Question

What is the product formed when acetyl chloride reacts with aniline in the presence of pyridine?
(A) Acetanilide
(B) Acetic anhydride
(C) Benzoylacetate
(D) Benzanilide

Answer 1

Hint: Reaction of acetyl chloride with aniline in presence of pyridine is Nucleophiles substitution of aromatic amines with acid. It forms N-substituted amides. In nucleophilic substitution nucleophile is substituted by other nucleophile

Step by step answer: Aniline is Aromatic amines in which amino group is attached to the benzene ring.

In the reaction of acetyl chloride $C{H_3}COCl$ with aniline, H-atom of $ - N{H_2}$ group is replaced by acetyl group $[C{H_3}C{O^ - }]$ therefore this is a type of nucleophilic substitution.

In this reaction $HCl$is lost which reacts with pyridine to form hydrochloride salt of pyridine.
This reaction is also known as Schotten Baumann reaction.
Role pyridine in reaction: Pyridine is a nucleophile for carbonyl groups. It is used as a catalyst in acylation reactions.
N-atoms in pyridine are nucleophilic because the lone pair of electrons on nitrogen cannot be delocalized around the ring.
Mechanism:
(i) Mechanism includes regular attack of nucleophile of an amine on a carbonyl.

In the final step proton is accepted from $ - N{H_2}$ and from acetanilide.

Therefore, from the above explanation the correct option is (A) Acetanilide.

Note: N-acetylation of aniline with acetyl chloride leads to production of one equivalent acid $(HCl),$ which is formed from salt with unreacted pyridine.
If we add an equivalent base to neutralize $HCl,$ catalytic amount of pyridine yields more in a shorter period.