In the reaction sequence shown below,determin | vedclass

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(D) $1$. The starting material is a chiral ketone with an alkyne group: $CH_3-CD(CH_3)-C \equiv C-CO-CH_3$. The chiral center at the $CD(CH_3)$ positi

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$8$

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(D) $1$. The starting material is a chiral ketone with an alkyne group: $CH_3-CD(CH_3)-C \equiv C-CO-CH_3$. The chiral center at the $CD(CH_3)$ position is fixed as $(R)$ or $(S)$.$2$. Reaction $A$ involves the reduction of the alkyne to a $cis$-alkene. This creates a new double bond,but since the geometry is fixed as $cis$,no new stereocenter is formed at the alkene carbons.$3$. Compound $C$ is the $cis$-alkene: $CH_3-CD(CH_3)-CH=CH-CO-CH_3$. It has one chiral center.$4$. The final step is the addition of $Br_2/CCl_4$ to the $cis$-alkene. This is an anti-addition reaction across the double bond.$5$. The addition of $Br_2$ to the $cis$-alkene creates two new chiral centers at the carbons where $Br$ atoms are added.$6$. Since the molecule already has one chiral center,and two new ones are formed,the total number of stereoisomers is $2^n$,where $n$ is the number of chiral centers. Here,$n=3$. Total stereoisomers = $2^3 = 8$.

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