Question

A bicyclic silver carboxylate is treated with Br2 and converts to the corresponding bicyclic bromide, with loss of CO2 and AgBr. The following reaction goes through which intermediate?

• A. Free radical intermediate
• B. Carbanion intermediate
• C. Carbocation intermediate
• D. Carbene intermediate

Hint:

Recall that the Hunsdiecker reaction is a radical chain decarboxylative halogenation.

Answer 1

The Correct Option is A

Step 1: This is the Hunsdiecker reaction, converting a carboxylic acid silver salt into an alkyl halide with one fewer carbon.

Step 2: Br2 reacts with the silver carboxylate to form an acyl hypobromite, releasing AgBr.

Step 3: The weak O-Br bond undergoes homolytic cleavage, generating a carboxyl radical and a bromine radical.

Step 4: The carboxyl radical rapidly loses CO2 by decarboxylation, giving an alkyl free radical.

Step 5: This radical abstracts bromine from another Br2, giving the product and continuing the chain.

\[\boxed{\text{Free radical intermediate}}\]