Why Alkenes Decolourise Bromine Water Faster Than Alkanes

Two clear liquids, hexane and hex-1-ene, are each mixed with orange bromine water. Which one will make the orange colour disappear first if both are kept out of direct sunlight?

Hex-1-ene will usually decolourise the bromine water rapidly, while hexane will not. The difference comes from their bonds. Alkenes contain a carbon-carbon double bond (\(\ce{C=C}\)) that readily reacts with bromine through an addition reaction, even at room temperature without UV light. Alkanes contain only single bonds and generally require UV light to react with bromine through a different process called substitution.

Both reactions can consume coloured bromine. The important difference is how readily they happen, what bonds change, and what products form.

01Why does a double bond make alkenes more reactive?

Compare ethane and ethene.

Ethane is an alkane:

\[ \ce{CH3-CH3} \]

Ethene is an alkene:

\[ \ce{CH2=CH2} \]

Ethane contains a carbon-carbon single bond. Ethene contains a carbon-carbon double bond.

You might predict that ethene would be harder to react because a double bond is stronger than a single bond.

That prediction sounds reasonable, but it misses something important: the two parts of a double bond are not identical.

A double bond contains a sigma bond and a pi bond

A carbon-carbon double bond consists of:

• One sigma (\(\sigma\)) bond, formed by direct overlap of orbitals between the carbon nuclei.
• One pi (\(\pi\)) bond, formed by sideways overlap of orbitals, producing electron density above and below the plane of the bonded atoms.

The complete double bond is stronger than a carbon-carbon single bond, but the additional pi bond is more readily involved in reactions than the sigma bond.

Its electron density is also more exposed to approaching molecules.

Bromine, \(\ce{Br2}\), is a non-polar molecule when isolated. However, as it approaches the electron-rich pi bond, its electron distribution becomes distorted.

The bromine atom closest to the double bond develops a partial positive charge.

This allows the electron-rich double bond to initiate a reaction with bromine. The process is called electrophilic addition.

The key idea is that the carbon atoms do not need to separate from one another. Their sigma bond remains intact while the pi bond is replaced by new bonds to other atoms.

What happens when bromine adds to ethene?

In the simplest bromine addition model:

\[ \ce{CH2=CH2 + Br2 -> BrCH2-CH2Br} \]

The product is 1,2-dibromoethane.

Follow the bond changes:

1. The carbon-carbon double bond becomes a single bond.
2. The bromine-bromine bond is broken.
3. A new carbon-bromine bond forms on each of the two carbon atoms.
4. No hydrogen atom is removed from ethene.

The product contains the same two carbon atoms and four hydrogen atoms as ethene, plus two bromine atoms.

This is an addition reaction because atoms are added across the double bond without replacing atoms already attached to the carbon skeleton.

Why does the orange colour disappear?

Bromine water contains dissolved bromine, which gives it a yellow-orange to orange-brown appearance, depending on concentration.

When bromine reacts with an alkene, the concentration of free bromine decreases.

The resulting organic addition products are generally colourless under these conditions.

Therefore, if enough alkene is present to consume the bromine, the orange colour disappears.

The colour change is evidence that bromine has been consumed. It is not caused by the alkene itself becoming colourless; the starting alkene was already colourless.

02Why don’t alkanes react the same way?

Now consider ethane:

\[ \ce{CH3-CH3} \]

Each carbon atom has four covalent bonds. There is no carbon-carbon pi bond for bromine to attack through the rapid addition pathway described above.

The carbon-carbon and carbon-hydrogen sigma bonds are relatively stable under ordinary conditions.

Consequently, ethane does not normally react rapidly with bromine water in the dark.

But this does not mean alkanes can never react with bromine.

UV light allows a different reaction

Under suitable UV illumination, bromine molecules can split to produce highly reactive bromine radicals.

These radicals can initiate a chain reaction in which a hydrogen atom on an alkane is replaced by a bromine atom.

For ethane, the overall equation for monosubstitution is:

\[ \ce{C2H6 + Br2 -> C2H5Br + HBr} \]

The reaction occurs under UV light, and the organic product is bromoethane.

Notice the difference from ethene:

• Ethene adds two bromine atoms across its double bond in the simplified addition reaction.
• Ethane replaces one hydrogen atom with one bromine atom and also forms hydrogen bromide, \(\ce{HBr}\).

This is free-radical substitution.

An alkane can therefore decolourise bromine under suitable photochemical conditions. It is incorrect to say that alkanes never decolourise bromine water.

