Esters

1. Overview

Esters give many fruits and flavourings their smell, and they are widely used as solvents. Every ester contains the group -COO-, and an ester can be made by joining a carboxylic acid to an alcohol, with a molecule of water lost. This section covers that reaction, esterification, and hydrolysis, which splits an ester again when it is heated with dilute acid or with dilute alkali. It starts with naming esters, which almost every question on these reactions needs.

Key Definitions

• Ester: an organic compound containing the ester group, -COO-, with the general formula RCOOR', where R is a hydrogen atom or a hydrocarbon group and R' is a hydrocarbon group.
• Esterification: the reaction of a carboxylic acid with an alcohol, catalysed by concentrated sulfuric acid, to form an ester and water; it is a reversible condensation reaction.
• Condensation reaction: a reaction in which two molecules join to form a larger molecule, with the loss of a small molecule such as water.
• Hydrolysis: a reaction in which a bond in a compound is broken by reaction with water, or with hydroxide ions in alkaline conditions; an ester is hydrolysed to a carboxylic acid (or its salt) and an alcohol.
• Acid hydrolysis of an ester: heating an ester under reflux with a dilute acid, such as dilute sulfuric acid, to form a carboxylic acid and an alcohol; the reaction is reversible, so it reaches equilibrium with some ester left.
• Alkaline hydrolysis of an ester: heating an ester under reflux with a dilute alkali, such as aqueous sodium hydroxide, to form an alcohol and the salt of the carboxylic acid (for example sodium ethanoate); with enough alkali, the reaction goes to completion.
• Heating under reflux: heating a mixture in a flask fitted with a vertical condenser, so that vapours condense and run back into the flask; volatile reactants and products are not lost, so the mixture can be heated for a long time.

Content

The ester group and its name

An ester has the general formula RCOOR'. R comes from the carboxylic acid and is a hydrogen atom or a hydrocarbon group. R' comes from the alcohol and is a hydrocarbon group (an alkyl group in AS examples), never a hydrogen atom: with H there, the compound would be the acid.

In the ester group, one carbon atom has a double bond to an oxygen atom and a single bond to a second oxygen atom, which is also bonded to a carbon atom of R'. So the group is C(=O)-O-C, with exactly two oxygen atoms.

The same ester can be written in several ways: CH₃COOCH₃, CH₃CO₂CH₃ and CH₃OOCCH₃ are all methyl ethanoate. To read any of them, find the oxygen with two single bonds. The side with the C=O carbon is the acid part; the other side is the alcohol part.

An ester's name has two words (the rules are in 13.1; at AS you name unbranched esters with up to six carbon atoms on each side):

1. First word, from the alcohol: the alkyl group, ending in -yl. Ethanol gives ethyl.
2. Second word, from the acid: -oic acid becomes -oate. Ethanoic acid gives ethanoate.
3. The C=O carbon counts as part of the acid.

Carbons	From the alcohol	From the acid
1	methyl	methanoate
2	ethyl	ethanoate
3	propyl	propanoate
4	butyl	butanoate
5	pentyl	pentanoate
6	hexyl	hexanoate

For example, HCOOCH₂CH₃ has a one-carbon acid part and an ethyl group, so it is ethyl methanoate, made from methanoic acid and ethanol. The name puts the alcohol part first, but most formulae put the acid part first, so do not read a formula from left to right as if it were the name.

An ester and a carboxylic acid with the same number of carbon atoms can be isomers: methyl ethanoate, ethyl methanoate and propanoic acid are all C₃H₆O₂.

Making esters: esterification

• Reagents: a carboxylic acid and an alcohol.
• Conditions: a few drops of concentrated sulfuric acid as catalyst, and heat (warm in a water bath for a test-tube sample; heat under reflux to make a larger amount).

Ethanoic acid and ethanol form ethyl ethanoate and water:

CH₃COOH(l) + CH₃CH₂OH(l) ⇌ CH₃COOCH₂CH₃(l) + H₂O(l)

The two molecules join and a water molecule is lost, so esterification is a condensation reaction. The water forms from the -OH of the acid and the H of the alcohol's -OH group. The alcohol's oxygen atom stays in the ester, linking the two parts.

• It is slow. Without a catalyst and heat, very little ester forms in a reasonable time.
• It is reversible. The mixture reaches equilibrium with acid, alcohol, ester and water all present, so the reaction is incomplete and the yield is below 100%. To convert more of one reactant, use an excess of the other, or remove a product as it forms. All four substances are in one liquid mixture, so water appears in the $K_{\text{c}}$ expression (see 7.1).
• Concentrated H₂SO₄ is the catalyst. It speeds up the forward and reverse reactions by the same factor, so equilibrium is reached sooner but its position does not move. In an answer, give its role in esterification as "catalyst". (It is in the same liquid as the reactants, so you may see it called a homogeneous catalyst. The AS outcomes do not ask you to use that term: see 8.3.)

Why concentrated acid and not dilute? Dilute acid is mostly water, and extra water pushes the equilibrium back towards the acid and alcohol. That is what you want for hydrolysis, below, but not for making an ester.

