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Organic Chemistry - Homologous series

Grade 11IGCSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A homologous series is a family of organic compounds that share the same general formula and have similar chemical properties.

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Members of a homologous series possess the same functional group, such as the hydroxyl group (−OH-OH) in alcohols or the carboxyl group (−COOH-COOH) in carboxylic acids.

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There is a gradual change in physical properties as the number of carbon atoms increases; for example, the boiling point increases due to stronger intermolecular forces (Van der Waals forces).

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Successive members of a homologous series differ by a −CH2−-CH_2- unit, which corresponds to a relative molecular mass (MrM_r) difference of 1414.

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All members of a series can usually be prepared by similar chemical methods.

📐Formulae

Alkanes: CnH2n+2\text{Alkanes: } C_nH_{2n+2}

Alkenes: CnH2n\text{Alkenes: } C_nH_{2n}

Alcohols: CnH2n+1OH\text{Alcohols: } C_nH_{2n+1}OH

Carboxylic Acids: CnH2n+1COOH (where n starts from 0)\text{Carboxylic Acids: } C_nH_{2n+1}COOH \text{ (where } n \text{ starts from 0)}

💡Examples

Problem 1:

Determine the molecular formula of the fourth member of the alkane homologous series and name it.

Solution:

C4H10C_4H_{10}, Butane

Explanation:

For alkanes, the general formula is CnH2n+2C_nH_{2n+2}. For the fourth member, n=4n=4. Substituting nn into the formula: C4H(2×4)+2=C4H10C_4H_{(2 \times 4) + 2} = C_4H_{10}. The prefix for 4 carbons is 'but-', and the suffix for alkanes is '-ane'.

Problem 2:

An unknown compound belongs to the alkene series and has a relative molecular mass (MrM_r) of 4242. Determine its molecular formula. (Given: C=12,H=1C=12, H=1)

Solution:

C3H6C_3H_6

Explanation:

The general formula for alkenes is CnH2nC_nH_{2n}. The relative molecular mass is calculated as (n×12)+(2n×1)=42(n \times 12) + (2n \times 1) = 42. This simplifies to 12n+2n=4212n + 2n = 42, which is 14n=4214n = 42. Solving for nn gives n=3n = 3. Therefore, the formula is C3H6C_3H_6 (Propene).

Problem 3:

Identify the functional group and the general formula for the compound Ethanoic acid (CH3COOHCH_3COOH).

Solution:

Functional group: Carboxyl group (−COOH-COOH); General formula: CnH2n+1COOHC_nH_{2n+1}COOH

Explanation:

Ethanoic acid is part of the carboxylic acid homologous series. It contains 22 carbon atoms in total. In the general formula CnH2n+1COOHC_nH_{2n+1}COOH, n=1n=1 gives C1H(2×1)+1COOHC_1H_{(2 \times 1) + 1}COOH, which results in CH3COOHCH_3COOH.

Homologous series - Revision Notes & Key Formulas | IGCSE Grade 11 Chemistry