NCERT Solutions for Class 12 Chemistry Chapter 13 Amines includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 12 exams must go through NCERT Solutions for Class 12 Chemistry Chapter 13 Amines. Going through the solutions provided on this page will help you to know how to approach and solve the problems.
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NCERT Solutions for Class 12 Chemistry Chapter 13 Amines
NCERT Solutions for Class 12 Chemistry Chapter 13 Amines are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.
Classify the following amines as primary, secondary and tertiary :
(a) Write structures of different isomeric amines corresponding to the molecular formula, C4H11N.
(b) Write IUPAC names of all the isomers.
(c) What type of isomerism is exhibited by different pairs of amines?
How will you convert: (i) Benzene into aniline, (ii) Benzene into N,N-dimethylaniline, (iii) Cl(CH2)4Cl into hexan-1,6-diamine?
Arrange the following in increasing order of their basic strength.
(a) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH
(b) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2
(c) CH3NH2< (CH3)2NH, (CH3)3N, C6H5NH2> C6H5CH2NH2.
(a) The order of basicity is C6H5NH2 < NH3 < C6H5CH2NH2 < C2H5NH2 < (C2H5)2NH
The basic nature of amine arises from their ability to donate the lone pair of electrons on N to electrophiles. The availability of this l.p. depends on two factors :
(i) Electron donating/withdrawing effect of the alkyl groups attached to the N atom.
(ii) Steric hindrance posed by alkyl groups around N.
The presence of alkyl groups on the N atom increases the electron density and makes the l.p. more available for electrophiles. This happens due to the +1 effect of the alkyl groups. But, if the alkyl groups are too bulky or too many in number, they tend to sterically hinder the incoming proton and the basic strength decreases. These two factors working in opposing directions, tend to balance out each other in 2° amines, making them most basic and the basic strength follows the order :
Complete the following acid-base reactions and name the products :
- CH3CH2CH2NH2 + HCl →
- (C2H5)3N + HCl →
Write the reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
Write the chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC name of the isomers which will liberate nitrogen gas on treatment with nitrous acid.
- 3-Methylaniline into 3-nitrotoluene
- Aniline into 1,3,5-tribromobenzene.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
- Propan-2-amine (primary),
- Propan-l-amine (primary),
- N-Methylpropan-2-amine (secondary),
- 2-Methylpropan-2-amine (primary),
- N-Methylbenzenamine or N-Methylaniline (secondary),
- N-Ethyl-N-methylethanamine (tertiary),
- 3-Bromobenzenamine or 3-Bromoaniline (primary).
Give one chemical test to distinguish between the following pairs of compounds.
- Methylamine and dimethylamine
- Secondary and tertiary amine
- Ethylamine and W-methylaniline
- Aniline and benzylamine
- Aniline and N-methylaniline
Account for the following :
(i) pKb of 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- and 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.
(i) If the pKb value of any base or compound is higher than that of another, it implies that the former is a weaker base than the latter. In aniline, the N-atom is attached to the benzene ring and therefore the lone pair on N is delocalised over the entire benzene ring. As a result, it cannot accept a proton or any other electrophile.
This is why it has a lower Kb value (lower basic strength) and high corresponding pKb value.
In methylamine, CH3NH2, the electron density on nitrogen is greater than that in case of aniline. This is because -CH3 group in methylamine, by virtue of its +1 effect, increases electron density on N, which is more available for protonation.
(ii) Any compound capable of forming hydrogen bonds with water, dissolves in it. Ethylamine is able to do the same and hence its solubility.
However, in aniline, the bulky hydrocarbon part – C6H5 prevents the formation of effective hydrogen bonding and therefore it is not soluble.
(iii) The formation of hydrated ferric oxide may be understood by taking into consideration the basic strength of CH3NH2. In presence of CH3NH2, water hydrolyses as
(iv) During nitration, the nitration mixture used (cone. HNO3 and cone. H2SO4) protonates the NH2 group to produce anilinium ion as
But when this reaction is carried out with aniline, no electrophile generation takes place. The reason being the presence of aniline as a base.
Aniline is a Lewis base, reacts with AlCl3 and hence deactivates it. The Lewis acid is therefore no more available for electrophile generation and hence reaction does not take place.
(vi) Diazonium salts carry a N atom with a positive charge. This positive charge is well dispersed in aromatic diazonium salts through resonance as shown below :
Such a charge delocalisation is not possible in aliphatic amines and hence they are less stable.
(vii) Gabriel phthalimide reaction gives pure primary amines without any contamination of secondary and tertiary amines. Therefore, it is preferred for synthesising primary amines.
Arrange the following :
(i) In decreasing order of the pKb values: 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.
The availability of l.p. on N of p-nitroaniline is drastically reduced by presence of electron withdrawing -NO2 group on it.
In contrast, presence of electron releasing -CH3 group increases the electron density on N atom and improves basicity in p-toluidine.
(b) C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2 Involvement of l.p. of N in resonance causes aniline to have low basicity. In II, the -Me group through its +I effect improves the electron density on N and therefore its basic strength increases. In III, the -NH2 is farther off from benzene ring and hence l.p. is localized on it and hence the basic strength is highest.
