Complete combustion of alkanes, alkenes and alcohols

1. Alkanes contain only strong single bonds, so apart from combustion they are fairly unreactive.
2. In complete combustion, every carbon atom ends up in carbon dioxide and every hydrogen atom ends up in water.
3. The same two products form from any hydrocarbon or alcohol that burns in plenty of oxygen.
   1. For methane: $${CH}_4+2O_2\to {CO}_2+2H_{2}O$$
   2. For ethene: $$C_2{H}_4+3O_2\to 2{CO}_2+2H_{2}O$$
   3. For ethanol: $$C_2{H}_{5}OH+3O_2\to 2{CO}_2+3H_{2}O$$
4. To balance a combustion equation, balance carbon first, then hydrogen, and oxygen last.

Alkenes react by addition across the double bond

1. The $C=C$ double bond makes alkenes unsaturated and much more reactive than alkanes.
2. In an addition reaction, one bond of the double bond breaks and a new atom joins each of the two carbon atoms.
   1. Two molecules join to make one product, which is saturated.
3. With hydrogen and a nickel catalyst at about $150{}^{\circ }C$, an alkene forms the alkane with the same carbon chain: $$C_2{H}_4+H_2\to C_2{H}_6$$
4. With bromine, an alkene forms a compound with a bromine atom on each carbon of the old double bond: $$C_2{H}_4+{Br}_2\to C_2{H}_4{Br}_2$$
   1. The product from ethene is 1,2-dibromoethane: $\begin{array}{ccccccc}& & \mathrm{H}& & \mathrm{H}& & \\ & & |& & |& & \\ \mathrm{H}& -& \mathrm{C}& -& \mathrm{C}& -& \mathrm{H}\\ & & |& & |& & \\ & & \mathrm{Br}& & \mathrm{Br}& & \end{array}$
5. Shaking an alkene with bromine water turns it from orange to colourless, while an alkane leaves it orange.
   1. Acidified potassium manganate(VII) also turns from purple to colourless with an alkene, but it reacts with alcohols too, so bromine water is the clearer test for a $C=C$ bond.

Steam adds to alkenes to form alcohols

1. Ethene reacts with steam at about $300{}^{\circ }C$ and high pressure, with a phosphoric acid catalyst, to form ethanol: $$C_2{H}_4+H_{2}O\to C_2{H}_{5}OH$$
   1. A hydrogen atom joins one carbon of the double bond and the $-OH$ group joins the other.
2. For any alkene, keep the whole carbon chain and change only the double bond to a single bond.
3. Then add the atoms from the other reactant, one to each carbon atom that was in the double bond.
4. In general formulae, an alkene $C_n{H}_{2n}$ gives $C_n{H}_{2n+2}$ with hydrogen, $C_n{H}_{2n}{Br}_2$ with bromine and $C_n{H}_{2n+1}OH$ with steam.
   1. Propene with hydrogen gives propane, $C_3{H}_8$.
   2. Propene with steam gives propanol, $C_3{H}_{7}OH$.

Alcohols burn and are oxidised to carboxylic acids

1. Alcohols burn in plenty of oxygen to form carbon dioxide and water, which is why ethanol is blended into UK petrol.
   1. For methanol: $$2{CH}_{3}OH+3O_2\to 2{CO}_2+4H_{2}O$$
2. Warming an alcohol with acidified potassium manganate(VII) oxidises it to a carboxylic acid.
   1. The solution changes from purple to colourless as the manganate(VII) reacts.
3. The $-{CH}_{2}OH$ end of the alcohol becomes $-COOH$, so the carbon chain keeps the same length.
   1. Ethanol is oxidised to ethanoic acid: $$\begin{array}{ccccccccc}& & \mathrm{H}& & \mathrm{H}& & & & \\ & & |& & |& & & & \\ \mathrm{H}& -& \mathrm{C}& -& \mathrm{C}& -& \mathrm{O}& -& \mathrm{H}\\ & & |& & |& & & & \\ & & \mathrm{H}& & \mathrm{H}& & & & \end{array}\to \begin{array}{ccccccccc}& & \mathrm{H}& & \mathrm{O}& & & & \\ & & |& & \Vert & & & & \\ \mathrm{H}& -& \mathrm{C}& -& \mathrm{C}& -& \mathrm{O}& -& \mathrm{H}\\ & & |& & & & & & \\ & & \mathrm{H}& & & & & & \end{array}$$
4. Wine or beer left open to the air turns to vinegar, because bacteria oxidise its ethanol to ethanoic acid.

Predicting products from the functional group

1. Start by finding the functional group: only single bonds for an alkane, $C=C$ for an alkene or $-OH$ for an alcohol.
2. Then pick the reaction: combustion for all three, addition for an alkene and oxidation for an alcohol.
3. Copy the carbon chain unchanged, then change only the atoms at the functional group.
   1. Pentan-1-ol, $C_5{H}_{11}OH$, is oxidised to pentanoic acid, $C_4{H}_{9}COOH$.
   2. Hex-1-ene, $C_6{H}_{12}$, adds hydrogen to form hexane, $C_6{H}_{14}$.
4. Finish by checking that the product has the same number of carbon atoms as the starting compound, unless it has burned.