How to Use a Reaction Observation to Distinguish an Alkane from an Alkene
Two clear hydrocarbon samples sit in separate test tubes. You add a few drops of orange-brown bromine water to each.
Two clear hydrocarbon samples sit in separate test tubes. You add a few drops of orange-brown bromine water to each. One sample rapidly loses its colour. The other remains orange-brown.
If the two possible substances are an alkane and an alkene, the sample that decolourises the bromine water is the alkene. The sample that shows no overall colour change is the alkane.
This is the bromine water test. It works because an alkene undergoes an addition reaction with bromine, while an alkane does not react under normal test conditions.
There are two important traps, though. An alkane can react with bromine under UV light, and bromine can move between water and an organic liquid without actually reacting. Both can lead to an incorrect conclusion.
Let’s work through how to perform the test, interpret the observations, and explain the chemistry accurately.
01What does the bromine water test show?
Bromine water is an aqueous solution of bromine, \(\ce{Br2}\). It has a characteristic orange-brown colour.
When you add it to a hydrocarbon, the observation depends on the bonds present in that hydrocarbon.
- An alkene contains a carbon-carbon double bond, \(\ce{C=C}\), which reacts with bromine.
- An alkane contains only carbon-carbon and carbon-hydrogen single bonds. It does not react with bromine under the normal test conditions.
The essential comparison is:
| Observation | Interpretation |
|---|---|
| Orange-brown bromine colour disappears rapidly | Consistent with an alkene |
| Orange-brown colour remains | Consistent with an alkane |
| Colour disappears only after exposure to UV light | Could indicate alkane substitution, not necessarily an alkene |
| Water layer becomes pale, but organic layer remains orange | Bromine may have moved between layers rather than reacted |
The test assumes that the unknown is either an alkane or an alkene. A positive result establishes the presence of a reactive unsaturated bond, but it does not uniquely identify an alkene if other substances are possible.
02Why does an alkene decolourise bromine water?
The answer comes down to the carbon-carbon double bond.
A double bond consists of:
- One sigma (\(\sigma\)) bond.
- One pi (\(\pi\)) bond.
The pi bond is more exposed and its electrons can interact with an incoming electrophile, such as bromine.
This makes the double bond reactive towards addition reactions.
Addition: what happens to the bonds?
Consider ethene, \(\ce{CH2=CH2}\), reacting with bromine.
Before the reaction, the two carbon atoms are connected by a double bond.
During the reaction:
- The pi bond is broken.
- The carbon-carbon sigma bond remains.
- Each carbon forms a new bond to a bromine-containing species.
The overall effect is that atoms are added across the original double bond.
For the simplified reaction with bromine in an inert organic solvent:
\[ \ce{CH2=CH2 + Br2 -> BrCH2-CH2Br} \]
The product is 1,2-dibromoethane.
This reaction is called an addition reaction because atoms from the reacting substance are added to the original molecule.
More specifically, it is an electrophilic addition reaction.
Why does the colour disappear?
Bromine itself is orange-brown.
As the alkene reacts, bromine is consumed to form new chemical products. The concentration of coloured bromine decreases, so the solution loses its colour.
This is the chemical reason behind the positive result.
However, there is a detail worth understanding if a question asks about the actual products of bromine water.
A note about bromine water and reaction products
Pure bromine in an inert organic solvent gives the familiar dibromoalkane product.
Bromine water is slightly different because bromine reacts with water:
\[ \ce{Br2 + H2O <=> HOBr + HBr} \]
The resulting hypobromous acid, \(\ce{HOBr}\), and hydrobromic acid, \(\ce{HBr}\), can also react with an alkene.
For ethene, simplified representations include:
\[ \ce{CH2=CH2 + HOBr -> BrCH2-CH2OH} \]
\[ \ce{CH2=CH2 + HBr -> CH3-CH2Br} \]
This means aqueous bromine can produce a mixture of products, including bromohydrins and bromoalkanes. Some dibromoalkane can also form.
