NCERT Exercise Solutions – Chemistry Chapter 13 Amines
13.1 Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3 )2CHNH2
(ii) CH3 (CH2 )2NH2
(iii) CH3NHCH(CH3 )2
(iv) (CH3 )3CNH2
(vi) (CH3CH2 )2NCH3
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)
13.2 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.
13.3 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.
Ans – (i) The N-single atom’s electron pair is delocalized over the benzene ring in aniline.
The nitrogen atom’s electron density therefore drops. The electron density on the N-atom is increased by the -CH3 group in CH3NH2, in contrast. As a result, aniline has a greater pKb value than methylamine since it is a weaker base than the latter.
(ii) Intermolecular hydrogen bonds cause ethylamine to dissolve in water. However, aniline is insoluble in water because of the significant hydrophobic component, or hydrocarbon portion, which makes H-bonding extremely insignificant.
(iii) Methylamine takes a proton from water since it is more basic than water, releasing OH- ions.
(iv) Nitration is typically done using a cone HNO3 + cone H2SO4 mixture. The majority of aniline protonates to generate an ahilinium ion when these acids are present. As a result, aniline and anilinium ion make up the reaction mixture when acids are present. Currently, the aniline’s -NH2 group is activating and o, p-directing, whereas the anilinium ion’s -+NH3 group is deactivating and rw-directing: P-nitroaniline is the major product of aniline nitration (caused by steric hindrance at the o-position), whereas m-nitroaniline is produced by anilinium ion nitration. In actuality, p-nitroaniline and m-nitroaniline are obtained in around a 1:1 combination. As a result of the protonation of the amino group, nitration of aniline results in a significant amount of m-nitroaniline.
13.4 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 .
Ans – (a) The decreasing order of pKb values or increasing order of basic strength is:
C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH
(b) C6H5NHCH3 is more basic than C6H5NH2 due to the presence of electron-donating −CH3 group in C6H5NHCH3.
The decreasing order of basic strength is:
(C2H5)2NH > CH3NH2 > C6H5N(CH3)2 > QH5NH2
(c) The increasing order of basic strength is:
p-nitroaniline < Aniline < p-Toluidine
(d) The decreasing order of basic strength in gaseous phase.
(C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3
(e) Increasing order of boiling point is:
(C2H3)2NH < C2H5NH2 < C2H5OH
(f) Increasing order of solubility in water is :
C6H5NH2 < (C2H5)2NH < C2H5NH2.
13.5 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?
13.6 Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
Ans – The Hinsberg test allows for the identification and distinction of primary, secondary, and tertiary amines. In this experiment, the amines are given the chance to interact with benzenesulphonyl chloride, Hinsberg’s reagent (C6H5SO2Cl). With Hinsberg’s reagent, the three types of amines respond in distinct ways. As a result, they are simple to recognise using Hinsberg’s reagent.
Benzenesulphonyl chloride and primary amines combine to generate N-alkylbenzenesulphonyl amide, which is soluble in alkali.
A secondary amine forms N, N – dialkylbenzene suiphonamide which remains insoluble in aqueous KOH and even after acidification with dilute HCl
A tertiary amine is insoluble in aqueous KOH and does not react with benzene suiphonyl chloride.
However, due to the production of the ammonium salt, acidification with weak HCI results in a clear solution.
13.7 Write short notes on the following:
(i) Carbylamine reaction
(iii) Hofmann’s bromamide reaction
(iv) Coupling reaction
(vii) Gabriel phthalimide synthesis.
Ans – (i)The carboxylamine reaction
The carboxylamine reaction is a test used to identify primary amines. Carbylamines (or isocyanides) are created when aliphatic and aromatic primary amines are heated with chloroform and ethanolic potassium hydroxide. The smell of these carbylamines is very awful. The results of this test are negative for secondary and tertiary amines.
Low temperatures (273-278 K) are required for the reaction between aromatic primary amines and nitrous acid, which is created on-site from NaNO2 and a mineral acid like HCl, to produce diazonium salts. Diazotization is the process of turning aromatic primary amines into diazonium ions.
For instance, aniline creates benzenediazonium chloride when treated with NaNO2 and HCl at 273–278 K, with NaCl and H2O as byproducts.
(iii) Hoffmann bromamide reaction
A primary amine with one less carbon atom than the original amide is created when a sodium hydroxide amide is treated with bromine in an aqueous or ethanolic solution. Hoffmann bromamide reaction is the name of this breakdown process. An alkyl or aryl group moves from the amide’s carbonyl carbon atom to the nitrogen atom during this reaction.
