See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the starting material. The starting material is 1-methyl-2-(acetyl)cyclohexan-1-one, i.e., a cyclohexanone where the carbon alpha to the ring ketone (C1) carries both a CH3 group and a –COCH3 (acetyl) group. Step 2 – Haloform reaction on the acetyl group. Treatment with Br2/NaOH (alkaline hypohalite conditions) selectively performs the iodoform/haloform reaction on the methyl ketone (–COCH3). The –COCH3 group undergoes trihalogenation at the methyl carbon and then base-promoted cleavage: –COCH3 → –COO⁻ (carboxylate) + CHBr3. So product (A) is the sodium carboxylate salt: 1-methyl-2-oxocyclohexane-1-carboxylate (sodium 1-methyl-2-oxocyclohexane-1-carboxylate), and CHBr3 (bromoform) is the other product. Step 3 – Acidification and decarboxylation of (A). When the carboxylate salt (A) is treated with H+/heat (Delta), it is first protonated to give the beta-keto acid (1-methyl-2-oxocyclohexane-1-carboxylic acid). Beta-keto acids readily undergo decarboxylation upon heating. Loss of CO2 from the alpha position gives 2-methylcyclohexan-1-one (a cyclohexanone with a methyl group at C2). Step 4 – Match to answer choices. 2-methylcyclohexan-1-one corresponds to option (b). Step 5 – Why other options fail: (a) 3-methylcyclohexan-1-one would require migration of the methyl group, which does not occur. (c) 2-bromo-2-methylcyclohexan-1-one would require retention of bromine on the ring, but halogenation occurs only on the acetyl methyl in the haloform step. (d) 2-(tribromomethyl)cyclohexan-1-one would require no cleavage of the C–C bond, which contradicts the haloform mechanism. Therefore, the correct answer is B.