0
$\begingroup$

In Organic Chemistry mechanisms, I've commonly seen water getting protonated to complete a mechanism, and I'm getting a bit confused on how water can easily get protonated.

From the autoionization of water, I've learned that water is in equilibrium with $\ce{H3O+}$ and $\ce{OH-}$, but it has an equilibrium constant of $K_\text{eq} = 10^{-14}$. So more often than not I suppose, water should not be protonated and the $\Delta G^\circ$ at standard state for protonating/deprotonating water would be rather large and positive.

In the mechanisms I've seen, the protonation of water happens in some sort of acidic conditions. For example, the mechanism of the oxidation of alcohols with $\ce{NaOCl}$ (from Wade "Organic Chemistry" 9th Edition pg 508; I made some annotations to illustrate my question)

Mechanism of oxidation of alcohols with NaOCl taken from Wade "Organic Chemistry" 9th edition pg 508

In the second step of "Formation of an alkyl hypochlorite derivative," I see water come in and deprotonate the oxygen. The way I'm seeing it, such acid/base reactions can only happen if they result in the formation of weaker conjugate acids and weaker conjugate bases. So does that mean the intermediate they show which gets deprotonated is a stronger acid than the $\ce{H3O+}$ that is formed (and I believe $\ce{H3O+}$ has a $\mathrm{p}K_\mathrm{a}=0$ so that intermediate would have a $\mathrm{p}K_\mathrm{a} < 0$)? And the alkyl hypochlorite that is formed is a weaker base than $\ce{H2O}$ (and I think $\ce{H2O}$ has a $\mathrm{p}K_\mathrm{b} = 14$ so that alkyl hypochlorite would have a $\mathrm{p}K_\mathrm{b} > 14$)?

Something bothers me about thinking that the intermediate is a stronger acid than $\ce{H3O+}$ and the alkyl hypochlorite is more basic than water, and I'm not sure I'm thinking about this in the right way. If acidic conditions make water more basic and more likely to be protonated into $\ce{H3O+}$, how come it starts taking protons from the intermediate in this mechanism instead of just taking protons from the acid that induced acidic conditions (which I think is $\ce{CH3COOH}$ in this case)? Is that intermediate in the mechanism really more acidic than the $\ce{H3O+}$ that forms?

And why do the acidic conditions of lots of protons dissolved in solution lead to more protonation of water -- wouldn't the dissociation of the acid (that's causing acidic conditions in the mechanism) lead to a greater amount of $\ce{H3O+}$? So shouldn't that shift the equilibrium in (2) to the left, leading to less protonated water?

$$\ce{CH_3COOH + H_2O <<=> CH_3COO^- + H_3O^+ (1)}$$ $$\ce{2H_2O <<=> H_3O^+ + OH^- (2)}$$

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
2
$\begingroup$

From the autoionization of water, I've learned that water is in equilibrium with $\ce{H3O+}$ and $\ce{OH-}$, but it has an equilibrium constant of $K_\text{eq} = 10^{-14}$.

The value of $K_\text{w} = [\ce{H+}][\ce{OH-]} = 10^{-14}$ is, on its own, meaningless. You could obtain that by setting $[\ce{H+}] = 1$ and $[\ce{OH-}] = 10^{-14}$, which actually corresponds to a case where quite a lot of water is protonated. You could also obtain that by setting $[\ce{H+}] = 10^{-14}$ and $[\ce{OH-}] = 1$, which means that very little water is protonated at all. The product being $10^{-14}$ isn't what matters; it's the individual concentrations which matter. An 'acidic' solution would lean far more towards the former than the latter.

The way I'm seeing it, such acid/base reactions can only happen if they result in the formation of weaker conjugate acids and weaker conjugate bases.

This is true, but only from a thermodynamic perspective, i.e., it only applies to the net reaction, or the equilibrium position. At any single point in time, protons are flying around all over the place. If it's the only way the mechanism can go forward, it's not illogical to protonate something, even if the equilibrium position doesn't favour it outright.

The point is, even if you only have a tiny concentration of a reactive intermediate, it's more relevant than a large concentration of unreactive stuff.

If acidic conditions make water more basic [...]

It doesn't make water 'more basic'; you're confusing the basicity of water with the position of the equilibrium.

If you're sitting on a seesaw, and someone heavy comes and sits on the other side, sending you flying upwards, it's not because you got lighter.

And why do the acidic conditions of lots of protons dissolved in solution lead to more protonation of water -- wouldn't the dissociation of the acid (that's causing acidic conditions in the mechanism) lead to a greater amount of $\ce{H3O+}$? So shouldn't that shift the equilibrium in (2) to the left, leading to less protonated water?

