How to Distinguish an Ester from a Carboxylic Acid

Two organic compounds can contain a carbonyl group and two oxygen atoms, yet one is a carboxylic acid and the other is an ester. The difference comes down to one bond: find the carbonyl carbon, then follow the oxygen attached to it by a single bond. If that oxygen is bonded to hydrogen, you have a carboxylic acid. If it is bonded to another carbon group, you have an ester.

Consider these two compounds:

• Butanoic acid: \(\ce{CH3CH2CH2C(=O)OH}\)
• Ethyl ethanoate: \(\ce{CH3C(=O)OCH2CH3}\)

Both have the molecular formula \(\ce{C4H8O2}\). They contain exactly the same numbers of carbon, hydrogen, and oxygen atoms, but their atoms are connected differently.

The structural formula tells you which compound you have. The molecular formula alone cannot.

01The one-bond rule: follow the oxygen

Both functional groups contain a carbonyl group, which is a carbon atom double-bonded to oxygen:

\[
\ce{C=O}
\]

The important difference is what else is attached to that carbon.

Carboxylic acids: the \(-COOH\) group

A carboxylic acid contains the carboxyl functional group:

\[
\ce{R-C(=O)-OH}
\]

Here, \(R\) represents the rest of the carbon-containing molecule.

Look carefully at the structure. The carbonyl carbon is connected to:

• One oxygen through a double bond.
• Another oxygen through a single bond.
• A hydrogen is attached to that single-bonded oxygen.

That final O-H bond is the giveaway.

For example, ethanoic acid has the structure:

\[
\ce{CH3-C(=O)-OH}
\]

The hydroxyl group (\(\ce{-OH}\)) is part of the carboxyl group. It is not a separate alcohol group.

Esters: the \(-COO-\) linkage

An ester contains the functional group:

\[
\ce{R-C(=O)-O-R’}
\]

Here, both \(R\) and \(R’\) represent carbon-containing groups.

Notice what has changed. The oxygen attached to the carbonyl carbon is now connected to another carbon rather than hydrogen.

For example, methyl ethanoate has the structure:

\[
\ce{CH3-C(=O)-O-CH3}
\]

The single-bonded oxygen acts as a bridge between two carbon-containing sections of the molecule.

The structural difference at a glance

The most reliable method is to trace the second bond from the single-bonded oxygen.

1. Find the \(\ce{C=O}\) group.
2. Locate the oxygen single-bonded to the same carbon.
3. Follow the other bond from that oxygen.
4. If it leads to H, identify a carboxylic acid.
5. If it leads to another carbon, identify an ester.

This works even when the structural formula is drawn in an unfamiliar orientation.

For example:

\[
\ce{CH3CH2-O-C(=O)-CH3}
\]

and

\[
\ce{CH3-C(=O)-O-CH2CH3}
\]

represent the same ester. The direction in which the formula is written does not change the connectivity.

A common trap: finding an O-H bond somewhere in the molecule

Suppose you see:

\[
\ce{CH3-C(=O)-CH2-OH}
\]

You might be tempted to call this a carboxylic acid because it contains both a carbonyl group and an O-H bond.

That conclusion is incorrect.

The carbonyl carbon is attached to carbon atoms on both sides. The hydroxyl group is attached to the neighbouring \(\ce{CH2}\) carbon, not to the carbonyl carbon.

This molecule contains a ketone and an alcohol group. It is a hydroxyketone, not a carboxylic acid.

There is another useful refinement: a molecule can contain an ester group and a separate alcohol O-H bond. You must identify the functional group from its local connectivity, rather than simply counting O-H bonds across the whole molecule.

What about a carboxylate ion?

You may also encounter a structure such as:

\[
\mathrm{CH_{3}COO}^{-}
\]

This is a carboxylate ion, not an ester or a neutral carboxylic acid.

It forms when a carboxylic acid loses its acidic hydrogen. For example:

\[
\ce{CH3COOH -> CH3COO^- + H^+}
\]

In a sodium carboxylate salt, the negative ion is paired with a sodium ion.

This distinction becomes important when interpreting ester hydrolysis and acid-base reactions later in the topic.

