Published by Rankguru Technology Solutions Private Limited
IL Ranker Series Chemistry for JEE Grade 12 Module 5
ISBN 978-81-983876-5-3 [SECOND
EDITION]
This book is intended for educational purposes only. The information contained herein is provided on an “as-is” and “as-available” basis without any representations or warranties, express or implied. The authors (including any affiliated organizations) and publishers make no representations or warranties in relation to the accuracy, completeness, or suitability of the information contained in this book for any purpose.
The authors (including any affiliated organizations) and publishers of the book have made reasonable efforts to ensure the accuracy and completeness of the content and information contained in this book. However, the authors (including any affiliated organizations) and publishers make no warranties or representations regarding the accuracy, completeness, or suitability for any purpose of the information contained in this book, including without limitation, any implied warranties of merchantability and fitness for a particular purpose, and non-infringement. The authors (including any affiliated organizations) and publishers disclaim any liability or responsibility for any errors, omissions, or inaccuracies in the content or information provided in this book.
This book does not constitute legal, professional, or academic advice, and readers are encouraged to seek appropriate professional and academic advice before making any decisions based on the information contained in this book. The authors (including any affiliated organizations) and publishers disclaim any liability or responsibility for any decisions made based on the information provided in this book.
The authors (including any affiliated organizations) and publishers disclaim any and all liability, loss, or risk incurred as a consequence, directly or indirectly, of the use and/or application of any of the contents or information contained in this book. The inclusion of any references or links to external sources does not imply endorsement or validation by the authors (including any affiliated organizations) and publishers of the same.
All trademarks, service marks, trade names, and product names mentioned in this book are the property of their respective owners and are used for identification purposes only.
No part of this publication may be reproduced, stored, or transmitted in any form or by any means, including without limitation, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the authors (including any affiliated organizations) and publishers.
The authors (including any affiliated organizations) and publishers shall make commercially reasonable efforts to rectify any errors or omissions in the future editions of the book that may be brought to their notice from time to time.
All disputes subject to Hyderabad jurisdiction only.
Dr. B. S. Rao, the visionary behind Sri Chaitanya Educational Institutions, is widely recognised for his significant contributions to education. His focus on providing high-quality education, especially in preparing students for JEE and NEET entrance exams, has positively impacted numerous lives. The creation of the IL Ranker Series is inspired by Dr. Rao’s vision. It aims to assist aspirants in realising their ambitions.
Dr. Rao’s influence transcends physical institutions; his efforts have sparked intellectual curiosity, highlighting that education is a journey of empowerment and pursuit of excellence. His adoption of modern teaching techniques and technology has empowered students, breaking through traditional educational constraints.
As we pay homage to Dr. B. S. Rao’s enduring legacy, we acknowledge the privilege of contributing to the continuation of his vision. His remarkable journey serves as a poignant reminder of the profound impact education can have on individuals and societies.
With gratitude and inspiration
Team Infinity Learn by Sri Chaitanya
Key Features of the Book
Chapter Outline
1.1 Types of Solutions
1.2 Methods of Concentration
1.3 Solubility
This outlines topics or learning outcomes students can gain from studying the chapter. It sets a framework for study and a roadmap for learning.
Specific problems are presented along with their solutions, explaining the application of principles covered in the textbook. Solved Examples
1. What is the molality of a solution of H2SO4 having 9.8% by mass of the acid?
Sol. 9.8% by mass of H2SO4 contains 9.8 g of H2SO4 per 100 g of solution.
Therefore, if mass of solution = 100 g, mass of solute, H2SO4 = 9.8 g,
Try yourself:
1. In a solution of H 2 SO 4 and water, mole fraction of H2SO4 is 0.9. How many grams of H2SO4 is present per 100 g of the solution?
Ans: 98
Try Yourself enables the student to practice the concept learned immediately.
This comprehensive set of questions enables students to assess their learning. It helps them to identify areas for improvement and consolidate their mastery of the topic through active recall and practical application.
CHAPTER REVIEW
Types of Solutions
■ A solution is a homogeneous mixture of two or more non–reacting components. Formation of solution is a physical process.
TEST YOURSELF
1. The mole fraction of a solvent in aqueous solution of a solute is 0.6. The molality of the aqueous solution is (1) 83.25 (2) 13.88 (3) 37 (4) 73
It offers a concise overview of the chapter’s key points, acting as a quick revision tool before tests.
Organised as per the topics covered in the chapter and divided into three levels, this series of questions enables rigorous practice and application of learning.
These questions deepen the understanding of concepts and strengthen the interpretation of theoretical learning. These complex questions combining fun and critical thinking are aimed at fostering higher-order thinking skills and encouraging analytical reasoning.
Exercises
JEE MAIN LEVEL
LEVEL 1, 2, and 3
Single Option Correct MCQs
Numerical Value Questions
THEORY-BASED QUESTIONS
Single Option Correct MCQs
Statement Type Questions
Assertion and Reason Questions
JEE ADVANCED LEVEL
BRAIN TEASERS
FLASHBACK
CHAPTER TEST
This comprehensive test is modelled after the JEE examination format to evaluate students’ proficiency across all topics covered, replicating the structure and rigour of the JEE examination. By taking this chapter test, students undergo a final evaluation, identifying their strengths and areas of improvement.
Level 1 questions test the fundamentals and help fortify the basics of concepts. Level 2 questions are higher in complexity and require deeper understanding of concepts. Level 3 questions perk up the rigour further with more complex and multi-concept questions.
This section contains special question types that focus on in-depth knowledge of concepts, analytical reasoning, and problem-solving skills needed to succeed in JEE Advanced.
Handpicked previous JEE questions familiarise students with the various question types, styles, and recent trends in JEE examinations, enhancing students’ overall preparedness for JEE.
Chapter Outline
11.1 Structure of Amines
11.2 Classification of Amines
11.3 Nomenclature of Amines
11.4 Preparation of Amines
11.5 Physical Properties of Amines
11.6 Chemical Reactions of Amines
11.7 Diazonium Salts
11.8 Cyanides and Isocyanides
11.9 Conversions
Amines constitute an important class of organic compounds derived by replaced one or more hydrogen atoms of ammonia molecule by alkyl/aryl group(s). In nature, they occur among proteins, vitamins, alkaloids, and hormones. Synthetic examples include polymers, dyestuffs, and drugs. Two biologically active compounds, namely adrenaline and ephedrine, both containing secondary amino group, are used to increase blood pressure. Novocain, a synthetic amino compound, is used as an anaesthetic in dentistry. Benadryl, a well known antihistaminic drug, also contains tertiary amino group. Quaternary ammonium salts are used as surfactants. Diazonium salts are intermediates in the preparation of a variety of aromatic compounds, including dyes
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.
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.
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 isomerism. 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 isomeris m. 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
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 isomerism: 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.
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 name 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.
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.
Nitriles, on reduction, produce primary amines.
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 NH 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 i s the most convenient method for preparing primary amines. This method gives an amine containing one carbon atom less than that in amide.
Schmidt reaction: By treating a mixture of a carboxylic acid and hydrazoic acid with cold conc. H2SO4, a primary amine is formed.
HSO24 23 2222 RCHCOOHNHRCHNHNCO +→−−++
Mechanism: + 3 RCNRNCO || O + −−→−==
3 NaCl RCOHNaN || O −−+→
Hydrolysis 2 23 RNHCO→−+
Curtius reaction: In this method, acid chlorides are treated with sodium azide to produce acyl azide, which on heating followed by alkaline hydrolysis, gives primary amines.
HSO24 23 2222 RCHCOClNHRCHNHNCO −+→−−++
Mechanism: + 3 RCNRNCO || O + −−→−==
3 NaCl RCClNaN || O −−+→
Donating group increases reactivity and withdrawing group decreases reactivity.
Hydrolysis 2 23 RNHCO→−+
Lossen reaction: In this method, acid chlorides are treated with hydroxyl amine to produce acyl azide, which on heating followed by alkaline hydrolysis, gives primary amines.
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 (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 similar molecular masses
S. NO Compound
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)
(3)
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
Answer key
(1) 4 (2) 4 (3) 1 (4) 3
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.
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.
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
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:
R–N–R > R–NH–R >R–NH2> NH3 | (3°) (2°) (1°)
R
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 hy drogen bonding is
maxim um 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 a s follows:
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.
T her 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.
NH2 NH2
NH2 NH2
NH2
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.
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 + + →
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 + NaOH653 2 65 2 C O NH+NaCl+ HNHCC H lH →
A cylation: 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
42
C2H5
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.
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.
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.
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.
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, benzene sulphonyl chloride is replaced by p-toluene sulphonyl chloride.
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.
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.
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.
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.
Nitrat ion: Direct nitration yields tarry oxidation products in addition to a mixture 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
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.
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.
N-alkyl benzene sulphonamide formed dissolves in NaOH.
RNH2+C6H5SO2Cl → RNHSO2C6H5+HCl
2. Hofmann mustard oil test: Treated with CS2 and then with HgCl2 A black precipitate of HgS is formed along with alkyl iso-thiocyanate.
3. Action of HNO2: Aromatic primary amines give a stable diazonium salt.
N, N-dialkyl benzene sulphonamide is formed, but it is insoluble in NaOH. R2NH+C6H5SO2Cl → R2NSO2C6H5+HCl
No reaction
4. Carbyl amine test: Treated with CHCl3 and alc. KOH.
5. Oxidation with KMNO4
6. Caro’s acid test
RNH2+CS2→ black
No balck precipitate is formed. R2NH+CS2→ 2 || S RNCSH
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. 2 RNHRHNORNR NO +→−− ↓
No reaction
Tertiary amines give a salt. −−+→ 2 | R RNRHNO + | 2 | H R RNRNO
Primary amines give carbyl amine with unpleasant odour. AlcKOH 23 RNHCHClRNC +→
Aldehydes and ketones are formed.
RCH2NH2→ RCHO 2 RCHRRCOR | NH −−→
Aldoximes and hydroxamic acid are formed. () 25 O 22 HSO RCHNH → NOH OH RC
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 → →
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
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 05C
6522 CHNHNaNO2HCl −° ++→
652 2 CHNClNaCl2HO. + ++
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.
232 CHNNClHOHPO D =++→
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
CHNNClCHClN =→+
Sandmeyer reaction: Benzene diazonium chloride is converted to chlorobenzene when treated with cuprous chloride and hydrogen chloride. 22 HCl
65 652 CuCl
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.
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) HO
Aniline gives p–amino azobenzene when it undergoes coupling with benzene diazonium chloride, at a pH value of 5 to 7.
N=NCl
p-Amino azobenzene (yellow) NH2 NH2 H+ +
N=N
TEST YOURSELF
1. Aniline NaNO22HO HCl,OC B A oil B , AC ° →→→
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.
PO25 22 RCONHRCNHO →+
From oximes: Oximes undergo dehydration on heating with phosphorus pentoxide and give nitriles.
PO25 2 RCHNOHRCNHO −=→+
Table 11.5 Names of some cyanides and isocynaides Common name IUPAC name
The reaction can also be run in pyridine, which takes up the two acids in the form of their pyridine salts.
CH3CH2C(OH)=NH(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 + →+
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 HO RCNR'MgXRCR'RCR' || || NMg O X
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. alcohol
RXAgCNRNCAgX +→+
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. ()()() Ni/H2 Aceticanhydride NaCN 32 CHCHClXY Z →→→
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
5.
The above reaction is (1) Mendius reaction (2) Schmidt reaction (3) Rosemund reaction (4) Stephen’s reaction
■ 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.
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/ HCl, Cu 2Br 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
CHONOCHCHONO −−=−−−=
3 32 methylnitrite ethylnitrite
■ Benzene derivatives in which –NO2 group is bonded to benzene carbon (C-N bond) are called nitrobenzenes.
nitrobenzene NO2 o-dinitrobenzene NO2 NO2
■ 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. 22 0 RCHNO 1 2 0 RCHNO | R 2 2
Na3AsO3+NaOH(or) Gloucose+NaOH
C6H5NO2 +6(H)
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 + D +→+
o HSO
66 2 6622 60C CH HONOCHNO HO < +→+
■ Nitrobenzene is prepared by heating benzene diazonium fluoroborate with sodium nitrite in presence of copper.
NaNO2
654 65242 Cu,heat CHNNBFCHNO+NaBF+N =→
■ 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:
■ 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
■ 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)
NH2 NO2 +6NH3+2H2O+3S
■ Electrophilic substitution reaction in nitrobenzene:
■ 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.
Alcohol 25 2 252 CHClAgNOCHNOAgCl D +→+
■ Decarboxylation of a-nitro acetic acid gives nitromethane.
■ 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, 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.
■ On reduction with tin and hydrochloric acid, ethyl nitrite forms ethyl alcohol and ammonia. A small quantity of hydroxylamine is also formed.
■ Direct nitration of ethane with fuming nitric acid at 470°C gives nitroethane.
0 470C 263 2522 CHHNOCHNOHO +→+
■ 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 1° and 2° 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. 2 Nitroform acinitroform RCHNORCHNOH | | O O ++ −−=−=−
NaOHRCHNONa | O + →−=−
■ 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 + −+→+
HCl
2R-CHNO2RCONOHO →++
2222
■ Alkyl cyanides on reduction with Na/ C 2 H 5 OH or H 2 /Raney Ni or LiAlH or catalytic reduction, amine are formed
Reduction RCNRCHNH22 →−
■ Oxidati on of aniline with acidified dichromate forms p-benzoquinone.
NH2
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.
