See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Reaction of nitrile with Grignard reagent: The starting material is cyclobutanecarbonitrile where the carbon of the C≡N is labeled as C-14 (radioactive/isotope label). When this reacts with CH3MgBr (a Grignard reagent), the methyl group attacks the electrophilic carbon of the nitrile (C-14), forming an imine salt intermediate. Upon hydrolysis with H3O+, the imine hydrolyzes to give a ketone. The product (A) is cyclobutyl methyl ketone where the carbonyl carbon is C-14 (the original nitrile carbon). This gives cyclobutane-C(14)(=O)-CH3. Step 2 - Identification of product (A): Since the nitrile carbon (C-14) becomes the carbonyl carbon upon hydrolysis, product (A) is cyclobutane ring with 14C(=O)-CH3. This matches option (c). Step 3 - Iodoform reaction with NaOI: Product (A), cyclobutyl methyl ketone (a methyl ketone), undergoes the iodoform reaction with NaOI (sodium hypoiodite). In the iodoform reaction, the methyl group adjacent to the carbonyl is triiodomethylated and then cleaved. The reaction cleaves the bond between the carbonyl carbon (C-14) and the CH3 group. The products are: - (B): cyclobutanecarboxylate salt (cyclobutane-14COO- Na+), i.e., the carboxylate containing C-14 - (C): CHI3 (iodoform, which comes from the CH3 group — normal carbon, not labeled) Step 4 - Why option (c) is correct for (A) and (C): - (A) = cyclobutane ring with 14C(=O)-CH3 (the labeled carbon is the carbonyl carbon) - (C) = CHI3 (iodoform from the unlabeled CH3 group) This matches option (c): cyclobutane ring with (14)C(=O)-CH3; CHI3 Step 5 - Why other options fail: - Option (a): Shows unlabeled carbonyl carbon — incorrect, the C-14 label from the nitrile becomes the carbonyl carbon. - Option (b): Shows unlabeled carbonyl carbon and 14CHI3 — both labels are wrong. - Option (d): Shows CH2-CHO product which would require a different reaction pathway (reduction, not Grignard addition to nitrile followed by hydrolysis). Therefore, the correct answer is C.