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).
