Question

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.

Answer #1

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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