AMINES CHAPTER 11
11.1 STRUCTURE OF AMINES
In all the amines, nitrogen atom assumes sp 3 hybridised state. Three of the hybrid orbitals of nitrogen are used in the formation of three sigma bonds with hydrogen atoms or alkyl groups, whereas the fourth hybrid orbital contains a lone pair of electrons. Thus, all the three amines acquire pyramidal shape like ammonia, as shown in Fig.11.1.
Fig.11.1 Pyramidal structure of amines
The bond angle around nitrogen atom, however, is close to tetrahedral angle 109°28'. The actual value of bond angle depends on the number and also the size of alkyl groups directly bonded to nitrogen atom. The bond angle in trimethyl amine is 108°.
TEST YOURSELF
1. The hybridisation of nitrogen atom in amines is (1) sp (2) sp3 (3) sp2 (4) dsp2
2. The general formula of amines is (1) C nH2n+1 N (2) C nH2n+2 N (3) C n H 2n+3 N (4) C nH2n N
3. C-N-C bond angle in trimethyl amine is (1) 111.7° (2) 109.5° (3) 108° (4) 117°
Answer key
(1) 2 (2) 3 (3) 3
11.2 CLASSIFICATION OF AMINES
Amines may be regarded as the alkyl or aryl derivatives of ammonia, formed by the replacement of one or more hydrogen atoms by corresponding number of alkyl or aryl or both groups. Amines are of three types.
A primary amine has only one alkyl group directly attached to nitrogen atom, a secondary amine has two alkyl groups, and a tertiary amine has three alkyl groups directly attached to nitrogen atom. Thus, characteristic functional groups for primary, secondary, and tertiary amines are:
Primary (1°)
Secondary (2°)
Tertiary (3°)
If the two alkyl groups in secondary amine and the three alkyl groups, in tertiary amine are same, they are known as simple amines, and if the groups are different, they are mixed amines.
Simple amines can be represented as:
Mixed amines can be represented as:
In addition to the above amines, tetra-alkyl derivatives, similar to ammonium salts, also exist, which are called quaternary ammonium salts. They are regarded as derivatives of ammonium salts, in which all the four hydrogen atoms are replaced by alkyl or aryl groups.
Ammonium halide Tetraalkyl ammonium halide
Example: [(CH3)4N]+Cl– [(C 6 H 5 ) 4 N] + I –Tetramethyl Tetraphenyl ammonium ammonium chloride iodide 25 +
N–ethyl–N–methyl–N–propyl ammonium bromide
Depending on the nature of group attached to nitrogen of amino group, amines are classified as aliphatic, aromatic, or arylalkyl amines.
Aliphatic amines can be represented as: CH3NH2, (CH3)2NH etc
Aromatic amines can be represented as:
C6H5NH2,(C6H5)2NH, (C6H5)3N, etc.
Arylalkyl amines can be represented as: C6H5CH2NH2, C6H5CH2NHCH2C6H5, etc.
TEST YOURSELF
1. Which of the following is a mixed 2° amine?
(1) Toludine
(2) N-methylaniline
(3) Dimethylamine
(4) Diethyl amine
2. Which of the following is a secondary amine?
(1) Diethylamine
(2) Aniline
(3) Isobutylamine
(4) Secondary butylamine
3. How many primary amines are drawn for the formula C4H11N?
(1) 1 (2) 2
(3) 3 (4) 5
4. Tertiary butylamine is (1) primary (2) secondary (3) tertiary (4) quaternary
Answer key
(1) 2 (2) 1 (3) 4 (4) 1
11.3 NOMENCLATURE OF AMINES
Amines are called alkylamines in common system. In IUPAC system, they are called alkanamines. The last letter ‘e’ in the name of the alkane is omitted.
Table 11.1 Common and IUPAC names of some amines
1. CH3NH2
2. CH3CH2NH2
3. CH3CH2CH2NH2 n–oropyl amine
4. 33 2 | CHCHCH NH Isopropyl amine
5. (CH3)3C–NH2 tert–butyl amine
If more than one –NH 2 group is present, suitable prefixes, like di, tri, etc., are used to indicate the number of such groups. Their positions in the chain are indicated and the letter ‘e’ in the name of parent alkane is retained. Secondary and tertiary amines are named N–substituted derivatives of primary amines, like N–alkylalkanamine and N,N–dialkylalkanamine, respectively.
Aryl amine, aniline, is named benzenamine. Toluidine is named aminotoluene. Common and IUPAC names of some amines are listed in Table 11.1.
11.3.1 Isomerism
Amines exhibit the following types of isomerism.
Chain isomerism: Aliphatic amines containing four or more carbon atoms exhibit chain iso merism. e.g., CH 3 CH 2 CH 2 CH 2 NH 2 and 322 3 CHCHCHNH | CH
Position isomerism: Alkyl amines with three or more carbon atoms exhibit position isomerism. e.g., CH 3 CH 2 CH 2 CH 2 NH 2 and 323 2 CHCHCHCH | NH
Metamerism: Secondary amines may exhibit metamerism, e.g., C2H5–NH–C2H5 and CH3–NH–C3H7
6. 323 2 CHCHCHCH | NH sec–butyl amine Butan–2–amine
7. CH2=CH–CH2NH2 Allylamine Prop-2-en-1-amine
8. H2N–(CH2)4–NH2 Tetramethylene diamine Butane-1, 4-diamine
9. C6H5NH2 Aniline Benzenamine
10. CH3 NH2 ortho–toluidine 2–methyl benzenamine
11. N(CH3)2 N,N–dimethylaniline N,N–dimethylbenzenamine
12. CH2NH2 Benzyl amine Phenylmethanamine
13. HO–CH2–CH2NH2 2–amino ethyl alcohol 2–aminoethanol
14. 223 2 HNCHCHCH | NH vic-propylenediamine Propane–1,2–diamine
15. CH3–NH–CH3 Dimethyl amine N–methylmethanamine
16. CH3–NH–C2H5 Ethyl methyl amine N–methylethanamine
17. 33 3 CHNCH | CH Trimethyl amine N, N–dimethylmethanamine
18. 2537 3 CHNCH | CH Ethyl methyl propyl amine N–ethyl–N–methylpropanamine–1
Functional isomerism: Primary, secondary, and tertiary amines having same molecular formula are functional isomers of one another, e.g.,CH 3 CH 2 CH 2 NH 2 ; CH 3 –NH–C 2 H 5 and 33 3 CHNCH | NH
Optical is omerism: Amines having chiral structures may show enantiomerism. C H H3C C2H5 NH2 C H H2N C2H5 CH3
Cyclic tertiary amines and quaternary salts with different groups also exhibit optical isomerism.
H7C3–N–C2H5X–CH3 C4H9
TEST YOURSELF
1. IUPAC name of (CH3)2CHNH2 is (1) trimethyl amine
(2) 2-methyl butanamine
(3) 2-propanamine
(4) 2-methyl propanamine
2. The number of structural isomers possible from the molecular formula C 3 H 9N is (1) 4 (2) 5
(3) 2 (4) 3
3. The correct IUPAC name of C2H5–N(CH3)–CH2CH2CH3 is
(1) N, N–diethyl butylamine
(2) N–methyl–N–methyl butylamine
(3) N–ethyl–N–methyl butanamine
(4) N–ethyl–N–methyl Propan –1–amine
4. IUPAC name of (C2H5)3C–NH2 is
(1) 3-ethyl propanamine-1
(2) 3-ethyl pentanamine-2
(3) 3-ethyl pentanamine-3
(4) 2-ethyl pentanamine-3
5. What is the IUPAC of aniline?
(1) amino benzene
(2) benzenamine
(3) phenyl amine
(4) benzyl amine Answer key
(1) 3 (2) 1 (3) 4 (4) 3
(5) 2
11.4
PREPARATION OF AMINES
Aniline was prepared by Unverborden for the first time by the destructive distillation of indigo. Hence, it is named aniline, as ‘anil’ means indigo.
From nitro benzene: Aniline is prepared by the reduction of nitro benzene with tin(Sn) and hydrochloric acid or H 2 – Pd in ethanol.
Iron, water, and a small quantity of hydrochloric acid are used for this reduction on a commercial scale.
This is preferred because FeCl 2 formed gets hydrolysed to release HCl so that a small amount of HCl is required to initiate the reaction.
Ammonolysis of alkyl halides: Heating alkyl halide with alcoholic ammonia results in replacement of halogen atom by an amino group (SN reaction).
0 100C 33 NHRXRNHX
Cleavage of C-X bond by ammonia molecule is known as ammonolysis. Free amine can be obtained from the ammonium salt by treatment with a strong base.
3 22 RNHXNaOHRNHHONaX
Order of reactivity of alklyl halides with ammonia is RI > RBr > RCl. Primary amine formed reacts with excess alkyl halide to form secondary, and tertiary amines, and finally, quaternary ammonium salt.
NHRNHRNH
RX 34 3amineQuaternaryamm.salt RNRNX
Primary amine is obtained as a major product by taking large excess of ammonia. Disadvantage of ammonolysis is formation of mixture of 1°, 2°, and 3° amines and also quaternary ammonium salt.
Amides, on reduction with LiAlH 4 , yield corresponding primary amine.
4 2 (1)LiAlH (2)HO 2 22 RCNH RCHNH || O
Nitriles, on reduction, produce primary amines.
24 25 H/NiorLiAlH NaHg/CHOH 22 RCN RCHNH
Note: This reaction is used for ascent of amine series.
Gabriel phthalimide synthesis : Treating phthalimide with ethanolic KOH forms potassium salt of phthalimide. This, on heating with alkyl halide, followed by alkaline hydrolysis, forms the corresponding primary amine. CO CO
Pthalimide
CO CO NR NaOH Aq COONa COONa
N-alkyl phthalimide + RNH2 (10 amine)
It is an exclusive method for the preparation of pure aliphatic primary amines. Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
Hoffmann bromamide reaction: Amide gives amines, when treated with bromine in the presence of alkali (Hofmann bromamide degradation).
RCONH2 + Br2 + 4NaOH →
RNH2 + Na2CO3 + 2NaBr + 2H2O
Mechanism: Br 2 +KOH → KOBr+HBr KOBr → K++OBr–
This is the most convenient method for preparing primary amines. This method gives an amine containing one carbon atom less than that in amide.
TEST YOURSELF
1. CH 3 CONH 2 +Br 2 + x NaOH → CH 3 NH 2 +Bi products; what is the value of x.
(1) 8
(2) 3
(3) 4
(4) 1
2. From Gabriel phthalimide synthesis, which amine is not formed?
(1) CH3–NH2
(2) C6H5NH2
(3) CH3CH2NH2
(4) (CH3)2CH–NH2
3. Treatment of ammonia with excess of ethyl chloride will yield
(1) diethylamine
(2) methylamine
(3) tetraethylammoniumchloride
(4) ethane
4. In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of amine produced are (1) two moles of NaOH and two moles of Br2
(2) four moles of NaOH and one mole of Br2
(3) one mole of NaOH and one mole of Br2
(4) four moles of NaOH and two moles of Br2
5. When acetamide is treated with NaOBr, the product formed is (1) CH3CN
(2) CH3COBr
Donating group increases reactivity and withdrawing group decreases reactivity.
(3) CH3NH2
(4) CH3OH
6. Among the following, which one does not act as an intermediate in Hofmann rearrangement? (1) RNCO (2) RCON (3) RCONHBr (4) RNC Answer key
(1) 3 (2) 2 (3) 3 (4) 2
(5) 3 (6) 4
11.5 PHYSICAL PROPERTIES OF AMINES
Lower aliphatic amines are soluble in water because they can form hydrogen bonds with water molecules. However, solubility decreases with increase in molar mass of amines due to increase in size of the hydrophobic alkyl part. Higher amines are essentially insoluble in water. The intermolecular hydrogen bonding in amines is weaker than that in alcohols. Boiling point of ethanamine is less than that of ethanol.
Intermolecular association is more in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in primary amine. Tertiary amines do not have intermolecular association due to the absence of hydrogen atom. Therefore, the order of boiling points of isomeric amines is as follows: Primary > Secondary > Tertiary.
Aniline is a colourless, oily liquid but turns brown on exposure to air due to oxidation. It has a characteristic unpleasant odour. It is slightly soluble in water but soluble in organic solvents like ether, acetone, etc. It is steam volatile and is purified by steam distillation. The boiling point of aniline is 183°C. Comparision of boiling points of mines, alcohols, and alkanes of similar molecular masses is shown in Table 11.2.
Table 11.2 Comparison of boiling points of mines, alcohols and alkanes of
TEST YOURSELF
1. The boiling points of amines and their corresponding alcohols and acids vary in the order
(1) RCH2NH2 > RCOOH > RCH2OH
(2) RCH2NH2 > RCH2OH > RCOOH
(3) RCH2NH2 < RCOOH < RCH2OH
(4) RCH2NH2 < RCH2OH <RCOOH
2. Most non-volatile among the following is (1) 1 – butanamine
(2) N – ethylethanamine
(3) N, N – dimethylethanamine
(4) 1 – butanol
3. The descending order of boiling points of the following compounds is
a) 1–butanamine
b) N–ethylethanamine
c) N, N–dimethyl ethanamine
d) n–butyl alcohol
e) iso pentane
(1) d
(3) a > d > b > c > e (4) a
4. Solubility of amines in water decreases with increase in molar mass of amines due to (1) Formation of hydrogen bonds with water molecule
(2) Increase in hydrophilic part
(3) Increase in size of the hydrophobic alkyl part
(4) Decrease in size of hydrophobic alkyl part
11.6 CHEMICAL REACTIONS OF AMINES
Difference in electronegativity between nitrogen and hydrogen atoms and the presence of unshared pair of electrons over the nitrogen atom makes amines reactive. The number of hydrogen atoms attached to nitrogen atom also decides the course of reaction of amines; that is, why primary (R-NH 2), secondary (R2 N-H), and tertiary amines (R 3 N) differ in many reactions. Moreover, amines behave as nucleophiles due to the presence of unshared electron pair. Some of the reactions of amines are described below.
11.6.1 Basic Nature of Amines
Due to the presence of a lone pair of electrons on nitrogen, amines exhibit basic character and behave as Lewis bases. They are stronger bases than water and are, therefore, protonated by water in their aqueous solutions. Amines are Lewis bases.
The basic nature of amines is compared in terms of the equilibrium constant K b.
Larger the value of Kb or smaller the value of pKb, more is the basic strength of amine. The Kb and pKb values for some amines are given in Table 11.3.
Table 11.3 pKb values of some amines
Her e, K b is dissociation constant of the base. As the value of Kb increases, strength of the base increases. A more convenient method of expressing the basic strength of amines is in terms of their pKb values. As the value of pKb decreases, strength of the base increases.
The strength of a base depends on the stability of its conjugate acid. As the stability of the conjugate acid of a base increases, it will be less acidic and, correspondingly, becomes a strong base.
Alkanaminies versus ammonia: Stability of conjugate acid of a base decides strength of a base. The stability of the conjugate acid of a base depends on charge spreading. +I effect of alkyl groups connected to the nitrogen of a base or +M effect of a group present may decrease the +ve charge formed on nitrogen of a base in its conjugate acid. It is stabilised and the base becomes a strong base.
Alkyl amines are thus, more basic than ammonia, as their conjugate acids become more stable due to the +I effect of the alkyl groups. Stability further increases with the number of alkyl groups connected to nitrogen increasing.
In gaseous state, order of basic strength:
In aqueous solution, in addition to +I effect of –R groups, hydrogen bonding with water also stabilises the conjugate acid.
The greater the extent of hydrogen bonding in protonated amine, more will be its stabilisation and, consequently, greater will be the basic strength of the corresponding amine. The hydration due to hydrogen bonding is maximum in monoalkyl ammonium ion. It is less in dialkyl ammonium ion and still less in trialkyl ammonium ion.
Therefore, basic strength should decrease in the order : 1° > 2° > 3°. Secondly, when the alkyl group is small, like –CH3 group, there is no steric hindrance to H– bonding. In case the alkyl group is bigger than CH3 group, there will be steric hindrance to H-bonding. Therefore, the change of nature of the alkyl group, e.g., from–CH 3 to C2H 5, results in change of the order of basic strength. Thus, there is subtle interplay of the inductive effects, solvation effect, and steric hindrance of the alkyl group, which decides the basic strength of alkyl amines in the aqueous state. The order of basic strength in case of methyl substituted amines and ethtyl substituted amines in aqueous solution is as follows:
CH3–NH–CH3 >CH3NH2>(CH3)3N>NH3.
C2H5–NH–C2H5>(C2H5)3N> C2H5NH2>NH3.
Arylamines versus ammonia: Availability of lone pair for donation decides the strength of base. Aliphatic amines are more basic than aromatic amines as the lone pair of aromatic amines is involved in resonance with the benzene ring and not in free state. Ethyl amine is more basic than aniline. In aniline, the lone pair on nitrogen is involved in resonance.
Ther e are five resonating structur es for aniline, whereas for anilinium, ion obtained by accepting a proton can have only two resonating Kekule structures. Greater the number of resonating structures, greater is the stability, i.e., aniline is more stable than anilinium ion. Hence, aniline or other aromatic amines would be less basic then than ammonia.
Presence of an electron withdrawing group in aromatic amines generally decreases the basic strength, whereas electron releasing group increases the strength of a base. Decreasing order of basic strengths of some amines is given below.
NH2 NO2 NH2 NH2 NH2 > > > NO2 NO2
O CH3 CH3 Cl NO2 NH2 NH2 NH2 NH2 NH2 > > > >
(+ M (hyper- (–M and –I effect) conjugation) effects)
H3C–N–CH3 NHCH3 NH2 > > NO2
CH2NH2 > > CH2NH2 CH2NH2
Amides are less basic than amines, as the lone pair in amides is in conjugation with p bond of C O – group.
R–C–NH2 O R–C=NH2 O
11.6.2
Reactions of Amines
Amines are involved in different chemical reactions.
Basic Nature: Aniline is soluble in dil HCl. It forms salts with acids. Aniline is regenerated when alkali is added to aniline hydrochloride. Aqueous solution of aniline hydrochloride is acidic due to cationic hydrolysis. + 23 RNH+HXRNHX → + 652 653 CHNH+HClCHNHCl +
Alkylation: Amines undergo alkylation with alkyl halides to give secondary and tertiary amines and, finally, quaternary ammonium salts. Aniline reacts with alkyl halide to give secondary and tertiary amines. Quaternary ammonium salt is finally formed. The byproduct in this reaction is neutralised by adding carbonate.
C6H5NH2 + CH3Cl → C6H5NHCH3 + HCl
C6H5NHCH3+CH3Cl → C6H5N(CH3)2+HCl ( ) ( ) 6533653 23 CHNCHCHClCHNCHCl + + →
Acylation: When aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides and esters undergo nucleophilic substitution reaction. It helps to calculate the number of NH2 groups in amines.
Number of NH2 groups = GMWofamidesGMWofamines
This reaction is carried out in the presence of a base stronger than the amine, like pyridine, which removes HCl so formed and shifts the equilibrium to the right hand side.
Aniline reacts with acetic anhydride or acetyl chloride to give acetanilide.
C6H5NH2+CH3COCl → C6H5NHCOCH3 + HCl
C6H5NH2+(CH3CO)2O → C6H5NHCOCH3+ CH3COOH
Order of reactivity in acetylation:
CH3COCl>CH3COOCOCH3> CH3COOC2H5
With benzoyl chloride, amines give anilides.
CH3NH2+C6H5COCl→CH3NH2+ C6H5COCl
C6H5NH2+C6H5COCl→C6H5NH2+ C6H5COCl
Introduction of RCO– or C6H5CO– group in place of –H atom in general is known as acylation. The former is acetylation and the latter is benzoylation.
The lone pair of electrons on nitrogen of acetanilide interacts with oxygen atom due to resonance.
Hence, the lone pair of electrons on nitrogen is less available for donation to benzene ring by resonance. Therefore, activating effect of – NHCOCH3 group is less than that of amino group.
Carbylamine reaction (Isocyanide test): On warming with chloroform in presence of alcoholic potash, aliphatic and aromatic primary amines form carbylamine isocyanide which has a foul smell. Only primary amines give this test.
Reaction with nitrous acid: Nitrous acid is prepared in situ from a mineral acid and sodium nitrite. Different types of amines react differently with nitrous acid. Primary aliphatic amines react to form unstable diazonium salts, liberating nitrogen gas quantitatively, and alcohols. Due to evolution of nitrogen gas
quantitatively, this reaction is used in the estimation of amino acids and proteins.
benzene sulphonyl chloride is replaced by p-toluene sulphonyl chloride.
Aromatic primary amines react with nitrous acid at very low temperature (0–5°C) to form stable diazonium salts, which are used for the synthesis of a variety of aromatic compounds.
The reaction of secondary and tertiary amines with nitrous acid is different.
Reaction with benzene sulphonyl chloride: Benzene sulphonyl chloride (C 6 H 5 SO 2 Cl) is known as Hinsberg’s reagent. It forms sulphonamides with primary and secondary amines.
With primary amine, it forms N-ethyl benzene sulphonyl amide.
Electrophilic substitution in aniline: The amino group in aniline releases electrons by exerting +M effect and increases the electron density at the ring carbon atoms. Electrophilic substitution takes place fast in aniline than in benzene due to this activation of the ring. Amino group is ortho and para orienting group as electron density increases at ortho and para positions compared to meta position.
Due to the presence of a strong electron withdrawing sulphonyl group, the hydrogen attached to nitrogen in sulphonamide is strongly acidic. The product is soluble in alkali. With secondary amine, it forms N, N–diethylbenzenesulphonamide.
Bromi nation: Aniline forms a white precipitate of symmetrical tribromoaniline (2, 4, 6-tribromoaniline) instantaneously when treated with bromine water at room temperature. NH2 + 3Br2
If we have to prepare monosubstituted derivative, the activating effect of – NH2 group is to be controlled by protecting the – NH 2 group by acetylation with acetic anhydride and then followed by hydrolysis. NH2
The p roduct formed does not contain any hydrogen atom attached to nitrogen atom. The product is not acidic in nature and, hence, insoluble in alkali.
Tertiary amines do not react with benzenesulphonyl chloride.
Benzene sulphonyl chloride can be used as a reagent for distinguishing and also separating the mixture of amines. Now,
Nitrat ion: Direct nitration yields tarry oxidation products in addition to a mixture
NH2 NH2 NH2 NH2 NH2
of o–, m– and p–nitro anilines. If acetanilide is nitrated, only o– and p–nitro acetanilides are formed. These are converted to o– and p–nitroanilines by hydrolysis.
Due to resonance, the lone pair of electrons on nitrogen of acetanilide interacts with oxygen atom. Thus, the lone pair of electrons on nitrogen is less available for donation to benzene ring. Hence, activating effect of amino group is lessened by converting into – NHCOCH3
Sulp honation: Aniline is sulphonated by heating with conc. H 2 SO 4 . Anilinium hydrogen sulphate formed rearranges to p–aminobenzene sulphonic acid called sulphanilic acid at 453 - 473K which exists as zwitter ion in solution state.
(major product)
In th e strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. That is why besides the ortho and para derivatives, significant amount of meta derivative is also formed.
Table 11.4 tests for three types of amines
1. Hinsberg’s test: The amine is treated with benzene sulphonyl chloride, and then with NaOH.
2. Hofmann mustard oil test: Treated with CS2 and then with HgCl2
N-alkyl benzene sulphonamide formed dissolves in NaOH.
RNH2+C6H5SO2Cl →
RNHSO2C6H5+HCl
Ani line does not undergo Friedel–Crafts reaction. With the catalyst aluminium chloride, the Lewis acid, aniline, forms salt. Hence, nitrogen of amino group acquires positive charges and acts as a strong deactivating group for further reaction.
11.6.3 Distinguishing Amines
The three types of amines can be differently identified by following certain tests. These tests are summarised in Table 11.4.
N, N-dialkyl benzene sulphonamide is formed, but it is insoluble in NaOH.
R2NH+C6H5SO2Cl → R2NSO2C6H5+HCl
No reaction
A black precipitate of HgS is formed along with alkyl iso-thiocyanate.
RNH2+CS → black
No balck precipitate is formed. R2NH+CS2→ 2 || S RNCSH
RNCSHgS2HCl −==+↓
No reaction
3. Action of HNO2: Aromatic primary amines give a stable diazonium salt. RCH2NH2+HNO2→ 2 22 2 2 RCHNNRCHN HO RCHOH ++ ≡→+ ↓
Yellow, oily, nitroso amine is formed. It gives a green colour on heating with phenol and H2SO4, which changes to red on dilution and then blue on adding alkali.