(iv) In gas phase, basicity follows the order : (C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3 In gas phase, the stabilization by solvation is not present and hence basic strength follows the expected order based on +I effect of alkyl groups.
(v) (CH3)2NH < C2H5NH2 < C2H5OH
(vi) C6H5NH2 < (C2H5)2NH < C2H5NH2
Amines can form hydrogen bonds with water and are therefore soluble in it. However, the solubility decreases if the mass of the hydrocarbon part increases.
How will you convert :
- Ethanoic acid into methanamine
- Hexanenitrile into 1-aminopentane
- Methanol to ethanoic acid
- Ethanamine into methanamine
- Ethanoic acid into porpanoic acid
- Methanamine into ethanamine
- Nitromethane into dimethylamine
- Propanoic acid into ethanoic acid?
Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
1°,2° and 3° amines can be distinguished by Hinsberg’s reagent.
Write short notes on the following :
- Carbylamine reaction
- Hofmann’s bromamide reaction
- Coupling reaction
- Gabriel phthalimide synthesis
(i) Carbylamine reaction : Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul smelling substances. Secondary and tertiary amines do not show this reaction. This reaction is known as carbylamine reaction or isocyanide test and is used as a test for primary amines.
(ii) Diazotisation : The conversion of primary aromatic amines into diazonium salts is known as diazotisation.
The conversion is brought about by reacting the amine with HNO2 which is prepared in situ.
(iii) Hofmann’s bromamide reaction : Primary amides when heated with Br2 and (aqueous or ethanoic solution of) NaOH lose a carbon atom and are converted to the corresponding amines. It is an example of step-down reaction.
(iv) Coupling reaction : The reaction of diazonium salts with phenols and aromatic amines to form azo compounds having an extended conjugate system with both aromatic rings joined through the — N = N — bond, is called coupling reaction. In this reaction, the nitrogen atoms of the diazo group are retained in the product. The coupling with phenols takes place in mildly alkaline medium while that with amines occurs under faintly acidic conditions. For example,
Coupling generally occurs at the p-position with respect to the hydroxyl or the amino group, if free, otherwise it takes place at the o-position.
(v) Ammonolysis : The process of cleavage of the C — X bond in alkyl halides by ammonia molecule is called ammonolysis. 1° amine thus obtained behaves as a nucleophile and further reacts with alkyl halide to form 2°, 3° and finally quaternary ammonium salt.
(vi) Gabriel phthalimide synthesis : In this reaction phthalimide is converted into its potassium salt by treating it with alcoholic potassium hydroxide. Then potassium phthalimide is heated with an alkyl halide to yield an N-alkylpthalimide which is hydrolysed to phthalic acid and primary amine by alkaline hydrolysis.
This synthesis is very useful for the preparation of pure aralkyl and aliphatic primary amines. However, aromatic primary amines cannot be prepared by this method.
Accomplish the following conversions :
- Nitrobenzene to benzoic acid
- Benzene to m-bromophenol
- Benzoic acid to aniline
- Aniline to 2,4,6-tribromofluorobenzene
- Benzyl chloride to 2-phenylethanamine
- Chlorobenzene to p-chloroaniline
- Aniline to p-bromoaniline
- Benzamide to toluene
- Aniline to benzyl alcohol.
Give the structures of A, B and C in the following reactions :
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.
Complete the following reactions :
Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
(i) Gabriel phthalimide reaction involves the nucleophilic attack of the phthalimide on the alkyl halide.
(ii) Such a nucleophilic substitution reaction is not possible if the substrate is an aryl halide.
(iii) The reason for it can be explained on the basis of
(a) Partial double bond character of C — X bond in aryl halide. Consider the following structures :
From the resonance structures we see that the C — X bond has a double bond character in structures II, III and IV and this makes cleavage of C — X bond difficult.
(b) Also the steric hindrance by the bulky aryl group prevents the incoming nucleophile.
Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
Amines do not directly react with nitrous acid, rather they react with a mixture of dil. HCl and NaNO2 and HNO2 is produced in situ.
The reactions are :
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?
(i) Loss of a proton from an amine gives RNH– ion while loss of a proton from alcohol gives RO- ion as shown below :
As O is more electronegative than N,RO– can accommodate the negative charge more easily than the RNH– can.
As, RO– is more stable than RNH– the former is formed more. As a result, amines are less acidic than alcohols.
(ii) (a) At boiling point, the molecules in a compound break free from their inter molecular forces and escape into the vapour phase.
Weaker the inter-molecular forces, lower will be the boiling point.
(b) In 1 ° amines, there is strong H-bonding that binds the amine molecules together. Whereas in 3° amine absence of H on N atom prevents hydrogen bonding completely.
This is why 1° amines have higher boiling point.
(iii) (a) The basic nature of amines is a result of the presence of l.p. of electron on the N atom. Also the electron density is increased on N due to the +I effect of alkyl group.
(b) In aryl amines the l.p. on N is involved in resonance with the benzene ring and hence less available for protonation.
(c) In aliphatic amines there is no such delocalisation and hence it is more basic.
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