For the qualitative test, the important point is that the reactive carbon-carbon double bond consumes bromine species, causing the orange-brown colour to fade.
The NSW Department of Education’s Chemistry Module 8 guide specifically highlights this distinction between bromine in an organic solvent and bromine water. It also emphasises understanding the reaction and its evidence rather than memorising just one product.
03Why doesn’t an alkane normally decolourise bromine water?
Now consider ethane, \(\ce{CH3CH3}\).
Unlike an alkene, ethane contains no carbon-carbon double bond. It has only single covalent bonds.
These bonds do not undergo the same addition reaction with bromine under normal test conditions.
Therefore:
\[ \ce{CH3CH3 + Br2 -> No\ reaction} \]
For a test performed at room temperature without UV exposure, the orange-brown bromine colour remains.
But there’s an important exception.
Substitution: what happens under UV light?
An alkane can react with bromine when exposed to UV light.
Instead of adding bromine across a double bond, the reaction replaces a hydrogen atom with a bromine atom.
For ethane:
\[ \ce{CH3CH3 + Br2 ->[UV] CH3CH2Br + HBr} \]
The products are bromoethane and hydrogen bromide.
Notice what has happened:
- One carbon-hydrogen bond is broken.
- One carbon-bromine bond is formed.
- The carbon skeleton remains unchanged.
This is a substitution reaction.
UV light provides the energy needed to initiate a free-radical substitution reaction.
The distinction is important because both reactions consume bromine.
If an alkane is exposed to UV light, its bromine colour can eventually fade as substitution occurs.
A student who ignores the reaction conditions might incorrectly identify that alkane as an alkene.
The bromine water test must therefore be performed away from UV light.
Addition versus substitution: the comparison
| Feature | Alkene + bromine | Alkane + bromine |
|---|---|---|
| Reaction type | Addition | Substitution |
| Conditions | Room temperature, away from UV | UV light required |
| Bond change | C=C becomes C-C | C-H replaced by C-Br |
| Main simplified products | Dibromoalkane with Br2 in an inert solvent | Haloalkane and HBr |
| Bromine colour | Disappears rapidly | Can fade under UV, generally more slowly |
| Result in the standard test | Positive | Negative |
This is why stating only that bromine water changes colour is not enough in an HSC response. You must explain the reaction responsible for the observation and identify the conditions.
04How to perform the test correctly
The original experimental question is straightforward:
Does the unknown hydrocarbon contain a carbon-carbon double bond?
Bromine water provides a visible way to investigate this because the reagent is coloured. You can observe whether it is consumed without needing to identify the products directly.
The experimental setup
For a comparison, prepare separate labelled test tubes containing small, equal quantities of the hydrocarbon samples.
For a school investigation, cyclohexane and cyclohexene are useful reference substances.
- Cyclohexane is the alkane.
- Cyclohexene is the alkene.
Add a few drops of bromine water to each tube using a dropper.
Gently shake each tube to mix the substances, then allow the layers to settle. Keep the tubes away from direct sunlight and UV sources.
What should you observe?
The following is an illustrative set of expected observations.
| Sample | Conditions | Observation | Conclusion |
|---|---|---|---|
| Cyclohexane | Room temperature, no UV | Orange-brown colour remains overall | No alkene reaction |
| Cyclohexene | Room temperature, no UV | Orange-brown colour disappears | Addition reaction |
| Cyclohexane | Exposed to UV | Colour gradually fades | Substitution reaction |
The cyclohexene sample gives the positive result because its carbon-carbon double bond reacts with bromine.
The cyclohexane sample gives the negative result because it does not undergo this reaction in the absence of UV light.
The UV-exposed cyclohexane sample demonstrates why the experimental conditions matter.
An important complication: two liquid layers
There is another possible source of confusion.
Cyclohexane and cyclohexene are both non-polar hydrocarbons. They do not mix completely with water.