02Using sodium carbonate to distinguish the compounds

Structural formulas provide the most direct identification method. However, suppose someone gives you two unlabelled liquid samples and asks you to identify which contains a carboxylic acid.

You need an experimental test.

A useful reagent is aqueous sodium carbonate, \(\ce{Na2CO3}\).

Before looking at the reaction, predict what will happen when you add sodium carbonate separately to:

• A sample of butanoic acid.
• A sample of ethyl ethanoate.

The butanoic acid produces bubbles. The ethyl ethanoate does not produce characteristic effervescence under ordinary room-temperature aqueous test conditions.

Why?

The reaction of a carboxylic acid with carbonate

Carboxylic acids can donate hydrogen ions in acid-base reactions.

When a carboxylic acid reacts with sodium carbonate, carbon dioxide gas is released.

The general balanced equation is:

\[
\ce{2RCOOH + Na2CO3 ->
2RCOONa + H2O + CO2}
\]

The products are:

• A sodium carboxylate salt.
• Water.
• Carbon dioxide gas.

The visible bubbles are carbon dioxide.

For butanoic acid, the equation is:

\[
\ce{2CH3CH2CH2COOH + Na2CO3 ->
2CH3CH2CH2COONa + H2O + CO2}
\]

The acid is converted into sodium butanoate.

Notice that carboxylic acids are weak acids. This means they only partially ionise in water. It does not mean they cannot react with sodium carbonate.

What happens with an ester?

An ester does not contain the carboxylic acid O-H bond.

Consequently, it does not undergo the characteristic acid-carbonate reaction that releases carbon dioxide.

For a pure sample of ethyl ethanoate, adding aqueous sodium carbonate at room temperature produces no characteristic effervescence.

Experimental comparison

There are two important limitations.

First, a negative carbonate test does not automatically prove that a substance is an ester. Alcohols, aldehydes, and many other organic compounds also do not produce this characteristic reaction. If the question limits the possibilities to a pure carboxylic acid or a pure ester, the test becomes much more useful.

Second, the test detects acidic substances, not exclusively carboxylic acids. Another acid in the sample can also release carbon dioxide.

This second limitation becomes particularly important during esterification.

For a broader comparison of the physical properties and characteristic reactions of these two functional groups, see Carboxylic Acids vs Esters: Structure, Properties and Reactions.

03Worked examples: applying the identification method

04Esterification: why equilibrium matters when testing

There is an important connection between carboxylic acids and esters.

A carboxylic acid reacts with an alcohol to form an ester and water. This reaction is called esterification.

For example, butanoic acid reacts with ethanol to form ethyl butanoate:

\[
\ce{CH3CH2CH2COOH + CH3CH2OH
<=> CH3CH2CH2COOCH2CH3 + H2O}
\]

Notice the structural change.

The carboxylic acid contains:

\[
\ce{-C(=O)-OH}
\]

The ester product contains:

\[
\ce{-C(=O)-O-CH2CH3}
\]

The oxygen is now connected to an ethyl group rather than hydrogen.

This is the same structural distinction you used earlier to identify an unknown compound.

Why doesn’t all the acid necessarily become ester?

The esterification reaction is reversible.

The forward reaction produces ester and water. The reverse reaction allows the ester and water to reform the carboxylic acid and alcohol.

Eventually, the system can reach dynamic equilibrium.

At equilibrium:

• The forward reaction continues.
• The reverse reaction continues.
• Both reactions occur at equal rates.
• The concentrations of reactants and products remain constant.

This means a reaction mixture can contain all four substances:

1. Carboxylic acid.
2. Alcohol.
3. Ester.
4. Water.

You should not assume that heating the mixture converts every acid molecule into an ester molecule.

What do the reagents actually do?

Esterification commonly uses concentrated sulfuric acid and heating under reflux.

Each component has a different role.

The catalyst supplies hydrogen ions that increase the rate of esterification.

It also increases the rate of the reverse reaction. At a fixed temperature, a catalyst does not change the equilibrium constant or equilibrium composition. It simply allows equilibrium to be reached faster.

