How to Distinguish Homologues from Isomers in HSC Chemistry

Two compounds can contain the same number and types of atoms but be different molecules. Equally, two compounds can have different molecular formulas yet belong to the same chemical family. The trick is knowing which evidence to compare.

Consider but-1-ene, \(\ce{CH2=CHCH2CH3}\), and but-2-ene, \(\ce{CH3CH=CHCH3}\). Both have the molecular formula \(\ce{C4H8}\), but the carbon-carbon double bond is in a different position. They are structural isomers.

Now compare but-1-ene with pent-1-ene, \(\ce{CH2=CHCH2CH2CH3}\). Both are alkenes, but their molecular formulas are \(\ce{C4H8}\) and \(\ce{C5H10}\). They differ by one \(\ce{CH2}\) unit, making them members of the same homologous series.

That gives you the essential distinction:

• Homologous compounds: Different molecular formulas, the same homologous series, and successive members differ by \(\ce{CH2}\).
• Isomers: The same molecular formula but different arrangements of atoms, either through different connectivity or different spatial arrangements.

The rest of this guide develops a reliable way to recognise both, including the carbon-valence checks that prevent you from accepting impossible structures.

01Why molecular formula alone cannot settle the question

Imagine receiving two samples of an unknown hydrocarbon. You want to establish whether they contain the same compound, different isomers, or members of a homologous series.

An elemental analysis can help, but it has an important limitation: knowing how many carbon and hydrogen atoms are present does not tell you how those atoms are connected.

What combustion analysis actually measures

In a classical combustion-analysis arrangement, a measured sample is burned in excess oxygen inside a heated combustion tube.

The products travel along a gas pathway:

1. The sample enters the heated tube, where complete combustion produces carbon dioxide and water.
2. The gas mixture passes through a water absorber, which captures the water.
3. It then passes through a carbon dioxide absorber, which captures the carbon dioxide.
4. The mass increases of the absorbers are measured to determine the amounts of hydrogen and carbon in the original sample.

Consider but-1-ene and but-2-ene.

Both undergo the same complete combustion reaction:

\[ \ce{C4H8 + 6O2 -> 4CO2 + 4H2O} \]

Predict what happens if equal amounts, measured in moles, of each compound are combusted.

They produce exactly the same amounts of carbon dioxide and water.

This is not a failure of the experiment. It is a limitation of what the experiment measures. Both compounds have the same elemental composition, so elemental analysis cannot distinguish their different carbon arrangements.

A small worked measurement

Illustrative calculation: Assume complete combustion of 1.00 mmol of each alkene.

From the balanced equation:

\[ \begin{aligned} n(\ce{CO2}) &= 4(1.00)\\ &=4.00\text{ mmol} \end{aligned} \]

Similarly:

\[ n(\ce{H2O})=4.00\text{ mmol} \]

The corresponding absorber mass increases would be:

• Water: \(0.00400\times18.0=0.0720\text{ g}\).
• Carbon dioxide: \(0.00400\times44.0=0.176\text{ g}\).

Both isomers give these same calculated results.

Now compare 1.00 mmol of pent-1-ene, \(\ce{C5H10}\):

\[ \ce{2C5H10 + 15O2 -> 10CO2 + 10H2O} \]

It produces 5.00 mmol of carbon dioxide and 5.00 mmol of water.

Illustrative stoichiometric predictions, not experimental measurements. All samples contain 1.00 mmol of hydrocarbon and undergo complete combustion.

The measurement distinguishes different elemental compositions, but it still cannot establish whether two compounds belong to the same homologous series.

For that, you need additional structural information. A structural formula tells you which atoms are connected, while a stereochemical representation can show their spatial arrangement.

02The exact distinction: homologues versus isomers

Think of a molecular formula as a parts list and a structural formula as the assembly instructions.

Isomers have the same parts list but arrange those atoms differently. Homologues belong to the same structural family, but contain different numbers of repeating units.

The analogy has a limit: atoms cannot be connected arbitrarily. Their bonding must obey chemical valence, and some bonds restrict how molecules can rotate.

1. Homologous pairs

A homologous series is a family of compounds with the same characteristic structural feature or functional group, a common general formula, and similar chemical properties.

