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An enantiomerically pure acid is treated with a racemic mixture of an alcohol having one chiral carbon. The ester formed will be a. Optically active mixture b. Pure enantiomer c. Meso compound d. Racemic mixture
a) The optically active acid will react with d and l forms of alcohol present in the racemie mixture at different rates to form two diastereomers in unequal amounts leading to optical activity of the product.
Sorry, but above solution is incorrect.We hope to get a correct solution by iitians,!¡!!!!!!!!!!!!!¡!!!!!!!!!!!!!!!!!!!!?!!!??!+(+;!;!!!;;:;!;:;;!!!!!!!!!!!!¡
The answer above is correct as the acid form will be same nd have equal chiral carbon nd will be optically active
The enatiomerically pure acid or optically active acid will react with both the forms (d and l) of the alcohol in the racemic mixture to form isomeric pair of esters. In each isomer the configuration of the chiral center(carbon) remains the same , so the mixture will be optically active.
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