See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Concept: Reaction of Grignard reagent with diethyl carbonate (Et-O-C(=O)-O-Et). Step 1: Identify the reagents. CH3MgBr (methylmagnesium bromide) is used in excess. Diethyl carbonate (Et-O-C(=O)-O-Et) is the substrate. The reaction is followed by acid workup (H+). Step 2: First addition. CH3MgBr attacks the carbonyl carbon of diethyl carbonate. One ethoxy group is displaced, giving an ester intermediate: Et-O-C(=O)-CH3 (ethyl acetate) after loss of EtO- and before workup. Step 3: Second addition. Since CH3MgBr is in excess, the Grignard reagent attacks the ester (ethyl acetate) intermediate again. The first addition to the ester gives a tetrahedral intermediate, which collapses by expelling the ethoxy group to give a ketone intermediate: CH3-C(=O)-CH3 (acetone). However, because CH3MgBr is still in excess, it immediately attacks the acetone. Step 4: Third addition. CH3MgBr adds to acetone to give (CH3)3C-OMgBr. Step 5: Acid workup (H+). The magnesium alkoxide is protonated to give (CH3)3COH, which is 2-methylpropan-2-ol (tert-butanol). Summary: Diethyl carbonate + 2 equivalents CH3MgBr (excess) followed by H+ workup gives tert-butanol, (CH3)3COH. Why other options fail: - (a) 2-methylpropan-1-ol: Would require a different carbon skeleton buildup not produced here. - (c) Acetone: Is only an intermediate; since CH3MgBr is in excess, it reacts further with acetone. - (d) Propan-2-ol: Would result from a single Grignard addition to an aldehyde or formaldehyde ester; not consistent with diethyl carbonate and excess CH3MgBr. Therefore, the correct answer is B.