How to Identify Functional-Group Isomers of Acids and Esters

Two compounds can have the same molecular formula, contain the same number of oxygen atoms, and both contain a carbonyl group, yet belong to different functional groups.

For example, consider these two compounds:

• Propanoic acid: \(\ce{CH3CH2COOH}\)
• Methyl ethanoate: \(\ce{CH3COOCH3}\)

Both have the molecular formula \(\ce{C3H6O2}\). However, one is a carboxylic acid and the other is an ester.

They are functional-group isomers: compounds with the same molecular formula but different functional groups because their atoms are connected differently.

The quickest way to identify functional-group isomers of acids and esters is to find the carbonyl group, \(\ce{C=O}\), and then inspect the oxygen attached to the carbonyl carbon by a single bond.

• In a carboxylic acid, that oxygen is bonded to hydrogen: \(\ce{-C(=O)-OH}\).
• In an ester, that oxygen is bonded to another carbon: \(\ce{-C(=O)-O-C-}\).

That single difference is the key to the entire identification method.

01Identify the functional group by tracing the oxygen

Both carboxylic acids and esters contain two oxygen atoms.

One oxygen is double-bonded to carbon, forming the carbonyl group. The second oxygen is connected to the carbonyl carbon by a single bond.

The important question is what that second oxygen is connected to.

Carboxylic acids: the O-H connection

A carboxylic acid contains the carboxyl functional group:

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

Here, \(R\) represents a hydrogen atom or a carbon-containing group.

The carbonyl carbon is connected to:

• An oxygen by a double bond.
• Another oxygen by a single bond.
• A carbon-containing group, or hydrogen in methanoic acid.

The singly bonded oxygen is also connected to hydrogen.

This O-H connection is what distinguishes the carboxylic acid functional group.

For example, ethanoic acid has the structure:

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

The \(\ce{-OH}\) is directly attached to the carbonyl carbon. It is therefore part of a carboxylic acid group, not an ordinary alcohol group.

Esters: the O-C connection

An ester has the general structure:

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

Here, both \(R\) and \(R'\) represent carbon-containing groups, although \(R\) can be hydrogen in a methanoate ester.

The carbonyl carbon is connected to:

• An oxygen by a double bond.
• Another oxygen by a single bond.
• A carbon-containing group, or hydrogen.

The important difference is that the singly bonded oxygen connects to another carbon rather than hydrogen.

For example, methyl ethanoate has the structure:

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

The oxygen between the two carbon-containing sections acts as a link.

The diagnostic comparison

A useful rule is to follow the single bond from the carbonyl carbon to the second oxygen.

If the oxygen connects to hydrogen, you have a carboxylic acid.

If the oxygen connects to another carbon, you have an ester.

For a broader explanation of recognising organic functional groups, see HSC Chemistry: How to Identify Organic Functional Groups.

02Use carbon valency to check whether a structure is possible

Recognising the functional group is only half the job. You also need to check that the proposed structure is chemically valid.

This is particularly important when constructing functional-group isomers from a molecular formula.

Remember the usual covalent valencies:

A double bond counts as two bonds when checking valency.

Checking a carboxylic acid

Consider:

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

Focus on the carbonyl carbon.

It has:

• Two bonds to oxygen through the double bond.
• One bond to the methyl carbon.
• One bond to the hydroxyl oxygen.

Its total bond-order count is:

\[ 2+1+1=4 \]

Carbon has exactly four bonds, so the structure is valid.

The hydroxyl oxygen also has two single bonds: one to carbon and one to hydrogen.

Checking an ester

Now consider:

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

The carbonyl carbon again has:

\[ 2+1+1=4 \]

The oxygen between the two carbon-containing groups has two single bonds:

• One to the carbonyl carbon.
• One to the methyl carbon.

It also satisfies oxygen’s usual valency of two.

Both structures are therefore valid, despite having different connectivity.

A common mistake: giving carbon five bonds

Imagine a student tries to draw a carboxylic acid with the carbonyl carbon attached to two methyl groups as well as the hydroxyl oxygen:

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

This structure is not valid as an ordinary neutral organic molecule.

The carbonyl carbon would have:

\[ 2+1+1+1=5 \]

That exceeds carbon’s usual valency of four.

