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Carbon and its Compounds - Homologous series

Grade 10CBSE

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

🔑Concepts

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A homologous series is a group of organic compounds having the same functional group and similar chemical properties, in which the successive members differ by a −CH2−-CH_2- unit.

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The difference in the molecular mass between any two successive members of a homologous series is 14 u14 \text{ u}.

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All members of a specific homologous series can be represented by the same general formula, such as CnH2n+2C_nH_{2n+2} for alkanes.

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As the molecular mass increases in a series, the physical properties like melting point and boiling point show a gradual increase (gradation) due to the increase in van der Waals forces.

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The chemical properties of members in a homologous series remain similar because they possess the same functional group.

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The first member of the Alkyne series is Ethyne (C2H2C_2H_2), and the first member of the Alkene series is Ethene (C2H4C_2H_4); n=1n=1 is not possible for these series as a double or triple bond requires at least two carbon atoms.

📐Formulae

CnH2n+2 (General formula for Alkanes)C_nH_{2n+2} \text{ (General formula for Alkanes)}

CnH2n (General formula for Alkenes)C_nH_{2n} \text{ (General formula for Alkenes)}

CnH2n−2 (General formula for Alkynes)C_nH_{2n-2} \text{ (General formula for Alkynes)}

CnH2n+1OH (General formula for Alcohols)C_nH_{2n+1}OH \text{ (General formula for Alcohols)}

Mass of −CH2− unit=12+(2×1)=14 u\text{Mass of } -CH_2- \text{ unit} = 12 + (2 \times 1) = 14 \text{ u}

💡Examples

Problem 1:

Determine the molecular formula and molecular mass of the third member of the Alkyne series.

Solution:

C4H6C_4H_6 and mass is 54 u54 \text{ u}.

Explanation:

The general formula for alkynes is CnH2n−2C_nH_{2n-2}. The series starts from n=2n=2 (Ethyne). Thus, the third member corresponds to n=4n=4. Formula: C4H(2×4)−2=C4H6C_4H_{(2 \times 4) - 2} = C_4H_6. Molecular mass: (4×12)+(6×1)=48+6=54 u(4 \times 12) + (6 \times 1) = 48 + 6 = 54 \text{ u}.

Problem 2:

Identify which of the following belong to the same homologous series: C2H6C_2H_6, C3H6C_3H_6, C4H10C_4H_{10}, C2H4C_2H_4.

Solution:

Pair 1: C2H6C_2H_6 and C4H10C_4H_{10} (Alkanes); Pair 2: C3H6C_3H_6 and C2H4C_2H_4 (Alkenes).

Explanation:

C2H6C_2H_6 and C4H10C_4H_{10} follow the general formula CnH2n+2C_nH_{2n+2} (Alkanes). C3H6C_3H_6 and C2H4C_2H_4 follow the general formula CnH2nC_nH_{2n} (Alkenes).

Problem 3:

What is the difference in the number of carbon and hydrogen atoms between Butanol (C4H9OHC_4H_9OH) and Pentanol (C5H11OHC_5H_{11}OH)?

Solution:

One Carbon atom and two Hydrogen atoms (−CH2−-CH_2- group).

Explanation:

Subtracting the atoms: (C5−C4)=C1(C_5 - C_4) = C_1 and (H11−H9)=H2(H_{11} - H_9) = H_2. This confirms that successive members of the alcohol homologous series differ by a −CH2−-CH_2- unit.

Homologous series - Revision Notes & Key Formulas | CBSE Class 10 Science