The total number of sp^2 hybridised carbon at | vedclass

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(B) The reaction proceeds in two steps:$1$. Reduction of the tetranitrile compound with excess $LiAlH_4$ followed by $H_2O$ yields a tetraamine: $(H_2

Vedclass · Accepted Answer

$28$

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(B) The reaction proceeds in two steps:$1$. Reduction of the tetranitrile compound with excess $LiAlH_4$ followed by $H_2O$ yields a tetraamine: $(H_2N-CH_2)_2CH-CH(CH_2-NH_2)_2$.$2$. The tetraamine reacts with excess acetophenone $(CH_3-CO-Ph)$ to form an imine (Schiff base) product $P$.In the product $P$,each of the four acetophenone-derived units contains one $C=N$ carbon and six aromatic ring carbons,totaling $7$ $sp^2$ carbons per unit.Since there are $4$ such units,the total number of $sp^2$ hybridised carbon atoms is $4 	imes 7 = 28$.

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

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