[sharbeldam]

First we put these 2 reactants (A/B in the picture) in NaOCH3/CH3OH with heat (E2) and they ask me which one reacts faster, I drew chair conformation of both (With Br in axial position, so it can undergo E2), in molecule A i got t-Bu in axial position and in molecule B i got t-Bu in equatiorial position.

* Would it be right to say that because B is more stable because the big group in eq position, it would react slower in E2?

If you can bear with me please, in the second part of the question, they say they put these 2 reactants in CH3OH only without the strong base, what would i expect different, I said SN1 but then how would i know which would be faster in that case? since the Br does not have to be axial.

Thank you in advance!

[Babcock_Hall]

How easy is it for a tertiary butyl group to go into the axial position?

[sharbeldam]

so like are you saying im right in my conclusion, since i did say it is better to be EQ hence more stable, hence it would react slower? so B would react slower in e2 because of that?
And what about sn1?

[Babcock_Hall]

It seems to me that molecule B will spend most of its time with Br in the axial position.  I am not sure about what will happen in the E1 case.

[hollytara]

T-Bu conformationally "locks" a cyclohexane ring with the t-Bu equatorial.

so for the cis t-Bu/Br cyclohexane, when t-Bu is eq, Br is axial and E2 is fast.  For trans t-Bu/Br cyclohexane with t-Bu eq the Br is eq and elimination is slow.

For SN1....   there may be a difference in rate based on whether an axial or equatorial Br dissociates more easily.   SN2 is known to prefer axial leaving groups (equatorial LG the path of nucleophilic attack must be through the ring).  So it may be that dissociation of the Br- in SN1 is better assisted by the methanol solvent when Br is axial.  But I couldn't find anything directly on point.  I am pretty sure that sometime between 1950 and 1975 someone did the experiment.

[sharbeldam]

" so for the cis t-Bu/Br cyclohexane, when t-Bu is eq, Br is axial and E2 is fast.  For trans t-Bu/Br cyclohexane with t-Bu eq the Br is eq and elimination is slow. "

Thank you so my assumption was wrong, i cant draw both molecules with Br in axial since that means that molecule A would have a big group in the axial position, so basically A molecule is very confused " Do I want my Br axial so E2 can occur or do I want my big T-bu group in equatorial because its more stable " and t-bu in equatorial would win the argument and hence Br would be in equatorial too and E2 would be slow.

B>A

I GET IT , THANKS A LOT!!