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NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Page No.262 - Intext Questions

Q9.1: Classify the following amines as primary, secondary or tertiary:

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(iii) (C2H5)2CHNH2                                                                              

(iv) (C2H5)2NH

Ans: Primary: (i) and (iii)

Secondary: (iv)

Tertiary: (ii)


Q9.2: (i) Write structures of different isomeric amines corresponding to the molecular formula, C4H11N.

(ii) Write IUPAC names of all the isomers.

(iii) What type of isomerism is exhibited by different pairs of amines?

Ans: (i), (ii) The structures and their IUPAC names of different isomeric amines corresponding to the molecular formula, C4H11N are given below:

(a) CH3-CH2-CH2-CH2-NH2

Butanamine (10)

(b) 

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Butan-2-amine (10)

(c) 

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

2-Methylpropanamine (10)

(d)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

2-Methylpropan-2-amine (10)

(e) CH3-CH2-CH2-NH-CH3

N-Methylpropanamine (20)

(f) CH3-CH2-NH-CH2-CH3

N-Ethylethanamine (20)

(g)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

N-Methylpropan-2-amine (20)

(h)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

N,N-Dimethylethanamine (3°)

(iii) The pairs (a) and (b) and (e) and (g) exhibit position isomerism.

The pairs (a) and (c); (a) and (d); (b) and (c); (b) and (d) exhibit chain isomerism.

The pairs (e) and (f) and (f) and (g) exhibit metamerism.

All primary amines exhibit functional isomerism with secondary and tertiary amines and vice-versa.

Page No.265 - Intext Questions

Q9.3: How will you convert?

(i) Benzene into aniline
 (ii) Benzene into N, N-dimethylaniline  
 (iii) Cl−(CH2)4−Cl into hexan-1, 6-diamine?

Ans: (i)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(ii)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(iii)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Page No.274 - Intext Questions

Q9.4: Arrange the following in increasing order of their basic strength:

(i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH

(ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2

(iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2.

Ans: (i) Considering the inductive effect of alkyl groups, NH3, C2H5NH2, and (C2H5)2NH can be arranged in the increasing order of their basic strengths as:

NH< C2H5NH2 < (C2H5)2NH

Again, C6H5NH2 has proton acceptability less than NH3. Thus, we have:

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Due to the −I effect of C6H5 group, the electron density on the N-atom in C6H5CH2NH2 is lower than that on the N-atom in C2H5NH2, but more than that in NH3. Therefore, the given compounds can be arranged in the order of their basic strengths as:

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(ii) Considering the inductive effect and the steric hindrance of the alkyl groups, C2H5NH2, (CH5)2NH2, and their basic strengths as follows:

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Again, due to the −R effect of C6H5 group, the electron density on the N atom in C6HNH2 is lower than that on the N atom in C2H5NH2. Therefore, the basicity of C6H5NH2is lower than that of C2H5NH2. Hence, the given compounds can be arranged in the increasing order of their basic strengths as follows:

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(iii) Considering the inductive effect and the steric hindrance of alkyl groups, CH3NH2, (CH3)2NH, and (CH3)3N can be arranged in the increasing order of their basic strengths as:

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

In C6H5NH2, N is directly attached to the benzene ring. Thus, the lone pair of electrons on the N−atom is delocalized over the benzene ring. In C6H5CH2NH2, N is not directly attached to the benzene ring. Thus, its lone pair is not delocalized over the benzene ring. Therefore, the electrons on the N atom are more easily available for protonation in C6H5CH2NH2 than in C6H5NH2 i.e., C6H5CHNH2 is more basic than C6H5NH2.

Again, due to the −I effect of C6H5 group, the electron density on the N−atom in C6H5CH2NH2 is lower than that on the N−atom in (CH3)3N. Therefore, (CH3)3N is more basic than C6H5CH2NH2. Thus, the given compounds can be arranged in the increasing order of their basic strengths as follows.

