In a mixture, two enantiomers are found to be present in 85 % and 15 % respectively. The enantiomeric excess (e, e) is

Question

In a mixture, two enantiomers are found to be present in 85 % and 15 % respectively. The enantiomeric excess (e, e) is (A) 85 % (B) 15 % (C) 70 % (D) 60 %

Tardigrade Admin · Accepted Answer

15 % will form racemic mixture with another 15 %. Hence, the enantiomeric excess is =(85-15)=70 %

Q. In a mixture, two enantiomers are found to be present in $85%$ and $15%$ respectively. The enantiomeric excess $(e, e)$ is

$85%$

$15%$

$70%$

$60%$

$15%$ will form racemic mixture with another $15%$ . Hence, the enantiomeric excess is $= (85 - 15) = 70%$

Q. In a mixture, two enantiomers are found to be present in 85% and 15% respectively. The enantiomeric excess (e,e) is

85%

15%

70%

60%