See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is (bromomethylene)cyclopropane bearing a phenyl group on the sp2 carbon, i.e., cyclopropylidene(phenyl)methyl bromide - specifically it is a vinylic/allylic bromide on a cyclopropylidene system: the structure is a cyclopropane ring with an exocyclic double bond, where the exocyclic sp2 carbon bears both Ph and Br. This is an alkenyl (vinyl) bromide of the type cyclopropyl=C(Ph)Br. Step 2 - Formation of Grignard reagent A: Treatment with Mg in ether converts the C-Br bond to a C-MgBr bond. Since this is a vinyl bromide attached to the cyclopropylidene system, the Grignard reagent A is cyclopropylidene(phenyl)methyl magnesium bromide: the sp2 carbon bearing Ph now bears MgBr instead of Br. So A = [cyclopropylidene=C(Ph)]MgBr. Step 3 - Reaction with HCHO (formaldehyde): The Grignard reagent attacks formaldehyde (HCHO) in a nucleophilic addition. The carbon bearing MgBr attacks the carbonyl carbon of HCHO, adding a -CH2-OMgBr group to the sp2 carbon. Step 4 - Workup with H+: Protonation of the alkoxide gives -CH2-OH. Step 5 - Product B: The result is the cyclopropylidene compound where the sp2 carbon bears Ph and CH2-OH (replacing Br with CH2OH). This corresponds to option (a): the cyclopropane ring with an exocyclic double bond, the terminal carbon of which bears Ph and CH2-OH. Step 6 - Why other options fail: Options (b), (c), and (d) contain alkyne (C≡C) units, which would require ring opening of the cyclopropane ring and rearrangement. While cyclopropylmethyl systems can rearrange, the direct Grignard addition to HCHO without ring-opening conditions simply gives the primary alcohol product by straightforward 1,2-addition. The cyclopropylidene Grignard does not ring-open under these mild conditions to give an alkyne. Options (b), (c), (d) all incorrectly invoke ring-opening to an acetylenic system. Therefore, the correct answer is B.