
Silver benzoate reacts with bromine to form
A.

B.

C.

D.

Answer
161.7k+ views
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.
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.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main Mock Test Series Class 12 Chemistry for FREE

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Classification of Drugs

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Solutions Class 12 Notes: CBSE Chemistry Chapter 1

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

Electrochemistry Class 12 Notes: CBSE Chemistry Chapter 2
