Question

Silver benzoate reacts with bromine to form
A.

B.

C.

D.

Answer 1

Hint: Silver benzoate reacts with bromine in the presence of carbon tetrachloride under reflux conditions. An alkyl halide is formed as the final product. Whereas silver halide and carbon dioxide are the side products. The overall process can be explained by the Hunsdiecker reaction.

Complete step by step solution:
Alkyl carboxylic salt here is silver benzoate, reacts with bromine and first produces a very unstable intermediate. Under the refluxing condition, unstable reaction intermediates further decarboxylate to desired alkyl halides. According to the given question, silver benzoate reacts with bromine to give bromobenzene.
Here the overall reaction proceeds via a radical mechanism. It is one of the important pathways to get a product with one carbon less. The chain initiation consists of homolytic cleavage of carbon-bromine bond forming aryl carbonyl radical. The overall driving force of this reaction is the participation of silver bromide which is less soluble and stable. Silver benzoyate is transformed into Aryl hypobromite in the presence of Br atom. Then by radical mechanism, a carboxyl radical and a bromine atom is formed. Aryl carboxyl radical decomposes into carbon dioxide by decarboxylation to produce Aryl radical, which further reacts with Br radical to form aryl bromide or bromobenzene.
Mechanism:

Thus, Option (B) is correct.

Note: Hunsdiecker reaction is a very important reaction to synthesise higher bromoalkane compounds in case of long chain fatty acids. The overall yield of alkyl halide of this reaction follows the order: ${{1}^{{\mathrm O}}}>{{2}^{{\mathrm O}}}>{{3}^{{\mathrm O}}}$. If this reaction starts with ${{3}^{{\mathrm O}}}$ alkyl carboxylate, it gives alkene as a major product instead of alkyl halide.