Esters

AS · 11 min

Esters are the compounds that give fruit its smell: ethyl butanoate smells of pineapple, 3-methylbutyl ethanoate of pear drops and bananas. They are made by joining a carboxylic acid to an alcohol, with the loss of water, and they can be split back into the acid and alcohol by hydrolysis. This note covers the structure and naming of esters (a frequent source of lost marks), their formation by condensation, their hydrolysis by dilute acid and by dilute alkali, and how to work backwards from hydrolysis products to the ester. Esterification is also the standard example of a homogeneous equilibrium in Paper 2.

Structure and naming

An ester contains the group $\ce{-COO-}$ : a carbonyl carbon bonded to a second oxygen, which is bonded to a carbon group. The general formula is $\ce{RCOOR'}$ , where R (which can be H) comes from the acid and R′ (a carbon group) comes from the alcohol.

Method

Naming an ester

Find the C=O. The part of the molecule containing the C=O, including the carbonyl carbon, comes from the acid: name it as "-oate" (ethanoate, propanoate).
The carbon group attached to the single-bonded oxygen comes from the alcohol: name it as an alkyl group (methyl, ethyl, propyl).
Write the alkyl name first, as a separate word: alkyl alkanoate.

ester	made from	name
$\ce{HCOOCH3}$	methanoic acid + methanol	methyl methanoate
$\ce{CH3COOCH3}$	ethanoic acid + methanol	methyl ethanoate
$\ce{CH3COOCH2CH3}$	ethanoic acid + ethanol	ethyl ethanoate
$\ce{CH3CH2COOCH3}$	propanoic acid + methanol	methyl propanoate
$\ce{HCOOCH2CH2CH3}$	methanoic acid + propan-1-ol	propyl methanoate
$\ce{CH3CH2CH2COOCH2CH3}$	butanoic acid + ethanol	ethyl butanoate

The formula is often written with the acid part first ( $\ce{CH3COOCH2CH3}$ ) but the name has the alcohol part first (ethyl ethanoate). The formula can also be written the other way round: $\ce{CH3CH2OOCCH3}$ is the same ethyl ethanoate. Always find the C=O before naming.

Esters with the formula $\ce{C_{n}H_{2n}O2}$ are functional group isomers of carboxylic acids. $\ce{C3H6O2}$ , for example, is propanoic acid, methyl ethanoate and ethyl methanoate.

Physical properties and uses

Esters have polar C=O and C–O bonds, so they have permanent dipole–permanent dipole forces, but they have no O–H group, so they cannot hydrogen bond to each other.

compound ( $\ce{C4H8O2}$ )	boiling point / $^\circ\text{C}$
butanoic acid	164
ethyl ethanoate	77

Esters have much lower boiling points than the isomeric carboxylic acids, so they are volatile, which is why they have strong smells.
Small esters are slightly soluble in water (water molecules can hydrogen bond to their oxygen lone pairs), but much less soluble than acids.
Uses: flavourings and perfumes (sweet, fruity smells); solvents, for example ethyl ethanoate in glues and nail-varnish remover; plasticisers.

Making esters: condensation

Key result

Esterification: carboxylic acid + alcohol, heated with a few drops of concentrated sulfuric acid as catalyst.

\ce{RCOOH + R'OH <=> RCOOR' + H2O}

This is a condensation reaction (two molecules join, water is lost). It is reversible and reaches equilibrium slowly.

\ce{CH3COOH + CH3CH2OH <=> CH3COOCH2CH3 + H2O}

The water is formed from the OH of the acid and the H of the alcohol's OH; the alcohol's oxygen stays in the ester.

Concentrated sulfuric acid does two jobs:

It is a catalyst (provides $\ce{H+}$ ), speeding up a very slow reaction.
It is a dehydrating agent: it absorbs some of the water produced, which shifts the position of equilibrium to the right (Le Chatelier's principle) and increases the yield of ester.

