Carboxylic acids and esters

In a nutshell

Carboxylic acids contain the -COOH group. They are weak acids, they dissolve in water because they hydrogen bond with it, and they are the parent family of three useful relatives: esters, acid anhydrides and acyl chlorides.

This subtopic covers how carboxylic acids react as acids, how esters are made and broken down (hydrolysed), and why acyl chlorides are the go-to reagent for making esters and amides in synthesis.

Assumed knowledge: Acids, Bonding and structure, Alcohols.

Core content

Properties of carboxylic acids: the family at a glance

Every compound in this note has a C=O group with another electronegative atom on the same carbon. Only that second group changes.

In structural formulae, OCR writes the carboxyl group as COOH and the ester group as COO. Do the same.

Properties of carboxylic acids: solubility in water

Short-chain carboxylic acids are soluble in water because the -COOH group forms hydrogen bonds with water molecules.

The -COOH group can hydrogen bond to water in three ways:

• the δ+ H of the acid's O-H bonds to a lone pair on the O of a water molecule;
• a δ+ H of water bonds to a lone pair on the C=O oxygen;
• a δ+ H of water bonds to a lone pair on the O of the O-H group.

Chain length matters:

• The first four acids (methanoic to butanoic acid) mix with water in all proportions.
• As the chain gets longer, solubility decreases. The non-polar hydrocarbon chain cannot hydrogen bond with water, and it makes up more and more of the molecule.

A full answer names the hydrogen bonds, says they form between the COOH group and water molecules, and (if asked to draw them) shows the lone pairs, δ+/δ- and a dashed line from a δ+ H to a lone pair.

Properties of carboxylic acids: reactions with metals and bases

Carboxylic acids are weak acids: they partially dissociate in aqueous solution.

CH₃COOH(aq) ⇌ CH₃COO^-(aq) + H⁺(aq)

They still do everything a typical acid does, forming carboxylate salts. The salt is named from the acid: ethanoic acid gives ethanoates, propanoic acid gives propanoates.

Things that trip people up:

• Magnesium ethanoate is (CH₃COO)₂Mg. Mg²⁺ needs two ethanoate ions, so you need 2CH₃COOH.
• Sodium metal gives ½H₂ per acid: 2CH₃COOH + 2Na → 2CH₃COONa + H₂. Check the H count.
• As an ionic equation, the carbonate reaction is 2CH₃COOH(aq) + CO₃^2-(aq) → 2CH₃COO^-(aq) + H₂O(l) + CO₂(g).

You do not need to compare how acidic different carboxylic acids are.

PAG7: qualitative analysis of organic functional groups (testing for a carboxylic acid)

The reaction with a carbonate is OCR's test for a carboxylic acid.

1. Add aqueous sodium carbonate (or solid Na₂CO₃) to the unknown.
2. A carboxylic acid gives effervescence.
3. To confirm the gas is CO₂, bubble it through limewater: it turns milky (cloudy).

Record what you see: "effervescence" or "bubbles", not "CO₂ is made". The gas is the conclusion; the bubbles are the observation.

Esters: naming esters

An ester's name has two parts, and the order trips people up.

• The first word comes from the alcohol and ends in -yl (methyl, ethyl, propyl). It is the chain attached to the single-bonded O.
• The second word comes from the carboxylic acid and ends in -oate. It includes the C=O carbon.

So CH₃COOCH₂CH₃ is ethyl ethanoate, and CH₃CH₂COOCH₃ is methyl propanoate.

Notice the formula is written acid part first, but the name is said alcohol part first.

Esters: esterification of carboxylic acids with alcohols

A carboxylic acid and an alcohol react to form an ester and water, in the presence of an acid catalyst, usually concentrated sulfuric acid.

CH₃COOH + CH₃CH₂OH ⇌ CH₃COOCH₂CH₃ + H₂O

ethanoic acid + ethanol ⇌ ethyl ethanoate + water

The ester is then separated from the mixture (for example by distillation). Many small esters have sweet, fruity smells, which is why they are used in flavourings and perfumes.

Phenol is the exception. Phenol is not readily esterified by carboxylic acids. To make a phenyl ester you use an acyl chloride instead (see below).

