Describe, in three steps, how you would synthesise phenylethylamine (C6H5CH2CH2NH2) from methylbenzene, giving reagents and conditions for each step. For each step, state the type of reaction that occurs.

Methylbenzene (C₆H₅CH₃) can be reacted with chlorine in the presence of UV light. The type of reaction is free radical substitution, and this produces phenylchloromethane (C₆H₅CH₂Cl). The next step is a nucleophilic substitution reaction involving potassium cyanide (KCN) in ethanol and water. For this reaction to occur, the solutions must be heated under reflux. This produces phenylethanenitrile (C₆H₅CH₂CN). This reaction adds another carbon atom onto the reactants, allowing us to continue with the final step. Water on its own should not be used as a solvent, as this is likely to cause subtitution with -OH instead of -CN.The final step is a reduction to an amine using lithium tetrahydridoaluminate (LiAlH₄), which is a powerful reducing agent. Sodium tetrahydridoborate (NaBH₄) is not used, as this is too weak to reduce a nitrile. The conditions require an ethoxyethane (ether) solvent, and treatment with a dilute acid. Alternatively, hydrogen gas and a platinum or palladium catalyst may be used, in a similar way as is used to hydrogenate alkenes. The product is phenylethylamine (C₆H₅CH₂CH₂NH₂).