When bromine water is shaken with either substance, two liquid layers can form.
For cyclohexane and cyclohexene, the organic hydrocarbon layer sits above the aqueous layer because these liquids are less dense than water.
Bromine is more soluble in the organic layer than in the aqueous layer.
This creates a possible false positive.
Imagine that you add bromine water to cyclohexane. After shaking, the lower aqueous layer becomes much paler, while the upper organic layer becomes orange-brown.
Has the bromine reacted?
No. It has mainly moved from the water layer into the organic layer.
The bromine has changed where it is dissolved, not necessarily undergone a chemical reaction.
Think of the bromine colour as a visible tracer. If it moves from one layer to another, it has changed location, not disappeared through a reaction.
The analogy stops there: the actual movement is caused by differences in solubility and intermolecular interactions.
The correct observation is therefore the loss of bromine colour from the overall mixture, not simply the fading of the aqueous layer.
This phase-partitioning issue is discussed in the NSW Department of Education’s Chemistry Module 8 guide, which also warns about false positives caused by UV exposure.
Three mistakes to avoid
Mistake 1: Performing the test in sunlight
UV light can initiate substitution in an alkane.
Correction: Keep the test away from UV light and direct sunlight.
Mistake 2: Looking only at the aqueous layer
Bromine can dissolve into the organic layer, making the water layer appear paler without chemical consumption.
Correction: Allow the layers to settle and inspect both.
Mistake 3: Adding too much bromine water
If there is a large excess of bromine, its remaining orange-brown colour can mask the fact that some bromine has reacted.
Correction: Use a small amount of bromine water and compare the result with appropriate controls.
A reliable investigation should also keep sample volumes, temperature, shaking, and observation time consistent.
Bromine water is hazardous. Any practical investigation must use a teacher-approved microscale procedure, suitable protective equipment, and appropriate laboratory ventilation.
05Worked examples
Worked example: Identify two unknown hydrocarbons
Two unknown liquid hydrocarbons, A and B, are tested separately.
Equal quantities of each are mixed with a few drops of bromine water at room temperature, away from UV light.
The observations are:
- Sample A rapidly loses its orange-brown colour throughout the mixture.
- Sample B develops a pale aqueous layer, but the upper organic layer remains orange-brown.
Identify the likely hydrocarbon in each sample and explain the observations.
Step 1
Sample A loses the bromine colour throughout the mixture.
This indicates that bromine has been chemically consumed by a reaction with an unsaturated bond.
Step 2
Sample A contains a carbon-carbon double bond, which reacts with bromine through addition.
Therefore, A is an alkene.
Step 3
The aqueous layer becomes pale, but the organic layer remains orange-brown.
This indicates that bromine has moved into the organic layer rather than being consumed in an addition reaction.
Step 4
Assuming the only possible substances are alkanes and alkenes:
- A is an alkene.
- B is an alkane.
The important distinction is that a change in the colour of one layer does not necessarily mean a chemical reaction has occurred.
Worked example: Why does an alkane decolourise bromine under UV?
An unknown hydrocarbon does not decolourise bromine water after 30 seconds at room temperature in the absence of UV light.
A fresh portion of the same sample is then exposed to UV light in the presence of bromine. The orange-brown colour gradually fades.
A student concludes that the unknown must be an alkene because bromine has been consumed.
Explain why this conclusion is incorrect.
Step 1
There is no rapid bromine consumption in the absence of UV light.
This is consistent with an alkane because it does not undergo the normal bromine addition reaction.
Step 2
UV exposure allows the alkane to undergo free-radical substitution.
A hydrogen atom is replaced by bromine, forming a haloalkane and hydrogen bromide.
For example:
\[ \ce{CH4 + Br2 ->[UV] CH3Br + HBr} \]
Step 3
Bromine is consumed during substitution, so the orange-brown colour gradually fades.
Step 4
The unknown is consistent with an alkane, not an alkene.
