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Organic Chemistry - Functional group chemistry

Grade 11IBChemistry

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

🔑Concepts

A homologous series is a series of compounds that can be described by a general formula, such as CnH2n+2C_nH_{2n+2} for alkanes. Members show a gradation in physical properties and similar chemical properties.

Functional groups are specific atoms or groups of atoms within a molecule that determine its chemical reactivity. Examples include the hydroxyl group (OH-OH) in alcohols and the carboxyl group (COOH-COOH) in carboxylic acids.

Saturated compounds contain only single carbon-carbon bonds (e.g., CH3CH3CH_3CH_3), while unsaturated compounds contain at least one double or triple carbon-carbon bond (e.g., CH2=CH2CH_2=CH_2).

IUPAC nomenclature provides a systematic way to name organic compounds based on the longest carbon chain (stem), the primary functional group (suffix), and any side chains or substituents (prefix).

Structural isomers are molecules that share the same molecular formula but have different structural formulas, such as CH3CH2CH2OHCH_3CH_2CH_2OH (propan-1-ol) and CH3CH(OH)CH3CH_3CH(OH)CH_3 (propan-2-ol).

Primary (11^{\circ}), secondary (22^{\circ}), and tertiary (33^{\circ}) classifications describe the carbon atom attached to the functional group based on how many other carbon atoms are bonded to it.

📐Formulae

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

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

CnH2n2 (Alkynes)C_nH_{2n-2} \text{ (Alkynes)}

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

CnH2n+1CHO (Aldehydes)C_nH_{2n+1}CHO \text{ (Aldehydes)}

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

💡Examples

Problem 1:

Identify the IUPAC name and functional group for the molecule CH3COCH2CH3CH_3COCH_2CH_3.

Solution:

The IUPAC name is butan-2-one. The functional group is a ketone (carbonyl group).

Explanation:

The molecule has a four-carbon chain (but-), contains only single C-C bonds in the skeleton (-an-), and has a double-bonded oxygen on the second carbon, which is the ketone functional group (-one).

Problem 2:

Determine the molecular formula and draw the structural isomers for an alkane with n=4n=4.

Solution:

Molecular formula: C4H10C_4H_{10}. Isomers: Butane (CH3CH2CH2CH3CH_3CH_2CH_2CH_3) and 2-methylpropane (CH(CH3)3CH(CH_3)_3).

Explanation:

Using the general formula CnH2n+2C_nH_{2n+2}, for n=4n=4, we get C4H4(2)+2=C4H10C_4H_{4(2)+2} = C_4H_{10}. There are two ways to arrange four carbons in a saturated chain: a straight chain or a branched chain.

Problem 3:

Classify the alcohol CH3C(CH3)2OHCH_3C(CH_3)_2OH as primary, secondary, or tertiary.

Solution:

Tertiary (33^{\circ}) alcohol.

Explanation:

The carbon atom bonded to the hydroxyl group (OH-OH) is bonded to three other carbon atoms (CH3CH_3 groups), making it a tertiary carbon.