The mistake is easy to make if you remember to include the \(\ce{-OH}\) group but forget that the carbonyl carbon already has two bond-order units occupied by its double bond to oxygen.

Always check the carbonyl carbon before accepting a proposed acid or ester structure.

03Check the molecular formula before comparing structures

Functional-group isomers must have the same molecular formula.

For saturated, acyclic, monocarboxylic acids and esters, the general molecular formula is:

\[ \ce{C_nH_{2n}O2} \]

Here, \(n\) is the total number of carbon atoms, including the carbonyl carbon.

This formula assumes:

• One carboxylic acid or ester functional group.
• No carbon-carbon double or triple bonds.
• No rings.
• No additional functional groups.

It is not a formula for every possible acid or ester.

For example:

There is one interesting boundary case.

Methanoic acid has the formula \(\ce{CH2O2}\), but there is no ester with only one carbon atom. An ester needs at least two carbon atoms: one for the carbonyl carbon and at least one attached to the singly bonded oxygen.

Therefore, the first possible saturated, acyclic acid-ester functional-group isomer pair occurs at two carbon atoms.

Constructing ester isomers from a molecular formula

You can use carbon counting to work out which ester structures are possible.

Consider \(\ce{C3H6O2}\).

The ester must contain:

• One carbonyl carbon.
• Two additional carbon atoms.
• Two oxygen atoms.

The carbonyl carbon is fixed. The remaining two carbon atoms must be distributed between the carbon-containing group attached directly to the carbonyl carbon and the group attached through oxygen.

There are two possible distributions.

Arrangement 1: One carbon on each side

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

This is methyl ethanoate.

It has:

• One methyl group attached to the carbonyl carbon.
• One methyl group attached to the oxygen.

Arrangement 2: No carbon on the acid-derived side

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

This is ethyl methanoate.

The carbonyl carbon is attached to hydrogen, while the oxygen connects to an ethyl group.

Both esters have the molecular formula \(\ce{C3H6O2}\).

This gives three possible compounds within the acid and ester families:

All three have the formula \(\ce{C3H6O2}\).

Propanoic acid is a functional-group isomer of both esters.

However, methyl ethanoate and ethyl methanoate are not functional-group isomers of each other. They have the same functional group but different atom connectivity. They are structural isomers within the ester family.

This distinction becomes particularly useful when a question asks you to construct all possible acid and ester structures for a given molecular formula.

04Distinguish connectivity from spatial arrangement

Here is another trap.

Two structural formulas can look different without representing different compounds.

For example:

\[ \ce{CH3CH2C(=O)OH} \]

and

\[ \ce{HOOCCH2CH3} \]

Both represent propanoic acid.

The second formula is simply written from the opposite direction. Every atom is connected to exactly the same neighbouring atoms.

Now compare propanoic acid with methyl ethanoate:

\[ \ce{CH3CH2COOH} \]

\[ \ce{CH3COOCH3} \]

These are different compounds because their atom-to-atom connections have changed.

This is the distinction between connectivity and spatial arrangement.

Connectivity describes which atoms are bonded to which other atoms.

Spatial arrangement describes how those connected atoms are arranged in three-dimensional space.

Rotating a molecule, flipping a drawing, or writing a structural formula backwards does not change its connectivity.

However, some molecules have restricted spatial arrangements that produce different stereoisomers. For example, a carbon atom attached to four different groups can have two different configurations.

Those stereoisomers have the same connectivity but different three-dimensional arrangements.

For acid-ester functional-group isomerism, the key difference is the connectivity of the oxygen atom, not the orientation of the drawing.

A useful comparison is:

If you need more practice distinguishing identical structures from genuine structural isomers, see How to Check if Two Skeletal Formulas Show the Same Molecule.

05Worked examples

06Questions and solutions

For all five questions, assume ordinary neutral organic compounds unless stated otherwise. Unless stereochemistry is explicitly supplied, compare the constitutional structures only.

Question 1

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

Compound A:

\[ \ce{CH3CH2CH2COOH} \]

Compound B:

\[ \ce{CH3CH2COOCH3} \]

Identify the functional group in each compound and determine whether they are functional-group isomers. Justify your answer using their molecular formulas and structural features.

