13. AMINES – LONG ANSWER TYPE QUESTIONS

13. AMINES 

Q. 1. Explain the following:

(i) Aniline dissolves in aqueous HCl.
(ii) Amines are stronger bases than ammonia.
(iii) Aniline is a weak base than ethylamine.
(iv) The amino group in ethylamine is basic whereas that in acetamide is not basic.
(v) Cyclo hexylamine is a stronger base than aniline.

Ans⇒ (i) The solubility of C₆H₅NH₂ in dil. HCl is due to the formation of water soluble salt.

C₆H₅NH₂ + H₃⁺ + Cl^– → C₆H₅NH₃⁺Cl^– + H₂O Anilinium chloride

(ii) Amines have electron repelling group which increase the electron density on nitrogen, while ammonia has no such group.

(iii) In aniline the electron pair of nitrogen atom is delocalised due to resonance and hence lesser available for protonation while ethylamine does not undergo resonance.

(iv) In amides the lone pair of electrons on nitrogen atom is delocalised and hence less available for protonation than in amines where no resonance is possible and thus

acetamide is a weaker base than ethylamine.
(v) Aniline is weaker base than cyclohexylamine because of resonance while there is no resonance in cyclohexylamine.

Q. 2. (i) Write str amines corresponding to t C4H11N.
(ii) Write IUPAC names of
(iii) What type of isomeri different pairs of amines?

Ans⇒ Eight isomeric amines are Possible

Q. 3. Arrange the following in increasing order of their basic strength:

(i) C₂H₅NH₂, C₆H₅NH₂, NH₃, C₆H₅CH₂NH₂ and (C₂H₅)₂NH
(ii) C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅-NH₂
(iii) CH₃NH₂, (CH₃)₂ NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅ CH₂NH₂

Ans⇒ (i) (C₂H₅)₂NH > C₂H₅NH₂ > C₆H₅CH₂NH₂ > NH₃ > C₆H₅NH₂

(ii) (C₂H₃)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > C₆H₅NH₂
(iii) (CH₃)₂NH > CH₃NH2 > (CH₃)₃N > C₆H₅CH₂NH₂ > C₆H₅NH₂

Q. 4. Write common name of the following compounds :

Ans⇒ (i) o-amine benzoic acid.
(ii) p-methoxy aniline (or p-anisidine).
(iii) neopentylamine.
(iv) Sec-butyl dimethy lamine.

Q. 5. Explain the following reactions by taking one suitable example in each case :

(i) Hoffman’s bromamide reaction
(ii) Gattermann’s reaction
(iii) Carbylamine reaction
(iv) Diazotisation
(v) Coupling reaction
(vi) Ammonolysis
(vii) Acetylation
(viii) Gabriel phthalimide synthesis

Ans⇒ (i) Hoffman’s Bromamide Reaction : When an amide (aliphatic or aromatic) is treated with bromine in alkali solution, it is converted to a paimary amine that has one carbon atom less than the starting amide.

RCONH₂ RNH2 + Na2CO3 + 2NaBr + 2H2O Aliphatic Amide 
.                                           Amine
 C₆H₅CONH₂ C₆H₅NH₂ + Na₂CO₃ + 2NaBr + 2H₂O
Aromatic amide                  Aniline 
(Benzamide)
CH₃CONH₂           CH₃NH₂
Ethanamide                        Methanamine

This reaction is extremely useful for stepping down (or descending) homologous series.

(ii) Gattermann’s reaction: When benzene ring and diazonium salt is treated with Cu/HCl or Cu/HBr, chlorobenzene or bromobenzene is formed. The yield is about 40%.

(iii) carbylamine reaction : Both aliphatic and aromatic primary amines when warmed with chloroform and an alcoholic solution of KOH produce isocyanides or carylamines which have an unpleasant odour.

R-NH₂ + CHCl₃ + 3 KOH (alc.) R-N C
.1° amine                                                   +3KCl + 3 H₂O

In contrast, secondary and tertiery amines (both aliphatic and aromatic) do not give this test.

