The intermediate(s) formed during the reaction C 6 H 5 CH 2 COOAg + Br 2 CCl 4 C 6 H 5 CH 2 Br , is :

Question

The intermediate(s) formed during the reaction C 6 ​ H 5 ​ CH 2 ​ COOAg + Br 2 ​ CCl 4 ​ ​ C 6 ​ H 5 ​ CH 2 ​ Br , is :

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The reaction shown is the Hunsdiecker reaction, which proceeds via a free-radical mechanism.

Step 1: Formation of acyl hypobromite. The silver salt reacts with bromine to form an acyl hypobromite intermediate.

$C_{6} H_{5} C H_{2} COO A g + B r_{2} \to C_{6} H_{5} C H_{2} COOB r + A g B r$

This corresponds to option (A).

Step 2: Radical formation. The acyl hypobromite undergoes homolytic cleavage to form an acyloxy radical and a bromine radical.

$C_{6} H_{5} C H_{2} COOB r Δ C_{6} H_{5} C H_{2} COO \cdot + B r \cdot$

These correspond to options (B) and (D) respectively.

Step 3: Decarboxylation. The acyloxy radical quickly decarboxylates to form a benzyl radical and carbon dioxide.

$C_{6} H_{5} C H_{2} COO \cdot \to C_{6} H_{5} C H_{2} \cdot + C O_{2}$

This corresponds to option (C).

Step 4: Product formation. The benzyl radical combines with a bromine radical to form the final product, benzyl bromide.

$C_{6} H_{5} C H_{2} \cdot + B r \cdot \to C_{6} H_{5} C H_{2} B r$

Answer: The intermediates formed are (A) $C_{6} H_{5} C H_{2} COOB r$ , (B) $C_{6} H_{5} C H_{2} COO \cdot$ , (C) $C_{6} H_{5} C H_{2} \cdot$ , and (D) $B r \cdot$ . Therefore, the correct options are (A), (B), (C), and (D).

Chemistry - Carboxylic acid and their Derivatives Question with Solution | TestHub

Options:(select one or more)

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