6.1.3 Carboxylic Acids and Esters

The –COOH group can form hydrogen bonds with water, so the smaller carboxylic acids dissolve readily. The carbonyl oxygen can accept a hydrogen bond and the O–H group can donate one.

As the hydrocarbon part of the molecule gets larger, that non-polar part becomes a bigger fraction of the molecule and water solubility falls. OCR does not require a comparison of the acid strengths of different carboxylic acids here.

Carboxylic acids react with reactive metals in the familiar acid–metal pattern: carboxylate salt + hydrogen. The acidic H is the hydrogen on the –COOH group.

With sodium, for example: 2CH₃COOH + 2Na → 2CH₃COONa + H₂. Two acid molecules are needed to make one H₂ molecule.

A carboxylic acid reacts with a carbonate to form a carboxylate salt, water and carbon dioxide. The visible observation is effervescence as CO₂ is released.

For example: 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂. This is also useful in qualitative analysis because phenol is too weakly acidic to give this carbonate reaction.

A basic metal oxide neutralises a carboxylic acid to give a carboxylate salt and water. There is no hydrogen gas because this is an acid–base reaction, not an acid–metal reaction.

Using magnesium oxide as an example: 2CH₃COOH + MgO → (CH₃COO)₂Mg + H₂O. Check the metal charge when you write the salt formula.

With an alkali such as NaOH, the reaction is carboxylic acid + alkali → carboxylate salt + water. One mole of a monocarboxylic acid reacts with one mole of NaOH.

For ethanoic acid: CH₃COOH + NaOH → CH₃COONa + H₂O. Keep the COO part together when you write the salt; the acidic H has been replaced by the metal ion.

Heat the carboxylic acid with the alcohol under reflux using an acid catalyst, commonly concentrated H₂SO₄. The reaction is reversible: carboxylic acid + alcohol ⇌ ester + water.

For example, ethanoic acid and ethanol form ethyl ethanoate and water. Reflux lets you heat the mixture without losing volatile reactants; it does not make an equilibrium reaction go to completion.

An acid anhydride reacts with an alcohol to give an ester plus a carboxylic acid. One acyl group becomes part of the ester; the other becomes the acid.

For ethanoic anhydride and ethanol: (CH₃CO)₂O + CH₃CH₂OH → CH₃COOCH₂CH₃ + CH₃COOH. Read the alcohol side carefully when you build the ester name and structure.

For acid hydrolysis, heat the ester with dilute aqueous acid. The products are the carboxylic acid and alcohol, and this is the reverse of esterification, so show the reaction as an equilibrium.

Ethyl ethanoate gives ethanoic acid and ethanol: CH₃COOCH₂CH₃ + H₂O ⇌ CH₃COOH + CH₃CH₂OH. The acid is a catalyst, not a product in the equation.

Heat the ester with aqueous alkali such as NaOH. The products are an alcohol and a carboxylate salt, not the free carboxylic acid.

For ethyl ethanoate: CH₃COOCH₂CH₃ + NaOH → CH₃COONa + CH₃CH₂OH. If a later step acidifies the mixture, the carboxylate can then be converted to the carboxylic acid.

OCR uses SOCl₂ (thionyl chloride) to convert a carboxylic acid into an acyl chloride. Replace the –OH of –COOH by Cl while keeping the carbonyl oxygen in place.

The full equation is RCOOH + SOCl₂ → RCOCl + SO₂ + HCl. For ethanoic acid, the acyl chloride is CH₃COCl — not CH₃COOCl.

An acyl chloride reacts readily with an alcohol to form an ester and HCl: RCOCl + R′OH → RCOOR′ + HCl. You do not need an acid catalyst for this route.

Acyl chlorides can also esterify phenol, which is not readily esterified by a carboxylic acid. That makes acyl chlorides particularly useful when you need to put an acyl group onto an –OH group in synthesis.

Water hydrolyses an acyl chloride rapidly to give a carboxylic acid and HCl. The carbonyl carbon stays in the same place and Cl is replaced by OH.

The general equation is RCOCl + H₂O → RCOOH + HCl. This reaction is why acyl chlorides must be kept dry if you want them to remain as acyl chlorides.

With ammonia, an acyl chloride forms a primary amide. The –Cl is replaced by –NH₂ on the acyl carbon.

A useful overall equation with excess ammonia is RCOCl + 2NH₃ → RCONH₂ + NH₄Cl. One ammonia molecule becomes part of the amide and another removes the HCl formed.

A primary amine reacts with an acyl chloride to form a secondary amide. In the product, the acyl group is attached to the nitrogen and that nitrogen still carries one H.

You can show the core reaction as RCOCl + R′NH₂ → RCONHR′ + HCl. If excess amine is present, a second amine molecule can accept the HCl. The key structural check is the amide link –CONH–.