Why Alkane Halogen Substitution Usually Needs UV Light

Two liquids have the same molecular formula, \(\ce{C6H12}\). Add orange-brown bromine to each, and something strange happens: one loses its colour in the dark, while the other only reacts noticeably when exposed to ultraviolet light. Why should light make such a difference?

The usual reason alkane halogen substitution needs UV light is that the light can break the bond in a chlorine or bromine molecule, producing highly reactive particles called radicals. These radicals start a chain reaction that replaces hydrogen atoms in an alkane with halogen atoms. Without a suitable source of radicals, the reaction is normally very slow at room temperature.

But here’s the important distinction: UV light starts a substitution reaction in an alkane. An alkene can react with bromine by addition, even in the dark. Understanding why means looking at what happens to the bonds, rather than just memorising reaction conditions.

01An experiment that separates the effect of light from molecular structure

Imagine two possible structures for \(\ce{C6H12}\):

• Cyclohexane, a saturated hydrocarbon with six carbon atoms joined in a ring by single bonds.
• Hex-1-ene, an unsaturated hydrocarbon with a six-carbon chain and a double bond between its first two carbon atoms.

Both have the same molecular formula. Their different structures give them very different reactions with bromine.

The first comparison: bromine in the dark

Suppose we put each hydrocarbon in a separate sealed, transparent sample cell containing the same starting concentration of dissolved bromine.

Both mixtures initially appear orange-brown because of \(\ce{Br2}\).

Keep both away from light.

What should happen?

Hex-1-ene can react readily with bromine, so its orange-brown colour disappears as bromine is consumed.

Cyclohexane, however, normally shows little observable reaction over the same period. Its bromine colour remains.

It’s tempting to conclude that cyclohexane simply cannot react with bromine. But that experiment has only demonstrated that it reacts very slowly under those conditions.

It hasn’t tested whether cyclohexane needs something to start the reaction.

The improved comparison: change the light, not the hydrocarbon

To isolate the effect of light, compare two identical cyclohexane-bromine mixtures. Keep one in darkness and expose the other to suitable UV radiation.

Include a hex-1-ene-bromine mixture in darkness as a comparison.

The predicted results are:

These are illustrative predictions, not measured experimental results. The rate of colour loss depends on concentration, temperature, illumination, and other conditions.

The comparison supports two conclusions.

First, cyclohexane can react with bromine when light provides a way to initiate the reaction. It isn’t completely incapable of reacting.

Second, the presence of a carbon-carbon double bond allows hex-1-ene to react through a different pathway without photochemical initiation.

There’s also an experimental limitation worth recognising. If bromine water is mixed with a hydrocarbon, two liquid layers may form, and bromine can move between them. One layer becoming paler isn’t necessarily proof of a chemical reaction. A controlled comparison should account for this, and identifying the products provides stronger evidence than colour alone.

02Why alkanes undergo substitution instead of addition

Consider one carbon atom in cyclohexane. It forms four covalent bonds: two to neighbouring carbon atoms and two to hydrogen atoms.

There is no carbon-carbon double bond available for the familiar type of halogen addition.

The carbon doesn’t have room to form an extra ordinary covalent bond while keeping its existing four bonds.

Instead, one carbon-hydrogen bond can be replaced with a carbon-bromine bond.

For one substitution in cyclohexane:

\[ \ce{C6H12 + Br2 ->[UV] C6H11Br + HBr} \]

The organic product is bromocyclohexane, \(\ce{C6H11Br}\). Hydrogen bromide, \(\ce{HBr}\), is the other product.

Notice the atom accounting:

• One hydrogen atom leaves the hydrocarbon.
• One bromine atom takes its place.
• The other bromine atom combines with the removed hydrogen.
• The carbon ring remains intact.

This is a substitution reaction, specifically free-radical substitution.

The strong carbon-carbon and carbon-hydrogen single bonds help explain why alkanes are relatively unreactive. For a broader refresher on these bonds, see Bonding in Alkanes: Properties Explained for HSC Chemistry.

Why an alkene is different

Now consider hex-1-ene.

Its double bond contains one sigma bond and one pi bond. The pi bond provides electron density that can interact with and polarise an approaching bromine molecule.

In a straightforward addition reaction, the carbon-carbon double bond becomes a single bond, and each of the two carbon atoms gains a bond to bromine.

For bromine addition in a suitable non-aqueous medium:

\[ \ce{C6H12 + Br2 -> C6H12Br2} \]

The organic product is 1,2-dibromohexane.