Successive members differ by one \(\ce{CH2}\) unit.

For example, the alkene series includes:

\[ \ce{C2H4,\ C3H6,\ C4H8,\ C5H10,\ldots} \]

Each successive member contains one additional carbon atom and two additional hydrogen atoms.

A pair of homologues does not have to consist of consecutive members. Two members separated by several positions will differ by multiple \(\ce{CH2}\) units.

For example, ethene and but-1-ene differ by two \(\ce{CH2}\) units.

The important features are:

• Both compounds belong to the same homologous series.
• They have the same characteristic functional group or bonding pattern.
• Their molecular formulas differ by one or more \(\ce{CH2}\) units.

Important: Having the same functional group is not enough to establish that two compounds are homologues. You must also check their molecular formulas.

For example, but-1-ene and but-2-ene are both alkenes, but they have identical molecular formulas. They are isomers, not homologues.

2. Isomers: same formula, different arrangement

Isomers are compounds with the same molecular formula but different arrangements of atoms.

There are two important types to recognise.

Structural isomers (constitutional isomers)

Structural isomers have the same molecular formula but different atom-to-atom connectivity.

For example:

\[ \ce{CH3CH2CH2OH} \]

and

\[ \ce{CH3CH(OH)CH3} \]

Both have the molecular formula \(\ce{C3H8O}\).

However, the hydroxyl group, \(\ce{-OH}\), is attached to a different carbon atom. The first compound is propan-1-ol, while the second is propan-2-ol.

They are positional structural isomers.

Another possibility is functional-group isomerism, where the compounds have different functional groups despite having the same molecular formula.

For example, an alcohol and an ether can both have the formula \(\ce{C3H8O}\).

Stereoisomers

Stereoisomers have the same molecular formula and the same atom-to-atom connectivity, but differ in their three-dimensional spatial arrangement.

A useful HSC example is cis-trans isomerism in alkenes.

Consider but-2-ene:

• In cis-but-2-ene, the two methyl groups are on the same side of the double bond.
• In trans-but-2-ene, the two methyl groups are on opposite sides.

Why can’t these molecules simply rotate into one another?

A carbon-carbon double bond contains a sigma bond and a pi bond. Rotation around the double bond would disrupt the pi bonding, so the two arrangements cannot freely interconvert in the way groups can rotate around an ordinary carbon-carbon single bond.

This is why cis-trans forms are distinct stereoisomers.

A comparison worth remembering

One particularly important distinction is that matching molecular formulas are necessary for isomerism, but they do not prove that two structures are isomers.

If the structures are actually identical, they are simply two representations of the same molecule.

03Carbon valence: how to reject impossible structures

You might correctly identify two structures as having the same number of carbon atoms, but still accept an impossible molecular structure.

This is where carbon valence becomes essential.

In ordinary neutral organic compounds, carbon forms four bonds in total. A single bond contributes one bond, a double bond contributes two, and a triple bond contributes three.

You can calculate the number of implied hydrogen atoms attached to a carbon using:

\[ H_{\mathrm C}=4-\sum b \]

Where:

• \(H_{\mathrm C}\) is the number of hydrogen atoms attached to that carbon.
• \(\sum b\) is the sum of the bond orders around that carbon.

For example, consider a carbon attached to three other carbon atoms through single bonds.

\[ H_{\mathrm C}=4-(1+1+1)=1 \]

It has one hydrogen atom.

Now consider a carbon involved in a double bond and two additional single bonds.

\[ H_{\mathrm C}=4-(2+1+1)=0 \]

It has no hydrogen atoms.

This gives you a useful validity check:

• Bond-order sum of four: carbon has no additional hydrogen.
• Bond-order sum below four: carbon has enough remaining valence for the corresponding number of hydrogens.
• Bond-order sum above four: the proposed structure is invalid for an ordinary neutral carbon atom.

This rule assumes the ordinary neutral carbon structures used in introductory organic chemistry. Unusual charged or radical species require separate treatment.

Carbon valence is not itself a classification test. It is a check that the structures you are comparing are chemically possible.

04A reliable method for distinguishing the two

When an exam gives you two molecular or structural formulas, use this sequence.

