JEE Advanced 2023 · Paper 1 · Chemistry Q13

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ChemistryInteger+4Question 47 of 51

The total number of $sp^2$ hybridised carbon atoms in the major product P (a non-heterocyclic compound) of the following reaction is _____.

1,1,2,2-Tetracyanoethane, (NC)2CH-CH(CN)2 → [(i) LiAlH4 (excess), then H2O; (ii) acetophenone (excess)] → P.

Your answer

Answer and worked solution

Official answer: 28

Reduction. Excess $\ce{LiAlH4}$ reduces all four nitriles to primary amines:
$\ce{(H2NCH2)2CH-CH(CH2NH2)2}.$

Condensation. Each $\ce{-NH2}$ condenses with acetophenone to an imine, $\ce{-N=C(CH3)Ph}$ , giving a tetra-imine. It is non-heterocyclic, so no ring closure occurs.

Counting $sp^2$ carbons. Each imine unit has:

one $\ce{C=N}$ carbon;
six aromatic carbons.

$4\times7=28$ .

Topics:Cyanides and isocyanides Nucleophilic addition Hybridisation

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