To find the total number of sp2 hybridized carbon atoms in the major product P, we analyze the reaction step-by-step:
Step 1: Reduction with LiAlH4 (excess) followed by H2O
The starting compound is ethane-1,1,2,2-tetracarbonitrile:
(NC)2CH−CH(CN)2
Upon reduction with excess lithium aluminium hydride (LiAlH4), all four nitrile (−C≡N) groups are converted into primary amine (−CH2NH2) groups:
(NC)2CH−CH(CN)2(i) LiAlH4 (excess), H2O(H2N−CH2)2CH−CH(CH2−NH2)2
Step 2: Reaction with excess Acetophenone
Acetophenone is Ph−CO−CH3 (C6H5−CO−CH3).
Primary amines undergo nucleophilic addition-elimination with ketones to form imines (Schiff bases):
−CH2−NH2+O=C(CH3)C6H5−H2O−CH2−N=C(CH3)C6H5
Since excess acetophenone is used and the problem specifies that P is a non-heterocyclic compound, all four amine groups react to give the tetraimine product P:
P: [C6H5−C(CH3)=N−CH2]2CH−CH[CH2−N=C(CH3)−C6H5]2
Step 3: Determination of sp2 Hybridized Carbon Atoms
In each −N=C(CH3)C6H5 group:
Imine Carbon (−N=C−):1 carbon (sp2 hybridized)
Phenyl Ring (−C6H5):6 carbons (sp2 hybridized)
Methyl Carbon (−CH3):1 carbon (sp3 hybridized)
Thus, each acetophenone residue contributes:
1+6=7sp2 hybridized carbon atoms
Since there are 4 such groups in product P:
Total sp2 hybridized carbons from acetophenone groups=4×7=28
The remaining backbone carbons (2×−CH− and 4×−CH2−) and methyl carbons (4×−CH3) are all sp3 hybridized.