Step 1: Write both molecular formulas.

Count every carbon, hydrogen, and other explicitly shown atom.

For skeletal formulas, remember that each unlabeled endpoint and corner normally represents a carbon atom, while hydrogens attached to carbon are usually omitted.

Step 2: Compare the molecular formulas.

• Different molecular formulas: the compounds cannot be isomers.
• Identical molecular formulas: they could be isomers, but you need to check their structures.

Use molecular formulas, not empirical formulas. Two compounds can have the same empirical formula but different molecular formulas.

Step 3: If the formulas differ, check for homologous relationships.

Ask:

• Do both compounds belong to the same homologous series?
• Do they have the same characteristic functional group or bonding pattern?
• Do their molecular formulas differ by one or more \(\ce{CH2}\) units?

If all these conditions are satisfied, they are homologues.

Step 4: If the formulas match, compare the arrangements.

• Different atom-to-atom connections: structural isomers.
• Same connectivity but different spatial arrangement: stereoisomers.
• Same connectivity and same stereochemical arrangement, despite different-looking drawings: the same molecule.

Step 5: Check carbon valence.

Make sure every carbon has four bonds in total, counting double and triple bonds correctly.

If a proposed structure violates carbon valence, reject it before classifying the pair.

If you need more practice distinguishing different-looking structural drawings from genuinely different molecules, see How to Check if Two Skeletal Formulas Show the Same Molecule.

05Worked examples

06Questions and solutions

Try each question before reading its solution. The questions progress from straightforward classification to more demanding combinations of molecular formula, connectivity, spatial arrangement, and carbon valence.

Question 1

Butane has the molecular formula \(\ce{C4H10}\), while pentane has the molecular formula \(\ce{C5H12}\).

Determine whether these compounds are homologues or isomers. Justify your answer using their molecular formulas and homologous series.

Solution 1

Butane and pentane are homologues, not isomers.

Both are saturated, open-chain hydrocarbons belonging to the alkane homologous series.

The general formula for alkanes is:

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

Compare their molecular formulas:

\[ \begin{aligned} \text{Butane: }&\ce{C4H10}\\ \text{Pentane: }&\ce{C5H12} \end{aligned} \]

The difference is:

\[ \ce{CH2} \]

Therefore, pentane is the next member of the alkane homologous series after butane.

The tempting mistake is to call them isomers because they have similar structures and chemical properties. However, isomers must have identical molecular formulas. These compounds have different formulas, so they are homologues.

Question 2

Consider the following two compounds:

• Compound A: propan-1-ol, \(\ce{CH3CH2CH2OH}\).
• Compound B: propan-2-ol, \(\ce{CH3CH(OH)CH3}\).

A student claims that these compounds must be homologues because they both contain an alcohol functional group.

Assess this claim. Calculate their molecular formulas and explain the structural relationship between the compounds.

Solution 2

Propan-1-ol and propan-2-ol are positional structural isomers, not homologues.

Both compounds have the molecular formula:

\[ \ce{C3H8O} \]

However, their hydroxyl groups occupy different positions.

In propan-1-ol, the hydroxyl group is attached to the first carbon atom.

In propan-2-ol, it is attached to the second carbon atom.

The carbon skeleton is the same length, but the atom-to-atom connectivity involving oxygen is different.

The carbon valence check also confirms that both structures are valid.

For the central carbon in propan-2-ol, there are:

• One bond to the first carbon.
• One bond to the third carbon.
• One bond to oxygen.
• One bond to hydrogen.

Therefore:

\[ \sum b=1+1+1+1=4 \]

The carbon satisfies its normal valence.

The student’s mistake is assuming that sharing a functional group automatically makes two compounds homologues. The molecular formulas must also differ by one or more \(\ce{CH2}\) units.

Here, the formulas are identical, and the hydroxyl group changes position. These are positional isomers.

Question 3

Two compounds have the following structural formulas:

\[ \text{A: }\ce{CH2=CHCH2CH3} \]

\[ \text{B: }\ce{CH2=C(CH3)2} \]

Both compounds have the molecular formula \(\ce{C4H8}\).

Determine their relationship. In your answer:

(a) Explain how their carbon connectivity differs.

