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Are $\ce{CH2=CH-CH=O}$ and $\ce{CH2=C=CH-O-H}$ tautomers?

There is only a shift of a hydrogen atom and a pi bond. So why are these two not tautomers?

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    $\begingroup$ The carbons in H2C=C=CH(OH) have to be linear. In CH2=CH−C(=O)(H) they don't. $\endgroup$ – MaxW May 20 at 6:00
  • $\begingroup$ @MaxW That's irrelevant, it's an issue only in resonance. $\endgroup$ – Mithoron May 20 at 14:38
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The molecular geometry argument is pretty common-sensical, and is true to a certain extent. But I don't think it's the cleanest argument. If you look at a typical ketone and its enol, you will notice that the enol has a more constrained geometry than the ketone:

Geometry constraints in keto-enol tautomers

In the keto tautomer, the highlighted carbon is free to go wherever it likes, since it's part of an sp3 chain. But in the enol tautomer, by virtue of the 120° bond angles in the alkene, the highlighted carbon must be in the same plane as the double bond.

Furthermore, it's not possible to claim that the ketone and the enol have the same geometries in the carbon chain. Many of the bond lengths in the two tautomers will be different: after all one has a C–C (and C=O) bond and the other has C=C (and C–O). Unlike resonance structures, both of these tautomers are things which exist independently, and they have different bond lengths.

The safest way is to go back to the IUPAC definition. That's a bit boring, but when you read these, you can see the effort that has gone into coming up with a definition that is (as far as possible) consistent with edge cases such as these. Tautomerism is defined as (https://doi.org/10.1351/goldbook.T06252):

Isomerism of the general form $\ce{G-X-Y=Z <=> X=Y-Z-G}$ where the isomers (called tautomers) are readily interconvertible.

Acrolein and propadienol do match the bonding pattern required for tautomerism, but don't pass the second criterion of being "readily interconvertible".

The reason is fairly simple. The process of tautomerisation involves protonating the oxygen, and deprotonating an α C–H (which way round this happens doesn't matter). To deprotonate a proton that is α to C=O, the C–H bond must be oriented parallel to the C=O π system, because the new C=C π bond is going to be made from that C=O π bond and the C-H σ bond. This cannot be the case in acrolein, because of the extended conjugation between the C=C and C=O bonds. Indeed, the α C–H bond in acrolein is perpendicular to the C=O π system.

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One of the requirements of tautomerism is that the carbon skeleton of the molecule remains unchanged. Only the proton must change it's location with reference to the molecule. Acrolein and propadienol have very different molecular geometries and hence are not tautomers of each other. Here's a comparision of their structures. The first one is acrolein and second one is propadienol.

enter image description here enter image description here

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    $\begingroup$ You have a link to the definition of tautomerism, which is great, but it doesn't actually stipulate in there that the carbon skeleton geometry must be unchanged. $\endgroup$ – orthocresol May 20 at 12:16
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    $\begingroup$ That said, I don't entirely disagree with it. Just a little bit. There is probably a correlation between your geometry argument and my reactivity argument in that, in general, drastic changes in molecular geometry make reactions less likely to happen. $\endgroup$ – orthocresol May 20 at 12:26

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