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Doubtful E2 in cycloalkane CIS 2 months 1 day before #13990

good morning again :) I have some doubts about what was happening in this example case.

I have 1-phenyl-2-iodocyclohexane in with potassium tert butoxide. It is clear that the idea converges by an E2 mechanism. However, I have several doubts.

I have drawn conformations A and B, the first thing is that it is displaced towards conformation B since there is less 1,3 diaxial interaction since phenyl is much more voluminous than iodine and fits better in the equatorial zone.

At this point, in conformation B, there are 2 plausible elimination hydrogens, the germinal one with phenyl and the neighboring one with iodine, I hope I make myself clear, I have circled them in green.

Here my question arises, which is easier to withdraw? normally the more substituted alkene is formed, however I wonder if that phenyl group won't generate some steric hindrance and eventually the other one will be removed? to give the Hoffman product. This would be my main question

Then I have a secondary doubt, which had arisen at the beginning, but I have already realized by observing the conformations better. It is about the possibility of leaving the phenyl group instead of the iodine, since the phenyl in the A conformation also has an antiperiplanar hydrogen. But hey, I discard this point because the conformation is not favored.

However, the question would be if, in the right circumstances (not in this case) could phenyl be a leaving group? I have not yet had to study aromatic compounds or the stability of phenyl as a cation.

A greeting and a thousand thanks in advance

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Last Edition: by Germán Fernández .

Doubt E2 in cycloalkane CIS 1 month 3 weeks before #13992

Good morning!
When using a hindered base, the major alkene is the least substituted due to the steric hindrances of the phenyl.
If you use a less bulky base such as methoxide, you would mostly get the most substituted alkene (1-phenylcyclohexene). Phenyl, for its part, cannot behave as a leaving group since it does not have the capacity to stabilize the negative charge. As you say, the elimination requires anti hydrogens, which in turn requires the leaving group to be axial.
Conformation B is the one that meets these requirements and in it you have two anti hydrogens, giving rise to two elimination products. The proportion of these products depends on reaction conditions. Small bases at high temperatures give the thermodynamic product (1-phenylcyclohexene), bulky and very strong bases at moderate temperatures give the kinetic product (3-phenylcyclohexene) Thank you for proposing these issues so well explained.

All the best!!!
The following user said thanks: Pajaro13

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Doubtful E2 in cycloalkane CIS 1 month 3 weeks before #13993

Thank you very much for the clarification!

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