See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the starting material: The starting material is 1-ethoxy-1-hydroxycyclopropane (a cyclopropane bearing both an OH and an OEt group on the same carbon, C1). Step 2 - Reaction with MeMgBr (Step 1): MeMgBr is a Grignard reagent but at 0°C in THF its primary role here is to deprotonate the alcohol OH of 1-ethoxy-1-hydroxycyclopropane, forming the magnesium alkoxide. This converts the OH to OMgBr, activating the molecule. Alternatively, the Grignard can open the cyclopropane ring: the strained cyclopropane with a leaving-group-like OEt at C1 undergoes ring opening. The magnesium alkoxide of the cyclopropanol acts as an activated cyclopropyloxy anion. Under Grignard conditions, the ethoxy group can be displaced, generating a cyclopropanone equivalent (or the magnesium alkoxide of cyclopropanone). More precisely, MeMgBr deprotonates the OH to give the alkoxide, which then facilitates elimination of EtO- to give cyclopropanone (or its Grignard adduct). Actually, 1-ethoxy-1-hydroxycyclopropane is a cyclopropanone hemiketal; under basic Grignard conditions, it hydrolyzes/equilibrates to cyclopropanone in situ. Step 3 - Cyclopropanone formation: 1-Ethoxy-1-hydroxycyclopropane is the ethyl hemiketal of cyclopropanone. Treatment with MeMgBr (a base/nucleophile) in THF at 0°C generates cyclopropanone in situ (by elimination of EtOH after deprotonation). Step 4 - Reaction with lithium phenoxide / cyclohexadienyl OLi (Step 2): The reagent shown is the lithium enolate of cyclohexanone (2-cyclohexen-1-one reduced to enolate, or directly the lithium enolate of cyclohexanone depicted as cyclohexadienyl OLi). This is the lithium enolate of cyclohexanone. The enolate carbon of cyclohexanone acts as a nucleophile and attacks cyclopropanone at its carbonyl carbon, performing a nucleophilic addition. This gives a beta-hydroxy ketone intermediate (an aldol-type product): the cyclohexanone enolate adds to cyclopropanone to give a lithium alkoxide where C1 of the former cyclopropanone now bears an OLi, and is connected to the alpha-carbon of cyclohexanone. Step 5 - Acid workup (Step 3): H+ protonates the alkoxide to give the alcohol. The product is 2-(1-hydroxycyclopropyl)cyclohexan-1-one: a cyclohexanone ring with a 1-hydroxycyclopropyl group at the 2-position (alpha carbon), which corresponds to option (b). Step 6 - Why other options fail: - Option (a) shows a spiro compound where a cyclopropane is spiro-fused to cyclohexanone; this would require a different type of reaction (spiroannulation), not a simple addition. - Option (c) shows a cyclohexane (no ketone) with a cyclopropyl OH; this ignores the ketone of the cyclohexanone enolate. - Option (d) shows cyclohexane with 1-hydroxycyclopropyl; again no ketone retained, which is wrong. - Option (b) correctly shows the ketone retained at C1 of cyclohexane and the 1-hydroxycyclopropyl group at C2, consistent with enolate addition to cyclopropanone. Therefore, the correct answer is B.