See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the starting material. The starting material is a cyclopentanone (five-membered ring ketone) with a methyl group (Me) on one alpha carbon and a phenyl group (Ph) on the other alpha carbon. From the Newman/stereochemical drawing, the Me substituent is on a wedge and Ph is on a dash relative to the ring plane, establishing the relative stereochemistry of the starting ketone (trans arrangement of Me and Ph). Step 2 – Identify the reagent. MeMgBr is a Grignard reagent. It adds a methyl nucleophile to the carbonyl carbon of the ketone. Step 3 – Determine the reaction outcome. The Grignard reagent (Me–) attacks the carbonyl carbon (C1 of the cyclopentanone ring) to give, after aqueous workup, a tertiary alcohol. The carbonyl carbon becomes a new stereocenter bearing: two methyl groups (one original substituent Me from the ring substituent side, one from MeMgBr addition) — wait, re-examining: the carbonyl carbon in the starting cyclopentanone has no substituents other than the ring bonds and the C=O. After MeMgBr addition, the carbonyl carbon gains a new Me group and becomes C–OH (tertiary alcohol), with the ring carbons on either side still bearing their Me and Ph groups. Step 4 – Analyze stereochemistry. The carbonyl carbon is sp2 (planar, trigonal), so MeMgBr can attack from either face (re or si face) with equal probability in the absence of chiral influence, generating two diastereomers (since the ring already has existing stereocenters with Me and Ph). The two products correspond to attack from the two faces of the carbonyl, giving (a) one diastereomer (OH and Ph on same side / specific relative configuration) and (b) the other diastereomer (OH and Ph on opposite side). Both are formed. Step 5 – Why both (a) and (b) are correct. Because the carbonyl is prochiral and the attack is not stereospecific under normal Grignard conditions, both faces are accessible. The existing stereocenters (Me at one alpha carbon, Ph at other alpha carbon) make the two products diastereomers, not enantiomers. Both diastereomeric tertiary alcohols are produced in the reaction mixture. Step 6 – Eliminate other options. Option (a) alone or (b) alone would imply complete facial selectivity, which is not expected here. Option (d) is wrong because (a) and (b) are indeed the correct products. Therefore, the correct answer is C.