See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Step 1 - Identify compound B: Compound B is 1,3,5-trimethylidenecyclohexane, a cyclohexane ring bearing three exocyclic methylene (=CH2) groups at the 1, 3, and 5 positions. Step 2 - Retrosynthetic analysis of B: POCl3/pyridine is a dehydration reagent that converts tertiary or secondary alcohols to alkenes (elimination). So compound B is formed by dehydration of a triol intermediate. The triol intermediate must be a cyclohexane with three -C(CH3)(OH)- or -CH(OH)- groups that, upon elimination, give =CH2 groups. Specifically, each exocyclic =CH2 in B requires that the precursor alcohol had a -C(OH)(H)(CH3) group... Wait, let us reconsider: =CH2 exocyclic means a CH2 double bond exo to the ring. The carbon in the ring is sp2, and the exocyclic carbon is =CH2. So the alcohol precursor would be ring-C-CH2OH... no. Actually, POCl3/pyridine eliminates an OH and an adjacent H to form a double bond. For an exocyclic =CH2, the alcohol is ring-C(OH)-CH3 (a tertiary alcohol where CH3MgBr added to a carbonyl on the ring). So if A has a carbonyl group directly on the ring (e.g., -C(=O)-R), CH3MgBr adds to give ring-C(OH)(CH3)-R. Then POCl3 eliminates to give ring-C(=CH2) if R=H (from aldehyde) or ring-C(=CH2) if it's a ketone where the methyl from Grignard becomes the exo methylene after elimination of the ring-H... Actually for a ketone C(=O) in the ring: CH3MgBr adds to give a tertiary alcohol ring-C(OH)(CH3), then elimination (loss of H from adjacent ring carbon and OH) gives endocyclic alkene, NOT exocyclic =CH2. Step 3 - Reconsider: For exocyclic =CH2 formation, we need an ester or acid derivative. When an ester (-COOCH3) reacts with excess CH3MgBr (2 equivalents), the first equivalent adds to give a tetrahedral intermediate that collapses to a ketone (methanol leaves), then the second equivalent of CH3MgBr adds to give a tertiary alcohol: ring-C(OH)(CH3)2. Then POCl3/pyridine eliminates to give ring-C(=CH2) as an exocyclic double bond. This is because from -C(CH3)2OH, elimination of H from one CH3 and OH gives =CH2 (exocyclic). Step 4 - Match to compound B: B has three exocyclic =CH2 groups. So A must have three ester groups (-COOCH3) on the cyclohexane ring at 1, 3, 5 positions. With large excess CH3MgBr, each ester reacts with 2 equiv CH3MgBr to give -C(CH3)2OH at each position. POCl3/pyridine then dehydrates each tertiary alcohol to give =CH2 exocyclic double bonds, yielding 1,3,5-tris(methylidene)cyclohexane = compound B. Step 5 - Identify compound A: Compound (d) is 1,3,5-tris(methoxycarbonyl)cyclohexane (trimethyl cyclohexane-1,3,5-tricarboxylate), which has three methyl ester groups at C1, C3, C5. This perfectly matches the requirement. Step 6 - Why other options fail: - (a) Trialdehyde: CH3MgBr adds once to each CHO to give secondary alcohol ring-CH(OH)CH3; elimination gives endocyclic or vinyl alkene but not exocyclic =CH2 with the correct substitution pattern for B. - (b) Mixed ester/aldehyde/ketone: Would give mixed products; the ester portion could give =CH2 but aldehyde gives secondary alcohol (not giving =CH2 upon elimination without correct substitution), and overall would not cleanly give the symmetric trisubstituted B. - (c) Two ketones and one ester: Ketones with CH3MgBr give tertiary alcohols ring-C(OH)(CH3)- with adjacent ring carbons; elimination from these would give endocyclic double bonds, not exocyclic =CH2. Only the ester group gives =CH2. So this would not give three =CH2 groups. - (d) Three methyl esters: Each ester + 2 CH3MgBr → -C(CH3)2OH, then POCl3 → =CH2 exocyclic. All three positions give =CH2, producing compound B. This is correct. Therefore, the correct answer is D.