Concentrated sulfuric acid has an additional role. It can absorb water, helping remove a reaction product and shift the equilibrium towards ester formation.

Using an excess of one reactant or removing water can also favour the forward reaction.

Why is reflux necessary?

Imagine heating a mixture of an alcohol and a carboxylic acid in an open flask.

Some of the volatile substances may evaporate and escape. You would lose reactants, and the composition of your mixture would change.

Reflux addresses this problem.

The reaction mixture is heated in a round-bottom flask. A vertical water-cooled condenser sits directly above it.

As vapour rises from the flask, it enters the condenser. Cold water flows into the lower sidearm and leaves through the upper sidearm, keeping the condenser cool.

The vapour condenses into liquid and flows back into the original flask.

This allows the mixture to be heated for an extended period without losing significant quantities of volatile reactants and products. The apparatus is not sealed.

The testing problem: why a crude reaction mixture can fizz

Here is a particularly important experimental trap.

Suppose you prepare an ester using butanoic acid, ethanol, and concentrated sulfuric acid.

After reflux, you take some of the reaction mixture and add sodium carbonate.

You observe bubbles.

Does this mean esterification failed?

No.

There are two possible sources of the carbon dioxide:

1. Unreacted butanoic acid remaining in the equilibrium mixture.
2. Concentrated sulfuric acid used as the catalyst.

Sulfuric acid also reacts with sodium carbonate:

\[
\ce{H2SO4 + Na2CO3 ->
Na2SO4 + H2O + CO2}
\]

Therefore, bubbling in the crude reaction mixture cannot tell you whether ester has formed.

In fact, the mixture may contain a substantial amount of ester while still producing carbon dioxide during this test.

The correct approach is to separate and purify the ester, removing residual acids and catalyst before using the carbonate test on the isolated product.

Even then, no bubbles only indicates that there is no detectable acidic contamination under the test conditions. It does not independently prove that the isolated compound is an ester.

Don’t confuse hydrolysis with an identification test

Another common mistake is to heat an unknown ester with aqueous sodium hydroxide and then expect to recover the original ester.

Hot alkaline conditions can hydrolyse an ester.

The general equation is:

\[
\ce{RCOOR’ + OH^- ->
RCOO^- + R’OH}
\]

The products are a carboxylate ion and an alcohol.

Notice that the carboxyl-containing product is now a carboxylate ion, not a neutral carboxylic acid.

This is why the reaction conditions matter. An ester can be converted into an alcohol and a carboxylate through alkaline hydrolysis, while acid-catalysed hydrolysis produces a carboxylic acid and an alcohol:

\[
\ce{RCOOR’ + H2O <=>
RCOOH + R’OH}
\]

The second reaction is reversible.

If your goal is simply to distinguish an unknown acid from an ester, use the structural formula first and an appropriate carbonate test on a clean sample. Do not deliberately hydrolyse the compound before identifying it.

05Questions and solutions

The following questions progress from straightforward structural identification to more demanding problems involving equilibrium, acid-base reactions, and functional groups within the same molecule.

Question 1

Two organic compounds have the same molecular formula, \(\ce{C4H8O2}\).

Compound A:

\[
\ce{CH3CH2CH2C(=O)OH}
\]

Compound B:

\[
\ce{CH3C(=O)OCH2CH3}
\]

Identify each compound as either a carboxylic acid or an ester. Explain the structural feature that allows you to distinguish them, despite their identical molecular formulae.

Solution 1

Compound A is a carboxylic acid, while compound B is an ester.

In compound A, the carbonyl carbon is double-bonded to one oxygen and single-bonded to another oxygen. The single-bonded oxygen is connected to hydrogen, forming the carboxyl group:

\[
\ce{-C(=O)-OH}
\]

In compound B, the single-bonded oxygen is connected to an ethyl group:

\[
\ce{-C(=O)-O-CH2CH3}
\]

This forms an ester linkage.

The compounds have identical molecular formulae because they contain the same numbers of each type of atom. However, their connectivity differs.

They are functional-group isomers.

The important distinction is not the number of oxygen atoms, but whether the single-bonded oxygen is attached to hydrogen or carbon.

