Why Cracking Produces Alkanes and Alkenes | HSC Chemistry

A refinery can separate crude oil into different fractions, but separation does not shorten the molecules inside those fractions. What happens when there are plenty of long-chain hydrocarbons but more demand for shorter fuels and alkenes used to manufacture plastics?

The answer is cracking. It breaks carbon-carbon bonds in long-chain hydrocarbons, producing smaller molecules. Some of those molecules are alkanes, while others are alkenes.

But why do you get both types of hydrocarbon from the same starting molecule?

The key is that the hydrogen atoms are distributed differently between the fragments. One fragment can end up with enough hydrogen atoms to form an alkane. Another can have fewer hydrogen atoms relative to its carbon atoms, forming a carbon-carbon double bond and becoming an alkene.

Consider one possible overall cracking reaction:

\[ \ce{C12H26 -> C8H18 + C4H8} \]

A molecule of dodecane produces octane and butene.

Notice that the atoms balance:

• Carbon: \(12 = 8 + 4\)
• Hydrogen: \(26 = 18 + 8\)

The alkane and alkene are different because their carbon atoms have different numbers of hydrogen atoms attached to them.

This is an overall reaction equation, not a claim that one carbon-carbon bond simply snaps and immediately produces two stable molecules. Cracking involves further reactions and rearrangements before stable products form.

01Why cracking produces different hydrocarbons

Start with the structure of an alkane.

An alkane is a saturated hydrocarbon. It contains carbon-carbon single bonds and carbon-hydrogen single bonds, with no carbon-carbon double or triple bonds.

For an open-chain alkane, the general formula is:

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

An open-chain alkene containing one double bond has the general formula:

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

The difference is two hydrogen atoms.

That difference is important when a long-chain alkane breaks into smaller molecules.

What happens to the hydrogen atoms?

Imagine breaking a long carbon chain into two shorter sections.

It is tempting to think that each section simply becomes a smaller alkane. However, the hydrogen atoms must also be accounted for.

For example, suppose dodecane produces an eight-carbon alkane:

\[ \ce{C12H26 -> C8H18 + ?} \]

The octane product contains eight carbon atoms and 18 hydrogen atoms.

What remains?

• Four carbon atoms.
• Eight hydrogen atoms.

The remaining product therefore has the formula \(\ce{C4H8}\).

For an open-chain hydrocarbon containing one carbon-carbon double bond, this is the formula of an alkene.

The overall reaction becomes:

\[ \ce{C12H26 -> C8H18 + C4H8} \]

The hydrogen atoms have been redistributed between the products during the cracking reactions.

The four-carbon fragment does not have enough hydrogen atoms to form an open-chain alkane. Instead, it can form a carbon-carbon double bond.

Why can’t both products simply be alkanes?

Let’s test that prediction.

Suppose dodecane breaks into two open-chain alkane molecules, with eight carbon atoms in one and four in the other.

The eight-carbon alkane is:

\[ \ce{C8H18} \]

The four-carbon alkane would be:

\[ \ce{C4H10} \]

Together, these products contain:

\[ 18+10=28 \text{ hydrogen atoms} \]

But the original dodecane only contained 26 hydrogen atoms.

Where would the extra two hydrogen atoms come from?

They cannot appear from nowhere. If no external hydrogen is supplied, this proposed reaction does not conserve atoms.

This explains why an alkane and an alkene are such a common product combination in simplified cracking equations.

An alkene has two fewer hydrogen atoms than the corresponding open-chain alkane. That allows the original hydrogen count to be conserved when the carbon chain is divided into two fragments.

Important limitation: Real cracking produces a mixture of hydrocarbons through several reaction pathways. Not every cracking event produces exactly one alkane and one alkene. The equation above represents one possible balanced outcome.

If you need to refresh the formulas and structural differences between these hydrocarbon families, see HSC Chemistry: Alkane, Alkene and Alkyne Formulae.

02What actually happens during cracking?

Cracking is a decomposition reaction in which large hydrocarbon molecules break down into smaller hydrocarbons.

The process requires substantial energy, usually supplied through heating.

There are two important approaches.

Thermal cracking

Thermal cracking uses very high temperatures, often with high pressure.

The heat provides the energy needed for chemical bonds to break and for further reactions to occur.

Thermal cracking commonly produces a significant proportion of smaller alkenes.

Catalytic cracking

Catalytic cracking uses a catalyst, commonly a zeolite, to help break down large hydrocarbon molecules.

