How to Distinguish Saturated and Unsaturated Hydrocarbons

A hydrocarbon has the molecular formula \(\ce{C4H8}\). Is it saturated or unsaturated?

If you remember that alkenes have the general formula \(\ce{C_nH_{2n}}\), you might immediately call it unsaturated. But there’s a catch: cyclobutane also has the formula \(\ce{C4H8}\), and every carbon-carbon bond in cyclobutane is a single bond.

So the molecular formula alone doesn’t always give you the answer.

The reliable method is to look at the carbon-carbon bonds first. When you only have a molecular formula, you need to check whether rings or multiple bonds could explain the number of hydrogen atoms.

01The difference between saturated and unsaturated compounds

For hydrocarbons, the distinction comes down to the bonds between carbon atoms.

• Saturated hydrocarbons contain only single carbon-carbon bonds (\(\ce{C-C}\)).
• Unsaturated hydrocarbons contain at least one carbon-carbon double bond (\(\ce{C=C}\)) or triple bond (\(\ce{C\bond{#}C}\)).

Consider three simple hydrocarbons:

A double bond contains one sigma bond and one pi bond. A triple bond contains one sigma bond and two pi bonds.

Those additional pi bonds are important because they allow unsaturated hydrocarbons to undergo addition reactions. We’ll use this idea later to understand why bromine water behaves differently with alkenes and alkanes.

One important detail: a compound doesn’t need to be an open chain to be saturated. A ring can contain only single carbon-carbon bonds and still be saturated.

02How to distinguish them using a molecular formula

If you are given a molecular formula rather than a structural formula, the first thing to establish is whether the compound is an open-chain hydrocarbon.

For simple, open-chain hydrocarbons, the general formulae are:

Notice that alkynes and dienes share the same general formula. A molecular formula doesn’t necessarily tell you exactly which multiple bonds are present.

What happens when a compound forms a ring?

Here’s where students often make a mistake.

An open-chain alkane with \(n\) carbon atoms has the formula:

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

When a carbon chain forms one ring, two hydrogen atoms are lost. The resulting cycloalkane has the formula:

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

The same reduction of two hydrogen atoms occurs when an open-chain alkane forms one carbon-carbon double bond.

That means both an alkene and a cycloalkane can have the formula \(\ce{C_nH_{2n}}\).

For example:

• Pent-1-ene, \(\ce{C5H10}\), is unsaturated because it contains a C=C bond.
• Cyclopentane, \(\ce{C5H10}\), is saturated because it contains only C-C single bonds.

Both have the same molecular formula, but their structures are different.

The important rule: A shortage of hydrogen atoms compared with an open-chain alkane suggests unsaturation, but it doesn’t prove that a carbon-carbon multiple bond exists. A ring can account for the same difference.

A reliable decision process

When you see a formula in an HSC question, follow this order:

1. If you have a structural or displayed formula: Inspect every carbon-carbon bond. Only single bonds means saturated; any double or triple bond means unsaturated.
2. If you have a molecular formula and an open-chain hydrocarbon is specified: Compare the hydrogen count with the general formulae for alkanes, alkenes, and alkynes.
3. If rings are possible: Don’t automatically identify a compound as unsaturated just because it has fewer hydrogen atoms than an open-chain alkane.
4. If the molecular formula allows multiple structures: State that the formula alone is insufficient and give examples of the possible structures.

These hydrocarbon formula patterns should not be applied blindly to compounds containing other elements. The carbon-carbon bonding structure remains the deciding factor for the distinction being discussed here.

03Addition versus substitution reactions

Understanding saturated and unsaturated structures also helps you predict how hydrocarbons react with halogens such as bromine.

Two reactions are particularly important:

• Addition reactions occur readily with unsaturated hydrocarbons.
• Substitution reactions occur with saturated hydrocarbons under suitable conditions, such as UV light.

The key difference is what happens to the carbon-carbon bonds and the atoms attached to the carbon chain.

Addition reactions: Unsaturated hydrocarbons

An alkene contains a carbon-carbon double bond.

During an addition reaction, the pi bond is broken, and new atoms attach to the two carbon atoms that were originally double-bonded.

The carbon-carbon sigma bond remains, so the double bond becomes a single bond.

For example, ethene reacts with bromine:

\[
\ce{CH2=CH2 + Br2 -> CH2Br-CH2Br}
\]

The product is 1,2-dibromoethane.

