s catalyst, form alkenes. Lindlar’s catalyst is palladium supported over barium sulphate containing sulphur or quinoline. Partially deactivated palladised charcoal is also known as Lindlar’s catalyst.
R–C ≡ CH + H2 → R – CH = CH2
R–C ≡ C – R + H2 → R – CH = CH – R
Controlled hydrogenation of acetylene in the presence of Lindlar’s catalyst or Pd-silica gel at 200�C gives ethylene.
HC ≡ CH + H2 Pd CBaSO Quinoline / 4 CH2 = CH2
When Lindlar’s catalyst is used, the predominant product is cis -alkene (syn addition).
R-C ≡ C-R + H2 R C = C H R Alkyne cis - alkene H Lindlar ' s catalyst →
However, alkynes, on reduction with sodium in liquid ammonia, form trans alkenes (anti addition).
R-C ≡ C-R+H2 R C=C H AR lkyne trans-alkene H 3 Na liq.NH →
10.2.4 Physical Properties
The first three members are gases, the next fourteen are liquids, and the higher ones are solids. Ethylene is a colourless gas with a faint sweet smell. All other alkenes are colourless
10: Hydrocarbons
and odourless, and insoluble in water but fairly soluble in non -polar solvents like benzene, ether, etc. They show a regular increase in boiling point with increase in molecular weight. For every –CH 2–group, increase in boiling point is around 20–30�C. Like alkanes, straight chain alkenes have higher boiling point than isomeric branched chain compounds.
Alkenes are less volatile than alkanes. Their boiling and melting points are higher than corresponding alkanes. Like alkanes, density of alkenes also shows regular gradation with increase in molecular mass. All alkenes are also lighter than water.
10.2.5 Chemical Properties
Carbon-carbon double bond in alkenes consists of strong sigma bond and a relatively weaker pi bond. The typical reactions of alkenes are the reactions in which the pi bond breaks and two strong sigma bonds are formed in its place. Most of the addition reactions in alkenes are initiated by electrophilic species because the loosely held pi electrons in alkenes serve as source for the attacking species. Thus, the characteristic reactions are electrophilic addition reactions. Some reagents also add by free radical mechanism. Under some special conditions, alkenes also undergo free radical substitution reactions. Oxidation, ozonolysis, and polymerisation are also prominent in alkenes.
Addition Reactions
Addition of hydrogen: Alkenes combine with hydrogen under pressure and in presence of a catalyst to form alkanes. This process is known as catalytic hydrogenation. With nickel catalyst, temperature of about 200–300�C and in presence of Pt or Pd, even at room temperature, hydrogenation occurs.
Ethylene adds on hydrogen in presence of Ni or Pt or Pd catalyst fo rming ethane.
CH2 = CH2 + H2 Ni CH3–CH3
Addition of halogens: Halogens, especially chlori ne and bromine, add on to alkenes at ordinary temperature to form vicinal dihalides.
The reaction is carried out in an inert solvent, like CCl4. The reddish orange colour of bromine solution in CCl 4 is discharged when bromine adds on an unsaturation site. This reaction is used as a test for unsaturation.
Addition of halogens to alkenes is electrophilic addition reaction.
Ethylene adds with bromine or chlorine directly to form ethylene bromide or ethylene chloride, respectively.
CH2 = CH2 + Br2 CCl4 BrCH2–CH2Br
Addition product of alkene and chlorine or bromine is a trans product. This reaction proceeds through a three–membered cyclic halonium ion intermediate.
U nder the normal conditions, alkenes do not add on iodine. Addition of fluorine is explosive.
Order of reactivity of halogens:
Cl2 > Br2 > I2
10.2.6 Stereochemistry of Some Addition Reactions
Alkene gives syn addition reaction with
(i) 4 KMnO OH HOH
(ii) OsO4
(iii) BH 3
cis alkene gives meso product and trans alkene gives (dl) mixture in syn addition reaction.
Baeyers reagent C OH H C H OH
Meso form
It is a stereospecific reaction. Baeyers reagent CH3 H CH3 H C OH CH3 H HO C CH3 H (dl) mixture C OH CH3 H HO C CH3 H (i) OsO (ii) NaHSO 4 3 H COOH H COOH C OH H C COOH COOH H OH Meso tartaric acid
It is stereospecific reaction.
HOOC H COOH (i) OsO (ii) NaHSO 4 3 C OH H C COOH COOH HO H (dl) mixture C OH H C COOH COOH HO H
Alkene gives syn addition with BH3, which on oxidation, gives alcohol.
3BH → H BH2 2 2 H O / O H → + H OH
H and OH are syn to each other
Stereochemistry of the products of syn addition reactions of alkenes can be known by the following letters
CSM, i.e.,
C stands for symmetrical cis alkene.
S stands for syn addition.
M stands for meso form (or erythro form).
If cis alkene gives meso form, trans alkene gives (dl) mixture (i.e., threo form).
If the alkene is unsymmetrical, then irrespective of type of addition, product will be racemic mixture.
Anti - Addition Reactions of Alkenes
Alkene gives anti addition reaction with bromine. Symmetrical cis-alkene gives (d l ) mixture and trans alkene gives meso in trans addition reaction.
If alkene is unsymmetrical, always, racemic mixture is obtained.
Mechanism: C C = X + X + slow X X → C C == X + fast → C C X X
H3C H Br2 →
C C CH3 Br H H Br +
C C CH3 Br H H Br (dl) mixture
CH3 H Br2 → CH3 CH3 CH3 Meso form C C Br Br H H
Addition of bromine with alkene is a stereospecific reaction as well as a stereoselective reaction.
Stereochemistry of the product(s) of anti addition reactions of alkenes can be known by the following letters:
CAR, i.e.,
C stands for cis alkene
A stands for anti addition
R stands for racemic mixture (i.e., threo forms)
cis alkenes gives racemic mixture, and trans alkene gives meso form (i.e., erythro forms) in anti addition reaction.
St ereochemistry of the product of anti elimination reactions of the substrate can be known by the following letters:
CAR, i.e.,
C stands for cis alkene (as product)
A stands for anti elimination
R stands for racemic mixture (or threo form) of the substrate
Threo form gives cis alkene, and erythro form (i.e., meso form) gives trans (or E alkene) in elimination reactions.
Addition of hydrogen halides : Alkenes react with hydrogen halides (halogen acids) to form alkyl halides. The order of reactivity of hydrogen halides is : HI > HBr > HC l.
10.2.7 Stereochemistry of Some Elimination Reactions
E 2 reaction is anti–elimination reaction, in which both leaving groups are anti to each other.
Threo-1-bromo-1, 2- diphenylpropane gives cis alkene with alc. KOH. This elimination reaction is a stereospecific reaction.
C6H5 C H C C6H5 H
CH3 alc KOH/” → H C6H5 CH3 C6H5 Cis-alkene
Threo form
On the other hand, erythro form gives trans alkene in this reaction. This reaction is also a stereospecific reaction as well as a stereoselective reaction.
C6H5 C H C C6H5 H
CH3 alc KOH/” →
Erythro form Trans - alkene
Addition of hydrogen halides to alkenes is also electrophilic addition reaction.
Ethylene reacts with hydrogen halides and gives ethyl halides.
CH2 = CH2 + HX → CH3–CH2X
Addition reactions of HBr to symmetrical alkenes take place by electrophilic addition. In case, the alkene and the reagent are both unsymmetrical, then two products are theoretically possible. For example, the reaction between propene and hydrogen bromide gives isopropyl bromide as main product.
2-Bromopropane (90%) 1-Bromopropane (10%) 3 2 CH CH CH HBr = + 3 2 2 CH -CH -CH Br 3 3 CH -CHBr-CH
The products above are formed when the reaction takes place in the dark and in absence of peroxides. Such addition reactions are governed by an empirical rule, called Markovnikov rule. This rule is useful when
a polar reagent adds on to an unsymmetrical alkene. The negative part of the addendum adds on to the carbon atom having less number of hydrogen atoms. (or) Markovnikov rule may also be stated as when a polar reagent adds to an unsymmetrical alkene, the positive part of the attacking reagent attaches itself to a carbon of the double bond so as to yield the more stable carbocation as an intermediate.
Mechanism of the addition of hydrogen bromide to an unsymmetrical alkene is given as follows:
HBr ionises as H+ and Br–. Electrophile, H+ attacks the double bond to form carbocation. + + (a) (b)
H3C-CH=CH2 + H–Br
Less stable primary carbocation More stable secondary carbocation H+
H3C–CH2–CH2 + Br H3C–CH–CH3 + Br
Due to more hyperconjugation in secondary carbocation (b), it is more stable than primary carbocation (a) and forms predominantly at a faster rate. Then, the secondary carbocation (b) is attacked by Br – readily to form the major product, 2-bromopropane.
HC CH + Br 3 CH HC CH CH Br 33 3 |
Peroxide Effect
In the presence of peroxide, addition of HBr to unsymmetrical alkenes takes place against the Markovnikov’s rule. (This happens only with HBr but not with HC l and HI.) This reaction is also known as Kharasch effect or Anti Markovnikov’s rule.
CH3-CH=CH2+HBr Propene Peroxide→ CH3-CH2-CH2Br 1-Bromopropane (major product)
Mechanism of the addition of hydrogen bromide to an unsymmetrical alkene in presence of peroxide is given as follows:
■ Peroxide undergoes homolysis to give free radicals.
RO : O R homolysis 2RO
Peroxide Free radical
Here, R is (benzoyl group)
CH COOC HCO 65 65 2
■ HBr combines with phenyl free radical to form bromine free radical.
■ Br • free radical attacks the double bond of the alkene to form a more stable 2� free radical. (10 free radical) (less stable) (20 free radical) (more stable) 3 2 CH CH CH Br = + 2 3 Br | H CH CH C • 3 2 CH C H CH Br •
■ More stable free radical attacks the HBr to form n-propyl bromide as major product.
CH CH CH Br HBr
CH CH CH Br Br Homolysis 32 32 2 (Major) Br | CH CH CH HBr Homolysis
CH CHBr CH Br 33 (Major)
■ Steps (iii) and (iv) are repeated again and again.
Addition of sulphuric acid: Alkenes add on cold concentrated sulphuric acid to form the corresponding alkyl hydrogen sulphate. This is electrophilic addition and follows Markovnikov rule.
Concentrated sulphuric acid readily adds on to ethylene to form ethyl hydrogen sulphate.
CH2 = CH2 + HOSO3H → C2H5HSO4
Alkyl hydrogen sulphates, when boiled with water, forms alcohols. Hence, this reaction is used for the preparation of alcohols, as alkenes are readily available from cracking of petroleum.
C2H5HSO4 + HOH → C2H5OH + H2SO4
Addition of cold concentrated sulphuric acid to unsymmetrical alkenes takes place in accordance with Markovnikov rule by the electrophilic addition. Propene adds sulphuric acid to form isopropyl hydrogen sulphate.
CH CH=CH + HOSO H 33 2
CH3–CH–CH3
OSO3H
Addition of Water
Water adds on to alkenes in presence of a few drops of concentrated sulphuric acid to form alcohols. This also follows Markovnikov rule.
In presence of sulphuric acid, ethylene adds a molecule of water giving ethyl alcohol.
Examples:
CH CH HOH
CH CH OH HSO 22 32 24
CH CCHH O methylpropen H 32 2 2 | CH 3 (ee) OH | | CH 3
CH CCH 33 ( (. ) terbutylalcoho l
Oxidation Reactions
With 1% alkaline potassium permanganate, a mild oxidising agent called Baeyer’s reagent, alkenes are oxidised to glycols (dihydric alcohols). This is called hydroxylation of double
bond. Decolourisation of pink colour of KMnO4 solution is used as a test for unsaturation.
Baeyer’s reagent oxidises ethylene to ethylene glycol and propene to propylene glycol.
CH2 = CH2 + H2O + (O) → HO–CH2–CH2–OH
Ethane–1, 2–diol
CH3–CH=CH2 + H2O + (O) → CH3CH(OH)CH2OH
Propane–1, 2–diol
Under drastic conditions, acidic potassium permanganate or acidic potassium dichromate oxidises alkenes to ketones and/or carboxylic acids, depending upon the experimental conditions where the carbon-carbon multiple bond is completely cleaved.
(CH3)2C = CH2 + 4(O) →
Methylpropene
(CH3)2C = O + CO2 + H2O
Propanone
CH CH CH CH O KMnO H 33 4 4 But-2-ene 2 3 CH COOH Ethanoic acid
When mixed with oxygen or air and passed through silver catalyst at high temperature (200–400 0 C) and pressure, lower alkenes add an atom of oxygen to form epoxides (cyclic ethers).
Ethylene is oxidised to ethylene oxide with air or oxygen in presence of silver catalyst.
CH CH OCHCH Ag 22 1 2 22 2 O
When burnt in excess air or oxygen, alkenes are oxidised to carbon dioxide and water. The reaction is called combustion or complete oxidation and it is highly exothermic.
C2H4 + 3O2 → 2CO2 + 2H2O + Heat
C nH2n + 3 2 n O2 → nCO2 + nH2O + Heat
Ozonolysis
Alkenes add on ozone to form unstable alkene ozonides, which when hydrolysed in presence of zinc, forms carbonyl compounds are formed. The overall reaction is called ozonolysis or reductive ozonolysis.
Ozonolysis reaction is highly useful in detecting (or locating) the position of the double bond in alkenes or other unsaturated compounds.
Ethylene adds on ozone to form unstable ozonide, which, on hydrolysis in presence of zinc, gives formaldehyde.
CH2=CH2 + O3 → CH2 CH2 O O O Zn HO 2 2HCHO
C=CH2 H3C H3C + O3 C H3C O H3C O O CH2 Methylpropene Ozonide 2 Zn H O →
C=O + HCHO CH3 CH3 Propanone Methanal
CH3 – CH = CH2 + O3 Zn HO 2 CH3CHO + HCHO
Polymerisation
Polymerisation is a process in which a large number of simple molecules combine under suitable conditions to form a giant molecule, known as a macromolecule or a polymer. The simple molecules are known as monomers.
n CH2=CH2
(CH CH )
Polythene Ethene (monomer) (polymer)
n CH3-CH=CH2
Propene Polypropene → 2 n 3 ( CH CH ) | CH
Polymers formed in the polymerisation reaction of substituted ethylene are used in industries in many ways. H2C=CH– is called vinyl group.
Polymers are used for the manufacture of plastic bags, squeeze bottles, refrigerator dishes, toys, pipes, radio and TV cabinets, etc. Polypropene is used for the manufacture of milk crates, plastic buckets, and other moulded articles.
Though these materials have now become common, excessive use of polythene and polypropylene is a matter of great concern for all of us.
TEST YOURSELF
1. The major product in the acid catalysed dehydration of 2 – pentanol is
(1) Pent – 1 – ene
(2) Pent – 2 – ene
(3) Pent – 3 – ene
(4) Methylbut – 1 – ene
2. Which of the following is a Lindlar catalyst?
(1) Sodium and liquid NH3
(2) Zinc chloride and HCl
(3) Partially deactivated palladised charcoal
(4) Cold dilute solution of KMnO 4
3. In the given reaction, ( ) 3Na/liquid.NH 332 CHCCCHHX −≡−+→
(X) will be
(1) butane
(2) trans – 2 – butene
(3) cis-2-butene
(4) 1– butene
4. 0 2 Br,500c 23 CHCHCHA; =−→ , the major product. A is formed through (1) free radical substitution
(2) electrophilic addition
(3) free radical addition
(4) electrophilic substitution
5. The major product is
(1) CH3CH = CHBr
(2) CH2 = CHCH2Br
(3) CH3(Br) C=CH2
(4) CH ≡ C – CH3
6. 25 alcKOH ZnCu 25 CHOH BCHClA; ←→ Here, A and B are
(1) CH4, C2H4
(2) C2H4, C2H6
(3) C2H6, C2H4
(4) C2H6, CH4
7. trans-2-butene 2Br Addition → Major product. The product is
(1) dl-2-3-dibromo butane
(2) 1 – 2, 3 – dibromo butane
(3) A mixture of d, 1 isomers
(4) meso-2, 3-dibromo butane
8. Which of the following chlorides exclusively gives 2-methyl-1-butene on dehydrohalogenation by a strong base?
(1) 2-chloro-2-methylbutane
(2) 2-chloro-3-methylbutane
(3) 1-chloro-2-methylpentane
(4) 1-chloro-2-methylbutane
9. The major product in the dehydro halogenation of 3-gromo-2, 2-dimethyl butane is
(1) 3,3-di methyl but-1-ene
(2) 2,3-di methyl but-1-ene
(3) 2,3-dimethyl but-2-ene
(4) 4-methyl pent-2-ene
10. Ethylene chlorohydrin is obtained from ethylene by the action of (1) dry chlorine gas
(2) dry HCl gas
(3) dilute HCL
(4) a solution of chlorine in water
11. The product formed in the following reaction sequence is
CH3-CH2-CH=CH2 ( ) ( ) i,HBrBenzoylperoxide iiAlcoholicKOH
(1) CH3-CH=CH-CH3
(2) CH3-CH2-CH=CH2
(3) CH3-CH2-CH2-CH2-OH
(4) CH3CH2CH(OH)CH3
12. Reaction of trans-2-phenyl-1bromocyclopentane on reaction with alcoholic KOH produces
(1) 4-phenylcyclopentene
(2) 2-phenylcyclopentene
(3) 1-phenylcyclopentene
(4) 3-phenylcyclopentene
13. The major product of the following reaction 22 Cl/HO
32 CHCHCH−=→
(1) 32 | | OH CHCHCH Cl (2) 32 OH CHCHCH | Cl |
(3) (4)
14. Br/CCl24 alcoholic KOH Zn/ alcoholic 32 Ä CHCHCl A B C −→→→
C is (1) acetylene
(2) ethylene
(3) ethane
(4) methane
15. (1) CH4, C2H4 (2) C2H4, C2H6
(3) C2H6, C2H4
(4) C2H6, CH4
16. ( ) 2 alc.KOH 3222 E CHCHCHCHCl Xmajor. −−−−→
X in this reaction is
(1) CH3–CH2–CH = CH2
(2) CH3–CH = CH – CH3
(3)
(4) CH3–CH = CH2
17. When 2-methyl butyl bromide is treated with sodium ethoxide in ethanol, what will be the major product?
(1) 2-methyl but-2-ene
(2) 3-methyl but-1-ene
(3) 2-methyl but-1-ene
(4) 2-methyl sodium butoxide
18. The compound which decolourises bromine water but does not give white precipitate with Tollen's reagent is
(1) C2H2 (2) C2H4
(3) C6H6 (4) CH4
19. When HCl is passed through propene in the presence of benzoyl peroxide, it gives (1) n – propyl chloride
(2) iso proply chloride
(3) 2 chloro ethanol
(4) benzoyl chloride
20. But-2-ene can be obtained by electrolysis of an aqueous solution of
(1) 2, 3-dimethyl maleic acid
(2) 2, 2-dimethylbutanedioic acid
(3) 2-methylbutanedioic acid
(4) 2, 3-dimethylbutanedioic acid
Answer Key
(1) 2 (2) 3 (3) 2 (4) 1 (5) 1 (6) 3 (7) 4 (8) 4 (9) 3 (10) 4 (11) 2 (12) 4 (13) 2 (14) 2 (15) 3 (16) 1 (17) 1 (18) 2 (19) 2 (20) 4
10.3 ALKYNES
Alkynes are aliphatic unsaturated hydrocarbons containing at least one carboncarbon triple bond. Each member of alkyne series has four hydrogen atoms less than the corresponding alkane. The general formula of alkynes is Cn H2n–2. The first and the most important member of the series is acetylene and this family, as a whole, is also known as acetylenes.
