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The products A and B in the following reaction sequence are:

Detailed Solution
Styrene + HBr (Markovnikov) gives 1-bromo-1-phenylethane; with Mg in dry ether it forms the Grignard reagent, which on reaction with $CO_2$ followed by $H_3O^+$ gives 2-phenylpropanoic acid (A). $SOCl_2$ converts A to the acid chloride, which with $CH_3NH_2$ gives N-methyl-2-phenylpropanamide (B).