Notice what has happened:

• The C=C double bond becomes a C-C single bond.
• The two bromine atoms are added across the original double bond.
• No carbon atoms are removed from the molecule.

This reaction occurs readily at room temperature without requiring UV light.

The bromine solution changes from orange-brown to colourless as bromine is consumed.

Substitution reactions: Saturated hydrocarbons

Alkanes contain only single carbon-carbon bonds.

They do not readily undergo the same addition reaction because they don’t have a C=C or C≡C bond.

However, alkanes can react with halogens through substitution when exposed to UV light.

During substitution, a hydrogen atom is replaced by a halogen atom.

For example, ethane reacts with bromine under UV light:

\[
\ce{CH3CH3 + Br2 ->[UV] CH3CH2Br + HBr}
\]

The products are bromoethane and hydrogen bromide.

Notice the difference from addition:

• One C-H bond is broken.
• One Br-Br bond is broken.
• A C-Br bond and an H-Br bond are formed.
• The carbon-carbon single bond remains unchanged.

UV light provides the energy needed to initiate a free-radical substitution reaction.

The orange-brown bromine colour can also disappear during this reaction because bromine is being consumed.

Addition versus substitution: A comparison

Here’s the trap: Both reactions can cause bromine to lose its colour.

You cannot identify addition or substitution just by observing that bromine has decolourised. You must also consider the starting hydrocarbon and the reaction conditions.

04Using bromine water to test for unsaturation

Suppose you’re given an unknown hydrocarbon and want to determine whether it contains a carbon-carbon multiple bond.

A useful qualitative test is the bromine water test.

Bromine water is an orange-brown solution of bromine in water. If an unsaturated hydrocarbon reacts with the bromine, the bromine is consumed and the solution loses its colour.

How the experiment works

Imagine two test tubes, each containing 2.0 mL of orange bromine water.

A gas syringe is connected to a delivery tube that extends below the surface of the solution. The syringe pushes a measured volume of hydrocarbon through the bromine water, producing bubbles.

The two tubes are kept at the same temperature and protected from UV light.

What should happen if the first tube contains ethene and the second contains ethane?

Ethene is unsaturated, so it reacts with bromine through addition. Ethane is saturated, so it does not react under these dark conditions.

The expected observations are:

The important observation is that ethene decolourises bromine water, while ethane does not.

Why must the experiment be kept away from UV light?

This is an important experimental detail.

Imagine that you leave both test tubes in direct sunlight. The ethane sample may now react with bromine through substitution.

The bromine colour can disappear even though ethane contains only single carbon-carbon bonds.

A student who ignores the reaction conditions might incorrectly conclude that both samples are unsaturated.

The problem isn’t the observation. It’s the interpretation.

The original test needs to be controlled so that the two possible reaction pathways aren’t confused.

A better experimental design:

• Keep the bromine water and hydrocarbon samples away from UV light.
• Use equal volumes of hydrocarbon samples.
• Keep the bromine water concentration and volume constant.
• Use a known alkane as a negative control.
• Use a known alkene as a positive control.

What does the bromine water test actually prove?

A positive result supports the presence of a reactive carbon-carbon multiple bond in a pure hydrocarbon sample.

However, there are limitations.

• The test doesn’t distinguish an alkene from an alkyne. Both can decolourise bromine water.
• It doesn’t tell you where the multiple bond is located.
• Other reducing substances can also decolourise bromine, so the test isn’t uniquely specific to C=C or C≡C bonds in every possible sample.
• A negative result is only useful if the sample has had adequate contact with the bromine water and the experimental conditions are suitable.

Bromine is also hazardous, so this is a supervised laboratory test using appropriate small-scale quantities, not a home experiment.

The strongest approach is to combine structural information, molecular formulae, and chemical test results rather than relying on one observation.

05Questions and solutions

Question 1

An acyclic hydrocarbon has the molecular formula \(\ce{C4H8}\).

Determine whether it is saturated or unsaturated. Show how the molecular formula supports your answer, and explain why the word acyclic matters.

Solution 1

The hydrocarbon is unsaturated because it must contain a carbon-carbon double bond under the stated assumptions.

An open-chain alkane has the general formula:

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

For four carbon atoms:

\[
H=2(4)+2=10
\]

The corresponding saturated open-chain hydrocarbon would therefore be \(\ce{C4H10}\).

The given compound is \(\ce{C4H8}\), which has two fewer hydrogen atoms.

