See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material and first reaction. The starting material is 1-phenylethanol (Ph-CH(OH)-CH3), a secondary alcohol. PCC (pyridinium chlorochromate) oxidizes secondary alcohols to ketones without over-oxidizing to carboxylic acids. Therefore, product (A) is acetophenone: Ph-C(=O)-CH3. Step 2: Identify the second reagent. The reagent is semicarbazide: NH2-NH-C(=O)-NH2. Semicarbazide reacts with aldehydes and ketones via a condensation reaction (nucleophilic addition followed by elimination of water) to form semicarbazones. Step 3: Reaction of acetophenone with semicarbazide. In semicarbazone formation, the terminal NH2 group of semicarbazide (the one directly attached to the hydrazine nitrogen, i.e., NH2-NH-) acts as the nucleophile attacking the carbonyl carbon of the ketone. After addition and elimination of water, the C=N bond forms between the carbonyl carbon and the NH nitrogen of semicarbazide. The product is: Ph-C(CH3)=N-NH-C(=O)-NH2 This is the semicarbazone of acetophenone. Step 4: Match with options. - Option (a): Ph-C(CH3)=N-C(=O)-NH-NH2 — this has the connectivity reversed (N-C(=O)-NH-NH2 instead of N-NH-C(=O)-NH2), incorrect. - Option (b): Ph-C(CH3)=N-NH-C(=O)-NH2 — this correctly shows the semicarbazone of acetophenone with CH3 on the carbon bearing the C=N bond. This matches. - Option (c): Ph-CH=N-N(CH3)-C(=O)-NH2 — this suggests an aldehyde-derived product with N-methyl substitution, incorrect. - Option (d): Ph-CH=N-C(=O)-NH2 — this suggests an aldehyde-derived product with wrong connectivity, incorrect. Option (b) is the semicarbazone of acetophenone, correctly showing Ph-C(CH3)=N-NH-C(=O)-NH2. Therefore, the correct answer is B.