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Grade: 6
        
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
4 years ago

Answers : (4)

Kevin Nash
askIITians Faculty
332 Points
							

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.

4 years ago
Krishnapal
11 Points
							Sorry, but above solution is incorrect.We hope to get a correct solution by iitians,!¡!!!!!!!!!!!!!¡!!!!!!!!!!!!!!!!!!!!?!!!??!+(+;!;!!!;;:;!;:;;!!!!!!!!!!!!¡
						
one year ago
Prachi
13 Points
							The answer above is correct as the acid form will be same nd have equal chiral carbon nd will be optically active
						
one year ago
Kartik
15 Points
							
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.
3 months ago
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