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So I was wondering if what I had in my head is correct:
1) You determine whether or not a molecule is E or Z based on whether or not the two biggest substituents are on opposite sides or the same side, respectively.
2) If you are given a problem asking whether a certain E configuration or Z configuration is more stable, the most stable one would be the molecule that has the most sterically hindered groups trans to each other?
I'm pretty sure #1 is right, but #2 confuses me. I came across a Bootcamp problem where there was an alkene, and on C1 there was a methyl and a benzene attached, and then on C2 there was like a propyl group then a methyl. In the question, it asked which configuration would be more stable, E (with the benzene and propyl groups on opposite sides of the double bond) or Z (with the benzene and the propyl groups on the same side of the double bond). The answer ended up being that Z was more stable (even though normally E is more stable), because benzene is sp2 and less sterically hindered or something to that effect, so having the methyl and propyl on opposite sides was actually more favorable.
So this leads me to this question: If you have an alkene with a chlorine and a propyl on C1, then a butyl and a methyl on C2, would the Z conformation (Chlorine and butyl on same side) be more stable since the methyl is more sterically hindered than Chlorine? Or would the E conformation be more stable because Chlorine is still sp3 (the lone pairs count in hybridization.... I think?)?
Thanks in advance guys, and sorry if this is poorly worded!
1) You determine whether or not a molecule is E or Z based on whether or not the two biggest substituents are on opposite sides or the same side, respectively.
2) If you are given a problem asking whether a certain E configuration or Z configuration is more stable, the most stable one would be the molecule that has the most sterically hindered groups trans to each other?
I'm pretty sure #1 is right, but #2 confuses me. I came across a Bootcamp problem where there was an alkene, and on C1 there was a methyl and a benzene attached, and then on C2 there was like a propyl group then a methyl. In the question, it asked which configuration would be more stable, E (with the benzene and propyl groups on opposite sides of the double bond) or Z (with the benzene and the propyl groups on the same side of the double bond). The answer ended up being that Z was more stable (even though normally E is more stable), because benzene is sp2 and less sterically hindered or something to that effect, so having the methyl and propyl on opposite sides was actually more favorable.
So this leads me to this question: If you have an alkene with a chlorine and a propyl on C1, then a butyl and a methyl on C2, would the Z conformation (Chlorine and butyl on same side) be more stable since the methyl is more sterically hindered than Chlorine? Or would the E conformation be more stable because Chlorine is still sp3 (the lone pairs count in hybridization.... I think?)?
Thanks in advance guys, and sorry if this is poorly worded!