How to Draw Products of Bromine Addition to Alkenes

Imagine you have two colourless hydrocarbon samples. You add a few drops of orange-brown bromine water to each, swirl them, and keep them away from ultraviolet light.

In one test tube, the bromine colour quickly disappears. In the other, it remains.

The difference is the presence of a carbon-carbon double bond. An alkene reacts with bromine by addition, while an alkane does not react under these conditions.

But how do you actually draw the product?

The rule is simple: change the C=C double bond into a C-C single bond, then attach one bromine atom to each of the two carbon atoms that were originally double-bonded.

For example, ethene reacts with bromine to form 1,2-dibromoethane:

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

Notice that the two bromine atoms attach to different carbon atoms. The carbon skeleton stays intact, and the product no longer contains a carbon-carbon double bond.

This is the standard product-drawing method for bromine addition. There is one important detail about bromine water, which we’ll examine shortly.

01How to draw the product of bromine addition

Before drawing anything, remember what makes an alkene different from an alkane.

An alkene contains at least one carbon-carbon double bond, written as C=C. An alkane contains only carbon-carbon single bonds.

The double bond consists of two shared electron pairs:

• One forms a sigma (\(\sigma\)) bond.
• The other forms a pi (\(\pi\)) bond.

The pi bond is more exposed and provides a region of relatively high electron density. This makes the double bond reactive towards bromine.

During the addition reaction, the pi bond is broken and new carbon-bromine bonds form. The original sigma bond between the carbon atoms remains.

The carbon atoms therefore stay connected, but the bond between them changes from double to single.

If you need to revise the underlying bonding and geometry, see Carbon-Carbon Bond Shapes: Single, Double and Triple Bonds.

The three-step drawing method

Use this method whenever you are asked to draw the product of bromine addition to an alkene.

Step 1: Find the double bond.

Identify the two carbon atoms joined by the C=C bond.

These are the only two carbon atoms that receive bromine atoms in the standard addition reaction.

Step 2: Change the double bond into a single bond.

Replace C=C with C-C.

Do not remove either carbon atom. The carbon skeleton remains unchanged.

Step 3: Attach one Br atom to each carbon.

Each of the two original alkene carbons receives one bromine atom.

Check that every carbon has four bonds after the reaction.

A useful way to remember the process is:

\[ \ce{C=C + Br2 -> Br-C-C-Br} \]

The bromine atoms must be attached to adjacent carbon atoms.

Why does each carbon receive one bromine atom?

Consider ethene, \(\ce{CH2=CH2}\).

Before the reaction, each carbon has:

• Two C-H single bonds.
• One C=C double bond.

After addition, each carbon has:

• Two C-H single bonds.
• One C-C single bond.
• One C-Br single bond.

Each carbon still has four bonds. No hydrogen atoms have been removed.

This is an important check because a surprisingly common mistake is attaching both bromine atoms to the same carbon.

That gives a different structural isomer, not the standard product of bromine addition.

02Addition versus substitution: don’t confuse the two

Bromine reacts differently with alkenes and alkanes.

The key is to look at what happens to the original molecule.

Addition: atoms are added across a double bond. The double bond becomes a single bond, and the original molecule gains new atoms.

Substitution: an existing atom is replaced by another atom. The carbon skeleton and its original bonding pattern are generally retained.

Compare the reactions:

Example 1: Addition to an alkene

Ethene reacts with bromine at room temperature.

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

The product is 1,2-dibromoethane.

Both bromine atoms are incorporated into the organic product.

There is no hydrogen bromide by-product in this simplified addition equation.

Example 2: Substitution in an alkane

Ethane reacts with bromine when exposed to ultraviolet light.

\[ \ce{CH3-CH3 + Br2 ->[UV] CH3-CH2Br + HBr} \]

The product is bromoethane, along with hydrogen bromide.

Here, one hydrogen atom has been replaced by bromine.