(iv) Coupling reaction
The coupling reaction is the process by which two aromatic rings are joined by an N=N bond. When phenol or aromatic amines are combined with arenediazonium salts, such as benzene diazonium salts, colourful azo compounds result.
An amino (-NH2) group is substituted for the halogen atom in an alkyl or benzyl halide when the reaction is allowed to proceed with an ethanolic solution of ammonia. Ammonolysis is the term for this rupture of the carbon-halogen bond.
When exposed with acid chlorides, anhydrides, or esters, aliphatic and aromatic primary and secondary amines undergo acetylation reaction via nucleophilic substitution. In this reaction, the hydrogen atom of the -NH2or > NH group is swapped out for the acetyl group, which results in the creation of amides. The HCl produced during the reaction is eliminated as soon as it is created to change the equilibrium to the right. A base that is stronger than the amine, such pyridine, is present during this reaction.
(vii) Gabriel phthalimide synthesis
The creation of aliphatic primary amines can be done using the Gabriel phthalimide synthesis, which is highly helpful. To create the potassium salt of phthalimide, phthalimide is treated with ethanolic potassium hydroxide. Alkyl halide is used to heat this salt further, and then alkaline hydrolysis and corresponding primary amine are produced.
13.8 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
13.9 Give the structures of A, B and C in the following reactions:
13.10 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 – Compound ‘B’ must be an amide, and molecule ‘C’ must be an amine since the compound ‘C’ with the chemical formula C6H7N is produced from compound ‘B’ on treatment with Br2 KOH.
Aniline is the only amine with the chemical formula C6H7N, or C6H5NH2.
The die amide from which it is generated must be benzamide because “C” is aniline (C6H5CONH2). Thus, benzamide is chemical “B.” Compound “A” must be benzoic acid since compound “B” is created by heating compound “A” and adding aqueous ammonia.
13.11 Complete the following reactions:
13.12 Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Ans – Primary aliphatic amines are made via the Gabriel phthalimide synthesis. It involves phthalimide-derived anion nucleophilic substitution (SN2) of alkyl halides.
The phthalimide anion’s nucleophilic assault on the organic halogen molecule is essential for the Gabriel phthalimide reaction to be successful.
Gabriel’s phthalimide process cannot produce arylamines, or aromatic primary amines, since aryl halides are difficult to convert into nucleophilic substitution reactions.
13.13 Write the reactions of
(i) aromatic and
(ii) aliphatic primary amines with nitrous acid.
Ans – Reaction with nitrous acid. Both aliphatic and aromatic amines of all three types interact with nitrous acid in a variety of ways to produce distinct results. Due to its extreme instability, nitrous acid must be produced on-site through the reaction of sodium nitrite with diluted hydrochloric acid.
(i) Under cold conditions, primary aliphatic amines react with nitrous acid to produce primary alcohol, nitrogen gas, and a vigorous effervescence. In order to create nitrous acid, which is unstable in nature, sodium nitrite is combined with diluted hydrochloric acid on-site.
(ii) Nitrous acid, which is created in-situ from NaNO2 and a mineral acid like HCl, reacts with aliphatic primary amines to create unstable aliphatic diazonium salts, which then react with alcohol and HCl to produce N2 gas.
13.14 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)A proton lost from an amine results in an amide ion, whereas a proton lost from alcohol results in an alkoxide ion.
R—O —H—>R— O– +H+
O will attract positive species more strongly than N because O is more electronegative than N. Since RNH® is less stable than RO, it is. Alcohols therefore have a higher acidity than amines. Alcohols are more acidic than amines, on the other hand.
(ii) Primary amines experience substantial intermolecular H-bonding because they have two H-atoms on the N-atom, whereas tertiary amines do not since they have no H-atom on the N-atom. Primary amines have higher boiling temperatures than tertiary amines with similar molecular masses as a result.
(iii) The following factors make aromatic amines significantly less basic than ammonia and aliphatic amines:
(a) The lone pair of electrons on the nitrogen atom get delocalized over the benzene ring due to resonance in aniline and other aromatic amines, making it less accessible for protonation. As a result, compared to ammonia and aliphatic amines, aromatic amines are weaker bases.
(b) Compared to the comparable protonated ion, aromatic amines are more stable.