(b) Demonstrate that the central carbon in compound B satisfies carbon’s normal valence.

(c) Explain why the compounds are not homologues.

Solution 3

Compounds A and B are structural isomers, not homologues.

(a) Their connectivity differs.

Compound A is but-1-ene. Its carbon atoms form a continuous four-carbon chain.

Compound B is 2-methylprop-1-ene. It has a branched carbon arrangement, with two methyl groups attached to the central alkene carbon.

Both structures contain four carbon atoms, eight hydrogen atoms, and one carbon-carbon double bond.

However, their carbon-to-carbon connections are different.

(b) The central carbon in compound B has four bonds.

The central carbon is connected to:

• One carbon through a double bond.
• One methyl carbon through a single bond.
• Another methyl carbon through a single bond.

Its total bond-order sum is:

\[ \sum b=2+1+1=4 \]

It therefore has no additional hydrogen atoms.

The terminal \(\ce{CH2}\) carbon has a double bond and two hydrogen atoms, giving another total of four.

The proposed structure is chemically valid.

(c) They are not homologues.

Although both compounds are alkenes, they have identical molecular formulas.

There is no additional \(\ce{CH2}\) unit separating them.

They are therefore structural isomers, specifically chain or skeletal isomers.

The important point is that two compounds can share a functional group and molecular formula while having different carbon skeletons.

Question 4

Pent-1-ene has the structural formula:

\[ \ce{CH2=CHCH2CH2CH3} \]

and molecular formula \(\ce{C5H10}\).

A student proposes the following structure as a branched structural isomer:

\[ \ce{(CH3)3C=CH2} \]

Assess whether this proposed structure is valid.

(a) Count its carbon and hydrogen atoms.

(b) Calculate the bond-order sum around the central carbon.

(c) Write a valid branched alkene with the molecular formula \(\ce{C5H10}\), and classify it against pent-1-ene.

Solution 4

The proposed structure is invalid and cannot be an isomer of pent-1-ene.

(a) Count the atoms.

The proposed structure contains:

• Three methyl groups, contributing three carbon atoms and nine hydrogen atoms.
• One central carbon atom.
• One terminal \(\ce{CH2}\) group, contributing one carbon atom and two hydrogen atoms.

Therefore, its atom count as written is:

\[ \ce{C5H11} \]

It does not have the claimed formula \(\ce{C5H10}\).

(b) Check the central carbon’s valence.

The central carbon has three single bonds to methyl groups and one double bond to the terminal carbon.

Its bond-order sum is:

\[ \begin{aligned} \sum b &=1+1+1+2\\ &=5 \end{aligned} \]

This exceeds the normal valence of four for a neutral carbon atom.

The structure is therefore chemically invalid as an ordinary neutral organic molecule.

Simply counting the number of carbon atoms is not enough to establish that a proposed structure is possible.

(c) Construct a valid branched alkene.

A suitable alternative is 2-methylbut-1-ene:

\[ \ce{CH2=C(CH3)CH2CH3} \]

Count its atoms:

• Terminal \(\ce{CH2}\): two hydrogen atoms.
• Methyl branch: three hydrogen atoms.
• \(\ce{CH2}\) in the ethyl group: two hydrogen atoms.
• Terminal \(\ce{CH3}\): three hydrogen atoms.

Total:

\[ 2+3+2+3=10 \]

The molecular formula is:

\[ \ce{C5H10} \]

The central alkene carbon has a bond-order sum of:

\[ 2+1+1=4 \]

The structure is valid.

2-methylbut-1-ene and pent-1-ene are structural isomers because they have the same molecular formula but different carbon connectivity.

They are not homologues because their molecular formulas are identical.

The key lesson is that a proposed isomer must satisfy both requirements: the correct molecular formula and chemically valid bonding.

Question 5

An examiner provides five structural drawings:

• P: But-1-ene, \(\ce{CH2=CHCH2CH3}\), with the ethyl chain drawn to the right of the double bond.
• Q: But-1-ene, \(\ce{CH3CH2CH=CH2}\), with the same chain written from the opposite direction.
• R: Cis-but-2-ene, \(\ce{CH3CH=CHCH3}\), with both methyl groups on the same side of the double bond.
• S: Trans-but-2-ene, \(\ce{CH3CH=CHCH3}\), with the methyl groups on opposite sides.
• T: Pent-1-ene, \(\ce{CH2=CHCH2CH2CH3}\).