The student’s mistake was assuming that every reaction consuming bromine must be an addition reaction.
The reaction conditions distinguish the two possibilities.
Worked example: How much bromine is present in the test?
A teacher-prepared bromine solution has an illustrative concentration of \(0.0150\ \mathrm{mol\,L^{-1}}\).
A student adds \(0.40\ \mathrm{mL}\) of this solution to a hydrocarbon sample.
Calculate the amount of bromine added. Assuming the simplified 1:1 addition model, determine how many moles of carbon-carbon double bonds this quantity could react with.
Step 1
The amount of substance is calculated using:
\[ n=cV \]
where:
- \(n\) is the amount in moles.
- \(c\) is the concentration in \(\mathrm{mol\,L^{-1}}\).
- \(V\) is the volume in litres.
Step 2
\[ 0.40\ \mathrm{mL}=4.0\times10^{-4}\ \mathrm{L} \]
Step 3
\[ \begin{aligned} n &= cV\\ &=0.0150\times4.0\times10^{-4}\\ &=6.0\times10^{-6}\ \mathrm{mol} \end{aligned} \]
Step 4
In the simplified addition model, one mole of bromine reacts with one mole of carbon-carbon double bonds.
Therefore, this quantity of bromine could react with:
\[ 6.0\times10^{-6}\ \mathrm{mol} \]
of C=C bonds.
Final answer: \(6.0\times10^{-6}\ \mathrm{mol}\) of bromine is present, equivalent to \(6.0\times10^{-6}\ \mathrm{mol}\) of C=C bonds in the simplified 1:1 model.
This is a useful scale estimate, not a quantitative determination of an unknown sample’s alkene concentration. In an actual qualitative test, excess bromine can leave a residual orange-brown colour even when some alkene has reacted.
06Questions and solutions
The following questions progress from standard identification towards more challenging experimental reasoning. The final question combines reaction stoichiometry, experimental controls, and the effect of a limiting reagent.
Question 1
A student adds bromine water to two separate hydrocarbon samples at room temperature, away from UV light.
Sample X rapidly decolourises the bromine water. Sample Y shows no overall colour change.
Assuming both samples are pure hydrocarbons and the only possibilities are an alkane and an alkene:
a) Identify X and Y.
b) State the reaction occurring in sample X.
c) Explain why sample Y does not react under these conditions.
Solution 1
X is an alkene, and Y is an alkane. Sample X undergoes addition, while sample Y does not react with bromine under the stated conditions.
a) Identification
X is an alkene because it rapidly decolourises bromine water.
Y is an alkane because it shows no overall colour change.
b) Reaction in sample X
The carbon-carbon double bond reacts with bromine through an addition reaction.
The pi bond is broken, the carbon-carbon sigma bond remains, and new bonds form between the carbon atoms and the incoming bromine-containing species.
Bromine is consumed, causing the orange-brown colour to disappear.
c) Why sample Y does not react
Sample Y contains only single carbon-carbon and carbon-hydrogen bonds.
Under normal test conditions, it does not undergo addition with bromine. Substitution requires UV light to initiate the reaction.
Therefore, the bromine colour remains.
Question 2
A student performs three tests using equal quantities of three hydrocarbon samples.
The results are recorded below.
| Test | Hydrocarbon | Conditions | Observation |
|---|---|---|---|
| A | Unknown P | Room temperature, no UV | No overall colour change |
| B | Unknown P | UV exposure | Bromine colour gradually disappears |
| C | Known cyclohexene | Room temperature, no UV | Bromine colour disappears rapidly |
a) Explain what the observations suggest about unknown P.
b) Explain why test B does not establish that P is an alkene.
c) Explain why test C is useful when interpreting the results.
Solution 2
Unknown P is consistent with an alkane, provided the sample is one pure hydrocarbon from the two proposed classes. Its reaction under UV light is explained by substitution rather than addition.
a) Interpretation of unknown P
In test A, P does not consume bromine under normal test conditions.
