Alkenes are manufactured by cracking larger alkane molecules.
(a) State the source of the large alkane molecules used in cracking.
(b) State two conditions needed for cracking large alkane molecules.
(c) When one molecule of dodecane, \( C_{12}H_{26} \), is cracked, three molecules of but-1-ene and one other product are formed.
(i) Use molecular formulae to complete the symbol equation for this reaction.
\( C_{12}H_{26} \rightarrow \) ………………………………… + …………………………………
(ii) Suggest the type of chemical reaction which happens during cracking.
(d) Propene will undergo polymerisation.
(i) Suggest the name of the polymer formed from propene.
(ii) Draw part of this polymer molecule to show three repeat units.
(iii) State the type of polymerisation propene undergoes.
▶️ Answer/Explanation
(a) petroleum
The large alkane molecules used in cracking come from petroleum, which is the crude oil extracted from the earth. Petroleum contains a mixture of hydrocarbons of varying chain lengths.
(b) 1. high temperature
2. catalyst
Cracking requires two main conditions: high temperature (typically around 400-700°C) to provide the necessary energy to break the carbon-carbon bonds, and a catalyst (usually silica or alumina) to speed up the reaction by lowering the activation energy.
(c)(i) \( C_{12}H_{26} \rightarrow 3C_4H_8 + H_2 \)
When dodecane (\( C_{12}H_{26} \)) cracks to form three but-1-ene (\( C_4H_8 \)) molecules, the remaining two hydrogen atoms form a hydrogen gas (\( H_2 \)) molecule to balance the equation.
(c)(ii) thermal decomposition
Cracking is a type of thermal decomposition reaction where large hydrocarbon molecules are broken down into smaller molecules by heat energy.
(d)(i) poly(propene)
The polymer formed from propene is called poly(propene) or polypropylene, where many propene monomers join together.
(d)(ii)
Structure showing three repeat units of poly(propene):
\[ \text{–[CH}_2\text{–CH(CH}_3\text{)]–[CH}_2\text{–CH(CH}_3\text{)]–[CH}_2\text{–CH(CH}_3\text{)]–} \]
The diagram should show a chain of six carbon atoms with methyl groups (\( CH_3 \)) attached to every other carbon, representing three repeat units of the polymer.
(d)(iii) addition
Propene undergoes addition polymerisation, where the double bond in the propene monomer opens up to form single bonds with adjacent monomers, without the elimination of any small molecules.
This question is about alkanes and alkenes.
(a) Short-chain alkanes and alkenes can be formed from long-chain alkanes in a chemical reaction.
(i) Name the type of chemical reaction which forms short‑chain alkanes and alkenes from long‑chain alkanes.
(ii) Decane has 10 carbon atoms. It forms ethane and ethene as the only products in this type of chemical reaction.
Write the chemical equation for this reaction.
(b) Ethane reacts with chlorine at room temperature to form chloroethane, C2H5Cl, and one other product.
(i) Name the other product formed.
(ii) State the condition needed for this reaction to take place.
(c) Ethene reacts with chlorine at room temperature to form dichloroethane, C2H4Cl2.
C2H4 + Cl2 → C2H4Cl2
(i) State why this is an addition reaction.
(ii) The chemical equation for this reaction can be represented as shown.
The energy change for the reaction is –180kJ/mol.
Use the bond energies in the table to calculate the bond energy of a C–Cl bond, in kJ/mol.
Use the following steps.
step 1 Calculate the energy needed to break bonds.
energy needed to break bonds = kJ
step 2 Use your answer in step 1 and the energy change for the reaction to determine the energy released when bonds are formed.
energy released when bonds form = kJ
step 3 Use your answer in step 2 and bond energy values to determine the energy of a C–Cl bond.
bond energy of a C–Cl bond = kJ/mol
▶️ Answer/Explanation
(a) (i) The reaction is called cracking (breaking long-chain hydrocarbons into shorter ones).
(ii) Decane cracking equation: C10H22 → 4C2H4 + C2H6
– Balanced: 10 carbons → 4 ethene (C2H4) + 1 ethane (C2H6).
(b) (i) The other product is hydrogen chloride (HCl).
(ii) Condition: UV light (initiates free-radical substitution).
(c) (i) This is an addition reaction because chlorine adds across the C=C double bond, forming a single product (dichloroethane).
(ii) Bond energy calculation:
Step 1: Energy to break bonds = (C=C) + (Cl–Cl) = 610 + 240 = 850 kJ/mol.
Step 2: Energy released = Step 1 + ΔH = 850 + 180 = 1030 kJ/mol.
Step 3: Energy in C2H4Cl2 bonds = 4(C–H) + 2(C–Cl) + (C–C).
⇒ 1030 = 4(410) + 2(C–Cl) + 350 ⇒ C–Cl bond = 340 kJ/mol.