A chemist finds that the addition of (+)-epinephrine to the catalytic reduction of butan-2-one(Figure 5-16) gives a product that is slightly optically active, with a specific rotation of +0.45°.Calculate the percentages of (+)-butan-2-ol and (-)-butan-2-ol formed in this reaction.
Solution 13P :
Step 1:
Given :
Observed specific rotation of butan-2-ol = +0.45°.
Percentages of (+)-butan-2-ol and (-)-butan-2-ol formed in this reaction = ?
The rotation of pure butan-2-ol = +13.5 %
We know formula for optical purity :
Optical purity(o.p) = 100 %
Hence , o.p for butan-2-ol = 100 %
= 3.33 %
Therefore, the o.p of butan-2-ol is 3.33 % which also means that the enantiomeric excess(e.e) is 3.33 % excess (+) enantiomer over (-) enantiomer.
Step 2:
Now, let’s calculate the percentages of (+) and (-) enantiomer in the mixture :
Let’s put it in the equation from and solve it :
(+) + (-) = 100 % -------------- (1)
(+) - (-) = 3.33 % -------------- (2)
We can rewrite equation (1) as :
(-) = 100 % - (+) ----------- (3)
Using equation (3), we can rewrite equation (2) as :
(+) - (100 % - (+)) = 3.33 %
The above equation can be simplified as :
2 (+) = 3.33 % + 100 %
2(+) = 103.33 %
(+) =
(+) = 51.66 %
Hence, the percentages of (+)-butan-2-ol formed in the reaction is 51.66 %
