# What is incorrect about this depiction of the hydrolysis of an ester?

First of all, I'd like to apologize for again asking a question solely featuring the deficiencies of my textbook.

I have a few specific questions, and a general one about the image at the bottom of this question:

1. Is the product really an acid? It doesn't even have an $H^+.$
2. It's stated that hydrolysis can happen when esters are treated with an acid as well as a base. But acids only donate $H^+$ so where would the extra $O$ come from to produce an alcohol as a product?
3. It's stated that a molecule of water is produced. Is this statement incorrect, or is the diagram incorrect?
4. In general, how would we change the diagram/information below to make it correct?

If you can answer any/all of these questions, I would greatly appreciate it.

Thanks again for this professional community.

• In Figure 8, the alcohol oxygen should not be red.
– ron
Jul 21 '16 at 21:55
• Can I ask what the book is as it seems poorly written? Nov 1 '16 at 22:05
• @Beerhunter It's Nelson Chemistry 12, a terrible, terrible high school textbook from Ontario, Canada. Nov 3 '16 at 2:12

## 1 Answer

There are indeed two mechanisms for ester hydrolysis, one using base catalysis and one using acid catalysis. The mechanisms of both are provided below (as an aside, you mention that your textbook isn't great- you should consider getting a different one, even if your teacher gives assignments out of the other book, it'll give you a second opinion and something to cross check your book against).

Base catalysed ester hydrolysis

Source: Organic Chemistry (Wothers and Clayden)

Acid catalysed ester hydrolysis

Source: Organic Chemistry (Wothers and Clayden)

Is the product really an acid? It doesn't even have an H+.

Yes, it is an acid. The acid is actually initially formed however because of the basic conditions of the base catalysed mechanism it gets deprotonated to give the carboxylate. Upon workup the carboxylate salt (in your example the sodium salt of the carboxylate) would protonate to give the carboxylic acid (COOH).

It's stated that hydrolysis can happen when esters are treated with an acid as well as a base. But acids only donate H+ so where would the extra O come from to produce an alcohol as a product?

The product is a gain a carboxylic acid, and as you can see from the mechanism above, water is involved, which is where the additional oxygen comes from. Its worth noting that with the acid catalysed mechanism the whole thing is reversible.

It's stated that a molecule of water is produced (in the base catalysed mechanism). Is this statement incorrect, or is the diagram incorrect?

It could be water or an alcohol depending on whether hydroxide or the alkoxide does the deprotonation. Water is most likely however, as shown above. In the acid catalysed mechanism it is the alcohol that is formed.

In general, how would we change the diagram/information below to make it correct?

See the mechanisms above, taken from Clayden and Wothers. In general the diagram in your textbook looks essentially 'fine', though I dont entirely like the way the structures are drawn (its a very american textbook way of doing it.)

• Good answer! I'd reinforce that the book doesn't seem very didatic, when it doesn't show the mechanisms and write "acid" under the conjugate base of the acid. $\ce{ROO-}$ isn't an acid (actually it's a base), but it's a result from deprotonation of the carboxylic acid.
– IanC
Jul 21 '16 at 20:24
• I think the alcoxides are more nucleophilic than the base hydroxide though. At least when we conduct a base catalysed esterification reaction, we first react the base with alcohol, to form alcoxides, and the latter make the nucleophilic attack on the acylglycerides. If hydroxide was a stronger nucleophile, it would make more sense to use it directly and then react the formed carboxylic acid with the alcohol.
– IanC
Jul 21 '16 at 20:36
• @IanC, in what step are you referring to, and for the acid/the base? I don't quite understand Jul 21 '16 at 20:39
• Sorry, I was a little confusing in my comment. For the step of deprotonation of the base catalysed hydrolysis, you said that water was more likely to deprotonate the carboxylic acid instead of the alcoxide. If the concentration of hydroxide is much greater than the concentration of alcoxide this makes sense (which I think is the case in the reaction discussed), but it's good to point out that alcoxides are stronger bases than hydroxides and if present in the same ammounts will more readily attack the carboxylic acid.
– IanC
Jul 21 '16 at 20:49
• @NotEvans. Wouldn't be better to say that the first mechanism is base-induced and not base-catalyzed as the $\text{OH}^{-}$ isn't produced at the final step? Sep 28 '20 at 16:04