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The reactivity of the enolate ion is most frequent explained with reference to its system of $\pi$ molecular orbitals. The image below taken from Wikimedia Commons depicts this system of $\pi$ molecular orbitals. $\ce {C}$-alkylation is favoured when the orbital interaction with the $\ce {HOMO}$ of the $\pi$ system is dominant while $\ce {O}$-alkylation is favoured when the electrostatic interaction with the negative charge built up on the oxygen atom is dominant. This is the case because the coefficient of the $\ce {HOMO}$ is the largest on the terminal carbon atom while being more electronegative, oxygen naturally bears greater negative charge.

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The explanation provided by Fleming (2010) on p. 160 is that in the lowest-lying $\pi$ molecular orbital, the oxygen atom exerts the strongest polarising effect on the electron distribution while for the second highest $\pi$ molecular orbital, the polarisation is in the opposite direction. I understand that these MO diagrams are grounded in quantum chemistry theory but is there any way that we could intuitively understand why the coefficient of the $\ce {HOMO}$ is indeed the largest on the terminal carbon?

Update

After discussion with some friends, I have more thoughts I would like to share. There is this connection between resonance structures and MO diagrams. For example, the MO diagrams of the allyl cation and allyl anion feature a node in the second MO, informing us that there is no sharing of the charge with the central carbon atom. This information is also reflected when we draw the two resonance structures. Can we perhaps, extend this connection a little further and say that the first MO for this enolate $\pi$ system reflects the charge distribution in the resonance structure with negative charge localised on the oxygen atom while the second MO reflects the charge distribution in the resonance structure with the negative charge localised on the carbon atom?

enter image description here

Image source

Reference

Fleming, I. Molecular Orbitals and Organic Chemical Reactions (Reference ed.). United Kingdom : Wiley, 2010.

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  • $\begingroup$ Not directly relevant to the question, but regarding the issue of C vs O alkylation, conventional HSAB theory is not correct, doi.org/10.1002/anie.201007100 $\endgroup$ – orthocresol May 13 at 16:22
  • $\begingroup$ Not sure if I answer your question so I leave a comment: In $\pi_2$, there is a node between C and O due to the opposing signs of the wavefunctions. You won't find any electrons here and the oxygen can't pull electrons on the other side of the node due to the deconstructive interference of the waves, hence why the polarization is in the opposite direction towards the terminal carbon. $\endgroup$ – MasterYoda Jun 4 at 22:08
  • $\begingroup$ This is not directly related, but you might want to give my answer a read: How to rationalise with MO theory that CO is a two-electron donor through carbon? Similar reasoning can be applied explaining the coefficient of pi2 at carbon. $\endgroup$ – Martin - マーチン Jun 6 at 11:34
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One possible way to intuit the result is to remember that all of the orbitals are normalized. Thus, the contribution of a single AO to the three MOs must add up to 1, and the contribution of the three AOs in each MO must also add up to 1. (This is a bit of an oversimplification, since the normalization is of the square of the wave function, but conceptually it's the same idea). So if a particular AO contributes a lot to one MO, it is proportionally less represented in the others. Also, the other AO's have to contribute less to the first MO in order for it to have the right magnitude.

In MOs composed of AOs from atoms of differing electronegativity, the lower energy MOs have a larger contribution from the more electronegative atom (and therefore smaller contribution from the less electronegative atoms). The higher energy MOs must therefore have a smaller contribution from the electronegative atom.

As a very simple first example, consider the $\sigma$ and $\sigma^*$ MOs of a polar diatomic molecule such as HF. The valence F AO is much lower in energy than the hydrogen orbital, so the bonding MO has a large contribution from F and very small contribution from H. In fact, the MO is nearly indistinguishable from the fluorine AO, as evidenced by the bond being nearly completely ionic. In order for the normalization to be correct, the antibonding orbital must be the reverse - a very small contribution from F and large contribution from H.

A slightly more complicated example is a triatomic like H2O. Here is a drawing of the lowest three valence MO's for XH2, when X is more electronegative than H: MOs

(This is a portion of Figure 7.3 of Albright et al. Orbital Interactions in Chemistry).

You can see that again the electronegative atom (O) is overrepresented in the lowest energy MOs, at the expense of the H atomic orbitals. Therefore, in the high energy orbitals, we have to compensate with the reverse, an overrepresentation of the H orbitals (as seen in the 3a1).

Now getting to your question of C-C-O, again the O is most electronegative, so it contributes most to the lowest energy orbital. That means it has less to contribute to the higher energy orbitals, where the carbon orbitals are the bigger contributors.

This overly simplistic approach is harder to apply to the difference between the two carbons, for which you really have to incorporate the actual math of the atomic orbital contributions, but maybe it at least helps with the concepts.

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  • $\begingroup$ Thanks for the answer. Your point on the normalisation of the orbital coefficients is helpful and leads us closer to the answer. However, I am interested as to why exactly does the largest amplitude of the 2nd MO lie on the terminal carbon. Maybe, this simply cannot be figured out based on intuition... $\endgroup$ – Tan Yong Boon Jun 7 at 7:50
  • $\begingroup$ Is it intuitive to think about denisty decreasing near nodes? Consider that if it were three equivalent atoms, there would be a node through the central atom in $\pi_2$. Even though the node is shifted a bit here, the density still tapers off approaching the node. I'd also add that the basic math of combining AOs to make MOs is not super complicated. If it's something you're really interested in, it's worth learning it right. $\endgroup$ – Andrew Jun 7 at 11:08
  • $\begingroup$ Minor point of correction: The depicted orbital scheme uses four orbitals, there would be one missing. $\endgroup$ – Martin - マーチン Jun 7 at 17:51
  • $\begingroup$ @Martin-マーチン - Are you referring to the three MOs from H2O that I've shown? You're correct that four AOs are involved, but since I've only used the $\sigma_g$ combination of the H orbitals, there are really only three orbitals being combined - the oxygen $p_z$ and $s$ and the H2 $\sigma_g$. (To complete the MO diagram for H2O, there is one MO that is the oxygen $p_y$ and two that result from combination of oxygen $p_x$ with H2 $\sigma^*_u$) $\endgroup$ – Andrew Jun 7 at 21:59

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