Consider the following reaction sequence: Ethylbenzene [ (1 eq) ] Br ₂/h P KCN Q [ 2. H ₂ O ] 1. DIBAL-H, -78^ C R The major product R is:

Step-by-step solution

Step 1: Free radical bromination of ethylbenzene with Br ₂/h is highly regioselective. The more stable secondary benzylic radical is formed, yielding 1-bromo-1-phenylethane as product P . Step 2: Nucleophilic substitution of P with KCN gives 2-phenylpropanenitrile ( Q ). Step 3: DIBAL-H (Diisobutylaluminium hydride) at -78^ C partially reduces the nitrile to an imine intermediate, which upon hydrolysis with water yields an aldehyde. Thus, Q is converted to 2-phenylpropanal ( R ). Answer: 2-phenylpropanal