[billsauce]

   
How many alkenes are formed by E2 elimination of HBr from 3-bromo-3,4-dimethylhexane using a strong base such as sodium methoxide?

I think it's 4 because there are 3 beta carbons, one from the carbon with 3 hydrogens, one with 2 hydrogens and 1 with 1 hydrogen. The one with 2 hydrogens will form E/Z structures, the others will only form 1 structure. Is this right? I took this question off of an online quiz that said there was 5...

[Schrödinger]

3-bromo-3,4-dimethylhexane

First things first : Why such a weird name? Unless you're missing something like cyclohex"ene" or something.

[billsauce]

its not a weird name. its an elimination reaction that starts off as an alkane and forms alkenes.

[Schrödinger]

Oops sorry! I read that as cyclohexane. My bad

Hint : The beta-Carbon with just 1 H is a chiral center.

[billsauce]

no it doesnt. it has an ethyl group

[discodermolide]

The E2 elimination requires a trans antiperiplanar alignment of the H and the leaving group (Br).
Unless I am blind I see 3 possibilities here.
http://en.wikipedia.org/wiki/Elimination_reaction

[orgopete]

I have five also, E and Z for 3,4-dimethylhex-2-ene and hex-3-ene, that is four. The 3-methylene-4-methylhexane is five.

[Schrödinger]

no it doesnt. it has an ethyl group

Of course it does. As a matter of fact, it has 2 chiral centers. C-3 and C-4 both are chiral centers.

[discodermolide]

I guess I'm blind, I get 5 as well. Changed my glasses