RNHRHNORNR NO +→−− ↓
Tertiary amines give a salt. −−+→ 2 | R RNRHNO + | 2 | H R RNRNO
4. Carbyl amine test: Treated with CHCl3 and alc. KOH.
5. Oxidation with KMNO4
Primary amines give carbyl amine with unpleasant odour. AlcKOH 23 RNHCHClRNC +→
Aldehydes and ketones are formed. RCH2NH2→ RCHO 2 RCHRRCOR | NH −−→
6. Caro’s acid test
Aldoximes and hydroxamic acid are formed. ( ) 25 O 22 HSO RCHNH → NOH OH RC
TEST YOURSELF
1. Hinsberg’s reagent is (1) C6H5SO3H (2) C6H5OH (3) C6H5SO2Cl (4) C6H5SO2Na
2. Primary amines can be distinguished from secondary and tertiary amines by reacting with (1) chloroform and alcoholic KOH (2) methyl iodide (3) chloroform alone (4) zinc dust
No reaction No reaction
Tetra alkyl hydrazine is formed. ( ) O 22 || RR 2RNHRNNRHO →−−−+
Dialkyl hydroxylamine is formed. ( ) 25 O HSO RNHR → | R RNOH
No reaction
Tertiary amine oxide is formed ( ) 24 O 3 HSO 3 RN RNO → →
3. A compound with molecular mass 180 is acylated with CH 3 COCl to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is (1) 2
(2) 5 (3) 4
(4) 6
4. Secondary amines, on oxidation with Caro’s acid, give (1) dialkyl hydroxylamine (2) tetraalkyl hydrazine (3) amine oxide (4) ketones
5. Choose the correct order of basic strength.
I) Methanamine
II) N-methyl ethanamine
III) N,N-dimethyl methanamine
(1) III > II > I
(2) III > I > II
(3) II > I > III
(4) II > III > I
6. Among the amines given below, order of basic strength is (A) C2H5NH2 (B) C6H5NH2 (C) NH3
(D) C2H5CH2 NH (E) (C2H5)2NH
(1) E>A>D>C>B (2) E>A>C>B>D
(3) D>E>A>C>B (4) D>B>C>A>B
7. CH 3CH2NH2 contains a basic NH2 group, but CH3CONH2 does not, because (1) acetamide is amphoteric in character (2) in CH 3CH 2NH 2, the electron pair on N-atom is delocalised by resonance (3) In CH3CH2NH2 , there is no resonance, while in acetamide the lone pair of electron on N-atom is delocalised and, therefore, less available for protonation (4) acetamide is neutral in character
Answer key
(1) 3 (2) 1 (3) 2 (4) 1 (5) 1 (6) 3 (7) 3
11.7 DIAZONIUM SALTS
The ion is called diazonium ion. The compounds possessing this ion along with X are called diazonium salts. The general formula of diazonium salts is where R is + 2 RNX , an alkyl/aryl group,
It is alkyl diazonium salt. X – may be Cl–, Br–, HSO4–, BF4–, etc.
Diazonium compounds are named by suffixing diazonium to the name of the hydrocarbon corresponding to –R group, followed by the name of X –.
C 6 H 5 N = N + Cl – , Benzene diazonium chloride, C 6 H 5 N = N + HSO 4 – , Benzene diazonium hydrogen sulphate, CH3CH2N=N+Br–, Ethane diazonium bromide.
Stability: Aliphatic diazonium salts are unstable and dissociate forming a carbocation, releasing nitrogen gas. The carbocation may form alcohol, alkene, or alkyl halide. Aromatic diazonium salts are stable as diazonium ion is resonance stabilised. The conversion of aromatic primary amines into diazonium salts is known as diazotisation.
11.7.1
Preparation of Benzene Diazonium Chloride
Benzene diazonium chloride is prepared by dissolving aniline in dilute hydrochloric acid and adding a solution of sodium nitrite at 0–5°C.
NaNO2 + HCl → HNO2 + NaCl
11.7.2 Physical Properties
Benzene diazonium chloride is a colourless crystalline solid. It is soluble in water and stable at 0°C, but reacts with warm water. Benzene diazonium fluoroborate is insoluble in water and stable at room temperature. In solid state, diazonium chloride decomposes.
11.7.3 Chemical reactions
Diazonium chloride is converted to several compounds and becomes important in organic synthesis.
When reduced with hypophosphorous acid (phosphinic acid) or ethanol, it gives benzene.
65 232 CHNNClHOHPO ∆ =++→
C6H6+N2+H3PO3+HCl
C6H5N=NCl+C2H5OH →
C6H6+N2+CH3CHO+HCl
If the temperature of the diazonium salt solution is allowed to rise upto 10°C, the salt gets hydrolysed to phenol.
C6H5N=NCl+H2O → C6H5OH+N2+HCl
Sandmeyer reaction: Benzene diazonium chloride is converted to chlorobenzene when treated with cuprous chloride and hydrogen chloride.
22 HCl 65 652 CuCl CHNNClCHClN =→+
Similarly, bromobenzene is formed with cuprous bromide and hydrogen bromide.
Benzene diazonium chloride gives benzonitrile with potassium cyanide and cuprous cyanide. Benzonitrile, on acidification, gives benzoic acid.
CuCN
65 65 CHNNClKCNCHCN =+→
These reactions are called Sandmeyer reactions. The above products can also be prepared by treating benzene diazonium chloride with HCl and Cu or HBr and Cu.
This reaction is known as Gatterman reaction.
Cu
65 652 CHNNClHCl CHClNCuCl=+→++
Cu 65 652 CHNNClHBrCHBrNCuCl =+→++
The yield in Sandmeyer reaction is better than that in Gatterman reaction.
Iodobenzene is formed when benzene diazonium chloride is treated with potassium iodide in the presence of copper catalyst.
Cu
65 65 2 CHNNClKICHIKClN =+→++
Benzene diazonium chloride, on treating with fluoroboric acid followed by heating gives fluoro-benzene.
65 4 6524 HCl CHNNClHBFCHNBF =+→ Heat 6523 CHFNBF→++
Benzene diazonium chloride, on treating with HBF4, followed by heating with aqueous sodium nitrite solution in presence of copper, nitrobenzene.
2 65 4 654 HCl CHN=NCl+HBFCHNBF + → NaNO2 65224 Cu, CHNONNaBF ∆ →++
11.7.4 Importance of Diazonium Salts
The replacement of diazo group by other groups is helpful in preparing those substituted aromatic compounds which cannot be prepared by direct substitution in benzene or substituted benzene. These salts provide important link in many synthetic processes. Diazonium salts can be prepared from almost all primary aromatic amines. One of the significant routes in which diazonium salts provide link during synthesis processes is Ar–H → Ar–NO2→ Ar–NH2 → Ar– 2 N +
Coupling reactions: Benzene diazonium chloride undergoes coupling reactions. Coupling reaction is an electrophilic aromatic substitution reaction. Benzene diazonium ion substitutes phenol and aniline in para position.
This reaction is useful in the synthesis of azodyes, which are coloured.
Phenol gives p-hydroxy azobenzene when it undergoes coupling with benzene diazonium chloride, at a pH value of 7 to 10.
N=NCl OH + N=N
p-Hydroxy azobenzene (orange)
Aniline gives p–amino azobenzene when it undergoes coupling with benzene diazonium chloride, at a pH value of 5 to 7.
N=NCl N=N
p-Amino azobenzene (yellow)
TEST YOURSELF
1. Aniline NaNO22HO HCl,OC B A oil B , AC °
In the above reaction sequence, A, B, and C are
(1) phenol, benzene, phenolphthalein
(2) benzene diazomium chloride, phenol, p-hydroxyl azobenzene (3) 65 656565 CHNNCl,CHOH,CHNNCH ⊕ Θ −≡ = (4) 65 65653 CHCHNH,CHOH,CHCOCH =
2. Benzene diazonium chloride, on reaction with KCN in the presence of CuCN, yields X. X, on hydrolysis, yields Y. Y can also be obtained from
(1) toluene by the action of Cl 2/FeCl3
(2) toluene by oxidation by KMnO 4 (3) toluene by nitration (4) toluene by sulphonation
3. Which of the following is the correct order of ease of coupling with C 6H5N2Cl?
A) Benzene
B) Nitrobenzene
C) Phenol
D) Chlorobenzene
(1) A > D > B > C
(2) C > A > B > D
(3) C > A > D > B
(4) B > D > A > C
4. Action of HCl on benzene diazonium chloride in the presence of copper powder gives
(1) p-chlorobenzene diazonium chloride
(2) o-chlorobenzene diazonium chloride
(3) chlorobenzene
(4) o-dichlorobenzene
5. Which diazonium salt is stable at room temperature?
(1) Benzene diazonium chloride
(2) Benzene diazonium fluoroborate
(3) Benzene diazonium nitrate
(4) Benzene diazonium bromide
6. Which one of the following is water insoluble and stable at room temperature?
(1) C6H5N2Cl (2) C6H5N2HSO4
(3) C6H5N2Br (4) C6H5N2BF4
7. Benzenediazonium chloride, on reaction with water at 283 K gives
(1) phenol
(2) aniline
(3) benzylamine
(4) benzaldehyde
8. In the diazotisation of aniline, the reagent or reagents used is/are
(1) HNO3, HCl
(2) HNO2 only
(3) NaNO2, HCl at 0–5°C
(4) NaNO2, HNO2 at 0–5°C
Answer key
(1) 2 (2) 2 (3) 3 (4) 3
(5) 2 (6) 4 (7) 1 (8) 3
11.8 CYANIDES AND ISOCYANIDES
Hydrogen cyanide exists in two tautomeric forms. HCN:HNC: + −≡−≡
Alkyl or aryl derivatives of these two forms are called cyanides and isocyanides, respectively. These two are functional isomers.
R–C ≡ N is cyanide and R–N ≡ C is isocyanide.
Cyanides are named ‘nitriles’ based on the corresponding carboxylic acids obtained from them, by hydrolysis. The terminal ‘ic acid’ in the name of acid is replaced by nitrile. IUPAC names of cyanides are derived by adding the secondary suffix ‘nitrile’ to the name of the hydrocarbon containing the same number of carbon atoms. Isocyanides are named carbylamines. Common and IUPAC names of some important cyanides and isocyanides are given in Table 11.5
11.8.1 Preparation of Cyanides
From alkyl halides: Alkyl halides, on heating with potassium cyanide in alcoholic solution, form cyanides. This is S N2 reaction.
alcohol 100C RXKCNRCNKX ° +→+
From acid amides: Acid amides, on heating with phosphorus pentoxide, undergo dehydration and give cyanides.
Table 11.5 Names of some cyanides and isocynaides
PO25 22 RCONHRCNHO →+
From oximes: Oximes undergo dehydration on heating with phosphorus pentoxide and give nitriles.
PO25 2 RCHNOHRCNHO −=→+
The reaction can also be run in pyridine, which takes up the two acids in the form of their pyridine salts.
CH3–CH2–CH=NOH(CH3–CH2–CONH2)+ (CH3CO)2O → CH3CH2CN+2CH3COOH
From ammonium carboxylates: Ammonium carboxylates, on heating with phosphorus pentoxide, yield alkyl cyanides.
PO25 42 RCOONHRCNHO →+
From Grignard reagent: Grignard reagent gives cyanides on treating with cyanogen chloride (N ≡ C–Cl).
RMgCl + ClCN → RCN+MgCl2
11.8.2 Properties of Cyanides
Cyanides are neutral substances with pleasant odour. They are soluble in water and organic solvents. Cyanides form carboxylic acids on hydrolysis, either with dil. acid or with alkali.
2 HO 2 H RCNRCONH + −→ 2 HO 3 H RCOOHNH + →+
CH3CN Methyl cyanide (or) Acetonitrile Ethanenitrile
CH3CH2CN Ethyl cyanide (or) Propionitrile Propanenitrile
CH3CH2CH2CN Propyl cyanide (or) Butyronitrile Butanenitrile
C6H5CN Phenyl cyanide (or) Benzonitrile Benzenenitrile
CH3NC Methyl isocyanide Methylcarbylamine (isonitrile)
C2H5NC Ethyl isocyanide Ethylcarbylamine
C6H5NC Phenyl isocyanide Phenylcarbylamine
CH2 = CH – CN Vinyl cyanide (acrylonitrile) Prop-2-en-nitrile
Complete reduction of cyanides gives primary amines. LiAlH4, hydrogen, and nickel or sodium and ethyl alcohol can be used as reducing agents. Partial reduction of cyanides with SnCl2 and HCl, followed by hydrolysis, forms aldehydes. This is called Stephen’s reaction.
4 LiAlH RCNRCHNH22 →
2 SnCl HCl RCNRCHNH →=
2 HO 3 RCHONH→+
Alkyl or aryl cyanides undergo addition with Grignard reagent. The addition product, on hydrorysis, gives ketone.
2
11.8.3 Preparation of Isocyanides
From alkyl halides: Alkyl halides produce isocyanides when heated with silver cyanide. A small quantity of cyanide is also formed.
From primary amines : Primary amines, on warming with chloroform in presence of alcoholic KOH, form isocyanides.
2 warm RNHCHCl3KOH++→ RNC + 3KCl + 3H2O
Cyanides are important intermediates in multistep synthesis. They are produced in millions of pounds annually, and are mainly used for the production of rubbers and synthetc textiles.
11.8.4 Properties of Isocyanides
Isocyanides possess unpleasant odour. Isocyanides are more volatile than corresponding cyanides. These are insoluble in water and soluble in organic solvents. Isocyanides are more toxic than cyanides.
On hydrolysis with dilute acids, isocyanides produce primary amines. They are not hydrolysed by alkali.
2 HO 2 H RNCRNHHCOOH + →+
On reduction with hydrogen and platinum or sodium and ethanol, isocyanides form secondary methyl amines. 25 Na,CHOH 3 RNC RNHCH →
They are oxidised to isocyanates when treated with HgO.
R–NC + HgORNCO + Hg
TEST YOURSELF
1.
In the above reaction sequence, Z is (1) CH3CH2CH2NHCOCH3
(2) CH3CH2CH2NH2
(3) CH3CH2CH2CONHCH3
(4) CH3CH2CH2CONHCOCH3
2. The compounds A and B in the following reaction are, respectively,
(1) A= Benzyl alcohol, B= Benzyl isocyanide
(2) A= Benzyl alcohol, B= Benzyl cyanide
(3) A= Benzyl chloride, B= Benzyl cyanide (4) A=Benzyl chloride, B= Benzyl isocyanide
3. The major product in the reaction of ethyl chloride with ethanolic AgCN is (1) ethyl cyanide (2) ethyl isocyanide
(3) ethyl nitrate
(4) nitroethane
4. Hydrolysis of alkyl isocyanide yields (1) primary amine
(2) tert. amine
(3) alcohol (4) aldehyde
The above reaction is (1) Mendius reaction (2) Schmidt reaction (3) Rosemund reaction (4) Stephen’s reaction
Answer key
(1) 1 (2) 4 (3) 2 (4) 1 (5) 4
11.9 CONVERSIONS
■ Methanamine to ethanamine
■ Ethanamine to methanamine
■ Ethanoic acid to aminomethane
■ Pentyl cyanide to 1-aminopentane
■ Nitromethane to N-methylmethanamine
■ Benzoic acid to aniline
■ Aniline to p-bromoaniline
■ Aniline to benzyl alcohol
■ Benzamide to toluene
■ Benzylchloride to 2-phenilethanamine
■ Aniline to sym-tribromofluorobenzene
■ Chlorobenzene to p-chloroaniline
■ Benzene to N,N-Dimethylaniline
■ 1,4-dichlorobutane to hexane-1,6-diamine
■ Benzene to p-nitrobenzaldehyde
■ Benzoic acid to m-nitrobenzylalcohol
Answers:
■
NH3 (O) 3 34 CHCOOHCHCOONH→→
2 Br/KOH 32 32 (Methanamine) CHCONHCHNH ∆ → →
3 2 NH SoCl 33 (Acetic acid) (Acetylchloride) CHCOOHCHCoCl→→
CHCONHCHNH →
2 Br/KOH 32 32 (Acetamide) (Methanamine)
■ 2 HO 3232 Pentylcyanide
CH-(CH)-CH-CN →
2 Br/KOH 3232 2 CH(CH)CHCONH−−−→
CH-(CH)-CH-NH
32322 (1-pentanamine)
■ 3 CHCl Sn/HCl 32 32 Reduction (Nitromethane) CHNO CHNH →→
33 Dimethylamine CHNNCH
■ COOH (Benzoic acid) (1)NaOH (2)CaO/NaOH → Nitration → NO2 Sn/HCl → NH2 Aniline
■ NH2
32 (CHCO)O Pyridine → NHCOCH3 → Br2 NHCOCH3
Br H + → NH2 p-Bromo aniline Br
■ NH2
2 NaNO/HCl → N 2 Cl HPO32 → methylation → CH3
2 Cl hυ → CH2Cl
aq.KOH → CH2OH Benzyl alcohol
■ CONH2 2 Br/NaOH →
2 NaNO/HCl →
N 2 Cl HPO32 → 33 Alkylation CHCl/AlCl →
■ CH3Cl (Benzyl chloride) 33 Alkylation CHCl/AlCl → CH2CN 4 LiAlH →
CH2 - CH2 - NH2
2-phenyl ethanamine
■ NH2 Br2 → Br NH2 Br Br 2 NaNO/HBr →
Br N 2 Cl Br Br HBF4 ∆ → Br F 2, 4, 6 Tribromo fluoro benzene Br Br
■ (Benzene)
33 CHCl/AlCl → CH3 Nitration → CH3 NO2
CrOCl22 H + → CHO (p-nitro benzaldehyde) NO2
■ COOH (Benzoic acid) Nitration → COOH (m-nitro Benzoic acid) NO2
11.9.1 Write the structure of the major organic product in each of the following reactions
■ Hoffmann degradation → Propanamine
■ CONH2 Hoffmann bromamidedegradation →
■ NH2 65 CHCOCl →
■ (C2H5)3N + HCl →
■ N 2 Cl + 3 2 NH HO/H CuCN Ä →→→
■ NO2 + 3 2 0 HO NaNO/HCl Fe/HCl Ä 0-5C →→→
■ NO2 + 3 2 0 HO NaNO/HCl Fe/HCl Ä 0-5C →→→
■ Cl Nitration NO2 NH2
Chloro Benzene Reduction p-Chloro aniline Cl Cl
■ → Nitration NO2 Reduction → NH2
3 2CHX → NH(CH3)2
■ KCN 24 24 Cl(CH)ClCN(CH)CN −−→−− Reduction 2262 (Hexamethylenediamine) NH(CH)NH→−−
■ NO2 652 2 0 CHNH HNO 0-5C Fe/HCl → →→
■2 KCN OH NaOH 32 PartialhydrolysisBr CHCHI →→→
■ 42 0 LiAlHHNO NaCN 32 0-5C CHCHBr →→→
■ 3 2 0 NH HNO NaOBr 3 Ä 0C CHCOOH →→→
■ 2 2 Br 652 HO CHNH →
■ 4 2 Ä (i)HBF 652 (ii)NaNO/Cu CHNCl →
■ C6H5N2Cl+H3PO2+H2O →
■ C6H5NH2+Ac2O →
Answers
■ 2 Br.KOH 3222 (Butanamide) CHCHCHCONH−−−−→
CHCHCHNH
3222 (Propanamine)
■ CONH2 (Benzamide)
2 Br+4KOH ∆ + → NH2 (Aniline) +2KBr+K2CO3+H2O
CHNHCHNHCOCH →
■ 65 CHCOCl 652 6565 (Aniline) (Benzanilide)
CHNHCHN(CH)I +− →
3 excessCHI 652 6533 Quaternary ammonium salt (Aniline)
CHNHClCHNHCl+− +→
■ ( ) ( ) 25 25 33 (Triethylamine) (Triethylammoniumchloride)
■ 2 HO/H CuCN 652 65 (Benzenediazonium chloride)
CHNClCHCN + →→
CHCOOH CHCONH ∆ →
NH3 65 652 (Benzamide)
■ 2 0 NaNO/HCl Fe/HCl 652 652 05C (nitrobenzene)
CHNOCHNH→→
3 HO 652 652 (nitrobenzene) (Phenol)
CHNClCHOH N HCl + ∆ →++
CHCHCONH CHCHNH → →−−
■ NaCN 32 32 (ethylbromide)
CHCHBrCH CHCN−−→−−
42 0 LiAlH HNO 3222 05C
CHCHCHOH
CHCHCHNH →−−−→ 3 (1Propanol)
■ NH3 NaOBr 3 32 (Aceticacid)
CHCOOHCHCONH ∆ → →
2 0 HNO 32 3 0C (Methanol) CHNH CHOH →
■ ■ 652 4 (Benzenediazonium chloride)
CHNClHBF+→
CHNONNaBF ++
65224 (nitrobenzene)
Fe/HCl 652 652 05C (nitrobenzene)
CHNH652 652 CHNCl →
CHCHICHCHCN → 2 OH NaOH 322 322 Partialhydrolysis Br (ethanamine)
2 0 HNO
CHNOCHNH→→
N = N NH2 p-amino azobenzene (yellow dye)
■ KCN 32 32 (ethyl iodide)
CHAPTER REVIEW
Structure of Amines
■ Amines are alkyl or aryl derivatives of ammonia.
■ Amines are classified as primary (RNH 2), secondary (R 2 NH), and tertiary (R 3 N) amines.
CHNClHPOHO++→
■ 652322 (Benzenediazonium chloride)
6633 (Benzene)
CHNH CHNHCOCH →
CHHPOHCl ++ 32 (CHCO)O 652 65 3 (Aniline) (Acetanilide)
Preparation of Amines
■ Destructive distillation of indigo gives aniline.
■ Industrially, aniline is prepared by the reduction of nitrobenzene with iron, water, and a small amount of HCl.
■ Nitrobenzene, on reduction with Sn/HCl, Zn/HCl or H2/Ni, gives aniline.
■ Phenol, on heating with NH 3 at 300°C in presence of ZnCl2(catalyst), gives aniline.
■ Phenol, on heating with NH 3 at 200°C in presence of Cu2O (catalyst), gives aniline.
■ Aniline can be purified by steam distillation.
■ Among isomeric amines, boiling points are in the order: 1° > 2° > 3°. This is due to decreasing ability to form hydrogen bonds.
Chemical Reaction of Amines
■ Amines are basic in nature, as they can donate a lone pair and accept proton.
■ Basic nature of amines depends on availability of lone pair for donation and stability of conjugate acid.
■ Electron-releasing groups increase the basic strength and electron-withdrawing groups decrease the basic strength.
■ Basic nature of amines in gaseous state: 3° > 2° > 1° > NH 3 (CH3)3N>(CH3)2NH>CH3NH2>NH3
■ In aqueous solution, order of basic nature: 2°>3°>1°>NH3
■ Aliphatic amines are more basic compared to aromatic amines.
■ Cyclohexyl amine is more basic than aniline. (C 2 H 5 ) 2 NH>C 2 H 5 NH 2 >NH 3 > C6H5NH2
■ Aniline is a Lewis base. It is almost neutral to litmus.
■ Aniline, on treatment with HCl, forms a salt named aniline hydrochloride.
■ Primary, secondary, and tertiary amines form quaternary salts with alkyl halides.
■ Aniline, on treatment with acetyl chloride or acetic anhydride, forms acetanilide.
■ Reactivity order of acylating agents is CH3COCl > (CH3CO)2O > CH3COOC2H5
■ With benzoyl chloride, in the presence of base, aniline gives benzanilide.
■ Aniline, on treatment with benzene sulphonyl chloride, gives N-phenyl benzene sulphonamide.
■ Aniline, with benzaldehyde in the presence of conc.H2SO4, forms benzylidene aniline, known as Schiff’s base.
■ Primary amines give carbylamine reaction with chloroform and alcoholic potash. Phenyl isocyanide has unplasent odour.
■ Aniline, on treatment with nitrous acid (NaNO2 + HCl), undergoes diazotisation to give benzene diazonium chloride [C6H5N2+ Cl–].
■ Aniline is used in the preparation of dyes, Schiff’s base, and sulphonamide.
■ Primary, secondary, and tertiary amines are distinguished using carbylamine test, Hinsberg test, and Hoffmann mustard oil reaction.
■ Carbylamine reaction is given only by aliphatic primary amines and aromatic primary amines and not by secondary and tertiary amines.
■ Benzene sulphonyl chloride is called Hinsberg reagent.
■ Hofmann mustard oil reaction is given only by primary amines. In this reaction, alkyl isothiocyanate (smells like mustard oil) and black precipitate (HgS) are formed.
■ On oxidation with KMnO4, primary amines give aldehydes and ketones; secondary amines form tetra alkyl hydrazine while tertiary amines do not react.
■ On oxidation with Caro’s acid, primary amines form amine, secondary amines form dialkyl hydroxyl amine, and tertiary amines give amine oxide.
Diazonium Salts
■ Aromatic diazonium salts like C6H5N2Cl are more stable (due to resonance) than aliphatic diazonium salts like CH 3N2Cl.
■ Benzene diazonium chloride is a colourless solid; highly soluble in water in dry state and easily decomposes.
■ Benzene diazonium chloride, with Cu2Cl2/ Cl, Cu 2 Br 2 / HBr and Cu 2 (CN) 2 /KCN, forms chlorobenzene, bromobenzene, and cyano-benzene, respectively, in Sandmeyer reaction.
■ Benzene diazonium chloride with Cu / HCl, Cu /HBr gives, respectively, chlorobenzene and bromo benzene, in Gattermann reaction.
■ Iodobenzene is formed when benzene diazonium chloride is treated with aqueous KI.
■ Benzene diazonium chloride with HBF 4 (fluoboric acid) gives benzene diazonium fluoroborate, which on heating gives fluoro-benzene.
■ Benzene diazonium chloride, on boiling with steam, forms phenol.