A catalyst provides an alternative reaction pathway and changes the distribution of products. It does not supply extra carbon or hydrogen atoms.

Catalytic cracking is widely used in petroleum refining to produce valuable shorter hydrocarbons.

The exact product distribution depends on the starting hydrocarbons, temperature, catalyst, and reaction conditions.

Cracking is not fractional distillation

These two refinery processes solve different problems.

Fractional distillation separates hydrocarbons according to differences in their boiling points. The molecules themselves remain chemically unchanged.

Cracking changes the molecules by breaking carbon-carbon bonds and forming smaller hydrocarbons.

For example:

• Fractional distillation can separate dodecane from other hydrocarbons.
• Cracking can convert dodecane into smaller hydrocarbons such as octane and butene.

This distinction matters in exam questions. If a question asks how a refinery produces smaller hydrocarbons, fractional distillation is not the answer.

03How can an experiment show that cracking has occurred?

Consider a school demonstration using paraffin oil and a heated catalyst.

The purpose is to investigate whether heating a long-chain hydrocarbon in the presence of a catalyst produces smaller, unsaturated hydrocarbons.

Setting up the apparatus

A small amount of paraffin oil is held in mineral wool near the closed end of a hard-glass tube.

Further along the tube is a section containing heated aluminium oxide, which acts as a catalyst.

A delivery tube connects the outlet to a gas collection vessel.

The arrangement works from left to right:

1. The paraffin oil is heated and produces hydrocarbon vapour.
2. The vapour travels along the glass tube.
3. It passes over the hot catalyst.
4. Cracking reactions produce smaller hydrocarbon molecules.
5. The gaseous products leave through the delivery tube and are collected.

A small sample of the collected product mixture can then be tested separately.

What should happen if cracking occurs?

Before considering the observation, make a prediction.

If the paraffin oil simply evaporates without undergoing a chemical reaction, its molecules remain unchanged.

The original paraffin hydrocarbons are saturated. They do not rapidly react with bromine water in the absence of ultraviolet light.

However, if cracking produces alkenes, the product mixture should react with bromine water.

This gives us a way to investigate whether unsaturated hydrocarbons have formed.

The bromine water observation

Bromine water is orange-brown.

When an alkene reacts with bromine, an addition reaction occurs across its carbon-carbon double bond. The bromine is consumed, so the orange-brown colour disappears.

A rapid change from orange-brown to colourless therefore provides evidence that an alkene is present in the cracking products.

However, there is an important limitation.

The bromine water test does not prove that every product is an alkene. A cracking mixture can contain both alkanes and alkenes, and the alkane fraction can remain unreacted while the alkene fraction consumes the bromine.

It is a test for the presence of unsaturation, not a complete analysis of the mixture.

What if the alkane is exposed to ultraviolet light?

This is where the distinction between addition and substitution becomes important.

An alkane can also react with bromine, but it requires different conditions.

Under ultraviolet light, bromine molecules can split into reactive bromine radicals. These can initiate a substitution reaction with an alkane.

The bromine colour can therefore disappear even when the hydrocarbon is an alkane.

A useful comparison is:

What could the original experiment fail to establish?

Suppose the student observes that bromine water becomes colourless and concludes:

“The catalyst has produced alkenes, and all the products must be alkenes.”

There are two problems.

First, the experiment contains both heat and a catalyst. Heating alone can cause thermal cracking. Without a control experiment, we cannot establish how much of the reaction was caused by the catalyst.

A better investigation would compare two otherwise identical setups:

• One contains the catalyst.
• The other does not contain the catalyst.

Both should use the same feedstock, temperature, and reaction time.

Second, bromine water only confirms the presence of an alkene. It cannot establish that the mixture contains no alkanes.

To determine the individual product composition, the products would need to be separated and analysed using a more detailed method, such as gas chromatography coupled with mass spectrometry.

This is the difference between obtaining evidence that a reaction has occurred and identifying every product of that reaction.

04Addition versus substitution: two different reactions

Cracking, addition, and substitution are not interchangeable terms.

They describe different changes to hydrocarbon molecules.

Addition: atoms attach across a double bond

An alkene contains a carbon-carbon double bond.

That double bond consists of one sigma bond and one pi bond. The pi bond is more exposed and makes the alkene more reactive towards addition reactions.

Consider ethene reacting with bromine:

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

The reaction proceeds at room temperature without requiring ultraviolet light.

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

The product is 1,2-dibromoethane.

Notice what has happened:

• The original carbon skeleton is preserved.
• Both bromine atoms are incorporated into the product.
• The carbon-carbon double bond becomes a single bond.
• One main addition product is formed for this symmetrical alkene.

