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(B) $1$. The starting material is a hemiacetal of cyclopropanone. Treatment with $MeMgBr$ (a base) deprotonates the hydroxyl group to form an alkoxide

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(B) $1$. The starting material is a hemiacetal of cyclopropanone. Treatment with $MeMgBr$ (a base) deprotonates the hydroxyl group to form an alkoxide.$2$. This unstable alkoxide undergoes ring opening to generate cyclopropanone and ethoxide ion. Cyclopropanone then reacts with the lithium enolate of cyclohexanone (the nucleophile).$3$. The nucleophilic attack of the enolate on the carbonyl carbon of cyclopropanone forms a new carbon-carbon bond.$4$. Finally,acidic workup $(H^+)$ protonates the resulting alkoxide to yield the final product,which is $2-(1-hydroxycyclopropyl)cyclohexanone$.

Column-$I$ (Reaction)	Column-$II$ (Mechanism)
$(A)$ $OH^- + CH_3Cl \to HOCH_3 + Cl^-$	$(1)$ $S_N2$ mechanism
$(B)$ $(CH_3)_3CBr + OH^- \to (CH_3)_3COH + Br^-$	$(2)$ $S_E2$ aromatic
$(C)$ $CH_3CH_2Br \xrightarrow[ethanol]{KOH} CH_2 = CH_2$	$(3)$ $S_N1$ mechanism
$(D)$ $C_6H_6 + Br_2 \xrightarrow{Fe} C_6H_5Br + HBr$	$(4)$ $\beta$-elimination (dehydrohalogenation)

Product $(A)$ in this sequence is:

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