Alcohol 6522 652 2 KOH
2CHNHCS(CHNH)CSHS +→=+
HCl
652 65652 (CHNH)CSCHNCSCHNH =→+
■ With CS 2 in presence of HgCl 2 , aniline gives isothiocyanate.
2 HgCl 6522 65 CHNHCSCHNCSHgS2HCl +→==+↓+
■ The isothiocyanates have a characteristic smell like mustard oil; the reaction is called Hofmann’s mustard oil reaction.
Exercises
JEE MAIN LEVEL
Level - I
Structure of Amines
Single Option Correct MCQs
1. 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
Classification of Amines
Single Option Correct MCQs
2. Which of the following is not a tertiary amine ?
(1) triethyl amine
(2) trimethyl amine
(3) 2-methyl propanamine-2
(4) N-ethyl-N-methyl propanamine-1
Electronic Structure of Amines
Single Option Correct MCQs
3. Number of resonating structures of aniline is _______
(1) 5 (2) 2 (3) 1 (4) 4
Nomenclature of Amines
Single Option Correct MCQs
4. Common Name of N, N-dimethylpropan2-amine is ________.
(1) diethyl isopropyl amine
(2) dimethyl isopropyl amine
(3) N, N-dimethylisopropylamine
(4) N, N-dimethylaminepropane
5. 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
6. IUPAC name of (CH3)3CNH2 is ______.
(1) trimethyl amine
(2) 2-methyl butanamine
(3) 2-methyl-2-propanamine
(4) 2-methyl propanamine
7. Common name of Phenylmethanamine is ____________.
(1) Amino benzene (2) Benzylamine
(3) phenyl amine (4) benzyl amine
8. What is the IUPAC of aniline?
(1) Amino benzene
(2) Benzenamine
(3) phenyl amine
(4) benzyl amine
Preparation of Amines
Single Option Correct MCQs
9. Nitroalkane on reduction gives _____.
(1) 1o amine (2) 2o amine
(3) 3o amine (4) amide
10. Ethanamine can be obtained from methyl iodide by the action of alc. KCN followed by _______.
(1) hydrolysis
(2) oxidation
(3) reduction
(4) action of NH3/heat
11. Treatment of ammonia with an excess of ethyl chloride will yield ________.
(1) diethylamine
(2) methylamine
(3) tetraethylammoniumchloride
(4) ethane
12. Which compound on reduction with LiAlH4 gives n – butylamine of
(1) n-butyl cyanide
(2) n-propyl cyanide
(3) Propyl, methyl amine
(4) Ethyl cyanide
13. 3 2 CHCl/alc·KOH HO/NaOH 6522 CHCHNH X Y →→ . ‘Y’ is ________.
(1) C6H5CN
(2) C6H5NC
(3) C6H5CH2NH2
(4) C6H5CH2OH
14. N,N-dimethyl ethanamide on reduction with LiAlH 4 gives ______.
(1) N-methylethanamine
(2) N, N-dimethylethanamine
(3) ethanamine
(4) trimethyl amine
15. Gabriel synthesis is used for the preparation of _____.
(1) primary aromatic amines
(2) primary aliphatic amines
(3) secondary amines
(4) tertiary amines
16. Treatment of sodium azide with ethyl chloride yields _______.
(1) triethyl amine
(2) quaternary ammonium salt
(3) diethyl amine
(4) ethyl amine
17. Acetoneoxime on catalytic hydrogenation gives ________.
(1) 1-propanamine
(2) isopropylamine
(3) ethylmethylamine
(4) CH4 and ethanamine
18. What is the major product of the reaction given below () () 3 2 iNH 3 iiH/Ni CHCHO−→
(1) CH3−CH2−NH2
(2) CH3−CH2−CH2−NH2
(3) CH3−NH−NH−CH3
(4) H2N−CH2−CH2−NH2
19. RCONHR + N2
What is the name of the above reaction?
(1) Claisen-Schmidt reaction
(2) Kolbe-Schmidt reaction
(3) Schmidt reaction
(4) Kolbe’s reaction
Isomerism in Amines
Single Option Correct MCQs
20. Which of the following on Hoffmann’s bromamide reaction gives alkanamine?
(1) RCH2NH2
(2) RCONHR
(3) RCONH2
(4) RCOONH4
21. Which of the following shows optical activity?
(1) butanamine-1
(2) butanamine-2
(3) isopropylamine
(4) ethyl methyl amine
22. How many primary amines are possible for the formula C4H11N?
(1) 1 (2) 2 (3) 3 (4) 4
Physical Properties of Amines
Single Option Correct MCQs
23. Which of the following turns brown on exposure to air and light?
(1) Nitrobenzene
(2) m-Dinitrobenzene
(3) Aniline
(4) Benzene diazonium chloride
24. Primary amines have lower boiling points than _______.
(1) corresponding alkanes
(2) corresponding 2o and 3o amines
(3) corresponding esters
(4) corresponding alcohols
25. Aniline is soluble in ________.
(1) dil. HCl
(2) dil. NaOH
(3) Water
(4) Na2CO3 solution
Chemical Properties of Amines
Single Option Correct MCQs
26. The major product obtained in the following reaction is:
CH3–NH2 + H2O →
(1) 33 CHNHOH+−
(2) CH3-CN
(3) CH3-NH-CO-NH-CH3
(4) CH3-NH-CO-CH3
27. Aniline forms anilinium salt when it reacts with ______.
(1) an alkyl halide
(2) acetyl chloride
(3) sulphuric acid
(4) benzoyl chloride
28. Arrange the following in the correct order of their basic character
I) NH3
II) CH3NH2
III) C6H5NH2
(1) III > II > I
(2) II > III > I
(3) II > I > III
(4) I = II = III
29. What is the correct order of base strength of the given amines?
(1) o – nitroaniline > p – nitroaniline > m – nitroaniline
(2) m – nitroaniline > p – nitroaniline > o – nitroaniline
(3) m – nitroaniline > o – nitroaniline > p – nitroaniline
(4) o – nitroaniline > m – nitroaniline > p – nitroaniline
30. What is the final product when aniline is treated with excess of CH 3I?
(1) C6H5NHCH3 (2) C6H5N(CH3)2
(3) C6H5N+(CH3)3I (4) C6H5I
31. A compound with molar mass 180 is acylated with CH3COCl to give a compound having molar mass 390. The number of amino groups present per molecule of the former compound is ______.
(1) 4 (2) 6 (3) 2 (4) 5
32. The major product obtained in the following reaction is:
NH3 + CuCl2 →
(1) ()() 2 3 4 2 CuNHCl +
(2) () 2 3 4 CuNH + +2
(3) 3 NHCl2Cu +
(4) ()() 3 4 CuNHCl +
33. What is A?
(1) (2) (3) (4)
34. What are the reagent/s that are used in the diazotisation of aniline?
(1) HNO3 and HCl
(2) HNO2 only
(3) NaNO2 and HCl at 0–5°C
(4) NaNO2 and HNO2 at 0–5°C
35. What is the product form when ethylamine reacts with methyl magnesium bromide?
(1) CH3CH3
(2) CH4
(3) CH3CH2CH3
(4) CH3CH2CH2CH3
36. The major gaseous product obtained in the following reaction:
CH3 – NH2 + COCl2 →
(1) CH3 – N = C = O
(2) CH3 – C ≡ N
(3) CH3 – NH – CO – NH – CH3
(4) CH3NHCOCH3
37. Which of the following does not give nitroalkane?
(1)
(2) 2 alc. AgNO 25 CHI →
(3) 3 Fuming of HNO CHCH33 →
(4) Both (1) and (2)
38. The reagent that gets attached to the nucleus when added to aniline is ______.
(1) CH3I
(2) C6H5COCl
(3) CH3COCl
(4) Br2
39. Aniline on reaction with con.H 2SO 4 gives X. What product will form if X is heated?
(1) Sulphanilic acid
(2) Sulphonamide
(3) Benzene sulphonyl chloride
(4) m-Amino benzene sulphonic acid
40. A niline gives meta derivative as major product with ______.
(1) CH3COCl/pyridine
(2) HNO3+H2SO4
(3) Br2/water
(4) CH3Cl/pyridine
41. Aniline on acetylation gives _____.
(1) phenol
(2) acetamide
(3) acetanilide
(4) benzene
42. Treatment of ammonia on excess of alkyl halide yields ______.
(1) triethyl amine
(2) quaternary ammonium salt
(3) diethyl amine
(4) ethyl amine
43. What is the product form when aniline is heated CS2 in the presence of HgCl 2?
(1) Phenyl cyanide
(2) Phenyl isocyanide
(3) Phenyl isothiocyanate
(4) p-Aminobenzene sulphonic acid
44. Nitrosomines (R2N – N = O) are soluble in water. On heating with conc. H 2SO4, they give secondary amines. What is the name of the reaction?
(1) Perkin’s reaction
(2) Sandmeyer’s reaction
(3) Fittig reaction
(4) Liebermann nitroso reaction
45. Chemical formula of Hinsberg’s reagent is _____.
(1) C6H5SO3H
(2) C6H5NHSO2C6H5
(3) C6H5SO2Cl
(4) C6H5NHCOC6H5
46. For carbyl amine reaction, we need alcoholic KOH and ______.
(1) any primary amine along with chloroform
(2) aromatic primary amine along with chloroform
(3) aliphatic primary amine along with chloroform
(4) any amine along with chloroform
47. Ethyl amine can be distingished from aniline by ______.
(1) Tollens’ reagent
(2) Schiff’s reagent
(3) Azodye test
(4) Carbylamine test
Applications of Amines
48. Benadryl, an antihistamine drug contains ____.
(1) a primary amine group (2) secondary amine group (3) tertiary amine group (4) all the above
Cyanides and Isocyanides
49. CH 3 – CO – NH 2 + X pyridine → CH 3CN. Here ‘X’ is __________.
(1) C6H5SO2Cl
(2) Ethanolic AgCN
(3) (CH3CO)2O
(4) CHCl3
50. What is the major product when ethyl chloride treated with ethanolic AgCN?
(1) Ethyl cyanide
(2) Ethyl isocyanide
(3) Ethyl nitrate
(4) Nitroethane
51. Hydrolysis of acetonitrile in acidic medium gives ______.
(1) CH3CH2OH (2) CH3NC
(3) CH3CH2CHO
(4) CH3COOH
52. Alkyl isocyanide on reduction with Zn – Hg/ HCl gives ______.
(1) primary amine
(2) tertiary amine
(3) N – alkyl alkanamine
(4) N – methyl alkanamine
Organic Compounds Containing Nitrogen
53. Nitrobenzene is categorized as ________.
(1) Nitroalkanes
(2) Nitroarenes
(3) Nitroparaffins
(4) Both 1 and 3
54. What is the IUPAC nomenclature of TNT?
(1) 2,4,6-trinitrotoluene.
(2) 1,2,3-trinitrotoluene.
(3) 1,3,6-trinitrotoluene.
(4) 2,4,6-trinitrobenzene
55.
Determine the number of positional isomers for the compound?
(1) 4
(2) 3
(3) 6
(4) 7
56. Ethyl chloride on reaction with alcoholic AgNO3 forms _______.
(1) CH3CH2NO2
(2) CH3-CH3
(3) CH3CH2NH2
(4) CH3NO2
57. On reduction, primary amine is formed by ____.
(1) 1-nitroethane
(2) ethylnitrite
(3) azobenzene
(4) ethylcarbylamine
58. 24 663 conc. HSO CH+HNO 100C °
What is the main product of the above reaction?
(1) Nitrobenzene
(2) o-Dinitrobenzene
(3) m-Dinitrobenzene
(4) p-Dinitrobenzene
59. Choose the correct statements regarding nitro compounds.
(1) The boiling point of nitro-alkanes is greater than nitroarenes.
(2) The boiling point of nitro compounds is greater than their corresponding hydrocarbons.
(3) The boiling point of nitro compounds is lower than their corresponding alkyl nitrites.
(4) The high boiling point of nitro compounds is majorly due to strong van der Waal forces of attraction.
60. Which diazonium salt is stable at room temperature?
(1) Benzene diazonium chloride
(2) Benzene diazonium fluoroborate
(3) Benzene diazonium nitrate
(4) Benzene diazonium bromide
61. Identify X. 1. Fe/HCl
322 2. NaOH CHCHNOX →
(1) CH3CH2NH2
(2) CH3CH2OH
(3) CH3-CH3
(4) CH3CH2Cl
62. Identify X.
1. Zn/NaOH 652 2. CHNOX D →
(1) C6H5NH2
(2) C6H5-NH-NH-C6H5
(3) C6H6
(4) C6H5OH
63. Identify X.
42 1. Zn/NHCl, HO
32 2. CHNO X D →
(1) CH3NHOH (2) CH3NH2 (3) CH4 (4) CH3Cl
64. What is X?
(1) C6H5NH2
(2) C6H5NHOH
(3) p - amino phenol
(4) Hydrazobenzene
65. The conversion of m-dinitrobenzene into m-nitro aniline can be done using _____.
(1) (NH4)2S
(2) Zn + NaOH
(3) Sn + HCl
(4) Zn + NH4Cl
66. The number of moles of hydrogen atoms required to convert one mole of nitrobenzene to hydrazobenzene is _____.
(1) 5 (2) 10
(3) 4 (4) 8
67. Identify X.
2 HCl/HO 22 RCHNOX D →
(1) RCOOH (2) RCH2COOH
(3) RCH2NH2 (4) RCH2NHOH
68. The electrophile, E+ attacks the benzene ring to generate the intermediate σ-complex. Of the following, which σ-complex has the lowest energy?