10.3.1 Nomenclature and Isomerism
In common system, alkynes are named derivatives of acetylene. In IUPAC system, they are named derivatives of the corresponding alkanes replacing ‘ane’ by the suffix ‘yne’, as shown in Table 10.2.
Table 10.2 Common and IUPAC names of alkynes
Compound
HC ≡ CH Acetylene Ethyne
H3C–C ≡ CH Methyl acetylene Propyne
CH3CH2C ≡ CH Ethyl acetylene But–1–yne
H3C–C ≡ C–CH3 Dimethyl acetylene But–2–yne
C3H7–C ≡ CH n-Propylacetylene Pent-1-yne
name
(CH3)2CH–C ≡ CH Isopropylacetylene 3-Methyl but-1-yne
(CH3)3C–C ≡ CH Tertiarybutyl acetylene 3,3-Dimethyl but-1-yne
10.3.2 Structure of the Triple Bond (–C ≡ C–)
Acetylene is a linear molecule since the bonded electron cloud between two carbon atoms is cylindrically symmetrical about internuclear axis. Carbons at triple bond are sp hybridised. A triple bond consists of one sigma bond and two pi bonds.
Carbon-carbon sigma bond is formed by the head-on overlapping of the two sp hybridised orbitals of the two carbon atoms. C–H sigma bond is formed by the remaining sp hybridised orbital of carbon atom with 1s orbital of each hydrogen along the internuclear axis.
H–C ≡ C–H bond angle is 180°. The –C ≡ C– bond length in acetylene is 1.20 A°, which is less than that in alkenes and alkanes. The –C ≡ C– bond energy (823 kJ mol–1) is more when compared with the double and single bond energies.
Alkynes exhibit chain isomerism, position isomerism and also functional isomerism with alkadienes. To exhibit chain isomerism and positional isomerism by an alkyne, it must contain at least five and four carbons, respectively.
10.3.3 Preparation
A l kynes can be prepared from haloalkanes and polyhalogen compounds.
From Calcium Carbide
Acetylene is obtained by the hydrolysis of calcium carbide. It is industrial method as well as a laboratory method.
CaC2 + 2H2O C2H2 + Ca(OH)2
Calcium carbide is obtained from CaCO 3 and coke.
3 2 CaCO CaO CO 2 CaO 3C CaC CO
To prepare 1 mole of C2H2, 1 mole of CaCO3 and 3 moles of carbon are required.
Dehydrohalogenation of Dihalides
Alkynes are prepared by dehydrohalogenation of vicinal dihalides, using alcoholic solution of potassium hydroxide. Acetylene is prepared by the action of alcoholic caustic potash on ethylene bromide or ethylidene bromide.
For the second step, NaNH 2, mineral oil is a better reagent.
In vicinal dihalide, the two halogen atoms are on adjacent carbons, while in geminal dihalides, the two halogens are on same carbon.
Gem dihalides can also be dehydrohalogenated to get alkynes.
–C ≡ C– triple bond is less stable since it is highly strained.
Highly strained bridged carbocation
Dehalogenation of Tetrahalides
By heating 1, 1, 2, 2-tetra bromoethane (acetylene tetrabromide) with zinc dust in alcohol, acetylene is obtained.
Br2HC–CHBr2 + 2Zn alcohol / H–C ≡ C–H + 2ZnBr2
10.3.4 Physical Properties
The first three alkynes are gases, while those containing five to thirteen carbon atoms are liquids, and higher alkynes are solids. Alkynes are less volatile than the corresponding alkanes and alkenes. All alkynes are lighter than water. These are insoluble in water, but soluble in organic solvents like ether, acetone, alcohol, etc. Their densities increase regularly with increase in molecular mass. All alkynes are colourless and odourless.
Pure acetylene has a pleasant odour. The garlic odour exhibited by acetylene is due to the presence of impurities, like hydrogen sulphide and phosphine. It is poisonous, highly inflammable, and lighter than air.
10.3.5 Chemical Properties
Alkynes, like alkenes, undergo electrophilic addition reactions. However, alkynes are less reactive than alkenes towards electrophilic addition reactions.
Reason for Reactivity
The bridged intermediate carbocation formed by the initial attack of the electrophile on the
Also, in acetylenic carbon atoms, the electrons are held more tightly by carbon nuclei and, so they are less readily available for reaction with electrophiles. In addition to electrophilic addition reactions, alkynes also undergo nucleophilic addition reactions in the presence of heavy metal ions. Alkynes also undergo oxidation reactions and polymerisation reactions.
Addition Reactions
Formation of addition product takes place according to the following steps.
The addition product formed depends on stability of vinylic cation. Addition in unsymmetrical alkynes takes place according to Markovnikov rule.
Addition of hydrogen: Alkynes react readily with hydrogen in the presence of finely divided nickel, platinum, or palladium as catalysts. During catalytic hydrogenation, ultimate products are alkanes.
R–C ≡ C–R + 2H2 catalyst R–CH2–CH2–R
HC ≡ CH + H2 catalyst H2C = CH2
HC ≡ CH + 2H2 catalyst H3C – CH3
CH3 – C ≡ CH + 2H2 CH3 – CH2 – CH3
Addition of halogens: Halogens, especially chlorine and bromine add on alkynes readily,
producing a tetrahalogen derivative. This reaction is carried out in an inert solvent, like CCl4. Due to the presence of two pi bonds, each molecule of the alkyne can react with two molecules of the reagent.
HC ≡ CH + Cl2 4 CCl ClCH = CHCl 1,2-Dichloroethene
ClCH = CHCl + Cl2 4 CCl Cl2CH–CHCl2 1,1,2,2-Tetrachloroethane (Westron)
Due to very slow reaction of iodine, only diiodo derivative is formed. Decolourisation of Br2/CCl4 by alkynes is a test for unsaturation.
Addition of hydrogen halides: Addition of one molecule of halogen acid gives vinyl halide which then adds another molecule to form gem-dihalide. This addition follows Markovnikov rule.
Ad dition of hydrogen cyanide: Acetylene adds on one molecule of hydrogen cyanide in the presence of barium cyanide or cuprous cyanide catalyst to form vinyl cyanide, also called acrylonitrile.
In presence of peroxides, HBr adds on to acetylene according to anti-Markovnikov rule. The order of reactivity of halogen acids is:
> HBr > HCl
Addition of water: When acetylene gas is passed through dilute sulphuric acid at 60–65°C in presence of mercuric sulphate as catalyst, a water molecule adds up and, finally, acetaldehyde is formed. This is called hydration. Vinyl alcohol and acetaldehyde are tautomers.
Acr ylonitrile is used in the manufacture of an important polymer, polyacrylonitrile. This follows nucleophilic addition mechanism. Addition of acetic acid: Acetylene combines with acetic acid in presence of mercuric sulphate catalyst. It first foms vinylacetate and then ethylidene acetate.
2Hg HC CH HOCOCH 3 2 CH CHOCOCH
CH2=CHOCOCH3 + HOCOCH3 2Hg
CH3–CH(OCOCH3)2
Vinyl acetate is used in plastic industry and ethylidene acetate, on heating decomposes to give acetaldehyde and acetic anhydride.
Combustion: Acetylene burns in air or oxygen to form carbon dioxide and water with evolution of large amount of heat. The oxyacetylene flame is used for welding purposes and gives a temperature of about 3500°C.
2C2H2 + 5O2 4CO2 + 2H2O; ∆ H = –1300 kJ mol–1
W ith alkaline potassium permanganate, acetylene is oxidised to oxalic acid, during which pink colour is decolourised.
CH ≡ CH + 4(O) 4 KMnO OH , 25 C
With chromic acid, acetylene is oxidised to acetic acid.
≡ CH+H2O+(O)
Ozonolysis
Alkynes, on reductive ozonolysis, give diketones. Acetylene adds ozone to form unstable ozonide, which, on hydrolysis in presence of zinc, glyoxal is formed.
Under special conditions, this polymer conducts electricity and a thin film of polyacetylene can be used as electrodes in batteries.
Cyclic polymerisation: When acetylene is passed through a red hot metallic tube, cyclic polymerisation (trimerisation) takes place with the formation of benzene. Red hot F e tube 3CH C H
10.3.6 Acidic Character of Alkynes
In oxidative ozonolysis, alkynes yield carboxylic acids.
CH 3 2 3 (i) O ( ii ) H O CH COOH 2HCOOH Formic acid
CH3–C CH
2 ( i ) O ( ii ) H O acid
CH3COOH + HCOOH Acetic acid Formic acid
Polymerisation
Linear polymerisation: When acetylene is passed into cuprous chloride solution containing ammonium chloride, linear polymerisation occurs, forming monovinyl acetylene and divinyl acetylene.
2 4 Cu Cl NH Cl CH CH CH CH 2 CH CH C CH 2 2 4 CH CH Cu Cl , NH Cl CH2=CH–C ≡ C–CH=CH2
Under suitable conditions, acetylene is linearly polymerised to produce polyacetylene.
2n HC CH ( CH CH CH CH ) n
Acet yl e ne and terminal alkynes (1-alkynes) behave as very weak acids. This character can be explained in terms of sp-hybrid state of the carbon atom. sp-hybridised carbon is more electronegative than sp2 and sp3 hybrid carbon atoms. On account of acidic nature, acetylene forms salts, which are called acetylides. The acidic hydrogens of 1-alkynes can be replaced by copper (I) or silver (I) ions in ammoniacal solutions.
Alkanes, alkenes, and alkynes follow the following trend in their acidic behaviour:
HC ≡ CH > H2C = CH2 > H3C – CH3
Acidity of alkynes is in the order:
CH ≡ CH > CH3– C ≡ CH > CH3 – C ≡ C –CH3
Formation of sodium acetylides: Acetylene reacts with sodium in liquid ammonia or sodamide to form sodium acetylides or ethynides.
H–C ≡ C–H + Na → H–C ≡ C–Na + 1 2 H2
Monosodiumacetylide
H–C ≡ C–H + 2Na Na–C ≡ C–Na + H2
Disodiumacetylide
CH3 – C ≡ C–H + 2 N a N H
3 CH C C Na + NH3
Sodium propynide
Higher alkynes may be prepared by the action of alkyl halides on mono and disodium acetylides.
HC ≡ CNa + BrC2H5 →
HC ≡ C–C2H5 + NaBr
CH3Br + NaC ≡ CNa + BrCH3 → CH3 –C ≡ C–CH3 + 2NaBr
Formation of copper acetylide: Acetylene when passed through an ammoniacal cuprous chloride solution, a red precipitate of cuprous acetylide is formed. This is a test for acetylene.
HC ≡ CH + Cu2Cl2 + 2NH4OH → Cu–C ≡ C–Cu ↓ + 2NH4Cl + 2H2O
Formation of silver acetylide: Acetylene, when passed through an ammoniacal solution of silver nitrate (Tollen’s reagent), a white precipitate of silver acetylide forms. This is also a test for acetylene.
TEST YOURSELF
1. NaNH2 HBrHBr Ä HCCHABC ≡→→→ , C in the reaction is
(1) Ethyne (2) Ethanol (3) Ethane (4) Ethene
2. An unsaturated hydrocarbon, X, on ozonolysis, gives A. Compound A, when warmed with ammoniacal silver nitrate, forms a bright silver mirror along the sides of the test tube. The unsaturated hydrocarbon X is
(1) 33 CHCCCH −≡−
(2) 33 33 CHCCCH || CHCH −=− (3) C CH3 CH3
(4) HCCCHCH23 ≡−−
3. 3 2 O Zn/HO
CHCH≡→ Product will be
(1) oxalic acid (2) glyoxal (3) CHO–COOH (4) CO2 + H2O
4. The treatment of C 2 H 5 MgX with 3 CHCCH −≡− produces (1) C2H6 (2) CH3 – HC = CH2 (3) CH3–C ≡ C–CH3 (4) CH3–CH=CH–CH3
5. The product Q and R in the following reactions, respectively, are
3 2 's O 23HO HCPQR Lindlar catalyst Zn CHCH + +≡→→+
(1) ethanol, methanoic acid (2) ethanoic acid, methanol
(3) ethanal, methanol (4) ethanoic acid, methanoic acid
6. The homologue of ethyne is (1) C2H4 (2) C2H6 (3) C3H8 (4) C3H4
7. Predict the correct intermediate and product in the following reaction. ( ) ( ) , −≡→→ 24 4 HOHgSO 3 HgSO CHCCHAB A (Intermediate) B (Products) (1) 33 CHCCH || O CH3–C ≡ CH
(2) 32 CHCCH | OH −= 32 4 CHCCH | SO −=
(3) 32 CHCCH | OH −= 33 O || CHCCH
(4) 32 4 CHCCH | SO −= 33 O || CHCCH
8. The decreasing order of acidic character among ethene (I), ethyne (II) and ethane (III) is
(1) I > II > III (2) II > I > III
(3) III > II > I (4) II > III > I
9. Which of the following reactions will yield 2, 2-dibromopropane?
(1) 3 CHCCH2HBr−≡+→
(2) 3 CHCHCHBrHBr≡+→
(3) CHCH2HBr≡+→
(4) 32 CHCHCHHBr−=+→
10. Acetylene is prepared by the electrolysis of concentrated aqueous solution of (1) potassium fumarate
(2) potassium succinate
(3) potassium acetate
(4) sodium oxalate
Answer Key
(1) 1 (2) 4 (3) 2 (4) 1
(5) 5 (6) 4 (7) 3 (8) 2
(9) 1 (10) 1
10.4 AROMATIC HYDROCARBONS
The name 'aromatic' was given to compounds which possess aroma (Greek : aroma means pleasant smell). Aliphatic compounds were assumed to have, properties similar to fats (aliphatic means 'fat-like'). Aromatic compounds are also known as arenes.
10.4.1 Chemistry of Aromatic Nature
Most of the aromatic compounds were found to contain benzene ring. Compounds having benzene ring in them are classified as benzenoid compounds.
The homologues of benzene and all aromatic hydrocarbons are known as arenes.
Aromatic compounds that do not contain benzene ring in them are known as nonbenzenoid compounds, which contain other highly unsaturated rings.
Benzenoid compounds include benzene and its derivatives and polynuclear hydrocarbons, such as naphthalene, anthracene, biphenyl, etc.
Some important polynuclear compounds are given in Table 10.3
Common and IUPAC names of some important benzene derivatives are given in Table 10.4.
Aromatic hydrocarbons contain less number of hydrogens than corresponding aliphatic compounds, leading to the conclusion that these are highly unsaturated compounds. However, these are resistant to the addition reactions which is a general characteristic of unsaturated compounds. Further, aromatic hydrocarbons undergo electrophilic substitution reactions. These possess unusual stability.
Table 10.3 Important polynuclear hydrocarbons
Table 10.4 Common and IUPAC names of some important benzene derivatives
Structure Common name IUPAC name
CH3
CH3
Toluene
NH3 Aniline
Methylbenzene
Aminobenzene or Benzeneamine or aniline
OH phenol Hydroxybenzene (or) phenol
OCH3 Anisole Methoxybenzene
CHO Benzaldehyde Benzene carbaldehyde
COOH Benzoic acid Benzene carboxylic acid
CH = CH2 Styrene Phenylethene
CH2 Cl Benzyl chloride Chlorophenyl methane
CHCl2 Benzalchloride Dichlorophenyl methane
CCl3 Benzotrichloride Trichlorophenyl methane
Cl Cl Orthodichlorobenzene 1,2–Dichlorobenzene
CH3 m–Xylene 1,3–Dimethyl benzene
O2N COOH p-Nitrobenzoic acid 4–Nitrobenzoic acid
Aromatic hydrocarbons are cyclic, generally containing five, six, or seven membered rings and found to be planar molecules obeying Huckel’s rule. Huckel’s Rule
Th ough compounds were identified as aromatic on the basis of aroma exhibited by these compounds, the term is no longer valid as several compounds were prepared later were found not to possess aroma, yet behaving like aromatic compounds. Hence, a more general approach to account for aromatic character was found necessary.
Huckel proposed a modern concept to explain aromatic character, known as Huckel’s rule (or) (4n + 2) rule.
For a molecule or an ion to be aromatic, it should be cyclic, planar , conjugated systems and should contain (4n + 2) electrons (where n = 0, 1, 2, 3, ......).
These pi electrons are delocalised and hence these molecules possess high degree of stability and behave as aromatic compounds.
Aromatic compounds, according to their molecular formula look like highly unsaturated but show resistance towards oxidation and addition reactions under normal conditions.
Aromatic compounds need not necessarily be carbocyclic (homocyclic) compounds. Heterocyclic compounds also may be aromatic provided they obey Huckel’s rule. Example:
Illustrative Examples
Benzene
It has 6 electrons
4n + 2 = 6 (n=1.)
Naphthalene
It has 10 electrons
4n + 2 = 10 (n=2.)
Anthracene
It has 14 electrons
4n + 2 = 14 (n=3.)
Phenanthrene
It has 14 electrons
4n + 2 = 14 (n = 3.)
Cyclopentadienyl anion
It has 6 electrons
4n + 2 = 6 (n=1.)
Cycloheptatrienyl cation
(Tropylium ion)
It has 6 electrons
4n + 2 = 6 (n=1.)
Cyclopropenyl cation
It has 2electrons
4n + 2 = 2 (n=0.)
