The products of the following reaction of a sample of 2‐butanol (ee = X%) show two doublets in ¹H NMR spectrum in the ratio of 3 ∶ 2. The value of X is _______.

Concept:

Enantiomeric excess (ee%) or optical purity is a method to measure purity of a chiral substance. It indicates which enantiomer is in greater amount than the other. A pure enantiomer has an ee of 100% while a racemic mixture has an ee of 0%. 

Explanation:

• Enantiomeric excess  (ee%)

​\(ee\% = {| [R] - [S]|\over[R] + [S]} \)

• ¹H NMR spectrum shows two doublets in the 3:2 ratio of R and S enantiomers. Therefore, the molar ratio of the enantiomeric products should be also in the ratio 3:2

or, 

3R : 2S

R = 2S /3

given:  R% + S% = 100%

Calculation:

Putting the value of R in above eq.

 2S/3  +  S = 100

(2S + 3S)/ 3 = 100

5S/3 = 100

S = 60%

R = 40%

Therefore, ee% = 60% - 40% 

ee% = 20%

Conclusion:

Hence, the value of X is 20%.