 NCERT Solutions Class 12 Chemistry Chapter 9 - Amines


Q9.5: Complete the following acid-base reactions and name the products:
(i) CH3CH2CH2NH2  + HCl  
(ii) (C2H5)3N HCl 

Ans:

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines


Q9.6: Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
Ans: Aniline reacts with methyl iodide to produce N, N-dimethylaniline.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

With excess methyl iodide, in the presence of Na2CO3 solution, N, N-dimethylaniline produces N, N, N−trimethylanilinium carbonate. NCERT Solutions Class 12 Chemistry Chapter 9 - Amines


Q9.7: Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Ans:

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines


Q9.8: Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.
Ans: The structures of different isomers corresponding to the molecular formula, C3H9N are given below:

(a) CH3-CH2-CH2-NH2     

     Propan-1-amine (10)

(b)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Propan-2-amine (10)

(c)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(d)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

N,N-Dimethylmethanamine (30)

10amines, (a) propan-1-amine, and (b) Propan-2-amine will liberate nitrogen gas on treatment with nitrous acid.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Page No.277 - Intext Questions

Q9.9: Convert

(i) 3-Methylaniline into 3-nitrotoluene.                                  
 (ii) Aniline into 1,3,5-tribromobenzene.

Ans:- (i)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(ii)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Exercises

Q9.1: Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

(i) (CH3)CHNH2    

(ii) CH3(CH2)2NH2    

(iii) CH3NHCH(CH3)2  

(iv) (CH3)3CNH2

(v) C6H5NHCH3 

(vi) (CH3CH2)2NCH3                  

(vii) m−BrC6H4NH2

Ans:

(i) 1-Methylethanamine (10 amine)                                          

(ii) Propan-1-amine (10 amine)           

(iii) N−Methyl-2-methylethanamine (20 amine)                      

(iv) 2-Methylpropan-2-amine (10 amine)

(v) N−Methylbenzamine or N-methylaniline (20 amine)         

(vi) N-Ethyl-N-methylethanamine (30 amine)

(vii) 3-Bromobenzenamine or 3-bromoaniline (10 amine)


Q9.2: Give one chemical test to distinguish between the following pairs of compounds.

(i) Methylamine and dimethylamine     

(ii) Secondary and tertiary amines

(iii) Ethylamine and aniline    

(iv) Aniline and benzylamine

(v) Aniline and N-methylaniline.

Ans: (i) Methylamine and dimethylamine can be distinguished by the carbylamine test.

Carbylamine test: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form foul-smelling isocyanides or carbylamines. Methylamine (being an aliphatic primary amine) gives a positive carbylamine test, but dimethylamine does not.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(ii) Secondary and tertiary amines can be distinguished by allowing them to react with Hinsberg’s reagent (benzenesulphonyl chloride, C6H5SO2Cl).

Secondary amines react with Hinsberg’s reagent to form a product that is insoluble in an alkali. For example, N, N−diethylamine reacts with Hinsberg’s reagent to form N, N−diethylbenzenesulphonamide, which is insoluble in an alkali. Tertiary amines, however, do not react with Hinsberg’s reagent.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(iii) Ethylamine and aniline can be distinguished using the azo-dye test. A dye is obtained when aromatic amines react with HNO2 (NaNO2  dil.HCl) at 0-5°C, followed by a reaction with the alkaline solution of 2-naphthol. The dye is usually yellow, red, or orange in colour. Aliphatic amines give a brisk effervescence due (to the evolution of N2gas) under similar conditions.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(iv) Aniline and benzylamine can be distinguished by their reactions with the help of nitrous acid, which is prepared in situ from a mineral acid and sodium nitrite. Benzylamine reacts with nitrous acid to form unstable diazonium salt, which in turn gives alcohol with the evolution of nitrogen gas.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

On the other hand, aniline reacts with HNO2 at a low temperature to form stable diazonium salt. Thus, nitrogen gas is not evolved.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(v) Aniline and N-methylaniline can be distinguished using the Carbylamine test. Primary amines, on heating with chloroform and ethanolic potassium hydroxide, form foul-smelling isocyanides or carbylamines. Aniline, being an aromatic primary amine, gives positive carbylamine test. However, N-methylaniline, being a secondary amine does not.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines


Q9.3: Account for the following:

(i) pKof aniline is more than that of methylamine.

(ii) Ethylamine is soluble in water whereas aniline is not.

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

(iv) Although amino group is o, p− directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.

(v) Aniline does not undergo Friedel-Crafts reaction.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Ans: (i) pKof aniline is more than that of methylamine:

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Aniline undergoes resonance and as a result, the electrons on the N-atom are delocalized over the benzene ring. Therefore, the electrons on the N-atom are less available to donate.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

On the other hand, in case of methylamine (due to the I effect of methyl group), the electron density on the N-atom is increased. As a result, aniline is less basic than methylamine. Thus, pKb of aniline is more than that of methylamine.

