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If I understand everything correctly, carboxylic acids behave like acids because oxygen is much more electronegative than hydrogen, so it takes his electron and releases H+. The problem is that alcohols have an OH group too, so why don't they behave like acids?

Jan
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roymend
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1 Answers1

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The figure below illustrates these two ionization reactions. The key difference between the two is that when the carboxylic ionizes the carboxylate anion is formed. The carboxylate anion is much more stable than the alkoxide anion formed when the alcohol ionizes. The carboxylate anion is much more stable because we can draw resonance structures to describe it. This resonance stabilization of the carboxylate anion explains why carboxylic acids are so much more acidic than an alcohol.

enter image description here

ron
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  • Although the answer is fine, there is still something weird when we got into inorganic acids. How would we justify that $\ce{HClO4}$ is so acid when it has no resonance structures (hibridisation is $sp^3$)? –  Oct 12 '18 at 16:01
  • @santimirandarp The $\ce{ClO4}$ anion is highly resonance stabilized (just google for drawings). This enhanced stability of the perchlorate anion explains why the acid is so highly ionized. – ron Oct 15 '18 at 19:46
  • There is no resonance without using $d$ orbitals. That's my point.. –  Oct 15 '18 at 19:47
  • @santimirandarp Sorry, perhaps I'm not following you. You asked why $\ce{HClO4}$ is so acidic and I said because of resonance stabilization of the perchlorate anion. Hybridization $\ce{sp^3}$ for chlorine is OK, no need for d-orbitals. – ron Oct 15 '18 at 20:02
  • No problem. Perchlorate anion has four simple bonds, so it is $sp^3$ and all oxygens are with same partial charge $-1$ there is no resonance...just one structure. Am I wrong? If we say there is a double bond, the only way possible is by means of $d$ orbitals... –  Oct 15 '18 at 20:06
  • @santimirandarp I think you are asking how is the tetrahedral geometry found in the perchlorate anion consistent with $\ce{sp^3}$ hybridization and resonance stabilizaton. Permeakra's answer in this earlier Q&A might be helpful. If not you might want to post a new question. It is a very different question than the one above. – ron Oct 15 '18 at 21:32
  • thanks, yes it was. So there is no resonance, at least it doesnt follows from that answer. In any case, it should be interesting to see some calculations probably... –  Oct 15 '18 at 23:14
  • @santimirandarp Yes, I suspect this is one of those cases were traditional hybridization approaches may break down and calculations would be informative. – ron Oct 15 '18 at 23:46
  • hmm... yes but not that strange. $\ce{H2SO4}$ is another example...i.e no stabilization by resonance. (it is not a downside of your answer, I'm just trying to understand the whole picture). –  Oct 20 '18 at 19:27