Esterification of the acid P with the alcohol | vedclass

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(B) The acid $P$ is a single enantiomer with $(S)$ configuration.The alcohol $Q$ is a racemic mixture,i.e.,a mixture of $(R)$ and $(S)$ enantiomers.Du

Vedclass · Accepted Answer

a mixture of diastereomers

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(B) The acid $P$ is a single enantiomer with $(S)$ configuration.The alcohol $Q$ is a racemic mixture,i.e.,a mixture of $(R)$ and $(S)$ enantiomers.During esterification,the chiral center of the acid remains unchanged,while the chiral center of the alcohol reacts.Reaction: $(S)	ext{-acid} + (R/S)	ext{-alcohol} ightarrow (S,R)	ext{-ester} + (S,S)	ext{-ester}$.The products $(S,R)$ and $(S,S)$ are diastereomers of each other because they have different configurations at one chiral center while the other remains the same.Therefore,a mixture of diastereomers is produced.

Column-$I$ (Reaction)	Column-$II$ (Electrophile)
$A$. $C_6H_6 + CH_3Cl \xrightarrow{AlCl_3} C_6H_5CH_3$	$P$. $CO_2$
$B$. $C_6H_5OH + CO_2 \xrightarrow[H^+]{NaOH} \text{Salicylic acid}$	$Q$. $:CCl_2$
$C$. $C_6H_5OH + CHCl_3 \xrightarrow{KOH} \text{Salicylaldehyde}$	$R$. $\stackrel{+}{C}H_3$
$D$. $C_6H_6 + SO_3 \xrightarrow{H_2SO_4} C_6H_5SO_3H$	$S$. $\stackrel{+}{S}O_3H$
	$T$. $SO_3$

Esterification of the acid $P$ with the alcohol $Q$ will give:

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