Recall Hofmann degradation (Hofmann bromamide reaction)

A primary amide on treatment with $Br_2/NaOH$ gives a primary amine with one less carbon: $RCONH_2 \xrightarrow{Br_2/NaOH} RNH_2$

The key step involves rearrangement of an intermediate (isocyanate formation), where the group attached to the carbonyl carbon migrates from the carbonyl carbon to nitrogen.

Examine Statement I

Statement I says:

In Hofmann degradation reaction, the migration of only an alkyl group takes place from carbonyl carbon of the amide to the nitrogen atom.

This is incorrect because the migrating group need not be only an alkyl group. In Hofmann rearrangement, the substituent attached to the carbonyl carbon migrates, and this can be alkyl or aryl.

Example: $C_6H_5CONH_2 \xrightarrow{Br_2/NaOH} C_6H_5NH_2$ Here, a phenyl group migrates, not an alkyl group.

Therefore, Statement I is false.

Examine Statement II

Statement II says:

The group is migrated in Hofmann degradation reaction to electron deficient atom.

In the rearrangement step, the migrating group shifts to nitrogen, which is part of an electron-deficient intermediate during rearrangement. Thus the migration is toward an electron-deficient atom/center.

Therefore, Statement II is true.

Choose the correct option

• Statement I: Incorrect
• Statement II: Correct

Hence the correct option is: $\boxed{D}$

Comparison with stored answer

Stored correct answer: $D$

My derived answer also is $D$, so they agree.