Topic 4 · page 11 of 17

Cracking and reforming

Cracking breaks long alkane chains into shorter alkanes and alkenes. Reforming rearranges straight chains into branched and cyclic molecules. Both change the molecules, so both need balanced equations.

13 min read4.12

Read first: Crude oil and fractional distillation

Cracking 4.12

There is more long-chain oil in crude oil than people need, and more demand for petrol-sized molecules. Cracking turns the surplus into more useful products. The alkenes made are the starting materials for polymers and other chemicals.

Cracking: skeletal formula of decane on the left, forming octane and ethene on the right. Ethene has a C=C double bond.
Decane, C10H22, cracks into octane, C8H18, and ethene, C2H4.
CX10HX22→CX8HX18+CX2HX4 \ce{C10H22 -> C8H18 + C2H4}

The conditions are a high temperature. A zeolite catalyst lets the reaction happen at a lower temperature, so less energy is needed.

Writing cracking equations

A cracking equation must balance for carbon and for hydrogen. At least one product is an alkene. The reason is that a long alkane, CₙH₂ₙ₊₂, breaks into pieces, and there are not enough hydrogen atoms for every piece to be an alkane.

Worked exampleCompleting a cracking equation

Complete: C₁₆H₃₄ → C₁₀H₂₂ + ……

  1. Carbon: 16 − 10 = 6 carbons are left.
  2. Hydrogen: 34 − 22 = 12 hydrogens are left.
  3. The missing product is C₆H₁₂. It has the formula CₙH₂ₙ, so it is an alkene (hexene).
CX16HX34→CX10HX22+CX6HX12 \ce{C16H34 -> C10H22 + C6H12}

Worked exampleA cracking equation with several products

Dodecane, C₁₂H₂₆, cracks to give two molecules of ethene and one other product. Find the other product.

  1. Two ethene molecules are 2 × C₂H₄ = C₄H₈.
  2. Carbon left: 12 − 4 = 8. Hydrogen left: 26 − 8 = 18.
  3. The other product is C₈H₁₈, octane.
CX12HX26→2 CX2HX4+CX8HX18 \ce{C12H26 -> 2C2H4 + C8H18}

Reforming 4.12

Straight-chain alkanes tend to burn too quickly in a petrol engine. Branched and ring molecules burn more smoothly, so the petrol performs better. The reaction uses a platinum catalyst.

Two reforming reactions with a platinum catalyst. Top: hexane forms cyclohexane and hydrogen. Bottom: hexane forms 2-methylpentane.
Reforming makes a ring (and hydrogen) or a branched isomer.

Forming a ring loses hydrogen. Forming a branched isomer does not change the formula.

CX6HX14→CX6HX12+HX2 \ce{C6H14 -> C6H12 + H2}
CX6HX14→CX6HX14 \ce{C6H14 -> C6H14}

The second equation looks odd, but it is correct: hexane and 2-methylpentane are isomers, so both sides have the formula C₆H₁₄. Write the structures, or the names, when you use it, so the reader can see the change.

Reforming can also make arenes such as benzene, which you will study later in the course.

CX6HX14→CX6HX6+4 HX2 \ce{C6H14 -> C6H6 + 4H2}

Three processes compared 4.12

Fractional distillation Cracking Reforming
Type of change physical chemical chemical
What happens separation by boiling temperature long chains break into shorter chains straight chains rearrange into branched or cyclic molecules
Products fractions (mixtures) shorter alkanes and alkenes branched alkanes, cyclic hydrocarbons, and hydrogen
Conditions heat; column hot at the bottom high temperature; zeolite catalyst platinum catalyst
Equation needed? no yes yes

Check yourself

  1. Define cracking and give two reasons why it is carried out.
  2. Complete the equation C₁₂H₂₆ → C₁₀H₂₂ + ……
  3. Complete the equation C₁₅H₃₂ → 2C₂H₄ + C₃H₆ + ……
  4. Explain why every cracking equation includes an alkene.
  5. Write an equation for the reforming of heptane, C₇H₁₆, into methylcyclohexane, C₇H₁₄.
  6. Explain why petrol manufacturers want branched and cyclic molecules in petrol.

Answers cover these until you have tried the questions

  1. Cracking breaks long-chain alkanes into shorter-chain alkanes and alkenes. It is done because there is more long-chain oil than is needed and more demand for petrol-sized molecules; the alkenes are useful for making polymers and other chemicals.
  2. Carbon 12 − 10 = 2; hydrogen 26 − 22 = 4. The missing product is C₂H₄ (ethene).
  3. Products so far: 2 × C₂H₄ = C₄H₈, plus C₃H₆, which gives C₇H₁₄. Carbon left: 15 − 7 = 8. Hydrogen left: 32 − 14 = 18. The missing product is C₈H₁₈ (octane).
  4. A long alkane has the formula CₙH₂ₙ₊₂. If it split into two alkanes with a and b carbons (a + b = n), they would need 2a + 2 + 2b + 2 = 2n + 4 hydrogens, but only 2n + 2 are available. There are not enough hydrogens, so one piece must have two fewer than an alkane would have. That piece has a C=C bond: it is an alkene.
  5. C₇H₁₆ → C₇H₁₄ + H₂. Check: carbon 7 = 7; hydrogen 16 = 14 + 2.
  6. Branched and cyclic molecules burn more smoothly in an engine than straight-chain alkanes, so the fuel performs better.