(ii) Ethylamine is soluble in water whereas aniline is not:

Ethylamine when added to water forms intermolecular H−bonds with water. Hence, it is soluble in water.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

But aniline does not undergo H−bonding with water to a very large extent due to the presence of a large hydrophobic −C6H5 group. Hence, aniline is insoluble in water.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide:

    NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Due to the I effect of −CH3 group, methylamine is more basic than water. Therefore, in water, methylamine produces OH ions by accepting H  ions from water.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Ferric chloride (FeCl3) dissociates in water to form Fe and Cl ions.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Then, OH ion reacts with Fe ion to form a precipitate of hydrated ferric oxide.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(iv) Although amino group is o,p− directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline:

Nitration is carried out in an acidic medium. In an acidic medium, aniline is protonated to give anilinium ion (which is meta-directing).

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

For this reason, aniline on nitration gives a substantial amount of m-nitroaniline.

(v) Aniline does not undergo Friedel-Crafts reaction:

A Friedel-Crafts reaction is carried out in the presence of AlCl3. But AlCl3 is acidic in nature, while aniline is a strong base. Thus, aniline reacts with AlCl3 to form a salt (as shown in the following equation).

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Due to the positive charge on the N-atom, electrophilic substitution in the benzene ring is deactivated. Hence, aniline does not undergo the Friedel-Crafts reaction.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines:

The diazonium ion undergoes resonance as shown below:

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

This resonance accounts for the stability of the diazonium ion. Hence, diazonium salts of aromatic amines are more stable than those of aliphatic amines.

(vii) Gabriel phthalimide synthesis is preferred for synthesizing primary amines:

Gabriel phthalimide synthesis results in the formation of 1° amine only. 2° or 3° amines are not formed in this synthesis. Thus, a pure 1° amine can be obtained. Therefore, Gabriel phthalimide synthesis is preferred for synthesizing primary amines.

Q9.4: Arrange the following:

(i) In decreasing order of the pKbvalues: C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2

(ii) In increasing order of basic strength: C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2

(iii) In increasing order of basic strength:

(a) Aniline, p-nitroaniline and p-toluidine  
 (b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2.

(iv) In decreasing order of basic strength in gas phase: C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3

(v) In increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2

(vi) In increasing order of solubility in water: C6H5NH2, (C2H5)2NH, C2H5NH2.

Ans: (i) In C2H5NH2, only one −C2H5 group is present while in (C2H5)2NH, two −C2H5 groups are present. Thus, the I effect is more in (C2H5)2NH than in C2H5NH2. Therefore, the electron density over the N-atom is more in (C2H5)2NH than in C2H5NH2. Hence, (C2H5)2NH is more basic than C2H5NH2.

Also, both C6H5NHCH3 and C6H5NHare less basic than (C2H5)2NH and C2H5NH2 due to the delocalization of the lone pair in the former two. Further, among C6H5NHCH3and C6H5NH2, the former will be more basic due to the T effect of −CH3 group. Hence, the order of increasing basicity of the given compounds is as follows:

C6H5NH2 < C6H5NHCH3 < C2H5NH2 < (C2H5)2NH

We know that the higher the basic strength, the lower is the pKb values.

C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH

(ii) C6H5N(CH3)2 is more basic than C6H5NH2 due to the presence of the I effect of two −CH3 groups in C6H5N(CH3)2. Further, CH3NH2 contains one −CH3 group while (C2H5)2NH contains two −C2H5 groups. Thus, (C2H5)2 NH is more basic than C2H5NH2.

Now, C6H5N(CH3)2 is less basic than CH3NH2 because of the−R effect of −C6H5 group.

Hence, the increasing order of the basic strengths of the given compounds is as follows:

C6H5NH2 < C6H5N(CH3)2 < CH3NH< (C2H5)2NH

(iii) (a)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

In p-toluidine, the presence of electron-donating −CH3 group increases the electron density on the N-atom.

Thus, p-toluidine is more basic than aniline.

On the other hand, the presence of electron-withdrawing

−NO2 group decreases the electron density over the N−atom in p-nitroaniline. Thus, p-nitroaniline is less basic than aniline.