The reaction is called a substitution reaction because bromine takes the place of hydrogen rather than adding across a carbon-carbon double bond.

Why does ultraviolet light matter?

The difference in conditions helps you distinguish the reactions.

For alkene addition, bromine reacts with the electron-rich double bond at room temperature. UV light is not needed.

For alkane substitution, UV light supplies energy to break the Br-Br bond and generate bromine radicals. These radicals can participate in a free-radical substitution reaction.

This means that an alkane can also consume bromine if UV light is present.

That creates a potential problem with the bromine test: if you expose an alkane sample to sunlight, it may decolourise bromine even though it contains no C=C bond.

Keep the test away from UV light if you want to distinguish alkenes from alkanes.

03The bromine water test: what is actually happening?

The bromine test is useful because alkenes and alkanes behave differently towards bromine under controlled conditions.

The original experimental problem is straightforward: two hydrocarbon samples may look almost identical, so how can you tell whether either contains a carbon-carbon double bond?

A chemical reaction provides an answer.

Setting up a fair comparison

Consider two common laboratory samples:

• Cyclohexane, an alkane.
• Cyclohexene, an alkene.

Both are colourless liquids. Their appearance alone does not identify the functional group.

A suitable comparison uses separate labelled test tubes containing equal small volumes of each hydrocarbon.

A few drops of bromine water are added to each tube. The tubes are gently mixed and protected from UV light.

The predicted outcomes are:

Cyclohexene: The bromine colour rapidly disappears because bromine reacts with the C=C bond.

Cyclohexane: The bromine colour remains because there is no double bond available for the addition reaction under these conditions.

The observation supports the prediction that the cyclohexene contains a reactive carbon-carbon double bond.

What does decolourisation tell you?

Bromine has an orange-brown colour in solution.

When an alkene reacts with bromine, the concentration of coloured bromine species decreases. The solution therefore loses its colour.

This is called decolourisation.

The observation is evidence that bromine has been consumed in a reaction with the sample.

It does not, by itself, prove the exact structure of the organic product.

That distinction matters. An experiment can provide strong evidence for a functional group without identifying every detail of the molecule.

Three experimental mistakes to avoid

1. Exposing the samples to UV light

This can cause an alkane to react with bromine through substitution.

The result may look like a positive test for an alkene.

2. Using too much bromine water

If you add a large excess of bromine water, some orange-brown colour may remain even after the alkene has reacted.

The remaining colour can mask the decolourisation.

Using a small amount of bromine reagent makes the observation easier to interpret.

3. Assuming that decolourisation always produces a dibromoalkane

This is a more subtle mistake.

The standard addition equation is useful, but the actual products depend on the reaction medium.

The distinction becomes important when the question specifically mentions bromine water.

04Important distinction: bromine in an inert solvent versus bromine water

There is a chemical detail worth knowing before you draw every bromination product as a dibromoalkane.

The standard reaction:

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

represents the addition of molecular bromine to ethene in an appropriate non-aqueous, inert solvent.

However, bromine water contains water, and water can participate in the reaction.

Bromine reacts reversibly with water:

\[ \ce{Br2 + H2O <=> HOBr + HBr} \]

The hypobromous acid, \(\ce{HOBr}\), can also add across an alkene’s double bond.

For ethene, this produces 2-bromoethanol:

\[ \ce{CH2=CH2 + HOBr -> HOCH2-CH2Br} \]

The product contains:

• One bromine atom attached to one carbon.
• One hydroxyl group (\(\ce{-OH}\)) attached to the adjacent carbon.

This type of product is called a bromohydrin.

In aqueous bromine, bromohydrin formation can be the major pathway rather than formation of the dibromoalkane. Some dibromoalkane can also form.

The NSW Department of Education’s Chemistry Module 8 guide discusses this distinction and explains why identifying the reaction and understanding its conditions matter more than simply memorising one product.

What should you draw in an HSC question?

Pay attention to the wording.

