[kaseli]

So I have the molecule 1-bromopentane reacting with potassium tert-butoxide in a solvent of tert-butanol (i asssume it's the solvent because it's written below the reaction arrow, isn't that the convention?)

I believe this reaction will proceed via an E2 mechanism because removal of the bromo group would leave a primary carbocation with no possibility of rearrangement, ruling out SN1 and E1.  Also, the nucleophile, tert-butoxide, is very bulky which deceases the likelihood of an SN2 reaction.

However, it is my understanding that in an E2 reaction, the hydrogen which leaves MUST be anti-coplanar to the leaving group.  Yet, there doesn't seem to be an anti-coplanar hydrogen in this molecule.  I've drawn Newman Projections but I just can't see it.

What am I missing?  Is it not E2?  Did I draw the projection wrong?  Any help would be appreciated.

[kaseli]

JK it looks like a hydride shift is possible to make a secondary carbocation.  Was thinking only of methyl shifts at first.

[Kate]

Unless I'm missing something, that's the hydrogen.

[kaseli]

I drew the projection differently.  I'm curious, for the back carbon, how come you drew one Hydrogen up and to the right but the other one you drew down?  How come they are not both up like how I drew them?

[Kate]

Single bonds can rotate and this rotation is much faster than the E2 reaction. So both projections are right, but if you want to show how the E2 mechanism can happen with your molecule, then it's best to draw the way I did because it really confirms that there's an hydrogen anti-coplanar to the Br atom.