The orange-brown bromine colour disappears because bromine is consumed.

Substitution: one atom replaces another

Now consider ethane.

Ethane is an alkane and contains only carbon-carbon and carbon-hydrogen single bonds.

It does not readily react with bromine in the dark at room temperature.

However, ultraviolet light supplies energy to initiate a reaction.

One hydrogen atom is replaced by a bromine atom:

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

The products are bromoethane and hydrogen bromide.

This is a substitution reaction because one atom has been replaced by another.

The important changes are:

• One carbon-hydrogen bond is broken.
• A carbon-bromine bond forms.
• The carbon-carbon skeleton remains unchanged.
• Hydrogen combines with the other bromine atom to form HBr.

The reaction can cause the orange-brown bromine colour to fade, but it is slower than the addition reaction of an alkene under comparable laboratory conditions.

Cracking: the carbon chain breaks

Cracking is different from both reactions.

Consider the three processes side by side:

The key distinction is the bond being changed.

Cracking breaks the carbon skeleton into smaller molecules.

Addition uses an existing double bond to attach new atoms or groups.

Substitution replaces an atom, such as hydrogen, with another atom, such as bromine.

Do not call cracking an addition reaction simply because an alkene appears in the products. The formation of a double bond does not mean atoms have been added across an existing double bond.

05Worked examples

06Questions and solutions

Question 1

A long-chain alkane with the molecular formula \(\ce{C9H20}\) undergoes cracking.

One of the products is pentane, \(\ce{C5H12}\). Assume the other product is an open-chain monoalkene.

Determine the molecular formula of the alkene and explain your reasoning.

Solution 1

The missing product has the molecular formula \(\ce{C4H8}\), which is an open-chain monoalkene.

The original molecule contains nine carbon atoms and 20 hydrogen atoms.

Pentane accounts for five carbon atoms and 12 hydrogen atoms.

The remaining carbon atoms are:

\[ 9-5=4 \]

The remaining hydrogen atoms are:

\[ 20-12=8 \]

Therefore, the missing product is:

\[ \boxed{\ce{C4H8}} \]

This fits the general formula of an open-chain monoalkene, \(\ce{C_nH_{2n}}\).

The key is to conserve both carbon and hydrogen atoms. Finding the carbon count alone is not enough to establish the molecular formula.

Question 2

Two separate samples are prepared:

• Sample A contains ethene.
• Sample B contains ethane.

Both samples are mixed with orange-brown bromine water at room temperature in the absence of ultraviolet light.

Sample A rapidly becomes colourless, while Sample B remains orange-brown.

The samples are then exposed to ultraviolet light. The bromine colour in Sample B gradually disappears.

Explain the reactions occurring in both samples. Include balanced equations, reaction types, bond changes, and the reason for the different observations.

Solution 2

Sample A undergoes addition, while Sample B can undergo substitution when exposed to ultraviolet light.

Sample A: Ethene

Ethene contains a carbon-carbon double bond.

Bromine adds across this double bond at room temperature:

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

The pi component of the double bond is broken, and a bromine atom attaches to each carbon.

The product is 1,2-dibromoethane.

The orange-brown bromine colour rapidly disappears because bromine is consumed in the addition reaction.

Sample B: Ethane

Ethane contains only carbon-carbon and carbon-hydrogen single bonds.

In the absence of ultraviolet light, it does not readily react with bromine under these conditions. Therefore, the orange-brown colour remains.

When ultraviolet light is introduced, bromine molecules can split into reactive bromine radicals, initiating a substitution reaction.

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

One hydrogen atom is replaced by bromine.

The products are bromoethane and hydrogen bromide.

The bromine colour gradually disappears as the reaction proceeds.

The important distinction is that both reactions can consume bromine, but they involve different structural changes. Ethene undergoes addition across a C=C bond, while ethane undergoes substitution of a hydrogen atom.

Question 3

A student investigates cracking using paraffin oil and heated aluminium oxide.

The paraffin vapour passes over the hot catalyst, and the gaseous products are collected. A sample of the products rapidly decolourises bromine water in the dark.

The student makes two conclusions:

1. The aluminium oxide catalyst must be responsible for all the cracking.
2. Every product formed must be an alkene because bromine water became colourless.

Evaluate both conclusions.

Suggest two improvements to the investigation that would provide stronger evidence.

Solution 3

Neither conclusion is justified. The observation confirms the presence of at least one alkene, but it does not establish that the catalyst caused all the cracking or that every product is an alkene.

