Can someone help me calculate the number of geometrical isomers in this compound? According to me there should be 3 isomers here but according to my teacher, there are 4.I believe that both EZ and ZE cases will be same here.
1
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2$\begingroup$ in EZ or ZE cases the double bond in the middle also becomes stereomeric hence bringing it back to 4 $\endgroup$– napstablookOct 8, 2021 at 3:29
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$\begingroup$
$\endgroup$
A possible reason you are being led astray is related to the way the substituents on the right side of the tri-substituted double bond are drawn in a linear fashion. However, when the hydrogen and n-propyl are drawn with the proper hybridization, only four isomers are possible, as napstablook has stated in a Comment.
In trienes 1 and 2 the tri-substituted double bond is stereogenic with the (Z)-designation in the former and the (E)-designation in the latter. The (Z)-propenyl group has priority over the (E)-propenyl group. The tri-substituted double bond in trienes 3 and 4 is non-stereogenic owing to the presence of identical groups at one end of the tri-substituted double bond.
answered Oct 15, 2021 at 23:25