There’s no hydrogen bromide by-product in this simple addition equation. Both bromine atoms become attached to the organic molecule.

The comparison is useful because the starting compounds have identical molecular formulas. Knowing only \(\ce{C6H12}\) cannot tell you whether the compound is an alkene or a saturated cycloalkane.

And adding UV doesn’t turn substitution into addition. The molecular structure and reaction pathway determine what bonds can change.

03What ultraviolet light actually does to a halogen molecule

We’ve established that light changes whether substitution happens readily.

But what absorbs the light?

A tempting explanation is that UV radiation simply breaks a carbon-hydrogen bond in the alkane, leaving a place for bromine.

That’s not the usual initiation mechanism.

The first important event is the splitting of a bond in the halogen molecule itself.

Step 1: UV light produces halogen radicals

A bromine molecule consists of two bromine atoms joined by a covalent bond.

When a bromine molecule absorbs a photon of suitable energy, its bond can break so that each bromine atom takes one electron from the shared pair.

This is called homolytic bond fission.

\[ \ce{Br2 ->[h\nu] 2Br^.} \]

The symbol \(h\nu\) indicates light energy, where \(h\) is Planck’s constant and \(\nu\) is the frequency of the radiation.

The dot in \(\ce{Br^.}\) represents an unpaired electron.

A particle with an unpaired electron is called a radical. Radicals are often highly reactive because of their electronic structure.

This first stage is initiation.

It explains why illumination can make such a dramatic difference. The light provides a route to generating reactive species that are scarce in an ordinary dark mixture.

Step 2: A bromine radical removes hydrogen from the alkane

A bromine radical can collide with a cyclohexane molecule and remove one of its hydrogen atoms.

\[ \ce{Br^. + C6H12 -> HBr + C6H11^.} \]

Two things happen.

A hydrogen-bromine bond forms, producing hydrogen bromide. The carbon atom that lost its hydrogen is left with an unpaired electron, producing a cyclohexyl radical, \(\ce{C6H11^.}\).

The original bromine radical has been consumed, but another radical has appeared.

This replacement of one radical by another is essential to the chain mechanism.

Step 3: The carbon radical reacts with another bromine molecule

The cyclohexyl radical reacts with \(\ce{Br2}\):

\[ \ce{C6H11^. + Br2 -> C6H11Br + Br^.} \]

Bromocyclohexane forms.

But look at the other product: another bromine radical.

That radical can remove hydrogen from a further cyclohexane molecule, creating another cyclohexyl radical, which can react with another bromine molecule.

The sequence can repeat.

These two repeating reactions are called propagation steps.

Think of the UV light as the starter motor in a car. It gets the reaction going, but the chain can continue through reactions of the radicals already present.

The analogy has a limit. A radical chain isn’t a machine that runs indefinitely. Radicals can disappear by reacting with one another or with substances that trap them, and the reaction also depends on available reactants and favourable collisions.

Step 4: Radicals can terminate the chain

Sometimes two radicals collide and combine without producing another radical.

For example:

\[ \ce{Br^. + C6H11^. -> C6H11Br} \]

This is a termination step.

Other radical combinations are possible. The important feature is that the reaction consumes radicals without replacing them.

As the radical population decreases, the chain reaction can slow or stop.

That gives us the three stages:

The light-controlled experiment is consistent with this mechanism, but colour changes alone don’t prove the existence or exact sequence of radical intermediates. The mechanism is a chemical explanation supported by wider evidence about radical reactions.

04Why UV is an effective initiator, but not an absolute requirement

There’s a useful energy comparison behind the mechanism.

The bond between two halogen atoms is generally easier to break than the strong carbon-hydrogen bonds in an alkane.

For an approximate comparison:

Bond dissociation energy is the energy needed to break a mole of specified bonds homolytically in the gas phase. Values vary with the particular molecule and bonding environment.

The key observation is that the halogen-halogen bonds have considerably lower bond dissociation energies.

So a suitable photon can help split a halogen molecule without needing to break a carbon-hydrogen bond directly.

Does substitution always need UV?

No. This is where the word usually in the question matters.

Ultraviolet radiation is a standard, effective condition for initiating alkane chlorination and bromination. But sufficiently energetic light at visible wavelengths can also initiate some halogen reactions when the halogen absorbs it. Heating or suitable radical initiators can also start radical reactions under appropriate conditions.

Chlorine and bromine are coloured partly because they absorb visible light. Some visible photons can therefore contribute to photochemical initiation.