The more accurate comparison is that alkenes usually react readily under ordinary room-temperature conditions without UV light, whereas alkanes generally require light to initiate substitution.

Addition versus substitution: the essential comparison

This is the distinction to explain in an HSC response. Saying that alkenes are simply “more reactive” is not enough. Identify the pi bond and the different reaction conditions.

03How the bromine-water experiment answers the question

The chemical explanation predicts a visible difference. A useful experiment must determine whether that difference really comes from the carbon-carbon double bond.

The apparatus and the prediction

Consider a comparison using cyclohexane and cyclohexene.

Cyclohexane is a saturated cyclic hydrocarbon with only single carbon-carbon bonds. Cyclohexene contains one carbon-carbon double bond.

Both are liquids that do not mix completely with water.

A controlled comparison would use two labelled, stoppered test tubes:

• Tube A contains cyclohexane and bromine water.
• Tube B contains cyclohexene and bromine water.

The bromine water initially forms the lower aqueous layer, with the less-dense organic liquid above it. The samples are kept away from UV light, mixed for comparable periods, and observed under similar conditions.

Predict the result before considering the observation.

If the double bond allows bromine to react readily, Tube B should lose its bromine colour much faster than Tube A, provided enough cyclohexene is present.

That is the characteristic behaviour.

What is actually observed?

The following table summarises expected qualitative behaviour. These are illustrative observations, not results from a particular recorded experiment.

The final row is important.

An alkene does not automatically make every amount of bromine water colourless. If there is more bromine than its double bonds can consume, some colour remains.

The flaw in a one-tube experiment

Suppose you mix an unknown hydrocarbon with bromine water and notice that the lower aqueous layer becomes paler.

Has bromine reacted?

Possibly. But there is another explanation.

Bromine can dissolve into the non-polar organic layer. Some bromine may have moved from the water into the hydrocarbon without undergoing a chemical reaction.

This is physical partitioning between two liquid layers, not addition or substitution.

The first design therefore has a weakness: observing only the aqueous layer cannot reliably distinguish chemical consumption from bromine moving into another phase.

Improving the design

A stronger investigation uses matched alkane and alkene controls, keeps the samples away from UV, uses comparable mixing and observation times, and examines the colour of both liquid layers.

It also includes a bromine-water-only control to identify changes unrelated to the hydrocarbons.

The investigator must distinguish three possible outcomes:

• Bromine is consumed in a chemical reaction.
• Bromine remains but moves between the water and organic layers.
• Bromine reacts through another pathway because the conditions have changed.

A rapid loss of bromine colour throughout a properly controlled mixture supports the presence of a reactive carbon-carbon multiple bond.

It is not absolute proof that an unknown substance is an alkene. Other substances can react with bromine, and a purely visual test has limitations.

This is why experimental conditions matter as much as the final colour.

04Does bromine water always produce a dibromoalkane?

There is one useful chemical detail behind the familiar HSC reaction equation.

The simple equation

\[ \ce{CH2=CH2 + Br2 -> BrCH2-CH2Br} \]

shows bromine addition correctly and describes the formation of 1,2-dibromoethane, particularly when bromine reacts in a suitable non-aqueous medium.

But bromine water contains water as well as bromine.

Water can participate in the addition reaction, producing a bromohydrin: a compound containing both a bromine atom and a hydroxyl group, \(\ce{-OH}\), on neighbouring carbon atoms.

For ethene, a possible overall equation is:

\[ \begin{aligned} &\ce{CH2=CH2 + Br2 + H2O}\\ &\ce{-> BrCH2-CH2OH + HBr} \end{aligned} \]

The organic product is 2-bromoethanol.

Under aqueous conditions, this can be a major product. Some 1,2-dibromoethane can also form, so a mixture of products is possible.

Both pathways consume bromine and can produce the same visible loss of colour.

The colour change alone therefore cannot establish the exact organic product.

For a standard HSC comparison, the simple dibromoalkane equation is a useful representation of bromine addition. When interpreting a real bromine-water experiment, remember that the solvent can also participate.

The detailed reaction mechanism is not needed to understand why alkenes decolourise bromine water more readily than alkanes.

05Worked examples: from reaction prediction to bromine quantities

06Questions and solutions

Question 1

Two test tubes contain equal amounts of cyclohexane and cyclohexene, respectively.

An equal volume of orange bromine water is added to each tube. The mixtures are kept in the dark at room temperature and mixed equally.

Predict which compound will decolourise the bromine water more readily. Explain your prediction in terms of bonding and reaction type.

Solution 1

Cyclohexene will decolourise the bromine water more readily because it contains a carbon-carbon double bond and can undergo electrophilic addition with bromine.

