How much of each enantiomer is present if a sample of (R)-2-bromopentane is 24% ee? Be sure to label each amount as being R or S.
Let amount of "R" enantiomer = R ad amount of "S" enantiomer = S
Assuming that 100 molecules of 2-bromopentane are given. Some of these 100 molecules some are R and other are S.
R + S = 100 ......... (1)
Given ee = 24%
R is in excess than S hence R > S
R-S = 24 ............. (2)
By (1) + (2) we get,
2R = 124
R = 62
Put R=60 in equation 1 we get
62 + S = 100
S = 100 - 62
S = 38
AMount of R is 62% and that of S is 38%.
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