■ Benzene diazonium chloride, on reduction with hypophosphorous acid solution or ethyl alcohol, forms benzene.
■ Benzene diazonium fluoroborate, on heating with NaNO 2 /Cu, forms nitrobenzene.
■ Benzene diazonium chloride, on coupling with phenol in weakly alkaline medium, forms an orange-coloured dye called p-hydroxy azobenzene.
■ Benzene diazonium chloride, on coupling with aniline in weakly alkaline medium, forms an yellow-coloured dye namely, p–aminoazobenzene.
■ The azoproducts have extended conjugate system with aromatic rings and –
N = N – bond. Hence, these compounds are coloured and are used as dyes.
Additional Information
■ The formula of nitroalkanes is RNO 2 and of alkyl nitrite is RONO. Nitroalkanes and alkyl nitrites can be considered the alkyl derivatives of nitrous acid.
■ The two tautomeric forms of nitrous acid are
RNO2 is the derivative of I while RONO is the derivative of II.
■ If the NO 2 group is bonded to the alkyl carbon atom through the oxygen atom, the derivatives are called alkyl nitrites
■ Benzene derivatives in which –NO2 group is bonded to benzene carbon (C-N bond) are called nitrobenzenes.
■ The aliphatic nitrocompounds may further be classified as primary, secondary, and tertiary, as the nitro group is attached to primary, secondary, and tertiary carbon atoms, respectively.
■ Nitrobenzene is prepared by heating benzene diazonium fluoroborate with sodium nitrite in presence of copper.
■ Benzene diazonium chloride reacts with nitrous acid in presence of cuprous oxide to give nitrobenzene.
2 CuO
652 2 6522 CHNCl HNOCHNO NHCl +→ ++
■ Nitrobenzene is a pale yellow, oily liquid (b.p 211°C) with odour like bitter almonds. It is heavier than water and almost insoluble in H2O but soluble in organic solvents.
■ Nitrobenzene is highly toxic, because it destroys RBC by forming complex with haemoglobin. It is steam volatile and can be purified by steam distillation.
■ Nitrobenzene undergoes reduction. The products formed depend on the nature of reducing agent as well as the pH or the medium.
■ Nitrobenzene, on reduction with active metals like Zn, Fe, or Sn, in presence of Conc. HCl form aniline.
SnHCl
652 6522 CHNO6(H)CHNH2HO + +→+
■ Reduction of nitrobenzene is in alkaline medium:
Na3AsO3+NaOH(or) Gloucose+NaOH
+6(H)
C6H5NO2
Zn/NaOH, CH3OH
Zn/NaOH,(or) NH2-NH2 +8(H) +10(H)
65 652 Azoxybenzene CHNNCH3HO ↑ −=−+
65 652 Azobenzene CHNNCH4HO −=−+
65 652 Hydrazobenzene CHNHNHCH4HO −−−+
■ Nitrobenzene, on reduction with Zn dust and NH 4 Cl, provides neutral medium, produces N-phenyl hydroxyl amine.
ZnNH4Cl 652 65 2 CHNO4(H)CHNHOHHO + ∆ +→ +
■ Nitrobenzene, on reduction with LiAlH 4, gives azo compounds but not primary amine.
4LiAlH 652 6565 Ether 2CHNO8(H) CHNNCH+→=
■ Nitrobenzene on reduction with H 2 in presence of Ni (or) Pd/C catalyst in alcohol,
produces aniline.
Pd/C 6522 6522 Ethanol CHNO3HCHNH2HO +→+
■ Nitrobenzene, on electrolytic reduction in weakly acidic medium, gives aniline, but in strongly acidic medium, it gives p-aminophenol
Electrlytic 652 Reduction CHNO → Weakly acidic
Strongly acidic medium CHNH652 NHOH NH2 OH
Rearrangement →
■ If more than one nitro group is present on the benzene ring, then it is possible to replace one of these groups without affecting other nitro group by using mild reducing agent like, NH 4HS, (NH4)2 S, or sodium polysulphide (Na 2Sx)
+3(NH4)2S →
+6NH3+2H2O+3S
■ Electrophilic substitution reaction in nitrobenzene: ∆ NO2 NO2 NO2 NO2 NO2 Cl SO3H H/FeCl23 24 HSOconc 100oC HNO/HSO324
■ Nitrobenzene is used as a solvent in Friedel–Crafts reaction, as an oxidising agent in organic synthesis.
■ Nitrobenzene is used in the manufacture of aniline, benzidine, azodyes, detergents,
IL RANKER SERIES NEET | 25 and pharmaceuticals. It is used in the manufacture of floor polishes, for cheap scents (under the name ‘oil of mirbane’), cheap soaps, and shoe polishes.
■ Ethyl nitrite is prepared by the action of HNO2 on ethyl alcohol.
CH3CH2OH+HNO2→CH3CH2ONO+H2O
■ By the action of potassium nitrite on ethyl chloride, ethyl nitrite is formed.
C2H5Cl+KONO → C2H5ONO+KCl
■ Ethyl nitrite is a gas with the smell of apples and it is insoluble in water. On hydrolysis, ethyl nitrite gives ethyl alcohol and nitrous acid.
H(or)
322 32 2 OH CHCHONOHO CHCHOHHNO + +→ +
■ On reduction with tin and hydrochloric acid, ethyl nitrite forms ethyl alcohol and ammonia. A small quantity of hydroxylamine is also formed. Sn 32 3232 HCl CHCHONO6[H]CHCHOHNHHO +→ ++
■ Ethyl nitrite increases pulse rate and decreases hypertension. It is used to cure heart disease and for the treatment of asthma. Its 4% alcoholic solution is used as diuretic under the name ‘sweet spirit of nitrite.’
■ Ethyl chloride forms nitroethane when treated with silver nitrite. A small quantity of ethyl nitrite is also formed.
∆ +→+
■ Decarboxylation of a-nitro acetic acid gives nitromethane.
2 Cl 32 RedP CHCOOHCHCOOH | Cl → NaNO22HO
■ Direct nitration of ethane with fuming nitric acid at 470°C gives nitroethane. 0 470C
■ Nitroalkanes are colourless liquids with pleasant odour. They are less volatile than isomeric alkyl nitrites and less soluble in water, but readily soluble in organic solvents.
■ The a -hydrogen atoms in aliphatic 10 and 20 nitroalkanes become acidic due to electron withdrawing nature of NO2 group. So, primary and secondary nitroalkanes form salts with alkali. The acidic nature is explained on the basis of tautomerism.
■ Pr imary nitroalkanes form nitrolic acid, which dissolves in sodium hydroxide and produces red colour. Secondary nitro alkanes give pseudo nitrols, which dissolve in ether or sodium hydroxide, giving blue colouration. Tertiary nitroalkanes do not react with nitrous acid. This reaction is the basis to distinguish three type of alcohols by Victor Meyer test.
■ Primary nitroalkanes, on hydrolysis with acids, give hydroxyl amine and carboxylic acids. Secondary nitroalkanes, on hydrolysis, give ketones. H 222 2 R-CHNOHORCOOHNHOH + −+→+
→++
■ Alkyl cyanides on reduction with Na/ C 2 H 5 OH or H 2 /Raney Ni or LiAlH or catalytic reduction, amine are formed
■ Oxidation of aniline with acidified dichromate forms p-benzoquinone.
22724 KCrO,HSO Controlled Oxidation → O O
■ Aniline forms diphenyl thiocarbamic acid with CS2 in alcoholic potash solution, which gives phenyl isothiocyanate with conc. HCl.
HCl 652 65652 (CHNH)CSCHNCSCHNH =→+
■ With CS 2 in presence of HgCl 2 , aniline gives isothiocyanate.
2 HgCl 6522 65 CHNHCSCHNCSHgS2HCl +→==+↓+
Alcohol 6522 652 2 KOH
2CHNHCS(CHNH)CSHS +→ =+
■ The isothiocyanates have a characteristic smell like mustard oil; the reaction is called Hofmann’s mustard oil reaction.
Exercises
NEET DRILL
Level 1
Structure of Amines
1. Which of the following is an enamine?
(1) N-CH3
(2) CH3 N-CH3
(3) NH N
(4) CH3 CH4 N OH
2. Anisidines are the compounds containing benzene ring in which (1) −CH3 & −NH2 are present
(2) −CH3 & −OCH3 are present
(3) −OCH3 & NO2 are present
(4) −OCH3 & −NH2 are present
3. Which of the following is a primary amine?
(1) CH3CH2CH2NH2
(2) CH3CH(NH2)CH3
(3) (CH3)3C−NH2
(4) All the above
4. Which of the following contains imino group [-NH-]?
(1) Aniline (2) O – Toludine
(3) Benzylamine (4) N-methyl aniline
5. The general formula of amines is (1) C nH2n+1N (2) C nH2n+2N (3) C n H2n+3N (4) C nH2nN
Classification of Amines
6. Which of the following is not a tertiary amine?
(1) Tri ethyl amine
(2) Tri methyl amine
(3) 2-methyl propanamine-2
(4) N-ethyl-N-methyl propanamine-1
7. Identify the compounds from the following that form primary amines under suitable reduction conditions(single step reaction).
1) C2H5NC
2) C2H6
3) C2H5CONH2
4) C6H5NO2
(1) 1, 4 (2) 3, 4
(3) 1, 3, 4 (4) 2, 3, 4
8. Primary amino group is absent in (1) p-amino phenol
(2) o-amino phenol
(3) N-methyl ethanamine
(4) phenyl amine
9. How many of the following are primary amines?
A) 3° butyl amine
B) 2° butyl amine
C) Isobutyl amine
D) Dimethyl amine
(1) 3 (2) 4 (3) 2 (4) 1
Nomenclature of Amines
10. The correct IUPAC name for CH = CH–CH2–NH–CH3 is (1) allylmethylamine
(2) 2-amino-4-pentene
(3) 4-amino pent-1-ene
(4) N-methyl prop-2-ene-1-amine
11. The IUPAC name of (CH3)2NCH(CH3)2 is (1) N, N dimethyl-2-propanamine (2) N, N dimethyl amino isopropane (3) N, N dimethyl-amino ethane (4) methyl-N-isopropyl methyl amine
12. The IUPAC name of (CH3)3CNH2 is (1) trimethyl amine (2) 2-methyl butanamine (3) 2-methyl propanamine-2 (4) 2-methyl propanamine
13. The IUPAC name of the compound N-(CH3)2 is
(1) N-phenyl-N-propene
(2) N, N-methyl benzene
(3) N, N-dimethyl benzenamine (4) N, N-dimethyl benzene
Preparation of Amines
14. Which sequence of reactions is best suitable method for the preparation of m-chloroaniline from benzene?
(1) (i) nitration (ii) Cl2/FeCl3 (iii) Sn + HCl
(2) (i) Cl2/FeCl3 (ii) nitration (iii) Sn + HCl
(3) (i) Cl2/FeCl3 (ii) NH3 & ZnCl2
(4) (i) nitration (ii) Cl2, acetic acid
(iii) Sn + HCl
15. Which sequence of reactions is useful to prepare 1 – propanamine? (1)
17. Treatment of ammonia with excess of ethyl chloride will yield
(1) diethyl amine
(2) methyl amine
(3) tetra ethyl ammonium chloride
(4) ethane
18. One mole of NH3 is made to react with one mole of CH3Cl. Which of the following is/ are formed?
(1) CH3NH2
(2) (CH3)2NH
(3) (CH3)3NH
(4) All the above
19. The conversion of propan-1-ol to n-butylamine involves a sequential addition of reagents. The correct sequential order of reagents is
(1) i) SOCl 2 ii) KCN iii) H 2 /Ni, NaHg/ C2H5OH
(2) i) HCl2 ii) H2/Ni, Na(Hg)/C2H5OH
(3) i) SOCl2 ii) KCN iii) CH3NH2
(4) i) HCl ii) CH3NH2
20. Which of the following gives primary amine on reduction?
(1) CH3CH2NO2
(2) CH3CH2–O–N–O
(3) (CH3)2CNO2
(4) (CH3)2CHNO2
21. Ethylamine (C2H5NH2) can be obtained from N-ethyl phthalimide on treatment with
(1) CaH2 (2) NaBH4
(3) H2O (4) NH2NH2
16. Which of the following species is not formed in Hofmann bromamide reaction?
(1) R−CONHBr (2) RCONBr
(3) RNCO (4)
RCONBr2
22. The major product formed in the reaction given below is 1.NaOH3Br2 2.H3Oo o NH
(1)
NH2 COOH
(2) Br CONH2
(3) COOH CONH2
(4) o o o
Physical Properties of Amines
23. Arrange the following in the increasing order of their boiling points.
N-Ethylethanamine Butanamine N, Ndimethyl ethanamine
I II III
(1) III > II > I
(2) III > I > II
(3) II > III > I
(4) II > I > III
24. Identify the correct statements from the following.
(A) Lower aliphatic amines are gases.
(B) Lower aliphatic amines have fishy odour.
(C) Lower aliphatic amines are soluble in water.
(D) Pure aniline is a reddish brown liquid.
(1) A, B Only
(2) A, B, C Only
(3) A, C, D Only
(4) B, C Only
Chemical Reactions of Amines
25. Which of the following is highly basic?
(1) Diphenylamine (2) Benzylamine
(3) Aniline (4) Triphenylamine
26. The order of basicity of the compounds o N H H H N N N is
(1) IV > I > III > II
(2) I > III > II > IV
(3) III > I > IV > II
(4) II > I > III > IV
The percentages of X, Y, and Z formed are respectively
(1) 2%, 36%, 72% (2) 51%, 47%, 2% (3) 38%, 34%, 28% (4) 71%, 27%, 2% 28.
(major). The compound ‘C’ in the reaction is
29. Identify the product formed in the given reaction.
30. Which of the following cannot be detected by isocyanide test?
(1) C6H5−NH2 (2) C6H5−NH–CH3
(3) CH3-CH-CH3 NH2 (4) CH3-C-CH3
31. Aniline does not undergo (1) nitration
(2) sulphonation
(3) Friedel-Crafts reaction
(4) bromination
32.
33. Nitrosoamines R 2 N–N=O are soluble in water. On heating with conc. H 2SO4, they give secondary amines. The reaction is called as
(1) Perkin’s reaction
(2) Sandmeyer’s reaction
(3) Fittig reaction
(4) Liebermann nitroso reaction
34. A colourless, odourless, and non-combustible gas is liberated when ethyl amine reacts with (1) NaOH (2) CH3COCl (3) NaNO2+HCl (4) H2SO4
35. The major product formed in the reaction given below will be NH2 NaNO2-2HCl H2O
(1) OH (2) OH
(3) OH (4) OH
36. Identify X and Y of the following reaction sequence. i) H2/Pd i) Br2/CH3COOH ii) (CH3CO)2O, pyridine ii) H3OI iii) CHCl3, KOH,D X Y
X- Y(2) NHOCOCH3 NHOH X= Y= (3) X- Y= Br NHOCOCH3 NC (4) Br NHOCOCH3 NC X= Y-
37. Which among the following will not liberate nitrogen on reaction with nitrous acid?
(1) dimethylamine (2) 2-aminopropane (3) ethylamine (4) methylamine
38. The chemical formula of Hinsberg reagent is
(1) C6H5SO3H (2) C6H5NHSO2C6H5 (3) C6H5SO2Cl (4) C6H5NHCOC6H5
39. Which of the following does not give diazonium salt with nitrous acid at 273 K?
(1) Benzenamine
(2) Benzyl amine
(3) p-Hydroxy aniline
(4) o-Hydroxy aniline
40. Which of the following amines will give carbylamine test? (1) NHCH3 (2) N(CH3)2 (3) NHC2H5 (4) NH2
41. Benzene sulphonyl chloride forms a soluble salt in alkali, when it reacts with (1) primary amine (2) secondary amine (3) tertiary amine (4) nitrous acid
42. When aniline is heated with conc.H 2SO4 at 455-475 K, it forms (1) aniline hydrogen sulphate (2) sulphanilic acid (3) amino benzene sulphonic acid (4) benzene sulphonic acid
43. Which of the following organic compounds react with benzene suphonyl chloride (Hinsberg reagent) to give sulphonamide product which is insoluble in alkali (1) C2H5NH2 (2) (C2H5)2NH (3) (C2H5)3N (4) (C2H5)2NCH3
44. Aniline on treatment with HCl and NaNO2 at low temperature gives (1) aminophenol (2) chloroaniline (3) diazonium salt (4) nitroaniline
45. NH2
47. Para-nitro aniline in good yield is obtained by the following method.
(1) Aniline is treated with nitration mixture.
(2) Aniline is subjected to hydrolysis, acetylation, and nitration in succession.
(3) Aniline is treated with HNO 2 at low temperature.
(4) Aniline is subjected to acetylation, nitration and hydrolysis in succession.
48. The product formed from the following reaction sequence is NH2
(1) NaNO2+HCL
(1) H2O/water
(3) NaOH
(4)(i) CO2 ii)H+
49. Which of the following are examples of electrophilic aromatic substitution reactions?
A) Bromination of acetanilide
B) Coupling reaction of aryl diazonium salts
C) Diazotisation of aniline
D) Acylation of aniline
The correct answer is
(1) A, B
(2) A, C
(3) A, C, D
(4) A, B, D
50. Select the amine which provides insoluble compound in NaOH with Hinsberg reagent.
(1) Aniline
(2) N, N-dimethyl aniline
(3) N-Methyl aniline
(4) O-Toluidine
Diazonium Salts
51. 2 NaNO+HCl ¢ 252 0-5C CHNH(1 mole ) X →
(Organic compound). The weight of X obtained is
(1) one gram more than the weight of amine reacted
(2) 47.5 grams more than the weight of amine reacted
(3) 30 gram more than the weight of amine reacted
(4) 17.5 gram more than the weight of amine reacted
52. The following compound is stable at 298 K N2X+ .Then X is
(1) Cl (2) Br (3) BF 4 (4) HSO4
53. Which of the following will be the most stable diazonium salt RN 2+X ?
(1) CH3N2+X (2) C6H5N2+X (3) CH3CH2N2+X (4) C6H5CH2N2+X
54. Among the following, incorrect resonance structure of benzene diazonium ion is (1) N N:
(2) N N: :
(4)
55. The electrophile, E+ attacks the benzene ring to generate the intermediate σ -complex. Of the following, which σ -complex is of lowest energy? (1)
56. Which one of the following gives the mos t stable diazonium salt? (1) CH3CH2CH2NH2 (2) NHCH3
57. In a set of reactions, p-nitro toluene yielded a produce E.
Cyanides and isocyanides
58. Identify A, B and C in the sequence.
(1) CH3CH2CN, CH3CH2CH2NH2, CH3CH2CH2OH (2) CH3CH2CN, CH3CH2OH, CH3CH2CH2NH2 (3) CH3CH2CN, C2H5OH, C2H5N2Cl (4) CH3CH2CN, CH3CH2NH2, C2H5OH
59. The isomer of the end product (B) in the following sequence of reactions is NH2 CHCI3 LIA/H4 A B KOH H2O (1) NC (2) CH2NH2
(3) NH C H O (4) NH-CH3
60. The IUPAC name of is
(1) 1,2, 3-propanetrinitrile
(2) 1, 2, 3-tricyanopropane
(3) 3-cyano-1,5-dinitrilepentane
(4) 1, 2, 3-propanetricarbonitrile
61. 2 H/Ni Ethanolic 25 KCN A)CHClXY →→ 253 (P°Amine) (Excess) B)CHCl+NH®Z
The incorrect statement about X, Y and Z is
(1) co-valency of carbon of functional group in ‘X’ is 4.
(2) Y is propanamine
(3) compound X is C2H5NC
(4) Y and Z are homologues
62. 3 2 NH Br 3 Ä NaOH CHCOOHAB →→ 3 25 CHCl/KOH Na-EtOH CHOH, Ä CD;
D is (1) CH3NHCH3
(2) CH3N(CH3)2
(3) C2H5NHCH3
(4) (C2H5)2NH
63. Ethyl isocyanide on acidic hydrolysis gives (1) ethylamine and methanoic acid. (2) propanoic acid and ammonium salt. (3) ethanoic acid and ammonium salt.
(4) methylamine and ethanoic acid.
Level 2
Structure of Amines
1. The number of saturated isomeric primary amines possible for the molecular formula C3H5N is (1) zero
(2) 3
(3) 2
(4) 4
2. Which of the following contains imino group?
(1) Aniline
(2) O-Toluidine
(3) Benzylamine
(4) N-phenyl methanimine
3. Which of the following is a carbamide?
(1) CH3CONH2
(2) NH2CONH2
(3) CH2(NH2)CONH2
(4) CO(OH)NH2
4. The diamide of carbonic acid is (1) acetamide
(2) formamide
(3) benzamide
(4) urea
5. Which of the following compounds is a cyclopentyldiethylamine?
(3) N(C2H5)2
(4) NHC2H5 NHC2H5
Classification of Amines
6. Which of the following is a 1° amine?
(1) Tert. butylamine
(2) Dimethyl amine
(3) N–Methylaniline
(4) N, N–Dimethyl aniline
7. Which of the following is a secondary amine?
(1) Dimethylamine
(2) Aniline
(3) Isobutylamine
(4) Secondary butyl amine
8. Which of the following is not a tertiary amine?
(1) Tri ethyl amine
(2) Tri methyl amine
(3) 2-methyl propanamine-2
(4) N-ethyl-N-methyl propanamine-1
9. Which type of isomerism is exhibited by primary, secondary and tertiary amine having the same molecular formula (C3H9N)?
(1) chain (2) metamerism
(3) geometrical (4) functional
Nomenclature of Amines
10. IUPAC Name of (CH3)3CNH2 is (1) trimethyl amine
(2) 2-methyl butanamine
(3) 2-methyl butanamine
(4) 2-methyl prop-2-amine
11. The IUPAC name of (C2H5)2NCH(CH3)2
(1) N, N diethyl-2-propanamine
(2) N, N dimethyl amino isopropane
(3) N,N dimethyl-amino ethane
(4) Methyl-N-isopropyl methyl amine
12. IUPAC name of the following is N
(1) 3-dimethyl amino-3-methyl pentanamine
(2) 3(N,N trimethyl)-3-pentanamine
(3) 3-N,N-trimethyl pentanamine
(4) N, N dimethyl-3 methyl 3-pentanamine
13. The IUPAC name of is (1) Tricyclo propanamine
(2) N, N-di-cyclopropylamine
(3) N, N di-cyclopropyl cyclopropanamine
(4) N, N, N tricyclo propanamine
Preparation of Amines
14. Which of the following amines cannot be prepared by Gabriel phthalimide synthesis?
(1) CH3CH2−NH2 (2) (CH3)2CH−NH2
(3) Ph−NH2 (4) C2H5−NH2
15. Excess alkyl halide on heating with alc.NH3 in a sealed tube results in the formation of (1) 1° amine (2) 2° amine
(3) 3° amine (4) all of these
16. Me−Cl+NH3(Excess) → major product. The major product will be (1) MeNH2 (2) Me2NH (3) Me3N (4) Me4NCl
17. Which of the following reactions will NOT give primary amine as a product?
(1) 4 3 (i) LiAlH 32 (ii) HO CHCONH Product ⊕ →
(2) 2 Br/KOH 32 CHCONH Product →
(3) 4 3 (i) LiAlH 3 (ii) HO CHCN Product ⊕ →
(4) 4 3 (i) LiAlH 3 (ii) HO CHCN Product ⊕ →
18. Which of the following reactions does not yield an amine?
(1) R – X + NH3 →
(2) 25 Na RCHNOH[H] CHOH −=+
(3) H 2 RCNHO + + → (4) 4 LiAlH 2 RCONH4[H]−+→
19. KOBr NH2 Br o A(Major product) lAH4 H2O NH2 o Br B(Major product)
In the above reactions, products A and B respectively are
20. Which of the following amines can be prepared by Gabriel phthalimide reaction? (1) neo-pentylamine (2) n-butylamine (3) t-butylamine (4) triethylamine
21. A(C8C8Cl20) NH3 NaOH Br2 C8H8CINO H2N Cl CH3
Consider the above reaction. The compound ‘A’ is
22. Which of the following reaction does not involve Hoffmann bromamide degradation?
(1) CH2-C-NH3 i)Br2,NaOH/H+ ii)NH3/D iii)LiAH4/H20
(2) i)NH3 NaOH ii)Br2, NaOH
(3)
Br2, NaOH KOH, H20
23. Which rearrangement takes place by the formation of isocyanate as an intermediate compound?
(1) Gabriel phthalimide reaction
(2) Hoffmann ammonolysis
(3) Hoffmann-bromamide rearrangement
(4) Hinsberg’s reaction
Physical Properties of Amines
24. The incorrect statement about aniline is (1) It is less basic than ethyl amine.
(2) It is steam volatile.
(3) On reaction with Na, it gives H 2.
(4) It is highly soluble in water.
