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

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

WBJEEWBJEE 2015Haloalkanes and Haloarenes

Solution:

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