Given this compound in an acidic medium, what would get protonated first? Would the double bond not get protonated preferentially, since oxygen is more electronegative than carbon and less willing to donate an electron pair for protonation?

I checked the $\mathrm{p}K_\mathrm{a}$ value for the conjugate acid of an aldehyde, and it was around -10, which is pretty low; this suggests that the aldehyde would have a very low basicity.

Also carbocation had a $\mathrm{p}K_\mathrm{a}$ value of around -3.

Going by this information does it not make sense for a double bond to get protonated first?

On the other hand, the oxonium ion is resonance stabilised, which would indicate that it is protonated in preference to the double bonds.

Which reasoning is correct? And more importantly, in these situations how are we supposed to predict which factor dominates?

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    $\begingroup$ You'd need pKa of carbocation not alkene. $\endgroup$ – Mithoron May 9 '16 at 22:08
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    $\begingroup$ please don't spam bold+italics, it doesn't help the readability $\endgroup$ – orthocresol May 9 '16 at 22:18
  • $\begingroup$ @Mithoron sorry about that,corrected. $\endgroup$ – Daipayan Mukherjee May 9 '16 at 22:35
  • $\begingroup$ @orthocresol oh,I thought they did, but sure. $\endgroup$ – Daipayan Mukherjee May 9 '16 at 22:37
  • $\begingroup$ Draw the mesomerics forms and select the part of the molecule in which this effect is the lowest then you'll have more electrons there to attack the proton. If you have nothing else it is a good way to have a good idea. $\endgroup$ – Hexacoordinate-C May 9 '16 at 22:38

In the scheme below I have shown the three most probable protonations of the molecule in question. I have considered that a tertiary carbocation is typically more stable than a secondary one and that a resonance-stabilised carbocation is typically more stable than a non-stabilised one.

protonations of the monoterpene

From left to right, let’s discuss the structures.

  • The one on the left is not good. The carbocation is only stabilised by being tertiary. It would take a very strong acid to sufficiently protonate in this way, or a good nucleophile that can capture the positive charge on the tertiary carbocation.

  • The one in the middle doesn’t look as bad initially, because it is resonance-stabilised. However, if you draw the other principal mesomeric structure, you arrive at a positively charged oxygen with a sextet — a very unstable thing to have. It is very improbable that protonation would occur like this.

  • There is another (undrawn) structure that would protonate the same double bond but on the other side. However, that carbocation would destroy the unsaturated ketone’s resonance and is therefore unfavored.

  • Finally, the protonated carbonyl. This is clearly the best cation, since it is an all-octet structure and the positive charge can be stabilised by resonance. Futhermore, if you draw the resonance structure, you realise that it localises the cation on a tertiary carbon — the best possible case.

Thus there should be no discussion; protonation of the carbonyl oxygen leads to the most stable cation and is therefore expected to prevail.

(Your lower $\mathrm{p}K_\mathrm{a}$ of carbocations is likely due to the fact that they can often deprotonate to form a double bond giving additional stabilisation.)

  • $\begingroup$ @ron I was sure that I had re-removed them, but ChemDraw again was more ‘intelligent’ than I was. Bugger. (Hopefully fixed.) $\endgroup$ – Jan Jun 9 '16 at 1:02

I am not saying it is a valuable answer because organic chemistry is not the field of chemistry in which I am the best. But doing what I did can help a bit as I said.

If you try a mesomeric form with the left bouble bond, which is possible in theory (everything is possible on a sheet of paper) you see that you need to make a carbon with a negative charge.

If you do the same in the second part of your molecule, you see that the gamma carbon take a positive charge and the oxygen the negative charge, which is correct about their electronegativity.

So the mesomeric forms I drew in the left of my sheet are more probable than those on the right.

So on your molecule you will have more a bouble bond between the two carbons on the left and more a "1.5" bond in the right and in the right the electrons are travelling between atoms and then they look less reactive (obviously none of them is static just these ones will move more than others). To get your proton you need electrons, and there are more of them in the left side then (which are not doing anything else).

So I would answer it is the bouble bond with react first.

I hope it can help you to think about that. And as I said I am not a specialist, I just propose a way to think.

And sorry in advance for my possible English mistakes ! :)



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