Chemical reactions of Amines
25. Which one of the following is the strongest base in liquid phase?
(1) (C2H5)2NH (2) C2H5NH2
(3) (C2H5)3N (4) NH3
26. The correct order of basic strength of the following are (I) NH2 (II) NH2
(1) I > IV > III > II (2) III > IV > I > II
(3) IV > III > I > II (4) II > IV > I > III
27. Arrange the following in the decreasing order of their basic strength.
(I) C6H5NH2 (II) C2H5NH2
(III) (C2H5)2NH (IV) NH3
(1) I > IV > II > III (2) III > II > I > IV
(3) II > III > IV > I (4) III > II > IV > I
28. Aniline on treatment with excess of bromine water gives
(1) Aniline bromide
(2) Ortho bromoaniline
(3) Para bromoaniline
(4) 2, 4, 6 tribromoaniline
29. Nitration of aniline in strong acidic medium also gives m – nitro aniline because (1) In absence of substituents, nitrogroup always goes to m – position.
(2) In electrophilic substitution reactions amino group is meta directive.
(3) inspite of substituents nitro group always goes to only m-position.
(4) In a strong acidic medium, aniline is present as anilinium ion.
30. NH2 NH2 HNO3 288K H2SO4
HNO2/H2O 2O alcohal
CHCl3/KOH(alc)
Consider the given reaction, percentage yield of
(1) (C) > (A) > (B) (2) (B) > (C) > (A)
(3) (A) > (C) > (B) (4) (C) > (B) > (A)
31. An organic compound ‘A’ having molecular formula C2H3N on reduction gave another compound ‘B’ Upon treatment with nitrous acid, ‘B’ gave ethyl alcohol. On warming with chloroform and alcoholic KOH, B formed an offensive smelling compound ‘C’. The compound ‘C’ is
(1) CH3CH2NH2
(2) CHCH32 NC =
(3) CH3–C ≡ N
(4) CH3CH2OH
32.
Propanenitrile
The conversion of A to B is an example of the reaction called as
(1) Reimer-Tiemann reaction
(2) carbyl amine reaction
(3) Stephen reaction
(4) Sandmeyer reaction
33. Which amine amongst the following will give positive carbylamine test?
(1) C6H5 – NH – CH3
(2) Me NH2
(3) C6H5 – NH – C4H9
(4) C6H5 – N(C2H5)2
isocvanide
Then C3H9N is (1) CH3– CH2 – CH2 – NH2 (2) (3) CH3 – NH – CH2 – CH3 (4) (CH3)3N
35. The following reaction is carried out. CH2NH2 HNO2 OoC
Which of the following alcohols is formed as a major product?
34. C3H9N
36. The incorrect statement regarding the reaction given below is Me-N-Me
NaNO2 + +
(1) The reaction occurs at low temperature.
(2) ‘B’ is N-nitroso ammonium compound.
(3) ‘B’ is N-nitro ammonium compound.
(4) The product ‘B’ formed in the above reaction is p-nitroso compound at low temperature.
37. HNO2 MajorProduct o-Phenylenediamine X → X is (1) NH (2) N N N H (3) N = N N2 + + (4) N2 NH2 +
38. CH2 - NH2 HNO 2 A A is (1) OH (2) OH (3) (4)
39. The product P (major) of the following reaction is HO CH2 NH2 * (1) * O (2) * O (3) O * (4) * CH = 0
40. Which of the following reactions take place by the formation and rearrangement of carbocation? (1) 22 4 CH3 Hg(OACI/HO) 2 NaBH I R-CH-CH=CH → (2) 2 22 CH3 BH HO I R-CH-CH →
CH3 HC 32 I CH-CH=CH →
HNO2 NH 2 OH
41. CH2 NH 2 HNO2 → The major product formed is ____.
(1) CH2 OH (2) CH 2
46. Which one of the following gives foul smell in carbylamine test but does not give a stable diazonium salt with ice cold nitrous acid?
(1) Aniline
(2) 4-aminobenzene carbaldehyde
(3) O-methoxy aniline
(3) OH (4)
42. Which reagent distinguishes aniline and ethylamine?
(1) CHCl3/KOH (2) CS2, HgCl2
(3) C6H5SO2Cl (4) HNO2
43. Aniline in a set of reactions yielded a product D.
The structure of the product D would be
(1) C6H5CH2NH2 (2) C6H5NHCH2CH3 (3) C6H5NHOH (4) C6H5CH2OH
44. How many of the following compounds give foul smell when heated with chloroform and alcoholic caustic potash? n-butyl amine, 3° butyl amine, isopropyl amine, benzylamine, aniline, and dimethylamine (1) 5 (2) 2 (3) 4 (4) 3
45. Which one of the following compounds form an alkali insoluble derivative with Hinsberg reagent?
(1) NH 2
(2) NH 2
(3) N (4) NH
(4) Benzylamine 47. N + 2 / ClCu/HCl X ........(I) Cl2 /FeCl 3 X ........(II) dark The attacking species in (I) and (II) are (1) Cl , Cl (2) Cl , Cl+ (3) Cl , Cl (4) Cl , Cl
48. The number of resonance structures possible for 2 3653 CHCHNH and CHNH ++ =−− are respectively (1) 2, 0 (2) 2, 2 (3) 2, 5 (4) 0, 2
49. Benzene sulphonyl chloride is (1) Collins reagent (2) Grignard reagent
(3) Hinsberg reagent
(4) Hoffmann reagent
50. Consider the following sequence of reactions NO 2 2 CHCl/KOH Sn/HCl AB→→=
The final product B is
(1)
3 (2)
(3)
(4)
Salts
51. Stable diazotisation means the conversion of (1) any primary amine into diazonium salt using NaNO 3 + HCl at ice cold temperature (2) aromatic primary amine into diazonium salt using NaNO2 + HCl at 60-70°C (3) aromatic primary amine into diazonium salt using NaNO 2 + HCl at ice cold temperature
(4) any primary amine into diazonium salt using NaNO 3 + HCl at ice cold temperature
52. In the diazotisation of aniline, the reagent or reagents used is/are (1) HNO3, HCl (2) HNO2 only (3) NaNO2, HCl at 0–5°C (4) NaNO2, HNO2 at 0–5°C
53. In benzene diazonium chloride, the functional group is (1) −N = N=Cl (2) −N = N+=Cl–(3) −N+ = N=Cl– (4) +•• -NN ≡
54. Reactivity in coupling with diazonium salt is maximum when the substrate is (1) (2) Cl
(4)
55. 1-naphthylamine and sulphanilic acid in acetic acid is used for the detection of (1) NO3 (2) NO (3) N2O (4) NO2–
56. Which one of the following nitro compounds does not react with nitrous acid? (1) H3 C C H2 NO2 C H 2 (2) H3 C C H 2 NO2 CH H3 C (3) H 3C - C - NO 2 H 3C H 3C (4)
Cyanides and Isocyanides
57. Benzyl isocyanide can be obtained by (A) CH 2Br AgCN → (B) . CH 2NH 2 3 CHCl AgKOH → (C) CH 2 - NH - CH 3 3 CHCl AqKOH →
Diazonium
(3)
Choose the correct answer from the options given below:
(1) Only B (2) A and C
(3) A and B (4) A and D
58. The number of mono chloro derivatives that the compound ‘D’ can give
(1) 3
(2) 4
(3) 2
(4) 1
FURTHER EXPLORATION
1. Identify the most probable product (M) in the following reaction.
59. The compounds A and B in the following reaction are, respectively:
2. The increasing order of the pK b of the following compounds is
(1) A = Benzyl alcohol, B = Benzyl isocyanide
(2) A = Benzyl alcohol, B = Benzyl cyanide
(3) A = Benzyl chloride, B= Benzyl cyanide
(4) A = Benzyl chloride, B= Benzyl isocyanide
+ 1 3 + 3 HO RMgX HO QR–CNP
. P and Q respectively are (1) 1°−alcohol and 1°−amine
(2) Ketone and acid
(3) Aldehyde and 1°−alcohol
(4) Acid and ketone
(1) (C) > (A) > (D) > (B)
(2) (B) > (D) > (A) > (C)
(3) (A) > (C) > (D) > (B)
(4) (B) > (D) > (C) > (A)
3. The product of the following reaction sequence is NH 3 3 + 3 2
(4) Br Br Br Br
4. The major product obtained in the following reaction is CN O NH 2 3 2 (i) CHCl/KOH (ii) Pd/C/H
CN OH NCHCl 2 H
O NCH 3 H
The possible products of the reaction are (p) NH 2 CH3 (q) NH 2 15
(r) 15 CH 3 NH 2 (s) NH 2
(1) p and q only
(2) p, q, and r are possible
(3) p, q, r, and s are possible
(4) q, r and s are possible
Matching Type Questions
1. In the reduction of nitro benzene, match the List-I with List-II.
List –I (Medium) List –II (Product formed)
(A) Neutral (p) Aniline
(B) Acidic (q) Azobenzene
(C) Alkaline (r) N-phenyl hydroxylamine
(A) (B) (C)
(1) r q p
(2) q p r
(3) p p r
(4) r p q
2. Match the List-I with List-II.
List-I List-II
(A) Amide (p) Carbylamine reaction
(B) Nitrile (q) Hinsberg reagent
(C) C6H5SO2Cl (r) Hoffmann bromamide reaction
(D) 1°-Amine (s) LiAlH 4
(A) (B) (C) (D)
(1) p q r s
(2) r s q p
(3) r q s p
(4) p r q s
3. Match column-I(Reaction) with columnII(products).
Column I Column II
(A) Gabriel synthesis (p) Benzaldehyde (B) Kolbe reaction (q) Ethers
(C) Williamson synthesis (r) Primary amines
(D) Etard reaction (s) Salicylic acid
(A) (B) (C) (D)
(1) r p q s
(2) q r p s
(3) s s p r
(4) r s q p
4. Match column-I with column-II.
Column I
(compound)
Column-II (pKb)
(A) NH3 (p) 4.75
(B) CH3–NH2 (q) 3.38
(C) NH 2 (r) 9.38
(A) (B) (C)
(1) p q r
(2) q r p
(3) r p q
(4) r q p
Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II)
(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 : The bond angle in trimethyl amine is 108°.
S-II : In trimethyl amine, nitrogen is sp 3 hybridised.
2. S-I : In Hoffmann degradation reaction, the migration of only an alkyl group takes place from carbonyl carbon of the amide to the nitrogen atom.
S-II : The group is migrated in Hoffmann degradation reaction to an electron deficient atom.
3. S-I : Aniline cannot be prepared by Gabriel phthalimide synthesis.
S-II : Aniline is a primary aryl amine.
4. S-I : N-phenyl ethanamide is more basic than N-methyl benzene carboxamide.
S-II : Aniline reacts with acetic acid to form acetanilide.
5. S-I : Functional group in acetanilide is an amide .
S-II : In aniline – NH2 group is an ortho, para directing group, but ring deactivating group.
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
1. (A) : Aniline is more stable than anilinium ion.
(R) : Aniline is more resonance stabilised than anilinium ion.
2. (A) : In a Hoffmann bromamide reaction, the amine formed has one carbon atom less than the parent 1° amide.
(R) : N-methyl acetamide undergoes Hoffmann bromamide reaction.
3. (A) : Gabriel phthalimide synthesis cannot be used for the preparation of aromatic primary amines.
(R) : The C–X bond in aryl halides possesses partial double bond character and hence X–cannot be replaced by a pthalimide ion.
4. (A) : When acetamide reacts with NaOH and Br2, methyl amine is formed.
BRAIN TEASERS
1. IUPAC name of
is
(1) N, N–dimethyl-2–Pentanamine
(2) N–ethyl–3–methyl–2–Pentanamine
(3) N(1–methylethyl)–3–methyl–2–Pentanamine
(4) N–ethyl–3–methyl-1-Pentanamine
2. The structure of the product (A) is OH
(R) : The reaction occurs through the intermediate formation of an intermediate called isocyanate.
5. (A) : Order of basicity of the following amines in the gaseous phase is NH 3 > primary amine > secondary amine > tertiary amine.
(R) : In the gaseous phase, the basic nature of aliphatic amine decreases with an increase in the number of alkyl groups.
6 (A) : When aniline is subjected to nitration by conc. HNO 3 & H 2SO4 meta nitro aniline is formed in considerable amount.
(R): NH 2 is o/p directing but ring deactivating group.
7. (A) : Aniline does not undergo Friedel Crafts alkylation and acylation.
(R) : Aniline forms salt with AlCl 3 the lewis acid which is used as a catalyst in Friedel Crafts reaction.
3. Major product of the following reaction is
(4) N O
4. Match the column-I with column-II
Column-I (Reactant) Column-II (Reaction)
(A) CH3CH2CN (p) Undergoes electrophilic substitution
(B) CH3CH2NC (q) Hydrolysis give primary amine
(C) NC (r) Reduction give primary amine
(D) CN (s) Hydrolysis give formic acid
(t) Reduction give a secondary amine
(A) (B) (C) (D)
(1) r qst pr pqst
(2) r qst pqst pr
(3) pr r qst pqst
(4) pr r pqst qst
5. The major product of the following reaction is
6. Repeated Hoffmann elimination reaction (exhaustive methylation followed by heating with AgOH) will often remove a nitrogen atom from an amine molecule. Which of the following compounds is most likely to be the product in this case?
Haffmann Elimination
FLASH BACK (Previous NEET Questions)
1. Given below are two statements:
Statement I : Primary aliphatic amines react with HNO2 to give unstable diazonium salts.
Statement II : Primary aromatic amines react with HNO2 to form diazonium salts, which are stable even above 300 K.
In light of the above statements, choose the
CHAPTER TEST
Section-A
1. The primary amino group is absent in
(1) p-amino phenol
(2) o-amino phenol
(3) N-methyl ethanolamine
(4) phenylamine
2. N, N-dimethylbutanamine-2 contains (1) six sp3 hybridised carbon atoms. (2) seven sp3 hybridised atoms. (3) two sp3 hybridised nitrogen atoms. (4) both (1) and (2) are correct.
3. One mole of an amine (A) consumes two moles of methyl bromide to give a quaternary ammonium salt. The amine (A) is
(1) (CH3)3CCH2NH2
(2) (CH3)2NCH2CH3
(3) CH3NCH2CH3
(4) N(CH2CH3)3
4. The number of primary, secondary and tertiary amine isomers possible (structural isomers) for the compound with a formula C4H11N are
(1) 4, 3, 1
(2) 4, 3, 2
(3) 3, 2, 1
(4) 4, 2, 1
most appropriate answer from the options given below:
(1) Both Statement I and Statement II are incorrect
(2) Statement I is correct but Statement II is incorrect
(3) Statement I is incorrect but Statement II is correct
(4) Both Statement I and Statement II are correct
5. Which of the following is not a primary amine?
(1) Dimethylamine
(2) Aniline
(3) Isobutylamine
(4) Secondary butyl amine
6. Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
(A) : Pyrrole is an aromatic heterocyclic compound.
(R) : It has cyclic delocalised 6π electrons. 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 and (R) is not the correct explanation of (A)
(3) (A) is true but (R) is false
(4) Both (A) and (R) are false
7. For a carbyl amine reaction, we need alcoholic KOH,
(1) any primary amine and chloroform
(2) aromatic primary amine and chloroform
(3) aliphatic primary amine and chloroform
(4) any amine and chloroform
8. Given below are two statements. One is labelled as Assertion (A) and the other is
labelled as reason (R).
(A): Mono bromination of aniline is done by protecting the amine group by acylation and followed by hydrolysis of a substituted amine.
(R) : Activating effect of -NHCOR is less than that of –NH2 group.
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.
9. Nitrogen containing organic compounds find many applications in daily life. Which of the following is a wrong match about them?
(1) Novocain….Anaesthetic in dentistry
(2) Benadryl…..Antihistamine
(3) Adrenaline …..Emergency hormone
(4) Cetyl trimethylammonium bromide…. Anionic soap
10. 4 LiAlH 252 CH-NOX →
4 LiAlH 652 CH-NOY → . X and Y are respectively
(1) primary amine, azo compound
(2) primary amine, primary amine
(3) azo compound, azo compound
(4) azo compound, primary amine
11. IUPAC name of (CH3)3C.NH2 is (1) trimethyl amine
(2) 2-methyl butanamine-1
(3) 2-methyl propanamine-2
(4) 2-methyl propanamine-1
12. Correct IUPAC name of ortho-toluidine is (1) 2-methyl anisole
(2) 2-methoxy aniline
(3) 2-methyl benzene amine
(4) 2 methyl benzyl amine
13. The structure of N, N-dimethyl - 3hexanamine is (1)
(2)
(3)
(4)
14. IUPAC name of NH2 is
(1) Pent-4-en-1-amine
(2) Pent-1-en-5-amine
(3) 4-Pentenamine
(4) 1-en-5-pentanamine
15. Nitrobenzene on reduction with hydrogen in the presence of nickel gives
(1) azobenzene
(2) hydrazobenzene
(3) phenyl hydroxyl amine
(4) aniline
16. Which of the following is hydrolysed to give secondary amine?
(1) Alkyl cyanide
(2)
(3) Nitro paraffins
(4) Acid amide
17. Industrial method and most economic method to prepare aniline from nitrobenzene is (1)
(2)
Fe/CH3COOH (3)
H2/Pt (4)
Fe + little HCl H2O
18. Reaction of propanamide with ethanolic sodium hydroxide and bromine will give (1) aniline (2) ethylamine (3) methylamine (4) propylamine
19. Treatment of ammonia with excess of ethyl iodide will yield (1) diethylamine (2) ethylamine (3) triethylamine (4) tetraethylammonium iodide
20. Nitrobenzene on reduction with Sn and HCl gives (1) 1° amine (2) 2° amine (3) 3° amine (4) amide
21. Acetonitrile on reduction gives (1) propanamine (2) methanamine (3) ethanamine (4) propane nitrile
22. On reduction, secondary amine is given by the compounds (1) Nitrobenzene (2) Methyl Cyanide (3) Nitroethane (4) Methyl Isocyanide
23. Identify the product (D) in the following reaction sequence.
(1) N(CH)32 332 32 I (CH)C-CH-CH(OH)-N(CH)
(2) (CH3)3C−CH2−CH(OH)−N(CH3)2
(3) (CH3)3C−CH2CN
(4) (CH3)3C−CH2−CH2N(CH3)2
24. Amongst the compounds given, the one that would form a brilliant coloured dye on treatment with NaNO2 in dil.HCl followed by the addition to an alkaline solution of β-naphthol is (1) N(CH3)2 (2) NHCH3 (3) H3C NH2 (4) CH2 NH2
25. In the given set of reactions, what is the compound D is?
dc.KOH B-Br NaOH(Br) + D (amine) COONa COONa Phthailmide B C
(1) CH3−NH2
(2) CH3CH2NH2
(3) CH3–CH2–NH–CH3
(4) (CH3)2NH
26. Gabriel’s synthesis is used for the preparation of
(1) primary aromatic amines
(2) primary aliphatic amines
(3) primary aliphatic amines
(4) tertiary amines
27. The major product of the following reaction is
28. In a set of reactions, m-bromobenzoic acid gave a product D. Identify the product D. COOH Br A
29. Predict the compound ‘X’ the following reaction.
22 23 2 Ph-CO-NH+Br+NaOHX+NaCO+2NaBr+2HO ∆ → 22 23 2 Ph-CO-NH+Br+NaOHX+NaCO+2NaBr+2HO ∆ → (1) (2) NH2 (3) N H (4) CH3NH2
30. Mixture of primary, secondary, and tertiary amines are separated by Hinsberg test. The reagent used is
(4) SOCl2
31. Wh ich of the following reaction does not involve Hoffmann bromamide degradation?
(1)
CH3-C-CH3 CH3-CH3 i)Br2, NaOH/H+ ii)NH3/ D iii)LiAH4/H2O 0 (2) Cl NH2 i)NH3-NaOH i)Br2-NaOH
(3) CH2-C-NH2 CH2-CH2 0 i)Br2, NaOH (4) CN CH2-CH2 Br2, NaOH KOH3-H2O
32. The product (P) of the following reaction is CH2 CONH2 COOCH3 i) Br2/NaOH ii) P is
33. Which of the following reactions is appropriate for converting acetamide to methanamine?
(1) Hoffmann hypobromamide reaction
(2) Stephen’s reaction
(3) Gabriel phthalimide synthesis
(4) Carbylamine reaction
34. The correct decreasing order of the basic nature of the following compounds is
(1) P > Q > R > S
(2) S > R > Q > P
(3) S > P > R > Q
(4) R > Q > P > S
35. Write the basicity order for the following compounds.
(1) III > II > I (2) II > I > III
(3) II > III > I (4) III > I > II
Section-B
36. The correct order in the aqueous medium of basic strength in case of methyl substituted amines is
(1) Me2NH > MeNH2 > Me3N > NH3
(2) Me2NH > Me3N > MeNH2 > NH3
(3) NH3 > Me3N > MeNH2 > Me2NH
(4) Me3N > Me2NH > MeNH2 > NH3
37. Arrange the following amines in the decreasing order of basicity
39. Which of the following is most basic in the aqueous solution?
(1) III > II > I (2) III > I > II (3) I > III > II (4) I > II > III
38. The correct order of the basic strength of following compounds is
40. The product formed from the following reaction sequence is NH2
(i) (CH3CO)2O, pyridine (ii) LiAIH4 (iii) H2O
42. Most reactive towards electrophilic substitution is (1) aniline
(2) aniline hydrochloride (3) nitrobenzene (4) N-acetyl aniline
43. A positive carbylamine test is given by (1) N, N – dimethyl aniline (2) N-methyl ethanamine
(3) N – methyl–2–methyl aniline
(4) P – methyl benzylamine
44. The compound C5H13N is optically active. The compound is
(1) N-methylbutanamine
(2) 2-aminopentane
(3) 1-amino pentane
(4) N,N1-dimethyl propanamine
45. The question has two statements. Statement I and statement II.
Statement I : Coupling of N2Cl with aniline is faster than with phenol.
Statement II : Aniline is more electron donating than phenol.
In light of the above statements, choose the correct answer from the options given below.
(1) Both statement I and statement II are correct.
(2) Both statement I and statement II are incorrect.
(3) Statement I is correct but statement II is incorrect.
(4) Statement I is incorrect but statement II is correct.
46. Aniline can be separated from phenol using (1) NaHCO3
(2) NaNO2 + HCl at 0°C
(3) NaCl
(4) Acidified KMnO4
47. The major product X in the reaction is
NH2
1) CHCl3 / KOH 2) Pd / C / H2 X
48. The product (B) in the reaction
The correct structure of B is (1) CH3−CH2−CH2−NH2
(2) O II 322 CH-CH-C-NH (3) CH3−CH2−COOH (4) CH3−CH2−NH2
50. Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
(A) : Aniline cannot be prepared by Gabriel phthalimide synthesis.
(R) : –NH 2 group in aniline is highly activating towards electrophilic substitution.
In the light of the given statements, choose the correct answer from the options given.
(1) Both (A) and (R) true, and R is 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
ANSWER KEY
Further Exploration
Matching Type Questions
Statement Type Questions
Assertion and Reason Type Questions
Brain Teasers
Flashback (1) 2
Chapter Test
Chapter Outline
12.1 Carbohydrates
12.2 Proteins
12.3 Enzymes
12.4 Vitamins
12.5 Nucleic Acids
12.6 Hormones
Living systems are made up of various complex biomolecules, like carbohydrates, nucleic acids, proteins, lipids, etc. Carbohydrates and proteins are essential constituents of our food. These biomolecules interact with each other and constitute the molecular logic of life processes. In addition, some simple molecules like vitamins and mineral salts also play an important role in the functions of organisms.
12.1 CARBOHYDRATES
Carbohydrates are mainly the compounds of C, H, and O. Initially, the carbohydrates were considered as the hydrates of carbon with formula, Cx(H2O)y
Examples: Glucose: C 6 H 12 O 6 or C 6 (H 2 O) 6 ; Sucrose: C12H22O11 or C12(H2O)11
But all the compounds with formula Cx(H2O)y are not necessarily carbohydrates.
Examples: Formaldehyde: HCHO or C(H2O); Acetic acid: CH3COOH or C2(H2O)2.
A few carbohydrates may not have the formula, Cx(H2O)y. e.g., rhamnose, C6H12O5.
Most of the carbohydrates are sweet to
BIOMOLECULES
taste. Hence, these are also called saccharides. In Greek; saccharon means sugar.
Carbohydrates are now defined as optically active polyhydroxy aldehydes or polyhydroxy ketones or the compounds that produce such units on hydrolysis.
12.1.1 Classification of Carbohydrates
Based on the hydrolysis products of carbohydrates, these are classified into 3 types.
Monosaccharides: These are single unit carbohydrates and cannot be broken into lower sugars during hydrolysis. About 20 monosaccharides occur in nature. Glucose and fructose are the most important members of this class.
Disaccharides and oligosaccharides: The disaccharides, on hydrolysis, give two monosaccharides of same or different kind.
Examples: Sucrose, maltose, lactose, etc.
Oligosaccharides on hydrolysis give two to ten monosaccharides. Disaccharides may also be treated as oligo saccharides.
Examples: hydrolysis Raffinosefructose glucose galactose →++ hydrolysis Gentianosefructoseglucoseglucose →++
Polysaccharides: Polysaccharides on hydrolysis give a large number of same or different monosaccharides, generally more than ten. The general formula is (C 6H10O5)n
Examples: Starch, cellulose, glycogen, dextrin, gums, etc.
Depending on the solubility in water, carbohydrates can be classified into two types.
Natural mono, di and oligosaccharides are crystalline solids, soluble in water, sweet to taste, and are called sugars. Polysaccharides are colourless amorphous solids and are generally insoluble in cold water. They are tasteless, and are generally called non-sugars.