The double bond contains a pi bond whose exposed electron density can react with an approaching bromine molecule.

Bromine is consumed as atoms add across the double bond, causing the orange colour to disappear if enough cyclohexene is present.

Cyclohexane contains only carbon-carbon and carbon-hydrogen sigma bonds. It does not undergo the same rapid addition reaction and generally requires UV light for free-radical substitution.

The tempting explanation that cyclohexane can never react with bromine is incorrect. The difference is the reaction pathway and the conditions required.

The observation should also consider both liquid layers, since bromine can move from the water into the organic layer without reacting.

Question 2

Two sealed tubes, P and Q, contain ethene and ethane, one gas in each. Each also contains bromine water.

In a controlled dark trial, Tube P rapidly loses its bromine colour throughout the mixture, while Tube Q retains visible bromine colour.

In a separate trial using fresh reactants, the tubes are exposed to suitable UV light for a prolonged period, and both eventually lose their visible bromine colour.

A student concludes that both gases must undergo the same addition reaction because both can decolourise bromine water.

Identify P and Q, write the simplified overall bromination equation for each gas, and explain why the student’s conclusion is incorrect. State one experimental control that would strengthen the comparison.

Solution 2

Tube P contains ethene, and Tube Q contains ethane. The student’s conclusion is incorrect because the two gases consume bromine through different reactions.

Ethene undergoes addition readily in the dark:

\[ \ce{C2H4 + Br2 -> C2H4Br2} \]

This simplified equation represents the formation of 1,2-dibromoethane.

The double bond becomes a single bond, and one bromine atom attaches to each carbon. No hydrogen atom is removed.

Ethane does not contain a C=C bond, but under UV light it can undergo free-radical substitution:

\[ \ce{C2H6 + Br2 -> C2H5Br + HBr} \]

This produces bromoethane and hydrogen bromide during monosubstitution.

The two reaction equations account for bromine atoms differently.

In addition, both bromine atoms become attached to the organic molecule.

In substitution, one bromine atom replaces hydrogen in the organic molecule, while the other forms HBr.

Actual aqueous addition can also form 2-bromoethanol, but it remains an addition process rather than alkane-style substitution.

A useful control would be bromine water exposed to the same lighting conditions without either hydrocarbon. This helps identify changes caused by the reagent or illumination rather than by the test substance.

The trap is assuming that identical visible endpoints require identical reaction mechanisms.

They do not. Colour disappearance indicates that the concentration of coloured bromine has fallen, not that a particular organic product has formed.

Question 3

An unknown hydrocarbon sample contains heptane and possibly hept-1-ene.

Two equal portions of the unknown are tested separately in the dark. The bromine solution has a concentration of \(0.0200\ \mathrm{mol\,L^{-1}}\).

• Portion A receives \(4.00\ \mathrm{mL}\) of bromine solution. All bromine colour disappears.
• Portion B receives \(6.00\ \mathrm{mL}\) of the same solution. Some orange colour remains.

Assume complete mixing and reaction, a 1:1 ratio between bromine and C=C bonds, no physical loss or redistribution affecting the observations, no competing reactions, and perfect detection of residual bromine.

Determine the possible range for the amount of hept-1-ene in each portion.

Can the exact amount of hept-1-ene be determined from these observations? Explain.

Solution 3

Each portion contains at least \(0.0800\ \mathrm{mmol}\), but less than \(0.120\ \mathrm{mmol}\), of hept-1-ene under the stated idealised assumptions. The exact amount cannot be determined from the two colour observations.

First calculate the bromine added to Portion A.

Using \(n=cV\):

\[ \begin{aligned} n_A&=(0.0200)(0.00400)\\ &=8.00\times10^{-5}\ \mathrm{mol}\\ &=0.0800\ \mathrm{mmol} \end{aligned} \]

All the bromine is consumed in Portion A.

Since one mole of hept-1-ene can consume one mole of bromine, at least \(0.0800\ \mathrm{mmol}\) of hept-1-ene must be present.

Now calculate the bromine added to Portion B:

\[ \begin{aligned} n_B&=(0.0200)(0.00600)\\ &=1.20\times10^{-4}\ \mathrm{mol}\\ &=0.120\ \mathrm{mmol} \end{aligned} \]

Some bromine remains in Portion B.

Therefore, its hept-1-ene content must be less than \(0.120\ \mathrm{mmol}\).

Writing \(n_{\mathrm{ene}}\) for the amount of hept-1-ene in each portion:

\[ 0.0800\leq n_{\mathrm{ene}}<0.120 \]

All amounts in this inequality are in millimoles.