Hence, the increasing order of the basic strengths of the given compounds is as follows:

p-Nitroaniline < Aniline < p-Toluidine

(b) C6H5NHCH3 is more basic than C6H5NH2 due to the presence of electron-donating −CH3 group in C6H5NHCH3.

Again, in C6H5NHCH3, −C6H5 group is directly attached to the N-atom. However, it is not so in C6H5CH2NH2. Thus, in C6H5NHCH3, the −R effect of −C6H5 group decreases the electron density over the N-atom. Therefore, C6H5CH2NH2 is more basic than C6H5NHCH3.

Hence, the increasing order of the basic strengths of the given compounds is as follows:

C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2.

(iv) In the gas phase, there is no solvation effect. As a result, the basic strength mainly depends upon the I effect. The higher the I effect, the stronger is the base. Also, the greater the number of alkyl groups, the higher is the I effect. Therefore, the given compounds can be arranged in the decreasing order of their basic strengths in the gas phase as follows:

(C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3

(v) The boiling points of compounds depend on the extent of H-bonding present in that compound. The more extensive the H-bonding in the compound, the higher is the boiling point. (CH3)2NH contains only one H−atom whereas C2H5NH2 contains two H-atoms. Then, C2H5NH2 undergoes more extensive H-bonding than (CH3)2NH. Hence, the boiling point of C2H5NH2 is higher than that of (CH3)2NH.

Further, O is more electronegative than N. Thus, C2H5OH forms stronger H−bonds than C2H5NH2. As a result, the boiling point of C2H5OH is higher than that of C2H5NH2and (CH3)2NH.

Now, the given compounds can be arranged in the increasing order of their boiling points as follows:

(CH3)2NH < C2H5NH2 < C2H5OH

(vi) The more extensive the H−bonding, the higher is the solubility. C2H5NH2 contains two H-atoms whereas (C2H5)2NH contains only one H-atom. Thus, C2H5NH2undergoes more extensive H−bonding than (C2H5)2NH. Hence, the solubility in water of C2H5NH2 is more than that of (C2H5)2NH.

Further, the solubility of amines decreases with increase in the molecular mass. This is because the molecular mass of amines increases with an increase in the size of the hydrophobic part. The molecular mass of C6H5NH2 is greater than that of C2H5NH2 and (C2H5)2NH.

Hence, the increasing order of their solubility in water is as follows:

C6H5NH2 < (C2H5)2NH < C2H5NH2


Q9.5: How will you convert:

(i) Ethanoic acid into methanamine                

(ii) Hexanenitrile into 1-aminopentane

(iii) Methanol to ethanoic acid                        

(iv) Ethanamine into methanamine

(v) Ethanoic acid into propanoic acid             

(vi) Methanamine into ethanamine

(vii) Nitromethane into dimethylamine           

(viii) Propanoic acid into ethanoic acid

Ans: (i)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(ii)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(iii)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(iv)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(v)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(vi)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(vii)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(viii)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

 

Q9.6: Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
Ans: Primary, secondary and tertiary amines can be identified and distinguished by Hinsberg’s test. In this test, the amines are allowed to react with Hinsberg’s reagent, benzenesulphonyl chloride (C6H5SO2Cl). The three types of amines react differently with Hinsberg’s reagent. Therefore, they can be easily identified using Hinsberg’s reagent.

Primary amines react with benzenesulphonyl chloride to form N-alkylbenzenesulphonyl amide which is soluble in alkali.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Due to the presence of a strong electron-withdrawing sulphonyl group in the sulphonamide, the H−atom attached to nitrogen can be easily released as proton. So, it is acidic and dissolves in alkali.

Secondary amines react with Hinsberg’s reagent to give a sulphonamide which is insoluble in alkali.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

There is no H−atom attached to the N-atom in the sulphonamide.Therefore, it is not acidic and insoluble in alkali

On the other hand, tertiary amines do not react with Hinsberg’s reagent at all.


Q9.7: Write short notes on the following:

(i) Carbylamine reaction                                  

(ii) Diazotisation

(iii) Hofmann’s bromamide reaction                

(iv) Coupling reaction

(v) Ammonolysis                                              

(vi) Acetylation

(vii) Gabriel phthalimide synthesis.