Based on the reducing nature of carbohydrates, these are again classified into reducing sugars and non-reducing sugars.
Carbohydrates which reduce Tollens’ reagent and Fehling’s solution are called reducing sugars. They form silver mirror with Tollens’ reagent and give red precipitate with Fehling’s solution.
All monosaccharides and disaccharides, except sucrose, are reducing sugars. Saccharides which cannot reduce Fehling’s solution and Tollens’ reagent are called nonreducing sugars. In disaccharides, if the reducing groups of monosaccharides are bonded, then they are non-reducing sugars.
Example: Sucrose
Sugars can be classified into D– and L– forms based on their configuration. The enantiomer which rotates the monochromatic light to right is written as (+) or ‘d’ and the other which rotates the monochromatic light to the left is written as (–) or ‘l’. The direction of the rotation of monochromatic light can be denoted by (+) and (–), but cannot indicate the arrangement of –OH and –H around chiral carbon atom.
Rosanoff proposed a system to designate the stereo chemistry of carbohydrates by
considering the simplest sugar, glyceraldehyde, as standard. The sugars having the same configuration as D–glyceraldehyde at the least priority chiral carbon adjacent to primary alcoholic group(–CH2OH) are called D–sugars and those having the configuration as L–glyceraldehyde are called L–sugars.
Practically, D–sugars may be D–(+) or D–(–) and L–sugars may be L–(+) L–(–). The symbol (+) or ‘d’ is used for dextro and (–) or ‘l’ is used for laevorotatory compound. It is observed that natural glucose, ribose, and fructose are in D–form.
12.1.2 Monosaccharides
Monosaccharides are two types. The optically active polyhydroxy aldehydes are called aldoses and optically active polyhydroxy ketones are called ketoses. Based on the number of carbon atoms, carbohydrates are called trioses, tetroses, pentoses, hexoses, etc., when the number of carbon atoms present is 3, 4, 5, 6, etc. This classification is given in Table 12.1
Glucose
Glucose is called grape sugar because grapes contain nearly 20% glucose. It is also present in fruits and honey.
It occurs in free form and in combined form as starch, cellulose, and sucrose. It is probably the most abundant organic compound on earth.
Table 12.1 Examples of aldoses and ketoses
Glucose occurs as dextrorotatory compound in nature, so it is called dextrose. Human blood contains a normal range of 65–110 mg glucose per 100 ml. It is also called blood sugar.
Preparation
of Glucose
In the laboratory, glucose is prepared by the acid hydrolysis of sucrose in the presence of alcohol. Glucose and fructose are formed in equal amounts. Glucose, being less soluble in ethyl alcohol than fructose, crystallises out. C12H22O11
6H12O6 + C6H12O6
Glucose is prepared on a large scale by acid hydrolysis of starch or cellulose at 120°C under a pressure of 2–3 atm.
(C6H10O5)n + n H2O
Structure of Glucose
The molecular formula of glucose was found to be C6H12O6.
Glucose forms glucose pentaacetate with acetic anhydride, which exists as a stable compound. It indicates the presence of five –OH groups in glucose molecule and the five hydroxyl groups should be attached to different carbon atoms. Out of these five –OH groups, one was found to be primary and the remaining four are secondary.
612632 34 23 CHO | CHO+(CHCO)(CHOCOCH) | CHOCOCH →
Glucose forms glucose oxime with NH2–OH and glucose cyanohydrin with HCN.
61262 4 2 CHNOH | CHOHNOH(CHOH) | CHOH = +→
6126 4 2 HOCHCN | CHOHCN(CHOH) | CHOH +→
These reactions indicate that glucose has one carbonyl group.
Glucose gets oxidised to six carbon monocarboxylic acid (gluconic acid) on reaction with mild oxidising agent, like bromine water or an alkaline solution of iodine. Glucose reduces Tollens’ reagent to metallic silver and also Fehling’s solution to reddish brown cuprous oxide. All these are mild oxidising agents. These reactions indicate that the carbonyl group present in glucose is an aldehydic group.
6126 4 COOH | CHOmildoxidant(CHOH) | Gluconcacid +→
On oxidation with strong oxidising agents like nitric acid, glucose as well as gluconic acid both yield a dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic group in glucose.
6126 4 4 2 COOHCOOH || CHO(CHOH)(CHOH) || COOHCHOH
GlucoseSaccharicacidGluconicacid →→
Glucose is reduced to sorbitol, a hexahydric alcohol, with hydrogen in presence of nickel.
C6H12O6+H2 Ni → HOH2C(CHOH)4CH2OH
Glucose, on prolonged heating with HI gives n–hexane.
C6H12O6 HI heat → CH3CH2CH2CH2CH2CH3
The above two reactions suggest that the six carbon atoms in glucose are linked in an unbranched linear chain.
When glucose is treated with dilute sodium hydroxide solution, it undergoes reversible isomerisation, resulting in the formation of a mixture of D–glucose, D–fructose, and D–mannose. This reaction is known as Lobry de Bruyn van Ekenstein’s rearrangement.
■ Glucose does not react with Schiff’s reagent, NaHSO3, and NH3, even though it contains an aldehydic group.
■ Glucose pentaacetate does not react with hydroxylamine, indicating the absence of free aldehydic group.
■ Glucose forms two isomeric methyl glucosides (α– and β–) on heating with methyl alcohol in the presence of dry HCl gas.
Mechanism:
Glucose, when heated with excess of phenyl hydrazine, a dihydrazone, known as glucosazone, is formed. On the basis of these experimental observations, glucose was given the open chain structure by Baeyer.
■ Glucose exists in two stereo isomeric forms: α– and β– forms. These two forms differ from each other in the stereo chemistry at C–1. The aqueous solution of glucose shows mutarotation.
Glucose crystallised from concentrated solution at 30°C gives α–D(+) glucose with melting point 146°C and [α]D = +111°. When pure crystals of α–D(+) glucose are dissolved in water, the specific rotation gradually decreases from +111° to +52.5°.
Glucose is (2R, 3S, 4R, 5R)–2, 3, 4, 5, 6-pentahydroxyhexanal.
D–glucose and D–mannose are diastereomers. They differ in the configuration only at C–2 carbon and are known as epimers. Similarly, D-glucose and D-galactose are C–4 epimers.
Glucose crystallised from hot saturated aqueous solution above 98°C gives β–D(+) glucose with melting point 150°C and [α]D = +19.2°. When pure crystals of β–D(+) glucose are dissolved in water, the specific rotation gradually increases from +19.2° to +52.5°.
The change in optical rotation is made fast by the addition of traces of acid or base. The spontaneous change of specific rotation of an optically active compound in solution with time, to an equilibrium level, is called mutarotation.
α–D(+) equilibrium β–D(+) glucose mixture glucose
[α]D=+111° (36% α + 64% β) [α]D = +19.2°
[α]D = +52.5°
Cyclic Structure of Glucose
The open chain formula of glucose accounts for most of the reactions satisfactorily but fails to explain the following reactions.
This mutarotation cannot be explained by open chain structure of glucose, but can be explained by cyclic structure.
Generally, alcoholic groups undergo rapid and reversible addition to aldehyde group to form hemiacetals. The alcoholic group bonded
to C–5 of glucose reacts intramolecularly with –CHO, forming a 6-membered hemiacetal ring. The asymmetric carbon, now at C–1, gives two optical isomers. They are not mirror images of each other and, hence, they are diastereomers. They differ in the configuration only at C–1 and are called anomers. The two cyclic forms exist in equilibrium with Fischer chain structure, as shown below.
Pyran
Furan
The Haworth horizontal structure of glucopyranose is identical to the Fischer vertical projection structure. The groups present on the right side in Fischer formula are written below the plane of the ring and those on the left side are written above the plane.
α–D–(+)–
The cyclic structure of glucose explains the presence of α–and b – forms, mutarotation. It explains the inability of glucose to form aldehyde ammonia and bisulphite compound. In the presence of other carbonyl reagents, the ring is opened and free aldehyde group is produced, which can react with those reagents.
Glucose is represented by Fischer, as shown below:
b –D–(+)– glucose
Glucose forms a six-membered ring pyranose containing 5 carbon atoms and one oxygen atom, like pyran. The five-membered ring formed like furan is called furanose. Glucose is present in pyranose form only, as shown in Fig. 12.1
Fructose
Fructose is a ketohexose. It is called fruit sugar since it is more abundant in ripe fruits. The naturally occurring fructose is laevorotatory, so it is called laevulose. It is present in honey and cane sugar along with glucose in combined form. It is the sweetest of all the sugars. Inulin, on acidic hydrolysis, gives only fructose. Fructose belongs to D–series and is laevorotatory. Hence, it is written as D–(–)–fructose.
Sucrose, on hydrolysis in the presence of acid, gives glucose and fructose. C12H22O11+H2O
Structure of Fructose
The molecular formula of fructose was found to be C6H12O6
Fig 12.1. a -D-(+) and b -D-(+) Glucopyranoses
Fructose contains five hydroxyl groups, out of which two are primary and three are secondary.
Fructose contains a carbonyl group and it was found to be ketonic from its oxidation products with a strong oxidising agent.
Fructose was found to contain ketonic functional group at second carbon atom and all the six carbon atoms are in unbranched chain, as in the case of glucose.
Since fructose and glucose form identical osazones when heated with excess of phenyl hydrazine, it was found that both glucose and fructose have the same configuration at C–3; C–4 and C–5.
Like glucose, fructose also shows mutarotation.
Though fructose does not contain an aldehydic group, it behaves as a reducing sugar due to Lobry de Bruyn van Ekenstein rearrangement.
Unlike glucose, fructose has cyclic furanose structure.
The a–and b–forms of fructose are anomers at C–2.
The open chain structure of fructose is as shown below.
The Fischer and Haworth structures of –D–fructose and b–D–fructose are shown in Fig. 12.2.
–D–(–) Fructofuranose
1 2 3 4 5 6 OH O
b –D–(–) Fructofuranose
Fig. 12.2. Cyclic structures of fructose
12.1.3 Disaccharides
A disaccharide, on hydrolysis, gives two monosaccharides. Similarly, a trisaccharide, on hydrolysis, gives three monosaccharides. Oligosaccharides, on hydrolysis, give 2 to 10 monosaccharide molecules of same or different kinds.
In oligo and polysaccharides, monosaccharide units are linked through oxygen atom by the loss of a water molecule. This linkage is called glycosidic linkage. For example, in sucrose, C–1 of α–D–glucose and C–2 of b –D–fructose are linked through glycosidic linkage.
Sucrose
Molecular formula of sucrose is C 12H 22O 11. Sucrose is present in all photosynthetic plants. It is mainly present in sugar cane, beet root, etc. It is a disaccharide consisting of glucose and fructose. Naturally available sucrose is a dextrorotatory substance but does not show mutarotation. It is a colourless and odourless crystalline substance, which is highly soluble in water.
Inversion of Cane Sugar
The acid hydrolysis of sucrose produces equimolar mixture of D–(+)–glucose and D–(–)–fructose.
Sucrose HCl → D–(+)–glucose + D–(–)–fructose
[a]D = +66.5° [a]D = +52.5° [a]D = –92.4°
Sucrose in an aqueous solution is dextrorotatory with [ a ] D = +66.50. The net
specific rotation of equimolar mixture of D–glucose and D–fructose is 0 52.592.4 20 2 +−
The laevorotation of fructose (–92.4°) is more than dextrorotation of glucose (+52.5°). Hence, its aqueous solution is laevorotatory and sign changes from ‘d’ (+66.5) to ‘l’ (–20°). This process is called inversion of cane sugar and the product is named invert sugar.
In sucrose, glucose is present in pyranose form and fructose is present in furanose form with α–glycosidic linkage. So, sucrose is known as α–D–glucopyranose and b –D–fructofuranose. The Haworth structure of sucrose is given in Fig. 12.3. 1
Fig. 12.4. Structure of lactose
Lactose emulsin → D-galactose + D-glucose
Maltose
Maltose is present in sprouted barley seeds. Its molecular formula is C12H22O11. It is obtained from starch by hydrolysis, using the enzyme diastase. Diastase is present in malt. Maltose, on hydrolysis, gives two glucose units.
(C6H10O 5)n+H2O diastase → (Starch)
C12H22O 11 maltase → 2 C6H12O 6 (Maltose) (Glucose)
Fig. 12.3. Structure of sucrose
A glycosidic linkage is formed between C–1 of a –glucose and C–2 of b –fructose. Sucrose is a non-reducing sugar, because hemiacetal hydroxyl groups of both glucose and fructose are not free. They are involved in the formation of glycosidic linkage.
Lactose
Lactose is called milk sugar because it is mainly present in milk. Its molecular formula is C12H22O11. Lactose is a disaccharide of b-Dglucose and b -D-galactose. The two units are linked by b -glycosidic linkage with C-1 of galactose and C-4 of glucose molecule. The enzyme emulsin catalyse hydrolysis b -glycosidic linkage. It is a reducing sugar, because it has one hemiacetal hydroxyl group in glucose unit. The structure of lactose is given in Fig. 12.4.
C–1 of first glucose is linked to C–4 of the second glucose through a-glycosidic linkage. Both the glucose units are present in pyranose forms. Maltose is also a reducing sugar, because it has one hemiacetal hydroxyl group in one glucose unit. Its structure is shown in Fig. 12.5.
Fig. 12.5. Structure of maltose
12.1.4 Polysaccharides
Polysaccharides act as structural materials for higher plants and reserve food for plants as well as animals. Polysaccharides are also called glycans, consisting of large number of monosaccharide units joined through glycosidic linkages. The general formula of polysaccharides is (C 6H10O5)n.
Examples: Starch, cellulose, dextrin, glycogen, etc.
Starch
Starch is a polysaccharide of a–D–glucose. It is also called amylum with formula (C6H10O5)n It is widely present in vegetables. It is present in leaves, stems, fruits, roots, and seeds also. Its important sources are wheat, maize, rice, potatoes, barley, sorghum, nuts, etc. It is the major food material and easily hydrolyses with enzyme amylase, present in saliva, to give glucose.
Starch is a white amorphous substance with no taste and smell. It is almost insoluble in cold water but soluble relatively more in boiling water. Starch solution gives a blue colour iodine, which disappears on heating and reappears on cooling. Starch on complete hydrolysis gives D–glucose units.
(C6H10O5)n hydrolysis diastase →
Starch
C12H22O11 hydrolysis maltase
C6H12O6
Maltose D–glucose
Starch cannot reduce Tollen’s reagent or Fehling’s solution. It does not form osazone, because the hemiacetal hydroxyl groups at C–1 of all glucose units are involved in glycosidic linkages. Starch is a mixture of two polysaccharides, amylose and amylopectin. The exact chemical nature of starch varies from source to source. Natural starch contains
15–20% amylose and 80–85% amylopectin.
Amylose is a linear polymer of a –D–glucose. It is water–soluble and gives blue colour with iodine solution. Chemically, amylose is a long, unbranched chain with 200–1000 a –D–(+) glucose units joined by a-glycosidic linkage between C1 of one glucose and C4 of the next glucose.
Its molecular mass may be from 10,000 –50,000 u. The structure of amylose is given in Fig. 12.6.
Fig. 12.6. Structure
Amylopectin is a branched chain polymer of a–D–glucose. Amylopectin is water–insoluble and does not give blue colour with iodine solution. It indicates that starch turns iodine solution blue due to the presence of amylose only. Amylopectin contains 25–30 D-glucose units in each chain.
The C 1 of one glucose and C 4 of other glucose linked through a –glycosidic linkage to form chains. But the branchings are due to glycosidic linkage between C 1 of a glucose in one chain and C 6 of a glucose in the adjacent chain. These linkages are as shown in Fig. 12.7.
of amylose
Fig. 12.7. Structure of amylopectin
Cellulose
Cellulose is a polysaccharide of b–D-glucose. Cellulose is the principal structural component of vegetable matter. Higher percentage of cellulose is present in the natural plant polymer, cotton. Cotton contains 90% of cellulose. Wood contains 40–50% cellulose.
The molecular formula of cellulose is (C 6H 10O 5) n. Photosynthesis in the plants is responsible for the formation of cellulose. Cellulose is a colourless amorphous solid. It is insoluble in cold water. Cellulose contains a large number of D-glucose units joined by b (1, 4)– glycosidic linkages. Cellulose is mainly linear chain polymer and the individual strands will be connected through a number of hydrogen bonds. So, it becomes rigid and acts as cell wall material. But in amylose of starch, hydrogen bonds are not formed, so it contains soft helical structure. Cellulose does not reduce Tollen’s reagent and Fehling’s solution because no free hemiacetal hydroxyl group is present in it. It does not form osazone.
The molecular mass of cellulose is nearly 50,000–5,00,000u. It contains 300 to 2500 D-glucose units. The structure of cellulose is given in Fig. 12.8.
Cellulotic bacteria is present in the stomach of ruminant animals like cattle and sheep. It breaks down the cellulose using the enzyme cellulase to get digested and to convert finally into glucose. Cellulase enzyme is not present in the intestine of human beings. Hence, human beings cannot digest cellulose.
Glycogen
The carbohydrates are stored in animal body as glycogen. Glycogen is also called animal starch because its structure is similar to amylopectin and is rather highly branched. It is present in liver, muscles, and brain. When the body needs glucose, enzymes break down glycogen to glucose. Glycogen is also found in yeast and fungi.
12.1.5 Importance of Carbohydrates
Carbohydrates play an important role in the life of both plants and animals. Some of them are used as food reserves of animals, like glycogen. Honey is an instant source of energy. Glucose is used as food for patients and children. Glucose may be used in the preparation of jams and jellies. In the treatment of calcium deficiency, calcium glucosate is used as medicine. Vitamin C can be prepared industrially, using glucose. The carbohydrate antibiotic is streptomycin. Kenamycins, neomycins, and gentamycins are used against bacteria that are resistant to penicillin.
Starch is the most valuable constituent of foods like rice, potatoes, etc. It is also used in the manufacture of dextrin, adhesives, and explosives. Cell walls of bacteria and plants are made up of cellulose. Cotton fibre, paper, and wood contain cellulose. Explosives like gunpowder, medicines, paints andlacquers are manufactured using cellulose nitrate. Cellulose acetate is used in the manufacture of rayon and plastics. D-ribose and 2-deoxy-D-ribose are present in nucleic acids.
Fig. 12.8 Structure of cellulose
Glycogen is produced from glucose, which is absorbed from the intestine into the blood, transported to liver, muscles, etc., and is polymerised enzymatically. Similarly, when the body needs glucose, the enzymes break down glycogen to glucose.
Carbohydrates act as biofuels to provide energy for functioning of living organisms. In human system, all the carbohydrates, except cellulose, can serve as a source of energy.
TEST YOURSELF
1. The monosaccharides of maltose is (1) α-D-glucose and α-D-glucose (2) β-D-glucose and α-D-glucose (3) α-D-glucose and α-D-fructose (4) α-D-glucose and β-D-fructose
2. The term ‘anomers of glucose’ refers to (1) isomers of glucose that differ in configurations at carbons one and four (C–1 and C–4) (2) a mixture of (D)–glucose and (L)–glucose (3) enantiomers of glucose (4) isomers of glucose that difference in configuration at carbon one (C–1)
3. The number of chiral carbon atoms in glucose hemiacetal is (1) 5 (2) 2 (3) 4 (4) 1
4. Glycosidic linkage between C1 of α−glucose and C2 of β−fructose is found in (1) maltose (2) sucrose (3) lactose (4) amylose
5. Acetylation of glucose with acetic anhydride gives product (X). The number of sp 2 hybridised carbon atoms present in one molecule of a compound (X) is (1) 5 (2) 6 (3) 7 (4) 4
6. Which one of the following compounds contains β-C1-C4 glycosidic linkage? (1) Lactose (2) Sucrose (3) Maltose (4) Amylose
7. A disaccharide X cannot be oxidised by bromine water. The acid hydrolysis of X leads to a laevorotatory solution. The disaccharide X is
8. Amylopectin is composed of (1) α-D-glucose, C1-C4and C1-C6 linkages
(2) β-D-glucose, C1-C4 and C2-C6 linkages
(3) β-D-glucose,C1-C4 and C1-C6 linkages (4) α-D-glucose,C1-C4 and C2-C6 linkages
9. The presence of primary alcoholic group in glucose can be explained by (1) Oxidation to gluconic acid with Br2 water (2) glucose reduces Tollens’ reagent
(3) glucose reaction with HI (4) oxidation to saccharic acid with conc. HNO3
10. Regarding cellulose, correct statements are
A) It is a linear polysaccharide.
B) It is most abundant organic compound in plant kingdom.
C) It is rigid.
D) It is a predominant constituent of cell wall of plant.
E) It involves in C1–C4 glycosidic linkage between a–glucose units.
(1) A, B, C, D, E (2) A, B, D only
(3) A, B, C, D only (4) B, C, D, E only 11. Lactose involves a glycosidic linkage between
(1) C-1 of β-D-glucopyranose and C-4 of β–D-galactopyranose
(2) C-1 of β-D-glycofuranose and C-4 of β-D-galactopyranose
(3) C-1 of β-D-galactopyrancse and C-4 of β-D-glucofuranose
(4) C-1 of β-D-galactopyranose and C-4 of β-D-glucopyranose
Answer key
(1) 1 (2) 4 (3) 1 (4) 2
(5) 2 (6) 1 (7) 1 (8) 1
(9) 4 (10) 3 (11) 4
12.2 PROTEINS
Proteins are the most abundant biomolecules of the living system. Chief sources of proteins are milk, cheese, pulses, peanuts, fish, meat, etc. They occur in every part of the body and form the fundamental basis of structure and functions of life. They are also required for growth and maintenace of body. The word protein is drived from Greek word, ‘proteios’ which means primary or of prime importance. All proteins are polymers of a amino acids.
12.2.1 Amino Acids
Amino acids are the bifunctional organic molecules containing both amino group and carboxyl group in the same compound.
Based on the position of amino group, the simple amino acids are classified as a -, b -, g -, d -, etc.
The general formula of a –amino acids is 2 HNCHCOOH | R
Examples: for a –amino acids: 22 HNCHCOOH
Glycine α 3 2 CH | HNCHCOOH
Alamine α
b -amino propionic acid b –amino butyric acid
Examples: for g –amino acids:
Examples: for g -amino butyric acid:
Examples: for g -amino pentanoic acid:
Out of numerous amino acids, a-amino acids contain primary amino group, except proline, which contains secondary amino group.
H COOH (Proline)
The amino acids containing equal number of –NH2 and –COOH groups are called neutral amino acids. If amino groups are more, it is basic and if carboxyl groups are more, it is acidic in nature. Twenty a -amino acids are classified into four types and are listed in Table 12.2. Structural formulae of all the twenty a –amino acids are given in Fig. 12.9.
12.2.2
Classification of Amino Acids
The amino acids which are synthesised in the body are called non-essential amino acids. The amino acids which are not synthesised by the body but must be supplied through the food are called essential amino acids.
The essential amino acids are 10 in number. They are valine, leucine, isoleucine, arginine, lysine, threonine, histidine, methionine, phenylalanine and tryptophan.
12.2.3 Properties of a Amino Acids
Amino acids are colourless crystalline solids which are water soluble. They have high melting points. They behave like salts.
The simplest amino acid is glycine. Glycine does not contain a chiral carbon. Except glycine, all other naturally occurring a-amino acids are optically active due to chiral carbon. Most of the naturally occurring amino acids have L–configuration.
Note: Glycine is so named because it is sweet to taste. Tyrosine is obtained from cheese.
Amino acids are amphoteric, since they exist as cations or as anions. The proton from –COOH group transfers to –NH 2 group to form both anion and cation within the same molecule, which is called zwitter ion (or) dipolar ion. It is also called inner salt.
At a particular pH, the zwitter ion behaves like neutral species and will not migrate towards any electrode, called isoelectric point. Its value depends on the groups present in the side chain of a-amino acid. For the neutral amino acids, the pH range is 5.5–6.3. Generally, the solubility of a-amino acids is least at isoelectric point. So, they can be separated easily at this point during hydrolysis of proteins.
At a particular pH, the dipolar ion acts as neutral ion (isoelectric point). At a particular pH, the dipolar ion of amines acid (zwitterion) acts as neutral ion and does not migrate to cathode or anode in electric point of the amino acid.
The isoelectric point of neutral amino acids is calculated by
Amino acids exist as cations in acidic medium and anions in basic medium.
All amino acids do not have same isoelectric point.
