Question

Identify the major product of the following reaction: 
 

 

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Hofmann bromamide degradation converts a primary amide into a primary amine containing one carbon atom less: \[ RCONH_2 \xrightarrow{Br_2+NaOH} RNH_2 \]

Answer 1

The Correct Option is D

Step 1: Identify the reactant from the text.
The substrate is benzamide $C_6H_5CONH_2$, an aromatic primary amide carrying the $-CONH_2$ group.
Step 2: Identify the reagent.
The reagent is bromine with sodium hydroxide, $Br_2 + NaOH$, which is the reagent for the Hofmann bromamide degradation.
Step 3: Recall the Hofmann bromamide reaction.
In this reaction a primary amide is converted into a primary amine that has one carbon atom fewer than the amide: \[ RCONH_2 \xrightarrow{Br_2 + NaOH} RNH_2. \]
Step 4: Apply it to benzamide.
Here $R = C_6H_5$, so the carbon of the carbonyl is lost and the product is aniline $C_6H_5NH_2$.
Step 5: Note the loss of one carbon.
Benzamide has seven carbons (six in the ring plus the carbonyl carbon); the product aniline has only the six ring carbons, confirming the one carbon shortening.
Step 6: Conclude.
The major product of the reaction is aniline, matching option 4. So the answer is
\[ \boxed{C_6H_5NH_2} \]