Why positively charged intermediates are preferred over negative charged intermediates, when formed during a reaction taking place in acidic conditions?

For example:

enter image description here Image source: Organic Chemistry by David R. Klein

Here, per Organic Chemistry by David R. Klein says,

That is, the mechanism does not exhibit any negative charges, which is consistent with acidic conditions.


  • 1
    $\begingroup$ What do you think happens to negatively charged species under acidic conditions? $\endgroup$
    – orthocresol
    Jan 2, 2017 at 17:42
  • $\begingroup$ What happens to a positively charged one? I am missing somewhere. $\endgroup$ Jan 2, 2017 at 17:55
  • 1
    $\begingroup$ Well, cations containing hydrogen are usually quite acidic and anions basic, aren't they? $\endgroup$
    – Mithoron
    Jan 2, 2017 at 19:55

1 Answer 1


Acidic conditions — whether you are using the Brønsted or the Arrhenius definition of acidity — is characterised by readily available protons. Remember the general dissolution ionisation of a Brønsted acid in water as shown below:

$$\ce{HA + H2O <=> H3O+ + A-}\tag{1}$$

The proton is readily liberated and transferred to species which are more happy to accept it — for strong acids, that means it is transferred to water, a really weak base. A proton typically gladly attaches to anything that has a negative charge, because those species are typically more basic than water in this example. Therefore, postulating an enolate or any other oxygen-centred anion in acidic solution is not considered correct since there will be enough protons to quench it.

However, this does not mean that no anions can exist in aquaeous solution. For example, a typical method to transform alcohols into iodides is refluxing in aquaeous hydrogen iodide. Here, $\ce{HI}$ is a very strong acid that very readily dissociates to give iodide anions, $\ce{I-}$. These, even though negatively charged, do exist in acidic conditions and are able to attack nucleophilicly. That’s all due to the high acidity of $\ce{HI}$.

$$\ce{R-OH + H3O+ + I- <<=> R-OH2+ + H2O + I- -> R-I + 2 H2O}\tag{2}$$


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