An amino acid having more COOH groups will have pI < 7
An amino acid having more NH2 group will have pI > 7
Table 12.2 List of twenty a -amino acids
Glycine
Asparagine Asn (or) N
Alanine
(or) A
Glutamine Gln (or) Q
Valine*
Serine Ser (or) S
Leucine*
(or) L
Threonine* Thr (or) T
Isoleucine* Ile (or) I
Tyrosine Tyr (or) Y
Methionine* Met (M)
Cysteine Cys (or) C
Proline
Phenylalanine*
Tryptophan*
C) With acid side chain
1. Aspartic acid Aspartic acid D) With basic side chain 1. Lysine* Lys (or) K
2. Aspartic acid Glu (or) E
* denotes essential amino acid
CO2H H NH2 H
Glycine
CO2H H NH2 CH
Isoleucine* CH3 CH2CH3
CO2H H NH2
CH3
Alanine
CO2H H NH2
CH CH3 CH3
CO2H H NH2
Methionine* (CH2)2SCH3
Valine* CO2H H HN
Proline
CO2H H NH2
CH2CHCH3
Leucine* CH3
CO2H H NH2
CO2H H NH2
Tryptophan* CH2 N H
CO2H H NH2
CH2COH
Aspartic acid O
CO2H H NH2
CH2CH2COH
Glutamic acid O
CO2H H NH2
2. Arginine* Arg (or) R
3. Histidine* His (or) H
CH2CH2CNH2 O Glutamine
CO2H H NH2
CH2OH
Serine
CO2H H NH2
Threonine* CHCH3 OH
CO2H H NH2
Phenylalanine* CH2
12.2.3 Structural Formation of Proteins
The amide bond formed between the amino group of one amino acid and the carboxylic group of another amino acid by the loss of water is called a peptide bond (–CO–NH–). Dipeptides are made from two amino acids linked by one peptide linkage. Tripeptides are made from three amino acids which may be same or different amino acids. If four to ten amino acid residues are present, the peptide is called oligopeptide. Polypeptides are called proteins. Generally, proteins are naturally
CH2CNH2
Asparagine O
CO2H H NH2 CH2 OH
Tyrosine
CO2H H NH2
CH2(CH2)3NH2
Lysine*
CO2H H NH2 (CH2)3NHCNH2
Arginine* NH
CO2H H NH2
CH2SH Cysteine
CO2H H NH2 CH2
Histidine* HN N
occurring polypeptides containing more than 100 amino acids having molecular mass higher than 10,000 u.
Examples: silk, hair, skin, enzymes, hormones, etc.
In writing the formula of the peptides, N–terminal amino acid containing free –NH 2 group must be written on the left side and the C–terminal amino acid containing free –COOH group must be written on the right side. The amino acids in a peptide chain are named from left to right by replacing –ine
Fig 12.9. Structure of a - amino acids
with –yl, except for the C– terminal amino acid. Instead of writing full name of amino acid in a polypeptide, it is most convenient to use three-letter abbreviation.
H2N–CH(CH3)–C–NH–CH2–C–NH–CH2–C–OH O O O
N–terminal
C–terminal
alanine glycine glycine
The above tripeptide is named alanylglycylglycine (or) ala–gly–gly.
The number of peptides possible for using different amino acids = xy,
Here, x is number of amino acids, y is 2 for dipeptide, 3 for tripeptide, 4 for tetrapeptide, etc. For example, the number of tripeptides possible with 3 amino acids is 3 3 = 27. A protein containing 50 amino acid units can be produced using 20 amino acids in 2050 ways.
Classification of Proteins
Proteins can be classified into two types on the basis of their molecular shape.
Fibrous proteins: When the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds, then fibrelike structure is formed. Such proteins are generally insoluble in water. Some common examples are keratin (present in hair, wool, silk) and myosin (present in muscles), etc.
Globular proteins:This structure results when the chains of polypeptides coil around to give a spherical shape. These are usually soluble in water. Insulin and albumins are the common examples of globular proteins.
Structure of Proteins
Protein structure can be explained by considering primary, secondary, tertiary and quaternary structures with different modes of stabilisation.
Primary structure: Each protein has one or more polypetide chains. Each polypetide in a
protein has amino acid linked with each other in a specific sequence. Primary structure of protein gives this sequence of amino acids. Change in the primary structure gives a different protein.
Secondary structure :The shape in which a long polypetide chain can exist is denoted by secondary structure. Two different secondary structures of proteins are: a –helix and b –pleated sheet structure. The free rotation of peptide chain is possible only around C–C bond joining amide group to a–carbon atom.
In secondary structure of RNA, single stranded helixes, sometimes fold back on themselves like a hairpin, thus acquiring double helix structure, possessing double stranded characteristics, as shown in Fig. 12.10.
Hydrogen bonds exist between –NH– and –CO– groups of polypetide chain. Maximum possible stability of a –helix is due to the number of hydrogen bonds in a polypetide chain. a–Helix is also known as 3.6 a helix. In each turn of a–helix, 3.6 amino acid residues, on an average, are present and a 13–membered chelate ring is formed by hydrogen bonding with L–configuration of polypeptide.
Fig. 12.10 Harpin structure for RNA
b –pleated sheet has the peptide chains completely stretched and then put together. Hydrogen bonds are present between polypeptide chains. The stretched peptides may be arranged parallel to one another, like in keratin of hair, or antiparallel, like in silk fibroin. a –helix and conformations of b –pleated structures are shown in Fig.12.11.
Tertiary structure: The tertiary structure of proteins represents overall folding of the polypeptide chains, i.e., further folding of the secondary structure. Based on their molecular shape proteins are of two types: fibrous proteins and globular proteins. Tertiary structure gives rise to these two molecular shapes. Overall folding of the polypeptide chain leads to globular structure. The extent
of folding depends upon the ionic bonds, hydrogen bonds, disulphide linkages, and hydrophobic interactions. Hydrophobic interactions are the attraction forces between alkyl groups of potyplides. These forces stabilise the tertiary structure of proteins.
Quaternary structure: Some proteins are composed of two or more polypeptide chains. These chains are called subunits and the proteins are called oligomers. These subunits form aggregates. The spatial arrangement of the subunits relative to each other is known as quaternary structure. The four structures of amino acids associated with the shape of proteins are diagramatically distinguished in Fig. 12.12 and Fig. 12.13
Fig 12.11 Secondary structure of polypeptide chains in proteins
Fig 12.12. Diagrammatic representation of protein structure
12.13. Representation of four structures of a protein
12.2.5 Denaturation of Proteins
The highly organised tertiary structure of natural proteins is responsible for their biological activity.
These structures are maintained by various attractive forces between different parts of the polypeptide chains. The breakdown of highly organised tertiary structure of protein and loss of its biological activity is called denaturation or unfold. Proteins undergo denaturation most readily at its isoelectric point. During denaturation, secondary and tertiary structures are destroyed but primary structure remains intact. Small changes in the environment can cause a chemical or conformational change, resulting in denaturation.
Proteins can be denaturated usually by:
■ Change in pH, which breaks down hydrogen bonds and electrostatic attractions
■ Adding detergents which break down normal hydrophobic interactions among non-polar groups
■ Adding reagents like urea which forms stronger hydrogen bonds with proteins than the hydrogen bonds between the groups
■ Heating or by agitation, which breaks down attractive forces
Addition of organic solvents, addition of heavy metal ions, and exposing to ultraviolet radiation are some other methods of protein denaturation. Denaturation may be reversible or irreversible. In irreversible denaturation, the denatured protein does not return to its original shape. Cooking of an egg white gives a hard and rubbery insoluble mass due to irreversible denaturation. The reversal of denaturation is called renaturation or refolding.
TEST YOURSELF
1. Which one of the following amino acids cannot be synthesized in the body?
(1) Leucine (2) Glycine (3) Alanine (4) Glutamine
Fig
2. Which structure(s) of proteins remain(s) intact during denaturation process?
(1) Both secondary and tertiary structures
(2) Primary structure only
(3) Secondary structure only
(4) Tertiary structure only
3. All the following amino acids contain benzene ring in their structure, except
(1) histidine
(2) phenyl alanine
(3) tryptophan
(4) tyrosine
4. Shape of polypeptide chain is explained by _____structure of protein.
(1) 10
(2) 20
(3) 30
(4) 40
5. Which of the following α-amino acids is not optically active?
(1) Proline
(2) Serine
(3) Leucine
(4) Glycine
6. The pk a1 and pk a2 value of alanine are 2.3 and 9.7, respectively. The isoelectric point of alanine is
(1) 7.4
(2) 6.0
(3) 5.7
(4) 9.7
7. Select the essential amino acid that has heterocyclic ring.
(1) Valine
(2) Glutamic acid
(3) Tryptophan
(4) Phenyl alanine
8. Select the essential amino acid with hetero aromatic character.
(1) Tyrosine
(2) Arginine
(3) Tryptophan
(4) Proline
9. Tertiary structure of protein will lead the polypeptide chains to get the following shapes
(1) Linear, octahedral
(2) Angular, tetrahedral
(3) Fibrous, globular
(4) Fibrous, planar
10. β-pleated structure of proteins is
(1) primary structure
(2) secondary structure
(3) tertiary structure
(4) quaternary structure
Answer key
(1) 1 (2) 2 (3) 1 (4) 2
(5) 4 (6) 2 (7) 3 (8) 3
(9) 3 (10) 2
12.3 ENZYMES
Almost all the enzymes are globular and are conjugated proteins. The enzymes act as specific catalysts in biological reactions. If once they are utilised in the reaction, they get deactivated in the further reaction, such that they must be replaced by synthesis in the body.
12.3.1 Mechanism of Enzyme Action
The mechanism of an enzyme as catalyst will be:
E + S → ES → EI → EP → E + P,
Here E is enzyme and S is substrate. They form a complex ES. ES transforms to an intermediate EI. EI changes to product EP, which finally gives the product P.
For the progress of a reaction, enzymes are needed only in small quantities. Enzymes reduce the magnitude of activation energy. For example, the activation energy for acid
hydrolysis of sucrose is 6.22 kJ/mole, while it is only 2.15 kj/mole when hydrolysed by the enzyme, sucrose. The enzymes which catalyse the oxidation of one substrate with simultaneous reduction of another substrate are named as oxidoreductase enzymes. Example: The enzyme zymase converts glucose to ethyl alcohol.
Poisons : The plant venoms and most of poisonous substances in animals are proteins which are called poisons.
Artificial sweeteners : Aspartame is a dipeptide and is 160 times sweeter than sucrose. Aspartame is named aspartyl phenyl alanine methyl ester. Structure of aspartame is given as,
CH CH NH CO
TEST YOURSELF
1. Which one of the following statements is not true about enzymes?
(1) Almost all enzymes are proteins.
(2) Enzymes work as catalysts by lowering the activation energy of a biochemical reaction.
(3) Enzymes are non-specific for a reaction and substrate.
(4) The action of enzymes is temperature and pH specific.
2. Enzymes in the living systems
(1) provide energy
(2) provide immunity
(3) transport oxygen
(4) catalyse biological reactions
3. Enzymes are associated with same nonprotein component called (1) co-factor
(2) promoter
(3) catalytic poison
(4) all
4. Regarding enzymatic reactions, the 4 steps are shown below
A) E+S→E−S B) E−P→E+P
C) E−I→E−P D) E−S→E−I
The correct sequence of the steps is
(1) A, D, C, B (2) A, B, C, D
(3) D, C, B, A (4) A, C, B, D
6. Regarding enzymes, incorrect statement is (1) an enzyme is generally a protein (2) an enzyme may be a conjugated protein (3) enzyme gets deactivated during reactions (4) enzyme gets activated during reactions
Answer key
(1) 3 (2) 4 (3) 1 (4) 1 (5) 4
12.4 VITAMINS
Vitamins are organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health of the organism. Plants can synthesise all vitamins.
Animals can synthesise a few but not all vitamins. Human body can synthesise vitamin A from carotene. Some members of vitamin B–complex and vitamin K are synthesised by microorganisms present in intestinal tract of human beings. Remaining vitamins are supplied to the organism through food.
The term vitamin was introduced by Funk. Vitamins are essential dietary factors. Vitamins are not utilised in cell building or as energy source but can act as catalysts in biological processes.
Their deficiency cause serious diseases, known as avitaminosis.
Hence, vitamins are essential constituents of our diet. They are necessary for the growth of children and or pregnant women. Vitamins are partly destroyed and are partly excreted.
Vitamins can be stored in the body to some extent, for example, the fat-soluble vitamins are stored in the liver and subcutaneous tissue.
Vitamins can perform their work in very small quantities. Hence, the total daily requirement of vitamins is usually very small.
12.4.1 Classification of Vitamins
Vitamins are classified into water–soluble vitamins and water–insoluble vitamins (or) fat soluble vitamins. Vitamins soluble in water are listed in Table 12.3 and insoluble ones are listed in Table 12.4.
1. Vitamin B1 (Thiamine)
2. Vitamin B2
3. Vitamin B6 (Pyridoxine)
Yeast, milk, green vegetables, and cereals Beri beri (loss of appetite, retarded growth)
Milk, eggwhite, liver, kidney Cheilosis (fissuring at corners of mouth and lips), degestive disorders, and burning sensation of the skin
Yeast, milk, egg yolk, cereals and grams Convulsions
4. Vitamin B12 Meat, fish, egg and curd Pernicius anaemia (RBC deficient in haemoglobin)
5. Vitamin C (Ascorbic acid)
Table 12.4 Fat–soluble vitamins
Citrus fruits amla, and green, leafy vegetables Scurvy (bleeding gums)
1. Vitamin A Fish liver oil, carrots, butter, and milk Xerophthalamia (hardening of cornea of eye) night blindness
2. Vitamin D Exposure to sunlight, fish and egg yolk Rickets (bone deformities in children) osteo-malacia (soft bones and joint pain in adults)
3. Vitamin E Vegetable oils like wheat germ oil, sunflower oil, etc. Increased fragility of RBCs and muscular weakness
4. Vitamin K Green, leafy vegetables Increased blood clotting time
TEST YOURSELF
1. Vitamin C is (1) ascorbic acid (2) lactic acid
(3) citric acid (4) paracetamol
2. Which of the following vitamins is responsible for beri-beri disease?
(1) A (2) B1
(3) K (4) D
3. Role of DNA in heredity was discovered for the first time through experiments involving (1) transformation using E.coli
(2) transduction involving bacteriophage (3) the use of enzymes protease and nuclease (4) transformation using heat killed virulent and living nonvirulent
Table 12.3 Water soluble vitamins
4. The RBC deficiency is deficiency disease of
(1) Vitamin B2
(2) Vitamin B12
(3) Vitamin B6
(4) Vitamin B1
5. Which of the vitamins given below is water soluble?
(1) Vitamin D
(2) Vitamin E
(3) Vitamin K
(4) Vitamin C
Answer key
(1) 2 (2) 2 (3) 4 (4) 4
(5) 3
12.5 NUCLEIC ACIDS
Chromosomes that are made up of proteins and another type of biomolecules, called nucleic acids, are the particles in nucleus of the cell, which are responsible for heredity.
Nucleic acids are biopolymers of nucleotides with a polyphosphate ester chain. Nucleic acids combine with proteins to give nucleoproteins, which are primary substances in living cells. There are two kinds of nucleic acids: Ribonucleic acid (RNA) and deoxyribonucleic acid (DNA).
The sequence of formation of nucleic acids is:
(base) + (sugar) → (nucleoside) → (nucleotide) → (nucleic acid)
Sugars: The two sugars present in nucleic acids are ribose and deoxyribose. These are aldopentose sugars and are present in furanose form. Ribose is present in RNA and deoxyribose is present in DNA. Ribose and deoxyribose differ structurally in terms of one oxygen atom on carbon at position -2.
Nitrogenous bases: The nitrogenous bases present in nucleic acids are the derivatives of pyrimidine or purine. The pyrimidine bases are cytosine, uracil, and thymine. The purine bases are adenine and guanine. It is important to note that cytosine, adenine, and guanine occur in both RNA and DNA. Uracil occurs only in RNA and thymine occurs only in DNA.
The purine bases are adenine and guanine. These bases are present both in RNA and DNA. Pyrimidine and purine bases and their structures are listed in Table 12.5.
Nucleosides: Nucleosides are N–glycosides in which nitrogen of purine or pyrimidine is bonded to the anomeric carbon of the sugar molecule.
Nitrogenous base + sugar → nucleoside
Nucleosides are named adenosime, guanosine, cytidine, thymidine, and widine when they contain adenine, guanine, cytosine, thymine, and uracil, respectively. In purine nucleosides, the C–1 of sugar is attached to N–9 of purines, but in pyrimidine nucleosides, the C–1 of sugar is attached to N–1 of the pyrimidines.
Nucleotides: Phosphoric acid esters of nucleosides are called nucleotides.
Nucleoside + phosphate → Nucleotide.
During the formation of nucleotide, the esterification takes place between –5 ’ ––OH group of sugar moiety and –OH group of phosphoric acid.
Table 12.5 Structures and systematic names of pyrimidine and purine bases
Example: Adenosine can form adenosine–
5 ’ –monophosphate (AMP), adenosine –5 ’ – diphosphate (ADP), and adenosine –5’ –triphosphate (ATP).
4' 3' 2' 1' + 2HO
Adenosine 5'-monophosphate (AMP)
Major species at pH 7
Adenosine triphosphate -D-Glucopyranose α
6-phosphate -D-Glucopyranose α
The naturally occurring nucleotides are ATP, ADP, AMP (adenosine derivatives), GTP (guanosine derivative), and UTP (uracil derivative).
12.5.1 Structures of DNA and RNA Nucleic acids are polynucleotides. In RNA, the repeating units are ribonucleotides and in DNA, the repeating units are
deoxyribonucleotides. Nucleotides are joined by phosphodiester linkage between the C–3'–oxygen of one nucleotide and the C–5'–oxygen of the next nucleotide.
Always, nucleotide sequences are written with the free 5'– end at the left and free 3'–end at the right. A trinucleotide sequence is written as ATG in Fig.12.14.
The structure of nucleic acids is discussed in the following two levels.
Primary structure: In the primary structure, the nucleotides are joined through phosphodiester bonds and should be written with free 5'–end on the left and free –3'–end on the right. It should be abbreviated by one letter code. It gives information regarding the sequence of nucleotides in the chain of a nucleic acid.
Secondary structure or helical structure: In the secondary structure of RNA, it has single stranded helix, but in DNA, it has double stranded helix. RNA molecules are of three types, called messenger RNA (mRNA), ribosomal RNA (rRNA) and transfer RNA(tRNA).
12.14. ATG, a trinucleotide
Main structural differences between DNA and RNA are compared in Table 12.6.
Table 12.6 Main structural difference between DNA RNA
1. Ribonucleotides are repeating units. Deoxyribonucleotides are repeating units.
2. Base thymine is presents. Base thymine is absent.
3. Base uracil is absent.
4. Double stranded helical structure
Base uracil is present.
Single stranded helical structure
Watson and Crick proposed double helical structure for DNA. According to X-ray crystallographic studies, DNA is composed of two polynucleotide chains running in opposite directions, such that they are held by hydrogen bonding between nitrogenous base pairs.
Proffessor Har Gobind Khorana shared the Nobel Prize for medicine and physiologes with Marshall Nirenberg and Robert Holley for cracking the genetic code.
Adenine forms two hydrogen bonds with thymine but no hydrogen bond with cytosine. Guanine forms three hydrogen bonds with cytosine but only one hydrogen bond with thymine, as shown in Fig.12.15.
Fig.
Fig. 12.15. Hydrogen bonds of adenine with thymine and of guanine with cytosine
Thus, adenine always pairs up with thymine and guanine pairs with cytosine to get maximum stability. All the hydrogen bonds are almost of same length.
DNA finger printing: Each and every human has unique fingerprints, but they can be altered by surgery. The sequence of bases on DNA is unique for a person. It cannot be changed at all for a person.
So, DNA fingerprinting is used for identifying criminals, identifying the dead bodies by comparing their DNA with parents or children, identifying racial groups to rewrite biological evolution, studying evolution of new biological species, determination of paternity of individual humans, and identifying viruses.
12.5.2 Biological Functions of Nucleic Acids
DNA may be regarded as the reserve of genetic information since it is the chemical basis of heredity. DNA is exclusively responsible for maintaining the identity of different species of organisms over millions of years. DNA have coded message for proteins to be synthesised in the cell. A DNA molecule is capable of self duplication during cell division and identical DNA strands are transferred to daughter cells. The proteins are synthesised by various RNA molecules in the cell but the message for the synthesis of a particular protein is present in DNA.
TEST YOURSELF
1. Name the particles in the nucleus of the cell responsible for heredity.
(1) Chromosomes
(2) Mitrochondria
(3) Ribosomes
(4) None of these
2. Which type of linkage is present in nucleotide between base and sugar?
(1) Peptide linkage
(2) Glycosidic linkage
(3) N-glycosidic linkage
(4) Amide linkage
3. DNA and RNA contain four bases each. Which of the following bases is not present in RNA?
(1) Adenine
(2) Uracil
(3) Thymine
(4) Cytosine
4. Which of the following nitrogenous bases is not present in DNA?
(1) Adenine
(2) Cytosine
(3) Thymine
(4) Uracil
5. If one strand of DNA has the sequence ATGCTTGA, the sequence in the complementary strand would be
(1) TCCGAACT
(2) TACGTAGT
(3) TACGAATC
(4) TACGAACT
6. Sugar moiety in DNA and RNA molecules, respectively, are
(1) β-D-2-deoxyribose, β-D-deoxyribose
(2) β-D-2-deoxyribose, β-D-ribose
(3) β-D-ribose, β-D-2-deoxyribose
(4) β-D-deoxyribose, β-D-2-deoxyribose
7. The base which is present in DNA but not in RNA, is :
(1) cytosine
(2) guanine
(3) adenine
(4) thymine
8. Nucleosides are composed of
(1) a pentose sugar and phosphoric acid
(2) a nitrogenous base and phosphoric acid
(3) a nitrogenous base and a pentose sugar
(4) a nitrogenous base, a pentose sugar, and phosphoric acid
9. The bases that are common in both RNA and DNA are (1) adenine, guanine, cytosine (2) adenine, guanine, thymine (3) adenine, uracil, cytosine (4) guanine, uracil, thymine
10. The helical structure or a secondary structure of proteins is stabilised by (1) peptide bonds (2) dipeptide bonds (3) H-bond (4) ether bonds
11. Each codon consists of _____ nitrogen bases. (1) four (2) twenty (3) three (4) sixty four
Answer key
(1) 1 (2) 3 (3) 3 (4) 4
(5) 4 (6) 2 (7) 4 (8) 3
(9) 1 (10) 3 (11) 3
12.6 HORMONES
The name hormone is due to its stimulating action. Like vitamins and enzymes, hormones are also effective in minute amounts. Hormones are referred to as chemical messengers because they transfer biological information from one group of cells to distant tissues or organs. These biomolecules are secreted by the ductless (endocrine) glands and can regulate various biological processes. Hormones are carried to different parts of the body by bloodstream to control metabolic reactions. These are not stored in the body, like fats and carbohydrates, but are continuously produced.
The plant hormones, unlike animal hormones, involve in the growth so, they are called growth hormones. Hormones generally act on tissues distant from the gland. But in few exceptions, these act on adjacent cells or on the cells producing them. Hormones are all generally proteins but not all of them are proteins. Hormones, in many cases, act by influencing the enzymes. The cells in the target
tissue distinguish and pick up selectively the hormone molecules with the help of receptors present in the cells. The hormone receptors are all proteins. The hormones are classified mainly into two types, steroidal and nonsteroidal hormones.
Steroidal hormones: Steroidal hormones possess four ring networks. Three of these are six-membered carbon rings and one is a five-carbon ring. Steroidal hormones are two types sex hormones and corticosteroids.
Androgens are secreted by testis. The principal male sex hormone is testosterone. It promotes muscle strength, deepening of voice, the growth of body hair and other male secondary sex characteristics.
Oestrogens are secreted by the ovary. Oestradiol is the principal female sex hormone. It regulates the menstrual cycle and the reproductive process. It is responsible for the development of female secondary sex characteristics, like breast development, shrill voice, and long hair.
Progesterone is a female sex hormone secreted by the corpus luteum. It is useful for preparing the uterus for the implantation of the fertilised egg. These are also useful as birth control agents.
Corticosteroids are secreted by adrenal glands. These are of two types. Mineralocorticoids control the NaCl content in the blood and can balance the water–salt ratio in the body. These can excrete the potassium through urine. Glucocorticoids influence some metabolic reactions like anti inflammatory effect.
Non–steroidal hormones: Non-steroidal hormones are of two types peptide hormones and amino acid derivative hormones. Peptide hormones are insulin, oxytosin, vasopressin, etc. Among them, insulin is the most important, which promotes anabolic reactions and inhibits catabolic reactions. It is secreted by islets of Langerhans. It is responsible for the entry of glucose and sugars into the living cells.
This is achieved by increasing the penetrating ability of cell membranes and by augmenting phosphorylation of glucose. This helps in the decrease of glucose in the blood.
This is, therefore, commonly called hypoglycemic factor. Its deficiency in a human being causes diabetes mellitus. It maintains constant sugar level in blood.
Insulin isolated from islets tissue of pancreas was the first hormone identified as protein. Sanger was awarded Nobel Prize for determining the structure of insulin.
Insulin has 51 amino acids, which are divided between two peptide chains. Chain ‘A’ has 21 amino acids and ‘B’ has 30 amino acids. A and B are joined by disulphide bonds between cystine residues.
Amino acid hormones are thyroidal hormones like thyroxin and tri-iodo thyronine. Thyroxin is secreted by thyroxin gland and can control metabolism of carbohydrates, lipids, and proteins.
Thimman introduced the name phytohormones for plant hormones. These are also called growth hormones, since these can regulate the growth and physiological functions in higher plants.
TEST YOURSELF
1. The organic compound that transfers biological information from one group of
CHAPTER REVIEW
Carbohydrates
■ Amino acids contain both carboxylic acid group and amine group.
