Ok, I think I've got an answer - you can probably disregard my first post.
My second answer was that in 2,3-dihydrofuran (1,2 elimination) the symmetry of p orbitals does not match and there is only C#2 available to eliminate hydrogens from.
I don't quite understand what you're saying here, can you draw it?
I think this is easier to approach from the opposite direction, hydrogenation of furan. Consider the HOMO-LUMO interaction of furan and H
2.
Here is the LUMO for furan:
Now if you combine that with the HOMO of H
2, how would you expect H
2 to add to furan?
It follows from the principle of microscopic reversibility that the favoured mode of addition of H
2 to furan will be the same as the favoured mode of elimination of H
2 (because the forwads and backward reactions proceed via the same transition states). In other words, if it is easier to do a 1,4-addition of H
2 to furan than a 1,2 addition, then it follows that the transition state for 1,4-addition to furan (= transition state for 1,4 elimination from 2,5-DHP) is favoured over the transition state for 1,2-addition to furan (= transition state for 1,2 elimination from 2,3-DHP).
But this is perhaps the HOMO/LUMO explanation you alluded to in your previous post that your professor didn't like?