In the lab. Put the acid, the alcohol, a few drops of concentrated H₂SO₄ and anti-bumping granules in a round-bottomed flask, and heat under reflux. The condenser is vertical and open at the top, with cooling water in at the bottom and out at the top. Heat with a water bath or an electric heater, as alcohols and esters are flammable. To purify the crude ester, shake it with sodium carbonate solution to remove acids (it fizzes while acid remains), separate the layers, dry it and distil it. Each step loses some ester, so the yield falls further.

Worked example 1: naming and making an ester

Question: An ester has the structural formula CH₃(CH₂)₄COOCH₃. (a) Name it. (b) Name the carboxylic acid and the alcohol it is made from, and give the conditions. (c) Write the equation, with state symbols, and state the type of reaction.

Step 1: split the formula at the oxygen with two single bonds. One side is CH₃(CH₂)₄CO-, with 1 + 4 + 1 = 6 carbon atoms including the C=O carbon: hexanoate. The other side is CH₃-, a methyl group.

Step 2: name it, alcohol part first. Methyl hexanoate.

Step 3: parent compounds and conditions. Hexanoic acid, CH₃(CH₂)₄COOH, and methanol, CH₃OH, heated with a few drops of concentrated H₂SO₄ as catalyst.

Step 4: equation.

CH₃(CH₂)₄COOH(l) + CH₃OH(l) ⇌ CH₃(CH₂)₄COOCH₃(l) + H₂O(l)

Check the atoms: both sides contain C₇H₁₆O₃ in total. Two molecules join and water is lost, so this is a condensation reaction.

Answer: (a) methyl hexanoate, C₇H₁₄O₂; (b) hexanoic acid and methanol, heated with concentrated H₂SO₄ as catalyst; (c) the equation in Step 4; the type of reaction is condensation.

Hydrolysis of esters

Hydrolysis splits an ester back into its two parts. With water alone the reaction is very slow, so the ester is heated under reflux with a dilute acid or a dilute alkali. Ester hydrolysis is also a way to make alcohols (16.1) and carboxylic acids (18.1).

With dilute acid (for example, dilute sulfuric acid or dilute hydrochloric acid), heated under reflux:

CH₃COOCH₂CH₃(l) + H₂O(l) ⇌ CH₃COOH(aq) + CH₃CH₂OH(aq)

This is the reverse of esterification, with H⁺ ions as the catalyst. It is reversible, so it reaches equilibrium with some ester left. The large excess of water in the dilute acid moves the position of equilibrium towards the acid and the alcohol.

With dilute alkali (usually aqueous sodium hydroxide), heated under reflux:

CH₃COOCH₂CH₃(l) + NaOH(aq) → CH₃COONa(aq) + CH₃CH₂OH(aq)

As an ionic equation:

CH₃COOCH₂CH₃(l) + OH⁻(aq) → CH₃COO⁻(aq) + CH₃CH₂OH(aq)

• The products are the alcohol and the salt of the carboxylic acid (here sodium ethanoate), not the acid itself.
• With enough alkali (at least one OH⁻ for each ester group), the reaction goes to completion. Any carboxylic acid that forms reacts at once with OH⁻ to give the carboxylate ion, RCOO⁻, which does not react with the alcohol, so the ester cannot re-form.
• OH⁻ is a reactant and is used up. It is not a catalyst.

To obtain the carboxylic acid, add a dilute strong acid, such as dilute hydrochloric acid, after the hydrolysis:

CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq)

	Dilute acid, heat	Dilute alkali, heat
Organic products	carboxylic acid + alcohol	carboxylate salt + alcohol
Extent	reversible, incomplete	complete
H⁺ or OH⁻ is	a catalyst	a reactant

More than one ester group. Every ester group in a molecule can be hydrolysed, so complete alkaline hydrolysis needs one OH⁻ per ester group. The diester made from ethane-1,2-diol and ethanoic acid needs 2 mol of NaOH for each mole of diester:

CH₃COOCH₂CH₂OOCCH₃(l) + 2NaOH(aq) → HOCH₂CH₂OH(aq) + 2CH₃COONa(aq)

Worked example 2: identifying an ester from its hydrolysis

Question: Ester X, C₇H₁₄O₂, has an unbranched chain on each side of the ester group. Heating X under reflux with aqueous sodium hydroxide gives propan-1-ol as one product. (a) Identify X. (b) Write the equation for this reaction. (c) X is also heated under reflux with dilute sulfuric acid. Name the organic products and explain why some X is left, while the reaction in (b) is complete.

Step 1: the alcohol part. Propan-1-ol gives the propyl group, CH₃CH₂CH₂-, which is C₃H₇.

Step 2: the acid part. C₇H₁₄O₂ minus C₃H₇ leaves C₄H₇O₂. This is the acid part RCOO- with R = C₃H₇: four carbon atoms, unbranched, so butanoate, CH₃CH₂CH₂COO-.

Step 3: the ester. X is propyl butanoate, CH₃CH₂CH₂COOCH₂CH₂CH₃, an isomer of methyl hexanoate from Worked example 1 (which would give methanol instead).