■ Amide linkages between amino acids are known as peptide bonds (– CO – NH –).
■ The product obtained from two amino acid molecules through a peptide bond is called a dipeptide.
cells to distant tissues or organs are called as
(1) Vitamins (2) Proteins
(3) Hormones (4) Carbohydrates
2. Which one of the following hormones modulate inflammatory reactions and are involved in the reactions to stress
(1) Mineralocorticoids
(2) Glucocorticoids
(3) Throxine
(4) Glucagon
3. In insulin molecule S-S linkage is in between (1) cystine - glycine (2) Cystine - cystine (3) Cystine-valanine (4) Proline-cystine
4. Which of the following hormones contains iodine?
(1) Insulin (2) Thyroxine (3) Adrenaline (4) Testosterone
5. Number of six-membered rings present in a steroid nucleus is (1) 1 (2) 2 (3) 3 (4) 4
6. Hormones are secreted by ductless glands of human body. Iodine-containing hormone is (1) adrenaline (2) thyroxine (3) testosterone (4) insulin
Answer key (1) 3 (2) 2 (3) 2 (4) 2 (5) 3 (6) 2
■ The product from three, four, and many amino acid molecules through peptide bond are called tri, tetra, and polypeptides, respectively.
■ A polypeptide chain formed from ‘n’ amino acids contains ‘n-1’ peptide bonds.
■ The numerical prefix (di, tri, tetra) of peptide is derived from the number of
amino acid molecules involved in bonding but not from number of peptide bonds.
■ A tripeptide has two peptide bonds between three amino acid molecules.
■ Generally a - aminoacids form proteins. All a-aminoacids contain a primary amino group, except proline.
■ The aminoacids that can be synthesised in the body are called non-essential amino acids and the amino acids that cannot be synthesised in the body but must be supplied through diet are called essential amino acids.
■ Amino acids containing equal number of –NH 2 and –COOH groups are called neutral amino acids.
■ Aminoacids containing more number of –NH 2 groups than –COOH groups are known as basic amino acid and more number of –COOH groups than –NH 2 groups are known as acidic aminoacids.
■ Amino acids containing –OH groups are tyrosine, serine, and threonine.
■ Amino acids containing benzene ring ar e phenylalanine, tyrosine, and tryptophan.
■ Amino acids are generally colourless crystalline solids and are highly polar.
■ In aqueous solution, the carboxyl group of amino acid transfers a proton to –NH 2 group to give zwitter ion or inner salt.
■ In acidic solution, amino acid exists as positive and in basic solution, as negative ion.
■ The pH at which dipolar ion acts as neutral ion and does not migrate either towards cathode or anode is known as isoelectric point of the amino acid.
■ The isoelectric point depends on different groups in the molecule of the amino acid. For neutral amino acids, the pH range is 5.5–6.3.
■ At isoelectric point, aminoacids have least solubility, which helps in the separation of mixture of aminoacids obtained from the hydrolysis of proteins.
■ Except glycine, all other naturally occurring aminoacids are optically active.
■ In Fisher projection formulae, all carbon atoms must be placed vertically with –COOH group at the top and amino group is kept horizontally.
■ If –NH 2 group is on left-hand side, it is L–form. If it is on right-hand side, it is D–form.
■ Most of the naturally occurring amino acids have L–configuration.
■ In a polypeptide, free amino group N–terminal residue is written on the left hand side and the free carboxyl group on the right hand side of the chain.
■ 22 33 O O || || HNCHCNHCHCNHCHCOOH || H CH C
N-terminal residue C-terminal residue Alanine Glycine Alanine This is read as alanyl glycylalanine.
■ Shorter peptides are called oligopeptides and longer peptides are called polypeptides. Polypeptides are amphoteric.
■ Most of the toxins (poisonous substances) in animal and plant venoms are proteins. Oligopeptides are effective hormones.
■ A dipeptide, called aspartame, being 160 times sweeter to sucrose, is used as a substitute of sugar.
■ Proteins are naturally occurring polypeptides containing more than 100 amino acid units. e.g., wool, nail, silk, hair, skin, connective tissues, and many hormones and enzymes.
■ Proteins are usually two types—fibrous proteins and globular proteins.
■ Primary structure of proteins refers to the sequence in which amino acids are arranged in protein and also the location of disulphide bridges.
■ Any two proteins will never have the same primary structure.
■ If a protein is made up of ‘m’ amino acids of ‘n’ types, the possible different types of protiens are ‘nm’.
■ Secondary structure of protein refers to the shape of polypeptide chains.
■ The shape of protein can be either a-helix, -pleated sheet or coil conformation.
■ Tertiary structure of protein gives threedimensional folding of protein. It includes both primary and secondary structures.
■ The three-dimensional folding of protein leads to fibrous or globular shapes.
■ Quaternary structure of protein explains the arrangement of different protein chains (sub units). It is possible only in oligomers.
■ Between protein chains, hydrogen bonding, electrostatic attractions, and hydrophobic interactions are present.
■ The processes such as heating, and treatment with acids, that bring about changes in the physical as well as biological properties of the proteins are called denaturation.
■ Denaturation changes the secondary and tertiary structure of proteins but has no effect on the primary structure.
■ Denaturation may be reversible or irreversible. Coagulation of eggwhite on boiling is an example of irreversible denaturation.
■ Reverse process of denaturation is called renaturation, which is possible in deoxyribonucleic acid.
Enzymes
■ Most enzymes are naturally occurring simple or conjugate proteins. They act as specific catalysts in biological reactions.
■ Enzymatic reaction may proceed through the following four stages.
E + S ES → complex EI complex EP→ E + P
Vitamins
■ Vitamins are certain organic compounds, required in small quantities in diet, but their deficiency causes specific diseases.
■ Vitamins are designated by alphabets A, B, C, D, etc and some of them are further categorised into subgroups, like B1, B2, B6, B12, etc.
■ Vitamins are classified into two groups depending on their solubility in water or in fat.
■ Vitamins of B group and vitamin C are water-soluble.
■ Vitamins A, D, E, and K are fat and oil soluble. These are stored in liver and adipose tissues.
■ Deficiency of vitamin A causes night blindness deficiency of vitamin C causes scurvy. Deficiency of vitamin D causes rickets.
■ Deficiency of vitamin B1 causes beri beri, B2 causes cheilosis, B6 causes convulsions, and B12 causes pernicious anaemia.
■ Vitamin E is found in oils and helps in fragility of RBCs and muscular strengthening.
■ Vitamin K is called green leafy vitamin and its deficiency leads to increased blood clotting time.
Nucleic Acids
■ Nucleic acids are polymers present in living cells and viruses. They are of two
types: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA).
■ The DNA stores and transmits genetic information and the RNA is responsible for the synthesis of proteins in living cells.
■ Nucleic acids are polymers whose repeating units are nucleotides.
■ The nitrogen base in nucleic acids is of two kinds: pyrimidines and purines.
■ Purine bases, adenine, and guenine are found in both RNA and DNA. Cytosine is found in both RNA and DNA.
■ Uracil is present only in RNA and thymine only in DNA.
■ Naturally occurring nucleic acids have -D-ribose in RNA and -D-deoxyribose in DNA.
■ A nitrogen base attached to a sugar molecule forms a nucleoside. A nucleoside joined to a phosphate group is called nucleotide.
■ Nucleic acids contain a chain of five membered ring sugars linked through phosphate groups and each sugar molecule is bonded to nitrogen atom of heterocyclic amine by a – N glycosidic bond.
■ Watson and Crick, based on X-ray diffraction studies of DNA, proposed a double helical structure for DNA.
■ The number of hydrogen bonds between thymine and adenine is 2, but between the
complementary bases cytosine and guanine it is 3.
■ Adenine pairs with thymine but not with cytosine, because adenine forms two Hbonds with thymine but no bonds with cytosines.
■ Primary structure of DNA gives sequence of bases in the strands. Secondary structure of DNA is double helix.
■ The synthesis of identical copies of DNA is called replication. A nucleic acid can be synthesised only in the 5–3 direction.
■ The amino acid specified by each three base sequence is called the genetic code. It is universal, commaless, and degenerate.
Hormones
■ Hormones transfer biological information from one group of cells to distant tissues or organs.
■ Hormones control metabolic activities and are effective in minute amounts.
■ Secretin, produced by intestinal mucosa, was first named hormone. Hormones are all generally proteins, but not all of them are proteins.
■ Hormones are classified into two types steroid hormones and non—steroid hormones.
■ Steroid hormones are produced by the adrenal cortex, testis, and ovary.
Exercises
NEET DRILL
LEVEL I
Carbohydrates
1. A non-reducing sugar A hydrolyses to give two reducing mono saccharides. Sugar A is: (1) fructose (2) galactose
(3) sucrose (4) glucose
2. Which of the following statements is/are correct?
I) Dextrose has one C O group II) Saccharic acid has two – COOH groups. III) Gluconic acid has one – COOH group (1) Only I (2) I and III (3) II and III (4) I, II, and III
3. The number of chiral centres in α– D(+)–glucopyranose and β–D(–) –fructofuranose respectively are (1) 3, 2 (2) 4, 4 (3) 4, 3 (4) 5, 4
4. Which of the following statement is not true about glucose?
(1) It is an aldopentose. (2) It is an aldohexose. (3) It contains five hydroxyl groups. (4) It is a reducing sugar.
5. Glucose cannot react with (1) HCN
(2) NH2-OH (3) [Ag(NH3)2]+ (4) NaHSO3
6. The letter ‘D’ in D – glucose signifies (1) configuration at all chiral carbons (2) dextrorotatory
(3) that it is a monosaccharide
(4) configuration at the penultimate chiral carbon
7. Which among the following is a reducing saccharide?
(1) Starch (2) Sucrose
(3) Cellulose (4) Fructose
8. Sucrose on hydrolysis gives :
(1) α -D-Glucose + β -D-Glucose
(2) α -D-Glucose + β -D-Fructose (3) α -D-Fructose + β -D-Fructose (4) β -D-Glucose + α -D-Fructose
9. Which of the following glycosidic linkages between galactose and glucose is present in lactose?
(1) C-1 of galactose and C-4 of glucose (2) C-1 of galactose and C-6 of glucose (3) C-1 of glucose and C-4 of galactose (4) C-1 of glucose and C-6 of galactose
10. Which reagent can be used to distinguish glucose and fructose? (I) Bromine water (II) Tollens reagent (III) Schiff’s reagent (1) (I), (II), and (III) (2) (II) and (III) (3) Only (I) (4) Only (III)
11. The number of chiral carbon atoms present in open chain structure of glucose and fructose ,respectively is (1) 3, 3 (2) 4, 4 (3) 3, 4 (4) 4, 3
12. Anomers have different (1) physical properties (2) melting points (3) specific rotation (4) all of these
13. Two monomers in maltose are: (1) α–D–glucose and α–D–Fructose (2) α–D–glucose and –D–glucose (3) α–D–glucose and –D–galactose (4) α–D–glucose and α–D–glucose
14. α–D-glucopyranose and β–D–glucopyranose can be called as all the following except (1) enantiomers
(2) anomers
(3) diastereomers
(4) epimers
15. HNO3 HI X Glucose Y. ←→ What are X and Y?
X Y
(1) n-Hexane Saccharic acid
(2) Gluconic acid Saccharic acid
(3) n-Hexanol Saccharic acid
(4) n-Hexanol Saccharic acid
Proteins
16. Which of the following statements is not correct about a zwitter ion?
(1) It is neutral and does not contain any charges.
(2) + –NH3 group is the acidic group.
(3) It is dipolar.
(4) It has more than one pKa value
17. Which of the following is a basic amino acid?
(1) Serine
(2) Alanine
(3) Tyrosine
(4) Lysine
18. The pH at which an amino acid carries no net charge is called its
(1) isoelectric point
(2) inversion point
(3) neutralisation point
(4) triple point
19. In glycine, the basic group is (1) –NH2
(2) –NH3
(3) –COOH
(4) –COO
20. The number of peptide bonds in the compound below is H3C N N H H H3C H3C NHNH2 CH3 O O
(1) 0 (2) 2
(3) 3 (4) 4
21. The correct structure of the dipeptide gly-ala is
(1) H H O O H2N – C – C – NH – C – C – OH CH3
(2) CH2SH O O H2N – CH – C – NH – CH2 – C – OH
(3) H H O O H2N – C – C – NH – CH – C – OH CH3
(4) H H O O H2N – C – C – NH – CH – C – OH CH2SH
22. The restriction of the biological nature and activity of proteins by heat or chemical agent is called
(1) dehydration (2) denaturation (3) deamination (4) denitrogenation
23. β-pleated structure of proteins is associated with (1) primary structure
(2) secondary structure
(3) tertiary structure (4) quaternary strucutre
24. Which of the following is a derivative of amino acid?
(1) Thyroxin (2) Estradiol
(3) Estrogene (4) Progesterone
25. Protein is composed of (1) amino acids (2) carbohydrates (3) vitamins (4) mineral salt
26. The tertiary structure of a protein is due to the involvement of which type of interactions?
(1) Hydrophobic interactions
(2) Hydrogen bonding
(3) Di sulphide bridges
(4) All of the above
27. Secondary structure of proteins refers to
(1) mainly denatured proteins and structure of prosthetic groups.
(2) three- dimensional structure, especially the bond between amino acid residues that are distant from each other in the polypeptide chain.
(3) linear sequence of amino acid residues in the polypeptide chain.
(4) regular folding patterns of continuous portions of the polypeptide chain.
28. Which of the following amino acids has an amide side chain?
(1) Asparagine acid (2) Glutamic acid
(3) Methionine (4) Asparagine
29. Biuret test is not given by
(1) proteins (2) starches
(3) poly peptides (4) urea
30. The bonds in protein structure, which are not broken on denaturaton are ____
(1) hydrogen bonds (2) peptide bonds
(3) ionic bonds (4) di sulphide bonds
Enzymes
31. Mark the wrong statement about enzymes.
(1) Enzymes are highly specific both in binding with substrates and in catalysing their reactions.
(2) Each enzyme can catalyse a number of similar reactions.
(3) Enzymes catalyse chemical reactions by lowering the energy of activation.
(4) Enzymes are needed only in very small amounts for their action.
32. Enzymes belong to which class of compounds?
(1) Polysaccharides (2) Polypeptides
(3) Heterocyclics (4) Hydrocarbons
33. Enzymes are made up of
(1) edible proteins
(2) proteins with specific structure
(3) nitrogen containing carbohydrates
(4) carbohydrates
Vitamins
34. Night blindness is associated with the (1) Vitamin A (2)Vitamin B
(3) Vitamin C (4) Vitamin D
35. Deficiency of pyridoxine leads to (1) convulsions (2) beri beri
(3) night blindness (4) rickets
36. The deficiency of which vitamin causes pernicious anaemia?
(1) Vitamin B2 (2) Vitamin B6
(3) Vitamin E (4) Vitamin B12
37. Hypothyroidism is due to
A. high level of iodine in the diet
B. enlargement of thyroid gland
C. low levels of iodine in the diet
D. increased levels of thyroxine
(1) A, B (2) B, C
(3) A, D (4) C, D
38. Thiamine and pyridoxine are also known respectively as:
(1) Vitamin B2 and Vitamin E
(2) Vitamin E and Vitamin B2
(3) Vitamin B6 and Vitamin B2
(4) Vitamin B1 and Vitamin B6
39. Vitamin B12 contains (1) Cu (2) Fe (3) Mg (4) Co
40. Which of the following B group vitamins can be stored in our body?
(1) Vitamin B1 (2) Vitamin B2
(3) Vitamin B6 (4) Vitamin B12
41. Thiamine and pyridoxine are also known respectively as
(1) Vitamin B2 and Vitamin E
(2) Vitamin E and Vitamin B2
(3) Vitamin B6 and Vitamin B2
(4) Vitamin B1 and Vitamin B6
42. Water soluble vitamins are (1) A, D (2) E, K
(3) D, E (4) C, B
43. Which of the vitamins given below is water soluble?
(1) Vitamin C
(2) Vitamin D
(3) Vitamin E
(4) Vitamin K
Nucleic Acids
44. Which of the following structures represent thymine?
(1)
(3)
45. Bases common to RNA and DNA are
(1) adenine, guanine, and cytosine
(2) adenine, uracil, and cytosine
(3) adenine, guanine, and thymine
(4) guanine, uracil, and thymine
46. Which of the following statements is/are true?
(1) Every individual has unique fingerprints and they occur at the tips of the fingers.
(2) A sequence of bases on DNA is also unique for a person and information regarding this is called fingerprinting.
(3) Fingerprints can be altered by surgery.
(4) All of the above.
47. Name the particles in the nucleus of the cell responsible for heredity.
(1) Chromosomes (2) Mitrochondria
(3) Ribosomes (4) None of these
48. When a nucleotide forms a dimer, the linkage present between them is (1) a glycosidic linkage.
(2) a disulphide linkage.
(3) a peptide linkage.
(4) a phosphodiester linkage.
49. Which of the following is not present in a nucleotide?
(1) Guanine (2) Cytosine
(3) Adenine (4) Tyrosine
50. Nucleoside on hydrolysis gives (1) pentose sugar and purine base (2) pentose sugar, phosphoric acid, purine or pyrimidine base (3) pentose sugar and a heterocyclic base (4) heterocyclic base and phosphoric acid
51. In a nucleoside, the base is attached to which position of the sugar molecule?
(1) C-1 (2) C-2
(3) C-3 (4) C-5
52. In a nucleotide, the sequence is represented as
(1) phosphate–base sugar (2) sugar–base–phosphate (3) base–sugar–phosphate (4) base–phosphate–sugar
53. Which base is present in RNA, but not present in DNA?
(1) Thymine (2) Uracil (3) Guanine (4) Cytosine
54. In nucleic acids, the sequence is represented as
(1) phosphate – base – sugar.
(2) sugar – base – phosphate.
(3) base – sugar – phosphate. (4) base – phosphate – sugar.
55. Which one of the following bases is not present in DNA?
(1) Cytosine (2) Thymine
(3) Quinoline (4) Adenine
56. The type of bond connecting two nucleotides is a
(1) peptide bond
(2) hydrogen bond
(3) phosphodiester bond
(4) glycosidic bond
57. Which of the following statements is not true about RNA?
(1) It always has a double stranded α-helix structure.
(2) It is present in the nucleus of the cell.
(3) It controls the synthesis of protein.
(4) It usually does not replicate.
58. Which one of the following is the correct structure for cytosine?
(1) N N H O H3C (2) N H O O H3C N–H
(3) N N H O NH2 (4) N N H O H2N
Hormones
59. Which of the following hormones contain iodine?
(1) Insulin (2) Thyroxine
(3) Adrenaline (4) Testosterone
60. Which of the followings is the more scientific definition of hormones?
(1) They are extracellular messengers.
(2) They always act at distantly located target organ.
(3) They are the products of well organised endocrine glands.
(4) They are the non-nutrient chemicals that act as intercellular messengers.
61. Insulin is a hormone that (1) releases glucose from glycogen (2) stimulates metabolism (3) increase blood sugar (4) decreases blood sugar
62. Substances synthesised in ductless glands are (1) proteins (2) hormones
(3) vitamins
(4) amino acids
Level 2
Carbohydrates
1. Arabinose is an example of (1) pentose sugar. (2) hexose sugar. (3) tetrose sugar. (4) heptose sugar.
2. Select a wrong statement about fructose. (1) It is a ketohexose.
(2) It can reduce Tollens reagent. (3) It can reduce Fehlings reagent. (4) It is tasteless.
3. Which of the following is NOT a disaccharide?
(1) Maltose (2) Xylose
(3) Lactose (4) Cellobiose
4. Which of the following is not a monosaccharide?
(1) Glucose (2) Arabinose
(3) Mannose (4) Lactose
5. Glucose has the feature that it (1) shows mutarotation
(2) is a reducing sugar
(3) is tasteless in nature
(4) All of the above
6. Glucose and fructose are (1) optical isomers. (2) tautomers.
(3) functional isomers. (4) chain isomers.
7. The biomolecule that cannot be hydrolysed further is (1) starch (2) glycogen (3) cellulose (4) glucose
8. In an aqueous solution of D-glucose, the percentages of α and ß anomers at equilibrium condition are, respectively (1) 80 and 20 (2) 20 and 80 (3) 36 and 64 (4) 64 and 36
9. Select the incorrect statement about glucose.
(1) It is a reducing sugar.
(2) It is a monosaccharide.
(3) It is a synthetic product.
(4) It is an aldohexose.
10. Hydrolysis of sucrose is carried out in the presence of
a) dilute HCl (or) H2SO4
b) invertase (sucrase)
Rate of hydrolysis in
(1) a > b
(2) a < b
(3) a = b
(4) cannot be compared
11. The question has two statements. Statement I and statement II.
Statement I: Sucrose is a non-reducing sugar.
Statement II : Fructose is a reducing sugar.
In light of the given statements, choose the most appropriate answer from the options given below.
(1) Both statement I and statement II are correct.
(2) Both statement I and statement II are incorrect.
(3) Statement I is correct but statement II is incorrect.
(4) Statement I is incorrect but statement II is correct.
12. Which one of the following is a reducing sugar?
(1) Starch
(2) Cellulose
(3) Glycogen
(4) Fructose
13. 3 Conc. HNO Glucose X; → ( ) + 3 2 AgNH Glucose Y; → X and Y respectively, are
(1) HOOC−(CH2)4−CHO, HOOC−(CH2)4 COOH
(2) HOOC−(CHOH) 4−CHO–(CHOH) 4 CH2OH
(3) HOOC−(CHOH)4−COOH–(CHOH)4 CH2OH
(4) HOOC−(CH 2 ) 4 −COOH, HOOC–(CH2)4−CH2OH
14. . The biomolecule that can not be hydrolysed further is
(1) starch
(2) glycogen
(3) cellulose
(4) glucose
15. Sucrose on hydrolysis gives
(1) α -D-glucose + β -D-glucose
(2) α -D-glucose + β -D-fructose
(3) α -D-fructose + β -D-fructose
(4) β -D-glucose + α -D-fructose
16. Which one of the following given below is a non-reducing sugar?
(1) Glucose
(2) Sucrose
(3) Maltose
(4) Lactose
17. Which of the following tests cannot be used for identifying amino acids?
(1) Biuret test
(2) Barfoed test
(3) Ninhydrin test
(4) Xanthoproteic
Proteins
18. The structure of Gly-Ala is
(1) N CO2H H O H2N
(2) N CO2H H O H2N
(3) N CO2H OH H O H2N
(4) N CO2H OH H O H2N
19. The functional groups present in asparagine, a non-essential aminoacid, are (1) −NH2, −COOH, = NH
(2) O –NH2 , – COOH , – C – NH2
(3) O O –NH – C – , – COOH , – C – Cl
(4) −NH2, −COOH, – OH
20. An α- amino acid exists as
NH–CH–COOH + at pH = 2 and its isoelectric point is 6. The amino acid at pH = 10.97 will exist as
R 3 I
(1) R –3 I NH–CH–COO +
(2) R –2 I NH–CH–COO
(3) R 2 I NH–CH–COOH
(4) R 2 I NH–CH–COOH +
21. A peptide made up of n number of amino acids contain how many peptide linkages?
(1) n (2) n + 1
(3) 2n – 1 (4) n – 1
22. Which of the following amino acids possess a nonpolar side chain?
(1) Isoleucine (2) Serine
(3) Cysteine (4) Glutamic acid
23. What is the number of chiral carbon atoms in alanine?
(1) 1 (2) 2 (3) 3 (4) 4
24. Which of the following is not an essential amino acid?
(1) Leucine (2) Isoleucine
(3) Valine (4) Alanine
25. The number of amino acid residues in human insulin is (1) 100 (2) 51 (3) 99 (4) 206
26. Biuret test is not given by (1) proteins (2) urea (3) polypeptide (4) carbohydrates
27. The helical structure of a protein is stabilized by (1) dipeptide bonds. (2) hydrogen bonds. (3) ether bonds. (4) peptide bonds.
28. The one that does not stabilise 2° and 3° structures of proteins is (1) van der Waals forces (2) –S–S– linkage (3) Hydrogen-bonding (4) –O–O–linkage
29. Hydrolysis of proteins give (1) β−amino acids (2) γ−amino acids (3) δ−amino acids (4) α−amino acids
30. Lysine is least soluble in water in the pH range of (1) 3 to 4 (2) 5 to 6 (3) 6 to 7 (4) 9 to 10
31. The given structure of an α -amino acid will exist at which pH?
37. Identify the correct statement among the following.
(1) Vitamin C is water insoluble.
(2) Vitamin A is water soluble.
(3) Vitamin C cannot be stored in our body.
(1) 7 (2) 14 (3) 12 (4) 6
32. Which one of the following statements is incorrect about enzyme catalysts?
(1) Enzymes are mostly protein as in nature.
(2) Enzymatic action is specific.
(3) Enzymes are denatured by ultraviolet rays and at high temperature.
(4) Enzymes are least reactive at optimum temperature.
Enzymes
33. Enzymes are made up of
(1) edible proteins
(2) proteins with specific structure
(3) nitrogen containing carbohydrates
(4) carbohydrates
34. Which of the following biomolecules act as specific catalysts in a biological reaction?
(1) Carbohydrates
(2) Lipids
(3) Vitamins
(4) Enzymes
35. Identify the correct statement for enzymes.
(1) Enzymes are active within a narrow range of pH and temperature.
