When optically pure (R)-2-bromobutane is heated with water, butan-2-ol is the product. The reaction forms twice as much (S)-butan-2-ol as (R)-butan-2-ol. Calculate the e.e. and the specific rotation expected for the product.

Solution 12P :

Step 1:

The enantiomeric excess (e.e.) is a method for expressing the relative amounts of enantiomers in a mixture. This can be calculated using the formula as shown below :

e.e. = 100% = 100%

The specific rotation is defined as the rotation found using a 10-cm (1-dm) sample cell and a concentration of 1 g/mL. It is denoted by []. It is calculated using the formula :

[] =

Where, = rotation observed in the polarimeter.

c = concentration in g/mL.

l = length of sample cell in dm(decimeter).

Step 2:

Given reaction :

The reaction produces twice as much as (S)-(+)butan-2-ol(d- enantiomer) as (R)-(-)butan-2-ol(l- enantiomer)

i.e , d = 2l

Now, let’s calculate the e.e :

e.e. = 100%

This can be rewritten as :

e.e = 100%

= 100%

e.e = 33.3 %.

Therefore, the e.e expected for the product is 33.3 %.