10.4.2 Structure of Benzene
The molecular formula of benzene is C 6H 6, containing eight hydrogen atoms less than the corresponding parent saturated hydrocarbon, i.e., hexane, C 6 H 14 . The carbon-hydrogen ratio indicates that it is highly unsaturated compound. Benzene was found to form a triozonide, which indicates the presence of three double bonds.
Benzene was found to produce one and only one mono substituted derivative, which indicated that all six carbons and six hydrogen atoms of benzene are identical.
Based on these observations, Kukule proposed cyclic hexagonal structure containing three double bonds alternatively.
The Kekule’s structure indicates the possibility of two isomeric 1,2-dibromo benzenes. In one of the isomers , the bromine atoms are attached to the doubly bonded carbon atoms, whereas in the other, they are attached to the singly bonded carbon atoms.
Br Br Br Br
Br Br Br Br
Br Br Br Br
Benzene was found to form only one ortho disubstituted product. To overcome this problem, Kekule suggested the concept of oscillating nature of double bonds in benzene.
Dewar proposed three more structures to benzene.
Dewar’s structures were accepted by the scientists to the extent of only 20%. Kekule’s structures were accepted to the extent of 80%.
10.4.3 Resonance in Benzene
Though Kekule’s dynamic structure was accepted as a reasonable structure for benzene, it could not explain certain experimental observations. In presence of platinum, cyclohexene adds one mole of hydrogen, where the heat of hydrogenation is –28.6 kcal mol–1
1,3–cyclohexadiene, under similar conditions, adds two moles of hydrogen where the heat of hydrogenation is –55.4 kcal mol–1, which is almost double to that of cyclohexene (one double bond).
+ 2H2 Pt H =–55.4 kcal mol–1
If benzene were a cyclohexatriene, then the heat of hydrogenation should be around 3 × 28.6= 85.8 kcal mol –1 , but actually, only 49.8 kcal mol –1 is liberated. To the extent of 36 kcal mol –1 (85.8–49.8), benzene is stabilised compared to cyclohexatriene. This energy is called resonance energy of benzene. The difference between expected heat of hydrogenation and observed heat of hydrogenation is called resonance energy.
X-ray diffraction studies showed that benzene is a planar molecule with all carboncarbon equal bond lengths. Had any one of the Kekule structures is correct, it should contain two types of bond lengths between carbon atoms. The measured bond length between carbon atoms was found to be intermediate between C – C single bond length (1.54 A°) and C = C double bond length (1.34 A°). This accounts for the partial double bond character between carbon atoms in benzene. The bond length was found to be 1.39 Ao. But benzene does not behave like unsaturated compound as it neither decolourises alkaline potassium permanganate solution nor bromine in carbon tetrachloride solution.
Benzene undergoes substitution reactions rather than addition reactions. Benzene was found to be a stable molecule.
To account for the experimental observations made, the phenomenon of resonance was used. These two structures differ in the positions of double bonds.
More the number of resonance structures, more will be the stability.
The actual structure of benzene is a resonance hybrid in which the contribution of Kekule structures is about 80%. (or) (or)
Resonance hybrid of benzene is represented as (or) (or)
The resonance hybrid structure of benzene has six delocalised pi electrons. This resonance hybrid structure for benzene explains the stability of benzene and its low reactivity. The resonance energy of benzene, which is the energy difference between energy of Kekule structures and that of the actual structure, is about 150.5 kJ mol–1 .
The difference between the energy of the most stable contributing structure and the energy of the resonance hybrid is known as resonance energy. Value of resonance energy has been determined by studying the enthalpies of combustion and hydrogenation.
10.4.4 Orbital Model of Benzene
The structure of benzene was also explained with the help of orbital theory. In benzene, all carbon atoms are sp2 hybridised. All carboncarbon single bonds are formed by sp2 – sp2 overlap. Each carbon atom has one pure ‘p’ orbital containing an unpaired electron and perpendicular to the plane of of three sp 2 hybrid orbitals. These ‘p’ orbitals overlap sideways to form three pi bonds. Overlapping of ‘p’ orbitals may take place in two ways to give two structures of benzene, which were given by Kekule.
(or) (or)
It is supposed that the pi electrons are distributed over all the carbon atoms to form a cloud-like structure above and below the plane of benzene ring. This type of distribution of electrons is called delocalisation and all six electrons are considered to belong to all carbon atoms. This can also be explained since all the carbon-carbon bond lengths in benzene are identical (1.39 A0). Due to delocalisation of pi-electrons, some energy is released and it is called resonance energy. The molecule becomes more stable due to delocalisation of pi-electrons.
10.4.5 Preparation of Benzene
Benzene is commercially isolated from coal tar (obtained by the destructive distillation of coal) by fractional distillation.
Polymerisation of acetylene: On passing acetylene gas through red hot iron or copper tube, benzene is formed.
3C2H2 0 600 C
Decarboxylation of benzoic acid: Heating sodium benzoate with soda lime gives benzene. It is a laboratory method of preparation.
COONa + NaOH CaO + Na2CO3
■ Reduction of phenol: Distillation of phenol in the presence of zinc dust gives benzene.
OH + Zn + ZnO
■ Hydrolysis of benzene sulphonic acid: Hydrolysis of benzene sulphonic acid with super heated steam gives benzene.
SO3H + H2O + H2SO4
10.4.6 Physical Properties
Aromatic hydrocarbons are generally colourless liquids or solids with characteristic aroma. Naphthalene balls which are used in toilets and for preservation of clothes because of unique smell of the compound and the repellent property.
Benzene is insoluble in water but soluble in organic solvents. Benzene and other aromatic hydrocarbons are highly inflammable and burn with a sooty flame. These are toxic and carcinogenic in nature. They possess higher boiling points than alkanes of comparable molecular weights. Boilingpoint of C 6H 6 is 80 C.
10.4.7 Chemical Properties
Benzene and its derivatives mainly undergo electrophilic substitution reactions. However, under special conditions, they also undergo addition reactions.
All electrophilic substitution reactions follow similar mechanism. The reaction was found to take place in three steps.
a) Generation of electrophile.
b) Formation of carbocation intermediate.
c) Removal of proton from the carbocation intermediate.
■ Generation of electrophile
Example:
In the case of nitration, the electrophile, nitronium ion is produced by transfer of proton from sulphuric acid to nitric acid in the following manner: HO3 SO H + H–O –NO 2 H–O–N O 2 + HSO4–H
H
nitric acid) (Nitrogen Ion)
Thus in nitration mixture, H2SO4 behaves as an acid and HNO3 as a base.
In chlorination, the electophile chloronium ion (Cl+) is formed by the reaction:
Cl2+ FeCl3 → Cl+ + FeCl4 –
In sulphonati on, the electrophile SO 3 is obtained by the reaction:
2H2SO4 SO3 + H3O+ + HSO4–
In Friedel–Crafts alkylation, the electophile, R+ is obtained by the reaction:
RCl + AlCl3 R+ + AlCl4–
In Friedel–Crafts acylation, the electrophile is formed by the reaction:
RCOClAlClRCOAlCl
CHCOClAlClCHCOAlCl
CHCOOAlClCHCOAlClCHCOO
Formation of carbocation:
sp3 hybridised carbon slow step where E is electrophile.
The carbocation formed in the above step is stabilised due to resonance.
R emoval of proton: In the last step, the carbocation loses a proton from sp3 hybridised carbon to regain its aromat ic character.
H H + AlCl4 E + HCl + AlCl3
H HH + HSO4–E + H2SO4
Nitration
Benzene when heated with a mixture of (1 : 1 by volume) concentrated nitric acid and concentrated sulphuric acid (nitration mixture) below 60�C gives nit robenzene. + HNO3 2 4 0 H SO 60 C NO2 + H2O
Halogenation
Benzene reacts with chlorine or bromine in presence of Lewis acids such as FeCl3 or FeBr3 or AlCl3 as catalyst to give halobenzene.
Since reaction with fluorine is very vigorous, direct fluorination of arenes is not possible. In case of iodination, hydriodic acid formed is a strong reducing agent, it will reduce iodobenzene (formed by the iodination of benzene) to benzene back, hence iodination should be carried out in presence of an oxidising agent like iodic acid, nitric acid or mercuric oxide.
C6H6 + I2 C6H5I + HI
5HI + HIO3 3I2 + 3H2O
The resonance hybrid structure is represented as:
5C6H6 + 5I2 + HIO3 5C6H5I + 3I2 + 3H2O + Cl2
FeCl Cl + HCl + Br2
FeBr Br + HBr
If e xcess of electrophilic reagent is used, further substitution reaction may take place in which other hydrogen atoms of benzene ring may also be successively replaced by the electrophile.
6Cl2 3
Sulphonation
6HCl
(C 6Cl6)
Benzene, when heated with fuming sulphuric acid at about 100�C, gives benzene sulphonic acid.
H2SO4
Table 10.5 Substitution products of benzene
+ H2O
Friedel-Crafts Alkylation
Benzene reacts with an alkyl halide in presence of anhydrous aluminium chloride to give alkyl benzene.
1 Chlorination Cl2 + FeCl3 Chloronium cation (Cl+)
2 Bromination Br2 + FeBr3 Bromonium cation (Br+)
3 Nitration conc. HNO3 + conc. H2SO4 Nitronium cation (NO2+)
4 Sulphonation conc. H2SO4 Sulphur trioxide (SO3)
5 FriedelCrafts ethylation C2H5Cl + AlCl3 Ethyl cation (C2H5+)
6 FriedelCrafts acetyltion
7. FriedelCrafts benzoylation
CH3COCl + AlCl3 Acetyl cation (CH3CO+)
C6H5COCl+ AlCl3 Benzoyl cation (C6H5CO+)
sulphonic acid
Friedel–Crafts Acylation
The reaction of benzene with an acyl halide (or) acid anhydride in presence of Lewis acid as catalyst to give acyl benzene is known as Friedel–Crafts acylation.
+ RCOCl 3AlCl
+ RCOCl 3AlCl
+ RCOCl 3AlCl
+CH3COCl 3AlCl
+CH3COCl 3AlCl
+CH3COCl 3AlCl
+(CH3CO)2O 3AlCl
+(CH3CO)2O 3AlCl
+(CH3CO)2O 3AlCl
COR + HCl
COR + HCl
COR + HCl
COCH3 + HCl
COCH3 + HCl
COCH3 + HCl
COCH3 + CH3COOH
COCH3 + CH3COOH
COCH3 + CH3COOH
Different electrophilic substitution reactions and the products are listed in Table 10.5.
Addition Reactions
Benzene does not undergo addition reactions under normal conditions. However, some addition reactions were found to take place under special conditions.
Hydrogenation: At high temperature and pressure, in presence of finely divided nickel benzene undergoes hydrogenation to give cyclohexane. + 3H2 Ni , h pressure
Addition of chlorine: Benzene reacts with chlorine in presence of sunlight at 500K to give benzene hexachloride (BHC) or hexachlorocyclohexane or Gammaxene or 666 or Lindane.
Three molecules of chlorine were added to benzene. + 3Cl2
Ozonolysis
One mole of benzene reacts with three moles of ozone to give a triozonide, which on hydrolysis in presence of zinc, gives three moles of glyoxal.
C6H6 + 3O3 → C6H6O9
C6H6O9 3 CHO | CHO 2 H O , Zn + 3H2O2
Combustion
When benzene is burnt in excess air, it burns with a sooty flame.
6 6 2 2 2 15 C H O 6 CO 3H O 2 +
10.4.8 Directive Influence of Functional Groups in Mono Substituted Benzene
In benzene, as all the six carbon atoms are equivalent. Monosubstitution gives only one compound. For example, monochlorination of benzene gives only chlorobenzene.
In the converting of a monosubstituted benzene into a disubstituted benzene derivative, the substituent already present in the benzene ring determines the position of the incoming group. This tendency of a group already present in the benzene ring to direct the incoming group towards a particular position is called the directive influence of group.
Depending upon the nature of the already present substituent, two types of substitutions are mainly observed.
Ortho and para directing groups: Groups like 3 2 O H; N H ; N HR; N HCOCH ; 3 O C H ; –CH3 and C2H5 release electrons to the benzene ring makes the benzene ring more susceptible to the attack of electrophiles. These groups direct the incoming substituent to enter into the ortho and para positions preferentially. Hence, these groups are called ortho and para
directing and ring activating groups. These groups activate benzene ring by increasing the stability of intermediate arenium ion formed by the attack of electrophile in rate determining step. Resonance in phenol is shown in Fig.10.3.
Example: NO2 group deactivates, benzene by 10–4 times. Nitro benzene gives meta dinitro benzene as major product on nitration.
Fig.10.3 Resonance in phenol
It can be observed from the resonance structures, that the negative charge is concentrated in ortho and para positions to phenolic –OH. However, it may be noted that –I effect of –OH group also operates, due to which the electron density on benzene ring is slightly reduced. But the overall electron density increases at these positions of the ring due to resonance.
The electrophile attacks at these positions giving rise to ortho and para disubstituted compounds.
For example, bromination of phenol gives 2,4,6-tribromophenol.
These groups deactivate benzene ring by decreasing the stability of intermediate arenium ion formed by the attack of electrophile in rate determining step. Resonance structures of nitrobenzene given in Fig.10.4.
This is due to the highly activating group –OH activates benzene ring to a great extent and, hence, substitution takes place in o- and p- positions.
Meta directing groups: Groups like –NO2, –CN, –COOH, –COOR, –COR, –CHO, –SO3H, –X, etc. withdraw electrons from the benzene ring specifically from ortho and para positions and makes the benzene ring less susceptible to the attack by electrophiles. These groups withdraw electrons particularly from ortho and para positions and thereby, meta position is relatively more electron dense.
Hence, these groups are called meta directing and ring deactivati ng groups.
Aryl halides – unusual behaviour:
In the case of aryl halides, halogens are moderately deactivating. In halobenzene, halogen decreases electron density on benzene ring, due to its strong –I-effect. Hence, further substitution on halobenzene is difficult. But electron density on ortho and para positons is relatively increased due to +M–effect of halogen. Therefore, halogen group is O, P–directing but ring deactivating group.
Resonance structures of chlorobenzene are given in Fig. 10.5.
Fig.10.4 Resonance in nitrobenzene
Fig.10.5 Resonance in chlorobenzene
13.4.9 Carcinogenicity and Toxicity
Benzene and polynuclear hydrocarbons containing more than two benzene rings fused together are toxic and said to possess carcinogenic (cancer producing) property. Such polynuclear hydrocarbons are formed on incomplete combustion of organic materials,
like tobacco, coal and petroleum.
They enter into human body and undergo various biochemical reactions and, finally, damage DNA and cause cancer.
Some of the carcinogenic hydrocarbons along with their structures are given in Fig.10.6
1,2,5,6–Dibe nzanthracene
Fig.10.6 Structure of same carcinogenic hydrocarbon
TEST YOURSELF
1. The number of s bonds, p bonds and lone pair of electrons in pyridine, respectively are (1) 12, 2, 1 (2) 11, 2, 0 (3) 12, 3, 0 (4) 11, 3, 1
2. Which of the following molecules is not aromatic? (1) (2) (3) (4)
3. Which of the following is the correct order of the rate of reaction of C 6H 6, C 6D 6 and
CH3 CH3 9,10–Dimethyl –1,2–benzanthracene
C6T6 towards sulphonation?
(1) C6H6 = C6D6 = C6T6
(2) C6H6 > C6D6 > C6T6
(3) C6T6 > C6D6 > C6H6
(4) C6H6 > C6D6 = C6T6
4. Benzophenone can be obtained by ____________.
(1) Benzoyl chloride + Benzene + AlCl 3
(2) Benzoyl chloride + Diphenyl amine
(3) Benzoyl chloride + Methyl magnesium chloride
(4) Benzene + Carbon monoxide+ ZnCl 2
5. Which one of the following compound is unstable at room temperature?
1,2-Benzanthracene
(1) (2) (3) (4)
6. In the nitration of benzene, the electrophile is
(1) NO2 ⊕ (2) NO2
(3) HNO3 (4) H2SO4
7. Regarding electrophilic substitution reactions of benzene, false statement is
(1) In sulphonation, SO3 is electrophile.
(2) In case of nitration, H2SO4 behaves as a base, HNO3 as an acid.
(3) In Friedal – Craft’s reactions catalyst is anhydrous AlCl3
(4) In second step of electrophilic substitution mechanism, carbocation is intermediate.
8. OH A B Zn dust CH3COCl AlCl3
In the above sequence, B is (1) Acetanilide
(2) Acetophenone
(3) Benzophenone
(4) N-phenyl acetamide
9. The major product formed in the reaction + CH3CH2CH2Cl anhydrous AlCl3
(1) anhydrous +CH3CH2CH2Cl AlCl3 (2)
10. So dium benzoate gives Benzene on being heated with ‘X’. Phenol gives Benzene on being heated with ‘Y’. ‘X’ and ‘Y’ are respectively.
(1) Sodalime and copper
(2) Zn dust and NaOH
(3) Soda lime and Zn dust
(4) NaOH and Zn dust
11. Most reactive towards aromatic electrophilic substitution is (1) Benzene
(2) nitrobenzene
(3) chlorobenzene
(4) Toluene
12. Benzene 3 0 24 .H;60C ConHNO ConSO <
B. compound "B" is (1) O-nitro chlorobenzene
(2) P-nitro chlorobenzene
A 23 / ClFeCl
(3) m - nitro chloro benzene
(4) O - Chloro nitro benzene
13. According to the Huckel rule, the number of p Electrons in anthracene is (1) 12 (2) 14 (3) 10 (4) 20
14. Electrophile in the sulphonation of benzene is
(1) 3 SOH + (2) 3 SOH
(3) SO3
(4) 3 SO +
15. Friedel-Crafts reaction of bromobenzene with methyl chloride gives (1) o-bromotoluene only (2) p-bromotoluene only (3) A mixture of o-and p-bromotoluenes (4) m-bromotoluene only
CHAPTER REVIEW
Alkanes
The acyclic saturated hydrocarbons are called alkanes.
Their general formula is CnH2n+2 where n = 1, 2, 3.....etc.
Hydrogenation of alkenes and alkynes in the presence of a catalyst such a Raney Ni, Pt or Pd at room temperature gives alkanes. (Sabatier and Sendern's reduction).
Grignard reagents react with substances containing active hydrogen to give alkanes R – MgX + H–OH →RH + Mg(OH)X with heavy water (D2O), deuterated alkanes are formed.
R–MgX + D – OD → RD + Mg(OD)X
Wurtz reaction.
