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vikas sir the question i posted was an iit 2009 question n the number of stereoisomers has been given as 6..... n cud u please tel me how do we get 2 enantiomers in the cis form as it is optically inactive.... please explain to me briefly how did u calculate the enantiomers in both. i am too confused wid it.... can u please refer me a way to slove such problems
dear shreyans here is the detailed solution , (d means dextro & L means leavo) left assymetric carbon double bond right assymetric carbon 1) d cis d 2) d trans d 3) L cis L 4) L trans L 5) d cis L 6) L trans L from this table , all d-L pairs are enantiomers & those paires which are not enantiomers are diasteriomers .... 1 - 3 & 2-4 are enantiomers , total enatiomers for cis are 2 , for trans 2 ... total steriomers are 6 ....
dear shreyans here is the detailed solution , (d means dextro & L means leavo)
left assymetric carbon double bond right assymetric carbon
1) d cis d
2) d trans d
3) L cis L
4) L trans L
5) d cis L
6) L trans L
from this table , all d-L pairs are enantiomers & those paires which are not enantiomers are diasteriomers ....
1 - 3 & 2-4 are enantiomers , total enatiomers for cis are 2 , for trans 2 ...
total steriomers are 6 ....
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