Anti-coplanar H in E2 reactions

[Toadesque]

Confused about this

Here's a reaction from Chad's quizzes

Here's the answer:

So here you have a strong Nuc and a strong base and a tertiary halide. This tells you it's an E2 reaction and not Sn2, I get that.

I don't really understand the concept of the whole co-planar hydrogen thing though and why the more substituted alkene (Zaitsev's) doesn't form instead, i.e this one

Should I just assume that a tertiary halide along with a strong base is going to be Anti-Zaitsev?

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[starnova]

just draw all the Alpha hydrogens, you'll see that all of them go backward out of the plane of the page except the one on CH3 group directly attached to Br. Basically, there's no hydrogen on the dash position for the base to pull from on both adjacent positions.

"Should I just assume that a tertiary halide along with a strong base is going to be Anti-Zaitsev?"
If you eliminate the CH3 group on the top, you're gonna have both hydrogens that are on wedge/dash position, so you can get a more substituted product (compared to the one in this example)