The precise behaviour depends on which halogen is present, the wavelength, absorption, temperature, and reaction conditions.

So the accurate HSC explanation is:

UV radiation can initiate alkane halogen substitution by producing halogen radicals through homolytic fission. It is not a universally indispensable ingredient, and it doesn’t act as a chemical catalyst.

05How to interpret a bromine colour change in an exam

An orange-brown bromine solution becoming colourless tells you that coloured bromine has been removed or altered in the observed sample.

It doesn’t, by itself, tell you whether the reaction was addition or substitution.

To decide, examine the structure, conditions, and products.

When addition is the likely explanation

If the organic compound contains a carbon-carbon double bond and reacts readily with bromine in the dark, addition is the usual explanation.

The product gains two bromine atoms across the original double bond.

For ethene:

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

The product is 1,2-dibromoethane.

When substitution is the likely explanation

If a saturated hydrocarbon shows little reaction in darkness but reacts with bromine under suitable illumination, free-radical substitution is a likely explanation.

For ethane:

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

The products of one substitution are bromoethane and hydrogen bromide.

A common mistake is to treat UV exposure plus bromine decolourisation as proof of substitution. That’s not enough. An alkene can still undergo addition under illumination.

Another mistake is assuming an alkane’s unchanged bromine colour in the dark proves it cannot react. The experiment only shows that the reaction is not occurring noticeably under those conditions.

The best evidence combines observations with structural information and product identification.

06Questions and solutions

Question 1

A sealed vessel contains ethane and chlorine at room temperature. The vessel is initially kept in darkness, then exposed to suitable UV radiation.

Assume monosubstitution is the only organic reaction.

a) State the expected effect of UV exposure on the reaction.

b) Write the balanced chemical equation and name both products.

c) Explain the specific role of UV radiation.

Solution 1

UV exposure can initiate substitution of chlorine for hydrogen in ethane. In darkness, very little substitution is normally observed over a comparable period at room temperature.

a) The reaction becomes possible at an appreciable rate when UV radiation generates reactive chlorine radicals.

b) The balanced equation is:

\[ \ce{C2H6 + Cl2 ->[UV] C2H5Cl + HCl} \]

The products are chloroethane, \(\ce{C2H5Cl}\), and hydrogen chloride, \(\ce{HCl}\).

One hydrogen atom in ethane has been replaced by chlorine, while the removed hydrogen forms hydrogen chloride.

c) UV light initiates the reaction by supplying photons that chlorine molecules can absorb, causing homolytic cleavage:

\[ \ce{Cl2 ->[h\nu] 2Cl^.} \]

The chlorine radicals can then remove hydrogen atoms from ethane and start the propagation steps.

The tempting mistake is to say that UV light adds chlorine across a carbon-carbon double bond. Ethane has no double bond. Its reaction is substitution, not addition.

Question 2

Two colourless hydrocarbons, P and Q, both have molecular formula \(\ce{C5H10}\).

One is cyclopentane and the other is pent-1-ene. Their identities are initially unknown.

Both are exposed to the same amount of dissolved bromine under controlled conditions.

The following observations are recorded:

• P decolourises bromine in darkness. Under the stated non-aqueous conditions, its single organic product contains two bromine atoms per molecule, and no hydrogen bromide is produced.
• Q shows no noticeable reaction in darkness. Under UV radiation, bromine is consumed, and hydrogen bromide is detected.

Assume each reacting hydrocarbon molecule undergoes only one addition or one substitution reaction, as appropriate, with no competing reactions.

a) Identify P and Q.

b) Write the balanced equation for each observed reaction.

c) Explain why the molecular formula alone cannot identify the reaction type.

d) A student argues that both compounds must undergo addition under UV because both can decolourise bromine. Explain the flaw.

Solution 2

P is pent-1-ene, which undergoes addition, and Q is cyclopentane, which undergoes UV-initiated substitution. Their identical molecular formulas do not imply identical bonding or reactions.

a) Pent-1-ene has a carbon-carbon double bond and can undergo addition with bromine in darkness.

Cyclopentane contains only single carbon-carbon bonds, so its usual reaction with bromine requires radical initiation.

b) For P:

\[ \ce{C5H10 + Br2 -> C5H10Br2} \]

The product is 1,2-dibromopentane.

The carbon-carbon double bond becomes a single bond, and two carbon-bromine bonds form.

For Q:

\[ \ce{C5H10 + Br2 ->[UV] C5H9Br + HBr} \]

The organic product is bromocyclopentane.

