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 %