Solution 1

Yes, the compounds are functional-group isomers. Compound A is butanoic acid, while compound B is methyl propanoate.

First, confirm their molecular formulas.

For compound A:

\[ N_{\mathrm H}=3+2+2+1=8 \]

It contains four carbon atoms, eight hydrogen atoms, and two oxygen atoms.

For compound B:

\[ N_{\mathrm H}=3+2+3=8 \]

It also contains four carbon atoms, eight hydrogen atoms, and two oxygen atoms.

Both therefore have the formula \(\ce{C4H8O2}\).

Now inspect the singly bonded oxygen.

In compound A, the oxygen is connected to hydrogen, forming the carboxylic acid group \(\ce{-COOH}\).

In compound B, the oxygen connects the carbonyl carbon to a methyl group, forming the ester group \(\ce{-COO-}\).

The molecular formulas match, but the functional groups differ.

The tempting mistake is to assume that two compounds with the same carbon chain and two oxygen atoms must belong to the same functional group. The oxygen connectivity proves otherwise.

Question 2

Three compounds have the molecular formula \(\ce{C3H6O2}\).

Determine:

(a) The functional group present in each compound.

(b) Which pairs are functional-group isomers.

(c) Whether compounds B and C are identical molecules.

(d) Explain why matching molecular formulas alone cannot establish that two compounds are functional-group isomers.

Solution 2

A is a carboxylic acid, while B and C are esters. Therefore, A and B are functional-group isomers, as are A and C. B and C are not functional-group isomers.

(a) Identify the functional groups.

Compound A:

\[ \ce{CH3CH2C(=O)OH} \]

The singly bonded oxygen connects to hydrogen, so A is propanoic acid.

Compound B:

\[ \ce{CH3C(=O)OCH3} \]

The singly bonded oxygen connects to a methyl carbon, so B is methyl ethanoate.

Compound C:

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

The singly bonded oxygen connects to an ethyl group, so C is ethyl methanoate.

(b) Identify the functional-group isomer pairs.

A and B are functional-group isomers because they have the same formula but different functional groups.

A and C are also functional-group isomers for the same reason.

B and C are not functional-group isomers because both contain ester groups.

(c) Are B and C identical?

No. They are different structural isomers within the ester family.

In B, the carbonyl carbon is connected to a methyl group, and the oxygen is connected to another methyl group.

In C, the carbonyl carbon is connected to hydrogen, while the oxygen is connected to an ethyl group.

The carbon atoms are distributed differently around the ester linkage.

(d) Explain why molecular formulas are insufficient.

A molecular formula only tells you how many atoms of each element are present. It does not tell you which atoms are bonded together.

All three compounds have the same molecular formula, but their connectivity and functional groups are not identical.

This is why you must check both molecular formula and structural connectivity before identifying functional-group isomers.

Question 3

A student is asked to draw a carboxylic acid with the molecular formula \(\ce{C3H6O2}\).

They propose:

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

The student claims that the structure is valid because it contains three carbon atoms and two oxygen atoms.

Evaluate the proposed structure.

In your response:

(a) Identify the specific bonding problem.

(b) Explain why the carbonyl carbon violates the usual valency of carbon.

(c) Draw one valid carboxylic acid and one valid ester with the formula \(\ce{C3H6O2}\).

Solution 3

The proposed structure is invalid because the carbonyl carbon has five bond-order units instead of four.

(a) Identify the bonding problem.

The carbonyl carbon is connected to:

• One oxygen through a double bond.
• One hydroxyl oxygen through a single bond.
• One methyl group through a single bond.
• A second methyl group through a single bond.

(b) Check carbon valency.

The total bond-order count is:

\[ 2+1+1+1=5 \]

Ordinary neutral carbon forms four covalent bonds.

The proposed structure therefore violates carbon’s usual valency.

There is also a hydrogen-counting problem: the two methyl groups and hydroxyl group imply seven hydrogen atoms, not six.

The student has not constructed a valid molecule with the required formula.

(c) Construct valid alternatives.

A valid carboxylic acid is propanoic acid:

\[ \ce{CH3CH2C(=O)OH} \]

A valid ester is methyl ethanoate:

\[ \ce{CH3C(=O)OCH3} \]

Both have the formula \(\ce{C3H6O2}\).