One hydrogen atom has been replaced by bromine, and hydrogen bromide is produced.

c) The molecular formula \(\ce{C5H10}\) is consistent with an open-chain alkene containing one double bond, but also with a saturated cycloalkane containing one ring.

The formula alone does not distinguish these arrangements.

That is why structural information is necessary.

d) The student’s conclusion is incorrect because colour loss indicates bromine consumption, not a unique mechanism.

Even when both reactions occur under UV, P can undergo addition while Q undergoes substitution.

The product evidence distinguishes them. P incorporates two bromine atoms without removing hydrogen. Q replaces one hydrogen and produces \(\ce{HBr}\).

The key trap is confusing a shared observation with a shared reaction pathway.

Question 3

A sealed reaction mixture contains excess methane and chlorine.

A UV pulse is absorbed by chlorine molecules. The amount of absorbed photons is equivalent to \(1.20\times10^{-5}\ \mathrm{mol}\) of photons.

After the reaction, analysis shows that \(4.80\times10^{-3}\ \mathrm{mol}\) of chloromethane has formed.

Assume the only net chemical reaction is monosubstitution producing chloromethane and hydrogen chloride, with no secondary substitution or side products.

a) Calculate the number of chloromethane molecules formed per absorbed photon, on average.

b) Calculate the net amount of chlorine consumed and the amount of hydrogen chloride produced.

c) Suppose each absorbed photon causes one chlorine molecule to split into two radicals. Does your answer to part (a) mean each chlorine radical formed exactly that many chloromethane molecules?

d) Explain how so much product can form from comparatively few absorbed photons without violating conservation of energy or atoms.

Solution 3

The reaction forms an average of 400 chloromethane molecules per absorbed photon. This is consistent with a radical chain reaction because one initiation event can lead to many propagation reactions.

a) The number of product molecules per photon is equal to the ratio of their amounts in moles:

\[ \begin{aligned} R&=\frac{n(\ce{CH3Cl})}{n(\text{photons})}\\ &=\frac{4.80\times10^{-3}}{1.20\times10^{-5}}\\ &=400 \end{aligned} \]

The ratio \(R\) is dimensionless.

Therefore, an average of 400 chloromethane molecules formed per absorbed photon.

b) The net reaction is:

\[ \ce{CH4 + Cl2 ->[UV] CH3Cl + HCl} \]

The mole ratio of chlorine consumed to chloromethane formed is 1:1.

Therefore:

\[ \begin{aligned} n(\ce{Cl2})_{\text{net}} &=4.80\times10^{-3}\ \mathrm{mol}\\ n(\ce{HCl}) &=4.80\times10^{-3}\ \mathrm{mol} \end{aligned} \]

These quantities follow from atom conservation and the assumed overall reaction.

c) No. The measured value is an average per absorbed photon, not the number of products formed by each radical individually.

Under the extra assumption that every photon produces two radicals and every radical successfully starts a chain, the average would be:

\[ \begin{aligned} n(\text{initial radicals}) &=2(1.20\times10^{-5})\\ &=2.40\times10^{-5}\ \mathrm{mol} \end{aligned} \]

Then:

\[ \begin{aligned} R_{\text{radical}} &=\frac{4.80\times10^{-3}}{2.40\times10^{-5}}\\ &=200 \end{aligned} \]

That would mean 200 product molecules per initially formed radical on average.

It still wouldn’t mean every chain had length 200. Some radicals could terminate quickly, while others might propagate much longer.

d) A chlorine radical can react with methane to produce a methyl radical:

\[ \ce{Cl^. + CH4 -> HCl + CH3^.} \]

The methyl radical can then react with chlorine:

\[ \ce{CH3^. + Cl2 -> CH3Cl + Cl^.} \]

The second reaction regenerates a chlorine radical.

Thus, the chain can continue without requiring another photon at every substitution step.

Energy conservation is not violated because the chemical reactions themselves involve bond breaking and bond formation. The UV pulse initiates the chain; it doesn’t have to provide all the energy involved in every subsequent reaction.

The tempting misconception is to treat photons as stoichiometric reactants. They are not counted in the balanced substitution equation in the way chlorine molecules are.

Question 4

Consider three conceptual experiments involving cyclohexane and bromine.

In experiment A, a sealed mixture is exposed to a short UV pulse. After the lamp is switched off, bromine continues to be consumed briefly in the dark.

In experiment B, the same mixture receives the same UV pulse, but a substance known to trap radicals is present as the lamp is switched off. Bromine consumption stops much sooner.

