When aniline is reacted with benzoyl chloride, the correct product is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → N-Phenylbenzamide
Primary and secondary amines react with acid chlorides by nucleophilic substitution at the acyl carbon: the lone pair on nitrogen attacks the carbonyl carbon, the chloride ion leaves, and one hydrogen on nitrogen is replaced by the acyl group. The product is an amide. A base such as pyridine or NaOH is added to remove the HCl formed.
\( \displaystyle \ce{C6H5NH2 + C6H5COCl} \xrightarrow{\text{base}} \ce{C6H5NHCOC6H5 + HCl} \)
The product has a phenyl group on the nitrogen of benzamide, so it is N-phenylbenzamide (benzanilide).
Why the other options are incorrect
Option (2) Biphenyl — Incorrect. Biphenyl, \( \ce{C6H5-C6H5} \), is made by joining two aryl groups, as in the Fittig reaction of chlorobenzene with sodium; no carbon–carbon bond is formed here.
Option (3) p-(Amino)phenylbenzoyl chloride — Incorrect. This would need acylation of the ring itself; the nitrogen is the most nucleophilic site of aniline, so the acyl group goes onto nitrogen, and the product no longer has an acid chloride group.
Option (4) N-Phenylaniline — Incorrect. N-Phenylaniline (diphenylamine) has a phenyl group, not a benzoyl group, on nitrogen; the product of this reaction keeps the C=O group.
NCERT (2026-27 edition):
Acylation – …