(iv) Diazotisation : The reaction of converting aromatic primary amines into diazonium salts with a cold (273-278K) solution of nitrous acid is called diazotisation.

(v) Coupling Reaction : Arenediazonium salts react with highly reactive (i.e., electron-rich) aromatic compounds such as phenols and amines to form brightly coloured azo compounds.
 Ar – N = N – Ar.

(vi) Ammonolysis : Reactions of an alkyl halide (R – X) with an alcoholic solution of ammonia giving a mixture of 1°, 2°, 3° amines and quaternary salt is called ammonolysis.

R – X + HNH₂ R – NH₂ + HX
            (excess)          (Chiefproduct)

If R – X is used in excess, a mixture of 1°, 2°, 3° amines along with some quarternary ammonium salts is obtained.

It is a typical nucleophilic substitution reaction.

(vii) Acetylation : The replacement of an active hydrogen of alcohols, phenols or amines with an acyl group (RCO) to form the corresponding esters or amides is called acetylation. Acetylation is carried out in the presence of a base like pyridine, dimethylaniline etc. by acetyl chloride or acetic anhydride.

CH₃COCl           +    C₂H₅OH CH₃COOC₂H₅+ HCl
Acetyl chloride          ethanol                   Ethyl acetate
.(CH₃CO)₂O + C₂H₅NH₂ →    CH₃CONHCH₂CH₃ + CH₃COOH
Acetic                 Ethylamine      N-Éthylacetamide         anhydride

(viii) Gabriel’s phthalmide synthesis : In this reaction phthalimide is converted into its potassium salt by treating it with alcoholic KOH. Then potassium phthalimide is heated with an alkyl halide to yield on N alkyl phthalimide which is hydrolysed to phthalic acid and a primary amine by heating with HCl or KOH solution. This synthesis is very useful for the preparation of pure aliphatic primary amines.

Q. 6. How is anilline prepared commercially ? Describe carbylamine test for primary amines.

Ans⇒ Aniline is prepared commercially by the reduction of nitrobenzene using iron and hydrochloric
acid.

            Fe + 2HCl → FeCl₂ + 2H] × 3
            C₆H₅NO₂ + 6H → C₆H₅NH₂ + 2H₂O
  C₆H₅NO₂ + ₃Fe + 6HCl + C₆H₅NH₂ + 3FeCl₂ + 2H₂O
(Nitrobenzene)                      (Aniline)

Because of the presence of hydrochloric acid in the reaction mixture actually aniline hydrochloride (C₆H₅NH₃ ⁺ Cl^–) is obtained. This on treatment with aqueous soudium carbonate gives aniline.

Q. 7. Write IUPAC name of following and classify them into primary, secondary and tertiary amines :

(i) (CH₃)₂ CHNH₂
(ii) CH₃(CH₂)₂NH₂
(iii) CH₃NHCH(CH₃)₂
(iv) (CH₃)₃ CNH₂
(v) C₆H₅NHCH₃
(vi) (CH₃CH₂)₂NCH₃
(vii) m-BrC₅H₄NH₂.

Ans⇒ (i) (CH₃)₂ CHNH₂ : 1-amino-1-methyl propane (primary)

(ii) CH₃ (CH₂)₂ NH₂
1-amino propane (primary)
(iii) CH₃NHCH(CH₃)₂
methyl isopropyl amine (secondary)
(iv) (CH₃)₃ CNH₂
tert-butylamine (primary)
(v) C₆H₅NHCH₃
N-Methyl benzenamine (secondary)
(vi) (CH₃CH₂)₂ NCH₃
Diethyl methyl amine (tertiary)
(vii) m-Br CGH NH₂
3-isobromo aniline (primary)

Q. 8. How will you prepare :

(i) Benzylamine by Gabriel Pthalimide synthesis.
(ii) Benzylamine from benzene ?

Ans⇒

Class 12th Chemistry Long Type Question