(1) (2)
(3) (4)
69. Which of the following is used in the preparation of floor polishes?
(1) Nitrobenzene
(2) Aniline
(3) Benzene diazonium chloride
(4) Phenyl hydroxyl amine
Diazonium Salts
Single Option Correct MCQs
70. In benzene diazonium chloride, the functional group is _______.
(1) – N = N = Cl (2) – N = N+ – Cl–(3) – N+ = N – Cl– (4) – N+= N–Cl
Structure of Diazonium Salts
Single Option Correct MCQs
71. Which is the incorrect resonance structure of benzene diazonium ion?
Physical Properties of Diazonium Salts
Single Option Correct MCQs
73. Which of the following is not a property of diazonium salts?
(1) Diazonium salts are colourless crystalline solids.
(2) Being ionic in nature they are soluble in water.
(3) Most of these salts decomposes when dried.
(4) The aqueous solutions of these salts are poor conductors of electricity.
29. The correct statement regarding the basicity of arylamines is
25. Number of isomeric aromatic amines with molecular formula C 8 H 11N, which can be synthesised by Gabriel Phthalimide synthesis is ___.
(1) Arylamines are generally more basic than alkylamines, because the nitrogen atom in arylamines is sp-hybridised
(2) Arylamines are generally less basic than alkylamines because the nitrogen lone pair electrons are delocalised by interaction with the aromatic ring πelectron system
(3) Arylamines are generally more basic than alkylamines because the nitrogen lonepair electron are not delocalised by interaction with the aromatic ring π electron system
(4) Arylamines are generally more basic than alkylamines because of aryl group
30. In the following compounds
The order of basicity is
(1) IV > I > III > II
(2) III > I > IV > II
(3) II > I > III > IV
(4) I > III > II > IV
31. In the following compounds the decreasing order of basic strength is
(1) C2H5NH2 > NH3 > (C2H5)2NH
(2) (C2H5)2NH > NH3 > C2H5NH2
(3) NH3 > C2H5NH2 > (C2H5)2NH
(4) (C2H5)2NH > C2H5NH2 > NH3
32. A Positive carbylamine test given by
(1) N, N-Dimethyl aniline
(2) N-Methyl aniline
(3) N-Methyl-O-methyl aniline
(4) p-methyl benzylamine
33. Aniline is oxidised with K2Cr2O7/H2SO4. The product formed is _______
(1) Nitrobenzene
(2) p-benzoquinone
(3) Nitrosobenzene
(4) Phenyl hydrozylamine
34. Carbyl amine is formed from aliphatic or Aromatic primary amine involving which of the following intermediate.
(1) Carbanion
(2) Carbene
(3) Carbocation
(4) Carbon radical
Numerical Value Questions
35. How many of the following are more basic than phenyl methanamine
i) Benzanamine
ii) N, N-dimethyl aniline
iii) N-Ethylethanamine
iv) N, N-dimethyl methanamine
v) N-Methyl phenyl methanamine
vi) 2,6-Dimethylaniline
vii) N-Phenyl ethanamide
36. How many of the following is/are correct order of the basic strength
1) Me3N > Me2NH > MeNH2 > NH3 (in gas phase)
2) Me 2NH > Me 3N > MeNH 2 (in protic solvent)
3) Et2NH > EtNH2 > Et3N (in protic solvent)
4) NH NH2 Me NH2 NH3 > > > (in protic solvent)
5) NH2 NH2 NO2
6) N N H
37. How many of the following compounds give carbylamine on reaction with CHCl 3 in presence of alcoholic KO H?
Diazonium Salts
Single Option Correct MCQs
38. In the diazotisation of aniline, which set of the reagents used
(1) HNO3, HCl
(2) NaNO2, HCl at 0−5°C
(3) NaNO2, HNO3 at 0−5°C
(4) HNO3 only
39. Which of the following statements is incorrect
(1) benzene diazonium fluoroborate is stable at room temperature
(2) benzene diazonium fluoroborate is almost insoluble in water
(3) diazonium salts are stabile due to +I effect of benzene
(4) cyanobenzene easily prepared from diazonium salt
40. Benzenediazonium chloride on reacting with phenol in weakly basic medium gives
(1) Benzene
(2) Chlorobenzene
(3) Diphenyl ether
(4) p-hydroxy azobenzene
41. The reaction: Cu/HCl 2 2 ArNClArCl+N +− → is known as :
(1) Sandmeyers reaction
(2) Finkelstein reaction
(3) Gattermann reaction
(4) Balz–Schiemann reaction
Numerical Value Questions
42. How many of the following compounds will not undergo azo coupling reaction with benzene diazonium chloride?
1) Phenol 2) Nitrobenzene
3) Anisole 4) Aniline
Multiple Concept Questions
Single Option Correct MCQs
43. An organic compound ‘A’ on reaction with NH3 followed by heating gives compound B. Which on further strong heating gives compound C(C 8 H 5 NO 2 ). Compound ‘C’ on sequential reaction with ethanolic KOH, alkyl chloride and hydrolysis with alkali gives a primary amine. The compound ‘A’ is: (1) CHO CHO (2) CHO COOH (3) COOH COOH (4) CHO OH
44. The major product obtained in the following reaction is CN O NH2 i) CHCl3/KOH ii)LiAIH4 (1) N CH3 CN O (2) CN OH NH CHCl2 (3) CN OH NHCH3 (4) OH NHCH3 NH2
45. W hi ch is the incorrect statement in the followin?
(1) Methylamine is more basic than ammonia.
(2) Amines form hydrogen bonds.
(3) Ethylamine has higher boiling point than propane.
(4) Dimethylamine is less basic than methylamine.
46. The correct order of basicity of the following compounds is
(a) CH3 C NH NH2 (c) CH3NH2
(b) CH3 C O NH2 (d) (CH3)2NH
(e) CH3CN
(1) b > a > c > d > e (2) a > c > b > d > e (3) c > a > b > e > d (4) a > c > b > e > d
47. Cyclohexylamine when treated with nitrous acid yields (P). On treating (P) with PCC results in (Q). When (Q) is heated with dil NaOH . we get (R) the final product (R) is (1)
49. The tests performed on compound ‘X’ and their inferences are :
Test
Observation
(a) 2,4-DNP test Yellow-orange coloured precipitate
(b) Iodoform test Yellow precipitate
(c) Azo-dye test No dye formation
based on above observations, the compound ‘X’ is
Numerical Value Questions
50. What is the value of x + y in the following reaction?
R–CONHBrNaOHR–NHNaCONaBr2HO ++→+++ xy
2 R–CONHBrNaOHR–NHNaCONaBr2HO ++→+++ xy
48. An organic compound [A]C 4 H 11N, shows optical activity and gives N2 gas on treatment with HNO2. The compound A reacts with PhSO2Cl producing a compound which is soluble in KOH. the structure of A is
NH2 0-5oC NaNO2-HCl B OH PH=8-9 II) NH4Cl A I)
If molar mass of compound B is x then the value of 2 x is ___.
Level - III
1. Which of the following is an enamine? (1) N
(2) N
2. The primary amino group is absent in (1) p-amino phenol
(2) o-amino phenol
(3) N-methyl ethanolamine
(4) phenylamine
3. Which of the following statements is wrong?
I) Amines possess a pyramidal shape
II) Amines acts as Bronsted–Lowry bases
III) 1° amines show metamerism
IV) 2° amines show metamerism
(1) I, II, and III (2) II, III, and IV
(3) III only (4) I, II, and IV
4. The number of amines that are aliphatic amines from the following list is ____.
5. 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
6. The number of cyclic primary amines possible for C 4 H 9 N (excluding stereo isomers)
7.
(Nitrogen containing compound) F cannot be prepared by (1) O || 322 CH-C-NH+Br+aq.NaOH → (2) 2 H/Ni Ethanolic NaCN 3 CHCl (i)(ii)→→ (3) Sn+HCl CHNO32 → (4)
14. In the reaction ‘C’ is NH2 NaNO2 HCl 0o-5oC NHCH3 C (A) B
(1) NH-NH N-CH3 H
(2) NHCH3 N=N
(3) CH3 -N N=N -CH2
(4) N=N NH2
17. The final product ‘C’ is the following series of reactions is NO2 NaNO2 Sn/HCl - Naphthol HCl NaOH A BC β →→→ is (1) OH N=N (2) N=N
(3) OH N=N (4)
18. A compound ‘X’ on treatment with Br 2 / NaOH, provided C 3 H 9N gives a product. This product gives positive carbylamine test. Compound ‘X’ is
(1) CH3COCH2NHCH3
(2) CH3CH2COCH2NH2
(3) CH3CH2CH2CONH2 NO2
(4) CH3CON(CH3)2
19. The order of basicity among the following compounds is:
(P) N H
(Q) N
(R) N N
(1) Q>R>P (2) R>Q>P
(3) R>P>Q (4) Q>P>R
20. Which of the given option has the correct basic strength order?
(1) (2)
THEORY-BASED QUESTIONS
Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II). Mark the correct answer as
(4) OH– < SH– < SeH–
21. The correct sequential order of the reagents for the given reaction is NO2
(1) HNO2, Fe/H+, HNO2, KI, H2O/H+
(2) HNO2, KI, Fe/H+, HNO2, H2O/warm (3) HNO2, KI, HNO2, Fe/H+, H2O/H+ (4) HNO2, Fe/H+, KI, HNO2, H2O/warm
22. In the following reaction sequence the amount of E (in gram) formed from 25 mole of aniline is _____.
(Atomic weight in g/mol; H=1, C=12, Br=80, N=14, O=16.The yield (%) corresponding to the product in each step is given in the parenthesis)
(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 Hofmann 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 Hofmann degradation reaction to electron deficient atom.
3. S-I : Pure aniline and other arylamines are usually colourless.
S-II : Arylamines get coloured on storage due to atmospheric reduction.
Assertion and Reason Questions
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A).
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A).
(3) if (A) is true but (R) is false.
(4) if both (A) and (R) are false.
4. (A) : Trityl amine (Triphenylmethylamine) is a tertiary amine.
(R) : N in Trityl amine is attached with no hydrogen.
5. (A) : Aniline is more stable than anilinium ion.
(R) : Aniline is more resonance stabilised than anilinium ion.
6. (A) : Pyrrole is an aromatic heterocyclic compound.
(R) : It has cyclic delocalised 6π electrons.
7. (A) : Tertiary amines have lower boiling point than secondary amine.
(R) : Hydrogen bonding is often responsible for an increase in boiling point.
8. (A) : Pyridine is more basic than pyrrole.
(R) : In pyridine nitrogen is sp2 hybridised whereas in pyrrole N is sp3 hybridised
9. (A) : N, N-dimethyl aniline does not react with nitrous acid, whereas aniline react with nitrous acid.
(R) : N, N-dimethyl aniline is less basic than aniline.
10. (A) : Coupling of aniline with diazonium chloride occurs in slightly acidic medium.
(R) : Aniline is less basic than aliphatic amines.
11. (A) : Primary amines with three (or) more carbon atoms are liquids and still higher ones are solids.
(R) : Lower aliphatic amines are soluble in water because they can form hydrogen bonds with water molecules.
12. (A) : The boiling point of nitro alkanes is higher than their corresponding alkyl nitrites.
(R) : Nitro alkanes show strong dipole–dipole interaction.
13. (A) : 3°-amines have lower boiling points than isomeric 1°and 2°-amines.
(R) : 1° and 2° amines can form intermolecular H-bonding but not 3° amines.
14. (A) : Higher aliphatic amines are less soluble in water than lower aliphatic amines.
(R) : Solubility of amines increases with increase in molecular mass of amines due to increase in size of hydrophobic alkyl part.
15. (A) : Aniline does not undergo FriedelCrafts reaction.
(R) : NH2 group of aniline reacts with AlCl3 (Lewis acid) to give acid–base reaction.
16. (A) : Aniline on nitration yields ortho, meta, and para nitro derivatives of aniline.
(R) : Nitrating mixture is a strong acidic mixture.
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1. Amongst the compounds given, the one that form bright colored dye on treatment with NaNO2/conc. HCl followed by addition of N, N-dimethylaniline is
4. Which of the following reactions gives benzylamine as product?
2. Consider the following amines:
Choose the correct statement(s):
(1) P is more basic than Q due to inductive effect.
(2) P is less basic than R due to resonance.
(3) Basic nature follows the order Q<P<S<R
(4) S is more basic than R due to hydrogen bonding.
3. Which of the following order for basic strength is/are correct?
5. Identify the compounds which will give clear solution with Hinsberg reagent (C6H5SO2Cl) followed by KOH
6. In which of the following reaction, secondary amine can be final product?
(3) + N H CHO NaBH2CN
(4) NH LiAlH4 H2O (1) (2)
7. Which of the following amines cannot be prepared by Gabriel Phthalimide synthesis?
(1) C6H5NH2
(2) H2C = CH – NH2
(3) (CH3)3N
(4) (CH3)2NH
8. Which of the following is true regarding Hoffmann bromamide reaction?
(1) It is an intramolecular rearrangement involving cyclic transition state.
(2) Reaction occurs through formation of alkyl isocyanate intermediate
(3) Reaction occurs through formation of alkyl isocyanide intermediate
(4) Aromatic amides containing deactivating groups are more reactive than aromatic amides containing activating groups
9. Which of the following given reaction/s is/ are correctly matched
(1)
10. Which of the following statements is/are correct
(1) In gas phase the basic strength order among the three types of amine is 3° > 2° > 1°
(2) Among the isomeric amines boiling points order is 3° > 2° > 1°
(3) Alcohols are more water soluble than amines of comparable molecular weight.