Conclusion 1: The catalyst caused all the cracking.

This cannot be established because the experiment uses both high temperature and a catalyst.

Thermal cracking can occur without a catalyst. Therefore, the student cannot distinguish the effect of heating from the effect of aluminium oxide.

An improvement would be to repeat the investigation without the catalyst while keeping the feedstock, temperature, and reaction time the same.

Comparing the product mixtures would help determine how the catalyst changes the reaction and product distribution.

Conclusion 2: Every product is an alkene.

This is also incorrect.

Bromine water rapidly decolourises when an alkene undergoes addition with bromine.

However, an alkane present in the same mixture does not need to react. It can remain unchanged while the alkene consumes the bromine.

Therefore, the colour change only demonstrates that at least one alkene is present.

A second improvement would be to separate and analyse the individual products rather than relying entirely on the bromine water test.

For example, gas chromatography can separate the mixture into individual components, and mass spectrometry can help identify them.

The investigation also needs to keep the bromine water test away from ultraviolet light. Otherwise, alkane substitution could also consume bromine and complicate the interpretation.

A good experimental conclusion must distinguish what the observation actually demonstrates from what the student wants it to demonstrate.

Question 4

A sample of cracking products contains only one open-chain alkane and one open-chain monoalkene.

The following illustrative experimental data are recorded:

Assume:

• Every alkene molecule contains one C=C bond.
• Each alkene reacts with one molecule of bromine.
• The alkane does not react with bromine in the dark.
• There are no other bromine-consuming substances.

(a) Calculate the amount of alkane and alkene in the mixture.

(b) Determine whether the molecular formulas of both products can be uniquely identified using the supplied data.

(c) If they cannot, give two different pairs of molecular formulas that satisfy the measurements.

Solution 4

The sample contains \(0.050\text{ mol}\) of alkane and \(0.030\text{ mol}\) of alkene. However, the molecular formulas cannot be uniquely determined from the supplied measurements.

Part (a): Calculate the amounts of each hydrocarbon.

Each alkene molecule reacts with one molecule of bromine.

Therefore:

\[ n(\text{alkene})=0.030\text{ mol} \]

The total sample contains \(0.080\text{ mol}\).

Subtract the amount of alkene:

\[ n(\text{alkane})=0.080-0.030 \]

\[ \boxed{n(\text{alkane})=0.050\text{ mol}} \]

Part (b): Determine whether the formulas are unique.

Let the alkane have the general formula:

\[ \ce{C_mH_{2m+2}} \]

Let the alkene have the general formula:

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

Using \(C=12.0\) and \(H=1.0\), their molar masses are:

\[ M_{\text{alkane}}=14m+2 \]

\[ M_{\text{alkene}}=14n \]

The total mass is the sum of the individual masses.

\[ 0.050(14m+2)+0.030(14n)=6.26 \]

Expand:

\[ 0.70m+0.10+0.42n=6.26 \]

Rearrange:

\[ 0.70m+0.42n=6.16 \]

Multiply by 50:

\[ 35m+21n=308 \]

Divide by seven:

\[ \boxed{5m+3n=44} \]

This equation does not have just one possible pair of integer solutions.

Part (c): Find two possible formula pairs.

One solution is:

\[ m=4,\qquad n=8 \]

This gives:

• Alkane: \(\ce{C4H10}\)
• Alkene: \(\ce{C8H16}\)

Check the mass:

\[ 0.050(58)+0.030(112) \]

\[ =2.90+3.36 \]

\[ =6.26\text{ g} \]

A second solution is:

\[ m=7,\qquad n=3 \]

This gives:

• Alkane: \(\ce{C7H16}\)
• Alkene: \(\ce{C3H6}\)

Check the mass:

\[ 0.050(100)+0.030(42) \]

\[ =5.00+1.26 \]

\[ =6.26\text{ g} \]

Both pairs satisfy the experimental measurements.

Therefore, the original data cannot distinguish between these possible product combinations.

An additional measurement, such as determining the molecular mass of each separated product, would be needed to identify the molecules uniquely.

This question demonstrates an important experimental principle: even several correct measurements may not provide enough information to identify a mixture uniquely.

Question 5

A refinery processes \(0.30\text{ mol}\) of dodecane using a simplified cracking pathway:

\[ \ce{C12H26 -> C8H18 + C4H8} \]

Assume that all the dodecane reacts according to this equation, without any hydrogen being supplied to the cracking reactor.