In experiment C, cyclohexane alone receives the UV pulse. Bromine is introduced afterwards, once the lamp is off. No appreciable substitution follows.

Assume the radiation is absorbed effectively by bromine but not appreciably by cyclohexane alone, and that other initiation pathways are negligible.

a) Explain the observations in A and B using initiation, propagation, and termination.

b) Why doesn’t A prove that UV is unnecessary for initiation?

c) Explain why C behaves differently from A.

d) Would A necessarily continue reacting indefinitely if plenty of bromine and cyclohexane remained? Explain.

Solution 4

The results are consistent with a radical chain reaction initiated by UV light but capable of continuing briefly after the light is removed. Radical trapping and the order of illumination explain the differences between A, B, and C.

a) In experiment A, UV exposure allows bromine molecules to undergo homolytic bond fission, producing bromine radicals.

These radicals start propagation:

\[ \ce{Br^. + C6H12 -> HBr + C6H11^.} \]

\[ \ce{C6H11^. + Br2 -> C6H11Br + Br^.} \]

Even after the lamp switches off, radicals already present may continue reacting.

In experiment B, the radical-trapping substance removes reactive radicals from the chain. Fewer radicals remain available to start further propagation steps, so bromine consumption stops sooner.

The trap’s chemical identity is not needed because the question specifies its relevant action.

b) Experiment A doesn’t prove that UV was unnecessary.

The UV pulse created the initial radicals.

The subsequent dark reaction is propagation, not fresh photochemical initiation.

This separates two ideas that are easily confused: light may be needed to start a radical chain without being required at every step of an already-running chain.

c) In experiment C, cyclohexane is illuminated without bromine present.

Under the stated assumptions, the alkane does not appreciably absorb the UV radiation or form radicals by direct photolysis.

When bromine is introduced afterwards in darkness, there is no significant population of radicals available to initiate the chain.

The critical condition is not merely that light was used at some earlier point. It is that an appropriate radical-forming species absorbed light at a time when initiation could occur.

d) No. Experiment A would not necessarily continue indefinitely.

Radicals can collide with one another and terminate:

\[ \ce{Br^. + Br^. -> Br2} \]

They can also be trapped by other substances.

Once the radical population becomes too small, substitution slows or stops, even if both molecular reactants remain.

The difficult distinction is between having enough reactants for the overall equation and having an active mechanism capable of converting them into products.

Question 5

A student argues:

“Bromine should always substitute hydrogen in an alkane faster than chlorine does, because the bromine-bromine bond is weaker and therefore easier for light to break.”

To test this reasoning, consider separate idealised reactions of ethane with chlorine and bromine.

Two light sources are available:

• Source V emits visible light at \(560\ \mathrm{nm}\).
• Source U emits ultraviolet light at \(365\ \mathrm{nm}\).

Use the following rounded gas-phase bond dissociation energies:

For this question, the energy of one mole of photons is:

\[ E_{\text{mol}}=\frac{119\,600}{\lambda} \]

where \(E_{\text{mol}}\) is in \(\mathrm{kJ\,mol^{-1}}\) and wavelength \(\lambda\) is entered numerically in nanometres.

Assume that:

• A halogen molecule can undergo photochemical initiation only when an absorbed photon has at least the energy needed to break its halogen-halogen bond.
• Both halogens absorb the wavelengths supplied, and thermal and multiphoton initiation can be neglected.
• Ethane is in excess.
• At \(365\ \mathrm{nm}\), equal numbers of radicals are initially generated in the separate chlorine and bromine experiments.

An illustrative observation is then proposed: under \(365\ \mathrm{nm}\) illumination, chloroethane forms faster than bromoethane.

a) Calculate the energy per mole of photons at both wavelengths. Under the assumptions, identify which halogen reactions can be initiated by each source.

b) Calculate the approximate enthalpy change for the hydrogen-abstraction propagation step with chlorine radicals and with bromine radicals. Use bond energies and the convention that endothermic reactions have positive enthalpy changes.

c) Explain why the proposed faster chlorination under UV is chemically plausible despite the stronger \(\ce{Cl-Cl}\) bond.

d) Can the bond energies alone prove that chlorination must be faster at \(365\ \mathrm{nm}\)? Explain what additional factors matter.

e) Explain why the result at \(560\ \mathrm{nm}\) does not contradict the usual teaching that alkane halogenation needs UV light.

