10.2.1 Formulae and structures of the alkanes

The first four alkanes and their formulae

1. Methane is ${CH}_4$, with a single carbon atom.
2. Ethane is $C_2{H}_6$.
3. Propane is $C_3{H}_8$.
4. Butane is $C_4{H}_{10}$.
5. Each fits the general formula $C_n{H}_{2n+2}$, so the count of hydrogens follows from the carbons.

Drawing the structure means showing every bond

1. A structure shows each atom and every covalent bond as a line between two atoms.
2. Every carbon atom has four lines drawn to it, because carbon forms four bonds.
3. Every hydrogen atom has exactly one line, because hydrogen forms one bond.
4. The carbons are drawn in a chain, joined to one another by single lines.
5. The remaining bonds on each carbon are filled with hydrogen atoms.

Counting the bonds as a check

1. The number of lines drawn to each carbon must come to four, every time.
2. A carbon with only three lines has a missing hydrogen.
3. The total number of hydrogen atoms drawn must match the molecular formula.
4. Propane has eight hydrogens, arranged three, two and three along the chain.
5. Butane has ten, arranged three, two, two and three.

Alkanes are saturated

1. Every bond between carbon atoms in an alkane is a single bond.
2. A molecule whose carbons are joined only by single bonds is saturated.
3. Saturated means the molecule holds as many hydrogen atoms as it can.
4. No more hydrogen can be added without breaking an existing bond.
5. This is what separates the alkanes from the alkenes, which have a double bond.

10.2.2 Alkenes and the C=C functional group

The alkenes and their formulae

1. Ethene is $C_2{H}_4$, the smallest alkene.
2. Propene is $C_3{H}_6$.
3. Butene is $C_4{H}_8$.
4. Each fits the general formula $C_n{H}_{2n}$.
5. An alkene has two fewer hydrogens than the alkane with the same number of carbons.

The double bond is the functional group

1. Every alkene contains a carbon to carbon double bond, written $C=C$.
2. That double bond is the alkene functional group.
3. Such a group is what gives a family its characteristic reactions.
4. The rest of the molecule is a chain of carbons carrying hydrogen atoms.
5. Every alkene reacts in a similar way because every one contains the same group.

Drawing the structures

1. Each carbon still forms four bonds in total, with the double bond counting as two of them.
2. Ethene is two carbons joined by a double bond, each carrying two hydrogens.
3. Propene is three carbons, with the double bond between the first and second.
4. The carbon at the end of the double bond carries two hydrogens, and the middle one carries a single hydrogen.
5. The third carbon in propene is joined by a single bond and carries three hydrogens.

Alkenes are unsaturated

1. A molecule containing a carbon to carbon double bond is unsaturated.
2. Unsaturated means the molecule could hold more hydrogen than it does.
3. The double bond can open so that two more atoms join on.
4. An alkane cannot do this, since it has no double bond to open.
5. This difference is what makes alkenes far more reactive than alkanes.

10.2.3 Addition reactions of alkenes and testing for unsaturation

An addition reaction opens the double bond

1. The double bond between the two carbons opens up.
2. Each of those carbons then has a spare bond to use.
3. An atom joins onto each of the two carbons.
4. A single product is formed, with nothing else made alongside it.
5. The product is saturated, since the double bond has gone.

Ethene reacts with bromine

1. Bromine, ${Br}_2$, adds across the double bond of ethene.
2. One bromine atom joins to each of the two carbon atoms.
3. The equation is: $$C_2{H}_4+{Br}_2\to C_2{H}_4{Br}_2$$
4. The product is 1,2-dibromoethane, with a bromine on each carbon.
5. Its structure is two carbons joined by a single bond, each carrying two hydrogens and one bromine.

The same reaction works for other alkenes

1. Every alkene has the same functional group, so every one reacts in the same way.
2. Propene adds bromine to give 1,2-dibromopropane.
3. The pattern extends directly: $$C_3{H}_6+{Br}_2\to C_3{H}_6{Br}_2$$
4. The two bromine atoms always end up on the carbons that held the double bond.
5. The rest of the molecule is unchanged by the reaction.

Bromine water distinguishes alkanes from alkenes

1. Bromine water is orange.
2. It is shaken with the hydrocarbon being tested.
3. With an alkene it turns colourless, because the bromine has been used up in addition.
4. With an alkane it stays orange, because there is no double bond to react with.
5. Orange to colourless is therefore the positive result for an alkene.

10.2.4 Combustion of alkanes and alkenes

Complete combustion oxidises the hydrocarbon

1. Burning an alkane or an alkene in plenty of oxygen is complete combustion.
2. Every carbon atom is oxidised to carbon dioxide.
3. Every hydrogen atom is oxidised to water.
4. The hydrocarbon has gained oxygen, which is what oxidation means here.
5. Energy is given out, so the reaction is exothermic.

Writing the equation for an alkane

1. Methane burns completely to give: $${CH}_4+2O_2\to {CO}_2+2H_{2}O$$
2. Ethane needs more oxygen for the same reason: $$2C_2{H}_6+7O_2\to 4{CO}_2+6H_{2}O$$
3. Each carbon in the formula gives one carbon dioxide molecule.
4. Each pair of hydrogens gives one water molecule.
5. Oxygen is balanced last, once the products are fixed.

Writing the equation for an alkene

1. Ethene burns completely to give: $$C_2{H}_4+3O_2\to 2{CO}_2+2H_{2}O$$
2. The products are the same two substances as for an alkane.
3. The double bond makes no difference to what is formed.
4. An alkene has fewer hydrogens, so it needs less oxygen per carbon than the matching alkane.
5. The same three-step balancing works for either family.

Burning in a limited supply

1. With too little oxygen the combustion is incomplete.
2. Carbon monoxide and soot can form, as with any hydrocarbon.
3. Alkenes tend to burn with a smokier flame than alkanes of similar size.
4. That smoke is unburnt carbon, released because the molecule has proportionally more of it.
5. A clean blue flame indicates that the supply of oxygen is plentiful.