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How many optically active stereoisomers are possible for butan-2,3-diol?

Asked by Arushi Juyal 12th November 2017, 4:07 PM
Answered by Expert
Answer:
When the molecule can be divided into equal halves, i.e., the molecule has symmetry and the number (n) of asymmetric carbon atoms is even, then
The no. of optically active forms a = 2(n-1) and the no.of meso forms, m = 2(n-2)-1
Therefore, the total no. of optically active isomers = a + m
                                                                         = 2(n-1) + 2(n/2)-1
 
For butan-2,3-diol , n= 2, a=2(2-1)  =21  =  2
                                      m= 2(n/2)-1 = 2(2/2-1) = 20 =1
 
So,the total optical isomers = 2+1 =3
                                              
Answered by Expert 13th November 2017, 2:45 PM
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