(2) They show low reaction selectivity.
(3) Work under harsh reaction conditions.
(4) They work only at high temperature.
36. The incorrect statement regarding enzymes is
(1) Enzymes are very specific for a particular reaction and substrate.
(2) Enzymes are biocatalysts.
(3) Like chemical catalysts, enzymes reduce the activation energy of a bioprocess.
(4) Enzymes are polyssaccharides.
(4) Riboflavin is Vitamin B6.
Vitamins
38. Deficiency of pyridoxine leads to (1) convulsions (2) beri beri
(3) night blindness (4) rickets
39. The vitamin that is also called as antihemorrhage vitamin is
(1) vitamin A (2) vitamin B12
(3) vitamin K. (4) vitamin B6.
40. Tocopherol is also called as (1) vitamin D (2) vitamin K
(3) vitamin E (4) vitamin A
41. Deficiency of which of the following causes convulsions?
(1) Vitamin A (2) Vitamin B6
(3) Vitamin B12 (4) Vitamin D
42. Cholecalciferol is also called as vitamin
(1)vitamin A (2)vitamin D
(3)vitamin E (4)vitamin K
43. Vitamin A is also called as (1) ascorbic acid (2) reionol
(3) retinol (4) phytol
44. Deficiency of the following vitamin leads to pellagra.
(1) A (2) B2 (3) B 3 (4) C
45. Increased fragility of RBC and muscular weakness is due to the deficiency of (1) vitamin B12 (2) vitamin D
(3) vitamin E (4) vitamin K
46. The night blindness is developed due to the deficiency of
(1) vitamin B6 (2) vitamin C
(3) vitamin B12 (4) vitamin A
Nucleic Acids
47. In a nucleotide, C1 of sugar is joined to (1) N1 of a pyrimidine base.
(2) N9 of a pyrimidine base.
(3) N9 of both pyrimidine and purine bases. (4) N1 of both pyrimidine and purine bases.
48. Which base is not present in RNA?
(1) Adenine (2) Guanine
(3) Cytosine (4) Thymine
49. Which of the following bases is not present in DNA?
(1) Adenine (2) Thymine
(3) Uracil (4) Cytosine
50. Select the incorrect statement about thymine.
(1) It is a purine base.
(2) It is present in DNA.
(3) It is 5-methyl uracil.
(4) It can form H-bonds with adenine.
51. Phosphate ester of X is called a nucleotide. X is
(1) adenine (2) guanine
(3) nucleoside (4) thymine
52. Dinucleotide is obtained by joining two nucleotides together by a phosphodiester linkage. Between which carbon atoms of pentose sugars of a nucleotide are these linkages present?
(1) 5’ and 3 ’ (2) 1 ’ and 5 ’
(3) 5’ and 5 ’ (4) 3 ’ and 3 ’
53. Synthesis of identical copies of DNA is called (1) transcription (2) replication (3) translation
(4) reverse transcription
54. Find the correct combination that can form a nucleotide of RNA.
(1) uracil + ribose + phosphate
(2) thymine + ribose + phosphate
(3) uracil + deoxyribose + phosphate
(4) adenine + deoxyribose + phosphate
55. The central dogma of molecular genetics states that the genetic information flows from
(1) Amino acids → Proteins → DNA
(2) DNA → Proteins → Proteins
(3) DNA → RNA → Proteins
(4) DNA → RNA → Carbohydrates
56. In A DNA, the linkage between different nitrogenous bases is
(1) a phosphate linkage.
(2) hydrogen-bonding.
(3) a glycosidic linkage.
(4) a peptide linkage.
57. The correct statement regarding RNA and DNA respectively is
(1) The sugar component in RNA is arabinose and the sugar component in DNA is ribose.
(2) The sugar component in RNA is 2-deoxyribose and the sugar component in DNA is arabinose.
(3) The sugar component in RNA is arabinose and the sugar component in DNA is 2–deoxyribose.
(4) The sugar component in RNA is ribose and the sugar component in DNA is 2–deoxyribose.
58. Which of the following is not present in RNA molecule?
(1) Uracil (2) Ribose
(3) Thymine (4) Phosphate
59. The stability of an α-helix structure of proteins depend upon
(1) dipolar interaction
(2) hydrogen bonding interaction
(3) van der Waals forces (4) π-stacking interaction
60. Which one of the following is not a protein?
(1) Wool (2) Nail (3) Hair (4) DNA
61. The presence or absence of hydroxy group on which carbon atom of sugar differentiates RNA and DNA?
(1) 2nd (2) 3rd (3) 4th (4) 1st
Hormones
62. Which one of the following hormones modulate inflammatory reactions and are involved in the reactions to stress?
(1) Mineralocorticoids
(2) Glucocorticoids
(3) Throxine
(4) Glucagon
63. The organic compound that transfer biological information from one group of cells to distant tissues or organs are called as
FURTHER EXPLORATION
1. The correct statement(s) about the following sugars X and Y is/are CH2OH CH2OH HOH2C O O O H H H H H H H H OH OH OH HO HO CH2OH CH2OH O O O H H H H H H H H H H OH OH OH OH HO HO
(1) X is a reducing sugar and Y is a nonreducing sugar.
(2) X is a non-reducing sugar and Y is a reducing sugar.
(3) The glucosidic linkages in X and Y are α and α, respectively.
(4) The glucosidic linkages in X and Y are β and α, respectively.
2. Match the column I with column II.
Column-I (compound)
Column-II (Linkage)
(A) Sucrose (p) (C1−α)of glucose–C4 of glucose
(1) vitamins (2) proteins
(3) hormones (4) carbohydrates
64. Steroid hormones are produced by the 1) adrenal cortex 2) pancreas
3) thyroid 4) testis
5) pituitary
(1) 1 and 4 (2) 1, 2, and 3
(3) 3, 4 (4) 4, 5
65. Receptors of hormones are generally (1) carbohydrates (2) vitamins
(3) lipids (4) proteins
(B) Maltose (q) (C1−α)of glucose–C4 of glucose and – C6 of glucose in another chain
(C) Lactose (r) (C1−β)of-glucose–C4 of glucose
(D) Starch (s) (C1−β)ofgalactose–C4 of glucose
(E) Cellulose (t) (C1−α)of-glucose–(C2−β) of fructose
(A) (B) (C) (D) (E)
(1) p q r s t
(2) q p s r t
(3) s q p t r
(4) t p s q r
3. Iso-electric point of the given amino acid will be
(1) 3.3 (2) 5.9 (3) 9.6 (4) 11.8
4. Identify the correct statements about amines. I) In Hoffman bromamide reaction, amine can be obtained as a final product. II) All disaccharides are reducing sugars.
III) Both aliphatic and aromatic secondary amines react with nitrous acid to yield N–nitroso-amines under suitable conditions.
(1) I, II only (2) II, III only
(3) I, III only (4) I, II, III only
5. Consider the following reactions:
(i)
MATCHING TYPE QUESTIONS
1. Match the Column-I with Column-II.
Column-I (Vitamin) Column-II (Deficiency leads to)
(A) Vitamin A (p) Muscular weakness
(B) Vitamin D (q) Increased blood clotting time
(C) Vitamin E (r) Night blindness
(D) Vitamin K (s) Osteomalacia
(A) (B) (C) (D)
(1) s r q p
(2) q p r s
(3) r q s p
(4) r s p q
2. Match the column-I with column-II.
Column-I Column-II
(A) RNA (p) Single stranded α− helix structure
(B) DNA (q) Forms the site for protein synthesis
(C) r–RNA (r) Double stranded α− helix structure
(D) m–RNA (s) Cytosine and uracil as pyrimidine base
(E) t–RNA (t) Carries the genetic information from DNA to ribosomes where the protein is synthesized
acetyl derivative
(ii) ( ) 2 3 2 Ni/H y eq.of CHCOO Glucose A
acetyl derivative
(iii) ( ) 3 2 z eq.of CHCOO Glucose acetyl derivative
‘x’, ‘y’, and ‘z’ in these reactions are respectively,
(1) 4, 6, and 5 (2) 5, 6, and 5
(3) 5, 4, and 5 (4) 4, 5, and 5
(u) Carrier of amino acid (v) Cytosine and thymine as pyrimidine base
(A) (B) (C) (D) (E)
(1) p,s r,v q t u
(2) p q r s t
(3) r s q t u
(4) p t,s q,r v q,r
3. Match List-I with List-II List-I (Vitamin) List-II (Deficiency disease)
(A) A (p) Pernicious anaemia
(B) B12 (q) Beri beri
(C) D (r) Rickets
(D) B6 (s) Xerophthalmia
(t) Convulsions
(A) (B) (C) (D)
(1) q p r t
(2) r s q t
(3) s p r t
(4) s p q t
4. Match the column-I and column-II.
Column I Column II
(A) COOH
H2N H H (p) Optically active amino acid
Glycine
(B) COH
Proline O N (q) Suitable for Van Slyke estimation
(C) Histidine COOH H2N CH2 N N H H (r) Neutral amino acid
(D) COOH
H2N
CH2COOH H (s) Basic amino acid
Aspartic acid
(A) (B) (C) (D)
(1) q,r p,r p,q,s pq
(2) p,s p,q p,q,r p,s
(3) q,r s,p r,s q,p
(4) r,q s,r s,r,q s,r
5. Match the column-I and column-II.
Column-I Column-II
(A) Glucose (p) Reduces Tollen’s reagent
(B) Fructose (q) Exhibits mutarotation in mild alkaline medium
STATEMENT TYPE QUESTIONS
Each question has two statements: statement I (S-I) and statement II (S-II)
(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
(C) Mannose (r) Produces tetra acetate derivative on treatment with anhydride and pyridine
(D) Glucopyranoside (s) Gets oxidised
(A) (B) (C) (D)
(1) p,q,s p,q,s p,q,s r
(2) p q r s
(3) q,p p,r s t
(4) q,p r q s,q
6. Match the column-I and column-II.
Column-I Column-II
(A) Sucrose (p) Amylose + Amylopectin
(B) Maltose (q) α–D-Glucose + β−D− Fructose
(C) Lactose (r) α-D-Glucose + α-DGlucose
(D) Starch (s) β−D−Galactose + β− D−Glucose
(A) (B) (C) (D)
(1) p r s q
(2) s p q r
(3) p p q r
(4) q r s p
(4) if statement I is incorrect, but statement II is correct
1. S-I : D – Glucose and D – Galactose are C – 4 epimers.
S-II : α−D fructose and β−D fructose are C-2 anomers.
2. S-I : Reduction of glucose with HI, Red P4/∆, gives n-hexane.
S-II : HI+Red P4/∆ can reduce –CHO and –OH group.
3. S-I : Primary structure of a protein tells about the sequence of amino acids in a polypeptide chain.
S-II : Quaternary structure of proteins tells about the shape of a polypeptide chain.
4. S-I : Glucagon is a hyperglycemic factor.
S-II : Addison’s disease is characterized by an increased susceptibility to stress.
5. S-I : α- helix is the most common secondary structure of a protein.
S-II : Insulin contains 51 amino acid residues, but it is classified as a protein.
6. S-I : A unit formed by the attachment of a base to the 1st position of sugar is known as nucleoside.
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
1. (A) : Rhamnose(C 6 H 12 O 5 ) is not a carbohydrate.
(R) : All compounds having the general formula Cx(H2O)y are carbohydrates.
2. (A) : Rate of hydrolysis of sucrose is higher in the presence of invertase than when carried out in the presence of dilute mineral acid.
S-II : When a nucleoside is linked to phosphorous acid at 5 ’ – position of sugar moiety, we get a nucleotide.
7. S-I : Carbohydrates having D-configuration are always dextro rotatory.
S-II : D-(+)-glucose and L-(-)-glucose are a pair of enantiomers.
8. S-I : Sucrose is a non reducing sugar.
S-II : In sucrose, glucose is in pyranose form and fructose is in furanose form.
9. S- I : Glucose and fructose give the same osazone.
S-II : Glucose and fructose have same configuration at C3, C4 and C5
10. S-I : Glucagon is a hyperglycemic factor.
S-II: Addison’s disease is characterised by increased susceptibility to stress.
(R) : Activation energy for enzyme catalysed hydrolysis of sucrose is less than that of acid catalysed hydrolysis.
3. (A) : β -D-glucopyranose is the most abundant naturally occurring aldohexose.
(R) : All the ring substituents in the chair conformation are equatorial.
4. (A) : In presence of an enzyme, substrate molecule can be attacked by the reagent effectively.
(R) : Active sites of an enzyme hold the substrate molecule in a suitable position.
5. (A) : Glycosides are hydrolysed in acidic conditions.
(R) : Glycosides are acetals.
BRAIN TEASERS
1. D-Aldopentose Optical active acid Meso acid ConHNO3 Br2 H2O (X)
X Y Z HCN NaCN H3O+ z can be?
(1)
(2)
2. DNA and RNA are composed of pentose sugar, nitrogen base and phosphoric acid. The nitrogen base that does not contain a keto(oxo) group in its structure is
(1) adenine (2) guanine
(3) cytosine (4) uracil
3. The enzyme present in human stomach is unable to break the following cleavage.
(1) α (1, 4) glycosidic linkage between β–D– glucose units.
(2) β (1, 4) glycosidic linkage between β–D– glucose units.
(3) α (1, 6) glycosidic linkage between α–D– glucose units.
(4) α, β (1, 2) glycosidic linkage btween β–D– glucose and β– D–fructose units.
4. For a neutral amino acid (X), isoelectric point is 5.8. Now, its solubility at this point in water is
(1) maximum
(2) minimum
(3) zero
(4) unpredictable
FLASHBACK (Previous NEET Questions)
1. The incorrect statement regarding enzymes is
(1) Like chemical catalysts, enzymes reduce the activation energy of bioprocesses.
(2) Enzymes are polysaccharides.
(3) Enzymes are very specific for a particular reaction and substrate.
(4) Enzymes are biocatalysts.
CHAPTER TEST
Section – A
1. Enzymes are
(1) complex substances produced in cells
(2) steroids
(3) living organisms
(4) dead organisms
2. The two functional groups present in a typical carbohydrate are
(1) –OH and –COH
(2) –CHO and –COOH
(3) > C = O and –OH
(4) –OH and –CHO
3. The ‘term anomers of glucose’ refers to (1) isomers of glucose that differ in configurations at carbons one and four (C-1 and C-4)
(2) a mixture of (D)-glucose and (L)-glucose
(3) enantiomers of glucose
(4) isomers of glucose that differ in configuration at carbon one (C-1)
4. Which class of carbohydrates cannot be hydrolysed further?
(1) Monosaccharides
(2) Disaccharides
(3) Polysaccharides
(4) Proteins
5. Which of the following monosaccharides is a pentose?
(1) Glucose
(2) Fructose
(3) Ribose
(4) Galactose
6. The difference between amylose and amylopectin is
(1) amylopectin has 1 → 4α–linkage & 1 → 6α–linkage.
(2) amylose have 1 → 4α–linkage & 1 → 6 b –linkage.
(3) amylopectin1 → 4α–linkage & 1 → 6 b –linkage.
(4) amylose is made up of glucose and galactose.
7. Reducing sugar among the following is (1) glycogen
(2) sucrose
(3) fructose
(4) amylose
8. Glucose gives silver mirror with tollen’s reagent. It shows the presence of
(1) an acidic group
(2) an alcoholic group
(3) an aldehyde group (4) a ketonic group
9. Match the List-I with List-II. List-I (Vitamin) List-II (Deficiency diseases)
(A) Thiamine (p) Pernicious anemia
(B) Cyanocobalamin (q) Convulsions
(C) Ascorbic acid (r) Cheilosis
(D) Pyridoxine (s) Beri-beri
(t) Scurvy (u) Osteomalacia
(A) (B) (C) (D)
(1) r p t u
(2) r p t q
(3) s p u r
(4) s p t q
10. Enzymes belong to which class of compounds?
(1) Polysaccharides
(2) Polypeptides
(3) Polynitro heterocyclic
(4) Hydrocarbons
11. What is the common name given to the enzyme that catalyses the oxidation of one substrate with the simultaneous reduction of another substrate?
(1) Reductioxidase (2) Oxidonductase (3) Oxidoreductase (4) Reductoxides
12. Which one of the following statements is incorrect about enzyme catalysts?
(1) Enzymes are mostly protein as in nature.
(2) Enzyme action is specific.
(3) Enzymes are denatured by ultraviolet rays and at high temperature.
(4) Enzymes are least reactive at optimum temperature.
13. The two functional groups present in a typical carbohydrate are
(1) −CHO and −COOH.
(2) −COOH and −C = O.
(3) −OH and −CHO.
(4) −OH and −COOH.
14. Read the following statements and choose the correct answer.
(i) All monosaccharides are reducing sugars.
(ii) All monosaccharides are not reducing sugars.
(iii) In disaccharides if aldehydic or ketonic or both aldehydic and ketonic groups are involved in glycosidic bond, these are non-reducing sugars.
(iv) In disaccharides, if aldehydic or ketonic groups are free, these are reducing sugars.
(1) (i), (iii) and (iv)
(2) (ii), (iii), and (iv)
(3) (i) and (iv)
(4) (ii) and (iv)
15. Which of the following is false?
(1) Sucrose is a non reducing sugar.
(2) Glucose is oxidised by bromine water.
(3) Glucose rotates plane polarised light in clock wise direction.
(4) Fructose is oxidised by bromine water.
16. α-D-glucose and β-D glucose are
(1) epimers
(2) anomers
(3) enantiomers
(4) both (1) & (3)
17. Glucose on reaction with Br 2 water gives
(1) saccharic acid
(2) hexanoic acid
(3) gluconic acid
(4) salicylic acid
18. Which of the following is wrong w.r.t proteins?
(1) Denaturation of protein destroys 1°, 2°, and 3° structures.
(2) Change in pH can bring denaturation of a protein.
(3) Heating a protein destroys its 2°and 3° structures.
(4) Denaturation can be reversible.
19. Number of peptide links in a tri peptide is (1) 1 (2) 2 (3) 3 (4) 4
20. Which of the following statements is not correct?
(1) Ovalbumin is a simple food reserve in egg-white.
(2) Blood proteins thrombin and fibrinogen are involved in blood clotting.
(3) Denaturation makes the proteins more active.
(4) Insulin maintains the sugar level in the blood of a human body.
21. Proper representation of the dipeptide, glycyl alanine is
(1) CH3 H H2N – CH – CO – NH – CH – COOH
(2) CH3 H HOOC – CH – NH – CO – CH – NH
(3) CH3 H2N – CH2 – CO – NH – CH – COOH
(4) CH3 HOOC – CH – NH – CO – CH2 – NH2
22. Select the correct statements about enzymes among the following:
(A) They have low molar mass.
(B) They are essentially fibrous proteins.
(C) They form colloidal sols with water.
(D) Many of them have been obtained in pure amorphous state from living cells.
(1) A, B, C, D
(2) A, B, C only
(3) B, C, D only
(4) C only
23. The name of a peptide is alanyl glycyl phenylalanyl aspargine. Select the correct statement about this peptide.
(1) It is used as an artificial sweetner.
(2) It contains three CO–NH–bonds.
(3) It contains two glycosidic linkages.
(4) It is a protein present in keratin.
24. Select the correct statements among the following.
A. Fibrous proteins are generally insoluble in water.
B. Insulin and albumins are examples of fibrous proteins.
C. Globular proteins are usually soluble in water.
D. Keratin and myosin are the examples of globular proteins.
(1) A and C (2) B, C, and D
(3) A, C and D (4) B and C
25. The stability of an α–helix structure of proteins depend upon
(1) dipolar interaction
(2) H–bonding interaction
(3) van der Waals forces
(4) π–stacking interaction
26. The secondary structure of a protein refers to
(1) fixed configuration of the polypeptide backbone
(2) α−helical backbone
(3) hydrophobic interactions
(4) sequence of α−amino acids
27. Regarding denaturation of a protein, choose the false statement.
(1) Denaturation of protein occurs on heating.
(2) Denaturation of protein occurs on changing the pH.
(3) Denaturation leads to the breaking of hydrogen bonds in proteins.
(4) Denaturation is always reversible.
28. Proteins and polypeptides are formed by the condensation of α-amino acids. The correct representation of the dipeptide, glycyl alanine is
(1) H2N–CH2–CO–NH–(CHCH3)–COOH
(2) H2N– (CHCH3)–CO–NH–CH2–COOH
(3) HOOC–CH2–NH–CO–CH(CH3)–NH2
(4) HOOC–CH(CH3)–NH–CO–CH2–NH2
29. Regarding denaturation of protein, false statement
(1) Denaturation of protein occurs on heating.
(2) Denaturation of protein occurs on changing the pH.
(3) Denaturation leads to the breaking of hydrogen bonds in proteins.
(4) Denaturation is always reversible.
30. Enzymes that utilize A.T.P in phosphate transfer require an alkaline earth metal (M) as a cofactor. M is
(1) Be (2) Mg (3) Ca (4) Sr
31. How many tripeptides can be prepared by linking the amino acids glycine, alanine and phenyl alanine?
(1) One (2) Three (3) Six (4) Twelve
32. –NH2 group is absent in (1) uracil (2) adenine (3) guanine (4) cytosine
33. The backbone of a nucleotide strand contains the following sequence of arrangement.
(1) Base–Sugar (2) Sugar–Phosphate (3) Base–Phosphate (4) Base1–Base2
34. During the formation of a nucleotide, esterification takes place between (1) 5 ’ OH of sugar and OH of phosphoric acid
(2) 31 OH of sugar and OH of phosphoric acid
(3) 51 OH of sugar and COOH of phosphoric acid
(4) 31 OH of sugar and COOH of phosphoric acid
Section-B
35. In nucleosides, the m th N-atom of the pyrimidine base is joined to the nth C-atom of the ribose sugar moiety. Here, m and n are
(1) 1, 1 (2) 1, 4 (3) 1, 2 (4) 1, 5
36. Which of the following is not present in DNA?
(1) Adenine
(2) Guanine
(3) Uracil
(4) Thymine
37. Given below are two statements.
Statement I: All enzymes are proteins but all proteins are not enzymes.
Statement II : Keratin is an enzyme.
In light of the above statements, choose the correct answer from the options given below.
(1) Both statement I and statement II are correct.
(2) Both statement I and statement II are incorrect.
(3) Statement I is correct but statement II is incorrect.
(4) Statement I is incorrect but statement II is correct.
38. Given below are two statements. One is labelled as Assertion (A) and the other is labelled as reason (R).
Assertion (A) : All naturally occurring α-amino acids except glycine are optically active.
Reason (R): Most naturally occurring amino acids have L-configuration.
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 and (R) is not the correct explanation of (A).
(3) (A) is true but (R) is false.
(4) Both (A) and (R) are false.
39. Among the following, achiral amino acid is (1) glycine. (2) alanine.
(3) proline. (4) tryptophan.
40. The amino acids that cannot be synthesized in the body but must be supplied through diet are (1) essential amino acids. (2) non-essential amino acids.
(3) α-amino acids.
(4) acidic amino acids.
41. The sex hormone which controls the development and maintenance of pregnancy is
(1) cortisone (2) thyroxine (3) progesterone (4) estrone
42. Number of stereocentres present in Linear and cyclic structures of glucose are, respectively (1) 4 and 4 (2) 4 and 5 (3) 5 and 5 (4) 5 and 4
43. The number of chiral centers in the openchain structure of Glucose is (1) 3 (2) 4 (3) 5 (4) 6
44. When glucose is heated with nitric acid, the product is______.
(1) lactic acid (2) saccharic acid
(3) glycolic acid (4) oxalic acid
45. Glucose is not (1) a hexose.
(2) a carbohydrate.
(3) an oligosaccharide.
(4) an aldose.
46. Given below are two statements..
Statement I : Primary structure of protein tells about the sequence of amino acids in a polypeptide chain.
Statement II : Quaternary structure of proteins tells about the shape of polypeptide chain.
In light of the given statements, choose the most appropriate answer from the options given below.
In light of the above statements, choose the correct answer from the options given below.
1) Both statement I and statement II are correct.
2) Both statement I and statement II are incorrect.
3) Statement I is correct but statement II is incorrect.
4) Statement I is incorrect but statement II is correct.
47. All the following are essential amino acids except (one letter symbols of amino acids are given)
(1) F (2) W
(3) L (4) G
ANSWER KEY
Neet Drill
LEVEL 2
48. From the following statements, identify the false statement regarding α–amino acids.
(1) Most of the naturally occurring amino acids belong to L–series.
(2) They are insoluble in water.
(3) They are the building blocks of proteins.
(4) They exist as zwitter ions.
49. Which one of the following does not contain −NH2 group?
(1) Proline (2) Histidine
(3) Arginine (4) Glutamine
50. Which one of the following is a neutral, essential amino acid?
(1) Lysine (2) Serine
(3) Valine (4) Glycine
Further Exploration (1) 2 (2) 4 (3) 2 (4) 3 (5) 1
Matching type Questions (1) 4 (2) 1 (3) 3 (4) 1 (5) 1 (6) 4
Statement Type Questions
Assertion and Reasoning Questions (1) 4 (2) 1 (3) 1 (4) 1 (5) 4
Brain Teasers
Chapter Test