(4) The C–N bond in aromatic amines is shorter than that of in aliphatic amines
11. Reduction of nitrobenzene by which of the following reagents gives aniline?
(1) Sn/HCl
(2) Fe/HCl
(3) H2–Pd
(4) Sn/NH4OH
12. How many of the following synthesis will fail to give following as final major produc t
(3) Tertiary amines show intermolecular hydrogen bonding.
(4) Amines have lower boiling points as compared to those of alcohols and carboxylic acids of comparable molecular mass.
16. In the following set of sequence of reactions, choose the correct statements. ()()() () + 2 HO/H 1515 79 882 P QR AminocompoundsAcid
A single mononitro derivative ()() () + 2 i)diazotisation ii)CuCN iii)HO/H 79 882 QR Aminocompounds
(1) The reaction of P giving Q and R is hydrolysis of an amide.
(2) The total number of sp 3 hybridised carbons in P are 2.
(3) Q giving R involves one of the reactions as sandmeyer reaction.
(4) S with ethylene glycol undergoes condensation polymerisation under suitable conditions.
Numerical Value Questions
17. The major product in the following reaction contains how many N-atoms N N N NH2 CH3I base + N
18. 1.86 g of aniline completely reacts to form acetanilide, 10% of the product is lost during purification. Amount of acetanilide obtained after purification (in grams ) is x × 10–2.The value of x is
19. 1, 4-dinitrobenzene 4 22 NHHSBr/HONaNO2PQR →→→ 22 Br/HO CuBr-HBrSn/HClST →→→ major end product.
Find the total number of halogen atoms present in the major end product?
20. What is the maximum number of compounds with the molecular formula C 4 H 11 N, (including stereoisomers) which give an alkali soluble precipitate with benzene sulfonyl chloride?
21. I F NaCN 60% x(major product) Δ NO2
Find out weight of the major product ‘X’. (given:C = 12, O = 16, H = 1, N = 14, I = 127, F = 19, Na = 23). (round off to nearest integer). ________
22. The number of mole of NaOH consumed in the following one mole of amide conversion:
Integer Value Questions
23. How many compounds are less basic than aniline.
26. How many of the following compounds are more basic than aniline?
24. Ho w many major contributing resonance structures are possible for resonance hybrid of phenyldiazonium ion ?
27. The number of mole of electrons transferred in the reduction of one mole of nitrobenzene by Sn, HCl is ______.
28. For the molecular formula C 8H11N (all are benzene derivatives), x is the number of structural isomers which give isocyanide test and y structural isomers which do not give isocyanide test, then find the value of (x-y).
29. How many of the following are correct statements
A) Aliphatic amides are weaker bases than aliphatic amines.
B) PhCONH 2 is stronger base than CH3CONH2
C) C6H5NH2 is weaker base than C2H5 C2H5 C2H5 C2H5 N
25. In how many of the following substrates nitrogen is intact with initial substrate after repeative exhaustive methylation followed by Hoffman eliminaton ?
G) Basic strength of para-CH3C6H4NH2 > C6H5NH2 > ortho-CH3C6H4NH2
H) Tertiary aromatic amines react with nitrous acid to yield p-nitroso derivatives
(E) (F) (G) (H) (I)
(C) (D)
30. T he number of hydrogen atoms required to convert 1 mole of nitrobenzene to hydrazobenene is
31. From among following pairs, number of pairs in which first compound is more basic than second compound is
34. Which is the rate determining step in Hofmann bromamide degradation?
(1) formation of (i) (2) formation of (ii) (3) formation of (iii) (4) formation of (iv)
Passage 2
32. How many structural isomeric primary amines are possible for the formula C4H11N?
Passage-Based Questions
Passage 1
The conversion of an amide to an amine with one carbon atom less by the action of alkaline hypohalite is known as Hoffmann bromamide degradation.
Two aromatic compounds (benzene derivatives) A and B of formula C 8 H 11N on heating with CHCl3 in presence of alc. KOH forms an unpleasant smelling compound. Compound ‘A’ on reaction with NaNO2 + HCl at 0°C followed by treatment with phenol forms an intense red coloured azodye. But compound ‘B’ does not form an azodye on reaction with NaNO 2 + HCl followed by treatment with phenol.
35. The number of isomers possible for compound ‘A’(including it self) are _____
36. The number of isomers possible for compound ‘B’ (including it self) are _____
Passage 3
Reaction of benzene with a mixture of conc. HNO 3 and conc. H 2SO 4 at about 60°C give nitrobenzene. Nitration of substituted benzenes depends on the nature of the group already present in the ring. Increase of temperature may result in multiple nitration.
37. Which of the following compounds undergo nitration at a rate faster than the other (1) Toluene (2) Phenol (3) Benzene (4) Chlorobenzene
In this reaction RCONHBr is formed from which the reaction has derived its name. Hoffmann reaction is accelerated if the migrating group is more electron releasing.
33. How can the converstion of (i) to (ii) be brought about
38. CO Nitration O Product in the given reaction is (1) CO NO2 O (2) CO O NO2 (3) CO O O2N
(4) CO O NO2
Passage 4
A,B,C are with molecular formula C8H11N three isomeric amines. A and B react with CHCl3 + KOH but not C. C reacts with NaNO2 + HCl to give yellow oily liquid.A on treatment with (1) acetic anhydride (2) Br2 and acetic acid (3) aqueous NaOH , heat produced compound D (C8H10NBr)
39. Compound C may be
(3)
40. Compound A is (I) CH3 NH2 CH3
(II) NH2 CH3 CH3
(III) NH2 CH3 CH3
(1) I and II only (2) I and III only (3) II and III only (4) I, II, III
Matrix Matching Questions
41. Match the Column-I with Column-II. Column I Column II
43. Match column-I(Reaction) with columnII(products).
Column-I Column-II
(A) Gabriel synthesis (p) Benzaldehyde
(B) Kolbe synthesis (q) Ethers
(C) Williamsonsynthesis (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
BRAIN TEASERS
1. Identify the most probable product (M) in the following reaction.
(2)
2. Basicity order of N in following compound is
(4)
5. The major product of the following reaction is
(1) b > d > a > c (2) a > b > d > c (3) a > b > c > d (4) a > c > b > d
3. The correct decreasing order of basicity of the following is,
(1) Q > R > S > P
(2) P > Q > R > S
(3) S > R > P > Q
(4) Q > R > P > S
4. The product P (major) of the following reaction is
6. Aliphatic amines on diazotization gives alcohol via carbocation intermediate NH2 NaNO2+HCl 2 1
(Carbons are labelled with numbers)
In the above reaction how many of the following products are possible (a) OH 3 4 1 2 (b) OH 3 4 1 2
stronger than para nitro phenol in List-I and y = Total number of bases which are stronger than aniline in list –II. Find the value of 1 xy xy + −+ is
List-I:
List-II:
7. If x = Total number of acids which are
FLASHBACK (Previous JEE Questions)
JEE Main
1. Compound (M,F,C14H13ON) HCl,Δ Filter Residue Q
Filtrate
Oily Liquid R NaOH + P
Compound P is neutral Q gives effervescence with NaHCO3 while R reacts with Hinsberg’s reagent to give solid soluble in NaOH. Compound P is (2023) (1)
2. The major product formed in the following reaction is (2023)
Br2/NaOH
3. The major product A and B from the following reactions are (2023) H
(1)
A
(2) A
A
A
4. Match List-I with List-II. (2023) List-I (Reagents used) List-II (Compound with functional group detected)
(A) Alkaline solution of copper sulphate and sodium citrate (p) HO
(B) Neutral FeCl3 solution (q) NH2
(C) Alkaline chloroform solution (r) CHO
(D)Potassium iodide and sodium Hypochlorite (s) CH
(A) (B) (C) (D)
(1) r s p q
(2) q s r p
(3) s p q r (4) r s q p
5. Match List-I (reagent used where Diazonium Chloride) with List-II (product formed) (2023)
List(Reagent) List-II (Product)
(A) NH2 (p) F
(B) HBF4, D (q) CN
(C) Cu, HCl (r) N N NH2
(D) CuCN/KCN (s) Cl
(A) (B) (C) (D) (1) r p q s
(2) r p s q (3) p r s q (4) s r q p
6. The product (P) formed from the following multistep reaction is (2023) NO2 Br2 (P) Product H2/Pd H3PO2 NaNO2,HCl,O0C H3C
(1) Br H3C (2) Br H3C OH
(3) Br H3C OH (4) Br H3C
7. Isomeric amines with molecular formula C8H11N give the following tests Isomer (P) can be prepared by Gabriel phthalimide synthesis. Isomer (Q) Reacts with Hinsberg’s reagent to give solid insoluble in NaOH. Isomer (R) Reacts with HONO followed
by β- naphthol in NaOH to give red dye. Isomer (P), (Q) and (R), respectively are (2023)
Statement-I : Pure Aniline and other arylamines are usually colourless.
Statement-II : Arylamines get coloured on storage due to atmospheric reduction (2023)
In light of the above statements, choose the most appropriate answer from the options given below:
(1) Statement I is incorrect but Statement II is correct.
(2) Both Statement I and Statement II are correct.
(3) Statement I is correct but Statement II is incorrect.
(4) Both Statement I and Statement II are incorrect.
17. Choose the correct colour of the product for the following reaction. (2023)
N=N OOCCH3
+ 1-Naphthyl amine
(1) Red
(2) Yellow
(3) Blue
(4) White
18. The correct order in aqueous medium of basic strength in case of methyl substituted amines is (2023)
(1) NH3 > Me3N > MeNH2 > Me2NH
(2) Me2NH > Me3N > MeNH2 > NH3
(3) Me2NH > MeNH2 > Me3N > NH3
(4) Me3N > Me2NH > MeNH2 > NH3
19. Identify the product formed (A and E)
JEE Advanced
Passage
A trinitro compound, 1,3,5-tris-(4-nitrophenyl) benzene, on complete reaction with an excess of Sn/HCl gives a major product, which on treatment with an excess of NaNO2/HCl at 0°C provides P as the product. P, upon treatment with excess of H 2 O at room temperature, gives the product Q. Bromination of Q in aqueous medium furnishes the product R. The compound P upon treatment with an excess of phenol under basic conditions gives the product S. The molar mass difference between compounds Q and R is 474 g mol–1 and between compounds P and S is 172.5 g mol –1 . (2023)
20. The total number of carbon atoms and heteroatoms present in one molecule of S is _____. [Use: Molar mass (in g mol–1): H = 1, C = 12, N = 14, O = 16, Br = 80, Cl = 35.5 Atoms other than C and H are considered as heteroatoms]
21. The number of heteroatoms present in one molecule of R is ______ . [Use: Molar mass (in g mol–1): H = 1, C = 12, N = 14, O = 16, Br = 80, Cl = 35.5 Atoms other than C and H are considered as heteroatoms]
22. Considering the reaction sequence given below, the correct statement(s) is(are) (2022)
(1) P can be reduced to a primary alcohol using NaBH4
(2) Treating P with conc. NH4OH solution followed by acidification gives Q.
(3) Treating Q with a solution of NaNO2 in aq. HCl liberates N2
(4) P is more acidic than CH 3CH2COOH.
23. Considering the following reaction sequence,
the correct option(s) is(are) (2022)
(1) P = H2/Pd, ethanol, R = NaNO2/HCl
U = 1.H3PO2, 2. KMnO4−KOH, heat
(2) P = Sn/HCl R = HNO2
CHAPTER TEST – JEE MAIN
Section-A
1. Reaction of propanamide with Br2/KOH(aq) produces:
(1) Ethylnitrile
(2) Propanenitrile
(3) Propylamine
(4) Ethylamine
2. Butanone oxime on reduction with Na/ C2H5OH gives (1) 3° amine (2) 2° amine (3) 1° amine (4) Amide
3. 3 2 1.NH Br 65 2. NaOH CHCOOHPQ.'Q'is D
(1) o-bromo Aniline
(2) Benzamide
(3) Aniline
(4) Benzene
4. In the given set of reactions
D + D is (amine) Phthalimide
KOH Et – Br B C
COONa
COONa
NaOH(aq)
= T =
U = 1.CH 3CH 2OH , 2. KMnO 4−KOH, heat
Q = , R = H2/Pd ethanol,
(1) CH3NH2
(2) CH3CH2NH2
(3) CH3CH2NHCH3
(4) (CH3)2NH
5. The reaction of RCONH2 with bromine and KOH gives RNH2 as the end product. Which one of the following is the intermediate product formed in this reaction ?
(1) RCONHBr (2) R–NH–Br
(3) R–N=C=O (4) RCONBr2
6. In the Hofmann bromamide degradation reaction, the number of mole of NaOH and Br2 used per mole of amine produced are (1) Two mole of NaOH and two mole of Br2.
(2) Four mole of NaOH and one mole of Br2.
(3) One mole of NaOH and one mole of Br2.
(4) Four mole of NaOH and two mole of Br2.
7. The total number of amines among the following which can be synthesised by Gabriel phthalimide synthesis is ________
(1) 2 (2) 2 (3) 3 (4) 4
8. Identify correct statement 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
9.
The correct basic strength of the given amines is
(1) P > Q > R > S
(2) S > R > Q > P
(3) S > P > R > Q
(4) R > Q > P > S
10. Which of the following statements is not correct regarding aniline?
(1) It is less basic than ethylamine.
