Formulae, functional groups and terminology

Worked example 13 marks

1. 1
   1. Recall the definition of a homologous series. The key features are what define the series.
2. 2
   1. Feature 1: All members have the same general formula (e.g., CₙH₂ₙ₊₂ for alkanes).
3. 3
   1. Feature 2: All members have the same functional group, which gives them similar chemical properties (e.g., the C=C bond in alkenes).
4. 4
   1. Feature 3: Successive members differ by a -CH₂- group.
5. 5
   1. Feature 4 (optional but good): There is a gradual trend in physical properties, such as boiling points, as the chain length increases.

Worked example 12 marks

1. 1
   1. Identify the number of carbon atoms in the longest chain. There are three carbon atoms.
2. 2
   1. The prefix for three carbons is 'prop-'.
3. 3
   1. Identify the functional group. The -OH group indicates it is an alcohol.
4. 4
   1. The suffix for an alcohol is '-ol'.
5. 5
   1. Combine the parts: 'prop' + 'an' + 'ol' gives 'propanol'. (Specifically, propan-1-ol as the OH is on the end carbon).

Worked example 22 marks

1. 1
   1. Identify the prefix. 'Pent-' means 5 carbon atoms in total.
2. 2
   1. Identify the suffix. '-oic acid' means it is a carboxylic acid, with the functional group -COOH.
3. 3
   1. The -COOH group contains one of the 5 carbon atoms. This leaves 4 other carbon atoms in the chain (C₄H₉).
4. 4
   1. The full formula is therefore C₄H₉COOH. To write a molecular formula, sum all atoms: 5 carbons, 10 hydrogens (9+1), and 2 oxygens.
5. 5
   1. The molecular formula is C₅H₁₀O₂.

Worked example 13 marks

1. 1
   1. For the displayed formula, 'eth-' means 2 carbons. Draw two C atoms joined by a single bond: C-C.
2. 2
   1. Alkanes are saturated, so each carbon must have 4 single bonds in total. Add H atoms to each carbon until it has 4 bonds. This results in 3 H atoms on each C.
3. 3
   1. The final structure is H₃C-CH₃, with all bonds shown.
4. 4
   1. For combustion, the reactants are ethane (C₂H₆) and oxygen (O₂). The products of complete combustion are carbon dioxide (CO₂) and water (H₂O).
5. 5
   1. Write the unbalanced equation: C₂H₆ + O₂ → CO₂ + H₂O.
6. 6
   1. Balance the carbons: 2 C on the left, so 2CO₂ on the right. C₂H₆ + O₂ → 2CO₂ + H₂O.
7. 7
   1. Balance the hydrogens: 6 H on the left, so 3H₂O on the right. C₂H₆ + O₂ → 2CO₂ + 3H₂O.
8. 8
   1. Balance the oxygens: On the right, there are (2×2) + (3×1) = 7 oxygen atoms. So we need 3.5 O₂ molecules. C₂H₆ + 3.5O₂ → 2CO₂ + 3H₂O.
9. 9
   1. Double the whole equation to avoid fractions: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O.

Worked example 13 marks

1. 1

   (a) 'Prop-' means 3 carbons. '-ene' means a C=C double bond. Draw three carbons in a chain with one double bond: C=C-C.

2. 2

   (a) Now, add hydrogen atoms so that each carbon has exactly 4 bonds in total. The first C gets 2 H's, the second C gets 1 H, and the third C gets 3 H's. The structure is CH₂=CH-CH₃.

3. 3

   (b) Alkenes are unsaturated due to the C=C bond.

4. 4

   (b) They undergo an addition reaction with bromine.

5. 5

   (b) Aqueous bromine is an orange-brown solution.

6. 6

   (b) The observation is that the orange-brown colour of the bromine water disappears; it becomes colourless.

Worked example 13 marks

1. 1

   (a) The formula is CH₃OH. Draw the carbon atom bonded to three hydrogen atoms and the oxygen atom. The oxygen atom is then bonded to the final hydrogen atom. Ensure C has 4 bonds and O has 2.

2. 2

   (b) Complete combustion produces carbon dioxide (CO₂) and water (H₂O).

3. 3

   (b) Write the unbalanced equation: CH₃OH + O₂ → CO₂ + H₂O.

4. 4

   (b) Balance C: 1 on left, 1 on right. Balanced.

5. 5

   (b) Balance H: 4 on left (3+1), so need 2H₂O on right. CH₃OH + O₂ → CO₂ + 2H₂O.

6. 6

   (b) Balance O: 4 on right (2 + 2). On left, 1 in CH₃OH, so need 3 more from O₂. This means 1.5 O₂ molecules. CH₃OH + 1.5O₂ → CO₂ + 2H₂O.

7. 7

   (b) Double the entire equation to remove the fraction: 2CH₃OH + 3O₂ → 2CO₂ + 4H₂O.

Worked example 13 marks

1. 1
   1. Identify the reactants: a carboxylic acid (propanoic acid) and a metal oxide (magnesium oxide). This is an acid + base reaction.
2. 2
   1. Recall that acid + base → salt + water.
3. 3
   1. The salt is formed from the metal cation (Mg²⁺) and the carboxylate anion (C₂H₅COO⁻). To balance the charges, two propanoate ions are needed for every one magnesium ion. The salt's formula is (C₂H₅COO)₂Mg.
4. 4
   1. The other product is water (H₂O).
5. 5
   1. Write the unbalanced equation: C₂H₅COOH + MgO → (C₂H₅COO)₂Mg + H₂O.
6. 6
   1. Balance the propanoate group: There are 2 on the right, so we need 2 C₂H₅COOH on the left. 2C₂H₅COOH + MgO → (C₂H₅COO)₂Mg + H₂O.
7. 7
   1. Check the other atoms. Mg is balanced (1 and 1). H and O in the acid/base reaction are balanced (2H from the acid's -COOH and 1 O from MgO form one H₂O). The equation is now balanced.

Worked example 12 marks

1. 1
   1. Isomers have the same molecular formula but different structures. We need to arrange 3 carbons, 8 hydrogens and 1 oxygen in two different ways.
2. 2
   1. Isomer 1 (Propan-1-ol): Arrange the 3 carbons in a chain (C-C-C). Place the -OH group on an end carbon (carbon 1).
3. 3
   1. Fill in the remaining bonds with hydrogen atoms, ensuring each carbon has 4 bonds. The structure is CH₃-CH₂-CH₂-OH.
4. 4
   1. Isomer 2 (Propan-2-ol): Arrange the 3 carbons in a chain (C-C-C). This time, place the -OH group on the middle carbon (carbon 2).
5. 5
   1. Fill in the remaining bonds with hydrogen atoms. The structure is CH₃-CH(OH)-CH₃.
6. 6
   1. Both structures have the formula C₃H₈O but are different compounds with different names and properties.