Question

How do anhydrides react with water?

Answer 1

At the carbonyl carbon that belongs to the bigger half.

A general anhydride looks like this:

The leaving group is highlighted, assuming `R'` is longer than `R`.

A general rule of thumb is that the longer the alkyl chain of the `R` group, the higher the pKa of the conjugate acid of the proposed leaving group and the better the leaving group it is. You can see this when comparing methanoic acid (pKa = `3.75`, one carbon) and pentanoic acid (pKa = `4.84`, five carbons).

Since we are assuming that `R'` is longer than `R`, let's attack the carbonyl right now and remove the `R'` carboxylate group.

An anhydride literally lacks water. (anhydrous = without water.)

Let's add water.

1. Water attacks the carbonyl carbon. Which one? Pick the one that makes the carboxylate with the `R'` leave, because that's what we're assuming will happen. This is reversible, and actually is slightly favored in the reverse direction, so we should add quite a bit of water to push the equilibrium forward.
2. Proton transfer via a suitable base. Naturally, the water we added. But this is still an equilibrium.
3. Something has to leave to stabilize this tetrahedral intermediate. We don't want the water we just added to gain a proton and then leave. That's the reverse reaction. And of course, `"O"^(2-)` is a much stronger base than `"R"'"OO"^(-)` simply by looking at its charge. So, the `\mathbf(R"'")` carboxylate group must leave. At this point it's irreversible.
4. The carboxylate would want a proton. Why? The pKa of a typical carboxylic acid is about `5`, but the pKa of hydronium is about `-1.7`. The weaker acid exists in its acidic form predominantly. Hence, hydronium gives up its proton nicely.

You can now see that two carboxylic acids form.

CHALLENGE: Can you draw the mechanism for the dehydration of two carboxylic acids into one anhydride? Do you think it requires heat? How about acid?