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Organic Chemistry - Nomenclature and homologous series

Grade 12IGCSEChemistry

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

🔑Concepts

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

Members of a homologous series show a gradation in physical properties. For example, as the number of carbon atoms increases, the boiling point increases due to stronger intermolecular forces.

Each successive member of a homologous series differs by a CH2CH_2 unit and has a relative molecular mass that increases by 1414.

IUPAC Nomenclature follows a specific hierarchy: Stem (number of carbon atoms in the longest chain), Suffix (primary functional group), and Prefix (substituents like halogens or alkyl groups).

Common stems for carbon chains: 11 = meth-, 22 = eth-, 33 = prop-, 44 = but-, 55 = pent-, 66 = hex-.

Functional Groups define the chemical reactivity: Alkanes (single bonds), Alkenes (C=CC=C double bonds), Alcohols (OH-OH), Carboxylic Acids (COOH-COOH), and Esters (COO-COO-).

Structural Isomers are molecules with the same molecular formula but different structural arrangements, such as chain isomers, position isomers, or functional group isomers.

📐Formulae

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

CnH2n (Alkenes)C_nH_{2n} \text{ (Alkenes)}

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

CnH2n+1COOH (Carboxylic Acids)C_nH_{2n+1}COOH \text{ (Carboxylic Acids)}

RCOOR (Esters)R-COO-R' \text{ (Esters)}

💡Examples

Problem 1:

Provide the IUPAC name for the compound with the structural formula CH3CH2CH2OHCH_3CH_2CH_2OH.

Solution:

Propan-1-ol

Explanation:

The longest carbon chain contains 33 carbons, giving the stem 'prop-'. All carbon-carbon bonds are single, indicating 'an'. The functional group is a hydroxyl group (OH-OH) attached to the first carbon, providing the suffix '-1-ol'.

Problem 2:

Identify the molecular formula of the fourth member of the alkene homologous series.

Solution:

C5H10C_5H_{10}

Explanation:

The alkene series starts with Ethene (n=2n=2). The second member is Propene (n=3n=3), the third is Butene (n=4n=4), and the fourth is Pentene (n=5n=5). Using the general formula CnH2nC_nH_{2n}, for n=5n=5, we get C5H10C_5H_{10}.

Problem 3:

Draw and name the ester formed by the reaction of ethanol (C2H5OHC_2H_5OH) and propanoic acid (C2H5COOHC_2H_5COOH).

Solution:

Ethyl propanoate (CH3CH2COOCH2CH3CH_3CH_2COOCH_2CH_3)

Explanation:

In esterification, the alcohol provides the alkyl prefix (ethanol \rightarrow ethyl) and the carboxylic acid provides the alkanoate suffix (propanoic acid \rightarrow propanoate).

Nomenclature and homologous series Revision - Grade 12 Chemistry IGCSE