Because the question specifies that the compound is acyclic, a ring cannot account for this difference. The structure must contain one C=C bond.

The compound is therefore unsaturated.

If the question had not specified that the compound was acyclic, cyclobutane would also have been a possible structure, and the formula alone would not have provided a unique answer.

Question 2

Two hydrocarbons, A and B, both have the molecular formula \(\ce{C5H10}\).

• A is an open-chain hydrocarbon.
• B is a cyclic hydrocarbon containing only single carbon-carbon bonds.

Classify both compounds as saturated or unsaturated. Explain how they can have the same molecular formula despite having different classifications.

Solution 2

A is unsaturated, while B is saturated.

For A, the open-chain alkane formula would be:

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

For five carbon atoms, this gives \(\ce{C5H12}\).

Since A has the formula \(\ce{C5H10}\) and is open-chain, it must contain one carbon-carbon double bond.

B has the same molecular formula, but its two-hydrogen reduction is explained by the formation of a ring rather than a double bond.

Because B contains only single carbon-carbon bonds, it is saturated.

The distinction is that a ring and a double bond can both reduce the hydrogen count by two relative to an open-chain alkane.

This is why molecular formulae cannot always be used to classify hydrocarbons without structural information.

Question 3

An unknown, acyclic hydrocarbon X has the molecular formula \(\ce{C6H10}\).

A student claims that X must be an alkyne because alkynes have the general formula \(\ce{C_nH_{2n-2}}\).

Evaluate the student’s claim by giving two different structural possibilities for X. Explain why both structures are consistent with the molecular formula.

Solution 3

X is definitely unsaturated, but the student cannot conclude that it must be an alkyne.

Start by calculating the formula of the corresponding open-chain alkane:

\[
H=2(6)+2=14
\]

The alkane would have the formula \(\ce{C6H14}\).

X has four fewer hydrogen atoms, meaning it has two degrees of unsaturation relative to the open-chain alkane.

There are at least two possible ways to account for this.

Possibility 1: One triple bond

Hex-1-yne has the structural formula:

\[
\ce{HC\bond{#}CCH2CH2CH2CH3}
\]

It contains one carbon-carbon triple bond. This accounts for the loss of four hydrogen atoms relative to hexane.

Possibility 2: Two double bonds

Hexa-1,3-diene has the structural formula:

\[
\ce{CH2=CHCH=CHCH2CH3}
\]

It contains two carbon-carbon double bonds. Each double bond accounts for a reduction of two hydrogen atoms.

Both structures have the molecular formula \(\ce{C6H10}\).

Therefore, the student is correct that X could be an alkyne, but the molecular formula does not prove it. X could also be a diene.

Question 4

A student investigates the reactions of ethene and ethane with bromine water.

Equal volumes of each gas are passed through separate test tubes containing equal volumes of bromine water. Both experiments are initially performed away from UV light.

The following illustrative observations are recorded:

(a) Classify ethene and ethane as saturated or unsaturated.

(b) Write a balanced chemical equation for the reaction occurring in experiment A.

(c) Write a balanced chemical equation for the reaction occurring in experiment C.

(d) Explain why the result in experiment C does not mean that ethane is unsaturated. Identify the reaction type and describe the relevant bond changes.

Solution 4

Ethene is unsaturated, while ethane is saturated. The different observations are explained by addition and substitution reactions.

(a) Classification

Ethene, \(\ce{C2H4}\), contains a carbon-carbon double bond, so it is unsaturated.

Ethane, \(\ce{C2H6}\), contains only carbon-carbon single bonds, so it is saturated.

(b) Reaction in experiment A

Ethene undergoes an addition reaction with bromine.

\[
\ce{CH2=CH2 + Br2 -> CH2Br-CH2Br}
\]

The product is 1,2-dibromoethane.

The C=C double bond becomes a C-C single bond, and one bromine atom attaches to each carbon atom.

The bromine is consumed, causing the orange-brown colour to disappear.

(c) Reaction in experiment C

Ethane undergoes a substitution reaction with bromine under UV light.

\[
\ce{CH3CH3 + Br2 ->[UV] CH3CH2Br + HBr}
\]

The products are bromoethane and hydrogen bromide.

(d) Explanation of the UV result

Experiment C does not show that ethane is unsaturated.

Instead, UV light initiates a free-radical substitution reaction.

One hydrogen atom in ethane is replaced by a bromine atom. The carbon-carbon single bond remains unchanged.