The product mixture then passes through a separate reactor containing a nickel catalyst. Excess hydrogen gas is supplied, and the butene is completely converted into butane.

The final mixture contains:

• \(0.30\text{ mol}\) octane
• \(0.30\text{ mol}\) butane

A bromine water test on the final mixture shows no decolourisation.

A student argues:

“The cracking reaction must have produced two alkanes because the final mixture contains only alkanes.”

Evaluate the student’s reasoning.

Calculate:

(a) The amount of hydrogen consumed.

(b) The mass of hydrogen incorporated into the products.

(c) The total mass of the final hydrocarbon mixture.

Explain why the final bromine water observation does not establish that cracking originally produced only alkanes.

Solution 5

The student’s conclusion is incorrect. The cracking reactor produces an alkane and an alkene, while the separate hydrogenation reactor converts the alkene into an alkane by adding hydrogen.

The distinction between the two stages is essential.

First, identify what happens during cracking.

The cracking reaction is:

\[ \ce{C12H26 -> C8H18 + C4H8} \]

For every mole of dodecane, one mole of octane and one mole of butene form.

Therefore, \(0.30\text{ mol}\) of dodecane produces:

\[ 0.30\text{ mol octane} \]

and

\[ 0.30\text{ mol butene} \]

The butene is unsaturated because it contains a C=C double bond.

Next, identify what happens during hydrogenation.

The butene reacts with hydrogen in the presence of a nickel catalyst:

\[ \ce{C4H8 + H2 ->[Ni,\ heat] C4H10} \]

This is an addition reaction.

The carbon-carbon double bond becomes a single bond, and one hydrogen atom attaches to each of the two carbon atoms that originally formed the double bond.

The carbon skeleton remains unchanged.

Part (a): Calculate the amount of hydrogen consumed.

The balanced equation gives a mole ratio of:

\[ \ce{C4H8:H2}=1:1 \]

Therefore:

\[ n(\ce{H2})=0.30\text{ mol} \]

\[ \boxed{0.30\text{ mol}} \]

Part (b): Calculate the mass of hydrogen incorporated.

The molar mass of hydrogen gas is:

\[ M(\ce{H2})=2.0\text{ g mol}^{-1} \]

Using:

\[ m=nM \]

\[ m=0.30\times2.0 \]

\[ \boxed{m=0.60\text{ g}} \]

The extra hydrogen atoms have come from the hydrogen gas supplied to the second reactor.

Part (c): Calculate the total mass of the final mixture.

The molar mass of octane is:

\[ M(\ce{C8H18})=8(12)+18=114\text{ g mol}^{-1} \]

Its mass is:

\[ m=0.30\times114=34.2\text{ g} \]

The molar mass of butane is:

\[ M(\ce{C4H10})=4(12)+10=58\text{ g mol}^{-1} \]

Its mass is:

\[ m=0.30\times58=17.4\text{ g} \]

The total final mass is:

\[ 34.2+17.4 \]

\[ \boxed{51.6\text{ g}} \]

We can check this against the starting material.

The molar mass of dodecane is:

\[ M(\ce{C12H26})=170\text{ g mol}^{-1} \]

Its starting mass was:

\[ 0.30\times170=51.0\text{ g} \]

The final mass is 51.6 g, which is 0.60 g greater.

That difference is exactly the mass of hydrogen added in the second reactor.

Why does the final bromine test not prove that cracking produced two alkanes?

The bromine water test is performed after both reactions have occurred.

Initially, the cracking reactor produces butene, which contains a C=C double bond. It would rapidly decolourise bromine water in the dark.

However, the subsequent hydrogenation reaction removes that double bond by adding hydrogen.

The final mixture contains octane and butane, both saturated alkanes. Therefore, it does not rapidly decolourise bromine water in the dark.

The negative result describes the final products, not necessarily the intermediate products formed during cracking.

If a sample were collected between the two reactors, it would contain butene and would give a positive bromine water test.

The deeper lesson is that the final composition of a reaction mixture does not always tell you which intermediate reactions produced it. You must account for every stage and every source of atoms.

07What to learn next

Cracking explains how large saturated hydrocarbons can produce smaller molecules, including alkenes.

Once an alkene has formed, its C=C bond opens up another set of possible reactions. Bromine can add across the double bond, hydrogen can convert the alkene into an alkane, and other reagents can produce different organic compounds.

The next useful skill is predicting exactly where the new atoms attach when an alkene undergoes addition.

Continue with How to Draw Products of Bromine Addition to Alkenes.