Topics

Hydrolysis of esters and amides

What hydrolysis is

Hydrolysis is splitting a molecule apart using water.
An ester hydrolyses back to the carboxylic acid and the alcohol it was made from.
An amide hydrolyses back to the carboxylic acid and the amine (or ammonia).
It is the reverse of the reaction that made them, which is why it always regenerates the two original halves.

Four cases to know

Starting material	Conditions	Products
ester	$\text{H}_2\text{O}/\text{H}^+$ , reflux	carboxylic acid + alcohol — reversible
ester	$\text{H}_2\text{O}/\text{OH}^-$ , reflux	carboxylate salt + alcohol — complete
amide	$\text{H}_2\text{O}/\text{H}^+$ , reflux	carboxylic acid + ammonium salt
amide	$\text{H}_2\text{O}/\text{OH}^-$ , reflux	carboxylate salt + free amine

All four need heating under reflux. Hydrolysis is slow at room temperature.

The rule that ties them together

Whichever product could react with the conditions, does.

In acid conditions, a basic product — an amine — is protonated to its salt: $\text{R}\text{NH}_2 \rightarrow \text{R}\text{NH}_3^+$ .
In basic conditions, an acidic product — a carboxylic acid — is deprotonated to its salt: $\text{RCOOH} \rightarrow \text{RCOO}^-$ .
So you never get a free acid in base, and you never get a free amine in acid. Show the charges.

Ester hydrolysis in detail

In acid — the exact reverse of esterification, so it is reversible:

$\text{RCOOR}' + \text{H}_2\text{O} \rightleftharpoons \text{RCOOH} + \text{R}'\text{OH}$

In base — the carboxylate salt forms, and the reaction goes to completion:

$\text{RCOOR}' + \text{NaOH} \rightarrow \text{RCOO}^-\text{Na}^+ + \text{R}'\text{OH}$

Why base hydrolysis goes to completion: the acid product is immediately converted to its carboxylate salt, which cannot react back with the alcohol. Removing a product this way drives the reaction forward.
To obtain the free acid afterwards, acidify the mixture with dilute HCl.
Base hydrolysis of an ester is therefore the preferred method when you want a complete conversion.

Amide hydrolysis in detail

In acid:

$\text{RCONH}_2 + \text{H}_2\text{O} + \text{H}^+ \rightarrow \text{RCOOH} + \text{NH}_4^+$

The ammonia or amine is released as its ammonium salt, because the acidic conditions protonate it.

In base:

$\text{RCONH}_2 + \text{NaOH} \rightarrow \text{RCOO}^-\text{Na}^+ + \text{NH}_3$

The free ammonia is released, and can be detected as a pungent gas turning damp red litmus blue.
The acid appears as its carboxylate salt.
Amides are harder to hydrolyse than esters and need more vigorous conditions — prolonged reflux with concentrated acid or alkali.

Tips and Tricks

Get the charges right. This is where most hydrolysis marks are lost. In base the organic product is $\text{RCOO}^-$ , not RCOOH. In acid the nitrogen product is $\text{RNH}_3^+$ , not $\text{RNH}_2$ .

Say heated under reflux. Hydrolysis at room temperature is far too slow, and the condition is worth a mark.

Use the right arrow. Acid hydrolysis of an ester is reversible — draw $\rightleftharpoons$ . Base hydrolysis goes to completion — draw $\rightarrow$ .

Be able to explain why base hydrolysis is complete: the acid is converted to its salt, which cannot react back with the alcohol, so the reaction cannot reach equilibrium.

Split the molecule in the right place. For an ester, break the bond between the carbonyl carbon and the single-bonded oxygen — the carbonyl carbon stays with the acid. For an amide, break the bond between the carbonyl carbon and the nitrogen.

Worked ExampleHydrolysing a dipeptide two ways

A dipeptide has the structure below, in which two amino acid units are joined by an amide (peptide) link.

$\text{H}_2\text{N}-\text{CH}(\text{CH}_3)-\text{CO}-\text{NH}-\text{CH}_2-\text{COOH}$

Draw the organic products formed when the dipeptide is heated under reflux with

(a) dilute hydrochloric acid, (b) aqueous sodium hydroxide.

Step 1 — Locate the bond that breaks

The amide (peptide) link is the $-\text{CO}-\text{NH}-$ in the middle. Hydrolysis breaks the bond between the carbonyl carbon and the nitrogen.

The carbonyl carbon stays with the left-hand fragment, and the nitrogen goes with the right-hand one.

Splitting there and adding the elements of water gives the two amino acids:

from the left: $\text{H}_2\text{N}-\text{CH}(\text{CH}_3)-\text{COOH}$
from the right: $\text{H}_2\text{N}-\text{CH}_2-\text{COOH}$

These are the neutral forms. Now apply the conditions.

Step 2 — Part (a), hydrolysis in acid

In acidic conditions any basic group is protonated. Each fragment carries an $-\text{NH}_2$ group, which is basic, so both are protonated to $-\text{NH}_3^+$ .

The $-\text{COOH}$ groups are not affected — the conditions are already acidic.

Step 3 — Part (b), hydrolysis in base

In basic conditions any acidic group is deprotonated. Each fragment carries a $-\text{COOH}$ group, which is acidic, so both are deprotonated to $-\text{COO}^-$ .

The $-\text{NH}_2$ groups are not protonated — there is no acid present to do it.

Step 4 — Check the pattern

The same bond breaks in both cases and the same two fragments are produced. Only the charges differ, and they differ in exactly the way the conditions dictate:

	$-\text{NH}_2$ group	$-\text{COOH}$ group
in acid	protonated to $-\text{NH}_3^+$	unchanged
in base	unchanged	deprotonated to $-\text{COO}^-$

An answer that draws the neutral amino acids in both cases has done the hard part — finding the bond — and then lost the marks on the easy part. Always ask what the conditions would do to each product.