Alkanes Worked ExamplesIGCSE Chemistry 0620

(a)(i) A hydrocarbon is a compound containing hydrogen and carbon only. (ii) Saturated means all carbon–carbon bonds are single bonds. The bonding in alkanes is single covalent.

(b) An alkane fits $\text{C}_n\text{H}_{2n+2}$ and contains only C and H. $\text{C}_6\text{H}_{14}$: $2(6)+2 = 14$ ✓. $\text{C}_5\text{H}_{10}$: needs 12 H ✗ (an alkene). $\text{C}_8\text{H}_{18}$ ✓. $\text{C}_3\text{H}_8\text{O}$ contains O ✗. $\text{C}_9\text{H}_{20}$ ✓.

(c) Pentane is an unbranched chain of 5 carbons. Carbon–carbon bonds: one fewer than the number of carbons, $5 - 1 = 4$. Every H is joined by one C–H bond: 12.

Write the equation. One H atom in butane is replaced by a Cl atom, and the H that is removed joins the other Cl atom.

One atom replaced by another = substitution (so not A: addition needs a C=C bond and gives one product). It is photochemical: ultraviolet light supplies the activation energy (so not C — nickel is the catalyst for adding hydrogen to alkenes). The other product is hydrogen chloride, not hydrogen (so not D).

(a) Ultraviolet (UV) light. It provides the activation energy, $E_a$, needed to start the reaction — that is what 'photochemical' means.

(b) One H is swapped for Cl; the removed H forms HCl.

(c) Butane has two kinds of carbon: the two end carbons (equivalent) and the two middle carbons (equivalent). Replacing an H on an end carbon gives 1-chlorobutane, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl}$:

Replacing an H on a middle carbon gives 2-chlorobutane, $\text{CH}_3\text{CHClCH}_2\text{CH}_3$. (Putting Cl on carbon 3 gives the same molecule, just turned round.)

(d) In a substitution reaction one atom or group of atoms is replaced by another atom or group of atoms.

(a) Boiling point increases as the number of carbon atoms (chain length) increases.

(b) Go up from 6 on the horizontal axis to the curve, then across to the boiling-point axis: about 68 °C (between the 5-carbon value, 36 °C, and the 7-carbon value, 98 °C).

(c) A substance is a liquid at 20 °C if it melts below 20 °C and boils above 20 °C. The plotted alkanes with boiling points above 20 °C have 5, 7 and 8 carbon atoms. Those with 1–4 carbons boil below 20 °C, so they are gases.

(d) Any one property with its correct trend: viscosity increases (the liquid becomes thicker and flows less easily), or volatility decreases (it evaporates less easily).

(a) Complete combustion of a hydrocarbon gives carbon dioxide and water only. Balance C first (6 $\text{CO}_2$), then H (7 $\text{H}_2\text{O}$), then O: $12 + 7 = 19$ O atoms, i.e. $9\tfrac{1}{2}$ $\text{O}_2$. Double everything to clear the half.

(b) With limited oxygen, carbon forms CO. O atoms needed: $6 + 7 = 13$, i.e. $6\tfrac{1}{2}$ $\text{O}_2$; double it.

(c) Carbon monoxide is toxic (poisonous) — and it is colourless and odourless, so it is not noticed.

(a) $M_r(\text{C}_4\text{H}_{10}) = 4(12) + 10(1) = 58$.

(b) Mole ratio butane : oxygen = 2 : 13.

(c) Butane : carbon dioxide = 2 : 8 = 1 : 4; butane : water = 2 : 10 = 1 : 5. $M_r(\text{H}_2\text{O}) = 18$.

(d) For gases at the same temperature and pressure, volume ratio = mole ratio, so no conversion to moles is needed.

(a) P. One hydrogen atom in ethane has been replaced by a chlorine atom, and a second product (HCl) forms.

(b) Q. Two reactants join to give only one product — the definition of an addition reaction. (R is cracking; S is combustion.)

(c) Ethane is saturated: all its carbon–carbon bonds are single, so there is no C=C bond for atoms to add across. Alkanes are generally unreactive apart from combustion and substitution by chlorine.

(a) Chlorine is a pale yellow-green gas. In jar B the yellow-green colour fades / disappears as chlorine is used up — the products (chloromethane and hydrogen chloride) are colourless.

(b) Jar A is a control. The foil keeps out the UV light, so its colour stays yellow-green — this shows the reaction needs ultraviolet light (the light supplies the activation energy).

(c) $\text{CH}_4 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{HCl}$. Hydrogen chloride dissolves in the water on the paper to form an acid, turning blue litmus red.

(d) Chlorine (and hydrogen chloride) are toxic, so a fume cupboard removes the gases. Methane is flammable — a naked flame could ignite it.

(a) Moles of gas from volume, then $M_r$ = mass ÷ moles.

(b) $M_r = 12n + (2n + 2) = 14n + 2$.

(c) Four carbons can be joined in an unbranched chain (butane) or as a three-carbon chain with a $\text{CH}_3$ branch on the middle carbon (methylpropane).

Group the carbons by their surroundings. The two $\text{CH}_3$ groups attached to the branching carbon (C1 and the branch) are identical. Then there is the CH on the branching carbon, the $\text{CH}_2$, and the $\text{CH}_3$ at the other end.

Replacing an H on each different type of carbon gives a different product: (1) Cl on one of the two identical $\text{CH}_3$ groups at the branched end; (2) Cl on the branching CH; (3) Cl on the $\text{CH}_2$; (4) Cl on the $\text{CH}_3$ at the far end. That is 4. C is correct.

D (5) counts the two identical $\text{CH}_3$ groups on the branching carbon as different. A and B miss carbons whose hydrogens can also be replaced.

(a) X, from the reactants up to the peak, is the activation energy, $E_a$. Y, from the reactants down to the products, is the enthalpy change, $\Delta H$ (negative — exothermic).

(b) Only the bonds that change need counting: one C–H and one Cl–Cl are broken (endothermic); one C–Cl and one H–Cl are made (exothermic).

$\Delta H$ = energy to break bonds − energy released making bonds.

(c) Colliding particles must have at least the activation energy to react. In the dark too few have enough energy; ultraviolet light supplies the activation energy, so the reaction starts only in UV light.

(a) $M_r(\text{C}_2\text{H}_6) = 30$. Ethane : chlorine = 1 : 1.

(b) Ethane : chloroethane = 1 : 1. $M_r(\text{C}_2\text{H}_5\text{Cl}) = 24 + 5 + 35.5 = 64.5$.

(c) Percentage yield = actual ÷ theoretical × 100.

(d) Substitution does not stop after one H: chlorine can go on to replace further hydrogen atoms, forming products such as $\text{C}_2\text{H}_4\text{Cl}_2$, so some ethane (and chloroethane) is turned into other compounds. Some product is also lost when it is separated from the mixture.