How many optically active isomers CH3-CH2-CH(Br) –CH3 can form
Answers
Answered by
0
Answer:
Hence, we have derived that, if 'n' (number of chiral centers) is odd for a compound with similar ends, then:
Number of meso isomers=2(n−1)/2.
Total number of optical isomers=2n−1.
Number of enantiomers=2n−1−2(n−1)/2.
Method is here ...
Try to solve...
Hope it helps....
Answered by
0
22j2v wgqh1b1n2h2v2vwg2h
Similar questions