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Question: Identify the product formed in the following reaction. $C_6H_5-CH_2-CH_3 \xrightarrow[(ii) \ H_3O^+...

Identify the product formed in the following reaction.

C6H5−CH2−CH3→(ii) H3O+(i) alk.KMnO4C_6H_5-CH_2-CH_3 \xrightarrow[(ii) \ H_3O^+]{(i) \ alk.KMnO_4} product

A

C6H5−CH2COOHC_6H_5 - CH_2COOH

B

C6H5−CH2−CH2−CH2−COOHC_6H_5 - CH_2 - CH_2 - CH_2 - COOH

C

C6H5−OHC_6H_5 - OH

D

C6H5−COOHC_6H_5 - COOH

Answer

C6H5−COOHC_6H_5 - COOH

Explanation

Solution

C6H5−CH2−CH3→(ii) H3O+(i) alk.KMnO4C6H5−COOHC_6H_5-CH_2-CH_3 \xrightarrow[(ii) \ H_3O^+]{(i) \ alk.KMnO_4} C_6H_5 - COOH

(Benzoic acid)

Alkaline KMnO4KMnO_4 oxidizes the benzylic side chain in ethylbenzene (which has at least one benzylic hydrogen) completely to a carboxyl group. Thus, after acid work-up with H3O+H_3O^+, the product is benzoic acid (C6H5–COOHC_6H_5–COOH).