Negative charged intermediate in acidic conditions - what's the problem?

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](https://i.stack.imgur.com/EnqIS.png)](https://i.stack.imgur.com/EnqIS.png)
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.
> 
> 
> 


**Why?**




*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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