Practical skills

Preparing an ester in the laboratory

Mix the carboxylic acid and alcohol in a round-bottomed flask, add a few drops of concentrated sulfuric acid carefully, and add anti-bumping granules.
Heat under reflux for about 30 minutes, so the volatile reactants are not lost while equilibrium is approached.
Rearrange the apparatus and distil off the ester (and some unreacted alcohol and acid) as an impure distillate.
Shake the distillate with sodium carbonate solution in a separating funnel to remove acidic impurities (effervescence of $\ce{CO2}$ ; release the pressure regularly). Run off and discard the aqueous layer.
Dry the ester layer with an anhydrous drying agent such as anhydrous calcium chloride or magnesium sulfate.
Redistil, collecting the fraction that boils at the ester's boiling point.

Small-scale tests: warming a few drops of acid and alcohol with a drop of concentrated $\ce{H2SO4}$ , then pouring into sodium hydrogencarbonate solution, releases the sweet smell of the ester.

Hydrolysis of esters

Hydrolysis is the reverse of esterification: water splits the ester into its acid and alcohol. With water alone it is extremely slow, so it is catalysed by dilute acid, or carried out with dilute alkali.

Key result

	acid hydrolysis	alkaline hydrolysis
reagent and conditions	dilute $\ce{H2SO4}$ or HCl, heat under reflux	dilute NaOH(aq), heat under reflux
products	carboxylic acid + alcohol	carboxylate salt + alcohol
extent	reversible: equilibrium, incomplete	goes to completion
to obtain the free acid	already present	acidify with dilute HCl afterwards

Acid hydrolysis of ethyl ethanoate:

\ce{CH3COOCH2CH3 + H2O <=>[H+] CH3COOH + CH3CH2OH}

Alkaline hydrolysis of ethyl ethanoate:

\ce{CH3COOCH2CH3 + NaOH -> CH3COONa + CH3CH2OH}

then

\ce{CH3COONa + HCl -> CH3COOH + NaCl}

Why alkaline hydrolysis goes to completion: the carboxylic acid formed reacts at once with the hydroxide ions to give the carboxylate ion. A carboxylate ion cannot react with the alcohol to re-form the ester, so the reverse reaction is removed and all the ester is used up. This is why alkaline hydrolysis is preferred when the products are wanted.

Tip

Alkaline hydrolysis of the esters in fats and oils (esters of propane-1,2,3-triol with long-chain acids) produces the sodium salts of those acids: soap. The process is called saponification. This is background, not required at AS.

Working back from hydrolysis products

Method

Identifying an ester from its hydrolysis products

Identify the carboxylic acid and the alcohol formed (from data, tests or $M_r$ ).
Remove the OH from the acid's COOH and the H from the alcohol's OH; join the acid's C=O carbon to the alcohol's oxygen.
Check the molecular formula: ester = acid + alcohol − $\ce{H2O}$ .
Name it: alkyl (from the alcohol) + alkanoate (from the acid).

Worked examples

Routine: naming esters

Name (a) $\ce{CH3CH2COOCH2CH3}$ , (b) $\ce{HCOOCH2CH2CH3}$ , (c) $\ce{CH3COOCH(CH3)2}$ , and give the structural formula of (d) methyl butanoate.

Solution

(a) The C=O side has three carbons (propanoate); the O side is ethyl: ethyl propanoate.

(b) One carbon on the C=O side (methanoate); propyl on the O side: propyl methanoate.

(c) Ethanoate; the alcohol part is $\ce{-CH(CH3)2}$ , from propan-2-ol: 1-methylethyl ethanoate (propan-2-yl ethanoate).

(d) Butanoic acid part and methyl: $\ce{CH3CH2CH2COOCH3}$ .

Routine: making an ester

Write the equation for the formation of propyl ethanoate, give the conditions, and name the type of reaction.

