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Carboxylic acids and their derivatives

Covers AS91391

The family

A carboxylic acid, $-\text{COOH}$ , has three derivatives in this standard: the acyl chloride, the ester and the amide.
All four have a carbonyl carbon bonded to a second atom, and that atom is what names the family.

Carboxylic acids are acidic

The O–H of the $-\text{COOH}$ group is polar enough for the hydrogen to be released as $\text{H}^+$ , so carboxylic acids are weak acids.
They show all the usual acid reactions.
With sodium carbonate — the standard test:

$2\text{RCOOH} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{RCOONa} + \text{H}_2\text{O} + \text{CO}_2$

The observation is bubbles of carbon dioxide. No other family on this course fizzes with carbonate.
With sodium hydroxide — a neutralisation giving a salt and water:

$\text{RCOOH} + \text{NaOH} \rightarrow \text{RCOONa} + \text{H}_2\text{O}$

The salt is ionic, so it is far more soluble in water than the acid was. This is the basis of separating an acid from a neutral compound.

Making an acyl chloride

Thionyl chloride, $\text{SOCl}_2$ , replaces the acid's –OH with a –Cl:

$\text{RCOOH} + \text{SOCl}_2 \rightarrow \text{RCOCl} + \text{SO}_2 + \text{HCl}$

Both by-products are gases, so they escape and the reaction goes to completion.
Note that $\text{SOCl}_2$ does the same job on an alcohol, giving a haloalkane — in both cases it swaps an –OH for a Cl.

Acyl chlorides are the reactive one

An acyl chloride reacts readily with anything carrying an –OH or an –NH group.

Reagent	Product	Also formed
water	carboxylic acid	HCl — steamy fumes
an alcohol	ester	HCl
conc $\text{NH}_3$	amide	HCl
a primary amine	N-substituted amide	HCl

$\text{RCOCl} + \text{R}'\text{OH} \rightarrow \text{RCOOR}' + \text{HCl}$
$\text{RCOCl} + \text{NH}_3 \rightarrow \text{RCONH}_2 + \text{HCl}$
$\text{RCOCl} + \text{R}'\text{NH}_2 \rightarrow \text{RCONHR}' + \text{HCl}$
The reaction with water is the test for an acyl chloride. Adding a few drops of water produces steamy fumes of hydrogen chloride immediately, at room temperature.

Two routes to an ester

From the acid: acid + alcohol, catalysed by concentrated $\text{H}_2\text{SO}_4$ , refluxed. Reversible, so the yield is limited.
From the acyl chloride: acyl chloride + alcohol. Not reversible, and it happens at room temperature, so the yield is much better.
If a question asks for the better route to an ester, the acyl chloride route is the answer — at the cost of one extra step.

Tips and Tricks

The carbonate test is specific to the carboxylic acid. Alcohols, esters, aldehydes and ketones all give no reaction, so fizzing identifies the acid unambiguously. Be careful with an acyl chloride, though — it hydrolyses in the aqueous carbonate to give the acid, which then fizzes too. Test for the acyl chloride with water alone first.

For the acyl chloride test say steamy fumes, and identify them as hydrogen chloride. "It reacts vigorously" is not an observation.

Include HCl as a product in every acyl chloride equation. Leaving it out unbalances the equation.

Remember $\text{SOCl}_2$ appears twice on the reaction map, converting an alcohol to a haloalkane and an acid to an acyl chloride. In both cases it replaces –OH with –Cl.

When asked to compare the two ester routes, the marks are for reversibility: esterification is an equilibrium and gives a limited yield, whereas the acyl chloride route goes to completion.

Worked ExampleChoosing and justifying a synthetic route

A chemist needs to prepare ethyl propanoate from propanoic acid in as high a yield as possible.

(a) Give the equation and conditions for the direct esterification. (b) Give a two-step route via an acyl chloride, with equations. (c) Justify which route gives the higher yield.

Part (a) — the direct route

Propanoic acid reacts with ethanol, catalysed by concentrated sulfuric acid, heated under reflux:

$\text{CH}_3\text{CH}_2\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O}$

Note the equilibrium arrows — this is the key feature of the route.

Part (b) — the two-step route

Step 1 — make the acyl chloride using thionyl chloride:

$\text{CH}_3\text{CH}_2\text{COOH} + \text{SOCl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{COCl} + \text{SO}_2 + \text{HCl}$

Both by-products are gases and escape, so this step goes essentially to completion.

Step 2 — react the acyl chloride with ethanol, at room temperature:

$\text{CH}_3\text{CH}_2\text{COCl} + \text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_3 + \text{HCl}$

Note the single arrow — this reaction is not reversible.

Part (c) — justifying the choice

The direct route is limited by equilibrium. Esterification is reversible: as the ester and water build up, the reverse reaction (hydrolysis of the ester) becomes significant, and the mixture settles at an equilibrium position containing substantial amounts of unreacted acid and alcohol.

Even with the sulfuric acid absorbing some of the water, the yield is typically only around 60–70 %, and the product must be separated from the leftover reactants.

The acyl chloride route goes to completion.

Step 1 is driven forward because $\text{SO}_2$ and HCl are gases that leave the reaction mixture, so they cannot react back.
Step 2 is not an equilibrium — the acyl chloride is far more reactive than the acid, and the reaction proceeds to completion at room temperature.

With no equilibrium limiting either step, the overall yield is considerably higher.