Passage
Preparation & Chemical reactions of amines
Preparation:
Nitro compounds are reduced to amines by passing hydrogen gas in the presence of finely divided nickel, palladium, or platinum as catalyst. Nitriles on reduction with lithium aluminum hydride or catalytic hydrogenation produce primary amines. The Gabriel synthesis is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide.
Chemical reactions:
Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides, and esters by nucleophilic substitution reaction. This reaction is known as acylation. Both aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide give corresponding isocyanide. Benzenesulphonyl chloride also known as Hinsberg’s reagent reacts with primary and secondary amines to form sulphonamide. Aniline reacts with bromine water at room temperature to give a white precipitate of 2,4,6-tribromoaniline. Aniline reacts with HNO3 in the presence of H2SO4 to give nitro derivatives. Aniline reacts with concentrated sulphuric acid to form anilinium hydrogensulphate which on heating with sulphuric acid at 453-473K produces p-aminobenzene sulphonic acid or sulphanilic acid.
Identify the major product (P) in the following chemical reaction.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → N-(3-methylphenyl)benzamide
Primary and secondary amines react with acid chlorides by nucleophilic substitution: the lone pair on nitrogen attacks the C=O carbon and Cl is replaced. A hydrogen on the nitrogen is thus replaced by an acyl group (acylation). With benzoyl chloride this is called benzoylation. It is carried out in the presence of a base (here 10% NaOH), which removes the HCl formed and drives the reaction forward:
\( \displaystyle \ce{C6H5COCl + H2N-C6H4-CH3} \xrightarrow{10\%\ \ce{NaOH}} \ce{C6H5CO-NH-C6H4-CH3 + HCl} \)
\( \displaystyle \ce{HCl + NaOH -> NaCl + H2O} \)
Only the N–H bond takes part; nothing on either ring changes. So the benzoyl group (with no methyl) is joined to the nitrogen, and the methyl group stays meta to the nitrogen on the other ring. The product is N-(3-methylphenyl)benzamide.
Why the other options are incorrect
Option (1) 3-methyl-N-(3-methylphenyl)benzamide — Incorrect. It has an extra methyl group on the benzoyl ring, which benzoyl chloride does not carry.
Option (3) 3-methyl-N-phenylbenzamide — Incorrect. The methyl group is on the benzoyl ring instead of the amine ring; this would come from 3-methylbenzoyl chloride and aniline.
Option (4) N-(4-methylphenyl)benzamide — Incorrect. The methyl group is para to the nitrogen; this would come from p-toluidine, not from the meta-substituted amine given.
NCERT (2026-27 edition):
Acylation
