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I was wondering why when acetaldehyde is treated with $\ce{OH-}$, the $\ce{OH-}$ will attack the methyl group instead of the carbonyl carbon, forming an enolate anion. I would have though that the $\ce{OH-}$ would have attacked the carbonyl carbon due to it having a relative positive charge.

Additionally, my professor said during lectures that form x is favored over form Y. Is this because a negative oxygen is able to delocalize the negative charge (and be protonated more easily) more efficently than a $\ce{CH2-}$ group? If so, is the x form so dramatically favored that it would not make sense to draw any mechanisms involving the y form?

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Enolate vs hydrate formation

Both the enolate and the hydrate form.

Aldehydes (and ketones) react with water in the presence of acid or base to form hydrates. However, for more aldehydes in aqueous solution (even dilute solution), the hydrate is only a minor component of the mixture (~1%).

The other 99% of the aldehyde molecules are perfectly able to form the enolate, but only a small number of them do because the difference in pKa between most alpha hydrogens on aldehydes and ketones (~21) and water (15.7).

So, in actuality most of the aldehyde molecules at any given time are neither enolates nor hydrates. Now, if a reaction should occur to consume enolate, then Le Châtelier will come out and nudge the system to make more enolate.

"Favored" resonance contibutors

The two resonance contributors $x$ and $y$ are not separate compounds. It is not possible for one structure or another to be "favored", though we often use that language because we can relate to comparing the structures of two distinct compounds, like the cis and trans isomers of an alkene. What we really mean when we say "favored" is that the resonance contributor is more important for describing the structure and behavior of the ion than the other. The true structure of the enolate ion is some linear combination of resonance contributors $x$ and $y$ (as well as some other minor things). Thus, the true structure might be like a weighted average over the contributors with $x$ having higher weight.

For example:

$$enolate = \frac{2}{3}x + \frac{1}{3}y$$

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  • $\begingroup$ I'm sorry if I am misintepreting resonance, but if they are resonance forms (ie all identical molecules), wouldn't their reaction with another molecule, say acetaldehyde, form always one product, in this case 3-hydroxybutanal $\endgroup$ – Matthieu Kratz Apr 23 '15 at 10:58

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