This suggests that P does not contain a readily reactive carbon-carbon double bond.
In test B, UV exposure allows bromine to react with an alkane through free-radical substitution.
A hydrogen atom is replaced by a bromine atom, consuming bromine and causing the colour to fade.
b) Why test B is not proof of an alkene
Both alkenes and UV-exposed alkanes can consume bromine.
The reaction type depends on the molecular structure and conditions:
- An alkene undergoes addition without UV.
- An alkane undergoes substitution when exposed to UV.
Therefore, test B alone cannot distinguish the two.
c) Purpose of test C
Cyclohexene is a known alkene, so it acts as a positive control.
Its rapid decolourisation confirms that the bromine reagent is working and that the experimental conditions allow the expected addition reaction.
This makes the lack of reaction in test A more meaningful.
Question 3
A student adds bromine water to cyclohexane and shakes the test tube.
After allowing the mixture to settle, the following observations are recorded:
- The lower aqueous layer is almost colourless.
- The upper organic layer is orange-brown.
The student reports that cyclohexane has decolourised bromine water and must therefore be an alkene.
Evaluate this conclusion and explain what has actually happened.
Solution 3
The student has incorrectly identified a physical redistribution of bromine as a chemical reaction. The observation is consistent with cyclohexane remaining unreacted.
Cyclohexane is non-polar and does not mix completely with water.
Bromine is more soluble in the organic cyclohexane layer than in the aqueous layer.
When the tube is shaken, bromine transfers from the water into the organic phase.
The lower aqueous layer becomes paler, but the upper organic layer becomes orange-brown.
The bromine has not necessarily been chemically consumed.
For an alkene, bromine reacts with the carbon-carbon double bond. This consumes the coloured bromine species rather than simply transferring them between layers.
The correct conclusion is that the student must inspect both layers before deciding whether a positive result has occurred.
A colour change in one phase is not sufficient evidence of an addition reaction.
Question 4
An unknown hydrocarbon has the molecular formula \(\ce{C4H6}\).
It rapidly decolourises bromine water at room temperature in the absence of UV light.
A student claims that the unknown must be an alkene.
Evaluate this claim by considering the following possible structures:
- Buta-1,3-diene: \(\ce{CH2=CH-CH=CH2}\)
- But-1-yne: \(\ce{HC\bond{#}C-CH2-CH3}\)
Explain why the bromine water test cannot uniquely identify the unknown as an alkene.
Solution 4
The student’s claim is not justified. The bromine water test indicates the presence of a reactive unsaturated bond, but it cannot distinguish an alkene from an alkyne.
Buta-1,3-diene contains two carbon-carbon double bonds.
These bonds can undergo addition reactions with bromine, consuming the coloured reagent.
But-1-yne contains a carbon-carbon triple bond.
It is also unsaturated and can react with bromine through addition reactions.
Both compounds have the molecular formula \(\ce{C4H6}\), and both can give a positive bromine test.
The test therefore establishes that the sample contains a reactive unsaturated bond, but it does not uniquely establish which type of bond is present.
A further test or structural analysis would be needed to distinguish the possible compounds.
This illustrates an important limitation of qualitative analysis: a test can provide strong evidence for a structural feature without identifying the entire molecule.
Question 5
An unknown hydrocarbon mixture is known to contain only cyclohexane and cyclohexene.
A chemist wants to determine how much of each substance is present.
Two equal aliquots of the mixture are tested separately. All bromine consumption is measured using an independent quantitative method.
The following illustrative data are obtained:
| Experimental tube | Conditions | Bromine consumed |
|---|---|---|
| A | Room temperature, protected from UV | 0.240 mmol |
| B | Exposed to UV under conditions allowing both reactions | 0.390 mmol |
The following assumptions apply:
- Both aliquots contain identical quantities of the original mixture.
- In tube A, all cyclohexene reacts, while cyclohexane remains unreacted.