R – X + Na + X – R Dryether → R – R + 2 NaX
Reduction of alkyl halides with Zn/HCl, Sn/HCl or Zn-Cu couple and alcohol or H2 in presence Pd – C gives alkanes.
Reduction of 1° and 2° alkyl halides with LiAlH 4 gives alkanes while 3° alkyl halides mainly undergo dehydrohalogenation to give alkenes.
On the other hand, NaBH4 reduces 2° and 3° alkyl halides but not 1° whereas Ph 3SnH reduces all the three types of alkyl halides to give alkanes.
Reduction of alkyl halides with HI and red P at 423 K gives alkanes.
CH3CH2–I + HI 423K,redP → CH3–CH3 + I2
Decarboxylation of sodium salts of fatty acids with soda-lime (NaOH + CaO) at 630 K gives alkanes
RCOONa + NaOH CaO,630K → R–H + Na2CO3
Kolbe's electrolysis of Sodium or Potassium salts of fatty acids gives alkanes.
Action of water on beryllium and aluminium carbides gives methane.
Be2C + 4 H2O→CH4 + 2 Be(OH)2; Al4C3 + 12 H2O→3 CH4 + 4 Al(OH)3
Boiling points. The boiling points of straight chain alkanes increases regularly with the increase in the number of carbon atoms.
Amongst isomeric alkanes, boiling points decreases with branching. For example, boiling points decrease in the order: n-pentane > isopentane > neopentane.
Melting point. The melting points of n-alkanes with even number of carbon atoms are much higher than those of the next lower alkanes with odd number of carbon atoms — Alternation effect.
Solubility. Being non-polar, all alkanes are insoluble in water but are highly soluble in organic solvents such as ether, benzene carbon tetrachloride, etc. All alkanes are lighter than water.
Chemical properties of Alkanes. Alkanes are inert, and are called paraffins. They, however, preferably undergo substitution reactions at high temperatures by free radical chain mechanism.
Halogenation: Replacement of one or more hydrogens of alkanes by halogens is called halogenation.
The order of reactivity of halogens is: F2 > Cl2 > Br2 > I2. The reaction with F2 is very violent but the reaction with I 2 is very slow and reversible. Therefore, iodination is carried out in presence of an oxidising agent such as HNO3, HIO3, etc.
The ease of substitution of different hydrogens follows the order: 3° > 2° > 1° but their relative rates vary with the nature of halogen. With Cl2, the relative rates of substitution of 3°, 2° and 1° hydrogens at 298 K is 5 : 3.8 : 1 while Br 2, it is 1600 : 82 : 1 at 400 K.
Nitration: Nitration of alkanes is carried out with conc. HNO 3 in the vapour phase at 423–673 K under pressure. Since HNO3 is a strong oxidising agent, during nitrogen, cleavage of C–C bonds also occurs. Therefore, nitration of alkanes usually gives a mixtures of nitroalkanes.
Sulphonation. It is carried out by heating an alkane with fuming sulphuric acid (conc. H2SO4 + SO3) at 675–725 K. Only branched chain and the higher n-alkanes (containing six or more carbon atoms) undergo sulphonation.
Combustion. Alkanes readily burn in excess of air or oxygen to form CO 2 and H 2O with liberation of large amount of heat and light.
Isomerization. n-Alkanes when heated with anhyd. AlCl 3 in presence of HCl at 573 K undergo isomerization to give branched chain alkanes.
Aromatization. n-Alkanes containing 6-8 carbon atoms when heated to about 773 K under 10-20 atm pressure in the presence of a catalyst containing of Cr2O 3, V 2O 5, Mo2O3 supported over AlCl3 get converted into aromatic hydrocarbons.
Pyrolysis or cracking of higher alkanes at high temperature (773-973 K) under a pressure of 6-7 atomspheres in presence or absence of a catalyst gives a mixture of lower alkanes, alkenes, etc.
C6H14 773K 67atm → C6H12 + H2 + C4H8+ C2H6 + C2H4 + CH4
Pyrolysis involves breaking of C–C and C–H bonds and occurs by a free radical mechanism.
Conformation of Ethane. The infinite number of momentary arrangement of the atoms in space which result through rotation about a single bond are called conformations or rotational isomers.
Ethane has infinite number of conformations, of which only two i.e., staggered and eclipsed are important. The dihedral angle between two conformations is 60°.
The staggered conformation of ethane is more stable than the eclipsed conformation by about 3.0 kcal or 12.55 kJ mol –1. Due to small difference in energy, the two conformations are readily inter convertible and it is not possible to separate the two conformations of ethane. However, at any given moment, most of the ethane molecules would exist in the staggered conformation due to its minimum energy and maximum stability.
Alkenes
The general formula of alkenes is CnH2n where n = 2,3,4.... etc. All alkenes contain a double bond which consists of one strong
s -bond and one weak p -bond. A p -bond is formed by sideways overlap of two p-orbitals. The carbon atoms of a double bond are sp2-hybridised and all the four atoms directly attached to the double bond lie in a plane. The H—C—C angle is 121.7° while H—C—H angle is 116.6. The C=C bond distance is 134 pm and C—H bond distance is 110 pm.
Stability of alkenes. Heat of hydrogenation can be used to determine the relative stability of alkenes. The more highly substituted the alkene, the more stable it is, i.e.,
R2C = CR2 > R2C = CHR > R2C = CH2 > RCH = CHR > RCH = CH2 > CH2 = CH2
This stability order can be explained on the basis of hyperconjugation.
Dehydrohalogenation (removal of a molecule of a halogen acid) of alkyl halides with hot alcoholic KOH solution gives alkenes. For example, 1-bromoethane gives ethene while 1-bromopropane gives propene. When two alkenes are possible, the more highly substituted alkene predominates (Saytzeff rule). For example, 2-bromobutane on dehydrohalogenation gives 80% of the more stable, but-2-ene and 20% of the less stable, but-1-ene.
The ease of dehydrohalogenation of alkyl halides having the same alkyl group is: iodides > bromides > chlorides> fluorides while the isomeric alkyl halides having the same halogen is : 3° > 2° > 1°.
Dehydrohalogenation is an example of b -elimination reaction since hydrogen is removed from the b -carbon and halogen from the a -carbon.
Dehalogenation (removal of a molecule of halogen) of 1, 1- and 1,2-dibromoethane with Zn dust in boiling methanol or ethanol gives ethene while that of 1, 1 and 1, 2-dibromopropane gives propene.
Dehydrati on of alcohols with conc. H 2 SO 4 at 433-443 K gives alkenes. The reaction occurs through a carbocation intermediate. For example, CH3CH2OH gives CH2=CH2
During dehydration of alcohols, sometimes rearranged products are formed. This is due to the rearrangement of the initially formed less stable carbocation to the more stable carbocation either by a 1, 2-hydride or 1, 2-methyl shift. For example, dehydration of 1-butanol gives the same mixture of products, i.e., but-2-ene (80%) and but-1-ene (20%) as obtained from dehydration of 2-butanol under similar conditions.
Kolbe's electrolysis of Sodium or Potassium salts of saturated dicarboxylic acids gives alkenes. For example, sodium succinate on electrolysis give ethylene.
Chemical Properties of Alkenes. (i) Electrophilic addition reactions. The typical reactions of alkenes are electrophilic addition reactions. In these reactions, an electrophile first attack the double bond forming a carbocation intermediate which then undergoes nuclephilic attack to form a 1, 2-addition product.
Addition of halogens. The order of reactivity is: F2 > Cl2 > Br2 > I2. Cl2 in CCl4 and Br 2 in CCl 4 add readily to alkenes to form 1, 2-haloalkanes. During the addition of Br2, orange red colour of Br2 is discharged. Therefore, this reaction is used as a test for unsaturation.
Addition of halogen acids (HX). The order of reactivity is: HI > HBr > HCl > HF.
Addition of HX to symmetrical alkenes gives only one product but to unsymmetrical alkenes can give two products. In absence of peroxides, the major product is governed by
Markovnikov's rule which states that negative part of the unsymmetrical reagents goes to that carbon atom of the double bond which has lesser number of H-atoms.
Mark.addn.
32 Propene CHCHCHHBr−=+→
33 2Bromopropane CHCHBrCH
Anti-markovnikov's addition or peroxide effect or Kharasch effect. In presence of peroxides such as benzoyl peroxide, the addition of HBr (but not of HCl or HI) to unsymmetrical alkenes occurs contrary to the Markovnikov's rule. For example.
Peroxide
32 Propene CHCHCHHBr=+→
322 1Bromopropane CHCHCHBr
This addition occurs by a free radical mechanism.
Addition of hypohalous acid (HOX).
Addition of HOCl (Cl 2 in H2O) or HOBr (Br 2 in H 2 O) gives halohydrins. For example, addition of HOCl to ethylene gives ethylene chlorohydrin (ClCH 2 CH2OH).
Addition of cold conc. H2SO4 to alkenes followed by hydrolysis with boiling H 2O gives alcohols.
24 Cold.conc.
32 HSO Propene CHCHCH=→
333 sulphate CHCHOSOHCH
33 Propan2ol CHCHOHCH
Combustion: Alkenes readily burn in air or oxygen to form CO 2 and H 2 O with liberation of large amount of heat.
Alkenes react with O2 in presence of silver catalyst at 575 K to giv e epoxyalkanes.
F or example, ethene gives epoxyethane or ethylene oxide or oxirane.
With cold dilute neutral or alkaline K MnO 4 , alkenes give 1, 2-glycols. During this reaction, pink colour of the KMnO 4 solution is discharged and a brown precipitate of MnO 2 is formed. Therefore, this reaction is used as a test for unsaturation under the name Baeyer's test and the cold dilute neutral or alkaline KMnO4 solution is called Baeyer's reagent.
With hot KMnO 4 solution, cleavage of C=C bond occurs leading to the formation of carboxylic acids, ketones and carbon dioxide depending upon the nature of the alkene. For example,
4 KMnO,KOH 32 373383K CHCHCH=→
CHCOOHCOHO ++
3 22 Ethanolicacid
( ) 4 KMnO,KOH
CHCCH=→
32 373383K 2 2Methylpropene
( ) 3 22 2 Propanone
CHCOCOHO =++
4 KMnO,KOH
CHCHCHCHCH=→
32 3 373383K Pent2ene
CHCHCOOHHOOCCH +
32 3 Propanoicacid Ethanoicacid
When O3 is passed through a solution of an alkene in some inert solvent such as CH2Cl2, CHCl3 or CCl4 at low temperature (196-200 K), ozonides are formed. These on reduction with Zn/H 2O or H2/Pd give aldehydes or ketones or a mixture of these depending upon the structure of the alkene ( ) ( ) 322 2 iO/CHCl,.196200K
CHCHCH=→
32 iiZn/HO Propene
32 Methanal Ethanal CHCHOOCH =+=
This two step process is called ozonolysis and is used to locate the position of a double bond in an unknown alkene.
Alkynes
The general formula of alky nes is C nH2n-2 where n = 2, 3, 4....etc. They contain one triple bond which consists of one strong s -bond and two weak p -bonds. The carbon atoms of a triple bond are sphybridized. Acetylene is a linear molecule with cylindrical electron cloud. The C=C bond length is 120 pm while C-H bond length is 107.9 pm.
Being linear, alkynes do not show geometrical isomerism.
Dehydrohalogenation of gem- and vic-dihalides first with alcoholic KOH followed by treatment with NaNH2 in liq. NH3 or preferably by the action of NaNH2 in liquid NH3 ( ) KOHalc 22 1,2Dibromoethane
BrCHCHBr ∆ →
23 NaNH/liq.NH
CHCHBr HCCH=→≡
2 196K Ethyne
1Bromoethene
Dehalogenation of tetrahalides with zinc dust in methanol yields alkynes. Dehalogenation of chloroform and iodoform by heating with Ag powder gives acetylene.
Kolbe's electrolysis of Na or K salts of maleic or fumaric acid gives acetylene.
Reaction of mono- and disodium acetylides with alkyl halides gives both symmetrical and unsymmetrical alkynes depending upon the nature of the alkyl halide. For example,
HC ≡ C– Na++ CH3Br→HC ≡ C – CH3 + NaBr Sod. acetylide Propyne
Na+ –C ≡ C– Na+ + 2 CH3Br → Disodium acetylide
CH3–C ≡ C–CH3 +2 NaBr But-2-yne
In the laboratory, acetylene is prepared by the action of H2O on CaC2.
CaC2 + 2 H2O → HC ≡ CH + Ca(OH)2 Acetylene
Acidic nature. Due to sp-hybridization of acetylenic carbon, H-atoms of terminal alkynes are acidic in nature. They react with Na at 475 K or Na in liq. NH3 at 196 K to form sodium alkynides.
With ammoniacal AgNO 3 solution or Tollens' reagent, terminal alkynes give white precipitate of silver alkynides.
With ammoniacal CuCI solution, terminal alkynes give red precipitate of copper alkynides.
The acidic character of ethane, ethene and ethyne follows the order:
CH ≡ CH > CH2 = CH2 > CH3 — CH3
The acidic character of carboxylic acids, phenols, alcohols, water, ammonia, acetylene and alkanes follows the order:
RCOOH > C 6 H 5 OH > H 2 O > ROH > HC ≡ CH > NH3 > RH
The basic character, however follows the reverse order :
R– > NH–2 > HC ≡ C– > RO– > HO– > C6H5O–> RCOO–
Electrophilic addition reactions. Like alkenes, alkynes also undergo electrophilic addition reactions. However, due to (i) greater electronegativity of sp-hybridized carbon as compared to sp 2 -hybridized carbon of ethene, and (ii) cylindrical nature of the pi-electron cloud, alkynes are less reactive than alkenes towards electrophilic addition reactions.
Addition of halogens. Cl 2 and Br 2 readily add to alkynes first forming 1, 2-dihaloalkenes and then 1, 1, 2, 2-tetrahaloalkanes. The order of reactivity decreases in the order: F2 > Cl2 > Br2 > I2.
Addition of halogen acids. The order of reactivity decreases in the order : HI > HBr > HCl > HF In absence of peroxides, addition occurs according to Markovnikov's rule but in presence of
peroxides, addition of HBr follows antiMarkovnikov's rule. For example,
CHCHCHBrCHCHCHBr =→
Addition of hypohalous acids to alkynes gives dihalocarbonyl compounds. With HOC1, acetylene gives 2, 2-dichloroethanal while propyne gives 1, 1-dichloropropanone.
Addition of water or hydration of alkynes occurs when treated with dil. H 2SO 4 in presence of HgSO 4 as catalyst to form carbonyl compounds. For example,
24 4 Dil.HSO HgSO
Dil.HSO
Addition of HCN to acetylene in presence of Ba(CN) 2 or CuCl/HCl gives vinyl cyanide or acrylonitrile.
Addition of hydrogen. Addition of H 2 to alkynes in presence of Ni at 523-573 K first gives alkenes and then alkanes. The reduction of alkynes can be stopped at the alkene stage. With Lindlar's catalyst, i.e., Pd supported over CaCO 3 or BaSO4 and partially poisoned by addition of S or quinoline, alkynes give cis alkenes but with Na in liquid NH3 (i.e., Birch reduction), trans-alkenes are formed.
Oxidation of terminal alkynes with cold KMnO 4 solutions gives a mixture of RCOOH + CO2 while oxidation of nonterminal alkynes gives diketones. Under these conditions. acetylene, however, gives oxilic acid. During this reaction, the pink colour of the KMnO 4 solution is discharged and a brown precipitate of MnO2 is .formed. Therefore, this reaction is used as a test for unsaturation under the name Baeyer's test.
Oxidation of non-terminal alkynes with hot KMnO4 solution gives only a mixture of dicarboxylic acids through cleavage of C ≡ C bond.
Reaction with ozone. Alkynes react with O 3 to form ozonides which upon reductive ozonolysis gives 1, 2-dicarbonyl compounds.
Aromatic Hydrocarbons
Hydrocarbons and their alkyl, alkenyl and alkynyl derivatives which contain one or more benzene rings either fused or isolated in their molecules are called aromatic hydrocarbons or arenes. Aromatic hydrocarbons which contain a benzene ring are also called benzenoids while those which do not contain a benzene ring are called non-benzenoids, e.g., azulene, tropolone, cyclopentadienyl anion, cycloheptatrienyl cation, etc.
All the C-atoms of benzene are sp 2hybridized, i.e., all the ∠CCC and ∠HCH = 120°. It is a planar molecule, i.e., all the six C—C and six C—H bonds lie in a plane.
Benzene is a resonance hybrid of two Kekule structures and has a resonance energy of 36.0 kcal mol–1 or 151 kJ mol–1. Due to resonance, all the carbon-carbon bond lengths are equal (139 pm) and lie in between those of carbon-carbon single bond lengths of 154 pm and carbon-carbon double bond lengths of 134 pm.
Huckel rule. Benzene is an ideal aromatic compound. The aromatic character of other cyclic conjugated polyenes can be determined on the basis of Huckel rule. The main points of Huckel rule are :
1. It should have a single cyclic cloud of delocalized p -electrons above and below the plane of the molecule.
2. It should be planar.
3. It should contain (4n + 2) p-electrons where, n = 0, 1, 2, 3.... etc.
A molecule which does not satisfy any one or more of the above conditions is said to be non-aromatic.
Methods of Preparation of Arenes. Benzene is obtained by coal tar distillation.
Benzene and its homologoues are obtained by (i) cyclic polymerization of alkynes, (ii) decarboxylation of aromatic acids with soda-lime, (iii) reduction of phenols with Zn dust, (iv) reduction of chlorobenzene with Ni-Al alloy/NaOH, (v) reduction of benzenediazonium chloride with H 3 PO 2 /Cu + , (vi) heating benzenesulphonic acid with super heated steam, (vii) aromatization of n-hexane and n-heptane, (viii) Wurtz-Fittig reaction, involving heating of an aryl halide with an alkyl halide in presence of metallic Na in dry ether, (ix) Friedel–Craft's reaction of benzene with alkyl halides in presence of anhyd. AlCl3 and (x) reaction of arylmagnesium halides (Grignard reagents) with alkyl halides in presence of dry ether.
Electrophilic substitution reactions. The typical reactions of arenes are electrophilic substitution reactions in which one or more H-atoms of benzene ring are replaced by the electrophile. The reaction occurs in two steps. In the first step, the electrophile attacks the benzene ring to form a resonance-stabilized carbocation. This step is slow and is the rate-determining
st ep of the reaction. In the second step, the carbocation loses a proton to form the substitution product. This step is fast and hence does not affect the rate of the reaction.
