Question

The major product S of the following reaction is 

• A. A
• B. B
• C. C
• D. D

Hint:

In anilides, the –NHCO– group activates the ring and directs electrophiles strongly to the para position.

Answer 1

The Correct Option is C

Step 1: Reaction with hydroxylamine (NH$_2$OH·HCl).
Benzil reacts with hydroxylamine to form the benzil monoxime. The oxime formation occurs at one of the carbonyl groups, giving a C=NOH functionality. This step does not alter aromatic rings.
Step 2: Beckmann rearrangement (H$_2$SO$_4$, heat).
The oxime undergoes Beckmann rearrangement.
In benzil monoxime, migration occurs of the phenyl group trans to the oxime OH. This produces a benzil-derived amide in which one phenyl migrates, forming: Ph–CO–NH–Ph (benzamide derivative).
Step 3: Bromination using Br$_2$/FeBr$_3$.
Electrophilic aromatic substitution occurs on the ring attached to nitrogen.
Anilide (Ar–NH–CO–Ph) strongly directs electrophiles to the para position because –NHCO– is an activating, ortho/para directing group.
Hence bromine enters para to the –NHCO– group on the anilide ring.
Step 4: Identify the correct structure.
Option (B) shows the amide with bromine at the para position relative to the –NHCO– group, which matches the expected regioselectivity.
Other options show ortho substitution or bromination on the wrong ring, which is not favored.
Step 5: Conclusion.
Thus the major product is structure (B).