Esters: esterification of acid anhydrides with alcohols

An acid anhydride also reacts with an alcohol to give an ester. The by-product is a carboxylic acid, not water.

(CH₃CO)₂O + CH₃CH₂OH → CH₃COOCH₂CH₃ + CH₃COOH

ethanoic anhydride + ethanol → ethyl ethanoate + ethanoic acid

Why use the anhydride instead of the acid:

• It is more reactive than the carboxylic acid.
• The reaction is not reversible, so the yield is higher.

One half of the anhydride becomes the ester; the other half leaves as the carboxylic acid.

Esters: hydrolysis of esters

Hydrolysis breaks the ester back into its two parts by reaction with water. Which products you get depends on whether you use acid or alkali.

Using methyl propanoate:

Acid: CH₃CH₂COOCH₃ + H₂O ⇌ CH₃CH₂COOH + CH₃OH

(methyl propanoate + water ⇌ propanoic acid + methanol; H⁺ is the catalyst)

Alkali: CH₃CH₂COOCH₃ + NaOH → CH₃CH₂COONa + CH₃OH

(methyl propanoate + sodium hydroxide → sodium propanoate + methanol)

If you want the free carboxylic acid after alkaline hydrolysis, add a strong acid to the salt: CH₃CH₂COONa + HCl → CH₃CH₂COOH + NaCl.

Still don't get it? · why acid and alkali give different products

Imagine swapping two stickers with a friend. If the swap is fair, either of you can swap back at any time, so you end up with a bit of both. Now imagine that after the swap, your friend puts their sticker in a locked album. They cannot swap back, so every swap is final.

Now the chemistry, one step at a time:

1. An ester plus water can split into a carboxylic acid plus an alcohol.
2. With acid, nothing stops the acid and alcohol joining back up into the ester. Both directions happen, so you get an equilibrium mixture: ester, water, acid and alcohol all present. That is the "fair swap".
3. With alkali, the carboxylic acid that forms reacts straight away with the OH^- ions to become a carboxylate ion (RCOO^-, as its sodium salt).
4. A carboxylate ion does not react with the alcohol to remake the ester. That is the "locked album": the backwards reaction cannot happen.
5. So in alkali the reaction keeps going until the ester is used up.

Exam wording: acid hydrolysis (hot aqueous acid) gives a carboxylic acid and an alcohol and is reversible; alkaline hydrolysis (hot aqueous alkali) gives a carboxylate salt and an alcohol and goes to completion.

Acyl chlorides: formation from carboxylic acids

A carboxylic acid is converted into an acyl chloride by reaction with thionyl chloride, SOCl₂. The -OH of the acid is replaced by -Cl.

CH₃COOH + SOCl₂ → CH₃COCl + SO₂ + HCl

ethanoic acid → ethanoyl chloride

Points worth knowing:

• Naming: -oic acid becomes -oyl chloride (propanoic acid → propanoyl chloride; benzoic acid → benzoyl chloride).
• Both by-products, SO₂ and HCl, are gases, so they leave the mixture easily. Both are toxic and acidic, so the reaction is done in a fume cupboard.
• Acyl chlorides react with water, so the reaction and the product must be kept dry.

Acyl chlorides: use in synthesis

Acyl chlorides are much more reactive than carboxylic acids. Their reactions are fast, happen at room temperature, and are not reversible, so they give high yields. That is why chemists convert an acid into its acyl chloride before making an ester or an amide.

Product names: ethyl ethanoate, phenyl ethanoate, ethanoic acid, ethanamide, N-methylethanamide.

The two kinds of amide:

• Primary amide, -CONH₂: the N is bonded to one carbon (the C=O carbon). Made from ammonia.
• Secondary amide, -CONH-: the N is bonded to two carbons. Made from a primary amine.

With water, alcohols and phenol, HCl is released and you see steamy (misty) white fumes. With ammonia and amines, the HCl reacts with a second molecule of the base to form a salt (NH₄Cl or CH₃NH₃Cl), which is why the equations need 2NH₃ or 2CH₃NH₂.

Esterifying phenol: phenol is not readily esterified by carboxylic acids, but it reacts readily with acyl chlorides. If a question asks how to make a phenyl ester, the answer is the acyl chloride route.

