Question

The major products formed in the following reaction sequence A and B are : 

• A. A
• B. B
• C. C
• D. D

Hint:

The haloform reaction converts a methyl ketone into a carboxylic acid salt (with one less carbon) and a haloform ($CHX_3$).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The reaction of a methyl ketone with a halogen ($Br_2$) in the presence of a strong base (KOH) is known as the Haloform Reaction. It is a diagnostic test for the $CH_3C=O$ group.
Step 2: Detailed Explanation:
The starting material is acetophenone ($C_6H_5COCH_3$).
1. In the presence of KOH and $Br_2$, the alpha-methyl group undergoes exhaustive bromination to form a trihalomethyl ketone intermediate: $C_6H_5COCBr_3$.
2. The hydroxide ion then attacks the carbonyl carbon, followed by the cleavage of the $C-C$ bond to release the $CBr_3^-$ ion.
3. Proton transfer occurs between the resulting benzoic acid and the $CBr_3^-$ ion to form a carboxylate salt and bromoform ($CHBr_3$).
Reaction: $C_6H_5COCH_3 + 3Br_2 + 4KOH \rightarrow C_6H_5COOK + CHBr_3 + 3KBr + 3H_2O$.
Product A is potassium benzoate ($C_6H_5COOK$) and product B is bromoform ($CHBr_3$).
Step 3: Final Answer:
The major products are A = Potassium benzoate and B = $CHBr_3$.