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Question

$(b)$ Optical punty or enanttomenc excess $= \frac{Excess \,of \,one \,enantiomer}{Total \,mixutre} 	imes 100$

Enantiomeric excess $(e. e.) = \fra

Vedclass · Accepted Answer

$20$

Vedclass · Answer

$(b)$ Optical punty or enanttomenc excess $= \frac{Excess \,of \,one \,enantiomer}{Total \,mixutre} 	imes 100$

Enantiomeric excess $(e. e.) = \frac{6g -4g}{(6+ 4)} 	imes 100 = 20\, \%$

Calculate enantiomeric excees of mixture containing $6g$ of $( +) - 2$ -butanol and $4g$ of $(-) - 2$ -butanol........$\%$

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