Question 2

Two compounds are represented by the following structural formulae:

Compound X:

\[
\ce{CH3CH2OC(=O)CH3}
\]

Compound Y:

\[
\ce{CH3CH2COO^- Na^+}
\]

A student claims that both compounds are esters because both contain a carbonyl group and two oxygen atoms.

Identify each compound, explain why the student’s classification is incorrect, and predict whether either compound will produce carbon dioxide when aqueous sodium carbonate is added at room temperature.

Solution 2

Compound X is an ester, while compound Y is sodium propanoate, a carboxylate salt. Neither produces characteristic carbon dioxide effervescence when tested with aqueous sodium carbonate under the stated conditions.

For compound X, trace the oxygen single-bonded to the carbonyl carbon. It is connected to an ethyl group:

\[
\ce{CH3CH2-O-C(=O)-CH3}
\]

This is an ester linkage.

For compound Y, the negative charge indicates that the carboxylic acid has lost its acidic hydrogen. The resulting carboxylate ion is paired with a sodium ion.

It is not an ester because the oxygen is not bridging the carbonyl carbon to another carbon-containing group in the ester arrangement.

The carbonate test gives different information from simply locating a carbonyl group.

• Compound X is an ester, so it does not undergo the characteristic acid-carbonate reaction.
• Compound Y is already a carboxylate ion rather than a neutral carboxylic acid, so it does not release carbon dioxide through this reaction.

The tempting mistake is to classify any structure containing a \(\ce{COO}\) arrangement as an ester. You must also check the charges and the atom connected to the single-bonded oxygen.

Question 3

A purified liquid sample contains only butanoic acid and ethyl ethanoate.

The total concentration of organic molecules in the sample is \(0.0800\ \text{mol L}^{-1}\).

A \(25.0\ \text{mL}\) aliquot is reacted with excess aqueous sodium carbonate. A gas syringe collects \(12.4\ \text{mL}\) of carbon dioxide at \(25.0^\circ\text{C}\) and \(100\ \text{kPa}\).

Calculate the mole percentage of butanoic acid in the mixture.

Then determine the mass percentage of butanoic acid.

Assume ideal gas behaviour, complete gas collection, and no other acidic substances.

Solution 3

The mixture contains approximately \(50.0\%\) butanoic acid by mole and \(50.0\%\) by mass.

The key is that only the carboxylic acid reacts with sodium carbonate to produce the measured carbon dioxide.

Step 1: Calculate the total amount of organic molecules in the aliquot.

Use:

\[
n=cV
\]

Convert the volume:

\[
25.0\ \text{mL}=0.0250\ \text{L}
\]

Substitute:

\[
\begin{aligned}
n_{\text{total}}&=0.0800\times0.0250\\
&=0.00200\ \text{mol}
\end{aligned}
\]

Step 2: Calculate the amount of carbon dioxide produced.

Use the ideal gas equation:

\[
PV=nRT
\]

Rearrange:

\[
n=\frac{PV}{RT}
\]

The gas volume is:

\[
12.4\ \text{mL}=0.0124\ \text{L}
\]

Using \(R=8.31\ \text{kPa L mol}^{-1}\text{K}^{-1}\):

\[
\begin{aligned}
n(\ce{CO2})
&=\frac{100\times0.0124}{8.31\times298}\\
&=5.01\times10^{-4}\ \text{mol}
\end{aligned}
\]

Step 3: Determine the amount of butanoic acid.

From the balanced equation:

\[
\ce{2RCOOH + Na2CO3 ->
2RCOONa + H2O + CO2}
\]

The mole ratio is:

\[
n(\text{acid}):n(\ce{CO2})=2:1
\]

Therefore:

\[
\begin{aligned}
n(\text{acid})
&=2(5.01\times10^{-4})\\
&=1.002\times10^{-3}\ \text{mol}
\end{aligned}
\]

Step 4: Calculate the mole percentage.

\[
\text{Mole percentage}
=
\frac{n(\text{acid})}{n(\text{total})}\times100
\]

Substitute:

\[
\begin{aligned}
&=\frac{0.001002}{0.00200}\times100\\
&=50.1\%
\end{aligned}
\]

The mixture is approximately \(50.0\%\) butanoic acid by mole.

