# How do anhydrides react with water?

Jan 12, 2016

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.)

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.
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, ${\text{O}}^{2 -}$ is a much stronger base than ${\text{R"'"OO}}^{-}$ simply by looking at its charge. So, the $\setminus m a t h b f \left(R \text{'}\right)$ 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.