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I am sorry in advance for posting an image in this question. I know it makes searchability difficult, but I don't see any other way

enter image description here

The above screenshot is from Claydens organic chemistry. We can see clearly from the screenshot that the pi electrons jump to the antibonding pi* molecular orbital. As far as my memory goes oxygen is more electronegative than carbon, which in molecular orbitals would mean that the energy of the atomic orbital of oxygen is lower i.e. the pi orbitals of oxygen would contribute more to the sigma molecualr orbital as there is less difference in energy.

We would also see that the pi orbitals of carbon would be make more contribution to antibonding pi molecular orbital* So my question is since the energy difference between the pi orbital of carbon and the anitbonding pi* molecular orbital is lesser in comparison with pi orbital of oxygen and the antibonding pi* molecular orbital, Shouldn't the negative charge be on carbon by virtue of its energy being comprable to (pi*) orbital?

PS- I am reffering to the molecular bond formation between oxygen and carbon. If my questions sounds unclear please let me know, I will try to frame it more clearly

Thanks in advance to the answerer

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  • $\begingroup$ Too much MO theory getting in the way. how would you draw the product with a negative charge on carbon? $\endgroup$
    – user55119
    Commented Sep 12, 2022 at 16:17
  • $\begingroup$ I know carbon doesn't want to have a negative charge on it, but the MO theory suggests just that. Maybe I am wrong somewhere $\endgroup$
    – Ipsy-Doopsy
    Commented Sep 12, 2022 at 16:43
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    $\begingroup$ When you form an additional bond, there is no more pi orbital. Or look at it this way. You add a hydride, two electrons and a proton. Two electrons (formally, and only for the purpose of visualising it) go to the pi* orbital, which would give 2- to the whole molecule (1- at oxygen and 1- at carbon). The proton then goes to the carbon and forms the bond. Where is the remaining charge? $\endgroup$
    – Martin - マーチン ♦
    Commented Sep 12, 2022 at 17:06
  • $\begingroup$ Maybe just repeating what others have said, but in a different way - when determining the charge distribution of the product, you forget about the $\pi*$ orbital, since it's no longer there. Start fresh with an MO diagram of an alkoxide (ie the product). What does that look like and where does the charge end up? $\endgroup$
    – Andrew
    Commented Sep 12, 2022 at 20:56

1 Answer 1

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It would seem the mechanism is a straight forward nucleophilic addition of hydride to the carbonyl group. O- would be the most stable intermediate. The reaction is a reduction of the carbonyl to an alcohol.

With carbonyl groups, the carbon is slightly more positive, the hydride adds there.

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