Solution

\ce{CH3COOH + CH3CH2CH2OH <=> CH3COOCH2CH2CH3 + H2O}

Heat under reflux with a few drops of concentrated sulfuric acid as catalyst. Condensation (esterification).

Standard: acid and alkaline hydrolysis

Methyl propanoate is heated under reflux (a) with dilute sulfuric acid, (b) with aqueous sodium hydroxide. Write an equation for each and explain why the yield of products is higher in (b).

Solution

(a) $\ce{CH3CH2COOCH3 + H2O <=> CH3CH2COOH + CH3OH}$

(b) $\ce{CH3CH2COOCH3 + NaOH -> CH3CH2COONa + CH3OH}$

In (a) the reaction is an equilibrium, so some ester always remains. In (b) the propanoic acid is converted to propanoate ions, which do not react with methanol, so the reverse reaction cannot occur and hydrolysis goes to completion.

Standard: identifying an ester from its products

An ester S, $\ce{C5H10O2}$ , is hydrolysed. One product is ethanoic acid. The other product, an alcohol, gives a pale yellow precipitate with alkaline aqueous iodine. Identify S.

Solution

Ester $\ce{C5H10O2}$ minus the ethanoate part leaves a three-carbon alcohol: $\ce{C5H10O2 + H2O -> C2H4O2 + C3H8O}$ .

The alcohol $\ce{C3H8O}$ is propan-1-ol or propan-2-ol. A positive tri-iodomethane test needs $\ce{CH3CH(OH)-}$ , so it is propan-2-ol.

S is $\ce{CH3COOCH(CH3)2}$ , 1-methylethyl ethanoate.

Exam-hard: the esterification equilibrium

$1.00\ \text{mol}$ of ethanoic acid and $1.00\ \text{mol}$ of ethanol are mixed with a little concentrated sulfuric acid and left to reach equilibrium. The equilibrium mixture contains $0.667\ \text{mol}$ of ethyl ethanoate.

(a) Calculate $K_c$ . (b) Calculate the amount of ester at equilibrium if $1.00\ \text{mol}$ of ethanoic acid is mixed with $2.00\ \text{mol}$ of ethanol at the same temperature. (c) Explain, using Le Chatelier's principle, why the yield increases.

Solution

(a) At equilibrium: ester $0.667$ , water $0.667$ , acid $1.00 - 0.667 = 0.333$ , ethanol $0.333\ \text{mol}$ . The volume $V$ cancels (equal numbers of moles on each side):

K_c = \frac{[\ce{CH3COOC2H5}][\ce{H2O}]}{[\ce{CH3COOH}][\ce{C2H5OH}]} = \frac{0.667 \times 0.667}{0.333 \times 0.333} = 4.0

$K_c$ has no units.

(b) Let $x$ mol of ester form: $\dfrac{x^2}{(1.00 - x)(2.00 - x)} = 4.0$ .

$x^2 = 4(2.00 - 3.00x + x^2)$ , so $3x^2 - 12x + 8 = 0$ .

$x = \dfrac{12 - \sqrt{144 - 96}}{6} = \dfrac{12 - 6.93}{6} = 0.845\ \text{mol}$ . (The other root, 3.15, is impossible since only 1.00 mol of acid is present.)

(c) Increasing the concentration of a reactant (ethanol) shifts the position of equilibrium to the right to oppose the change, so more ester forms: $0.845\ \text{mol}$ instead of $0.667\ \text{mol}$ . $K_c$ is unchanged because the temperature is unchanged. See Equilibrium constants.

Exam-hard: finding Mr by back titration

$2.20\ \text{g}$ of an ester R is heated under reflux with $50.0\ \text{cm}^3$ of $1.00\ \text{mol dm}^{-3}$ sodium hydroxide (an excess). After cooling, the unreacted sodium hydroxide requires $50.0\ \text{cm}^3$ of $0.500\ \text{mol dm}^{-3}$ hydrochloric acid for neutralisation.