(2) It can be steam distilled.
(3) It reacts with sodium and gives hydrogen.
(4) It is soluble in H2O
11. The amine that does not form hydrogen bonds
(1) Isopropyl amine
(2) Neopentyl amine
(3) Tertiary butyl amine
(4) N,N-Dimethyl aminoethane
12. Correct order of boiling point of isomeric amines of comparable molecular mass is
(1) Primary > secondary > tertiary
(2) Tertiary > secondary > primary
(3) Secondary > tertiary > primary
(4) Secondary > primary >tertiary
13. The decreasing order of basicity of the following amines is NH2
(1) (I) > (III) > (IV) > (II)
(2) (II) > (III) > (IV) > (I)
(3) (III) > (I) > (II) > (IV)
(4) (III) > (II) >(I) > (IV)
14. Carbylamine test is used to detect the presence of primary amino group in an organic compound. Which of the following compounds is formed when this test is performed with aniline?
15. Identify X and Y of the following reaction sequence
(i) H2/Pd X Y (i) Br2/CH3COOH (ii) H3O+ (ii) (CH3CO)2 O, pyridine iii) CHCl3, KOH,D
NH2 NC Br COCH3 COCH3 X = Y =
NHCOCH3 NHOH CHO X = Y =
NHCOCH3 NC Br X = Y =
NHCOCH3 NC Br X = Y =
16. Consider the below reaction, the compound ‘A’ is
Me
The product formed in the above reaction is (1) Me H Et NH2 (2) Me H Et H2N
(3) A 1:1 mixture of A and B
(4) A partially racemised mixture of A and B with predominant A
18. Which of the following is an example of electrophilic substitution?
(1) Diazotization
(2) Diazo coupling
(3) Sandmayer reaction
(4) Action of KCN on ArN2Cl
19. What is ‘C’ in the sequence of reactions? 1) CO2 MgCl NH3 Br2 A B 2) H+ 2) H+ NaOH (1) NH2 (2) CONH2 (3) NH2 (4) CONHBr
20 CH652NCl Sn+HCl 652 H yellow C HNOA B + + →→
Consider the above reaction, the product ‘B’ is
(1) N=N
(2) N=N–N H
(3) N=N
(4) N H
Sec tion-B
21. The total number of reagents from those given below, that can convert nitrobenzene into aniline is _________.( round off to nearest integer)
I. Sn−HCl II. Sn-NH4OH
III. Fe-HCl IV. Zn-HCl
V. H2-Pd VI. H2-Raney Nickel
22. How many of the following are more basic than CH3−CH2−NH3NH3, (C2H5)3N, (C2H5)2NH
23. Of the following reactions, how many reaction, are used for the preparation of amines.
(a) 4 LiAlH R-CN≡→
(b) R – C – NH2 LiAlH4 O
CHAPTER TEST – JEE ADVANCED 2022 P2 Model
Section-A
[Integer Value Question]
1. NO2
(i) Br2 / Fe
(ii) H2 / Ni
(iii) NaNO2 / HCl
(iv) CuCN / D
(vi) H2O (v) LiAlH4
(c) NaOH/Br2 2 RCONH →
(d) R – C – CH3 + H3C – NO2 NaOH O
(e) 1) R-X/NaOH 2) hydrolysis NH O O
(f) Hydrazine/NaOH 3 RCOCH →
(g) NaOH 2 RCONH →
(h) 2 H,Ni R-CH-NO22 →
24. The number of moles of NaOH required to convert 1 mole of benzamide into aniline (along with Na 2 CO 3 as by product) in Hoffman’s degradation
25. The diazonium ions give diazo coupling with phenol in the presence of dilute NaOH. How many of the following are more reactive than benzene diazonium chloride? N2 N2 N2 N2 + + + + NHMe NO2 CH3 CN
Total number of delocalised π electrons in the final product of the given reaction is
2. The number of amines which can be produced from Gabriel phthalimide synthesis is/are (from chloro derivatives)
(1) NH2 (2) NH2
(3) NH2 (4) NH
3. Compound (A) with molecular formula C5 H13N was treated with excess of methyl iodide to give a compound (B) which on heating with AgOH gave majorly a compound (C). Compound (C) on heating with alkaline KMnO4 gave isobutyric acid (salt) and CO 2 . The compound (A) on reaction with HNO 2 gave an alcohol (D). The number of possible amines of the given compound (A), which satisfy all the above given conditions are
4. The number of resonating structures of arylammonium ion is
5. How many isomers (including stereo) of the aromatic compound, (containing benzene ring) C8H11N on reaction with HNO2(0 – 5°) gives alcohol?
6. How many mole of NaOH are consumed in Hoffmann bromamide reaction of one mole of RCONH2?
7. The total number of sp2 hybridised carbon atoms in the major product P (a nonheterocyclic compound) of the following reaction is ___. CN CN NC NC
8. The reaction of 4-methyloct-1-ene (‘P’wt:, 2.52 g) with HBr in the presence of (C6H5CO)2O2 gives two isomeric bromides in a 9 : 1 ratio, with a combined yield of 50%. Of these, the entire amount of the primary alkyl bromide was reacted with an appropriate amount of diethylamine followed by treatment with aq. K 2CO 3 to give a non-ionic product S in 100% yield. The mass (in mg) of S obtained is _______ [Use molar mass (in g mol–1): H = 1, C = 12, N = 14, Br = 80]
Section-B
[Multiple Option Correct MCQs]
9. Considering the reaction sequence given below, the correct statement(s) is(are):
(1) [P] gives a foul-smelling compound on reaction with CHCl3/KOH
(2) [R] is salt that is stable at room temperature.
(3) [Q] on warming with water gives an organic compound that turns blue litmus red.
(4) [S] on reaction with H2/Pd in ethanol gives aniline.
10. Consider the following reaction sequence COCI
The correct option(s) is/are
(excess),
Acetophenone (excess)
11. Which of the following amides are more reactive than benzamide (C 6 H 5 CONH 2 ) towards Hoffmann bromamide reaction?
12. Which of the following method(s) is/are not used to prepare p-bromo aniline as major product? (1)
(3) 2,N-dimethylaniline
(4) p-methylbenzylamine
13. A positive carbylamine test is given by (1) N, N-dimethylaniline (2) 2,4-dimethylaniline
(II) (III) (IV)
Which of the following statement(s) is/are correct?
(1) I and II are aromatic, but I is more basic than III
(2) I is anti-aromatic, II is aromatic and II is more basic than I
(3) The order of basicity of above compounds is IV > III > II > I
(4) The conjugate acid of II is less stabilized than conjugate acid of IV
Section-C (Single Option Correct MCQs)
15. With respect to the following reaction, consider the given statements HNO3 H2SO4, 288 k Products
(A) o -Nitroaniline and p -nitroaniline are the predominant products.
(B) p-Nitroaniline and m-nitroaniline are the predomtinant products.
(C) HNO3 acts as an acid.
(D) H2SO4 acts as an acid.
choose the correct option.
(1) (A) and (C) are correct statements.
(2) (A) and (D) are correct statements.
(3) (B) and (D) are correct statements.
(4) (B) and (C) are correct statements.
16. In the following reactions, the major product W is
(4)
17.
Which statement is incorrect
(1) Reduced product of P and Q will be metamers to each other.
(2) By dry distillation of hydrolysed products of P with Ca(OH)2, gives benzophenone.
(3) Hydrolysed benzenoid product of Q, reacts with NaNO2 + HCl followed by reaction with phenol, gives orange red dye
(4) Electrophile involved in the formation of Q is dichlorocarbene
18. Aniline is reacted with bromine water and the resulting product is treated with an aqueous solution of sodium nitrite in presence of dilute hydrochloric acid. The compound so formed is converted into a tetrafluoroborate which is subsequently heated dry. The final product is
(1) 1,3,5-tribromobenzene
(2) p-bromofluorobenzene
(3) p-bromaniline
(4) 2,4,6-tribromofluorobenzene
ANSWER KEY
- II
Theory-based Questions
JEE Advanced Level
Brain Teasers
Flashback
Chapter Test – JEE Main
Chapter Test – JEE Advanced
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 t he 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.
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
Glucose forms glucose oxime with NH2–OH and glucose cyanohydrin with HCN.
61262 4
6126
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.
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.
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.
to explain the following reactions.
■ 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
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.
α–D–(+)– glucose
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.
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
Fischer cross formulation projection
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.
Fig 12.1. a -D-(+) and b -D-(+) Glucopyranoses
Structure of Fructose
The molecular formula of fructose was found to be C6H12O6.
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 α–and β–forms of fructose are anomers at C–2.
The open chain structure of fructose is as shown below.
a –D–(–) Fructofuranose
The Fischer and Haworth structures of α–D–fructose and β–D–fructose are shown in Fig. 12.2
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.
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.
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.
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)
12H22O 11
→ 2 C6H12O 6 (Maltose) (Glucose)
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.
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 of amylose
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 C6 of a glucose in the adjacent chain. These linkages are as shown in Fig. 12.7.
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 α-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 in ner 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
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 = mn,
Here, m is number of amino acids, n 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. W hich 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 onl y 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,
TEST YOURSELF
1. W hich o ne 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
5. 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 r eactions
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.
Table 12.3 Water soluble vitamins
Sl.No. Name of Vitamins
1. Vitamin B1 (Thiamine)
2. Vitamin B2
3. Vitamin B6 (Pyridoxine)
4. Vitamin B12
5. Vitamin C (Ascorbic acid)
Table 12.4 Fat–soluble vitamins
Sl.No.
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
Meat, fish, egg and curd Pernicius anaemia (RBC deficient in haemoglobin)
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
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 an other 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.
N itrogenous 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 adenosine, guanosine, cytidine, thymidine, and uridine 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
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?
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 us ually 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
JEE MAIN LEVEL
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
Amino Acids and proteins
16. The number of amino acids found in proteins that a human body can synthesize is ______.
(1) 20 (2) 10
(3) 5 (4) 14
17. Which of the following is an optically inactive amino acid:
(1) Glycine
(2) Lysine
(3) Isoleucine
(4) Aspartic acid
18. Determine which structural feature distinguishes proline from α–amino acids?
(1) It is optically inactive
(2) It contains aromatic group
(3) It is a dicarboxylic acid
(4) It is a secondary amine
19. Which of the following statements is not correct?
(1) Amino acid can exist as an inner salt.
(2) Each polypeptide has one C-terminal and one N-terminal.
(3) Enzymes are naturally occurring simple proteins.
(4) The union of two amino acids produces two peptide linkages.
20. The peptide linkage is _ ______.
(1) – CH – COO – NH
(2) – CH – CO – NH
(3) – CH – CH2 – CO – NH2
(4) CH NH NH CO
21. Enzymes belong to which class of compound?
(1) Polysaccharides
(2) Polypeptides
(3) Heterocyclics
(4) Hydrocarbons
22. The primary structure of a protein tells about (1) 3D arrangement of all atoms.
(2) shape of polypeptide chain.
(3) specific sequence of amino acids.
(4) 3D arrangement of oligo peptide chains.
23. Secondary structure of protein refers to (1) mainly denatured proteins and structure of prosthetic groups.
(2) three-dimensional structure, especially the bond between amino acid residues that are distinct from each other in the poly-peptide chain.
(3) linear sequence of amino acid residues in the polypeptide chain.
(4) regular folding patterns of continuous portions of the polypeptide chain.
24. ß -pleated structure of proteins has ______.
(1) primary structure
(2) secondary structure
(3) tertiary structure
(4) quaternary structure
25. Tertiary structure of a protein will lead the polypeptide chains to get which of the following shapes:
(1) linear, octahedral
(2) angular, tetrahedral
(3) fibrous, globular
(4) fibrous, planar
26. Wh ich of the following is an example of irreversible denaturation of a protein?
(1) Boiling of egg.
(2) Change of sequence of amino acids.
(3) Enzymatic action.
(4) Synthesis of proteins.
27. Enzymes are made up of _______.
(1) edible proteins
(2) proteins with specific structure
(3) nitrogen containing carbohydrates
(4) carbohydrates
28. Which of the following statements is incorrect about enzyme catalysis?
(1) Enzymes are mostly proteinous in nature.
(2) Enzyme action is specific.
(3) Enzymes are denatured by ultraviolet rays and at high temperature.
(4) Enzymes are least reactive at an optimum temperature.
29. The function of enzymes in the living system is to _______.
(1) transport oxygen
(2) provide immunity
(3) catalyse biochemical reactions
(4) provide energy
Vitamins
30. Match List I with List II.
List I List II
A. B1 I) Riboflavin
B. B2 II) Retinol
C. A III) Ascorbic acid
D. C IV) Thiamine
(A) (B) (C) (D)
(1) IV I III II
(2) IV III I II
(3) III IV II I
(4) IV I II III
31. Which are the water soluble vitamins?
(1) A, D
(2) E, K
(3) D, E
(4) C, B
32. Which one of the following sets of vitamins is fat soluble?
(1) C, D, B6, B12
(2) A, D, E, K
(3) A, D, B1, B2
(4) D, B1, B2, E
33. Convulsion is due to deficiency of vitamin ________.
(1) B12 (2) B2
(3) B 5 (4) B6
Nucleic acids
34. Which of the following constitutes the genetic material of the cell?
(1) Nucleic acids
(2) Proteins
(3) Lipids
(4) Carbohydrates
35. The pentose sugars in DNA and RNA have the structure of _________.
(1) furanose
(2) open chain
(3) pyranose
(4) four membered ring
36. Which one of the following is not present in DNA?
(1) Adenine
(2) Ribose
(3) Cytosine
(4) Guanine
37. Which of the following bases is not present in DNA?
(1) Adenine
(2) Guanine
(3) Cytosine
(4) Uracil
38. The phosphodiester linkage in a nucleotide is between _________.
(1) 5’ and 1’ carbons
(2) 5’ and 3’ carbons
(3) 1’ and 5’ carbons
(4) 2’ and 5’ carbons
39. Adenosine monophosphate (AMP) is a _____.
(1) nucleotide (2) nucleoside
(3) insecticide (4) antibacterial
40. Primary and secondary structures of nucleic acid reveals
(1) nucleotide sequence and Single or double helix structure.
