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 

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Correct option is C

In stereochemistry, enantiomeric excess (ee) is a measurement of purity used for chiral substances. It reflects the degree to which a sample contains one enantiomer in greater amounts than the other. A racemic mixture has an ee of 0%, while a single completely pure enantiomer has an ee of 100%. A sample with 70% of one enantiomer and 30% of the other has an ee of 40% (70% − 30%).

​If one knows the moles of each enantiomer produced then:

$$$$% ee = ((R–S)/(R+S)×100)

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%