Alkenes Worked ExamplesIGCSE Chemistry 0620

(a) Aqueous bromine is orange. In Y it changes from orange to colourless.

(b) Y is the alkene. Alkenes are unsaturated: their molecules contain a C=C double bond, and bromine adds across it, so the bromine is used up. X is an alkane (saturated — only single C–C bonds), so it does not react and the aqueous bromine stays orange.

(c) The aqueous bromine stays orange — ethanol has no C=C bond (all its carbon–carbon bonds are single).

(a) Compare each pair of bars: supply exceeds demand only for kerosene (12 v 10) and fuel oil and bitumen (36 v 19).

(b) Reason 1: there is more long-chain fuel oil than is needed, but not enough gasoline (demand 30, supply 18). Cracking breaks larger alkane molecules into smaller, more useful ones such as gasoline. Reason 2: cracking also produces alkenes (and hydrogen), which are not present in petroleum but are needed, e.g. to make polymers and ethanol.

(c) Balance atoms. Carbon: $16 - 8 - 4 = 4$ left for 2X, so X has 2 C. Hydrogen: $34 - 18 - 8 = 8$ left for 2X, so X has 4 H. X is $\text{C}_2\text{H}_4$ (ethene).

(d) Propane loses two H atoms, which form one hydrogen molecule.

(a) The porcelain chips are the catalyst for cracking.

(b) Cracking needs a high temperature (as well as a catalyst) to break the long alkane molecules; the paraffin vapour must pass over the hot catalyst.

(c) The gas contains an alkene — an unsaturated compound with a C=C bond — formed by cracking.

(d) As the tube cools, the gas inside contracts and cold water would be sucked back into the hot boiling tube, which could crack it.

(e) Octane is $\text{C}_8\text{H}_{18}$. Carbon left: $12 - 8 = 4$ → two ethene molecules. Check H: $18 + 2(4) = 26$ ✓.

Cracking breaks up a large alkane molecule. The products contain only the carbon and hydrogen atoms that were in it: smaller alkanes, alkenes and, in some reactions, hydrogen — e.g. $\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_8\text{H}_{18} + \text{C}_2\text{H}_4$ or $\text{C}_2\text{H}_6 \rightarrow \text{C}_2\text{H}_4 + \text{H}_2$.

Carbon dioxide contains oxygen, which is not involved in cracking — it is a combustion product, so D is wrong. A leaves out hydrogen, and C leaves out the smaller alkane and hydrogen and wrongly includes carbon dioxide. B is correct.

Bromine adds to alkenes at room temperature with no catalyst — that is why it can be used as a quick test. So A is wrong.

Hydrogen adds in the presence of a nickel catalyst, giving the alkane: $\text{C}_3\text{H}_6 + \text{H}_2 \rightarrow \text{C}_3\text{H}_8$ (propane). B is correct.

Steam adds using an acid catalyst (and high temperature and pressure) to give an alcohol, not a carboxylic acid. So C (wrong catalyst and product) and D (no catalyst) are wrong.

(a) In an addition reaction only one product is formed.

(b) Condition: a nickel catalyst. One H adds to each carbon of the C=C bond, giving butane.

(c) Condition: an acid catalyst (with high temperature and pressure). H adds to one carbon of the C=C and OH to the other. Because but-2-ene is symmetrical, the OH is always on carbon 2: butan-2-ol.

(a) Products of complete combustion: carbon dioxide and water. Balance C: 4 $\text{CO}_2$. Balance H: 8 H → 4 $\text{H}_2\text{O}$. Balance O: $8 + 4 = 12$ O atoms → 6 $\text{O}_2$.

(b) Now 4 CO and 4 $\text{H}_2\text{O}$: O atoms $4 + 4 = 8$ → 4 $\text{O}_2$.

(c) Carbon (soot) — unburnt carbon particles.

(a) $M_r(\text{C}_4\text{H}_8) = 48 + 8 = 56$.

(b) 1 mol of alkene reacts with 1 mol of $\text{H}_2$.

(c) $M_r(\text{Br}_2) = 2 \times 80 = 160$.

(d) Moles of but-1-ene = moles of bromine that react; then $V = n \div c$, and convert $\text{dm}^3$ to $\text{cm}^3$.

(a) Convert $\text{cm}^3$ to $\text{dm}^3$, find moles, then $M_r$ = mass ÷ moles.

(b) $12n + 2n = 14n = 42$, so $n = 3$: $\text{C}_3\text{H}_6$, propene.

(c) Propene is $\text{CH}_2{=}\text{CHCH}_3$. (i) Adding $\text{H}_2$ across C=C gives propane. (ii) Adding $\text{Br}_2$ puts one Br on each carbon of the former C=C bond.

(a) Divide each mass by its $M_r$; $M_r(\text{Br}_2) = 160$.

(b) One $\text{Br}_2$ molecule adds across each C=C bond. Ratio $\text{Br}_2 : \text{Z} = 0.0200 : 0.0100 = 2 : 1$, so Z has 2 C=C bonds.

(c) A saturated 6-carbon chain is $\text{C}_6\text{H}_{14}$. Each C=C bond removes 2 H atoms: $14 - 2 \times 2 = 10$, so $\text{C}_6\text{H}_{10}$.

(a) Q — a catalyst provides a pathway with a lower activation energy, so its peak is lower. Both curves start and finish at the same levels: $\Delta H$ is unchanged.

(b) X is the activation energy (uncatalysed); Y is the enthalpy change, $\Delta H$.

(c) Bonds broken: the C=C bond becomes C–C, and one H–H bond breaks. Bonds made: one C–C and two new C–H. (The four original C–H bonds are unchanged, so they can be left out.)

$\Delta H$ = broken − made.

(d) Bond making is exothermic (bond breaking is endothermic).

(a) In addition, each Br joins one carbon of the old C=C bond. The Br atoms are on carbons 1 and 2, so the C=C bond was between C1 and C2.

(b) Four-carbon chain, Br on carbons 1 and 2: 1,2-dibromobutane. (But-2-ene would give 2,3-dibromobutane instead.)

(c) Show every atom and bond; all carbon–carbon bonds are now single.