An answer sheet makes the following claims:

1. P and Q are geometric isomers because the ethyl chain appears on a different side of the double bond.
2. R and S are the same molecule because their atoms have identical connectivity.
3. P and R are structural isomers because their molecular formulas match but their double bonds occupy different positions.
4. P and T are isomers because both compounds are alkenes.

Evaluate all four claims.

Identify which pairs represent the same molecule, structural isomers, stereoisomers, or homologues. Explain the conditions that determine each classification, including why not every alkene can exhibit geometric isomerism.

Solution 5

The correct classifications are:

• P and Q: the same molecule.
• R and S: stereoisomers.
• P and R: structural isomers.
• P and T: homologues.

The four claims need to be considered separately.

Claim 1: P and Q are geometric isomers.

This claim is incorrect.

P and Q both represent but-1-ene. Their structural formulas are simply written in opposite directions.

More importantly, but-1-ene has the terminal structure:

\[ \ce{CH2=CH-} \]

The terminal carbon has two identical hydrogen substituents.

For geometric isomerism around a carbon-carbon double bond, each double-bonded carbon must have two different substituents.

This condition is not satisfied because one carbon has two identical hydrogen atoms.

Consequently, but-1-ene cannot exhibit cis-trans or E/Z isomerism.

The different positions of the ethyl chain in the drawings do not create different molecules.

P and Q are two representations of the same compound.

Claim 2: R and S are the same molecule.

This claim is incorrect.

R and S have identical molecular formulas and identical atom-to-atom connectivity.

However, their spatial arrangements differ.

In cis-but-2-ene, both methyl groups are on the same side of the double bond.

In trans-but-2-ene, they are on opposite sides.

Each double-bonded carbon has two different substituents: hydrogen and methyl.

The double bond restricts rotation, so the two configurations cannot freely interconvert.

R and S are therefore stereoisomers, specifically geometric isomers.

The important distinction is that identical connectivity does not always mean identical molecules. Spatial arrangement must also be considered.

Claim 3: P and R are structural isomers.

This claim is correct.

Both compounds have the molecular formula:

\[ \ce{C4H8} \]

However, their double bonds occupy different positions.

For P, the double bond is between the first and second carbon atoms.

For R, the double bond is between the second and third carbon atoms.

This changes the connectivity of the carbon atoms involved in the double bond.

Both compounds are valid alkenes, but their structures differ.

They are positional structural isomers.

Notice that P and R are not stereoisomers. Their difference is in connectivity, not simply spatial arrangement.

Claim 4: P and T are isomers.

This claim is incorrect.

Both compounds belong to the alkene homologous series, but their molecular formulas differ:

\[ \begin{aligned} P &: \ce{C4H8}\\ T &: \ce{C5H10} \end{aligned} \]

The difference is:

\[ \ce{CH2} \]

Pent-1-ene contains one additional methylene unit compared with but-1-ene.

Therefore, P and T are successive members of the same homologous series.

They are homologues, not isomers.

The deeper lesson

The most tempting mistake in this question is to rely on only one structural feature.

For example:

• Matching functional groups do not automatically establish homologous relationships.
• Matching molecular formulas do not automatically establish isomerism.
• Matching connectivity does not automatically mean two molecules have the same spatial arrangement.
• Different-looking drawings do not automatically represent different molecules.

You need to check molecular formula, connectivity, spatial arrangement, and carbon valence together.

07What to study next

Once you can reliably distinguish homologues from isomers, the next useful step is understanding how their structures influence their physical and chemical properties.

For homologous compounds, adding a \(\ce{CH2}\) unit changes molecular size and generally affects properties such as boiling point, volatility, and viscosity.

For isomers, different connectivity or spatial arrangement can produce different physical and chemical behaviour despite an identical molecular formula.

Continue with Properties Across a Homologous Series: HSC Chemistry Guide to connect the structural differences you’ve identified with the properties you are expected to explain in HSC Chemistry.