beta elimination

[atlanta213]

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can you explain why b is answer?

Picture is attached.

 

[Xtian]

i believe it's sterics. There's only one hydrogen on the tertiary carbon and there's a big methyl group there. However on the secondary carbon you have two H's ready for the picking...

Xtian

 

[Mista E]

Normally we'd expect it to leave the more substituted alkene, in accordance with the Zaitsef rule.

However, when the Br leaves, an H leaves with it as well.
In order for the Br and H to leave, the two must leave fulfilling two requirements: the leave AXIAL and ANTI.

In the picture, the BR is already axial, so that's fine. Now you need the adjacent carbon's H to leave anti. Since the carbon with the CH3 doesnt have an axial H anti to the BR (rather it has a CH3 anti to the BR), we have to go to the other adjacent carbon which does have an anti axial H.

 

[yakuza]

This is what I would think if I was doing the problem...

Since its in this configuration, the question will be about stability. You want to pull of the H on a beta carbon, BUT here theres 2 beta carbons. So now you pick the H thats the same axia/equatorial as Br.

Br is axial, and the only other beta-carbon's H thats axial is the one on the left.

 

[yakuza]

nvm he answered it

 

[Ferdowsi]

wats B elemination?

 

[condoleezarice]

i believe it's sterics. There's only one hydrogen on the tertiary carbon and there's a big methyl group there. However on the secondary carbon you have two H's ready for the picking...

Xtian

Just a warning, this answer is wrong. Methyl groups don't cause much steric hindrance. The base/nucleophile is unhindered as well. The other answers about being anti are correct.

 

[doc3232]

Just a warning, this answer is wrong. Methyl groups don't cause much steric hindrance. The base/nucleophile is unhindered as well. The other answers about being anti are correct.

I agree.
A more stable bond will much more greatly affect the product than a tiny methyl group.
Also,
This question is tough, it could go so many ways: SN1 SN2 E1 E2
Secondaries make everything complicated...

 

[Contach]

But the question says it's an elimination.. does that help you?
What's a "secondary"?

 

[futuredent22]

This question is tough, it could go so many ways: SN1 SN2 E1 E2
Secondaries make everything complicated...

i guess it kinda "could" go any of the 4 ways, but by far the most likely pathway is E2 because of that strong base. the reason the answer is B is because E2 is anti. The base pulls a hydrogen that is anti to the leaving group, and there is only 1 axial hydrogen anti to the Br.

 

[futuredent22]

But the question says it's an elimination.. does that help you?
What's a "secondary"?

he meant that it would form a secondary carbocation if it went sn1 or e1.

 

[Contach]

he meant that it would form a secondary carbocation if it went sn1 or e1.

i see

And what's a "beta"-elimination?

 

[orthodocs12]

yes. mista E and yakutsk put it well. u want di axial position for an E2 rxn, a rxn under which a base like this will partake.