The two observations give a range, not an exact composition.

Heptane does not consume bromine appreciably through substitution in the dark under the stated conditions, so it does not account for the disappearance.

A student might calculate \(0.0800\ \mathrm{mmol}\) and claim that it is the exact amount of hept-1-ene. That is unjustified because the first portion may contain more alkene than the amount of bromine supplied.

A more precise determination would require a measured endpoint at which a known added quantity of bromine just exceeds the amount the alkene can consume, with suitable corrections for real experimental limitations.

Question 4

Ethene is passed into bromine water in the dark. Independent chemical analysis establishes that \(0.0600\ \mathrm{mmol}\) of bromine has reacted.

A student writes:

“The orange colour disappeared, so exactly \(0.0600\ \mathrm{mmol}\) of 1,2-dibromoethane must have formed. Hydrogen bromide cannot be produced during an addition reaction.”

Evaluate both claims.

Write two balanced overall reaction equations that explain how bromine could react with ethene in this aqueous system. State what the colour observation can and cannot establish.

Solution 4

Neither claim is necessarily correct. Bromine-water decolourisation establishes that bromine has been consumed, but it does not prove that 1,2-dibromoethane is the only product or that no HBr forms.

One possible reaction is the simplified addition of bromine:

\[ \ce{C2H4 + Br2 -> C2H4Br2} \]

This produces 1,2-dibromoethane.

If this were the only reaction, the 1:1 mole ratio would give:

\[ \begin{aligned} n(\ce{C2H4Br2}) &=n(\ce{Br2})\\ &=0.0600\ \mathrm{mmol} \end{aligned} \]

However, the solvent is water.

Water can participate in the reaction, producing 2-bromoethanol:

\[ \begin{aligned} &\ce{CH2=CH2 + Br2 + H2O}\\ &\ce{-> BrCH2-CH2OH + HBr} \end{aligned} \]

This equation is balanced.

The reactants collectively supply two carbon atoms, six hydrogen atoms, two bromine atoms, and one oxygen atom. The products contain the same numbers of each atom.

In this pathway, one bromine atom attaches to the organic molecule, while the other appears in HBr. The HBr is present in aqueous solution rather than necessarily being released as a separate gas.

If this pathway alone consumed the bromine, the quantities would be:

\[ \begin{aligned} n(\text{2-bromoethanol}) &=0.0600\ \mathrm{mmol}\\ n(\ce{HBr}) &=0.0600\ \mathrm{mmol} \end{aligned} \]

In practice, both organic addition products can be formed, so their individual amounts cannot be deduced from the total bromine consumed without further evidence.

The original observation supports bromine consumption by an alkene. It does not uniquely identify the product distribution.

The tempting mistake is treating the simplified school reaction equation as an exhaustive description of every reaction occurring in bromine water.

Question 5

Two unknown colourless liquids, X and Y, are each either cyclohexane or hex-1-ene. They might be the same substance or different substances.

Each is tested separately in the dark using the following amounts:

• \(1.00\ \mathrm{mL}\) of bromine water containing \(0.0200\ \mathrm{mmol}\) of dissolved \(\ce{Br2}\).
• \(5.00\ \mathrm{mL}\) of the unknown organic liquid, forming the upper layer.

Before mixing, the bromine water is visibly orange.

After thorough mixing and separation, both the upper and lower layers appear colourless in both tests.

A student confidently concludes that X and Y must both be hex-1-ene.

However, a separate physical model predicts that bromine can distribute between the two liquid layers without reacting.

For this question, use the following hypothetical illustrative data, not measured bromine-water properties:

• The partition coefficient is \(K=10.0\), meaning that at equilibrium the bromine concentration in the organic layer is ten times its concentration in the aqueous layer.
• A layer only appears visibly coloured when its dissolved bromine concentration is at least \(5.00\ \mathrm{mmol\,L^{-1}}\).
• If an alkene is present, there is enough of it to consume all the bromine rapidly in the dark.
• Ignore bromine hydrolysis, evaporation, other reactions, and changes in liquid volumes.

(a) Calculate the bromine concentration in each layer if the unknown liquid is cyclohexane and no reaction occurs.

(b) Determine whether the two layers could both appear colourless without any bromine being consumed.

(c) Decide whether the student’s conclusion about X and Y is justified. Propose a measurement that could distinguish chemical consumption from physical redistribution, and state the predicted outcome for each possible identity.

(d) Explain why repeating the test under UV light would make the identification less reliable.