Chlorination or bromination occurs with Cl2 or Br2 in presence of a Lewis acid (also called halogen carrier) such an anhyd. AlCl3, FeBr3, etc. The effective electrophile here is either Cl + (chloronium ion) and Br+ (bromonium ion).
Iodination is carried out with I2 in presence of an oxidising agent such HNO3, HIO3 or HgO.
Nitration is carded out with a mixture of conc. HNO3 + conc. H2SO4 (nitrating mixture). The effective electrophile here is NO2+ (nitronium ion).
Sulphonation is carried out either with conc. H2SO4 or ClSO3H (chlorosulphonic acid). The effective electrophile here is SO3. Unlike, halogenation and nitration, sulphonation is a reversible reaction. That is why benzenesulphonic acid on heating with super heated steam undergoes desulphonation to give benzene.
Friedel–Craft's alkylation is carried out with an alkyl halide in presence of anhyd. AlCl3, FeBr3, etc. The effective electrophile here is R+ .
Friedel–Craft's acylation is carried out with an acid chloride or anhydride in presence of anhyd. AlCl3. The effective electrophile here is the acylium ion,
Addition reactions. Benzene normally does not undergo addition reactions but under drastic conditions, i.e., high temperature and pressure, they do undergo some addition reactions. With H2 in presence
of Ni at 473 – 523 K, benzene gives cyclohexane but with Cl 2 in presence of sunlight and in absence of halogen carrier, it gives benzene hexachloride (BHC).
Oxidation of benzene with O2 in presence of V2O5 at 773 K gives maleic anhydride.
Reductive ozonolysis of benzene gives glyoxal; that of o-xylene gives glyoxal, methylglyoxal and dimethylglyoxal while that of mesitylene gives only methylglyoxal.
Directive influence of groups. When a monosubstituted benzene derivative is converted into a disubstituted derivative, the substituent already present in the benzene ring determines the position of the incoming group. This is called directive influence of groups.
o,p-Directing groups. The substituents or groups which direct the incoming group to o-and p-positions are called o, p-directing groups. For example, CH 3–, CH 3CH 2–,
C 6 H 5 –, Cl–, Br–, I–, –OH, –OCH 3 , –OCOCH 3 , –NH 2 , –NHCH 3 , –N(CH 3 ) 2 , –NHCOCH3, etc.
m-Directing groups. The substituents or groups which direct the incoming group to the m-position are called m-directing groups. For example, (CH 3) 3 N + – NO 2, – CN, – CF3, – CHO, – COR, – COOH, –COOR, SO3H, etc.
Effect of substituents on Reactivity. o, p-Directing groups increase the electron density in the benzene ring and hence activate the nucleus towards further electrophilic substitution reactions. In contrast, m-directing groups decrease the electron density in the benzene ring and hence deactivate the nucleus towards further electrophilic substitution reactions.
Although halogens are little deactivating, yet they are o, p-directing.
Exercises
JEE MAIN LEVEL
Level - I
Alkanes
Single Option Correct MCQs
1. The fully eclipsed conformation of n-butane is least stable due to the presence of (1) bond opposition strain only
(2) steric strain only
(3) bond opposition strain as well as steric strain
(4) no strain is present in the molecule.
2. Energy barrier between staggered and eclipsed forms of ethane is
(1) 0.6 Kcal/mole
(2) 2.9 Kcal/mole
(3) 12 Kcal/mole
(4) 12 Kcal/mole
3. By the reduction of alkyl halides, we get alkanes. The best alkyl halide is
(1) Fluoride
(2) Bromide
(3) Chloride
(4) Iodide
4. The products formed when the sodium salt of a fatty acid is heated with soda lime are (1) Alkane & NaHCO3
(2) Alkane & NaOH
(3) Alkane & Na2CO3
(4) Alkene & Na2SO4
5. The gases liberated at anode in the electrolysis of aqueous sodium acetate are (1) CO2 and H2
(2) C2H6 and CO2
(3) H2 and C2H6
(4) H2 and O2
6. In Wurtz reaction, in the preparation of alkanes, metallic sodium acts as (1) Oxidising agent
(2) Reducing agent
(3) Dehydrogenating agent
(4) Dehydrohalogenating agent
7. Methyl magnesium iodide reacts with ethyl alcohol to produce (1) Ethane
(2) Methane
(3) Propane (4) Ether
8. Arrange the following compounds in the descending order of their melting points a) n-pentane b) iso-pentane
c ) neopentane
(1) c > b > a (2) c > b > a
(3) b > c > a (4) c > a > b
9. Methyl magnesium iodide reacts with ethyl alcohol to produce (1) Ethane (2) Methane
(3) Propane (4) Ether
10. The Kolbe’s electrolysis proceeds via (1) Nucleophilic substitution mechanism (2) Electrophilic addition mechanism (3) Free radical mechanism
(4) Electrophilic substitution mechanism
11. Alkanes having odd carbons cannot be prepared in
A) Wurtz reaction
B) Frankland reaction
C) Kolbe’s electrolysis
D) Sabatier-senderens reaction
(1) B, C and D (2) A, C and D
(3) A, B and D (4) A, B and C
12. When n-heptane is heated to about 773K under high pressure in presence of Cr 2O 3 deposited over Al 2 O 3 , the final product formed is
(1) Cyclohexane
(2) Methylcyclohexane
(3) Toluene
(4) Benzene
13. Which of the following is correct sequence of steps in a chain reaction?
(1) Propagation, initiation, termination
(2) Initiation, termination, propagation
(3) Propagation, termination, initiation
(4) Initiation, propagation, termination
14. Here the major product X is (1)
(2)
(3)
(4)
15. Aromatization of n-hexane gives
(1) Benzene
(2) Toluene
(3) Methane
(4) A mixture of octanes
16. Cracking is a process in which
(1) Petrol is produced by cracks on the surface of wax
(2) Combustion of petrol is carried out
(3) Compounds of high molecular mass are converted into compounds of lower molecule
(4) None of the statements is correct
17. Iodination of alkane is carried out in the presence of (1) Alcohol
(2) HNO3 and HIO3
(3) Any reducing agent
(4) Benzene
Numerical Value Questions
18. The number of mono brominated products (including stereoisomers) is_____
19. 2323 0 CrO/AIO 614 600C,35atm CH X → number of π electrons in compound ‘X’ is
20. 2molesofaqueoussodiumacetateon electrolysisgives“X” litres of gaseous mixture at STP. Find X 22.4
21. What volume of methane (NTP) is formed from 16.4 g of sodium acetate by fusion with soda lime?
Alkenes
Single Option Correct MCQs
22. Identify the correct match from the following lists
List-I
List-II
(A) Dehydrohalogenation (I) Alc. KOH
(B) Dehydration (II) Con. H2SO4/170°C
(C) Unsaturation (III) Zn dust
(D) Dehalogenation (IV) Br2 water
(A) (B) (C) (D)
(1) II IV III I
(2) I II IV III
(3) IV I II II
(4) I III IV II
23. In which of the following reactions, only single isomer of alkene is formed?
(1) KOH 65222 Ethanol CH–CH–CH– CH – Br →
(2)
(3) (4)
24. The metal used for the debromination reaction of 1,2-dibromoethane
(1) Na (2) Zn (3) Mg
(4) Li
25. alc.KOHdil.HSO24 25 CHClAB. →→ Here A and B are (1) C2H4, C2H5OH (2) C2H6, C2H5OH (3) C3H8, C2H5OH (4) C2H2, C2H5OH Ans.(1)
26. Baeyer’s reagent oxidizes ethylene to (1) Ethylene chlorohydrin (2) Ethyl alcohol (3) CO2 and H2O (4) Ethanel-1,2-diol
27. The products of ozonolysis of an alkene are ethanal and methanal, then that alkene is (1) CH3–C≡C–CH3 (2) CH3–CH=CH–CH3 (3) CH3–CH=CH2 (4) CH3–CH=CH2
28. Ethylene on hydration gives (1) Glycol (2) Ethanol
(3) Ethane
(4) Ethanoic acid
29. Ethyl hydrogen sulphate is obtained by reaction of Conc. H2SO4 on (1) Ethylene (2) Ethane
(3) Ethyl chloride (4) Ethanal
30. The product obtained when propene undergoes addition reaction with HBr in the presence of benzoyl peroxide is
(1) 1-bromopropane
(2) 2-bromopropane
(3) 1,2-dibromopropane
(4) 2,2-dibromopropane
32. Markovnikov’s rule governs the addition of (1) Unsymmetrical reagents to symmetrical alkenes
(2) Symmetrical reagents to unsymmetrical alkenes
(3) Unsymmetrical reagents to unsymmetrical alkenes
(4) Symmetrical reagents to symmetrical alkenes 33. 2 alcoholicKOH Br
(1) Acetylene
(2) Ethane
(3) Ethene
(4) Methane
34. In which of the following reactions, anti Markownikoff’s rule is observed?
(1) peroxide
32 CHCHCHHCl−=+→
(2) peroxide
32 CHCHCHHBr−=+→
(3) peroxide
32 CHCHCHHI−=+→
(4) 3224 CHCCHHSO −= → +
35. Arrange the following alkenes in the descending order of their reactivity with HBr
a) Ethene b) Propene
c) 2-Butene d) 2-Methyl-2-butene
(1) a > b > c > d (2) d > c > b > a
(3) d > c > a > b (4) a > b > d > c
36. Ozonolysis products of 2-ethylbut-1-ene are
(1) Methanal and 3-pentanone
(2) Methanal and 2-pentanone
(3) Ethanal and 2-butanone
(4) Ethanal and 3-pentanone
37. Cis 4 KOH 273K 2buteneKMnOproduct → −+
(1) Meso 2,3–butane diol
(2) Dextro 2,3–butane diol
(3) Laevo 2,3–butanediol
(4) Racemic mixture of 2 and 3
(3)
(4) Mixture of 2 and 3
Numerical Value Questions
39. The number of Hydrocarbons, involved in Reductive Ozonolysis to gives methanol
40. 3-methylpent 2-ene on reaction with HBr in presence of peroxide forms and addition product (A). The number of possible stereoisomers for ‘A’ is_____
41. 3-methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is _____
42. KMnO4 2222 2 KOH, 373.383K CHCHCHCHCHCH A2CO=−−−=−→+
43. Number of molecules of alkene required for the reaction of one molecule of B 2H6 to carry out hydroboration
38. products
(1)
(2)
44. How many alkenes of formula C 6 H 12 excluding stereoisomers, can form saturated straight chain hydrocarbons Alkynes
45. Hybridization of triply bonded carbons, carbon–carbon triple bond length and bond strength of carbon-carbon triple bond are
(1) sp; 1.20 A° and 600 kJ mol –1
(2) sp2; 1.20 A° and 825 kJ mol –1
(3) sp; 1.20 A° and 825 kJ mol –1
(4) sp, 1.340 A° and 825 kJ mol –1
46. NaNH2 Ä Alcoholic 22 KOH CHBrCHBrX−→→
Acetylene. Here X is (1) Vinyl bromide
(2) 1, 1–Dibromoethane
(3) Ethyl bromide
(4) Ethylene dibromide
47. Acetylene is produced by the action of water on
(1) Be2C
(2) Al4C3
(3) CaC2
(4) Mg2C3
48. 2
22 CHBrCHBrAB.BC.
The compound ‘C’ is (1) Vicinal dihalide
(2) Geminal dihalide
(3) Benzyl halide
(4) Allyl halide
49. The major organic compound formed by the reaction of 1, 1, 1-trichloroethane with silver powder is (1) acetylene (2) ethane (3) 2-butyne (4) 2-butene
50. Number of linear atoms in acetylene molecules is a maximum of (1) 1 (2) 2 (3) 3 (4) 4
51. 32 CaCOCaOCO ∆ + →
CaC2HOCaOHCH +→+
2 CaO3CCaCCO +→+ ( ) 22 22 2
The weight of CaCO 3 needed for the preparation of 2.24 lit of Acetylene gas under S.T.P conditions is
(1) 10 g (2) 20 g
(3) 5 g (4) 2 g
52. The final product formed when acetylene reacts with hydrogen bromide is (1) CH2Br–CH2Br (2) CHBr2–CHBr2 (3) CH3–CHBr2
(4) CH3–CHBr2
53. The intermediate compound formed when acetylene is hydrated in presence of dil. H2SO4 and HgSO4 is (1) Acetaldehyde (2) Ethanol (3) Vinyl chloride (4) Ethanal
54. The reductive ozonolysis product of acetylene is (1) (2) (3) (4)
55. ExcessNaNH23ExcessCHCl HCCH A B. ≡→→ (1) 2-Butyne (2) 1-Butyne (3) Propyne (4) Monosodium acetylide
56. Which of the following cannot give precipitate with cuprous ammonium chloride?
(1) CH3 – C ≡ C − CH3
(2) CH3 – C ≡ CH
(3) HC ≡ CH
(4) CH3 − CH2 – C ≡ CH
57. Tollen`sreagent HCCH A ≡→ (1) 133 (2) 240 (3) 120 (4) 373
58. When alkyl substituted acetylene undergoes addition with hydrogen in presence of Lindlar’s catalyst, the alkene formed is (1) A mixture of cis and trans (2) Trans (3) Cis
(4) In presence of Lindlar’s catalyst, addition does not take place
59. The reagent used for getting trans alkene from alkyl substituted acetylene with hydrogen is
(1) Na in liq. NH3
(2) Li in liq. NH3 (3) Both (1) or (2) (4) H2 in presence of Ni (or) Pt (or) Pd
60. The acidic nature of hydrogens in acetylene can be explained by the reaction with (1) Sodium metal
(2) Ammonical cuprous chloride solution (3) Ammonical silver nitrate solution (4) All the above
61. The color of the precipitate formed when acetylene is passed through ammonical cuprous chloride solution is (1) White (2) Red (3) Blue (4) Green
62. In the following reaction, what is X?
(1) CH3CH2OH
(2) CH3−O–CH3
(3) CH3CH2CHO
(4) H2C = CHOH
63. Acetylenic hydrogens are acidic because (1) sigma electron density of C–H bond in acetylene is nearer to carbon which has 50% s-character
(2) acetylene has only one hydrogen on each carbon
(3) acetylene contains least number of hydrogens among the possible hydrocarbons having two carbons
(4) acetylene resembles acetic acid
64. Which of the following type of reaction occurs when 1-butyne reacts with HBr?
(1) An electrophilic addition
(2) Markownikoff’s electrophilic addition
(3) Anti-Markownikoff’s radical addition
(4) None of the above
65. Which of the following statements is correct?
(1) alkynes are more reactive than alkenes towards halogen addition
(2) alkynes are less reactive than alkenes towards halogen addition
(3) both alkynes and alkenes are equally reactive towards halogen addition
(4) primary vinylic RCHCH + =
cation is more reactive than secondary vinylic cation 2 RCHCH +
66. alc.KOH 322 CHCHCHClX −−→ PdBaSO4 2 CHCHH Y + → ≡+
Here, in the above sequences, the final products ‘X’ and ‘Y’
(1) are a pair of homologues
(2) have the same percentage composition
(3) have the same empirical formula
(4) all the above
67. Which of these will not react with acetylene?
(1) NaOH(aq)
(2) Na
(3) AgNO3, OH–
(4) HCl
68. 24 2 0 4 alcKOH ConHSO Br CCl Ä 170C XYQR →→→
statement (s) about R is/ are
A) It decolorizes Baeyer’s reagent
B) It gives benzene when subjected to polymerization
C) It gives red precipitate with ammonical Cu2Cl2
D) It gives ethylene on hydrogenation in the presence of Lindal’s catalyst
(1) A, B and C
(2) B, C and D
(3) A, B, C and D
(4) A, B and D
69. 25 2 CHBr NaNH 32 CHCHCCHXY ≡→→ the above reactions are
(1) 32 322 CHCHCOONa,CHCHCHCH = (2) 32 3225 CHCHCCNa,CHCHCCCH ≡≡
(3) 32223223 CHCHCHCHNa,CHCHCHCH
(4) ( ) 32 25 3 2 CHCHCCNa,CHCHCH ≡−
Numerical Value Questions
70. How many reagents from the list below can be used to remove hydrogen as proton from Terminal alkyne?
3 2 3232 CHMgBr,NaNH,Na,NaHCO,KNH,CHCHLi
71. 0 RedhotCu 3 2000C CHCCHX, −≡→ the no. of π bonds in X is
72. In bromination of propyne with Bromine (excess), the number of σ bonds formed in the product are x and π bonds are y, find sum of x and y
73. Sum of pure and hybrid orbitals present in the hydrocarbon obtained by hydrolysis of Mg2C3 is ______
74. What is the ratio of 2° and 3° hydrogen atoms in the given structure
75. The number of moles of oxygen required for the combustion of heptyne is
Single Option Correct MCQs
76. Formation of benzene from acetylene is
(1) Trimerization
(2) Tetramerization
(3) Dimerization
(4) Condensation
77. The carbon-hydrogen bonds in benzene are (1) σSP2−S
(2) σSP3−S
(3) Both σSP2−S and σSP3−S
(4) σSP−S
78. The structure of benzene can also be explained by (1) Valence bond theory
(2) Crystal field theory
(3) molecular orbital theory
(4) Both 1 and 3
79. Huckel’s rule of aromaticity is (1) Having 6 pi electrons
(2) Having 3 double bonds
(3) Having (4n+2) pi electrons
(4) Having alternate double bonds
80. The dipole moment of benzene is (1) Zero
(2) Less than p- dichloro benzene
(3) Greater than p-dichloro benzene
(4) Equal to that of chloro benzene
81. Which of the following statement is not true for benzene?
Benzene
(1) It is planar molecule
(2) All C-C bond lengths are equal
(3) The resonance energy is 36 kcal/ mole (4) It contains three localized pi bonds
82. In benzene molecule there are three double bonds and (1) 10 sigma bonds
(2) 12 sigma bonds
(3) 6 sigma bonds
(4) 3 sigma bonds
83. Sodium salt of benzoic acid when heated with which of the following gives benzene
(1) Sodamide
(2) Soda lime
(3) Sodium chloride
(4) Sada ash
84. The two structures of benzene proposed by Kekule differ in
(1) The position of carbon nuclei
(2) The position of hydrogen nuclei
(3) The position of the single bonds
(4) The position of the double bonds
85. Resonance energy of benzene is
(1) 150 Kcals Mol−1
(2) 36 Kcals Mol−1
(3) 6 KJ Mol−1
(4) 200 Kcals Mol−1
86. The C-C bond lengths in benzene are
(1) 1.54 A° and 1.34A°
(2) 1.34A° and 1.20A°
(3) 1.39 A° only
(4) 1.20 A° only
87. Maximum number of atoms present in same plane of benzene molecule (1) 6 (2) 10
(3) 12 (4) 4
88. Why does 1,3- cyclohexadiene undergo
dehydrogenation readily?