- In tube B, all cyclohexene reacts and each cyclohexane molecule undergoes exactly one substitution.
- Each cyclohexene molecule consumes one mole of bromine in the simplified addition model.
- Each cyclohexane molecule undergoing its first substitution consumes one mole of bromine.
- No other reactions occur.
a) Calculate the amount of cyclohexene in each aliquot.
b) Calculate the amount of cyclohexane in each aliquot.
c) A student proposes testing the same aliquot in tube A and then exposing it to UV light. The bromine colour has already disappeared, and no further colour change is observed. The student concludes that the mixture contains no cyclohexane.
Explain why this conclusion is incorrect.
Solution 5
Each aliquot contains 0.240 mmol of cyclohexene and 0.150 mmol of cyclohexane. The student’s conclusion is incorrect because the first experiment has exhausted its bromine reagent, leaving nothing available for the UV-initiated substitution reaction.
a) Calculate the amount of cyclohexene
In tube A, the reaction occurs away from UV light.
Under these conditions, cyclohexene undergoes addition, while cyclohexane does not react.
Therefore, the bromine consumed in tube A corresponds entirely to the cyclohexene present.
Using the 1:1 stoichiometric relationship:
\[ \begin{aligned} n(\text{cyclohexene}) &=n(\ce{Br2})\\ &=0.240\ \mathrm{mmol} \end{aligned} \]
The amount of cyclohexene in each aliquot is:
\[ \boxed{0.240\ \mathrm{mmol}} \]
b) Calculate the amount of cyclohexane
Tube B contains both hydrocarbons and is exposed to UV light.
Under the stated assumptions, both reactions occur:
- Cyclohexene consumes bromine through addition.
- Cyclohexane consumes bromine through substitution.
The total bromine consumed is therefore the sum of both contributions.
\[ n(\text{total Br}_2) = n(\text{cyclohexene}) + n(\text{cyclohexane}) \]
Rearranging:
\[ \begin{aligned} n(\text{cyclohexane}) &=0.390-0.240\\ &=0.150\ \mathrm{mmol} \end{aligned} \]
The amount of cyclohexane in each aliquot is:
\[ \boxed{0.150\ \mathrm{mmol}} \]
c) Evaluate the student’s proposed experiment
The student has confused the absence of a reaction with the absence of a reactant.
Tube A has already consumed all its bromine during the addition reaction.
Even if cyclohexane is present, it cannot undergo substitution without bromine being available.
UV light initiates the substitution reaction, but it does not replace the need for a reactant.
Therefore, exposing the already-decolourised tube to UV cannot establish whether cyclohexane is present.
The correct experimental approach is to use a fresh aliquot containing bromine for the UV test.
Comparing the bromine consumption in the two experiments allows the two contributions to be separated.
This is the deeper principle: a negative observation is only meaningful when the conditions allow the reaction you are trying to detect.
06What should you remember for an HSC response?
When a question asks how to distinguish an alkane from an alkene using a reaction observation, structure your answer around three things:
1. The observation
An alkene rapidly decolourises bromine water, while an alkane does not under normal test conditions.
2. The reaction
The alkene undergoes addition across its carbon-carbon double bond. The alkane does not react in the absence of UV light.
3. The conditions
The test must be protected from UV light because an alkane can undergo substitution with bromine when exposed to UV.
Also remember to inspect the entire mixture if two liquid layers form.
A strong exam response connects the visible observation to the chemical reaction. Simply writing that bromine water changes from orange-brown to colourless does not fully explain why the test distinguishes the two hydrocarbons.
Once you can interpret this reaction correctly, the next useful step is connecting the observation to molecular structure. Understanding the alkane, alkene, and alkyne formulae helps you recognise which bonds are available for addition, while understanding bonding in alkanes explains why their single-bond structures behave differently.
The key distinction to carry forward is simple: addition identifies reactive unsaturation; substitution under UV is a different reaction, even though both can consume bromine.