If (pm) 2-methylbutanoic acid is esterified b | vedclass

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Question

(C) The reaction is: $(\pm)$ $2$-methylbutanoic acid + $(\pm)$ $2$-butanol $\longleftrightarrow$ ester + $H_2O$. $(\pm)$ $2$-methylbutanoic acid exist

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) The reaction is: $(\pm)$ $2$-methylbutanoic acid + $(\pm)$ $2$-butanol $\longleftrightarrow$ ester + $H_2O$. $(\pm)$ $2$-methylbutanoic acid exists as a pair of enantiomers: $(R)$ and $(S)$. $(\pm)$ $2$-butanol exists as a pair of enantiomers: $(R)$ and $(S)$. The esterification reaction involves the formation of an ester from these two chiral components. The possible combinations of the acid and alcohol enantiomers are: $1$. $(R)$-acid + $(R)$-alcohol $\rightarrow$ $(R,R)$-ester $2$. $(R)$-acid + $(S)$-alcohol $\rightarrow$ $(R,S)$-ester $3$. $(S)$-acid + $(R)$-alcohol $\rightarrow$ $(S,R)$-ester $4$. $(S)$-acid + $(S)$-alcohol $\rightarrow$ $(S,S)$-ester Since all four resulting esters contain two chiral centers and are not meso compounds,all $4$ of them are optically active.

If $(\pm)$ $2$-methylbutanoic acid is esterified by reaction with $(\pm)$ $2$-butanol,how many optically active compounds will be present in the final equilibrium reaction mixture?

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