Choose the incorrect statement for N-methylbenzamide.

The correct answer is Option (1) → It is formed by benzoylation of aniline.

Benzoylation of an amine is its reaction with benzoyl chloride (\( \ce{C6H5COCl} \)): a hydrogen on the nitrogen is replaced by the benzoyl group, \( \ce{C6H5CO} \)–, and the product is an amide. The group already on the nitrogen stays in the product.

N-Methylbenzamide is \( \ce{C6H5CONHCH3} \): a benzoyl group and a methyl group on nitrogen. It is therefore formed from methanamine:

    \( \displaystyle \ce{C6H5COCl + CH3NH2 -> C6H5CONHCH3 + HCl} \)

Benzoylation of aniline puts a phenyl group on the nitrogen instead, giving benzanilide (N-phenylbenzamide):

    \( \displaystyle \ce{C6H5COCl + C6H5NH2 -> C6H5CONHC6H5 + HCl} \)

• Option (1) It is formed by benzoylation of aniline. — An incorrect statement about N-methylbenzamide. Benzoylation of aniline gives benzanilide (N-phenylbenzamide), not N-methylbenzamide.

• Option (2) It has one benzene ring. — A correct statement about N-methylbenzamide. The only ring is in the benzoyl group; the N-substituent is a methyl group.

• Option (3) It is an amide. — A correct statement about N-methylbenzamide. It contains the –CONH– group.

• Option (4) It is prepared through the benzoylation of methanamine. — A correct statement about N-methylbenzamide. This is the reaction of benzoyl chloride with \( \ce{CH3NH2} \) written above.

NCERT (2026-27 edition):
Acylation – …