See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is a para-substituted benzene ring bearing a ketone group C(=O)-CH3 (acetyl group) at one position and a -CH2-CH2-Br (2-bromoethyl) group at the para position. This compound is 4'-(2-bromoethyl)acetophenone. Step 2 - Wolff-Kishner reduction on the ketone: The Wolff-Kishner reduction (N2H4, HO-, heat) converts a ketone (C=O) to a methylene (CH2) group. The C(=O)-CH3 group is reduced to CH2-CH3 (an ethyl group). This reaction is selective for carbonyl groups and does not affect alkyl halides under these conditions. Step 3 - Effect on the bromoethyl group: The -CH2-CH2-Br group is an alkyl halide, not a carbonyl. However, under strongly basic conditions (HO-, heat) required for Wolff-Kishner reduction, the primary alkyl bromide -CH2-CH2-Br can undergo E2 elimination (base-induced dehydrohalogenation). The hydroxide ion acts as a base, abstracting a beta hydrogen from -CH2-CH2-Br to give a vinyl group -CH=CH2 (styrene-type side chain). Step 4 - Combined result: After Wolff-Kishner reduction and base-induced elimination, the product has an ethyl group (-CH2-CH3) at one para position (from reduction of C=O-CH3) and a vinyl group (-CH=CH2) at the other para position (from elimination of HBr from -CH2-CH2-Br). This gives 1-ethyl-4-vinylbenzene (p-ethylstyrene). Step 5 - Match to options: Option (b) shows a benzene ring with an ethyl group at one para position and a vinyl group at the other para position, which matches exactly. Step 6 - Why other options fail: - (a) would require substitution of Br by OH, not reduction/elimination - (c) would require only reduction of the ketone with no reaction at the bromide, but the strongly basic conditions cause elimination - (d) involves rearrangement of the bromide which is not expected here Therefore, the correct answer is B.