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(B) The reaction shown is the Gabriel phthalimide synthesis. In the first step,benzyl bromide reacts with potassium phthalimide to form $N$-benzylphth

Vedclass · Accepted Answer

$Ph-CH_2-NH_2$

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(B) The reaction shown is the Gabriel phthalimide synthesis. In the first step,benzyl bromide reacts with potassium phthalimide to form $N$-benzylphthalimide. In the second step,alkaline hydrolysis $(HO^-/H_2O)$ of $N$-benzylphthalimide yields benzylamine $(Ph-CH_2-NH_2)$ and phthalic acid. Therefore,the product $(A)$ is benzylamine.

The product $(A)$ of the given reaction is:

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