Ans: (i) Carbylamine reaction

Carbylamine reaction is used as a test for the identification of primary amines. When aliphatic and aromatic primary amines are heated with chloroform and ethanolic potassium hydroxide, carbylamines (or isocyanides) are formed. These carbylamines have very unpleasant odours. Secondary and tertiary amines do not respond to this test.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

For example,

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(ii) Diazotisation

Aromatic primary amines react with nitrous acid (prepared in situ from NaNO2 and a mineral acid such as HCl) at low temperatures (273-278 K) to form diazonium salts. This conversion of aromatic primary amines into diazonium salts is known as diazotization.

For example, on treatment with NaNO2 and HCl at 273−278 K, aniline produces benzenediazonium chloride, with NaCl and H2O as by-products.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(iii) Hoffmann bromamide reaction

When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, a primary amine with one carbon atom less than the original amide is produced. This degradation reaction is known as Hoffmann bromamide reaction. This reaction involves the migration of an alkyl or aryl group from the carbonyl carbon atom of the amide to the nitrogen atom.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

For example,

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(iv) Coupling reaction

The reaction of joining two aromatic rings through the −N=N−bond is known as coupling reaction. Arenediazonium salts such as benzene diazonium salts react with phenol or aromatic amines to form coloured azo compounds.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

It can be observed that, the para-positions of phenol and aniline are coupled with the diazonium salt. This reaction proceeds through electrophilic substitution.

(v) Ammonolysis

When an alkyl or benzyl halide is allowed to react with an ethanolic solution of ammonia, it undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino (−NH2) group. This process of cleavage of the carbon-halogen bond is known as ammonolysis.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

When this substituted ammonium salt is treated with a strong base such as sodium hydroxide, amine is obtained.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Though primary amine is produced as the major product, this process produces a mixture of primary, secondary and tertiary amines, and also a quaternary ammonium salt as shown.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(vi) Acetylation

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Acetylation (or ethanoylation) is the process of introducing an acetyl group into a molecule.

Aliphatic and aromatic primary and secondary amines undergo acetylation reaction by nucleophilic substitution when treated with acid chlorides, anhydrides or esters. This reaction involves the replacement of the hydrogen atom of −NH2 or > NH group by the acetyl group, which in turn leads to the production of amides. To shift the equilibrium to the right hand side, the HCl formed during the reaction is removed as soon as it is formed. This reaction is carried out in the presence of a base (such as pyridine) which is stronger than the amine.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

When amines react with benzoyl chloride, the reaction is also known as benzoylation.

For example,

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(vii) Gabriel phthalimide synthesis

Gabriel phthalimide synthesis is a very useful method for the preparation of aliphatic primary amines. It involves the treatment of phthalimide with ethanolic potassium hydroxide to form potassium salt of phthalimide. This salt is further heated with alkyl halide, followed by alkaline hydrolysis to yield the corresponding primary amine.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

 

Q9.8: Accomplish the following conversions:

(i) Nitrobenzene to benzoic acid                                 

(ii) Benzene to m-bromophenol

(iii) Benzoic acid to aniline                                         

(iv) Aniline to 2,4,6-tribromofluorobenzene

(v) Benzyl chloride to 2-phenylethanamine                

(vi) Chlorobenzene to p-chloroaniline

(vii) Aniline to p-bromoaniline                                    

(viii) Benzamide to toluene

(ix) Aniline to benzyl alcohol.
Ans: (i)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(ii)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(iii)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(iv)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(v)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(vi)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

         NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

 

(vii)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(viii)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(ix)

 NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

 

Q9.9: Give the structures of A, B and C in the following reactions:

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines
Ans: (i)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

 

(ii)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

 

(iii)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

 

(iv)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(v)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

 

(vi)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines


Q9.10: An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N. Write the structures and IUPAC names of compounds A, B and C.
Ans: It is given that compound ‘C’ having the molecular formula, C6H7N is formed by heating compound ‘B’ with Br2 and KOH. This is a Hoffmann bromamide degradation reaction. Therefore, compound ‘B’ is an amide and compound ‘C’ is an amine. The only amine having the molecular formula, C6H7N is aniline, (C6H5NH2).

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Therefore, compound ‘B’ (from which ’C’ is formed) must be benzamide, (C6H5CONH2).