(a) Calculate the $M_r$ of R. (b) R has a fruity smell and gives methanol on hydrolysis. Identify R.

Solution

(a) NaOH added: $50.0 \times 1.00 / 1000 = 0.0500\ \text{mol}$ . NaOH left over $= n(\ce{HCl}) = 50.0 \times 0.500 / 1000 = 0.0250\ \text{mol}$ . NaOH used by the ester $= 0.0250\ \text{mol}$ .

One ester group reacts with one NaOH, so $n(\textbf{R}) = 0.0250\ \text{mol}$ and $M_r = 2.20 / 0.0250 = 88.0$ .

(b) Esters $\ce{C_{n}H_{2n}O2}$ : $14n + 32 = 88$ , $n = 4$ : $\ce{C4H8O2}$ . The alcohol is methanol, so the acid part has three carbons: $\ce{CH3CH2COOCH3}$ , methyl propanoate.

Watch out

Ester names backwards. $\ce{CH3COOCH2CH3}$ is ethyl ethanoate; $\ce{CH3CH2COOCH3}$ is methyl propanoate. The alkyl group from the alcohol comes first in the name, even though it is written last in the usual formula.
Dilute sulfuric acid for esterification. Esterification needs concentrated sulfuric acid (catalyst and dehydrating agent). Hydrolysis uses dilute acid.
Alkaline hydrolysis products. The products are the salt (sodium carboxylate) and the alcohol, not the carboxylic acid, until the mixture is acidified.
"Esterification goes to completion". It is an equilibrium. Yields are typically about two-thirds unless the conditions are adjusted.
Ester boiling points. Esters cannot hydrogen bond to each other; do not explain their boiling points with hydrogen bonding between ester molecules.

Exam tip

Name-and-draw marks are common: practise converting both ways, and draw the displayed formula with the C=O and the C–O–C clearly shown when asked.
Conditions: esterification "heat with concentrated $\ce{H2SO4}$ "; hydrolysis "heat under reflux with dilute acid or dilute NaOH(aq)".
Equilibrium questions about esterification are standard Paper 2 material: $K_c$ for this reaction has no units, and volume cancels.
In identification questions, the hydrolysis products are the starting point; identify the alcohol (using oxidation or iodoform tests) and the acid (using $M_r$ or tests), then join them.

Summary

Ester group $\ce{-COO-}$ ; named alkyl (from the alcohol) + alkanoate (from the acid).
No hydrogen bonding between ester molecules: lower boiling points than isomeric acids; fruity smells; used as flavourings, perfumes and solvents.
Formed by condensation of a carboxylic acid and an alcohol with concentrated $\ce{H2SO4}$ catalyst, heat; reversible.
Acid hydrolysis (dilute acid, reflux): reversible, gives acid + alcohol.
Alkaline hydrolysis (dilute NaOH, reflux): complete, gives carboxylate salt + alcohol; acidify to get the acid.
Ester formula = acid + alcohol − water.

Practice

Question

Name: (a) $\ce{CH3COOCH2CH2CH3}$ , (b) $\ce{HCOOCH2CH3}$ , (c) $\ce{CH3CH2CH2COOCH2CH3}$ .
Give the structural formula of (a) methyl butanoate, (b) ethyl propanoate.
Write the equation and conditions for the formation of ethyl methanoate.
Give the products of the alkaline hydrolysis of methyl ethanoate with sodium hydroxide, and explain how ethanoic acid could be obtained from the products.
Explain why ethyl ethanoate boils at a lower temperature than its isomer butanoic acid.
Give the names and structures of the three isomers of $\ce{C3H6O2}$ that are a carboxylic acid or an ester, and describe a chemical test that identifies the acid.
Give two roles of concentrated sulfuric acid in esterification.
$0.600\ \text{mol}$ of ethanoic acid and $0.600\ \text{mol}$ of ethanol reach equilibrium. The mixture contains $0.400\ \text{mol}$ of ethyl ethanoate. Calculate $K_c$ .
Ester X, $\ce{C6H12O2}$ , is hydrolysed by aqueous sodium hydroxide to sodium ethanoate and an alcohol Y. Y is oxidised by acidified dichromate(VI) to a ketone and gives a pale yellow precipitate with alkaline aqueous iodine. Identify Y and X, and state whether X is chiral.
$3.52\ \text{g}$ of ethyl ethanoate is heated under reflux with $60.0\ \text{cm}^3$ of $1.00\ \text{mol dm}^{-3}$ NaOH. Calculate the volume of $0.500\ \text{mol dm}^{-3}$ HCl needed to neutralise the excess NaOH afterwards, assuming complete hydrolysis. ( $A_r$ : H 1.0, C 12.0, O 16.0)