Still don't get it? · one pattern for every acyl chloride reaction

Think of a dance where one partner always leaves the floor holding hands with a newcomer's friend. Whoever cuts in, the same thing happens: the old partner walks off with a companion, and the newcomer takes their place.

Now the chemistry, step by step:

1. Write the acyl chloride as R-CO-Cl.
2. Write the other reactant as H-X: water is H-OH, ethanol is H-OCH₂CH₃, phenol is H-OC₆H₅, ammonia is H-NH₂, methylamine is H-NHCH₃.
3. The Cl leaves and takes the H with it: that makes HCl.
4. X takes the Cl's place on the C=O carbon: R-CO-X.
5. If X came from water you have an acid (R-COOH); from an alcohol or phenol, an ester (R-COOR'); from ammonia, a primary amide (R-CONH₂); from a primary amine, a secondary amide (R-CONHR').
6. Ammonia and amines are bases, so a second molecule of them reacts with the HCl to form a salt. That is why those equations need two of them.

Exam wording: acyl chlorides react with alcohols and phenols to form esters, with water to form carboxylic acids, with ammonia to form primary amides and with primary amines to form secondary amides, releasing HCl each time.

Worked examples

Example 1: naming esters and writing the esterification equation

(a) Write an equation for the reaction of butanoic acid with propan-1-ol, naming the ester. (b) Name the acid and the alcohol needed to make CH₃COOCH(CH₃)₂.

(a)

1. Acid part (keeps the C=O carbon): butanoic acid gives butanoate.
2. Alcohol part (on the single-bonded O): propan-1-ol gives propyl.
3. Join them, losing water, and use ⇌ because the reaction is reversible:

CH₃CH₂CH₂COOH + CH₃CH₂CH₂OH ⇌ CH₃CH₂CH₂COOCH₂CH₂CH₃ + H₂O

The ester is propyl butanoate. Catalyst: concentrated H₂SO₄.

Check by counting atoms: left C₄H₈O₂ + C₃H₈O = C₇H₁₆O₃; right C₇H₁₄O₂ + H₂O = C₇H₁₆O₃. Balanced.

(b)

1. Split at the single-bonded O: CH₃COO- | -CH(CH₃)₂.
2. CH₃COO- has 2 carbons including the C=O carbon: ethanoic acid.
3. -CH(CH₃)₂ is joined to the O through its middle carbon: propan-2-ol.

Where marks go: writing the name backwards ("butyl propanoate" in (a)) or leaving out the water in the equation each lose the mark.

Example 2: hydrolysing an ester in acid and in alkali

Ethyl propanoate, CH₃CH₂COOCH₂CH₃, is heated under reflux with (a) dilute sulfuric acid, (b) aqueous sodium hydroxide. Write an equation for each and name the organic products.

(a) Hot aqueous acid gives the carboxylic acid and the alcohol, reversibly:

CH₃CH₂COOCH₂CH₃ + H₂O ⇌ CH₃CH₂COOH + CH₃CH₂OH

Products: propanoic acid and ethanol.

(b) Hot aqueous alkali gives the carboxylate salt and the alcohol, and goes to completion:

CH₃CH₂COOCH₂CH₃ + NaOH → CH₃CH₂COONa + CH₃CH₂OH

Products: sodium propanoate and ethanol.

Where marks go: giving propanoic acid as the product in (b) loses the mark, because in alkali the acid ends up as its salt. Using → in (a), or ⇌ in (b), is also marked wrong.

Example 3: a two-step route through an acyl chloride

Show how propanoic acid can be converted into N-ethylpropanamide, CH₃CH₂CONHCH₂CH₃. Give reagents and equations.

Step 1: make the acyl chloride with SOCl₂:

CH₃CH₂COOH + SOCl₂ → CH₃CH₂COCl + SO₂ + HCl

Step 2: react propanoyl chloride with the primary amine ethylamine, CH₃CH₂NH₂:

CH₃CH₂COCl + 2CH₃CH₂NH₂ → CH₃CH₂CONHCH₂CH₃ + CH₃CH₂NH₃Cl

The product is a secondary amide: its N is bonded to two carbons.

