Solution Found!
When optically pure (R)-2-bromobutane is heated with water, butan-2-ol is the product
Chapter 5, Problem PROBLEM 5-12(choose chapter or problem)
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.
Questions & Answers
QUESTION:
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.
ANSWER:Step 1 of 2
When the (R)-2-bromobutane is heated with water, then 2-butanol is formed. The amount of (S)-2-butanol is twice the amount of (R)-2-butanol. Let us assume that the yield of (R)-2-butanol is 1. Then the yield of (S)-2-butanol becomes 2.
The formula that is used for calculation of enantiomeric excess is found as:
Hence, the enantiomeric excess is found to be 33.33%.