See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Concept: When an acid chloride (RCOCl) reacts with excess Grignard reagent (R'MgX), the first equivalent of Grignard adds to the acid chloride to give a ketone intermediate (after loss of MgXCl), and then a second equivalent of Grignard adds to the ketone to give a tertiary alcohol after acidification. The key constraint is that both R' groups added come from the SAME Grignard reagent, so the two groups added to the carbonyl carbon must be identical. The resulting alcohol has the structure R-C(OH)(R')(R'), where R comes from the acid chloride and both R' groups come from the Grignard reagent. Step 1: Identify the structural requirement. The product of RCOCl + 2 R'MgX → R-C(OH)(R')2. The two substituents introduced by the Grignard reagent must be the same group (R'), because only one Grignard reagent is used in excess. Step 2: Analyze each option to see if it can be made this way (two identical groups from Grignard, one group from acid chloride): (a) Ph-C(OH)(Ph)-CH3: Groups on the carbinol carbon are Ph, Ph, CH3. Two Ph groups (from PhMgBr) and one CH3 (from CH3COCl). This fits: CH3COCl + excess PhMgBr → Ph-C(OH)(Ph)-CH3. POSSIBLE. (b) Ph-C(OH)(CH3)-CH3: Groups are Ph, CH3, CH3. Two CH3 groups (from CH3MgBr) and one Ph (from PhCOCl). This fits: PhCOCl + excess CH3MgBr → Ph-C(OH)(CH3)2. POSSIBLE. (c) Et-C(OH)(Et)-CH3: Groups are Et, Et, CH3. Two Et groups (from EtMgBr) and one CH3 (from CH3COCl). This fits: CH3COCl + excess EtMgBr → Et-C(OH)(Et)-CH3. POSSIBLE. (d) CH3-C(OH)(Et)-Ph: Groups on the carbinol carbon are CH3, Et, Ph. All three substituents are DIFFERENT. For this to be made from RCOCl + excess R'MgX, two of the three groups must be identical (both from R'MgX). Since CH3, Et, and Ph are all different, no two are the same, so this cannot be prepared by this method. Step 3: Conclusion. Option (d) cannot be prepared because it requires three different substituents on the carbinol carbon, but the Grignard method with a single Grignard reagent necessarily introduces two identical groups. Therefore, the correct answer is D.