Where marks go: using ammonia in step 2 would give the primary amide, propanamide, instead. Writing 1 CH₃CH₂NH₂ with HCl as the by-product is not the full equation, since the HCl reacts with a second molecule of the amine.

Example 4: percentage yield of an esterification (synoptic)

6.00 g of ethanoic acid is heated with excess ethanol and a few drops of concentrated sulfuric acid. After purification, 5.28 g of ethyl ethanoate is obtained. Calculate the percentage yield. (A_r: C 12.0, H 1.0, O 16.0, given on the Data Sheet.)

Equation (1 : 1 ratio of acid to ester):

CH₃COOH + CH₃CH₂OH ⇌ CH₃COOCH₂CH₃ + H₂O

Step 1: molar masses.

$M(\text{CH}_3\text{COOH}) = (2 \times 12.0) + (4 \times 1.0) + (2 \times 16.0) = 60.0\ \text{g mol}^{-1}$

$M(\text{CH}_3\text{COOCH}_2\text{CH}_3) = (4 \times 12.0) + (8 \times 1.0) + (2 \times 16.0) = 88.0\ \text{g mol}^{-1}$

Step 2: moles of the limiting reagent (the acid, because ethanol is in excess).

$n(\text{acid}) = \frac{6.00\ \text{g}}{60.0\ \text{g mol}^{-1}} = 0.100\ \text{mol}$

Step 3: theoretical moles and mass of ester (1 : 1).

$n(\text{ester})_{\text{theoretical}} = 0.100\ \text{mol}$

$m(\text{ester})_{\text{theoretical}} = 0.100\ \text{mol} \times 88.0\ \text{g mol}^{-1} = 8.80\ \text{g}$

Step 4: percentage yield.

$\%\ \text{yield} = \frac{5.28\ \text{g}}{8.80\ \text{g}} \times 100 = 60.0\ \%$

Answer: 60.0 % (3 significant figures, matching the data).

Why it is well below 100 %: the reaction is reversible, so the mixture reaches equilibrium before all the acid has reacted, and some ester is lost during purification. Making the ester from ethanoyl chloride or ethanoic anhydride instead avoids the equilibrium problem.

Where marks go: working from the moles of ethanol (the reagent in excess) gives a meaningless answer; always use the limiting reagent. Dividing the theoretical mass by the actual mass gives a yield over 100 %, which should tell you the fraction is upside down.

Common exam mistakes

• Explaining solubility as "carboxylic acids are polar" with no hydrogen bonds. The mark needs hydrogen bonds between the COOH group and water molecules.
• Drawing a hydrogen bond to or from an H on carbon. Only the δ+ H on O-H (of the acid or of water) takes part, and the bond goes to a lone pair on an O.
• Leaving out CO₂ or water in the carbonate equation, or writing H₂ as a product. Acid + carbonate gives salt + water + carbon dioxide.
• Getting the salt formula wrong: CH₃COOMg instead of (CH₃COO)₂Mg, with no 2 in front of the acid.
• Not balancing the sodium equation. 2CH₃COOH + 2Na → 2CH₃COONa + H₂; one Na per acid gives only ½H₂.
• Writing "CO₂ produced" as the observation in the carbonate test. The observation is effervescence; CO₂ is the conclusion (confirmed by limewater turning milky).
• Naming esters backwards. CH₃CH₂COOCH₃ is methyl propanoate, not "propyl methanoate". The -oate part includes the C=O carbon.
• Forgetting the water in an esterification equation, or using → instead of ⇌.
• Calling the catalyst "dilute H₂SO₄", or writing H₂SO₄ into the equation as a reactant. The catalyst is concentrated H₂SO₄ (an acid catalyst), and it goes over the arrow because it is not used up.
• Giving the carboxylic acid as the product of alkaline hydrolysis. In hot aqueous alkali you get the carboxylate salt (e.g. sodium propanoate) and the alcohol.
• Saying acid hydrolysis goes to completion. It is reversible; only alkaline hydrolysis goes to completion.
• Wrong by-products for SOCl₂. The by-products are SO₂ and HCl, not H₂O or Cl₂.
• Trying to esterify phenol with a carboxylic acid. Phenol is not readily esterified by carboxylic acids; use an acyl chloride.
• Using one NH₃ with NH₄Cl as a product (unbalanced). With ammonia or amines, the equation needs two molecules of the base.
• Mixing up primary and secondary amides. A primary amine gives a secondary amide (N bonded to two carbons). Ammonia gives the primary amide.
• Using the reagent in excess for a yield calculation. Work from the moles of the limiting reagent.