(2) amino acid sequence and 3D-folding.
(3) amino acid sequence and shape of protein.
(4) single/double helix structure and Nucleotide sequence.
41. The base pairing occurs in double helix of DNA is
(1) A to T and G to C.
(2) A to G and T to C.
(3) A to C and G to T.
(4) G to T and A to C.
42. The purine base present in RNA is ______.
(1) Thymine (2) Thymine
(3) Cytosine (4) Uracil
43. The pentose sugar in DNA and RNA has
(1) open chain structure
(2) pyranose structure
(3) furanose structure
(4) none of the above
44. What is the main role of DNA in a living system?
(1) It is the structural material of cell walls.
(2) It is an enzyme.
(3) It carries the hereditary characteristics of the organism.
(4) It participates in cellular respiration.
45. Receptors of hormones are generally _____.
(1) Carbohydrates (2) Vitamins
(3) Lipids (4) Proteins
46. Hormones are secreted by ductless glands of human body. Iodine containing hormone is __________.
(1) Adrenoline (2) Thyroxine
(3) Testosterone (4) Insulin
Level - II
Carbohydrates
Single Option Correct MCQs
1. Which class of carbohydrates cannot be hydrolysed further?
(1) Monosaccharides
(2) Disaccharides
(3) Polysaccharides
(4) Proteins
2. Which of the following monosaccharides is a pentose?
(1) Glucose (2) Fructose
(3) Ribose (4) Galactose
3. Cane sugar is regarded as an example of (1) Monosaccharide
(2) Disaccharide
(3) Non-saccharide
(4) Polysaccharide
4. The two functional groups present in a typical carbohydrate are
(1) –CHO and – COOH
(2) >C=O and −OH
(3) – OH and – NH2
(4) – OH and – COOH
5. Compound represented by this structure is
(1) D-threose (2) L-threose
(3) D-erythrose (4) L-erythrose
6. Whic h of the following is not an oligosaccharide?
(1) Xylose (2) Maltose
(3) Lactose (4) Sucrose
7. The number of chiral carbon atoms in glucose hemiacetal is
(1) 5 (2) 2 (3) 4 (4) 1
8. Which one of the following is one of the objection to open chain structure of glucose?
(1) Glucose forms penta acetate.
(2) Glucose reacts with hydroxylamine to form an oxime.
(3) Penta acetate of glucose does not react with hydroxylamine.
(4) Glucose is oxidised by nitric acid to gluconic acid.
9. The number of mole of CH3OH required to react with 1 mole of glucose to form acetal is ___.
(1) 5 (2) 2 (3) 4 (4) 1
10. Which one of the following is the standard for giving D, L-configuration of sugars?
(1) Erythrose
(2) Arabinose
(3) Glyceraldehyde
(4) Glucose
11. Ribose and deoxy ribose are
(1) Pair of enantiomers
(2) Pair of epimers
(3) Pair of anomers
(4) Pair of pentose sugars
12. Which of the following does not show mutarotation?
(1) Glucose (2) Fructose
(3) Maltose (4) Sucrose
13. Glucose does not react with (1) Br2/H2O (2) H2NOH
(3) HI (4) NaHSO3
14. Among the following, the most suitable reducing agent to reduce aqueous solution of D-ribose to an optically inactive product, as shown below is
(1) LiAlH 4 (2) DIBAL-H
(3) NaBH4 (4) B2H6
15. Lactose is a reducing sugar. The reason is (1) Anomeric carbon of α-D(+) glucose is not involved in glycoside linkage.
(2) Anomeric carbon of β- D (+) galactose is not involved glycoside linkage.
(3) Anomeric carbon of β-D(+) glucose is not involved in glycoside linkage.
(4) Anomeric carbon of α-D(+) galactose is not involved in glycoside linkage.
16. The glycosidic linkage present in sucrose is between
(1) C-1 of β-D-glucose and C-4 of α-Dglucose
(2) C-1 of α-D-glucose and C-4 of β-Dglucose
(3) C-1 of α-D-glucose and C-4 of α -D-fructose
(4) C-1 of α-D-glucose and C-2 of β-Dfructose
17. According to CIP rules, the configuration of chiral carbon atoms in D (+) glucose are (1) 2S, 3S, 4R, 5R (2) 2S, 3R, 4S, 5R
(3) 2R, 3R, 4S, 5S (4) 2R, 3S, 4R, 5R
18. Which of the following on hydrolysis give similar monosaccharide units
A) Sucrose
B) Lactose
C) Maltose
D) Amylose
(1) A, B (2) A, C
(3) A, D (4) C, D
19. Common feature of starch and glycogen is, that both
(1) both are abundently present in animals (2) are storage polysaccharides (3) are heteropolymer of glucose and fructose (4) give red colour with I2 solution
20. Which of the following carbohydrates is the essential constituent of cell wall?
(1) Starch
(2) Maltose
(3) Cellulose
(4) Sucrose
21. Regarding amylopectin, the wrong statement is
(1) it is a branched chain polysaccharide.
(2) it is insoluble in water.
(3) it gives blue colour with iodine solution. (4) one molecule may contain 25-30 D-glucose units.
Numerical Value Questions
22. The number of non-reducing sugars of the following; starch, maltose, glucose, lactose, sucrose, cellulose, glycogen, and dextrin.
23. The number of stereogenic centres in α-Dglucose are _____.
24. When glucose is reacted with bromine water the major product is X. The number
of carboxylic groups present in compound ‘X’ are _____.
Proteins
Single Option Correct MCQs
25. –NH2 group is absent in (1) Tyrosine (2) Histidine (3) Isoleucine (4) Proline
26. Which of the following amino acid is not optically active?
(1) Proline (2) Serine
(3) Leucine (4) Glycine
27. The number of tripeptides formed with two amino acids A and B is: (1) 5 (2) 3 (3) 2 (4) 4
28. During the denaturation of protein, the structure of protein that remains intact and unaffected is (1) primary only (2) primary and secondary only (3) primary, secondary and tertiary (4) secondary and tertiary only
29 Which of the following is not an essential amino acid?
30. Biuret test is not given by (1) proteins (2) starch (3) poly peptides (4) urea
31. β-pleated structure of proteins is related to (1) primary structure (2) secondary structure (3) tertiary structure (4) quaternary structure
32. The naturally occurring amino acid that contains only one basic functional group in its chemical structure is (1) asparagine (2) histidine
(3) arginine
(4) lysine
33. Sulphur containing amino acids from the following are:
(a) isoleucine
(b) cysteine
(c) lysine
(d) methionine
(e) glutamic acid
(1) a, and b
(2) a, b, and c
(3) b, c, and e
(4) b, and d
Numerical Value Questions
34. How many of the given are acidic amino acids?
1) Leucine
2) Glutamic acid
3) Aspartic Acid
4) Aspargine
5) Proline
35. A short polypeptide on complete hydrolysis produces 3 mole of glycine (G), two mole of leucine (L) and two mole of valine (V) per mole of peptide. The number of peptide linkages in the given polypeptide are ____.
Enzymes
Single Option Correct MCQs
36. Which of the following biomolecules act as specific catalysis in biological reaction?
(1) Carbohydrates (2) Lipids
(3) Vitamins (4) Enzymes
37. Which of the following statements about enzymes is true?
(i) Enzymes lack in nucleophilic groups.
(ii) Enzymes are highly specific both in binding chiral substrates and in catalysing their reactions.
(iii) Enzymes catalyse chemical reactions by lowering the energy of activation.
(iv) Pepsin is a proteolytic enzyme.
(1) (i) and (iv) (2) (i), and (iii) (3) (ii), (iii), and (iv) (4) (i) only
38. Regarding enzymatic reactions, four 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
39. Enzymes are associated with same nonprotein component called (1) Co-factor (2) Promoter (3) Catalytic poison (4) Co-activator
40. x UreaCO+NH23 → . The enzyme ‘x’ is (1) Invertase (2) Lactase (3) Urease (4) α-Amylase
Vitamins
Single Option Correct MCQs
41. Which of the following vitamin is responsible for beri beri disease?
(1) Vitamin-A
(2) Vitamin-B1
(3) Vitamin-K
(4) Vitamin-D
42. Night blindness is caused by deficiency of (1) vitamin-B12
(2) vitamin-A
(3) vitamin-C (4) vitamin-E
43. Which of the following vitamin deficiency causes anaemia?
(1) Vitamin-B2
(2) Vitamin-B12
(3) Vitamin-B6
(4) Vitamin-B1
44. Cheilosis occurs due to deficiency of (1) thiamine (2) nicotinamide
(3) pyridoxamine (4) riboflavin
Numerical Value Questions
45. How many of the following statements are correct?
(i) Vitamins A, D, E, and K are insoluble in water.
(ii) Vitamins A, D, E, and K are stored in liver and adipose tissues.
(iii) Vitamin B, and vitamin C are water soluble.
(iv) Water soluble vitamins should not be supplied regularly in diet.
46. Convulsion is caused by the deficiency of vitamin B x . The value of ‘x’ is
Nucleic Acids
Single Option Correct MCQs
47. The sugar present in RNA is (1) α-D(−)-Ribofuranose (2) -D(−)-Ribofuranose
(3) -D(−)-Ribopyranose
(4) α-D(−)-Ribospyranose
48. –NH2 group is absent in
(1) Uracil (2) Adenine
(3) Guanine (4) Cytosine
49. Which base is not present in RNA
(1) Adenine (2) Guanine
(3) Cytosine (4) Thymine
50. 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
51. Which of the following is correct about H-bonding in DNA
(1) A = = = = T, G ≡≡≡≡ C
(2) G – – – – T, A – – – – C
(3) A = = = = G, T ≡≡≡≡ C
(4) A = = = = A, T ==== T
52. 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
53. Thymine is (1) 5-Methyluracil
(2) 4-Methyluracil
(3) 3-Methyluracil (4) 1-Methyluracil
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
Numerical Value Questions
55. The number of H-bonds present between adenine and thymine in DNA
56. The number of oxygen present in a nucleotide formed from a base, that is present only in RNA is _______.
57. The number of amido groups present in uracil are ____.
Hormones
Single Option Correct MCQs
58. Which of the followings is the more scientific definition of hormone?
(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 non-nutrient chemicals that act as intercellular messengers.
59. Insulin is a hormone that (1) releases glucose from glycogen (2) decrease metabolism (3) increase blood sugar (4) decrease blood sugar
60. The sex hormone that controls the development and maintenance of pregnancy is
11. Which of the statements about ‘denaturation’ given below are correct?
(A) Denaturation of proteins causes loss of secondary and tertiary structures of the protein.
(B) Denaturation leads to the conversion of double strand of DNA into single strand.
(C) Denaturation effects primary structure that gets disorted.
(1) (B) and (C)
(2) (A) and (C)
(3) (A) and (B)
(4) (A), (B), and (C)
12. Select the incorrect statements about globular proteins.
I) These have linear arrangement of polypeptide chains.
II) These are insoluble in water.
III) Kertain is a globular protein.
(1) Only I
(2) I and II
(3) II and III
(4) I, II and III
13. A dodecapeptide made of molecular weight 980 u, on complete hydrolysis gave alanine (molecular weight 89 u) as one of the products. If alanine makes up 52.9 % by weight of the hydrolysis products, how many alanine units are there in the dodecapeptide?
14. In amino acid the carboxylic group ionises at 1 2.40 Ka P = and ammonium ion 2 9.60 Ka P = then iso electric point is ___.
15. 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.
16. Which among the following pairs of vitamins is stored in our body relatively for longer duration?
(1) vitamin-A and vitamin-D
(2) ascorbic acid and vitamin-D
(3) thiamine and ascorbic acid
(4) thiamine and vitamin-A
17. A sugar ‘X’ dehydrates very slowly under acidic condition to give furfural which on further reaction with resorcinol gives the coloured product after sometime. Sugar ‘X’ is
(1) Aldopentose
(2) Aldotetrose
(3) Oxalic acid
(4) Ketotetrose
18. 3 CHCOCl/pyridine D-(+)-Glucose
Glucose pentaacetate?
Which statement is true about glucose pentaacetate?
(1) It will react with phenylhydrazine but not with Tollens’ reagent.
(2) It will react with hydroxylamine but not with phenylhydrazine.
(3) It will react with both hydroxylamine and Fehling’s solution.
(4) It will react neither with phenylhydrazine nor with hydroxylamine.
