Which of the following is the reagent for Hofmann degradation of amides?

Br₂/4KOH

The correct answer is option 1. Br₂/4KOH.

The Hofmann degradation, or Hofmann rearrangement, is a chemical reaction where a primary amide is converted to a primary amine with one fewer carbon atom. The reaction involves the use of bromine (Br\(_2\)) and a strong base, typically potassium hydroxide (KOH).

Here is the detailed mechanism and reasoning:

1. Reaction Setup: The reaction requires the presence of bromine and an excess of a strong base. The typical conditions involve using Br\(_2\) and KOH.

2. Formation of N-Bromoamide: The amide reacts with bromine to form an N-bromoamide intermediate.

\(RCONH_2 + Br_2 \rightarrow RCONHBr + HBr\)

3. Deprotonation: The strong base (KOH) deprotonates the N-bromoamide.

\(RCONHBr + KOH \rightarrow RCONBr^- + K^+\)

4. Rearrangement: The deprotonated N-bromoamide undergoes a rearrangement, where the carbonyl carbon is attacked by the nitrogen, leading to the formation of an isocyanate intermediate (R-N=C=O) and the elimination of bromide (Br\(^-\)).

5. Hydrolysis: The isocyanate intermediate is hydrolyzed by water to form a primary amine and carbon dioxide.

\(R-N=C=O + H_2O \rightarrow R-NH_2 + CO_2\)

For this reaction, the correct ratio of the reagents is typically important to drive the reaction to completion efficiently. The notation Br\(_2\)/4KOH suggests the stoichiometry used in some protocols to ensure the complete conversion of the amide to the amine.

Thus, the correct reagent for the Hofmann degradation of amides is: Option 1: Br\(_2\)/4KOH