For a standard question asking you to draw the product of bromine addition, the 1,2-dibromoalkane is the expected structural model.

If the question specifically tests the chemistry of aqueous bromine and asks for the major product, don’t ignore the water.

05Worked examples

06Questions and solutions

Question 1

Propene reacts with bromine in an inert solvent at room temperature, away from UV light.

(a) Draw the structural formula of the organic product.

(b) Give the product’s systematic name.

(c) Identify the type of reaction and describe the expected colour change.

Solution 1

The product is 1,2-dibromopropane, formed through an addition reaction. The orange-brown bromine solution decolourises.

(a) Structural formula

Start with:

\[ \ce{CH3-CH=CH2} \]

Change the C=C bond to C-C and attach one bromine atom to each of the original double-bonded carbons:

\[ \boxed{\ce{CH3-CHBr-CH2Br}} \]

(b) Systematic name

The longest carbon chain contains three carbon atoms, giving propane.

The bromine substituents occupy positions 1 and 2.

Therefore, the product is 1,2-dibromopropane.

(c) Reaction type and observation

This is an addition reaction because bromine atoms add across the original carbon-carbon double bond.

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

Notice that no hydrogen atoms have been removed from the original molecule.

Question 2

Two separate experiments are performed using bromine in a dry, inert solvent.

Experiment A

A sample contains 0.250 mmol of ethene. It is mixed with 0.400 mmol of bromine at room temperature, away from UV light.

Assume the reaction goes to completion.

(a) Write the balanced reaction equation.

(b) Calculate the amount of 1,2-dibromoethane formed.

(c) Calculate the amount of bromine remaining.

Experiment B

A fresh sample contains 0.250 mmol of ethane. It is mixed with the same amount of bromine.

The mixture is initially kept in the dark. It is then exposed to UV light.

(d) Explain the expected difference between the two observations.

(e) Write the balanced equation for the reaction occurring in Experiment B under UV light.

Solution 2

Experiment A produces 0.250 mmol of 1,2-dibromoethane and leaves 0.150 mmol of bromine. Experiment B undergoes substitution only after UV exposure.

(a) Balanced equation for Experiment A

Ethene undergoes addition:

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

The structural equation is:

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

(b) Amount of product formed

The mole ratio of ethene to 1,2-dibromoethane is 1:1.

Therefore:

\[ \begin{aligned} n(\text{product}) &=n(\text{ethene})\\ &=0.250\text{ mmol} \end{aligned} \]

The amount of 1,2-dibromoethane formed is 0.250 mmol.

(c) Amount of bromine remaining

One mole of bromine reacts with one mole of ethene.

Therefore, 0.250 mmol of bromine is consumed.

\[ \begin{aligned} n(\ce{Br2}\text{ remaining}) &=0.400-0.250\\ &=0.150\text{ mmol} \end{aligned} \]

The remaining amount is 0.150 mmol.

(d) Expected observations

In Experiment A, ethene reacts rapidly with bromine even without UV light. The bromine colour decreases as the reaction proceeds.

In Experiment B, ethane does not react visibly with bromine in the dark under these conditions.

When UV light is introduced, bromine radicals can form and initiate substitution. Bromine is then consumed, so its colour can fade.

The important point is that bromine decolourisation is not exclusive to alkenes if UV light is present.

(e) Balanced equation for Experiment B

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

One hydrogen atom in ethane has been replaced by bromine.

The production of HBr distinguishes substitution from the simple addition reaction in Experiment A.

Question 3

A student is asked to draw the product of propene reacting with bromine.

They propose:

\[ \ce{CH3-CBr2-CH3} \]

Explain why this is not the correct product of the standard bromine addition reaction.

Your answer should:

(a) Identify the structural problem with the proposed product.

(b) Draw the correct product.

(c) Explain why checking the molecular formula alone is not sufficient to verify the structure.