29. The correct structure of histidine in a strongly acidic solution (pH = 2) is (1) NH2 + N NH3 + COOH
(2) NH N+ NH2 COOH H
(3)
30. Suitable pH for the electrophoretic separation of a mixture of glutamic acid [pI = 3.22], threonine [pI = 5.6], and histidine [pI = 7.59 is
(1) 7.59
(2) 3.22
(3) 5.6
(4) any pH can be used
31. Correct statements are
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
32. Consider the given compounds: NH3 + N H3 + COOH (Y) (X) (Z)
T he correct order of acidic nature of the positions X, Y, Z is
(1) Z > X > Y (2) X > Y > Y
(3) X > Z > Y (4) Y > X > Z
33. Chargaff’s rule states that in a organism
(1) Amount of adenine (A) is equal to that of guanine (G) and the amount of thymine (T) is equal to that of cytosine (C)
(2) Amount of adenine (A) is equal to that of cytosine (C) and the amount of thymine (T) is equal to that of guanine (G)
(3) Amount of all bases are equal
(4) Amount of adenine (A) is equal to that of thymine (T) and the amount of guanine (G) is equal to that of cytosine
Numerical Value Questions
34. How many moles of HCOOH will be obtained when one mole of fructose is treated with excess HIO4
THEORY-BASED QUESTIONS
Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II). Mark the correct answer as
(1) if both statement I and statement II are correct.
(2) if both statement I and statement II are incorrect.
(3) if statement I correct but statement II is incorrect.
(4) if statement I incorrect but statement II is correct.
1. S-I : All monosaccharides, whether aldose or ketose, are reducing sugars.
S-II : Carbohydrate that cannot be hydrolysed further to give a simpler unit of polyhydroxy aldehyde or ketone is monosaccharide.
2. S-I : Carbohydrates having D-configuration are always dextro rotatory.
S-II : D(+)glucose and L(-)glucose are a pair of enantiomers
3. S-I : Aldo triose is optically active.
S-II : Keto triose is optically inactive.
4. S-I : Glucose give a reddish-brown precipitate with Fehling’s solution.
S-II : Reaction of glucose with Fehling’s solution gives CuO and gluconic acid.
5. S-I : Hydrolysis of sucrose is called inversion of cane sugar.
35. The optical rotation of α-D fructose is -21° and that of its β-form is -133°. The equilibrium mixture of these anomers has an optical rotation of -92°. What is the % of α-form (upto one decimal place) in its equilibrium mixture?
36. How many mole of acetyl chloride are used per mole of sucrose for esterification?
S-II : Sucrose is dextro rotatory whereas invert sugar formed due to hydrolysis of sucrose is leavo rotatory.
6. S-I : α-helix is the most common secondary structure of protein.
S-II : Insulin contains 51 amino acid residues but it is classified as a protein.
7. S-I : Histidine is a basic amino acid.
S-II : In the structure of histidine there are more number of basic groups than acid groups.
8. S-I : Proline is optically inactive.
S-II : Proline is imino acid.
9. S-I : The sum of the number of C o groups and the number of peptide linkages in given structure of peptide chain Asp-Gly-Lys is 6
S-II : The number of hetero atoms within the cyclic ring system of the monomer of Nylon-6 is one.
10. S-I : All enzymes are proteins but all proteins are not enzymes.
S-II : Keratin is an enzyme.
11. S-I : A unit formed by the attachment of a base to 1’-position of sugar is known as nucleoside.
S-II : When nucleoside is linked to photophorous acid at 5’-position of sugar moiety, we get nucleotide.
12. S-I : Glucagon is a hyperglycemic factor.
S-II : Addison’s disease is characterised by increased susceptibility to stress.
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.
13. (A) : Rhamnose(C 6 H 12 O 5 ) is not a carbohydrate.
(R) : All compounds having the general formula Cx(H2O)y are carbohydrates.
14. (A) : Sucrose is a non-reducing sugar.
(R) : In sucrose, glucose is in pyranose form and fructose is in furanose form.
15. (A) : Both glucose and fructose have the same D-configuration.
(R) : Both D-glucose and D-fructose are dextrorotatory.
16. (A) : Cellulose is a linear polymer of β-Dglucose.
(R) : Cellulose is formed by the condensation of β-D-glucose molecules.
17. (A) : Cellulose is harder than starch.
(R) : Hydrogen bond strength is more in cellulose than in starch.
18. (A) : In a sucrose molecule, glucose is present in the furanose form and
JEE ADVANCED LEVEL
Multi Option Correct MCQs
1. D-Mannose differs from D-glucose in its stereochemistry at C-2. Which is/are not the correct structure of pyranose form of D-Mannose.
fructose is present in the pyranose form.
(R) : Pyranose and furanose are heterocyclic ring compounds.
19. (A) : Amino acids are water-soluble, high melting solids and behave like salts rather than simple amines or carboxylic acids.
(R) : This behaviour is due to presence of both acidic (−COOH) & basic (−NH2) groups in the same molecule.
20. (A) : In aqueous solution α-amino acids exists as internal salt called as zwitter ion.
(R) : Proline is a natural amino acid having a secondary amino group.
21. (A) : Rate of hydrolysis of sucrose is higher in presence of invertase than when carried out in presence of dilute mineral acid.
(R) : Activation energy for enzyme catalysed hydrolysis of sucrose is less than that of acid catalysed hydrolysis.
22. (A) : Competitive inhibitors compete with natural substrate for their attachment on the active sites of enzymes.
(R) : In competitive inhibition, inhibitor binds to the allosteric site of the enzyme.
23. (A) : Enzymes are proteins but all proteins are not enzymes.
(R) : Enzymes are bio-catalyst and possess a stable configuration having a active site pocket.
(3)
(4)
2. Identify the compounds that gives same osazone. (1)
(2)
(3)
(4)
3. For the denaturation of protein, correct statement(s) is/are?
(1) Primary structure of protein is not destroyed.
(2) Secondary, tertiary structure of protein are destroyed.
(3) Entropy increases during denaturation.
(4) Boiling of egg is an example for irreversible denaturation of protein.
4. The correct statement(s) about the following sugars X and Y is/are
(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 glycosiding linkages in X and Y are α and β respectively
(4) The glycosiding linkages in X and Y are β and α respectively
5. A protein sequence in solution at temperature T folds from a denatured extended state to its native state. Choose the probable relation(s) for this process.
(1) ΔH < 0 (2) ΔH > 0
(3) ΔS < 0 (4) ΔS < 0
6. Find correct statements
(1) Activation energy for acid hydrolysis of sucrose is 6.22 kJ/mole.
(2) A, D, E, K are fat soluble vitamins.
(3) Beriberi is caused due to deficiency of “Thiamine”.
(4) Cytosine and Thymine are differed with molar mass difference of ‘14’.
7. What is are true regarding amylose?
(1) (+)-Maltose is the only disaccharide obtained on hydrolysis of amylose.
(2) Amylose on complete hydrolysis gives D-(+)-Glucose as only monosaccharide.
(3) Amylose has α-1, 4-glycosidic linkage.
(4) Amylose is a reducing carbohydrate.
8. Which of the following statement(s) is/are correct
(1) Above is a tripeptide with 2-peptide linkages.
(2) Peptide when subjected to Lassaigne’s test with FeCl3 gave blood red color.
(3) One of the amino acids formed on hydrolysis is basic amino acid.
(4) One of the amino acids formed on hydrolysis is acidic amino acid.
9. Which of the following statements are correct with reference to isoelectric point?
(1) It is the point at which amino acids bear no net charge.
(2) It corresponds to the pH at which concentration of zwitter ion is maximum
(3) At isoelectric point amino acid exists as a base.
(4) At isoelectric point amino acid exists as a acid.
Integer Value Questions
10. A tetrapeptide made up of natural amino acid has leucine as the N-terminal residue which is coupled to a chiral amino acid. Upon complete hydrolysis, the tetrapeptide gives alanine, phenylalanine, glycine and leucine. The total number of possible sequences of the tetrapeptide is _____.
11. How many tripeptides are possible when glycine, alanine, and phenylalanine are allowed to from peptide bonds?
12. Find iso-electric point of the given amino acid
Passage-based Questions
Passage 1
Carbohydrates are polyhydroxy aldehydes and ketones and those compounds on hydrolysis give such compounds that are also carbohydrates. The carbohydrates that are not hydrolysed are called monosaccharides. Other carbohydrates are oligosaccharides and polysaccharides. Monosaccharides with aldehydic group are called aldose and with free ketonic group are called ketose. All carbohydrates are optically active. the number of optical isomers = 2n. where, n=number of asymmetric carbons.
15. Number of chiral carbons present in glucose is x. The number of optical isomers are possible in glucose molecule y, th en ? 8 xy + =
16. maximum number of monosaccharide units present in oligosaccharides
Passage 2
COOH HO NH 2 CH 3
13. An unknown compound(A) (molarmass = 180) on acylation gives a product (molar mass = 390). Then find the number of hydroxyl group present in compound (A).
14. A decapeptide (mol. wt = 796) on complete hydrolysis gives glycine (mol wt =75) alanine and phenylalanine. Glycine contributes 47.0 % to the total weight of the hydrolysed products. The number of glycine units present in the decapeptide is ____.
17. The product (A) is (1) an achiral compound. (2) chiral compound (racemic mixture). (3) a chiral compound (optically pure). (4) a mixture of diasteremoers.
18. Compound (B) is (1) COOH OH2 + NH3 + CH 2Cl (2) COOH2 + O
Passage 3
19. The aqueous solution of lysine is neutral at (1) pH = 7 (2) pH = 9.8 (3) pH=2.2 (4) pH=5.6
20. At pH = 3 which of the following of aspartic acid is predominant?
(1)
Passage 4
When a solution of an α–amino acid is placed in an electric field depending on the pH of the medium, following three cases may happen.
(i) In alkaline solution, α-amino acids exist as anion II, and there is a net migration of amino acid towards the anode.
(ii) In acidic solution, α-amino acids exist as cation III, and there is a net migration of amino acid towards the cathode.
(iii) If II and III are exactly balanced there is no net migration; under such conditions any one molecule exists as a positive ion and as a negative ion for exactly
the same amount of time, and any small movement in the direction of one electrode is subsequently cancelled by an equal movement back toward the other electrode. The pH of the solution in which a particular amino acid does not migrate under the influence of an electric field is called the isoelectric point of that amino acid.
21. NH3 + N H3 + COOH (Z) (X) (Y)
Arrange in order of increasing acid strengths.
(1) X > Z > Y
(2) Z < X < Y
(3) X > Y > Z
(4) Z > X > Y
22. In aqueous solutions, amino acids mostly exist as
(1) NH2CHRCOOH
(2) NH2CHRCOO–
(3) + 3 HNCHRCOOH
(4) +3 H NCHRCOO
Matrix Matching Questions
23. Match list-I with list-II.
LIST-I Enzyme LIST-II
Conversion of
(A) Invertase (p) Starch into maltose
(B) Zymase (q) Maltose into glucose
(C) Diastase (r) Glucose into ethanol
(D) Maltase (s) Cane sugar into glucose
(A) (B) (C) (D)
(1) r s q p
(2) r q p s
(3) s r p q
(4) s q r p
24. Match List-I with List-II.
List-I
List-II
(A) Glucose+HI (p) Gluconic acid
(B) Glucose+Br2 water (q) Glucose pentacetate
(C) Glucose+acetic anhydride (r) Saccharic acid
(D) Glucose+HNO3 (s) Hexane
(A) (B) (C) (D)
(1) s p q r
(2) s r q p
(3) r p s q
(4) p r s q
25. Match List-I with List-II.
List-I (Sugar)
List-II (type)
(A) Glucose (p) Keto hexose
(B) Fructose (q) Aldo hexose
(C) Arabinose (r) Aldo tetrose
(D) Erythrose (s) Aldo pentose
(A) (B) (C) (D)
(1) r q p s
(2) q r p s
(3) q r s p
(4) q p s r
26. Match Column-I with Column-II.
Column I Column II
(A) Sucrose (p) Amylose + Amylopectin
(B) Maltose (q) α-D-Glucose+β-DFructose
(C) Lactose (r) α-D-Glucose+ α-D-Glucose
(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
27. Match Column-I with Column-II.
Column-I Column-II
(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
28. Match Column-I with Column-II.
Column I Column II
(A) Vitamin A (p) Calciferol
(B) Vitamin D (q) Tocopherol derivative
(C) Vitamin E (r) Retinol
(D) Vitamin K (s) Antihaemorrhagic (t) Thymine
(A) (B) (C) (D)
(1) r p q s
(2) r p q t
(3) r p s q
(4) p r q s
29. Match the vitamins given in Column-I with the deficiency diseases listed in Column-II.
Column)-I Column-II
(A) Vitamin A (p) Osteomalacia
(B) Vitamin C (q) Beriberi
(C) Vitamin D (r) Scurvy
(D) Vitamin B1 (s) Xerophthalmia
(A) (B) (C) (D)
(1) p q s r
(2) s r p q
(3) s p r q
(4) p r q s
BRAIN TEASERS
1. Which of the following compounds is a β-ketohexafuranose?
(1)
2. L-isomer of a compound ‘A’ C 4 H 8 O 4 gives a positive test with [Ag(NH 3 ) 2 ] + Treatment of ‘A’ with acetic anhydride yields triacetate derivative. Compound ‘A’ produces an optically active compounds (B) and an optically inactive compound (C) on treatment with bromine water and HNO3,
