- #1
Declan McKeown
- 4
- 0
While I'm fairly sure this question is pretty straight forward, I can't remember why (ii) is not considered a geometric isomer. (The answer is (c)) http://imgur.com/ObccNpv
Looking at ii I would have thought it to be trans-pent-2-ene due to the ethyl and methyl group either side of the double bond, however it is not and I am not sure why. Would anyone happen to know why this is?
Thanks :)