(1) It can be easily reduced
(2) It has no resonance energy
(3) It gains considerable stability by becoming benzene
(4) It cannot undergo dehydrogenation
89. In halogenation of aromatic compounds, catalyst is (1) Lewis base
(2) FeCl3
(3) AlCl3
(4) FeCl3 or AlCl3
90. Nitration mixture is (1) 1:1 of conc. HNO3 and conc. HCl
(2) 1:1 of conc. HNO3 and conc. H2SO4
(3) 1:1 of conc. HNO2 and conc. H2SO4
(4) 1:10 of conc. H2SO4 and conc. HNO3
91. In Friedel-craft’s alkylation, catalyst is (1) anhydrous NaCl
(2) anhydrous AlCl3
(3) Trialkyl carbonium ion
(4) Alkyl halide
92. Benzene reacts with ________ To yield acetophenone
(1) CH3COCl + AlCl3
(2) C6H5COCl + AlCl3
(3) RCOCl + AlCl3
(4) RCOCl + AlCl3
93. In the above sequence B is (1) Acetanilide
(2) Acetophenone
(3) Benzophenone
(4) N- Phenylacetamide
94. Which of the following behaves as a saturated compound?
(1) C2H4 (2) C2H2
(3) C3H6 (4) C6H6
95. Here the product B is
(1) Benzene
(2) Toluene
(3) Benzene sulphonic acid
(4) Chlorobenzene
96. What is the molecular formula of the product formed when benzene is reacted with ethyl chloride in presence of anhydrous aluminium chloride?
(1) C8H10
(2) C6H6
(3) C8H8
(4) C6H5Cl
97. Number of double bonds in gammaxene (BHC)is
(1) 3 (2) 2
(3) 1 (4) 0
98. Which of the following is represented incorrectly?
(1)
(2)
(3) Distillation of phenol with zinc dust gives benzene
(4) Benzene reacts with CH3COCl in the presence of AlCl3 to give C6H5COCl
99.The wrong statement in the following is
(1) sulphonation of benzene takes place only with hot concentrated sulphuric acid
(2) In the nitration mixture concentrated sulphuric acid is used for the formation of nitronium ion
(3) Because of unsaturation benzene easily undergoes addition reactions
(4) Benzene burns with a sooty flame
100.Which of the following is a wrong statement?
(1) The electrophile in sulphonation is H2SO4
(2) In Friedel-crafts reaction of benzene
with an acid chloride the electrophile reagent is RCO+
(3) In the nitration of toluene by electrophilic substitution the electrophile is nitronium ion
(4) Reaction of benzene with chlorine in the presence of sunlight is an addition reaction
101.Which of the following is not correct?
(1) Benzene can be obtained by the cyclisation of ethene
(2) Benzene reacts with H2 in the presence of Ni at 200°C and high pressure to give cyclohexane
(3) C6H6Cl6 can be obtained from C6H6 + Cl2 + light
(4) Benzene reacts with ozone to form benzene triozonide
102.Benzene does not undergo addition reaction easily because
(1) It has six hydrogen atoms
(2) It has a cyclic structure
(3) Double bonds present in benzene is strong
(4) Resonance stabilised system is to be preserved
103.Under pressure, (with hydrogen) in presence of nickel catalyst, benzene form s (1) (2) (3)
(4) C6H14
104.Carcinogenic pollutants are formed on incomplete combustion
(1) tobacco
(2) coal
(3) petroleum
(4) all the above
105.Benzene and polynuclear hydrocarbons containing more than two benzene rings fused together are (1) Toxic
(2) Causes cancer
(3) possess carcinogenic property
(4) All the above
106.Among the following which is carcinogenic pollutant?
A. 1,2 benzopyrene
B. 1,2,5,6-Dibenzanthracene
C. 3-methyl cholanthrene
(1) only A
(2) only B
(3) only C
(4) All
Numerical Value Questions
107. How many of the following groups are activating towards electrophilic aromatic substitution reaction.
-NHCOR, -OCOR, -COOR, -NR 2, -NH 2, -OH, -OR
108.How many of the following are Meta directing groups in electrophilic substitution in benzene nucleus is/are
109.How many of the following species are ortho para directing
−COOH,−CHO,−OH,−NO 2 ,−NH 2 ,− NHCOR, −COR,−CF3,−SO3H,−NR3⊕,−CONHCH3
110. 25 3 CHCl Zn PhenolX Y DustAnhydrousAlCl number of structural isomers of Y are
111.Number of reactions which gives benzene as major product?
(a) Fetube HCCH
(b) (c) (d) (e)
112.Benzene reacts with excess of Cl2 in presence of sun light gives product A. The no. of halogens in A is ‘x’ and no. of hydrogens in A is y then x y valueisequalto_________.
Level - II
Alkane
Single Option Correct MCQs
1. Which of the following liberates methane on treatment with water?
(1) Silicon carbide
(2) Calcium carbide
(3) Beryllium carbide
(4) Magnesium carbide
2. Which of the following hydrocarbon is not formed when Wurtz reaction takes place between ethyl iodide and propyl iodide?
(1) Butane (2) Propane
(3) Pentane (4) Hexane
3. Electrolysis of an aqueous solution of potassium, acetate the hydrocarbon obtained is
(1) Methane (2) Ethane
(3) propane
(4) A mixture of the above three alkanes
4. Propane can be best prepared by the reaction
(1) Na/ether 323 Ä CH-CH-I+CHI →
(2) ( ) Ether, Ä 323 2 CH-CHBr+CHCuLi →
(3) 2 HO 33 Electrolysis CHCOONa+CHCOONa →
(4) 630k 32 NaOH+CaO CHCHCOONa →
5. Sodium propionate on decarboxylation with sodalime gives
(1) Propane (2) Ethane
(3) Butane (4) Pentane
6. Which of the following is the most stable conformer of n-Butane?
(1)
(2)
(3)
(4)
7. Iodination of alkane is carried out in the presence of (1) alcohol
(2) HNO3 or HIO3
(3) any reducing agent
(4) benzene
8. The correct order of melting points is
(1) ethane < propane < butane (2) butane < propane < ethane (3) propane < ethane < butane (4) ethane < butane < propane
Numerical Value Questions
9. Total number of isomers(including stereo isomers) possible for heptane are
10. The number of primary hydrogens in isobutane is
11. How many different organic compounds obtained by the following reaction?
12. Maximum number of isomeric monochloro derivatives that can be obtained from 2,2,5,5-Tetramethyl hexane by chlorinations is________
Alkene
Single Option Correct MCQs
13. Which of the following compound will exhibit geometrical isomerism.
(1) 1-phenyl 2-butene
(2) 3- phenyl 1-butene (3) 2-phenyl 1-butene (4) 1,1-diphenly 1-propene
14. Which of the following is the major product obtained when sec-Butyl chloride is heated with alcoholic KOH?
(1) 1-Butene
(2) 2-Butene
(3) 1-Butanol
(4) 2-Butanol
15. Under normal conditions ethene does not show addition reaction when it reacts with (1) Cl2/CCl4
(2) Br2/CCl4
(3) I2/CCl4
(4) H2/Pt
16. Anti-Markownikoff’s addition of HBr is not observed in
(1) 1-butene
(2) 1-pentene
(3) Propene
(4) 2-butene
17. An alkene on ozonolysis yield OHC-CH2CH2- CH2-CHO only. The alkene is (1) CH2 = CHCH2CH2CH2CH2CH3 (2)
Alkyne
Single Option Correct MCQs
20. 2-Hexyne gives trans-2-Hexene on treatment with
(1) Na/Liq.NH3
(2) Pd/BaSO4
(3) LiAlH 4
(4) Pt/H2
21. The product(Z) of the following reaction is 2 3 HBr HCCCHZ −≡→
(1) H3CCH2CHBr2
(2) H3CCBr2CH3
(3) H3CCHBrCH2CHr2Br
(4) BrCH2CH2CH2Br
CHCHCCH oducts
22. The ozonolysis product (s) of the following reaction is (are) 3 23 ) 32 ) / Pr iO iiHOCHCOOH
(1) CH3COCH3
(2) CH3COCH3 + HCOOH
(3) CH3COCH3 + HCHO
(4) CH3CH2COOH + HCOOH
Numerical Value Questions
18. How many of the following alkenes on addition of HBr would give the same product in the presence or absence of peroxide?
Propene, 1-butene, 2-butene, 3-hexene, 2, 3-dimethyl-2-butene, 1, 2-dimethyl cyclohexene, 1,4-dimethyl-2-cyclohexene, 3, 4-dimethyl-3-hexene, cyclohexene.
19. How many types of 10 alcoholic groups can be obtained by hydroboration oxidation of the following alkene?
(Consider all double bonds are involve in hydroboration oxidation excluding stereo isomers) CH3
23. When propyne is treated with aqueous H2SO4 in the presence of HgSO4 the major product is
(1) Propanol
(2) Propyl hydrogen sulphate
(3) Acetone
(4) Propenol
24. Addition of HOCl to ethyne gives:
(1) ethyl chloride
(2) vinyl chloride
(3) dichloroacetaldehyde
(4) ethylidene chloride
Numerical Value Questions
25. Maximum number of atoms that can be possible in one plane in prop yne
Aromatic hydrocarbons
Single Option Correct MCQs
26. 322222 CHCHCHCHCHCH 7223 3 CrOAlOOO CH X −→ Find the product ‘X’
(1) (3) CH3
(2) CH3 (4) CH3
27. Which of the following is not the method of preparation of benzene?
(1) Redhot 22 Cutube 3CHCH = →
(2) Redhot Fetube 3CHCH ≡ →
(3) OH (Vapour) + Zn (dust)
(4) 2 1) 22 2) Red-Hat-CuTube HO CaCHO+→
28. Benzene on nitration gives nitrobenzene in presence of HNO 3 and H 2SO 4 mixture, where:
(1) Both H2SO4 and HNO3 act as a bases
(2) HNO3 acts as an acid and H2SO4 acts as a base
(3) Both H2SO4 and HNO3 act as an acids
(4) HNO3 acts as a base and H2SO4 acts as an acid
29. Amongst the following compounds the one which is most easily sulphonated is:
(1) Benzene
(2) Nitro benzene
(3) Toluene
(4) Chlorobenzene
30. The electrophile in Friedel-Craft acylation is
(1) R+ (2) RCO
(3) RC + O (4) R – COO
31. Ozonolysis of benzene gives (1) Methyl glyoxal (2) glyoxal
(3) Pyruvic acid
(4) Pyruvaldehyde
32. Benzene sulphonic acid is converted into benzene by (1) heating with dilute sulphuric acid
(2) heating with dilute NaOH (3) heating with soda lime
(4) heating with zinc dust
33. In which of the following reactions, aromatic character is retained?
(1) 2 H/Ni/ Ä 66 CHX →
(2) 3 2 O 66 Zn/HO CHY →
(3) 3 3 CHCOCI 66 AlCl CHQ →
(4) CI2 66 light CHR →
Numerical Value Questions
34. 25 3 Zndust (Monoalkylatedproduct) anhy.AlCl Phenol CHCl XY→→
The number of structural benzenoid isomers of ‘Y’ are
35. 322 33 CHOH3BrClZn-Cu BFÄAlClEtOHABCD →→→→
Molecular weight of D is [At wt of C=12, H = 1, Cl = 35.5, Br = 80]
Level - III
1. Compound ‘A’ undergoes following sequence of reactions to give compound ‘B’. The correct structure and chirality of compound ‘B’ is: [where Et is −C 2H5 group
CH3 CH3 CH3 Br 2 2 Mg,EtODOB→
(1) CH3 CH3 CH3 D , Achiral
(2) CH3 CH3 CH3 OD , chiral
(3) CH3 CH3 CH3 D , chiral (4) CH3 CH3 O , Achiral
2. The major product in the following reaction.
(1)
ColdKMnO/OHA→
Which is true about this reaction?
(1) A is meso-2, 3-butandiol formed by syn addition
(2) A is meso-2, 3-butandiol formed by anti addition
(3) A is racemic mixture of d and l 2, 3-butandiol formed by anti addition
(4) A is racemic mixture of d and l 2, 3-butandiol formed by syn addition
5. How many statements is/are correct?
33 2 Na in liquid NH X moles NaNH 32 CHCl CHCHBrCHBrPQR −−→→→
A) X value is 3
B) Q is 2-Butyne
C) R is trans-2-Butene
D) In the Reaction Q → R, reaction intermediates involved are free radical anion, free radical and carbanion.
6. How many structural alkenes are formed by E2 elimination of 3-Bromo-3,4-dimethyl hexane using strong base like sodium methoxide.
Ä Ä(A)gas+BC →→
If IUPAC name of compound ‘C’ is x -Methylpent- z -en- y -one.Then value of (x+y) is ___. (Where x ,y and z are integers)
8. The alkyne H 3 C−CH 2 −C ≡ CH and H 3 C− C ≡ C−CH 3 Can be distinguished by the following methods except
(1) Tollen’s reagent
(2) Ammoniacal Cu2Cl2 solution
(3) Na-metal
(4) Baeyer’s reagent
9. The mass of silver required to prepare one mole of acetylene from iodoform is ____g. (Atomic mass of Ag = 108 g.mol –1) 10. 3BF 22 +FCHCHBrProduct
The predominant product is:
Which of these reactions are possible?
(1) (a) and (b) (2) (b) and (d) (3) (a) and (d) (4) (b), (c) and (d)
12. The compound which will have the lowest rate towards nucleophilic aromatic substitution on treatment with OH – is
11. Consider the following reaction:
13. The Friedel-Crafts alkylation….
(1) Works well for primary chlorides
(2) works very well for tertiary chlorides
(3) Works very well for acyl chlorides
(4) Work very well without catalyst
14. Maximum number of products expected (ignore yield) in the monobromination of p -xylene, o -xylene, and m -xylene are x , y and z respectively. Then x + y + z is
THEORY-BASED QUESTIONS
Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II). Mark the correct answer as
(1) if both statement I and statement II are correct.
(2) if both statement I and statement II are incorrect.
(3) if statement I correct but statement II is incorrect.
(4) if statement I incorrect but statement II is correct.
1. S-I : cis-But-2-ene has less dipole moment than trans-but-2-ene.
S-II : R−CC−R’ reacts with H2 in presence of Na/liq.NH3 to form cis-alkene.
2. S-I : Of the two isomeric alkanes shown below, (B) is more stable than (A).
S-II : Hyperconjugation observed only in alkene (B).
3. S-I :
:
4. S-I : Acetylene is acidic in nature.
S-II : Acetylene reacts with alkalies like NaOH (or) KOH to form salts.
5. S-I : Generally deactivating groups are meta directing for EAS reaction.
S-II : Deactivating groups increase electron density at meta position.
6. S-I : Tropolone is an aromatic compound and has 8 π electrons
S-II : π electrons of > C = O group in tropolone is involved in aromaticity
Assertion and Reason Questions
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A).
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A).
(3) if (A) is true but (R) is false.
(4) if Both (A) and (R) are false.
7. (A) : Among isomeric pentanes, 2,2-dimethyl propane has highest boiling point.
(R) : Branching does not affect the boiling point.
8. (A) : Halogenation of methane gives different halo alkanes [mono, di, tri, tetra halo compounds]
(R) : Halogenation of alkanes is electrophilic substitution reaction.
9. (A) : Addition of Br2 to 1-butene gives two optical isomers.
(R) : The product contains one asymmetric carbon.
10. (A) : 2-bromo 3-methyl butane heated with potassium hydroxide in ethanol forms 2- methyl-2-butene as a major product.
(R) : It follows Saytzeff rule.
11. (A) : Isobutene in presence of conc. H2SO4 reacts with water to form tertiary alcohol.
(R) : It follows markovnikov’s rule.
12. (A) : Acetylene on treatment with Tollens’ reagent gives white precipitate and acetylene on treatment with ammonical Cu2Cl2 gives red precipitate
(R) : Acetylene is acidic in nature
13. (A) : The compound cyclooctatetraene has the following structural formula:
It is cyclic with 8π-electron system but it is not an aromatic compound.
(R) : (4n+2) π electron rule does not hold good and ring is not planar
14. (A) : Benzene is more stable than hypothetical cyclohexatriene.
(R) : The delocalized π electron cloud is attracted more strongly by nuclei of carbon atoms.
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1. Which of the following molecules of alkanes will give only one mono halogenated product on reaction with halogen in presence of sunlight?
(1) CH3–CH3 (2) H3C–CH2–CH3
(3) C(CH3)4 (4)
2. The possible monobromo products of above reaction.
(1) Cyclohexyl methyl bromide
(2) 1-Bromo-1-methyl cyclohexane
(3) 1-Bromo-2-methyl cyclohexane
(4) 1-Bromo-3-methyl cyclohexane
3. Which of the following is not correct?
15. (A) : Benzene on reaction with propyl chloride with AlCl 3 give isopropyl benzene as major product.
(R) : In Friedel-Crafts alkylation reaction rearrangement of carbocation will not occur if Lewis acid is AlCl 3
16. (A) : Halogens are deactivating but ortho and para directing for EAS reaction.
(R) : Halogens stabilise σ-complex at ortho and para by resonance.
17. (A) : Activating groups are always ortho and para directing for EAS reaction.
(R) : Activating groups increase electron density at ortho and para by resonance or hyperconjugation.
−−−→ CH3 – CH2 – CH3 – COOH (C) 3 Conc. HNO 323 670 K CHCHCH−−→mixture of nitro alkane (D) 42 3 523 K,100 atm 22 Cu CHO CHOH+→−
(1) A & C (2) A & B
(3) A, B & C (4) A & D
4. Identify the correct statements
(1) Hydrogenation of alkenes in the presence of Ni/200-300°C is syn addition
(2) Hydrogenation of cis 2, 3-diphenyl-2butene gives meso product
(3) Hydrogenation of trans 2, 3-diphenyl2-pentene gives racemic mixture
(4) Hydrogenation of cis 2, 3-diphenyl-2pentene gives meso compound
5. Correct reaction(s) among the following is/ are
9. In which of the following reactions, reactant and products are correctly matched?
Select correct statement(s):
(1) A is cold alkaline KMnO4
(2) B is I. RCO3H II. H3O+
(3) C is obtained when D is heated with conc H2SO4
(4) Reaction C to Z is stereo non-selective
7. Which of the following reactions give meso product? (1)
(4) phBr + (CH3)3CONa → phOC(CH3)3 10. 4 2 propene products CCl Br → + . The above reactions involves
(1) Formation of cyclic bromonium ion
(2) anti addition of Br2 to the double bond
(3) Attack of Br– on cyclic intermediate is fast reaction.
