2
$\begingroup$

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.

$\endgroup$

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
1
  • 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 $\endgroup$ Commented May 12, 2019 at 23:13

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