[umair javed]

Hi everyone.

I thought a lot about the answer of the following question but couldn't figure it out. If anyone can help.

Using molecular orbital theory, why thermal elimination of hydrogen to form furan is facile in 1,4-elimination (2,5-dihydrofuran) but difficult in 1,2-elimination (2,3-dihydrofuran)?

Thanks for sharing your knowledge and time.

[Dan]

Not 100% sure on this but I have an idea.

Hint: What does the difference in ease of oxidation tell you about how ΔG for these two reactions compares? Can you rationalise that?

[umair javed]

I haven't thought from this perspective though there will be difference between the ΔG values.

My first answer was in 2,3 dihydrofuran (1,2 elimination) we have to remove two hydrogens from the same carbon to make furan which is difficult as compared to 1,4 elimination in 2,5-dihydrofuran (remove hydrogens from different carbons). But that was wrong.

Then I was sorting it out by the concept of HOMO and LUMO (pericyclic reaction) but professor told me that think from molecular orbital theory.
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. for which we need more energy. By loosing two of its hydrogens it will share electron with with C#3 and make bond in disrotatory fashion which is difficult as compare to 1,4 elimination.

After this he said I am close.

i cant think anymore now. My chemistry is not that much good

[Dan]

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 H₂.

Here is the LUMO for furan:



Now if you combine that with the HOMO of H₂, how would you expect H₂ to add to furan?

It follows from the principle of microscopic reversibility that the favoured mode of addition of H₂ to furan will be the same as the favoured mode of elimination of H₂ (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 H₂ 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?

[AWK]

http://onlinelibrary.wiley.com/doi/10.1002/kin.1065/pdf

Molecular hydrogen elimination from 2,5-dihydrofuran, 2,3-dihydrofuran, and 2-methyl-2,5-dihydrofuran: Quantum chemical and kinetics calculations
Faina Dubnikova, Assa Lifshitz
Volume 33, Issue 11, pages 685–697, November 2001

[umair javed]

Thanks Dan that was the answer I was looking for.

Yesterday evening professor gave me the answer and is similar to yours.

If we consider the reverse reaction, the first reaction (1,4-elimination) can be considered as the reaction between butadiene and H₂. and the second one can be a reaction of ethylene and H₂

Therefore, possibility of the reaction can be examined by the interaction between HOMO of butadiene and LUMO of H2 or LUMO of butadiene and HOMO of H2 for the first reaction.

The symmetry of p-orbital matches in 1st reation and is symmetry allowed process so it is facile, while in the 2nd reaction the symmetry of p-orbitals mismatch and that is why reaction is difficult.

I don't know how to draw these orbitals on computer and copy paste here.