(4) Product is geminal dihalide.
11. Which of the following alkenes will give same alcohol on hydration with either dil. H 2 SO 4 or with OMDM (oxymercuration demercuration)
8. An organic compound of molecular formula C8H12 incapable of showing stereoisomerism gives a single product by the following reaction.
The compound may be (1) CH2 CH2
12. Heat of hydrogenation of CH3 CH3 is less than:
(1)
CH2 CH3 (2) CH3 CH3 (3) CH3 CH3 (4) CH3 CH3
13. In the reaction 2 222 NaCl HO CHCHBr → =+
Product(s) is/are
(1) CH2Br–CH2Br (2) CH2Br–CH2OH (3) CH2Br–CH2Cl (4) CH2OH–CH2OH
14. Identify the correct reaction(s) for which reactant and product are correctly matched?
(1) 4 dil. cold 22 22 alkKMnO CHCH CHOHCHOH =→ (2) 3 2 1 2 2) ) 2 O ZnHO CHCH HCOOH →
16. Some Statements regarding the reaction are given below.
3 Me-CC-EtNa/liq.NHP ≡→ 24Br/CCl →
Select the correct statement(s). (consider major products)
(1) P is a trans-alkene
(2) Q1 is a pure compound and optically inactive due to internal compensation
(3) In the P to Q 1 Conversion step the Br 2 adds on P in a syn manner and the intermediate formed is a cyclic bromenium ion.
(4) Q2 is a binary mixture and is optically inactive due to external compensation
17. 3 3 CHLiMeILi 3NH CHCCHABC −≡−→→→ 24 H/Pd-BaSOD→
C and D are:
(1) Enantiomers
(2) Diastereoisomers
(3) Geometrical isomers
(4) Structural isomers
15. In which cases major product will be a rearranged product?
(1) CH3 CH3 CH3 CH2 224 HO/HSO →
(2) CH3 CH3 CH3 CH2 22 4 Hg(OAc)/HO NaBH/NaOH →
(3) CH3 CH3 CH3 CH2 3 22 BH/THF HO/OH
(4) CH3 CH3 CH3 CH2 HI →
18. Which of the following reaction is/are correct (1) redhot CHCCOOHCu-tube≡−→ COOH HOOC COOH (2) 3() 22 1. , 2. / l NaNH BrHO MeMe−≡−→ meso compound (Major) (3) O CPBA m → (4) 4 NHCl,CuCl CHCH2CHCHCCH≡→=−≡ ↓ HCl Chloroprene
19. The products X and Y obtained in the following reactions ar 1.NBS,hv 2.alc.KOHX→ 1.NBS,hv 2.alc.KOHY→
(1) X= (2) y= Br Br
(3) x = Br (4) y =
20. Which of the following molecules in pure form is/are stable at room temperature?
(1) CH2 CH2 (2) (3) (4)
21. Which of the following reactions give an aromatic product? (1)
24. Find the total number of isomers of C 7H14 (consider only 5-membered ring and including stereo isomers)
25. 2323 600,15 n-Hexane CrOAlO Catm ° → . Find product A The sum of σ and π bonds present in ‘A’ is
26. The number of isomeric tetraenes (including those containing sp-hybridised carbon atoms) that can be formed in the following reaction sequence is ______. CH3 24 2 1.H/PdBaSO 2.Br(excess) 3.Alc.KOH,Ä →
27. An acyclic compound(A) contains six carbon atoms and the required number of hydrogen atoms. It can absorb one molecules of H2 in presence of metal catalyst. It forms one monochloro substituted product when reacts with Cl 2 in presence of light. What is the mass of all CH 3 groups present in compound(A) in g mol–1?
28. Consider the following reactions
3Anhyd.AlCl 3 +CHCl(T) →
3Anhyd.AlCl 22 +CHCl(U) →
3Anhyd.AlCl 3 +CHCl(V) →
Numerical Value Questions
22. Total number of monochloro derivatives will be
23. The maximum number of carbon atoms in the expected products of the following reaction. 3 Na/Ether 323 2 Product CH I CHCHBrCHCHCHBr+−−→
3Anhyd.AlCl +CCl4(W) →
The sum of the number of sp2 hybridised carbon atoms and chlorine atoms across all the chief organic products [T], [U], [V] and [W] is
29. 10 mL of saturated hydrocarbon is completely burnt in 95 mL of oxygen. All the volumes are measured under similar conditions, how many carbon atoms are present in one molecule of saturated hydrocarbon
30. Molecular mass of the product (in g/mol) obtained in the given reaction is:
(At. Wt. H=1; C=12; O=16; Al=27; S=32; Cl=35.5; Fe=56)
31. A saturated polyhalogen compound (A) on heating with zinc gives 2-Butyne. What should be the minimum number of halogens in one molecule of the reactant (A) to give the product?
Integer Value Questions
32. How many of following undergo Diels–Alder reaction
33. All the alkenes excluding terminal alkenes, including stereoisomers of the formula C5H10 are subjected to reduction by D2/Pd. The total number of isomeric products including stereoisomers formed is
34. If the number of hydrogen atoms in the final product of the following reaction sequence is x, the value of 2 x is (round off to nearest integer)
35. A hydrocarbon A(C 6 H 10 ) on reductive ozonolysis does not fragment into 2 separate units but forms a dialdehyde (B) which on exhaustive and normal Clemmensen reduction forms an alkane (C). The number of possible structures for A is ________. (structural)
36. In the alkane 2, 4, 4-trimethylhexane, i) Number of 1o-hydrogens are a ii) Number of 2o-hydrogens are b iii) Number of 3o-hydrogens are c
iv) Number of monochloro structural isomers are d
v) Maximum number of 2o carbons on a plane are e
vi) Number of alkynes possible to prepare given alkane from hydrogenation are f
Then | a b+c d+e+f | is ______.
37. The number of −CH2−(methylene) groups in the final product formed from the following reactions sequence is __________
C H3 CH3 CH3 C H3 2 32 24 1.Br/hv(mono) 2.alc.KOH/Ä 3.O/Zn/HO 4.Conc.HSOMajorProduct→
38. How many of the following alkynes on treatment with dilute H2SO4 in presence of HgSO4 give a methyl ketone
HC CH, CH3C CH, CH3CH2C CH, CH 3C CCH 3, C 6H 5C CH, C 6H 5C CCH3, CHCH2C CCH2CH3, CH3CH2CH2C CH
39. Number of sp2 carbon atoms in Y from the following reaction sequence is _______
Note: X is the one of the products with highest molecular mass
40. Isobutane reacts with chlorine in sunlight to produce 1-Chloro-2-methyl propane 68% and 2-Chloro-2-methyl propane 32% of the total yield. Under similar conditions propane reacts with chlorine in sunlight to produce 1-Chloropropane 45% and 2-Chloropropane 55% of total yield. The percentage yield of the major product obtained by monochlorination of 1,3,5 -trimethyl cyclohexane is x × 10%. The value of x will be
Passage-based Questions
Q(41-42)
Halogenation of alkanes is free radical substitution. In the chlorination, the ratio of relative reactivity of 1, 2 and 3-alkyl C-H is 1: 3.8 : 5. But in bromination this relative reactivity is 1: 82 :1600, Since the cleavage of C-H bond is slow step, isotopic effect observed in halogenations of alkanes.
41. In the mono chlorination(Cl 2 /hυ)of isopentane, mole percentage of the major product formed is
42. Chlorination (Cl 2/hυ) of C2H5D leads to a mixture containing 93% C2H4DCl and 7% C2H5Cl. Based on these mole percentages, relative reactivity of C-H when compared with C-D in chlorination is__.
Q(43-44)
Mono chlorination of compound[P] gives the mixture of products as shown below (consider starting material is com pletely consumed).
45. The number of α-hydrogens in the carbocation intermediate just before formation of ‘B’ is____.
46. The number of stereogenic carbons present in the product obtained by reduction of A with Pd/H2 is_.
Q(47-48)
In the following reaction sequence, the % yield corresponding to the product in each step is given in the parenthesis. (atomic wt. : H = 1, C = 12, N = 14, O = 16 & Br = 80 g.mol–1)
43. Number of fractions (consider only organic products) which come out in fraction distillation of the product mixture is (1) 2 (2) 3 (3) 4 (4) 5
44. Number of optically inactive compounds formed in the above reaction is?
(1) 0 (2) 1 (3) 2 (4) 3
Q(45-46)
Dehydration of 2,2,3,4,4-Pentamethyl-3Pentanol gave two alkenes (A) and (B). Ozonolysis of the lower boiling alkene (A) gave formaldehyde and 2,2,4,4- Tetramethyl–3–Pentanone. Ozonolysis of B gave formaldehyde and 3,3,4,4-Tetramethyl-2-Pentanone.
47. The amount of D (in grams) formed from 10 mole of benzene is ____.
48. The amount of E (in grams) formed from 10 mole of benzene _____.
Q(49-50)
Oxidation of alkenes by cleavage with acidic or alkaline KMnO4 or acidic K2Cr2O7 at higher temperature yields products depending upon the nature of alkene. A hot solution of is a strong oxidising agent which gives only ketones and carboxylic acids and not aldehydes
Oxidation of alkenes with OsO 4 followed by hydrolysis yields glycols
49. A n organic compound (X) on oxidation with hot KMnO4 gives only butanoic acid. Number of carbon atoms and hydrogen atoms present in one molecule of (X) is ( y) and (z) respectively. Find the value of z–y.
50. Compound(A) on treatment with OsO 4 following by Na2SO3 will give cis-2-methyl butane-2,3-diol.One mole of (A) on complete combustion gives x mole of CO2 and y mole water vapour. Find the sum of x and y
Q(51-52)
51. The maximum number of atoms present in the same plane in compound ‘G’ are
52. The amount of ‘E’ obtained when the reaction is carried out with 117 g of benzene is _____.
(Assume 100% yield in all the steps and ‘C’ is distributed equally for the paths ‘F’ and ‘D’).
Q(53-54)
Consider following reaction
[At wt.of C = 12, H = 1, D = 2.01 and Br = 79.90]
53. Molecular weight of C is ____.
54. Molecular weight of D is ____.
Q(55-56)
55. Number of gaseous hydrocarbons formed in above reaction sequence is ______.
56. If total number of hybridised orbitals in J is x and I is y then value of y x is ______.
Matrix Matching Questions
57. Match the column-I (reaction) with columnII(catalysit).
Column I Column II
(A) Kolbe’s electrolytic method (p) Anhy. AlCl3/ HCl
(B) Wurtz reaction (q) Cr2O3 or V2O3
(C) Isomerisation (r) R-COONa
(D) Aromatisation (s) R-X
(A) (B) (C) (D)
(1) r s p q
(2) p r s q
(3) p s q r
(4) r s q p
58. Match the following?
Reaction Nature of Reaction
(A) dry ether Na RBr → (p) Partial reduction (B) , NaOH CaO RCOONa−+ ∆ → (q) Electrophilic addition
(C) ( ) 2 Lindlars Catalyst H ′ ′ −≡−→RCCR (r) Wurtz reaction
(D) 24 / 22 BrCCl HCCH=→(s) Decarboxylation
(t) Polymerisation
(A) (B) (C) (D) (1) r t p q
(2) r s p q
(3) q r p q (4) t p r q
59. Match the column-I (reactant to get benzene) with column-II(catalyst).
Column I
Column II
(A) Sodium benzoate (p) Red hot iron tube
(B) Phenol (q) dil. HCl
(C) Acetylene (r) anhydrous AlCl3
(D) Benzene sulphonic acid (s) Zinc dust, ∆
(t) Soda lime, ∆
(A) (B) (C) (D)
(1) r s p t
(2) q t p r
(3) t s p q
(4) q s p r
60. Match the Column-I(aromaticity) with Column-II(condition).
Column I Column II
(A) Aromatic (p) Planar
(B) Anti-aromatic (q) Non-planar
(C) Huckel rule (r) 4nπ- delocalised electrons
(D) Cyclo octa tetraene (s) (4n+2)π delocalised electrons
(A) (B) (C) (D)
(1) p,s p,r s qr
(2) p,q p,s s q,r
(3) s p,r p,s q,r
(4) p,q p s q
61. Match the column-I (structure) with column-II (property).
Column I
Column II
(A) Θ Θ (p) Anti-aromatic
(B) (q) Non-aromatic
(C) (r) Aromatic
(D) 12-Annulene (s) Non-planar
(A) (B) (C) (D)
(1) r p r qs
(2) p r r q
(3) q s r p
(4) p q s r
62. Match the column-I (reaction) with columnII(products).
Column I Column II
(A) CH3 –CH = CH2 22 / ClHO → (p) Acetic acid and butanoic acid
(B) CH3 CH3 CH3 HBr Peroxide → (q) Product contains two chiral carbons
(C) 2-Hexene 3 22 () () iO iiHO → (r) Acetic acid and propanoic acid
(D) 2-pentene 4 24 , KMnO HSO ∆ → (s) Reaction proceeds through intermediate halonium ion
(A) (B) (C) (D)
(1) s q r p
(2) s q p r
(3) q s p r
(4) q p s r
63. Match the column-I (conversion) with column-II (reagent used).
Column-I (conversion)
Column-II (Reagent used)
(A) CH3CH = CHCH3 ? → 2CH3COOH (p) KMnO 4 /OH –; 273K
(B) (CH3)2C = CH2 ? → (CH3)2C = O + CO2 + H2O (q) O3; Zn; H2O
(C) CH3CH = CH2 ? → CH3CH(OH) CH2OH (r) KMnO4/H+/∆
(D) RC CR' Cold KMnO → cis- alkene (s) Na/Liq NH3 (t) pd/C
(A) (B) (C) (D)
(1) p r q s
(2) r q p t
(3) r r p s
(4) r t p t
64. Match the column-I (reaction) with columnII (steriochemistry of product).
Column-I
(A) C
Column-II
(p) Racemic mixture
(B)
(q) erythro product
(C)
(r) threo product
(D) CH3 H H CH3 24 / ClCCl → (s) meso product
(A) (B) (C) (D)
(1) p,r q,s r,s q,r
(2) q,s p,r q,s q,s
(3) p,s q,s q,s p,s (4) q,s q,s r,s p,q
65. Match the Column-I (reaction) with Column-II (steriochemistry of product).
Column I Column II
(A) cis -But-2-ene 2 4 B CC → (p) Four enantiomers are formed.
(B) CF3 – CH = CH2 HCl → (q) The major products are according to antiMarkovnikov’s path.
(C) C6H5 – CH = CH – CH3 + BrCl → (r) Two optical isomers are formed.
(D) CH3 – CH = CH – CH3 + HBr → (s) A pair of enantiomers in equal amounts.
(A) (B) (C) (D)
(1) s q p r
(2) p q r s
(3) s p q r
(4) s q r p
66. Math the column-I (type of reaction) with column-II (catalyst used).
Column I Column II
(A) Dehydrohalogenation (p) Alc.KOH
(B) Dehydration (q) Conc. H2SO4/170°C
(C) Unsaturation (r) Ag
(D) Epoxidation (s) Br2 water
BRAIN TEASERS
1. There are two paths to prepare compound (A) (2-methylhex-1-en-3-yne). which of the following statements is correct?
(1) Path I is feasible.
(2) Path II is feasible.
(3) Both paths are feasible.
(4) Both paths are not feasible.
2. Consider the major products of following reactions:
The sum of sp2 carbon atoms and the number of π electrons not involved in resonance in P, Q, R and S is
(A) (B) (C) (D)
(1) q s r p
(2) p q s r
(3) s p r q
(4) r p s q
Q(3-4)
Consider the following scheme of reactions CH2 CH2 33 twoeqv.HBrMgtwoeqv MeCOOCMeÄPC
m is total number of products including stereoisomers
3. The value of m is ------------------------
4. 0.48g of liquid ‘T’ when completely combusted in a bomb calorimeter, the rise in temperature of calorimeter system is found to be 0.6 oC. Thermal heat capacity of calorimeter including its contents is 8K.Cal/C. The standard molar enthalpy of combustion of liquid ‘T’ at 27°C is − x kcal. The value of x is ___.(Take molar gas constant, R = 2.0 calK–1mol–1)
5. How many of the statements are correct about the following reaction sequence.
3-Methyl-1-butene 2 22 4 3 22 HO/H Hg(OAc)/HO NaBH/OH BH/THF HO/OH (X)major (Y)major (Z)major + → → →
I) ‘Y’ is optically inactive due to external compensation II) X & Z are optically inactive
III) X & Z are optically active IV) X is optically active but Z is optically inactive
6. Identity the magnitude of x-y
7. Given below are two statements. One is labelled as Assertion (A) and the other is labelled as reason (R).
Assertion (A) : Corey-House reaction can be used to prepare both symmetrical and asymmetrical alkane.
Reason (R) : Lithium metal is involved in it.
In light of the given statements, choose the correct answer from the options given.
(1) Both (A) and (R) are true and (R) is the correct explanation of (A)
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(3) (A) is true but (R) is false
(4) Both (A) and (R) are false
8. Identify number of reactions that can be give benzene as major product.
Q(9-10)
An alkane having molecular weight 72 have 3 isomers x,y & z and their boiling point order is x > y > z then answer following questions
9. The number of isomeric monochloro derivatives are possible for y are ____
10. The total Number of Primary C–H bonds in x, y and z are ____.
Q(11-12)
When substituted benzenes undergo electrophilic attack, group already on the ring affect both, the rate of the reaction and the site of attack. We say, therefore, that substituent groups affect both reactivity and orientation in electrophilic aromatic substitutions. Find the reactions of Benzene or benzene derivatives and answer the following
11. In the reaction sequence given below is carried out with one mole of reactant (P), the amount of product (S) formed (in g) is
The Number of valence electrons in element [A] are
13. Consider the given reaction and identify incorrect statement from the given following. 3 / / 33 NaNH PdC BCHCCCHA ∆ ←−≡−→ A) A is more soluble than B. B) The boiling point of A is higher than
that of B and melting point of A is lower than that of B
C) A is more polar than B because dipole moment of A is zero.