Bromine is still consumed, so its colour fades.

This demonstrates why reaction conditions matter when interpreting a bromine test.

Ethene decolourises bromine through addition, while ethane can consume bromine through substitution when exposed to UV light.

The observation of colour loss alone cannot identify which reaction occurred.

Question 5

An unknown, acyclic hydrocarbon X contains only carbon and hydrogen and has the molecular formula \(\ce{C6H10}\).

Two experiments are performed.

Experiment 1: A sample containing 0.0200 mol of X undergoes complete catalytic hydrogenation over nickel. The reaction consumes 0.0400 mol of hydrogen gas and produces \(\ce{C6H14}\).

Experiment 2: A separate sample undergoes complete addition with excess bromine. It consumes 0.0400 mol of \(\ce{Br2}\) per 0.0200 mol of X.

Two students propose different structures:

• Student A claims X must contain two C=C double bonds because it reacts with two molecules of hydrogen per molecule of X.
• Student B claims X must contain one C≡C triple bond because its formula matches the general formula for an alkyne.

Determine whether either student is correct. Give two possible structural types for X and explain why both experiments produce the same stoichiometric results for these structures.

Solution 5

Neither student can uniquely identify the structure of X. Both one C≡C triple bond and two C=C double bonds are consistent with all the experimental results.

This question tests an important distinction: the number of hydrogen or bromine molecules consumed tells you about the total number of pi bonds available for addition, not necessarily how those bonds are arranged.

Step 1: Calculate the hydrogen-to-hydrocarbon mole ratio.

The experiment uses:

• \(0.0200\) mol of X.
• \(0.0400\) mol of \(\ce{H2}\).

Therefore:

\[
\frac{n(\ce{H2})}{n(X)}
=
\frac{0.0400}{0.0200}
\]

\[
=2.00
\]

Each mole of X consumes two moles of hydrogen.

Step 2: Determine the total degree of unsaturation from the formula.

For six carbon atoms, the corresponding open-chain alkane has:

\[
\ce{C6H14}
\]

The unknown is \(\ce{C6H10}\).

There are four fewer hydrogen atoms than in the corresponding alkane.

This represents two degrees of unsaturation.

Because X is specified as acyclic, these two degrees could be accounted for by:

• One carbon-carbon triple bond.
• Two carbon-carbon double bonds.

Step 3: Test Student A’s proposed structure.

A hydrocarbon containing two C=C bonds has two pi bonds in total.

Each double bond can add one molecule of hydrogen during complete hydrogenation.

Therefore:

\[
\ce{C=C + H2 -> C-C}
\]

For two double bonds, two molecules of hydrogen are consumed.

The final product is a saturated six-carbon alkane, \(\ce{C6H14}\).

Student A’s proposed structure is possible, but it is not the only possibility.

Step 4: Test Student B’s proposed structure.

A carbon-carbon triple bond contains two pi bonds.

Complete hydrogenation therefore requires two molecules of hydrogen:

\[
\ce{C\bond{#}C + 2H2 -> C-C}
\]

The triple bond is converted into a single bond, and the resulting hydrocarbon is also \(\ce{C6H14}\).

Student B’s proposed structure is also possible.

The fact that the molecular formula matches the alkyne general formula does not exclude a diene.

Step 5: Consider the bromine experiment.

The bromine experiment gives the same mole ratio:

\[
\frac{n(\ce{Br2})}{n(X)}
=
\frac{0.0400}{0.0200}
=2.00
\]

A hydrocarbon with two C=C bonds can add two molecules of bromine, one across each double bond.

A hydrocarbon with one C≡C bond can also add two molecules of bromine across the triple bond.

Both structures therefore consume two moles of bromine per mole of hydrocarbon during complete addition.

Final conclusion

Neither student has proved their proposed structure.

Both experiments measure the total number of available pi-bond equivalents. They do not reveal whether those two pi bonds are grouped together in one triple bond or distributed across two double bonds.

A structural formula or additional structural evidence would be needed to distinguish the possibilities.

This is an important limitation of molecular formulae and simple addition tests: they can reveal useful information about a molecule without necessarily identifying its exact structure.

Once you can distinguish saturated and unsaturated hydrocarbons, the next step is predicting their reaction products.

For more practice with the two pathways, see Addition Reactions of Unsaturated Hydrocarbons Explained and Halogen Substitution of Saturated Hydrocarbons for HSC Chemistry.