Step 5: Calculate the mass percentage.

Butanoic acid and ethyl ethanoate have the same molecular formula:

\[
\ce{C4H8O2}
\]

Therefore, they have the same molar mass.

Equal mole fractions consequently correspond to equal mass fractions.

The mass percentage of butanoic acid is also approximately \(50.0\%\).

This question demonstrates why a gas test can be used to analyse the composition of a mixture, provided the other substances present do not interfere with the reaction.

Question 4

A chemist mixes \(0.200\ \text{mol}\) of butanoic acid with \(0.200\ \text{mol}\) of ethanol.

Concentrated sulfuric acid is added, and the mixture is heated under reflux to produce ethyl butanoate and water.

At equilibrium, the mixture contains \(0.120\ \text{mol}\) of ethyl butanoate.

Assume:

• No ester or water was initially present.
• No side reactions occur.
• The reaction follows the balanced esterification equation.

(a) Calculate the equilibrium amounts of butanoic acid, ethanol, and water.

(b) A student adds sodium carbonate to the crude equilibrium mixture and observes effervescence. They conclude that esterification has failed. Evaluate this conclusion.

(c) Explain the separate roles of concentrated sulfuric acid and reflux. Suggest one way to increase the equilibrium yield of ester without simply adding more catalyst.

Solution 4

(a) The equilibrium mixture contains \(0.080\ \text{mol}\) of butanoic acid, \(0.080\ \text{mol}\) of ethanol, \(0.120\ \text{mol}\) of ester, and \(0.120\ \text{mol}\) of water.

The balanced equation is:

\[
\ce{CH3CH2CH2COOH + CH3CH2OH
<=> CH3CH2CH2COOCH2CH3 + H2O}
\]

All four substances have a mole ratio of \(1:1:1:1\).

Step 1: Determine the amount of reactants consumed.

Since \(0.120\ \text{mol}\) of ester is produced, \(0.120\ \text{mol}\) of each reactant must have been consumed.

Step 2: Calculate the remaining butanoic acid.

\[
\begin{aligned}
n(\text{acid})
&=0.200-0.120\\
&=0.080\ \text{mol}
\end{aligned}
\]

Step 3: Calculate the remaining ethanol.

\[
\begin{aligned}
n(\text{ethanol})
&=0.200-0.120\\
&=0.080\ \text{mol}
\end{aligned}
\]

Step 4: Determine the amount of water.

The mole ratio of ester to water is \(1:1\).

Therefore:

\[
n(\ce{H2O})=0.120\ \text{mol}
\]

The equilibrium composition is:

(b) The student’s conclusion is incorrect. Effervescence does not mean that esterification failed.

There is still \(0.080\ \text{mol}\) of unreacted butanoic acid in the mixture. This acid reacts with sodium carbonate to release carbon dioxide.

Additionally, concentrated sulfuric acid is present and can also react with sodium carbonate to release carbon dioxide.

The observed bubbles therefore cannot distinguish between residual carboxylic acid and the acid catalyst.

Importantly, the mixture contains \(0.120\ \text{mol}\) of ester, so esterification has clearly occurred.

The student has confused evidence of an acidic substance with evidence that no ester has formed.

(c) Concentrated sulfuric acid has two relevant roles.

First, it acts as an acid catalyst. It increases the rates of both the forward and reverse reactions, allowing equilibrium to be reached more quickly.

Second, concentrated sulfuric acid can absorb water. Removing a product helps shift the equilibrium towards ester formation.

Reflux allows the reaction mixture to be heated for an extended period while returning condensed vapour to the flask. It reduces the loss of volatile substances.

One way to increase the equilibrium yield is to use excess ethanol. Increasing the concentration of a reactant shifts the equilibrium towards the products.

Alternatively, removing water as it forms can favour ester production.

Simply adding more catalyst will not change the equilibrium position at a fixed temperature.

Question 5

A student prepares a compound, X, by reacting ethanoic acid with ethane-1,2-diol.

