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(D) Step $1$: Formation of product $A$.Cyclohexene reacts with $Br_2$ to form $1,2$-dibromocyclohexane. This intermediate then reacts with $HC \equiv

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$8$

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(D) Step $1$: Formation of product $A$.Cyclohexene reacts with $Br_2$ to form $1,2$-dibromocyclohexane. This intermediate then reacts with $HC \equiv C-CH_2-O^- Na^+$ (a nucleophile) via an $S_N2$ mechanism to substitute one $Br$ atom,yielding product $A$ ($1$-bromo-$2$-(prop-$2$-ynyloxy)cyclohexane).Product $A$ contains one triple bond ($2 \pi$ bonds = $4 \pi$ electrons).Step $2$: Formation of product $B$.$1,2$-dibromocyclohexane reacts with $3$ equivalents of alcoholic $KOH$ (a strong base) to undergo double dehydrohalogenation,resulting in the formation of $1,3$-cyclohexadiene (product $B$).Product $B$ contains two double bonds ($2 \pi$ bonds = $4 \pi$ electrons).Step $3$: Calculation of total $\pi$ electrons.Total $\pi$ electrons = ($\pi$ electrons in $A$) + ($\pi$ electrons in $B$)Total $\pi$ electrons = $4 + 4 = 8$.

The major products from the following reaction sequence are product $A$ and product $B$. The total sum of $\pi$ electrons in product $A$ and product $B$ are . . . . . . .(nearest integer)

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