30. Match Column-I with Column-II.
Column-I Column-II
(A) Guanine (p) Purine
(B) Thymine (q) Pyrimidine
(C) Uracil (r) RNA
(D) Ribose (s) DNA
(A) (B) (C) (D)
(1) q r p q
(2) p,r,s q,s q,r r
(3) q p q s
(4) q p r s
respectively. Compound (A) is: (1)
3. Reduction of hexose A (molecular formula C 6H 12O 6) with sodium borohydride gives compounds B and C. Compound B is optically inactive, whereas compound C is optically active. Which of the following is compound A?(D-psicose is C 3-epimer of D-fructose, D-mannose is the C2-epimer of D-glucose)
(1) D-fructose (2) D-glucose
(3) D-mannose (4) D-psicose
4. Which of the following disaccharides is the β-anomer of 4-O-( β-D-glucopyranosyl) D–glucopyranose? (1)
5. A mixture of methionine (A), glutamic acid (B) and lysine (C) is separated by paper electrophoresis in two experiments, one at pH=1 and one at pH=12. After the separation, the paper strips are treated with ninhydrin to reveal the location of amino acids (purple spots). Which of the following correct match is of strips with appropriate
FLASHBACK (Previous JEE Questions)
JEE Main
1. Match List-I with List-II (2023)
List- I (Vitamin)
List- II (Deficiency Disease)
(A) Vitamin A (p) Beriberi
(B) Thiamine (q) Cheilosis
(C) Ascorbic acid (r) Xeropthalmia
(D) Riboflavin (s) Scurvey
(A) (B) (C) (D)
(1) r q s p
(2) s q r p
(3) r p s q
(4) s p r q
2. Match Column-I with Column-II (2023)
Column-I (Natural amino acid)
Column-II (One letter code)
(A) Arginine (p) D
(B) Aspartic acid (q) N
(C) Asparagine (r) A
(D) Alanine (s) R
(A) (B) (C) (D)
(1) s p q r
(2) r p q s
(3) s p r q
(4) p r s q
3. Match Column-I with Column- II. (2023)
Column-I (Natural Amino acid)
Column-II (One Letter Code)
(A) Glutamic acid (p) Q
(B) Glutamine (q) W
(C) Tyrosine (r) E
(D) Tryptophan (s) Y
(A) (B) (C) (D)
(1) r s p q
(2) q p s r
(3) s r p q
(4) r p s q
4. T he one that does not stabilise 2° and 3° structures of proteins is (2023)
(1) –S–S– linkage
(2) H-bonding
(3) –O–O– linkage
(4) van der Waals forces
5. L-isomer of tetrose X(C4H8O4) gives positive Schiff’s test and has two chiral carbons. On acetylation, ‘X’ yields triacetate. ‘X’ also undergoes the following reactions 3 4 HNO NaBH
The compound ‘X’ is (2023) (1)
(2)
(4)
6. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R). (2023)
Assertion(A) : A solution of the product obtained by heating a mole of glycine with a mole of chlorine in presence of red phosphorous generates chiral carbon atom.
Reason (R) : A molecule with 2 chiral carbons is always optically active.
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. 7. () + 2 3 i)HCN ii) HO/H iii) HNO
The products formed in the above reaction are (2023)
(1) One optically active and one meso product
(2) Two optically active products
(3) Two optically inactive products
(4) One optically inactive and one meso product.
8. Which is not true for arginine? (2023)
(1) It has a fairly high melting point.
(2) It is crystalline solid.
(3) It is associated with more than one pK a values.
(4) It has high solubility in benzene.
9. Match items of Row I with those of Row II.
Row II:-
(i) α-D(–) Fructofuranose
(ii) β-D(–)Fructofuranose
(iii) α-D(–)Glucopyranose
(iv) β-D(–)Glucopyranose
Correct match is (2023)
(1) A→(iii); B→(iv); C→(ii); D→(i)
(2) A→(iv); B→(iii); C→(i); D→(ii)
(3) A→(iii); B→(iv); C→(i); D→(ii)
(4) A→(i); B→(ii); C→(iii); D→(iv)
10. Given below are two statements: one is la belled as Assertion (A) and the other is labelled as Reason(R).
Assertion(A) : ketoses give Seliwanoff ‘s test faster than aldoses.
Reason (R) : ketoses undergo β elimination followed by formation of furfural. (2023)
In light of the above statements, choose the correct answer from the options given below:
(1) Both (A) and (R) are true and (R) is the correct explanation of (A).
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(3) (A) is true but (R) is false.
(4) Both (A) and (R) are false.
11. Which of the following artificial sweeteners has the highest sweetness value in comparison to cane sugar? (2023)
(1) Sucralose (2) Saccharin
(3) Alitame (4) Aspartame
12. A protein ‘X’ with molecular weight of 70,000 u, hydrolysis gives amino acids, one of these amino acid is (2023)
(1) CH3 COOH CH3 NH2 (2) NH2 COOH CH3
(3) NH2 COOH CH3 C H3 (4) CH3 COOH
13. Match List-I with List-II. (2023) List-I List-II
(A) Molisch’s test (p) Peptide
(B) Biuret test (q) Carbohydrate
(C) Carbylamine test (r) Primary amine
(D) Schiffs test (s) Aldehyde
(A) (B) (C) (D)
(1) r s q p
(2) q p r s
(3) r s p q
(4) p q r s
14. The correct representation in six membered pyranose form for the following sugar [x]
(3)
(2)
(4)
15. All structures given below are of vitamin C. The most stable of them is (2023) (1)
is (2023)
JEE Advanced
16. A disaccharide X cannot be oxidised by bromine water. The acid hydrolysis of X leads to a laevorotatory solution. The disaccharide X is (2023) (1)
17 Which of the following statement(s) is(are) true? (2019)
(1) Monosaccharides cannot be hydrolysed to give polyhydroxy aldehydes and ketones.
(2) Oxidation of glucose with bromine water gives glutamic acid.
(3) Hydrolysis of sucrose gives dextrorotatory glucose and laevo rotatory fructose.
(4) The two six-membered cyclic hemiacetal forms of D-(+)-glucose are called anomers.
18. The Fischer projection of D-glucose is given below. (2018)
CHAPTER TEST – JEE MAIN
Section-A
1. Regarding denaturation of protein, false statement is
(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.
2. 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
3. Secondary structure of proteins refers to
(1) Mainly denatured proteins and structure of prosthetic groups.
(2) Three-dimensional structure, especially
The correct structure(s) of β-glucopyranose is (are)
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.
4. 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
5. 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
6. Which base is present in RNA but not present in DNA?
(1) Thymine
(2) Uracil
(3) Guanine
(4) Cytosine
7. Which of the following is a nucleotide in m-RNA?
(1)
(2)
(3)
(4)
8. Stability of α-Helix structure of proteins depends upon
(1) dipolar interaction
(2) H-bonding interaction
(3) van der Waals forces
(4) π-stacking interaction
9. Glucose on oxidation with nitric acid gives:
(1) gluconic acid
(2) saccharic acid
(3) sorbic acid
(4) aldonic acid
10. Which of the following statement is not true for glucose?
(1) Glucose exists in two crystalline forms α and β.
(2) Glucose gives Schiff’s test for aldehyde.
(3) Glucose reacts with hydroxylamine to form oxime.
(4) The penta-acetate of glucose does not react with hydroxylamine to give oxime.
11. Fructose reduces tollens reagent due to
(1) asymmetric carbons
(2) enolisation of fructose followed by conversion to aldehyde by base.
(3) secondary alcoholic group.
(4) enolisation of fructose followed by conversion to alcohol to aldehyde.
12. The correct statement(s) about the following sugars X and Y is/are
(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.
13. Which of the following saccharides are reducing in narure
(1) A, B only (2) A, D only (3) B only (4) A, C, D
14. Which of the following best describes the polysaccharide amylose?
(1) a 1, 4-O-α-linked poly-D-glucose
(2) a 1, 4-O-β-linked poly-D-glucose
(3) an alternating 1, 4-α/β-linked poly-Dglucose
(4) a 1, 4-O-α-linked poly-D-mannose
15. The correct observation in the following reactions is
+ AB
Gly cosidic bond Seliwanoffs Sucrose ? Cleavage reagent (Hydrolysis)
(1) Formation of blue colour.
(2) Gives no colour.
(3) Formation of red colour.
(4) Formation of violet colour.
16. Which of the following is an animal polysaccharide?
(1) Amylopectin (2) Glycogen
(3) Amylose (4) Cellulose
17. The wrong statement about glucose is (1) It has one 1°-alcoholic group.
(2) It has four 2°-alcoholic group.
(3) It has one aldehydic group.
(4) It has one 3°-alcoholic groups.
18. Which of the following has a branched chain structure?
(1) Amylopectin
(2) Amylose
(3) Cellulose
(4) Nylon
19. The incorrect statement among the following is
(1) α-D-glucose and β-D-glucose are anomers.
(2) The pentaacetate of glucose does not react with Hydroxyl amine.
(3) Cellulose is a straight chain polysaccharide made up of only β-D-glucose unit.
(4) α-D-glucose and β-D-glucose are enantiomers.
20. Which vitamin deficiency causes Xerophthalmia ?
(1) Vitamin-A
(2) Vitamin-C
(3) Vitamin-B
(4) Vitamin-B6
Section-B
21. For the complete acetylation of one mole of glucose, the number of mole of acetic anhydride required is
22. The number of acetal (or ketal) groups in the following saccharide is:
23. A tripeptide on complete hydrolysis gives isoleucine, histidine and valine. On partial hydrolysis it never gives Val-His and HisVal. Find the total number of such primary structures of tripeptide?
24. The number of essential amino acids among the following is ______ [histidine, glycine, valine, alanine, aspartic acid, lysine, methionine].
25. The number of dipeptides formed when alanine combines with valine, phenyl alanine and tryptophan is
CHAPTER TEST – JEE ADVANCED
2022 P2 Model
Section-A
[Integer Value Questions]
1. How many numbers of following statements are correct?
i) When pure α-D-glucopyranose is dissolved in water, its optical rotation slowly changes.
ii) α-D-glucopyranose and β-Dglucopyranose are anomers.
iii) Methyl glucosides do not react with Fehling’s or Tollens’ reagent.
iv) α-D-glucopyranose react with NaHSO3 to form white precipitate.
v) α-D-methyl glucoside and β-D-methyl glucosides are anomers.
vi) Reduction of Fructose with Na–Hg/H2O gives both sorbitol and mannitol.
2. Gly-Lys-Pro-Glu- His
Regarding above pentapeptide chain, pick the number of correct statements given below.
A) It is also represented as GKPGH.
B) On complete hydrolysis gives 2 essential amino acids.
C) On complete hydrolysis gives one acidic amino acid and two basic amino acid
D) At one place of chain, nitrogen atoms are arranged similar to nitrogen atoms in guanidine [guanidyl group of nitrogen atoms]
E) Glycine end of chain is called as C-terminal
F) On complete hydrolysis gives four primary amino acids & one secondary amino acid.
3. When the following aldohexose exists in its D-configuration, the total number of stereoisomers in its pyranose form is ‘ X ’.
Find the value of 2 X CH(OH)
4. How many mole of HCOOH will be obtained when one mole fructose is treated with HIO4?
5. In aqueous solution, glucose exist in how many isomeric forms?
6. How many mole of HIO4 is required to break down the given molecule?
7. Galactose reacts with periodate, producing HCOOH, HCHO and IO3 only. In a typical experiment, a10 mL solution of galactose required 25 mL of 0.75 M periodate solution to reach the equivalence point. The solution is made free from formic acid and iodate ion by extraction and then treated with H 2 O 2 , which oxidises all formaldehyde into formic acid which is titrated against 0.1 M NaOH solution. Titration required 37.5 mL of alkali to reach the equivalence point. Then the molar ratio of formic acid to formaldehyde produced in the original reaction is _________.
8. The number of hydrogen bonds possible between a molecule of cystosine and a molecule of guanine in DNA molecule is
Section-B
[Multiple Option Correct MCQs]
9. Which of the following statements are correct reference to amino acid?
(1) A carboxylic acid that contains an amino group.
(2) Amino acids are the building blocks of peptides and proteins.
(3) An amino acid may exist as a zwitter ion under suitable conditions.
(4) Amino acids are negatively charged in basic medium.
10. A protein can be denaturated by _________
(1) change in pH
(2) addition of detergents
(3) adding urea
(4) strong heating
11. In amylopectin, the glucose molecules are connected by
(1) β-1,4-glycosidic bonds
(2) α-1,6-glycosidic bonds
(3) α-1,4-glycosidic bonds
(4) β-1, 6-glycosidic bonds
12. What is/are true about cellulose?
(1) It is a bipolymer of D-(+)-glucose.
(2) Hydrolysis of completely methylated cellulose gives 2,3,6-tri-O-methy1-Dglucose.
(3) It reduces Fehling’s solution.
(4) It shows mutarotation.
13. The forces that stabilise secondary and tertiary structure of proteins is/are (1) hydrogen bonds.
(2) disulphide linkages.
(3) van der Waals forces of attraction.
(4) glycosidic linkages.
14. Which of the following is/are obtained on hydrolysis of sucrose?
(1) α-D-glucose
(2) β-D-glucose
(3) α-D-fructose
(4) β-D-fructose
Section-C
[Single Option Correct MCQs]
15. Incorrect with respect to Lobry de BruynVan Ekenstein rearrangement is
(1) Glucose undergoes reversible isomerisation.
(2) A mixture of D-glucose, D-mannose, and D-fructose is formed.
(3) This rearrangement explains how fructose reduces Tollens’ reagent.
(4) If the starting material is D-mannose, then D-glucose is not formed.
16. Glucose on prolonged heating with HI gives (1) 1-Hexene
(2) Hexanoic acid
(3) 6-Iodohexanal
(4) n-hexane
17. Benedict’s solution is used to identify (1) Lipids