The purified compound has the structural formula:

\[
\ce{CH3C(=O)OCH2CH2OH}
\]

The student makes the following observations:

• Aqueous sodium carbonate produces no visible bubbles when added to purified X.
• X is subsequently heated under reflux with excess aqueous sodium hydroxide.
• After the reaction, adding sodium carbonate to the alkaline mixture still produces no visible bubbles.

The student concludes that X cannot be an ester because it contains an O-H bond. They also conclude that the alkaline hydrolysis must have failed because no carboxylic acid was produced.

Evaluate both conclusions.

Write a balanced equation for the alkaline hydrolysis of X, identify the functional groups present in X, and explain why neither carbonate test necessarily produces carbon dioxide.

Solution 5

Both conclusions are incorrect. X is an ester containing a separate alcohol group, and the absence of carbon dioxide after alkaline hydrolysis is consistent with successful formation of a carboxylate ion.

This question tests whether you can identify functional groups by their connectivity rather than simply looking for particular atoms.

Step 1: Identify the ester linkage in X.

The structure is:

\[
\ce{CH3-C(=O)-O-CH2CH2OH}
\]

The oxygen single-bonded to the carbonyl carbon is connected to another carbon-containing group.

Therefore, X contains an ester linkage.

Step 2: Identify the separate O-H bond.

The terminal section of the molecule is:

\[
\ce{-CH2CH2OH}
\]

This is an alcohol group.

Its O-H bond is not attached to the carbonyl carbon. Therefore, it does not form a carboxylic acid functional group.

X is a hydroxy-functional ester, not a carboxylic acid.

Step 3: Explain the first carbonate test.

The purified compound contains an ester group and an alcohol group, but no carboxylic acid group.

Alcohols do not undergo the characteristic carbon dioxide-producing reaction with aqueous sodium carbonate under ordinary conditions.

Therefore, the absence of bubbles is expected.

It does not mean that X is not an ester.

Step 4: Write the alkaline hydrolysis equation.

The ester reacts with hydroxide ions to produce a carboxylate ion and an alcohol.

For X:

\[
\ce{CH3COOCH2CH2OH + OH^- ->
CH3COO^- + HOCH2CH2OH}
\]

The products are:

• Ethanoate ions.
• Ethane-1,2-diol.

The ester linkage is broken, allowing the carbon-containing alcohol section to form ethane-1,2-diol.

Step 5: Explain the second carbonate test.

The hydrolysis occurs in excess sodium hydroxide, so the carboxyl-containing product is predominantly present as the ethanoate ion:

\[
\mathrm{CH_{3}COO}^{-}
\]

It is not present mainly as neutral ethanoic acid.

The ethanoate ion does not produce carbon dioxide when sodium carbonate is added under the stated conditions.

Therefore, the absence of bubbles is consistent with successful alkaline hydrolysis.

If the mixture were subsequently acidified, ethanoate ions could be converted into ethanoic acid. However, any excess mineral acid used during acidification would also react with carbonate, so it would need to be removed before using that test to identify the carboxylic acid specifically.

The deeper lesson is that the presence of oxygen, a carbonyl group, or even an O-H bond does not independently identify a functional group. You must consider the exact connectivity and the chemical species present under the reaction conditions.

06Where to go next

Once you can reliably distinguish \(\ce{-COOH}\) from \(\ce{-COOR’}\), the next step is predicting how these structures react.

In particular, being able to identify the carboxylic acid and alcohol in an esterification reaction makes it much easier to construct the balanced equation, identify the limiting reactant, and calculate the theoretical ester yield.

Continue with How to Calculate Theoretical Ester Yield from Two Reactants.

NSW syllabus context: Year 12 students in 2026 continue to study the Chemistry Stage 6 Syllabus (2017). The Chemistry 11-12 Syllabus (2025) begins implementation for Year 11 in 2028 and Year 12 in Term 4 2028, with the first HSC examination under the new syllabus scheduled for 2029. This explanation is written for the current Stage 6 course and does not claim that every detail is a separately prescribed requirement of the 2025 syllabus. See NESA’s syllabus implementation timeline.