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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.

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    $\begingroup$ 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! :-) $\endgroup$
    – G M
    Jun 1, 2014 at 8:55
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    $\begingroup$ 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. $\endgroup$
    – Greg E.
    Jun 1, 2014 at 17:44
  • $\begingroup$ AJ, is there any additional explanation with the answer? This is a very surprising result. $\endgroup$
    – jerepierre
    Jun 2, 2014 at 19:21
  • $\begingroup$ No there isnt any. $\endgroup$
    – AJ_
    Jun 4, 2014 at 3:21

2 Answers 2

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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.

enter image description here

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.

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    $\begingroup$ 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. $\endgroup$
    – Greg E.
    Jun 1, 2014 at 19:29
  • $\begingroup$ Whoops, now I see. Thanks for pointing that out. I'll edit my answer. $\endgroup$
    – ron
    Jun 1, 2014 at 19:42
  • $\begingroup$ You edited the figure, but your answer currently still says the methanal conjugate base is resonance-stabilized. $\endgroup$
    – Nicolau Saker Neto
    Jun 1, 2014 at 21:47
  • $\begingroup$ 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. $\endgroup$
    – AJ_
    Jun 2, 2014 at 2:38
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    $\begingroup$ Your explanation of the $\ce {pK_a}$ values is likely incorrect as aldehydes form significant amounts of their geminal diols in aqueous solution. See here: chemistry.stackexchange.com/questions/107616/… $\endgroup$
    – Tan Yong Boon
    Aug 11, 2020 at 0:43
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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.

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  • $\begingroup$ But the conjugate base of ethanal has a resonance structure that places the negative charge on the oxygen. r=0 for ethanal! $\endgroup$
    – jerepierre
    Jun 2, 2014 at 19:20
  • $\begingroup$ 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. $\endgroup$
    – Dissenter
    Jun 2, 2014 at 19:21
  • $\begingroup$ 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/… $\endgroup$
    – jerepierre
    Jun 2, 2014 at 19:23
  • $\begingroup$ Perhaps. I don't know how to compute the electron density so I can't dispute this (or confirm it). $\endgroup$
    – Dissenter
    Jun 2, 2014 at 19:24

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