Picture 1 and Picture 2

Which will be the most acidic hydrogen in both cases? Please explain.

According to me, in the first compound 2 should be most acidic as in both 1 and 2, resonance occurs but 2’s carbon is closer to the oxygen, which can stabilize the negative charge on carbon.

In the second compound, 1 should be the most acidic as there is $+I$ at 2 by carbon 3, which destabilizes the negative charge on it after removing hydrogen from carbon 2.

Am I correct?


2 Answers 2


Resonance effect is a more dominating effect than inductive effect. Here I think in both the structures hydrogen attached to carbon 1 will be more acidic. The reason is the negative charge formed on the carbon after removal of hydrogen is resonance stabilised to a greater extent.

In the first structure though after removal of hydrogen from the second carbon, the negative charge formed will be resonance stabilised, but its delocalisation is less compared to the delocalisation of the negative charged which would have been formed from the removal of hydrogen from the first carbon. So, if the negative charge forms from the first carbon ( from the removal of hydrogen), then it is more stabilised. Hence the hydrogen attached to the first carbon is more acidic.

In the second structure also the hydrogen attached to the first carbon is more acidic. This is because if negative charge forms on the second carbon, there is no stabilisation effect. But if the negative charge forms on the first carbon, then again by resonance it will be stabilised. So in the second structure also the hydrogen attached to the first carbon is more acidic.


Somewhat surprisingly, given both the commonness of the compounds in question and the otherwise comprehensiveness of published $\mathrm{p}K_\mathrm{a}$ data, I was unable to find any published acidity data for any of the four protons. Most notably, even looking up the structures in SciFinder, ChemSpider and the NIST Chemistry Webbook returned no acidity results. I must conclude that vinylogous ketones (possibly even aldehydes) have not yet been a research topic and would assume a low-hanging physical chemistry paper to be up for grabs.

Therefore, we must use rationalisation and chemical intuition to determine which factors dominate in which conjugate base of molecule.

To state the exceedingly obvious first: both protons marked 1 should have practically the same acidity since the distal methyl group will have practically no steric and absolutely no electronic effects that we must account for. That doesn’t really help us, though, since we are supposed to compare 1 and 2 within each molecule and not across them.

Upon deprotonation of the left-hand structure, we will generate one of the following enolates:

$$\begin{align}\ce{H2C=CH-CH=C(-O^-)-CH3} && \ce{H3C-CH=CH-C(-O^-)=CH2}\end{align}$$

Both of the two feature a five-atom π system with one internal and one terminal double bond; but one π system is branched while the other is linear. Also, the internal double bond is triple-substituted in the left-hand anion while it is double-substituted in the right-hand one. These two features lead me to assume that proton 1 in your first compound is more acidic than proton 2.

The deprotonation of the right-hand structure will generate one of the following enolates:

$$\begin{align}\ce{H2C=CH-CH=C(-O^-)-CH2-CH3} && \ce{H3C-CH=CH-C(-O^-)=CH-CH3}\end{align}$$

(I have immediately disregarded proton 3 as it is purely aliphatic and unstabilised; the race is only between protons 1 and 2.) As I mentioned above, the left-hand enolate is practically identical to the one above but the right-hand enolate is now also trisubstituted. Therefore, when purely comparing double bonds the left-hand structure (terminal & trisubstituted) is now less favourable than the right-hand structure (internal & trisubstituted) — this would favour proton 2. On the other hand, the π system after proton 2’s removal is branched but that upon proton 1’s removal is linear. I would assume that these two effects work against each other — i.e. that a linear π is better than a branched one.

However, when comparing the two effects with each other I think the higher substituted double bonds should win the contest over linear vs. branched. Therefore, I assume that in your second compound proton 2 is more acidic.

As I mentioned above, I unfortunately do not have experimental data to back my claim.

To just quickly comment on your suggestions:

  • As I stated implicitly, neither of the anions will actually be carbon-centred; the predominant mesomeric structure will always be the oxygen-centred anion. Thus, which carbon is closer to oxygen is a moot discussion as they do not carry any charge.

  • Methyl groups do indeed destabilise anions by $+I$ effects. However, they also serve to stabilise double bonds by the same $+I$ effect. Since we are not dealing with a carbon-centred anion, the destabilising effect is practically non-existant.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.