The number of stereoisomers formed in a react | vedclass

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(D) The starting material is a racemic mixture of $(\pm) Ph(C=O)C(OH)(CN)Ph$. This means it contains two enantiomers,$(+)$ and $(-)$.When this reacts

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) The starting material is a racemic mixture of $(\pm) Ph(C=O)C(OH)(CN)Ph$. This means it contains two enantiomers,$(+)$ and $(-)$.When this reacts with $HCN$,the carbonyl group $(C=O)$ is converted into a cyanohydrin group $(C(OH)(CN))$.The original chiral center at the second carbon remains unchanged,while a new chiral center is created at the carbonyl carbon.Each enantiomer of the starting material will produce two diastereomers because the new chiral center can have either $R$ or $S$ configuration.Since we start with a racemic mixture (a $1:1$ mixture of two enantiomers),we get two pairs of diastereomers.Specifically,$(+) 	ext{ isomer} ightarrow (R,R) + (R,S)$ and $(-) 	ext{ isomer} ightarrow (S,S) + (S,R)$.Thus,a total of $4$ stereoisomers are formed. However,looking at the options provided,the question likely asks for the number of stereoisomers formed from one enantiomer or is based on a specific interpretation. Given the standard nature of this problem,the correct answer is $4$. Since $4$ is not an option,and based on common textbook problems of this type,if the question implies the number of diastereomeric products from the racemic mixture,the answer is $4$. If the question is flawed,we select the most logical outcome. Re-evaluating: the product is $Ph-C(OH)(CN)-C(OH)(CN)-Ph$. This molecule has two chiral centers. The $(R,R)$,$(S,S)$,$(R,S)$,and $(S,R)$ forms are possible. The $(R,S)$ and $(S,R)$ are identical (meso form). Thus,there are $3$ stereoisomers: $(R,R)$,$(S,S)$,and the meso $(R,S)$. Therefore,the correct option is $D$.

The number of stereoisomers formed in a reaction of $(\pm) Ph(C=O)C(OH)(CN)Ph$ with $HCN$ is $..........$

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