Answers

(a) Propyl ethanoate. (b) Ethyl methanoate. (c) Ethyl butanoate.
(a) $\ce{CH3CH2CH2COOCH3}$ (b) $\ce{CH3CH2COOCH2CH3}$
$\ce{HCOOH + CH3CH2OH <=> HCOOCH2CH3 + H2O}$ ; heat (under reflux) with a few drops of concentrated sulfuric acid.
$\ce{CH3COOCH3 + NaOH -> CH3COONa + CH3OH}$ : sodium ethanoate and methanol. Distil off the methanol, then add dilute hydrochloric (or sulfuric) acid to the sodium ethanoate: $\ce{CH3COONa + HCl -> CH3COOH + NaCl}$ ; distil off the ethanoic acid.
Butanoic acid molecules form hydrogen bonds with each other (through O–H), but ethyl ethanoate has no O–H and its molecules are held only by permanent dipole–dipole and instantaneous dipole–induced dipole forces. These are weaker, so less energy is needed to separate ester molecules.
Propanoic acid $\ce{CH3CH2COOH}$ ; methyl ethanoate $\ce{CH3COOCH3}$ ; ethyl methanoate $\ce{HCOOCH2CH3}$ . Add sodium carbonate (or hydrogencarbonate): only propanoic acid gives effervescence of $\ce{CO2}$ , which turns limewater milky.
Catalyst (provides $\ce{H+}$ to speed up the reaction); dehydrating agent (absorbs water, shifting the equilibrium to the right and increasing the yield).
At equilibrium: ester 0.400, water 0.400, acid 0.200, ethanol $0.200\ \text{mol}$ . $K_c = (0.400 \times 0.400) / (0.200 \times 0.200) = 4.00$ (no units).
X minus the ethanoate part: $\ce{C6H12O2 + H2O -> C2H4O2 + C4H10O}$ , so Y is $\ce{C4H10O}$ . Oxidised to a ketone: secondary. Positive iodoform: contains $\ce{CH3CH(OH)-}$ . Y is butan-2-ol, $\ce{CH3CH(OH)CH2CH3}$ . X is $\ce{CH3COOCH(CH3)CH2CH3}$ , 1-methylpropyl ethanoate (butan-2-yl ethanoate). X is chiral: the carbon attached to the ester oxygen carries H, $\ce{CH3}$ , $\ce{C2H5}$ and $\ce{OCOCH3}$ , four different groups.
$M_r(\ce{CH3COOC2H5}) = 88.0$ ; $n = 3.52 / 88.0 = 0.0400\ \text{mol}$ , which uses $0.0400\ \text{mol}$ NaOH. NaOH added $= 0.0600\ \text{mol}$ , so excess $= 0.0200\ \text{mol}$ . Volume of HCl $= 0.0200 / 0.500 = 0.0400\ \text{dm}^3 = 40.0\ \text{cm}^3$ .

Structure and naming

Physical properties and uses

Making esters: condensation

Hydrolysis of esters

Working back from hydrolysis products

Worked examples

Practice

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