D) catalytic hydrogenation of A and B gives different hydrocarbons
Identify the incorrect statements from the options given below (1) A, and B only (2) A, B, and D only (3) B, C, and D only (4) C, and D only
Q(14-15)
A stereoselective reaction forms one stereoisomer preferentially over another stereoisomer. In this reaction for reaction stereoisomerism may or may not possible. A reaction is stereospecific if the reactant can exist as stereoisomers and each stereoisomers of reactant forms a different stereoisomer or a different set of stereoisomers of the product. All stereospecific reaction are stereoselective but a stereoselective reaction may or may not stereospecific.
LIST-I (Reaction)
(A) CH3 CH3
(B) CH3 CH3 2 /,DNi
(C) CH3 CH2 HBr
(D) CH3 CH3
LIST-II (Nature of reaction)
LIST-III (major product)
(i) Stereospecific (p) Erythro isomer
(ii) Stereoselective (q) Threo isomer
(iii) Regioselective (r) Meso
(iv) Only syn addition (s) Racemic mixture
14. Incorrect match among the following is (1) I-B-P (2) I-A-R (3) II-A-S (4) II-C-Q
15. Incorrect match among the following is (1) III-C-S (2) IV-C-S (3) IV-D-P (4) IV-A-Q
16. Match the Column-I with Column-II.
Column I
Column II
(A) (p) Hg(OAc)2/ H2O followed by NaBH4
(B) (q) H2O/H2SO4
(C) (r) BH3/THF followed by H2O2/ NaOH
(D) (s) Carbocation intermediate (t) Four centered cyclic transition state
(A) (B) (C) (D) (1) q,s p r,t q,r (2) s,t p,r r,s p,t
(3) p,s q,r r,t p,q (4) p,r r,t q,r q,t
17. The number of possible isomeric products formed when 3-chloro-1-butene reacts with HCl through carbocation formation is ________.
FLASHBACK ( Previous JEE Questions )
JEE Main
1. Which of the following reaction is correct? (2024)
(1) 2 2 HNO,0C 3222 322 HO CHCHCHNH CHCHOHNHCl + + →
(2)
(3)
(4) C 2 H 5 CONH 2 + Br 2 + NaOH → C2H5CH2NH2 + Na2CO3 + NaBr + H2O
2. The correct order of reactivity in electrophilic substitution reaction of the following compounds is (2024)
(1) B > C > A > D
(2) D > C > B > A
(3) A > B > C > D
(4) B > A > C > D
3. Compound A formed in the following reaction reacts with B gives the product C. Find out A and B. (2024)
(1) 3 322 ACHCC,BCHCHCHBr + =−≡=−−−
(2) 32322 ACHCHCH,BCHCHCHBr =−==−−−
(3) 3233 ACHCHCH,BCHCCH =−−=−≡
(4) 3 323 ACHCCNa,BCHCHCH + =−≡=−−
4. IUPAC name of following hydrocarbon (X) is:
(1) 3,4,7 –Trimethyloctane
(2) 2-Ethyl-2,6-diethylheptane
(3) 2,5,6-Trimethyloctane
(4) 2-Ethyl-3,6-dimethylheptane
5. Molar mass of the hydrocarbon (X) which on ozonolysis consumes one mole of O 3 per mole of (X) and gives one mole each of ethanol and propanone is___ g mol–1 (molar mass of C: 12 g mol–1, H : 1gmol–1) (2023)
6. The major product ‘A’ of the following given reaction has _____ sp2 hybridised carbon atoms. 2,7-Dimethyl-2,6-octadiene ( ) H+ AMajor → (2022)
7. In the presence of sunlight, benzene reacts with Cl2 to give product, X. The number of hydrogens in product X is ___. (2022)
8. In bromination of propyne, with bromine 1, 1,2,2-Terabromopropane is obtained in 27% yield. The amount of 1,1,2,2-Tetrabromopropane obtained from 1 g of bromine in this reaction is ________________ × 10−1 g. (nearest integer) (Molar mass : Br = 80 g/mo l) (2022)
9. CH3 OH
The product ‘A’ and ‘B’ formed in above reactions are (2021)
(1) A= B=
(2) A= B=
(3) A= B=
(4) A= B=
JEE Advanced
12. The number of –CH2– (methylene) groups in the product formed from the following reaction sequence is ________. (2022)
In the above reaction, 3.9 g of benzene on nitration gives 4.92 g of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is _____%. (round off to the nearest Integer)
(Given atomic mass: C:12.0 u, H:1.0u, O:16.0 u, N:14.0 u) (2021)
11. Which of these will produce the highest yield in Friedel–Crafts reaction? (2020)
13. The number of isomeric tetraenes (not containing sp-hybridised carbon atoms) that can be formed from the following reaction sequence is ________. (2022)
24 2 1.H/Pd-BaSO 2.Brexcess 3alc.KOH/Ä
CHAPTER TEST – JEE MAIN
Section A
1. the given reactions sequence, the major product ‘C’ is:
22 423 Br (i) NaNH (i) alc. KOH CCl (ii) NaNH (ii) CH-Cl (A) (B)(C),
Product (C) is:
(1) Ph−C CNa
(2) Ph−CH2–C Na
(3) Ph−C C – CH3
(4) Ph−CH = C = CH2
4. Benzene on nitration gives nitrobenzene in presence of HNO3 and H2SO 4 mixture, where:
(1) Both H2SO4 and HNO3 act as a bases
(2) Both H2SO4 and HNO3 act as a acids
(3) HNO3 acts as a base and H2SO4 acts as an acid
(4) HNO3 acts as a acid and H2SO4 acts as an base
5. Correct statements for the given reaction are:
2eqv.B→
A. Compound ‘B’ is aromatic.
B. The completion of above reaction is very slow.
C. ‘A’ shows tautomerism.
D. The bond lengths of C–C in compound B are found to be same.
Choose the correct answer from the options
given below:
(1) A,B and D only
(2) A, B and C only
(3) A, C and D only
(4) B, C and D only
6. Wrongly matched structure and it’s property is
(1) N+ H H is anti-aromatic but not nonaromatic
(2) is non-aromatic but not antiaromatic
(3) rotation about the double bond connecting the rings is most feasible.
(4) 10 annulene
(10) - annulene is not aromatic
7. A compound ‘A’ on reaction with ‘X’ and ‘Y’ produces the same major products but different by products ‘a’ and ‘b’. Oxidation of ‘a’ gives a substance produced by ants.
‘X’ and ‘Y’, respectively are
(1) KMnO4/H+ and dil.KMnO4, 273 K
(2) dil.KMnO4, 273K and dil.KMnO4/H+
(3) KMnO4/H+ and dil. O3, H2O/Zn
(4) O3, H2O/Zn and KMnO4/H+
8. The product formed in the following multistep reaction is
22 3 (ii) HO,NaOH 32 (iii) PCC (iv) CHMgBr CH-CH=CH →
(1) CH3CH2CH(OH)CH3
(2) CH3CH2COOCH3
(3) CH3)3COH
(4) CH3–CH2–CH2–CH2–OH
9. In the following reaction ‘X’ is ( ) Anhy.3 AlCl 323HCl 4 CHCHCHX →
(1) (CH3)2CH(CH2)2CH3
(2) Cl–CH2–(CH2)4CH2−Cl
(3) Cyclohexane
(4) CH3(CH2)4CH2Cl
10. Match the structures in Column-I with their corresponding property with Column-II.
Column-I Column-II
(A)
(p) Aromatic (B)
(q) Anti-aromatic
(C) (r) Non-aromatic
(D) (s) (4n+2)πe
(A) (B) (C) (D)
(1) ps ps r q
(2) qs pr q r
(3) p qr r q
(4) ps r c q
11. Choose the correct set of reagents for the following conversion. trans-Ph − CH = CH−
CH3 → cis-Ph − CH = CH−CH3
(1) Br 2 , alc. KOH, NaNH 2 , H 2 Lindlar Catalyst
(2) Br2, alc. KOH, NaNH2, Na(liq.NH3)
(3) Br2, aq.KOH, NaNH2, Na(liq.NH3)
(4) Br2, aq.KOH, NaNH2, H2 Lindlar Catalyst
12. Consider the following structures.
(2) A and B only
(3) B, C & D only
(4) B and C only
I II III
Choose the correct statement regarding the above structures
(1) Dipole moment varies as II > III > I
(2) II is more stable than I
(3) I is the most reactive among three
(4) All of the above
13.
What is stereochemistry of product?
(1) Racemic mixture
(2) Optically inactive
(3) Diastereomers
(4) Meso product
14. 3-methyl-2-pentene on reaction with HBr in presence of peroxide forms an additional product the number of possible stereoisomers for the product
(1) Two (2) Four
(3) Six (4) Zero
15. But-2-yne is reacted separately with one mole of hydrogen as shown below:
A) A is more soluble than B.
B) The boiling point & melting point of A are higher and lower than B respectively.
C) A is more polar than B because dipole moment of A is Zero.
D) adds easily to B than A.
Identify the incorrect statement from the option given below:
(1) A, C & D only
A and B are related as
(1) A and B are mixture of diastereomers
(2) A and B are mixture of enantiomers
(3) A and B are optically active
(4) B is optically active and A is optically inactive
17. The major product in the following reaction sequence is
18. Decreasing order of reactivity towards electrophile substitution for the following compounds is:
(3) e > d > a > b > c
(4) c > b > a > d > e
19. The decreasing order of acidic characters of the following is:
(I) HC CH (II)
(III)
(1) I > II > III (2) II > I > III
(3) III > II > I (4) I > III > II
20. A hydrocarbon ‘X’ with formula C5 H 8 uses two mole of H2 on catalytic hydrogenation of its one mole. On ozonolysis, ‘X’ yields two mole of methane dicarbaldehyde. The hydrocarbon X is:
(1) Benzene
(2) 1-methylcyclopenta-1,4-diene
(3) cyclohex-1,3-diene
(4) cyclohex-1,4-diene
The value of (number of isomeric alkenes for P + number of isomeric dihalides for Q) is
CHAPTER TEST – JEE ADVANCED 2020 P1 Model Section-A [Single Option Correct MCQs]
1.
The number of carbon atoms in the product ‘Z’
2. An environmentally friendly chemical process would normally be expected to have a high % atom economy, indicating that a high proportion of the starting materials end up as part of the final product, hence
22. The number of non-bonding electrons present in the major product obtained in the following conversion is/are
23. In the following series of reactions the maximum number of atoms present in molecule ‘C’ in one plane is _______.
(A is a lowest molecular weight alkyne)
24. How many of the following on hydrolysis, give methane.
(i) Al4C3 (ii) Mg2C3 (iii) Be2C (iv) BeC2 (v) CaC2
25. red
the total number of carbons with sp 2 hybridisation in ‘A’ is _____.
reducing the amount of waste. Efforts are constantly being made to increase the % atom economy of chemical processes. As an example, the manufacture of ethene oxide (C2H4O) for many years was via the classical chlorohydrin route.
C2H4 + Cl2 + H2O → ClCH2CH2OH + HCl, ClCH2CH2OH + Ca(OH)2 + HCl → C2H4O + CaCl2 + 2H2O.
Calculate the % atom economy for overall reaction. (At wt : Ca = 40, Cl = 35.5, O = 16, H = 1, C = 12)
3. The relative reactivity of 1oH, 2oH, & 3oH in bromination reaction if found to be 1 : 82 : 1600, respectively.
What is the % yield of A?
4.
The number of double bond(s) present in the hydrocarbon is_____.
8. Consider the following sequence of reactions. 3 (aq) CFI+KOH(X)+(Ysalt) →
33 CHCOCHCa(OCl)Cl() other Products ∆ + →+ Z
33 CHCOCHCa(OCl)Cl() other Products ∆ + →+ Z (B) (Z)+Agpowder(A)+white→↓
The given reaction proceeds via formation of ___ membered cyclic intermediate compound. 5.
c
What is the numerical value of N × M?
6. If the reaction sequence given below is carried out with 5 mole of n -heptane, the amount (in g) of the product [G] formed is
(A) is also obtained by hydrolysis of calcium carbide (CaC2).
If p = maximum number of atoms in a plane in a molecule of (X)
q = maximum number of identical bond angles involving three different atoms in one molecule of (Z).
r = bond order between identical atoms in a molecule of (A), find the value of 2 pqr ++
Section-B
[Multiple Option Correct MCQs]
9. Which of the following alkynes on complete reduction yields optically inactive compound
(1) 1-Butyne
(2) 3-Methyl pent-1-yne
The yields of [A] to [G] are given in parentheses.Compound [B]has the highest dipolemoment among the three compounds that [A] can produce. [Given: Atomic mass of H = 1, C = 12, N = 14, O = 16, Br = 80]
7. 17 mg of a hydrocarbon (M.F. C10H16) takes up 8.40 mL of the H2 gas measured at 0°C and 760 mm of Hg. Ozonolysis of the same hydrocarbon yields.
(3) 3-Methyl pent-1,4-diyne
(4) 4-Methyl-1-hexyne
10. Which of the following reactions does occur as shown
(1) CH3C C– Na+ + CH3CO2H → CH3CO2–Na+ + CH3C CH
(2) CH3C C– Na+ + CH3OH → CH3O–Na+ + CH3C CH
(3) CH3C CH + NaNH2 → NH3 + CH3C C–Na+
(4) CH3C CH + NaOH → H2O + CH3C C–Na+
11. Electrolysis 32 (aq) (MajorProduct) (CH)CHCOOK X →
The incorrect statements about X are
(1) X is chain isomer of pentane
(2) X has higher boiling point than n-hexane
(3) X will give three mono halogenated isomers
(4) The IUPAC name of X is 2,2-dimethylbutane
12. Acetone is the major product in:
(I)
(2) CH3CH2CHBr2 2 (i) alc. KOH (ii) NaNH →
(3) R-X RCCNa−≡→
(4) Potassium maleate electrolysis →
Section-C [Matrix Matching Questions]
15. Match the Column-I with Column II.
Column I
(A) CH2
Column II
+ 3 HO
22 HC=C=CH →
(II) +2 Hg/HSO24
(III)
3 HCCCH−≡→
3 HO / /OHHCCCH−≡→
3 22 BHTHF
(IV) +2 Hg/HSO24
32 HCCHCHCH−−≡→
(1) I
(2) II
(3) III
(4) IV
13. In which of the following reactions optically inactive compound(s) is/are formed?
(1) cis-But-2-ene42 KMnO,OH;HO cold →
(2) 3,3-Dimethylbut-1-ene ( ) 32 24 1) CHCOOHg,HO 2) NaBH,OH →
(3) Prop-1-ene
3 22 1) BH/THF 2) HO,NaOH →
(4) Prop-1-ene
3 3 1) BH/THF 2) CHCOOH →
14. Which of the following reactions will give alkyne?
(1) 2 3 + NaNH 3 O 3 H || CHCHCHCH−−−→ Cl Cl
24dil.HSO → (p) Over all reaction involves Markonikov’s addition of water on alkene
(B) CH2
26 22 BH/THF HO/OH → (q) Over all reaction involves AntiMarkonikov’s addition of water molecule on alkene
(C) CH2 OMDM → (r) Reaction involves carbocation rearrangement.
(D) CH2 1.HBr/Peroxide 2.aq.KOH → (s) CH3 OH (t) OH
(A) (B) (C) (D)
(1) pr qt ps qt
(2) pr qt ps s
(3) pr qr p t
(4) ps qt s t
16. Match the Colu mn-I with Column-II
Column I Column II
(A) CH3 – C C –CH3 Pd-BaSO4 Quinoline → (p) Saytzev's elimination
(B) R – C C – R Na/Liq.HNH3 → (q) Hoffman elimination
(C) 32 3 Br CH-HC-HC-CH |2525 CHOH/CHO → (r) cis-alkene
(D) ( ) + Ä325 3 CHN-CHBr → (s) trans-alkene
(A) (B) (C) (D)
(1) r s p q
(2) p q r s
(3) s r q p
(4) r s p q
17. Match the Column-I and Column-II.
Column I Column II
(A) CH3 CH3 OH 24 Conc. HSO Ä → (p) Hofmann`s product
(B) CH3 CH3 Cl C/C2525 Ä O HHOH → (q) Saytzeff`s product
(C) CH3 CH3 CH3 CH3 Cl
(r) Transition state
(D) CH2 CH3 CH3 Br Alc. KOH → (s) Carbocation
(A) (B) (C) (D)
(1) q,r,s q,r p,r q,r (2) q,s q,r p,r p,r
(3) q,r,s q,r p,r p,r (4) p,r q p,r p,r
18. Match the Column-I with Column-II:
Column I Column II
(A) CH3 CH3 4 3 1) 2) OsO NaHCO → (p) Optically inactive due to internal compensation
(B) CH3 CH3 3 1) 2) mCPBA HO → (q) Optically inactive due to external compensation
(C) CH3 CH3 2D Ni → (r) Product have presence of two chiral center
(D) CH3 CH3 2 2 Br HO →
(s) Diastereomers will be formed
(A) (B) (C) (D)
(1) q,r p,r q,r q,r
(2) pq s s pqrt
(3) pq s s qt
(4) pq s s t
ANSWER
KEY JEE Main Level
- I
- II
Level - III
(11) 2 (12) 2 (13) 2 (14) 6
Theory-based Questions (1) 2 (2) 2 (3) 1 (4) 3 (5) 3 (6) 2 (7) 2 (8) 3 (9) 1 (10) 1 (11) 1 (12) 1 (13) 1 (14) 1 (15) 3 (16) 3 (17) 1
JEE Advanced Level (1) 1,,3,4 (2) 1,2,3,4 (3) 2 (4) 1,2,3 (5) 1,4 (6) 1,2 (7) 1,2 (8) 3,4 (9) 1,2,3 (10) 1,2,3 (11) 1,3 (12) 1,2,3,4 (13) 1,2,3 (14) 1,3,4 (15) 1,4
(48)
(49) 8 (50) 10 (51) 12 (52) 96 (53) 59 (54) 137 (55) 3 (56) 1 (57) 1 (58) 2 (59) 3 (60) 1 (61) 1 (62) 2 (63) 4 (64) 2 (65) 1 (66) 2
| CHAPTER 10: Hydrocarbons
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(11) 1 (12) 0 (13) 2
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4 (2) 23 (3) 99 (4) 4 (5) 20 (6)
(7) 3 (8) 4 (9) 1,2,3 (10) 1,2,3 (11) 1,2,4 (12) 1,2 (13) 1,2,3,4 (14) 1,2,3,4 (15) 1 (16) 1 (17) 2 (18) 1
