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(A) The reaction shown is the first step of the Gabriel phthalimide synthesis. Phthalimide reacts with $KOH$ to form potassium phthalimide,which acts

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$N$-($4$-bromobenzyl)phthalimide

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(A) The reaction shown is the first step of the Gabriel phthalimide synthesis. Phthalimide reacts with $KOH$ to form potassium phthalimide,which acts as a nucleophile. This nucleophile then undergoes an $S_N2$ reaction with the alkyl halide,$4-bromobenzyl  chloride$ $(Br-C_6H_4-CH_2Cl)$. In $S_N2$ reactions,the nucleophile attacks the carbon atom attached to the leaving group. Since the $CH_2Cl$ group is more reactive towards $S_N2$ than the aryl bromide ($C-Br$ bond),the nitrogen of the phthalimide attacks the $CH_2$ carbon,displacing the $Cl^-$ ion. Therefore,the product is $N-(4-bromobenzyl)phthalimide$.

The major product of the following reaction is:

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