[mikeylikesit182]

Hello! I have a question over the "correct" route for a reaction (ChemDraw included.) My teacher explained this to me and I still can't quite understand the why the "correct" route is preferred. The goal is to dehydrate an alcohol to Alkene. The E2 route with NaOH apparently will not work but the E1 with H2SO4 and excess H2O will. Any insight into this would be greatly appreciated, since I cannot understand why one is better. Thank you!

[Babcock_Hall]

When NaOH interacts with tert-butanol, what do you think will happen first?

[mikeylikesit182]

I think OH- will deprotonate the H attached the Oxygen because it is more acidic than any of the Hydrogens on the Methyls. I know it can't perform an SN2 because the molecule is a Tertiary alcohol. Thanks for the response!

[Babcock_Hall]

I agree, but then what is wrong with the intermediate in the bottom diagram?

[mikeylikesit182]

Woops. I drew it the wrong way! I redrew it and I think I see it now. Oxygen would make for a poor leaving group and it would also have -2 charge for this to happen. I think?

[Babcock_Hall]

I agree that O^2- would be an extremely poor leaving group.

[mikeylikesit182]

I really appreciate the help. Thanks for your time