Solution 3

The student’s structure is incorrect because both bromine atoms have attached to the same carbon. The correct addition product has one bromine atom attached to each of the two original double-bonded carbons.

(a) Structural problem

The proposed product is 2,2-dibromopropane.

Although its carbon atoms have valid bonding arrangements, it does not represent the standard addition of bromine across the original C=C bond.

The bromine atoms must attach to adjacent carbon atoms.

(b) Correct product

Starting with:

\[ \ce{CH3-CH=CH2} \]

Change the double bond to a single bond and attach one Br to each carbon:

\[ \boxed{\ce{CH3-CHBr-CH2Br}} \]

The correct product is 1,2-dibromopropane.

(c) Why the molecular formula is insufficient

Both the student’s proposed structure and the correct product have the molecular formula:

\[ \ce{C3H6Br2} \]

However, they have different arrangements of atoms.

The student’s structure has both bromine atoms on carbon 2. The correct product has bromine atoms on carbons 1 and 2.

This is an example of structural isomerism.

A molecular formula tells you how many atoms are present, but it does not tell you how those atoms are connected.

You must check the actual structural formula against the reaction that occurred.

Question 4

Ethene is separately reacted with bromine under two different conditions.

Reaction A: Ethene reacts with bromine in a dry, inert organic solvent at room temperature.

Reaction B: Ethene reacts with bromine water at room temperature.

Both mixtures lose their orange-brown colour.

(a) Draw the expected organic product of Reaction A.

(b) Explain why the major organic product in Reaction B can be different.

(c) Draw the major bromohydrin product of Reaction B.

(d) Explain why both reactions can produce the same visible observation even though their organic products differ.

Solution 4

Reaction A produces 1,2-dibromoethane, while the major organic product in aqueous bromine can be 2-bromoethanol. Both reactions decolourise the bromine because coloured bromine species are consumed.

(a) Product of Reaction A

In the dry, inert solvent, the standard addition reaction occurs.

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

The product is 1,2-dibromoethane.

One bromine atom attaches to each of the original alkene carbons.

(b) Why Reaction B differs

Bromine water contains water, which can participate in the reaction.

Bromine reacts with water to form hypobromous acid and hydrobromic acid:

\[ \ce{Br2 + H2O <=> HOBr + HBr} \]

The hypobromous acid can then add across the ethene double bond.

This introduces both a bromine atom and a hydroxyl group into the organic molecule.

(c) Bromohydrin product

The major bromohydrin product is 2-bromoethan-1-ol.

Its structural formula is:

\[ \boxed{\ce{HOCH2-CH2Br}} \]

One carbon receives the hydroxyl group, while the adjacent carbon receives bromine.

(d) Why the observations are similar

Both reactions consume bromine species responsible for the orange-brown colour.

The visible observation is therefore decolourisation in both cases.

However, the identity of the major organic product depends on the reaction medium.

This demonstrates why you must consider the conditions when predicting a reaction product. Identical observations do not necessarily mean identical chemical reactions.

Question 5

Two pure hydrocarbons, A and B, have the same molecular formula:

\[ \ce{C4H8} \]

They are investigated using bromine in a dry, inert solvent.

The following observations are recorded:

A student makes the following claim:

“Both compounds have the same molecular formula, so they must have the same carbon structure. A must be but-1-ene because it forms a dibromoalkane.”

Evaluate this claim.

Your answer must:

(a) Identify the most likely class of compounds represented by A and B.

(b) Explain why the two compounds can have the same molecular formula but different reactions.

(c) Give two possible structures for A and draw their corresponding bromine addition products.

(d) Explain what the experimental results establish and what they cannot establish about the exact structures of A and B.

Solution 5

The student’s claim is incorrect. The results are consistent with A being an alkene and B being a saturated cyclic hydrocarbon, but they do not identify A as but-1-ene.

The important distinction is that the molecular formula does not uniquely determine the arrangement of carbon atoms or the location of a double bond.

