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 :
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).
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%
e.e = 33.3 %.
Therefore, the e.e expected for the product is 33.3 %.