Phthalimide is made to undergo following sequence of reactions.
Phthalimide →((i) KOH (ii) Benzylchloride) ‘P’
Total number of π bonds present in product 'P' is/are ________.

Solution

Chemistry ma’am · 3 steps

Build P first, N-benzylphthalimide, then count each ring and each C=O separately.

1. Make the anion. KOH removes the acidic N–H proton of phthalimide. This gives potassium phthalimide, whose nitrogen is a good nucleophile.
2. Attach benzyl. The phthalimide anion attacks benzyl chloride by S_N2 and pushes out Cl⁻. So P is N-benzylphthalimide, C₆H₄(CO)₂N-CH₂C₆H₅. This is the first stage of the Gabriel synthesis.

   Part of P	π bonds
   Benzene ring of the phthalimide unit	3
   Two C=O groups	2
   Benzene ring of the benzyl group	3
   CH₂ and the N–C bonds	0

3. So. Total π bonds in P = 3 + 2 + 3 = 8.

The common slip: 5. Counting only the phthalimide part and leaving out the benzyl ring gives 5, not 8. Step (ii) adds a whole benzene ring.

The idea to keep. To count π bonds, draw the product completely first. Then count 3 for every benzene ring, 1 for every double bond and 2 for every triple bond.

Answer from the NTA final answer key. Solution by PrepWiser.