VI. a) What is e.e. of a mixture containing 25% (+)-2butanol and 75% (-)2 butanol if specific rotation of (+)-2. butanol is 13.5 ^ 0.
Answers
Answered by
0
Answer:
These enantiomers are 25 % of +2-butanol and 75 % of -2-butanol. ... Hence, the enantiomeric excess ee of this mixture is calculated to be 50 %.
Answered by
1
Answer:
These enantiomers are 25 % of +2-butanol and 75 % of -2-butanol. We have to calculate the enantiomeric excess, ee. Hence, the enantiomeric excess ee of this mixture is calculated to be 50 %.
Explanation:
These enantiomers are 25 % of +2-butanol and 75 % of -2-butanol. We have to calculate the enantiomeric excess, ee. Hence, the enantiomeric excess ee of this mixture is calculated to be 50 %.
Similar questions