(a) Identify the likely compound classes

Hydrocarbon A rapidly decolourises bromine in the dark.

This is evidence that A contains a reactive carbon-carbon double bond and undergoes addition.

A is therefore most likely an alkene.

Hydrocarbon B does not react visibly with bromine in the dark, but it reacts under UV light and produces a brominated organic compound and HBr.

This is consistent with a substitution reaction.

Because B has the formula \(\ce{C4H8}\), it is consistent with a saturated cyclic hydrocarbon, such as cyclobutane or methylcyclopropane.

(b) Explain why the molecular formulas can be identical

For an acyclic alkane, the general formula is:

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

For an alkene containing one double bond, the general formula is:

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

Cycloalkanes containing one ring also have the general formula:

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

Therefore, both alkenes and cycloalkanes can have the molecular formula \(\ce{C4H8}\).

The difference is structural:

• An alkene contains a C=C double bond.
• A saturated cycloalkane contains a carbon ring with only single bonds.

A has a reactive double bond, whereas B is consistent with a saturated ring.

The same molecular formula does not guarantee the same chemical properties.

(c) Two possible structures for A

There are several possible alkene structures with the formula \(\ce{C4H8}\).

Consider but-1-ene:

\[ \ce{CH2=CH-CH2-CH3} \]

It reacts with bromine to form:

\[ \ce{CH2=CH-CH2-CH3 + Br2} \]

\[ \ce{-> CH2Br-CHBr-CH2-CH3} \]

The product is 1,2-dibromobutane.

Now consider 2-methylpropene:

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

It reacts with bromine to form:

\[ \ce{CH2=C(CH3)2 + Br2} \]

\[ \ce{-> CH2Br-CBr(CH3)2} \]

The product is 1,2-dibromo-2-methylpropane.

Both starting compounds have the formula \(\ce{C4H8}\), and both addition products have the formula \(\ce{C4H8Br2}\).

However, their carbon skeletons and product structures are different.

Therefore, the experimental information does not establish that A is specifically but-1-ene.

(d) What can and cannot be concluded?

The experiments establish several important things.

For A:

• Rapid bromine decolourisation in the dark supports the presence of a carbon-carbon double bond.
• The 1:1 stoichiometric relationship between bromine and alkene is consistent with one C=C bond per molecule.
• The product formula is consistent with the addition of one \(\ce{Br2}\) molecule.

However, the experiments cannot identify the exact position of the double bond or the arrangement of the carbon skeleton.

For B:

• The lack of reaction in the dark is consistent with a saturated hydrocarbon.
• The UV-induced reaction supports free-radical substitution.
• Formation of \(\ce{C4H7Br}\) and HBr is consistent with one hydrogen being replaced by bromine.

However, these observations alone do not establish whether B is cyclobutane, methylcyclopropane, or another structure consistent with the evidence.

The tempting mistake is to assume that a molecular formula and one reaction product uniquely identify a compound. They do not.

To distinguish the possible structures, additional structural evidence is needed. Techniques such as proton and carbon-13 NMR can help identify the arrangement of atoms and distinguish positional and structural isomers.

07Where to go next

Once you can draw the product of bromine addition, the next challenge is working out exactly which alkene you started with.

The bromine test provides evidence for unsaturation, but it cannot tell you precisely where the double bond is located.

That is where structural identification becomes important. Understanding carbon bonding, recognising structural isomers, and interpreting NMR data allow you to move from identifying a functional group to identifying the actual molecule.

For further revision, revisit Carbon-Carbon Bond Shapes: Single, Double and Triple Bonds and How to Distinguish Saturated and Unsaturated Hydrocarbons.

Course note: This explanation follows the Chemistry Stage 6 (2017) course context, including the Module 8 investigation of carbon-carbon double bonds. The Chemistry 11-12 Syllabus (2025) is scheduled for implementation from 2028, with its first HSC examination in 2029. See NESA’s syllabus implementation information for the transition timetable.