Homologous Series Exam Questions

A homologous series is a family of organic compounds that:

• Have the same general formula [1]
• Have the same functional group [1]
• Show a gradual trend/gradation in physical properties and have similar chemical properties [1]

Examiner note: Each successive member differs by CH2 (a methylene group). “Same type of compound” is too vague — you must mention general formula and functional group. The gradual trend in physical properties (e.g., boiling point increases with chain length) distinguishes a homologous series from a random group of compounds.

(a) C5H12 [1] (n = 5, so 2(5) + 2 = 12 hydrogens)

(b) Pentane [1]

(c) C3H6 [1] (n = 3, so 2(3) = 6 hydrogens)

(d) Propene [1]

Examiner note: The naming convention uses prefixes: meth- (1C), eth- (2C), prop- (3C), but- (4C), pent- (5C). The suffix tells you the homologous series: -ane (alkane), -ene (alkene), -ol (alcohol), -anoic acid (carboxylic acid). Practise applying general formulae until they become automatic.

• As the chain length increases, the molecules become larger / have a greater relative molecular mass [1]
• Larger molecules have stronger intermolecular forces / van der Waals forces between them [1]
• More energy is needed to overcome these forces, so the boiling point is higher [1]

Examiner note: The key word is “intermolecular forces” — not “bonds.” The covalent bonds within the molecule do not break during boiling. Larger molecules have more electrons, leading to stronger London dispersion forces. This trend applies to all homologous series.

(a) H-C-C-H with 3 H atoms on each carbon (showing all bonds) [1]

(b) H2C=CH2 showing the C=C double bond and 2 H atoms on each carbon [1]

(c) H-C-C-O-H with 3 H atoms on the first carbon and 2 H atoms on the second carbon [1]

(d) H-C-C(=O)-O-H with 3 H atoms on the first carbon, a double bond to one O and a single bond to O-H on the second carbon [1]

Examiner note: In displayed formulae, every bond must be drawn as a line. Missing a single bond (e.g., not showing a C-H bond) loses the mark. For ethene, the double bond between the two carbons must be clearly drawn as two parallel lines. For ethanoic acid, the C=O and C-O-H must both be shown.

• All members of a homologous series contain the same functional group [1]
• The functional group is the reactive part of the molecule [1]
• Since they share the same functional group, they react in the same way / undergo the same types of reactions [1]

Examiner note: Alkanes have no functional group, which is why they are relatively unreactive (they undergo combustion and substitution but not addition). Alkenes are more reactive than alkanes because the C=C double bond acts as the functional group and is the site of addition reactions.

(a) Carboxylic acids [1]

(b) Propanoic acid [1] (CH3CH2COOH — the COOH group contains one carbon, so 3 additional carbons give propanoic acid, but the molecular formula C4H8O2 corresponds to butanoic acid: CH3CH2CH2COOH)

Corrected: Butanoic acid [1] — the total formula C4H8O2 means CnH2n+1COOH where n + 1 = 4, so n = 3 giving CH3CH2CH2COOH

(c) CH3CH2CH2COOH or the full displayed formula showing all bonds [1]

Examiner note: The acid turns litmus red, confirming it is a carboxylic acid. Count all carbons including the one in COOH. C4H8O2 fits CnH2n+1COOH with n = 3 (the COOH carbon is part of the 4 total carbons). This gives butanoic acid, not propanoic acid.

Differences (any two):

• Alkanes have only single C-C bonds; alkenes have a C=C double bond [1]
• Alkanes have the general formula CnH2n+2; alkenes have CnH2n [1]
• Alkenes are unsaturated; alkanes are saturated [1]
• Alkenes decolourise bromine water; alkanes do not [1]

Similarities (any two):

• Both are hydrocarbons (contain only carbon and hydrogen) [1]
• Both undergo combustion reactions [1]
• Physical properties (boiling point, state at room temperature) show a gradual increase with chain length in both series [1]

Examiner note: The most important distinction is saturated vs unsaturated. This is the basis of the bromine water test: if bromine water is decolourised, the compound is an alkene (unsaturated). This test is frequently examined.

Definition: Isomers are compounds with the same molecular formula but different structural formulae / different arrangements of atoms [1]

Isomer 1: Butane — a straight chain of 4 carbon atoms (CH3CH2CH2CH3) [1]

Isomer 2: Methylpropane / 2-methylpropane — a chain of 3 carbon atoms with a CH3 branch on the middle carbon (CH3CH(CH3)CH3) [1]

Both must be correctly drawn with all bonds or condensed formula clearly showing the different structures [1]

Examiner note: Both isomers have the formula C4H10 but different structures. The branched isomer (methylpropane) has a lower boiling point than the straight-chain isomer (butane) because branching reduces the surface area, weakening intermolecular forces. This is a supplement-level concept but commonly appears on Paper 4.