This is a common misconception. The issue is that in a neutral solution of water, there is not much room for that equilibrium to go to the left.

$$\begin{align} \ce{HA + H2O &<=> A- + H3O+} \tag{1} \\ \ce{2H2O &<=> H3O+ + OH-} \tag{2} \end{align}$$

At pH 7, before you add any acid, you have $[\ce{H3O+}] = [\ce{OH-}] = 10^{-7}$. Now, if you add $\pu{0.1 M}$ of some acid to the solution, sure, you can push this equilibrium of $(2)$ 'to the left'. But by how much? There's only $\pu{10^{-7} M}$ of hydroxide for this new acid to react with. Even if you assume that all of this hydroxide reacts up, it's barely enough to make a dent in all the extra acid you added.

For all intents and purposes, then, you can ignore the autodissociation of water. The concentrations involved in this reaction are completely insignificant when compared against the concentrations involved in the other reaction, namely the acid dissociation, $(1)$.

Improve this answer
$\endgroup$
5
  • $\begingroup$ This is why the rate of reactions often is exquisitely sensitive to pH - for every pH unit, a reaction step might be 10 times faster or slower. Here, the acidic pH helps with the protonation of the hypochlorous acid, but makes the deprotonation to the alkyl hypochlorite more difficult. In solution, you have to choose a single pH as a compromise. Enzyme, on the other hand, are capable of protonating one location while deprotonating another by immobilizing acids and bases in specific positions of the active site. $\endgroup$
    – Karsten ♦
    Commented Aug 31, 2022 at 9:36
  • $\begingroup$ So then what I'm getting is that we have an acidic solution from dissociation of $CH_3COOH$ but that doesn't really shift the equilibrium of water much, so those acidic conditions don't impact whether water can be protonated? And while protonating water at an individual level may be energetically unfavorable (just to confirm: is just the protonation of water step by itself energetically unfavorable?) or kinetically slow (as @Karsten explained), in the context of the mechanism as a whole it can happen because the entire reaction is favorable? $\endgroup$
    – Baseball Fan
    Commented Aug 31, 2022 at 14:11
  • $\begingroup$ And the fact that this reaction needs acidic conditions doesn't really have to do with the protonation of water/deprotonation to the alkyl hypochlorite step (if anything, I suppose having acidic conditions makes protonating the water more difficult)? But the fact that we have acidic conditions is really just helping to protonate to hypochlorous acid? $\endgroup$
    – Baseball Fan
    Commented Aug 31, 2022 at 14:11
  • $\begingroup$ Also, is the reason we need to use $H_2O$ as the base instead of $OH^-$ because we have an acidic solution and some of the very little $OH^-$ has been neutralized (although I guess the amount neutralized is very small)? Even if we didn’t have an acidic solution, would there even be enough $OH^-$ to use as a base at neutral pH? $\endgroup$
    – Baseball Fan
    Commented Aug 31, 2022 at 14:12
  • $\begingroup$ In general, the steps that lead to a successful reaction are rare. However, these are the only steps we show, not the many collisions that don't even give the first step, or those that almost make it to completion but then go back to the starting point. $\endgroup$
    – Karsten ♦
    Commented Aug 31, 2022 at 14:20
-1
$\begingroup$

Instead of comparing or even considering dissociation of water into H+ and OH-, the key is consider relative reactivity of the the components in the equation to each other.

Protonation on NaOCl creates a leaving group of H2O, allowing Cl+ to "oxidize" or "bleach".

Remember these are equilibrium reactions.

In the second equation, ClOH2 will favor binding to an alkyl group because the alkyl group contributes negative charge a bit more than H. Water simply acts as a leaving group, and actually CH3COO- can be the proton acceptor in both steps, though it is listed only in the second step where the ketone is formed and Cl- is the leaving group.

Finally, HClO is a stronger acid than H2O because Cl is more strongly electron withdrawing than H.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Would the downvoter please step up and explain why they feel this answer is incorrect. $\endgroup$
    – Robert DiGiovanni
    Commented Aug 31, 2022 at 4:23
  • 3
    $\begingroup$ When you hover over the down-vote button, it says "this answer is not useful", it does not say "incorrect". I think there is a lot of good stuff in your answer. One individual, with the right to vote anonymously, did not think what you had (or what you had written before the edit) was a useful answer. Wasn't me (you have to trust me on that because voting is anonymous).... $\endgroup$
    – Karsten ♦
    Commented Aug 31, 2022 at 9:39
  • $\begingroup$ Appreciate that. It would be really helpful if the downvoter would leave a comment so I can improve my answer. I only did this stuff for a living for more than 20 uears. $\endgroup$
    – Robert DiGiovanni
    Commented Aug 31, 2022 at 17:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.