Solution 5

The student’s conclusion is not justified. Even if neither unknown reacts with bromine, the stated partitioning and visibility limits allow both layers to look colourless while all the original bromine remains present.

The key is that a substance can become too dilute to see without being chemically consumed.

Part (a): Calculate bromine concentrations without reaction.

Let:

• \(C_w\) be the bromine concentration in the water layer, in \(\mathrm{mmol\,L^{-1}}\).
• \(C_o\) be the bromine concentration in the organic layer, in the same units.
• \(V_w\) and \(V_o\) be the water and organic volumes in litres.
• \(n_0\) be the original amount of bromine, in millimoles.

From the given partition coefficient:

\[ K=\frac{C_o}{C_w}=10.0 \]

Therefore:

\[ C_o=10.0C_w \]

If there is no reaction, conservation of bromine requires:

\[ n_0=C_wV_w+C_oV_o \]

Substitute the partition relationship:

\[ n_0=C_w(V_w+KV_o) \]

The volumes are:

\[ \begin{aligned} V_w&=0.00100\ \mathrm{L}\\ V_o&=0.00500\ \mathrm{L} \end{aligned} \]

Calculate the aqueous concentration:

\[ \begin{aligned} C_w&=\frac{n_0}{V_w+KV_o}\\ &=\frac{0.0200\ \mathrm{mmol}}{0.0510\ \mathrm{L}}\\ &=0.392\ \mathrm{mmol\,L^{-1}} \end{aligned} \]

Now calculate the organic concentration:

\[ \begin{aligned} C_o&=10.0C_w\\ &=10.0(0.392)\\ &=3.92\ \mathrm{mmol\,L^{-1}} \end{aligned} \]

Part (b): Apply the visibility limit.

The hypothetical visibility threshold is:

\[ 5.00\ \mathrm{mmol\,L^{-1}} \]

The calculated aqueous concentration is:

\[ 0.392\ \mathrm{mmol\,L^{-1}} \]

The calculated organic concentration is:

\[ 3.92\ \mathrm{mmol\,L^{-1}} \]

Both are below the visibility threshold.

Therefore, both layers would appear colourless even though no chemical reaction occurred.

All \(0.0200\ \mathrm{mmol}\) of bromine could still be present, distributed between two layers.

Part (c): Determine what can actually be concluded.

Neither X nor Y can be identified uniquely from the visual observations.

Two chemically different explanations fit the same result.

If the unknown is cyclohexane, the model predicts no rapid chemical consumption in the dark. Bromine may simply partition between the layers until both are too pale to detect visually.

If the unknown is hex-1-ene, its double bond can react with the bromine. Under the stated assumption of sufficient alkene and complete reaction, the remaining free bromine approaches zero.

The final visual observation is identical, but the chemical situations are different.

A suitable follow-up would be an analytical determination of the total unreacted bromine across both layers, using a validated method that accounts for the two phases.

Under the assumptions:

This measurement would distinguish chemical consumption from mere redistribution.

The visual result cannot do that because its detection threshold is too high.

Part (d): Explain the effect of UV light.

UV exposure would make the identification less reliable because cyclohexane could then undergo free-radical substitution with bromine.

The two competing explanations would no longer be restricted to:

• An alkene reacting while an alkane does not.
• Bromine physically moving between the two layers.

A third possibility would become important: an alkane chemically consuming bromine through substitution.

The best diagnostic conditions therefore minimise light-induced substitution.

The conceptual trap is assuming that a colourless observation always means a zero concentration.

It does not.

A visible colour is a measurement with a detection limit, and a chemical reaction must be distinguished from a physical change in where the coloured substance is located.

06What the bromine-water test really tells you

The bromine-water test is most useful when you interpret the reaction conditions and molecular structure together.

Rapid loss of bromine colour in the dark supports the presence of a reactive carbon-carbon multiple bond because its pi electrons allow addition. An alkane generally does not react through that pathway, but it can consume bromine under UV light through substitution.

Three limits prevent overconfident conclusions: bromine may be present in excess, it can partition between liquid layers, and water can participate in the products formed.

For a simple HSC reaction question, the central distinction remains clear: alkenes undergo addition across C=C, while alkanes undergo substitution under suitable conditions.

Once you can explain that difference, the next useful skill is predicting exactly where the incoming atoms attach. The guide to drawing bromine-addition products from alkene structures develops that skill.

If identifying the double bond from a structural formula is still uncertain, revise the alkane, alkene, and alkyne formulae first. Knowing where the pi bond is tells you where addition can occur.