Concept:The reaction of a ketone with HCN produces a cyanohydrin. The alkyl groups originally on the ketone remain firmly attached to the central carbon.
Solution:- The starting material is 2-butanone (ethyl methyl ketone), which has one methyl group (\( CH_3 \)) and one ethyl group (\( C_2H_5 \)) attached to the carbonyl carbon.
- The nucleophile \( CN^- \) attacks the carbonyl carbon, converting the \( C=O \) to \( C-OH \) after protonation.
- The resulting cyanohydrin will still have the methyl and ethyl groups attached to that central carbon: \( CH_3(C_2H_5)C(OH)CN \).
Why other options are incorrect:Option B is the cyanohydrin of acetone. Option C is the cyanohydrin of formaldehyde. Option D is the cyanohydrin of 3-pentanone (diethyl ketone).
Quality & Fidelity Assurance:
Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.