In each structure, the carbonyl carbon has a bond-order total of four, and both oxygen atoms satisfy their usual valency of two.

These compounds are functional-group isomers.

The important lesson is that matching the number of atoms is not enough. Every proposed structure must also satisfy the bonding requirements of its atoms.

Question 4

A student is constructing every possible saturated, acyclic ester with the molecular formula \(\ce{C4H8O2}\).

They identify two possibilities:

• Methyl propanoate: \(\ce{CH3CH2COOCH3}\)
• Ethyl ethanoate: \(\ce{CH3COOCH2CH3}\)

They conclude that only two ester structures are possible because the three carbon atoms remaining after the carbonyl carbon must be divided into a two-carbon group and a one-carbon group.

Explain why this conclusion is incomplete.

Determine every possible ester constitutional structure with the formula \(\ce{C4H8O2}\), including branched structures. Explain how your carbon-counting method establishes that your list is complete.

Solution 4

There are four possible ester structures, not two. The student has overlooked the possibility that the carbonyl-side group contains no carbon atoms beyond the carbonyl carbon itself. This produces two additional methanoate esters.

Start with the general ester structure:

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

The carbonyl carbon accounts for one of the four carbon atoms.

The remaining three must be distributed between \(R\) and \(R'\).

Importantly, \(R'\) must contain at least one carbon because the singly bonded oxygen must connect to another carbon.

Consider the possible distributions.

Arrangement 1: Two carbons in R and one in R’.

This gives:

\[ \ce{CH3CH2C(=O)OCH3} \]

Methyl propanoate.

Arrangement 2: One carbon in R and two in R’.

This gives:

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

Ethyl ethanoate.

These are the two structures identified by the student.

However, there is a third carbon distribution.

Arrangement 3: No carbons in R and three in R’.

When \(R\) contains no carbon atoms, the carbonyl carbon is bonded to hydrogen.

The structure is therefore a methanoate ester.

The three-carbon group attached to oxygen can have two different arrangements.

The first is a straight-chain propyl group:

\[ \ce{H-C(=O)-O-CH2CH2CH3} \]

This is propan-1-yl methanoate.

The second is a branched propan-2-yl group:

\[ \ce{H-C(=O)-O-CH(CH3)2} \]

This is propan-2-yl methanoate.

Both have the formula \(\ce{C4H8O2}\).

The complete list is:

No additional ester structures arise under the stated assumptions.

The carbonyl-side group can contain zero, one or two additional carbons. The oxygen-linked group must contain at least one carbon. When the oxygen-linked group contains all three remaining carbons, there are two possible carbon skeletons: straight-chain and branched.

This gives four valid ester structures.

For comparison, the two saturated, acyclic carboxylic acid structures with this formula are butanoic acid and 2-methylpropanoic acid.

All four esters are functional-group isomers of both acids because they share the molecular formula \(\ce{C4H8O2}\) but have a different functional group.

Question 5

Consider four compounds. All are neutral, saturated, acyclic organic compounds with the molecular formula \(\ce{C5H10O2}\).

Compound X: Butan-2-yl methanoate.

\[ \ce{HCOOCH(CH3)CH2CH3} \]

Compound X’: The same atom-to-atom connections as X, but with the opposite three-dimensional configuration at the carbon directly attached to the ester oxygen. The two structures are drawn with opposite wedge-and-dash arrangements at this carbon.

Compound Y: 2-methylbutanoic acid.

\[ \ce{CH3CH2CH(CH3)COOH} \]

Compound Z: Methyl butanoate.

\[ \ce{CH3CH2CH2COOCH3} \]

For Y, no particular stereochemical configuration is specified.

A student makes two claims:

1. X and X’ must be the same molecule because their molecular formulas and atom-to-atom connections are identical.
2. X and Y must also be the same molecule because they contain the same number of atoms and both contain a carbonyl group and two oxygen atoms.

Evaluate both claims.

Then classify every pair of compounds as one of the following:

• Identical molecules.
• Functional-group isomers.
• Structural isomers with the same functional group.
• Stereoisomers.

