Enantiomeric excess (ee) = 84 % of S-enantiomer
specific rotation of S-enantiomer = + 25.0
Optical rotation of R enantiomer and S enantiomer is same but in opposite direction.So, specific rotation of R-enantiomer = - 25.0
Supposing the yield of S isomer to be = x
Then the yield of R isomer is = 100 - x
Since, the enantiomeric excess is of the S isomer, it must be in more yields.
Therefore, enantiomeric excess = yield of S isomer - yield of R isomer
or,
84 % = x - (100 - x)
or,
2x = 84 + 100 = 184
or,
x = 184 / 2 = 92 %
The percentage of S isomer in the mixture = 92 %
The percentage of R isomer in the mixture = (100 - 92) % = 8 %
We know that ,specific rotation of the mixture = [yield of S isomer x rotation of S isomer] + [yield of R isomer x rotation of R isomer]
or, specific rotation of the mixture = [92 % x + 25.0] + [8 % x - 25.0] = [0.92 x + 25.0] + [0.08 x - 25.0]
Thus the specific rotation of the mixture = 23.0 + [- 2.0] = + 21.0
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