Solution 5

Bromine is easier to photolyse than chlorine according to the supplied bond energies, but this does not mean bromination must be faster. The crucial distinction is between generating the first radicals and the subsequent hydrogen-abstraction reactions that propagate the chain.

a) Compare photon energies and initiation.

For \(560\ \mathrm{nm}\):

\[ \begin{aligned} E_{\text{mol}} &=\frac{119\,600}{560}\\ &\approx214\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

This is greater than the \(\ce{Br-Br}\) bond dissociation energy of \(193\ \mathrm{kJ\,mol^{-1}}\), but less than the \(\ce{Cl-Cl}\) value of \(243\ \mathrm{kJ\,mol^{-1}}\).

Under the simplified assumptions, 560 nm light can initiate bromination but cannot initiate chlorination by direct single-photon cleavage of the halogen bond.

For \(365\ \mathrm{nm}\):

\[ \begin{aligned} E_{\text{mol}} &=\frac{119\,600}{365}\\ &\approx328\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

This exceeds both halogen-halogen bond energies.

Therefore, the 365 nm source can initiate both reactions under the given assumptions.

Notice the first surprise: UV isn’t necessary for every possible halogen photolysis. The visible source can also provide sufficient photon energy for bromine.

b) Compare the hydrogen-abstraction propagation steps.

For chlorine:

\[ \ce{Cl^. + C2H6 -> HCl + C2H5^.} \]

A carbon-hydrogen bond is broken, and a hydrogen-chlorine bond forms.

Use:

\[ \Delta H\approx \sum E_{\text{broken}} -\sum E_{\text{formed}} \]

Then:

\[ \begin{aligned} \Delta H_{\text{Cl}} &\approx423-431\\ &=-8\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The chlorine hydrogen-abstraction step is slightly exothermic according to these approximate data.

For bromine:

\[ \ce{Br^. + C2H6 -> HBr + C2H5^.} \]

The carbon-hydrogen bond is broken, but this time a hydrogen-bromine bond forms.

\[ \begin{aligned} \Delta H_{\text{Br}} &\approx423-366\\ &=+57\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The bromine hydrogen-abstraction step is endothermic.

This is the major difference hidden by the student’s original reasoning.

c) Why faster chlorination is plausible.

The \(\ce{Br-Br}\) bond is easier to break during initiation.

However, after radicals are formed, they must react with ethane to remove hydrogen.

Chlorine radicals form the relatively strong \(\ce{H-Cl}\) bond. This makes the hydrogen-abstraction step energetically favourable overall.

Bromine radicals form the weaker \(\ce{H-Br}\) bond. For ethane, hydrogen abstraction is endothermic and generally has a less favourable kinetic pathway.

This helps explain why bromination of alkanes is often slower and more selective than chlorination, despite the weaker bromine-bromine bond.

The student has used information about initiation to predict the speed of the entire chain reaction. That is the exact reasoning error.

d) What the bond energies cannot prove.

The energy calculations do not, by themselves, establish the relative reaction rates.

Reaction enthalpy tells us the energy difference between reactants and products. Reaction rate depends on activation energy and the kinetic pathway, not just the overall enthalpy change.

The rates also depend on light absorption, radical concentrations, propagation rates, termination, temperature, and competing reactions.

Even with equal numbers of initiating radicals, the bond enthalpies alone do not mathematically prove a particular product formation rate.

They make the proposed observation chemically plausible, and they reveal why the student’s prediction is not justified.

This limitation is especially important in challenging questions: a favourable energy change is not the same thing as a measured rate.

e) Why visible-light initiation does not contradict the usual UV condition.

The familiar UV condition is a reliable way to initiate halogen substitution, not an absolute rule excluding every other wavelength.

In the idealised model, 560 nm photons carry enough energy to split \(\ce{Br-Br}\) bonds but not enough to split \(\ce{Cl-Cl}\) bonds directly.

Real photochemistry also depends on whether those photons are absorbed and whether absorption produces radicals, so bond energy alone cannot guarantee an observable reaction.

The final conclusion is that three different questions must be kept separate:

• Can the light initiate the reaction?
• How readily do the resulting radicals propagate the reaction?
• How quickly are radicals removed or the reactants consumed?

The first question involves light absorption and radical formation. The other two control how far and how fast the reaction proceeds.

This is also the next useful step in understanding alkane halogenation: initiating substitution does not guarantee a single clean product. Different carbon-hydrogen positions may react at different rates, and a haloalkane may undergo further substitution. For product prediction and repeated substitution, continue with Halogen Substitution of Saturated Hydrocarbons for HSC Chemistry.