Explain the role of connectivity, functional groups, and spatial arrangement in reaching your conclusions.

Solution 5

Both student claims are incorrect. X and X’ are stereoisomers, while X and Y are functional-group isomers. Matching molecular formulas and atom connectivity do not always establish that two molecules are identical.

First, confirm that the four compounds have the same molecular formula.

For X:

\[ N_{\mathrm H}=1+1+3+2+3=10 \]

There are five carbon atoms, ten hydrogen atoms and two oxygen atoms.

The same formula applies to X’ because it has identical atom counts.

For Y:

\[ N_{\mathrm H}=3+2+1+3+1=10 \]

For Z:

\[ N_{\mathrm H}=3+2+2+3=10 \]

All four therefore have the formula \(\ce{C5H10O2}\).

Now examine the individual relationships.

X compared with X’: stereoisomers

The student is incorrect to call X and X’ the same molecule.

The carbon directly attached to the ester oxygen has four different groups attached:

• Oxygen.
• Hydrogen.
• Methyl, \(\ce{-CH3}\).
• Ethyl, \(\ce{-CH2CH3}\).

This carbon is a stereogenic centre.

Changing its three-dimensional configuration while keeping all atom-to-atom connections unchanged produces a different stereoisomer.

Because the configuration is reversed at the only stereogenic centre, X and X’ are enantiomers.

They have identical molecular formulas and connectivity but different spatial arrangements.

They are not functional-group isomers because both are esters.

X compared with Y: functional-group isomers

The second student claim is also incorrect.

X is an ester:

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

Y is a carboxylic acid:

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

Although they have the same molecular formula, the oxygen atoms are connected differently.

In X, the singly bonded oxygen connects the carbonyl carbon to another carbon.

In Y, the singly bonded oxygen connects the carbonyl carbon to hydrogen.

Therefore, X and Y are functional-group isomers.

X compared with Z: structural isomers with the same functional group

Both X and Z are esters, so they are not functional-group isomers.

However, their carbon arrangements differ.

In X, the carbonyl-side group is methanoate, while the oxygen-linked group is butan-2-yl.

In Z, the carbonyl-side group is butanoate, while the oxygen-linked group is methyl.

They have the same molecular formula but different connectivity within the ester family.

They are structural isomers.

X’ compared with Y and Z

The same reasoning applies to X’.

X’ is an ester with the same connectivity as X, so:

• X’ and Y are functional-group isomers.
• X’ and Z are structural isomers within the ester family.

Y compared with Z

Y is a carboxylic acid, while Z is an ester.

They have the same molecular formula but different functional groups.

They are therefore functional-group isomers.

The complete classification is:

The important distinction is that connectivity and spatial arrangement answer different questions.

X and X’ have the same connectivity but different spatial arrangements.

X and Y have different connectivity and different functional groups.

The molecular formula alone cannot distinguish either situation.

This is why a reliable identification method needs three checks: molecular formula, atom-to-atom connectivity, and any explicitly specified stereochemistry.

07Where this sits in the NSW HSC Chemistry syllabus

There is an important syllabus distinction for students studying in 2026 and 2027.

NESA’s Chemistry 11-12 Syllabus (2025) explicitly includes constructing representations of functional-group isomers of carboxylic acids and esters from C1 to C8.

However, the new syllabus is implemented from 2028, with the first HSC examination under it in 2029.

Students in Year 12 during 2026 and 2027 continue to study the Chemistry Stage 6 Syllabus (2017).

Therefore, the specific C1-C8 construction requirement discussed here is forward preparation for the 2025 syllabus, rather than a new requirement that should automatically be attributed to the 2017 syllabus.

08The next step: naming acids and esters

Once you can identify an acid-ester functional-group isomer pair, the next useful skill is naming each structure correctly.

For esters, remember that the name follows the two sides of the oxygen connection:

• The alkyl group attached to oxygen is named first.
• The acid-derived part is named second, using the alkanoate ending.

For example, \(\ce{CH3COOCH2CH3}\) is ethyl ethanoate.

The ethyl group is attached to oxygen, while the ethanoate portion contains the carbonyl carbon.

Understanding this carbon arrangement makes both ester identification and IUPAC naming much more reliable.