Key definitions

• Carboxylic acid: a compound containing the carboxyl group, -COOH.
• Weak acid: an acid that partially dissociates in aqueous solution.
• Hydrogen bond: an intermolecular attraction between a lone pair on an N, O or F atom and a δ+ hydrogen atom bonded to N, O or F in a neighbouring molecule.
• Salt: a compound formed when the H⁺ ion of an acid is replaced by a metal ion or an ammonium ion (for a carboxylic acid, a carboxylate salt such as sodium ethanoate).
• Ester: a compound containing the -COO- group, formed from a carboxylic acid (or its derivative) and an alcohol.
• Esterification: the reaction of a carboxylic acid with an alcohol, in the presence of an acid catalyst (e.g. concentrated H₂SO₄), to form an ester and water; also the reaction of an acid anhydride or an acyl chloride with an alcohol (or phenol) to form an ester.
• Hydrolysis (of an ester): the breaking of the ester by reaction with water; in hot aqueous acid it forms a carboxylic acid and an alcohol, and in hot aqueous alkali it forms a carboxylate salt and an alcohol.
• Acid anhydride: a compound containing the -COOCO- group, e.g. ethanoic anhydride, (CH₃CO)₂O.
• Acyl chloride: a compound containing the -COCl group, e.g. ethanoyl chloride, CH₃COCl; formed from a carboxylic acid with SOCl₂.
• Primary amide: an amide in which the nitrogen is bonded to one carbon atom, -CONH₂.
• Secondary amide: an amide in which the nitrogen is bonded to two carbon atoms, -CONH-.

Specification

• I can explain why carboxylic acids are soluble in water in terms of hydrogen bonding between the COOH group and water molecules. (6.1.3 a)
• I can write equations for the reactions of carboxylic acids in aqueous conditions with metals, carbonates, metal oxides and alkalis, naming the carboxylate salts formed. (6.1.3 b)
• I can use the reaction with a carbonate to identify a carboxylic acid (effervescence of CO₂). (6.1.3 b, PAG7, 6.3.1 c vii)
• I can describe and write equations for the esterification of carboxylic acids with alcohols using an acid catalyst (concentrated H₂SO₄), and name the ester. (6.1.3 c i)
• I can describe and write equations for the esterification of acid anhydrides with alcohols. (6.1.3 c ii)
• I can describe and write equations for the hydrolysis of esters in hot aqueous acid to form carboxylic acids and alcohols. (6.1.3 d i)
• I can describe and write equations for the hydrolysis of esters in hot aqueous alkali to form carboxylate salts and alcohols. (6.1.3 d ii)
• I can write an equation for the formation of an acyl chloride from a carboxylic acid using SOCl₂. (6.1.3 e)
• I can write equations for the reactions of acyl chlorides to form esters (including with phenol), carboxylic acids, and primary and secondary amides. (6.1.3 f)
• I can explain why phenol is esterified using an acyl chloride rather than a carboxylic acid. (6.1.3 f)

Related notes

• Acids: acids, bases, salts and weak acid dissociation
• Amount of substance: moles and percentage yield
• Bonding and structure: hydrogen bonding
• Alcohols: oxidising primary alcohols to carboxylic acids
• Organic synthesis: heating under reflux and distillation
• How far?: reversible reactions and K_c, which explain the limited yield of esterification
• Aromatic compounds: phenol and its weak acidity
• Carbonyl compounds: oxidising aldehydes to carboxylic acids
• Amines: the primary amines that react with acyl chlorides
• Amino acids, amides and chirality: structures of primary and secondary amides
• Polyesters and polyamides: ester and amide links in condensation polymers
• Carbon-carbon bond formation: acyl chlorides in Friedel-Crafts acylation
• Chromatography and qualitative analysis: the full set of functional group tests