I understand how the diagram below results in hyperconjugation stabilising alkenes. enter image description here

Essentially, the C-H sigma bonding orbital interacts with the pi bonding orbitals creating two molecular orbitals. It also interacts weakly with the pi* orbital which lowers the energy of psi 1 and 2 stabilising the system.

What I do not understand is the effect this has on bond lengths. We have done a computational investigation in the lab and have found that both the C-H and C-C bonds are weaker, while the C-C bond between the double bond and the alkyl substituent is stronger. I don't understand why.

I have received an explanation for this phenomenon using resonance structures, but I am really looking to understand how to reconcile these observations with the MO diagram above.


1 Answer 1


The MO interpretation is not really all that complicated: you are allowing the C–H σ orbital to overlap with the C=C π* orbital. This leads to a very slight population of the π* orbital, which is antibonding with respect to the doubly bonded carbons. Therefore, that C=C bond length is slightly lengthened. Likewise, the C–H σ orbital is slightly depopulated, which makes the bond length longer.

At the same time, you are forming a slightly stronger C(sp2)–C(H) bond because these new MOs have bonding interactions between the C–H σ orbital and the p orbital on C(sp2). This is not entirely made clear in the diagram above from Fleming's text, which I have addressed separately here. From the diagram, it seems that there are equal amounts of bonding character (from ψ1) and antibonding character (from ψ2) However, the main point which I wrote in that post is that the π* orbital mixes (very slightly) into both ψ1 and ψ2. This increases the C–C bonding character in ψ1 and decreases the antibonding character in ψ2, such that there is a (very slightly) shorter bond.

The actual text in the book makes this clear, but the diagram does not. I suggest reading the text carefully.

  • 1
    $\begingroup$ If an answer has helped solve your question, please consider upvoting and/or accepting one (by clicking on the tick next to the voting buttons). That's the Stack Exchange way of saying thank you. See also: chemistry.stackexchange.com/help/someone-answers $\endgroup$ May 12, 2019 at 23:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.