Phthalimide is made to undergo the following | vedclass

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(B) $1$. Phthalimide reacts with $KOH$ to form potassium phthalimide.$2$. Potassium phthalimide then undergoes an $S_N2$ reaction with benzyl chloride

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) $1$. Phthalimide reacts with $KOH$ to form potassium phthalimide.$2$. Potassium phthalimide then undergoes an $S_N2$ reaction with benzyl chloride to form $N$-benzylphthalimide (product $P$).$3$. The structure of $N$-benzylphthalimide contains a benzene ring ($3 \ \pi$ bonds),another benzene ring from the benzyl group ($3 \ \pi$ bonds),and two carbonyl groups ($C=O$,each having $1 \ \pi$ bond).$4$. Total $\pi$ bonds = $3$ (phthalimide ring) + $3$ (benzyl ring) + $2$ ($C=O$ bonds) = $8$.

Phthalimide is made to undergo the following sequence of reactions. Total number of $\pi$ bonds present in product '$P$' is/are:

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