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Currently studying high school organic reaction mechanisms and encountered a question. It goes like this:

The C=O bond and the C–O bond of an ethanoate ion is equal at 127 pm, while the C=O bond of ethanoic acid is shorter at 121 pm than its C–O bond, which has a length of 141 pm. By drawing resonance structures, explain why this is so.

So I know why the C=O and C–O bonds in the ethanoate ion is equal because both of its resonance structures have the same significance, but I'm not so sure about the difference in the bond lengths in ethanoic acid. When I tried to draw the resonance structure for ethanoic acid, I was thinking that the O atom of the hydroxyl group is $sp^3$ hybridised (since each of the two lone pairs will occupy two separate $sp^3$ hybridised orbitals), so that means there is no resonance effect.

After scouring through the internet, the O atom of the OH group is $sp^2$ instead. Why is this so?

(p.s. high school student here, so I might not be able to understand high-level explanations just yet, sorry)

Karsten
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warren
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2 Answers2

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You're absolutely right in saying that in ethanoate ions, both resonance structures are equivalent - as both oxygens are the "same" in environment and position. However, in ethanoic acid, because of the presence of the hydrogen atom, the second structure is considerably less stable than the first - notice that the oxygen atom bearing a positive charge is unstable.

Why sp2 and not sp3? Well, the hybridisation of an atom is something that remains fixed in a molecule of the compound. You can't have the hydroxyl O being sp3 hybridised and yet participating in resonance - because the double bond wouldn't form without a p orbital for conjugation!

So the only sensible explanation (and experimentally verified fact) - is that the hydroxyl O is sp2 hybridised - one lone pair in an sp2 orbital of its own, the hydrogen bonded to O via another sp2 orbital, the carbon bonded via a third sp2 orbital (sigma bonding), and the second lone pair delocalised from the 2p orbital to form the pi bond in the second resonance structure. And there you have it!

enter image description here

sidsr003
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  • I was going to post the same answer but sidsr003 got there first, well done this is the same as what I teach my undergraduates when I am teaching organic chemistry. – Nuclear Chemist Feb 12 '22 at 10:28
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    What is the "experimentally verified fact" you refer to? Note that, as discussed earlier at several places in Chem SE, photoelectron spectroscopic data suggests the $\ce{O-H}$ bonds in the water molecule are roughly $\ce{sp^3}$ hybridized while the oxygen lone pairs reside in more or less $\ce{p}$ and $\ce{sp}$ orbitals. – ron Feb 12 '22 at 18:13
  • Thank you so much for your answer, but I am still not convinced as of why the O atom of the hydroxyl group is sp2 hybridised, since your explanation seems to assume first that there is resonance in ethanoic acid. How would you know that the O atom is sp2 hybridised, without knowing that there is in fact the resonance effect in ethanoic acid? – warren Feb 13 '22 at 12:21
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The experimental evidence

The bond next to the $\ce{C=O}$ double bond in acetic acid and in esters is conventionally drawn as a single $\ce{C-O}$ bond, but does not show free rotation. For example in methyl formate, the most common arrangement of atoms is this:

enter image description here

Rotating around the single bond (marked with red dots) does not happen readily (the conformation with the $\ce{-CH3}$ pointing at us is estimated to occur roughly a billion times less than the one shown). Similarly, the $\ce{C-N}$ bond in amides does not behave like a single bond, and amides are planar as well.

The length of the $\ce{C=O}$ double bond is about the same as in other molecules, where there is no oxygen or nitrogen single bond with the carbon. The length of the $\ce{C-O}$ double bond is a bit shorter than in, for example, methanol ($\ce{H3C-OH}$). In general, bond lengths can tell us whether a bond is a single or a double bond or something in between.

The interpretation

Somehow, the bond next to the $\ce{C=O}$ double bond has some "double bond character". The easiest explanation when writing a Lewis structure is to show a second resonance contributor. For the acetate ion, the two resonance contributors are identical (with the role of the two oxygen atoms switched). For the acetic acid molecule, one resonance contributor has formal charges (the minor one) and the other does not (the major one).

The question of the hybridization is not so important because you only have two atoms attached to the atom in question. If there are three atoms attached (as in the formamide below), you would either have a trigonal planar or a trigonal pyramidal arrangement, telling you whether it makes more sense to invoke $\mathrm{sp^2}$ or $\mathrm{sp^3}$ hybridization.

enter image description here

Source: https://chemistry.stackexchange.com/a/10917/72973

In this case, however, it does not matter for the major resonance contributor. The minor conributor, however, has to be $\mathrm{sp^2}$ (because you need the unhybridized p-orbital for the double bond).

The C=O bond and the C–O bond of an ethanoate ion is equal at 127 pm [...]

If the bond lengths are equal, it makes little sense to call one a single bond and the other a double bond. It would be nicer if the question read "the two bonds between carbon and oxygen in the acetate ion are of the same length, while in acetic acid, the bond between carbon and the hydroxyl group is longer than the other bond with oxygen". In this way, you don't make an assumption of the type of bond (single, double, or somewhere inbetween).

Karsten
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  • The rotation barrier between s-trans and s-cis acetic acid is a bit hard to measure because there is hardly any s-cis acetic acid present. However, it has been isolated in an argon matrix at 8 K: 10.1021/ja038341a – Karsten Feb 12 '22 at 14:54
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    Thank you for the answer. Yes, I do understand that due to the resonance effect, the C=O and C–O bonds are 'partial' double bonds. But when I was drawing the resonance structure of acetic acid, I was confused as I thought it makes no sense that there is resonance in acetic acid in the first place, since I thought the O atom of the –OH group was sp3 hybridised, without an hybridised p orbital to overlap with adjacent unhybridised p orbitals. Hence, the question. – warren Feb 13 '22 at 12:27