# Is the hydrogen atom attached to the benzylic carbon in benzaldehyde acidic?

If benzaldehyde loses hydrogen atom at the benzylic carbon, the negative charge will disperse into the benzene ring via resonance. So shouldn't the conjugate base formed be stable, implying that the hydrogen was acidic?

• Its pKa is 14.90, which is not acidic, but is acidic than most non-carboxyl-containing organic compounds. – DHMO Jan 25 '17 at 8:51

Yes and no.

What you discuss is only possible if the $\ce{C-H}$ bond is perpendicular to the ring plane. However, since the carbonyl group is planar and better conjugation can be achieved with the $\ce{C=O}$ bond participating in resonance with the benzene ring the predominant configuration of unsubstituted benzaldehyde should be planar, i.e. the $\ce{C-H}$ bond being in the ring plane.

If you introduce steric hindrance, e.g. 2,6-di-tert-butylbenzaldehyde, the chance of the carbonyl group turning out-of-plane is much greater resulting in a better overlap of the $\ce{C-H}$ bond with the ring system and thus a lower $\mathrm{p}K_\mathrm{a}$.

• I think I like this answer better than mine. I'm out of votes for today, but in another 50-odd minutes I'll probably up-vote yours and delete mine. Nice job. – airhuff Jan 24 '17 at 23:09
• So, it is solely up to the OP. I found this chemistry.stackexchange.com/help/deleted-answers , that says "Answers can be deleted at any time by their authors, unless the answer has been accepted by the question asker. " – airhuff Jan 24 '17 at 23:22
• @Suprateek Das, I think was the better answer, good change ;) – airhuff Jan 29 '17 at 8:25

Acidity of a weak acid is relative by definition (e.g. Ka). The wording of your question may not be quite right; you really mean "proton" or $\ce{H+}$ rather than hydrogen atom leaving or we'd be talking about radicals, not ions. But your statement is correct in that if you had a cyclohexane ring for example rather than a benzene ring, then you would not get the stabilization as you discussed. Thus, that hydrogen is much more acidic (meaning it leaves as $\ce{H+}$) because of the aromatic functionality than it would be in a compound without such a negative-charge-stabilizing functional group.

• There's no mesomeric stabilisation in this case. OP is wrong. – Mithoron Jan 24 '17 at 12:40
• It looks right to me. But if you can show in an electron-pushing diagram how it's wrong I will consider deleting my answer. – airhuff Jan 24 '17 at 19:48
• Electron pushing isn't the complete story here - as long as the C=O remains coplanar with the ring, the negative charge will reside in a sp2 orbital that is orthogonal to the pi system of the ring, so it cannot enjoy the stabilisation by delocalisation into the ring. – orthocresol Jan 24 '17 at 22:27
• Good point @orthocresol. I think I may pull this answer in favor of Jan 's. I don't know the ramifications for the OP having already accepted this answer, but I assume it's no problem and I can just delete it. – airhuff Jan 24 '17 at 22:47
• @airhuff You can’t delete an accepted answer afaik. You’ll have to wait for OP to mark the other answer accepted. – Jan Jan 24 '17 at 23:11