Question

(i) Complete the following equations :

(I) \( \mathrm{CH_3NH_2 + C_6H_5COCl} \longrightarrow \)

(II) \( \mathrm{C_6H_5NH_2} \) + 3 \( \mathrm{Br_2} \) \( \xrightarrow{\mathrm{H_2O}} \)

(III) \( \mathrm{C_6H_5CONH_2} \) \( \xrightarrow{\mathrm{NaOBr}} \)

(ii) (I) Explain the carbylamine reaction.

(II) Why can aromatic primary amines not be prepared by the Gabriel phthalimide synthesis ?

Hint:

Acylation gives an amide, bromine water gives tribromoaniline, and Hofmann degradation gives aniline. Aryl halides do not undergo nucleophilic substitution, so Gabriel synthesis fails for aryl amines.

Answer 1

Step 1: Part (i)(I): Acylation of methylamine:
A primary amine reacts with an acid chloride by nucleophilic substitution. The NH2 group replaces Cl, and HCl is lost. This is called benzoylation.
\[ \mathrm{CH_3NH_2 + C_6H_5COCl} \rightarrow \mathrm{C_6H_5CONHCH_3 + HCl} \]
The product is N-methylbenzamide.

Step 2: Part (i)(II): Bromination of aniline:
The NH2 group strongly activates the ring at the ortho and para positions. In bromine water all three positions react at once.
\[ \mathrm{C_6H_5NH_2 + 3Br_2} \xrightarrow{\mathrm{H_2O}} \mathrm{2,4,6\text{-}C_6H_2Br_3NH_2 + 3HBr} \]
The product is 2,4,6-tribromoaniline, a white precipitate.

Step 3: Part (i)(III): Hofmann bromamide degradation:
An amide with NaOBr (Br2 and NaOH) loses its carbonyl carbon and forms a primary amine with one carbon less.
\[ \mathrm{C_6H_5CONH_2 + Br_2 + 4NaOH} \rightarrow \mathrm{C_6H_5NH_2 + Na_2CO_3 + 2NaBr + 2H_2O} \]
The product is aniline.

Step 4: Part (ii)(I): Carbylamine reaction:
Aliphatic and aromatic primary amines, on heating with chloroform and alcoholic KOH, form isocyanides (carbylamines). These have a very foul smell.
\[ \mathrm{R{-}NH_2 + CHCl_3 + 3KOH} \xrightarrow{\Delta} \mathrm{R{-}NC + 3KCl + 3H_2O} \]
The reaction is used as a test for primary amines. Secondary and tertiary amines do not give it.

Step 5: Part (ii)(II): Limit of Gabriel synthesis:
In the Gabriel method, potassium phthalimide must attack an alkyl halide by nucleophilic substitution.
Aryl halides do not undergo this substitution easily. The C-X bond has partial double bond character because of resonance, and the carbon is sp2 hybridised.
So the phthalimide anion cannot replace the halogen of an aryl halide, and aromatic primary amines cannot be made this way.

Final Answer:
(i)(I) N-methylbenzamide. (II) 2,4,6-tribromoaniline. (III) aniline. (ii) The carbylamine reaction gives a foul smelling isocyanide from a primary amine; aryl halides resist nucleophilic substitution by phthalimide.\[ \boxed{\mathrm{C_6H_5CONHCH_3;\ 2,4,6\text{-}tribromoaniline;\ C_6H_5NH_2}} \]