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Further, benzamide is formed by heating compound ‘A’ with aqueous ammonia. Therefore, compound ‘A’ must be benzoic acid.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

The given reactions can be explained with the help of the following equations:

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

 

Q9.11: Complete the following reactions:

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Ans: (i)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

 

(ii)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

 

(iii)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

 

(iv)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

 

(v)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(vi)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(vii)

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines


Q9.12: Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Ans: Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic substitution (SN2) of alkyl halides by the anion formed by the phthalimide.

                    NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

But aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Hence, aromatic primary amines cannot be prepared by this process.

 

Q9.13: Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
Ans: (i) Aromatic amines react with nitrous acid (prepared in situ from NaNO2 and a mineral acid such as HCl) at 273 − 278 K to form stable aromatic diazonium salts i.e., NaCl and H2O.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

(ii) Aliphatic primary amines react with nitrous acid (prepared in situ from NaNO2 and a mineral acid such as HCl) to form unstable aliphatic diazonium salts, which further produce alcohol and HCl with the evolution of N2 gas.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

 

Q9.14: Give plausible explanation for each of the following:

(i) Why are amines less acidic than alcohols of comparable molecular masses?

(ii) Why do primary amines have higher boiling point than tertiary amines?

(iii) Why are aliphatic amines stronger bases than aromatic amines?

Ans: (i) Amines undergo protonation to give amide ion.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

Similarly, alcohol loses a proton to give alkoxide ion.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

In an amide ion, the negative charge is on the N-atom whereas in alkoxide ion, the negative charge is on the O-atom. Since O is more electronegative than N, O can accommodate the negative charge more easily than N. As a result, the amide ion is less stable than the alkoxide ion. Hence, amines are less acidic than alcohols of comparable molecular masses.

(ii) In a molecule of tertiary amine, there are no H−atoms whereas in primary amines, two hydrogen atoms are present. Due to the presence of H−atoms, primary amines undergo extensive intermolecular H−bonding.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

As a result, extra energy is required to separate the molecules of primary amines. Hence, primary amines have higher boiling points than tertiary amines.

(iii) Due to the −R effect of the benzene ring, the electrons on the N- atom are less available in case of aromatic amines. Therefore, the electrons on the N-atom in aromatic amines cannot be donated easily. This explains why aliphatic amines are stronger bases than aromatic amines.

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FAQs on NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

1. What are amines and what are their properties?
Ans. Amines are organic compounds that contain nitrogen as the central atom. They are derived from ammonia (NH3) by replacing one or more hydrogen atoms with organic groups. Some properties of amines include their characteristic odor, basic nature, and ability to form hydrogen bonds.
2. How are amines classified?
Ans. Amines can be classified into three categories based on the number of organic groups attached to the nitrogen atom: primary amines (one organic group), secondary amines (two organic groups), and tertiary amines (three organic groups).
3. What are some common uses of amines?
Ans. Amines have various applications in different industries. Some common uses of amines include: - Pharmaceuticals: Amines are important components in many drugs and medicines. - Agriculture: Amines are used in the production of fertilizers, herbicides, and pesticides. - Dyes and Pigments: Amines are used to produce vibrant colors in dyes and pigments. - Rubber Industry: Amines are used as accelerators and anti-degradants in rubber manufacturing. - Water Treatment: Amines are used as corrosion inhibitors and pH adjusters in water treatment processes.
4. How are amines prepared?
Ans. Amines can be prepared through various methods, including: - Gabriel Synthesis: In this method, phthalimide is treated with alkyl halides to produce primary amines. - Reduction of Nitro Compounds: Nitro compounds can be reduced using reducing agents like hydrogen gas or metal hydrides to obtain amines. - Hoffmann Bromamide Reaction: Amides are treated with bromine and a strong base to produce amines. - Reactions of Cyanides: Cyanides can be hydrolyzed or reacted with Grignard reagents to yield amines.
5. What are some common reactions of amines?
Ans. Amines undergo various reactions, some of which include: - Alkylation: Amines can react with alkyl halides to form new amines with additional alkyl groups. - Acylation: Amines can react with acyl chlorides to form amides. - Hofmann Rearrangement: Primary amines can be converted into primary amides by reacting with chloroform and a strong base. - Gabriel Phthalimide Synthesis: Amines can be synthesized by reacting phthalimide with alkyl halides. - Reductive Amination: Amines can be prepared by the reduction of carbonyl compounds in the presence of reducing agents.
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