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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

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

Grade:11

1 Answers

vikas askiitian expert
509 Points
11 years ago

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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