Step 4: alkaline hydrolysis.

CH₃CH₂CH₂COOCH₂CH₂CH₃(l) + NaOH(aq) → CH₃CH₂CH₂COONa(aq) + CH₃CH₂CH₂OH(aq)

Step 5: acid hydrolysis.

CH₃CH₂CH₂COOCH₂CH₂CH₃(l) + H₂O(l) ⇌ CH₃CH₂CH₂COOH(aq) + CH₃CH₂CH₂OH(aq)

Butanoic acid and propan-1-ol can react to re-form X, so the mixture reaches equilibrium with some X left. In alkali, butanoic acid becomes butanoate ions as soon as it forms; these do not react with propan-1-ol, so X cannot re-form.

Answer: (a) propyl butanoate, CH₃CH₂CH₂COOCH₂CH₂CH₃. (b) The equation in Step 4; the products are sodium butanoate and propan-1-ol. (c) Butanoic acid and propan-1-ol; acid hydrolysis is reversible and reaches equilibrium, but in alkali the acid is removed as butanoate ions, so the reaction is complete.

Key Equations

R is a hydrogen atom or a hydrocarbon group from the acid; R' is the hydrocarbon group from the alcohol.

Esterification (concentrated H₂SO₄ catalyst, heat): $$\text{RCOOH} + \text{R}'\text{OH} \rightleftharpoons \text{RCOOR}' + \text{H}_2\text{O}$$

Acid hydrolysis (dilute acid, heat under reflux; H⁺ is the catalyst): $$\text{RCOOR}' + \text{H}_2\text{O} \rightleftharpoons \text{RCOOH} + \text{R}'\text{OH}$$

Alkaline hydrolysis (NaOH(aq), heat under reflux; goes to completion): $$\text{RCOOR}' + \text{OH}^- \rightarrow \text{RCOO}^- + \text{R}'\text{OH}$$

Acidification after alkaline hydrolysis (add a dilute strong acid): $$\text{RCOO}^- + \text{H}^+ \rightarrow \text{RCOOH}$$

Common Mistakes to Avoid

• Mistake: naming an ester back to front, or leaving the C=O carbon out of the acid part. Fix: alcohol part first (-yl), then acid part (-oate), counting the C=O carbon in the acid. CH₃CH₂COOCH₂CH₂CH₂CH₂CH₃ is pentyl propanoate, not "ethyl pentanoate" (read from left to right) or "pentyl ethanoate" (C=O carbon left out).
• Mistake: changing the alcohol's carbon skeleton in the ester. Fix: the ester oxygen joins the carbon that carried the -OH. Methanoic acid and butan-2-ol give HCOOCH(CH₃)CH₂CH₃, not HCOOCH₂CH₂CH₂CH₃.
• Mistake: writing "H₂SO₄" or "acid" with no concentration. Fix: esterification uses concentrated H₂SO₄ (catalyst) and heat; acid hydrolysis uses a dilute acid, such as dilute HCl(aq), heated under reflux.
• Mistake: giving the carboxylic acid as a product of alkaline hydrolysis. Fix: the product is the salt. In Worked example 2, NaOH(aq) gives sodium butanoate, CH₃CH₂CH₂COONa; butanoic acid forms only after you acidify.
• Mistake: calling the reaction that makes an ester substitution, hydrolysis or oxidation, or giving the role of concentrated H₂SO₄ as "dehydrating agent". Fix: when asked for the type of reaction, write condensation (two molecules join, water is lost), which is safer than just "esterification"; concentrated H₂SO₄ is the catalyst. Splitting an ester with water is hydrolysis.
• Mistake: hydrolysing only one ester group in a molecule that has several. Fix: each ester group reacts. CH₃COOCH₂CH₂OOCCH₃ needs 2NaOH and gives HOCH₂CH₂OH and two CH₃COONa.
• Mistake: writing -CH₃ in a displayed formula, or drawing the ester group with an extra oxygen. Fix: show every bond, including each C-H bond; the ester group has exactly two oxygen atoms, C(=O)-O-C.
• Mistake: drawing reflux apparatus that is sealed, or with water entering the condenser at the top. Fix: the condenser is vertical and open at the top, with water in at the bottom and out at the top.

Exam Tips

• State the reagents and conditions in full: "concentrated H₂SO₄ (catalyst), heat" for esterification; "dilute HCl(aq), heat under reflux" or "NaOH(aq), heat under reflux" for hydrolysis.
• Name an ester in two words, alcohol part first, then check the carbon count: propyl butanoate has 3 + 4 = 7 carbon atoms, C₇H₁₄O₂.
• Describe a hydrolysis by giving the reagent, the condition, both organic products and whether it goes to completion. For alkali, name the salt, and add "then acidify" if the acid is wanted.
• Deduce an ester from its hydrolysis products by joining the acid part, RCO-, to the alcohol's alkyl group through an oxygen atom. Check that ester + water has the same atoms as acid + alcohol.
• Write esterification and acid hydrolysis with ⇌ (both reversible) and alkaline hydrolysis with →, and balance every equation.