# Acidity of aldehydes

Which is more acidic between methanal ($\ce{HCHO}$) and ethanal ($\ce{CH3CHO}$). Please explain using General organic chemistry basic concepts.

My Effort:

I saw the stability of the conjugate base.

It would be $\ce{HCO-}$ and $\ce{CH2^{(-)}CHO}$. Due to resonance I can say that second one is more stabilized and thus a better acid.

But answer says that methanal is more acidic than ethanal.

• Hi AJ_ welcome to ChemistryS.E.! It looks like an homework question, can you please show us your efforts, thoughts and your attempts?Please take a look to our homework policy! Thanks a lot! :-) – G M Jun 1 '14 at 8:55
• Formaldehyde/methanal is particularly electrophilic, hence in aqueous solutions it actually exists predominantly as the hydrate (methanediol, $\ce{CH2(OH)2}$). If comparing $pK_a$ values with water as the solvent, this may at least partially account for the seemingly aberrant results. – Greg E. Jun 1 '14 at 17:44
• AJ, is there any additional explanation with the answer? This is a very surprising result. – jerepierre Jun 2 '14 at 19:21
• No there isnt any. – AJ_ Jun 4 '14 at 3:21

## 2 Answers

Honestly, I don't see how anyone could have predicted this outcome before seeing the actual $\ce{pK_a}$ data. If you look up the acidities of methanal and ethanal, you find $\ce{pK_a}$ = 13.27 and 13.57 respectively. A pretty small difference to begin with. As the picture below illustrates, in the case of methanal we are removing the proton directly bound to the carbonyl carbon to produce the anion pictured below.

Now if someone asked, "would pulling the same proton off from ethanal be easier or harder?", you could answer "harder" because that methyl group (being electron releasing) that is attached to the carbonyl carbon would act to slightly destabilize the anion with the negative charge on carbon. But it turns out that when ethanal is treated with base a different proton is removed! The proton that is on the methyl group (said to be alpha to the carbonyl carbon) is the proton that is removed. Without knowing the actual $\ce{pK_{a}{'s}}$ beforehand, how could you know that the anion involved in the methanal case would be ever so slightly more stable than the resonance structures involved in the ethanal case?

Edit: Thanks to Greg E and Nicolau Saker Neto for pointing out an error in the earlier version of the figure and the corresponding text respectively.

• I'm fairly sure that there is no resonance stabilization in the case of methanal. The triple-bonded oxygen violates the octet rule, and I believe the lone-pair would have to reside in a localized $sp^2$ orbital. – Greg E. Jun 1 '14 at 19:29
• Whoops, now I see. Thanks for pointing that out. I'll edit my answer. – ron Jun 1 '14 at 19:42
• You edited the figure, but your answer currently still says the methanal conjugate base is resonance-stabilized. – Nicolau Saker Neto Jun 1 '14 at 21:47
• But my doubt still remains. The anion formed from ethanal is clearly resonance stabilized with charge delocalisation thus making it more stable than the ion from methanal.Where did I go wrong? Also thanks for the replies. – AJ_ Jun 2 '14 at 2:38
• Yep, I was surprised too, but them's the facts (the pKa's that is). All you can say is that for some reason the methanal anion is more stable than the delocalized ethanal anion. Just because one structure has more resonance structures than a different structure, doesn't necessarily mean that the former is more stable. – ron Jun 2 '14 at 2:54

The answer lies with induction.

The oxygen is better able to stabilize the residual negative charge on the conjugate base of methanal by virtue of residing closer to the negative formal charge.

On the conjugate base of ethanal note that the electronegative oxygen is one carbon removed from the negative formal charge while on the conjugate base of methanal, the oxygen is on the same carbon with the negative formal charge. Thus, the oxygen in methanal will be better able to withdraw electron density from the spot where heterolytic bond clevage occurred. This follows from Coulomb's Law:

${F=k\frac{q_1q_2}{r^2}}$

As we can see from Coulomb's law, force and distance are related in an inverse-square fashion. In other words, increasing the distance of the oxygen nucleus from the site of the negative charge dramatically decreases the electron-withdrawing effects of oxygen.

• But the conjugate base of ethanal has a resonance structure that places the negative charge on the oxygen. r=0 for ethanal! – jerepierre Jun 2 '14 at 19:20
• That doesn't mean the negative charge is necessarily there; resonance structures are discrete while the actual molecule definitely does not conform to these discrete forms we may draw on paper. And no r =! zero because there is a distance between the electrons and the nucleus. – Dissenter Jun 2 '14 at 19:21
• If you calculate the electron density of this anion, the electron density will primarily be located on the oxygen. See: instruct.uwo.ca/chemistry/283g/Nifty%20Stuff/… – jerepierre Jun 2 '14 at 19:23
• Perhaps. I don't know how to compute the electron density so I can't dispute this (or confirm it). – Dissenter Jun 2 '14 at 19:24