Q.Identify A, B and C: benzyl bromide, treated with

Step 1. Benzyl bromide, C6H5CH2Br, treated with Mg/ether (i), forms the Grignard reagent on its own benzylic carbon: A = benzylmagnesium bromide, C6H5CH2MgBr.

Step 2. A + CO2 (ii, dry ice), then H3O+ (iii), carboxylates that same carbon directly (Section 12.10.4/Grignard concept: RMgX + CO2, then H3O+, gives R-COOH): B = phenylacetic acid, C6H5CH2COOH.

Step 3. Separately, benzyl bromide, treated with NaCN in THF, undergoes nucleophilic substitution (SN2, an unhindered primary/benzylic halide) to give the nitrile, benzyl cyanide, C6H5CH2CN, + NaBr.

Step 4. Benzyl cyanide, hydrolysed with H3O+ (Section 12.10.2), gives C = phenylacetic acid, C6H5CH2COOH, + NH3 -- the identical final product reached by B, but by an entirely independent synthetic route. …

Trace the two reagent paths separately -- Mg/ether then CO2/H3O+ is the Grignard-carboxylation route to a carboxylic acid; NaCN/THF then H3O+ is the nitrile-substitution-then-hydroly …

• Assuming B and C must be different compounds simply because the question gives them different letters -- the whole point of pairing these two reagent sequences is that they are two IN …