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Organic Chemistry: Basic Principles and Techniques - IUPAC Nomenclature

Grade 11ICSEChemistry

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

🔑Concepts

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IUPAC Nomenclature structure: SecondaryPrefix+PrimaryPrefix+WordRoot+PrimarySuffix+SecondarySuffixSecondary Prefix + Primary Prefix + Word Root + Primary Suffix + Secondary Suffix.

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Word Root: Denotes the number of carbon atoms in the longest continuous chain (e.g., C1C_1: Meth, C2C_2: Eth, C3C_3: Prop, C4C_4: But).

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Primary Suffix: Indicates saturation or unsaturation: −ane-ane for alkanes, −ene-ene for C=CC=C, and −yne-yne for C≡CC\equiv C.

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Secondary Suffix: Indicates the principal functional group (e.g., −ol-ol for Alcohols, −al-al for Aldehydes, −oic acid-oic \ acid for Carboxylic acids).

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Longest Chain Rule: Select the longest continuous carbon chain containing the principal functional group and maximum number of multiple bonds.

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Lowest Locant Rule: Number the parent chain such that the principal functional group gets the lowest possible locant, followed by multiple bonds and then substituents.

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Alphabetical Order: When multiple substituents are present, they are listed in alphabetical order (ignoring prefixes like di−di-, tri−tri-).

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Priority of Functional Groups: −COOH>−SO3H>−COOR>−COCl>−CONH2>−CN>−CHO>>C=O>−OH>−NH2>>C=C<>−C≡C−>−R,−X,−NO2-COOH > -SO_3H > -COOR > -COCl > -CONH_2 > -CN > -CHO > >C=O > -OH > -NH_2 > >C=C< > -C\equiv C- > -R, -X, -NO_2.

📐Formulae

CnH2n+2 (Alkanes)C_n H_{2n+2} \text{ (Alkanes)}

CnH2n (Alkenes)C_n H_{2n} \text{ (Alkenes)}

CnH2n−2 (Alkynes)C_n H_{2n-2} \text{ (Alkynes)}

R−OH (General Alcohol formula where R=CnH2n+1)R-OH \text{ (General Alcohol formula where } R = C_n H_{2n+1} \text{)}

R−COOH (General Carboxylic acid formula)R-COOH \text{ (General Carboxylic acid formula)}

💡Examples

Problem 1:

Give the IUPAC name for CH3−CH(CH3)−CH2−CH2−OHCH_3-CH(CH_3)-CH_2-CH_2-OH.

Solution:

3-Methylbutan-1-ol

Explanation:

The longest chain has 4 carbons (Butan-). The functional group −OH-OH is at C1C_1, giving it priority in numbering. A methyl group is attached to C3C_3.

Problem 2:

Write the IUPAC name of CH3−CO−CH2−CH2−COOHCH_3-CO-CH_2-CH_2-COOH.

Solution:

4-Oxopentanoic acid

Explanation:

The molecule contains both a ketone (>C=O>C=O) and a carboxylic acid (−COOH-COOH). Since −COOH-COOH has higher priority, it forms the secondary suffix (−oic acid-oic \ acid) and its carbon is C1C_1. The ketone group at C4C_4 is treated as a prefix (oxo−oxo-).

Problem 3:

Name the following compound: CH2=CH−CH2−C≡CHCH_2=CH-CH_2-C\equiv CH.

Solution:

Pent-1-en-4-yne

Explanation:

The chain has 5 carbons (Pent). Numbering starts from the end that gives lower locants to the multiple bonds. In this case, numbering from the left gives locants 1 and 4. Since the double bond and triple bond are at terminal positions, the double bond (−ene-ene) is given numerical priority over the triple bond (−yne-yne).

IUPAC Nomenclature Class 11 Notes & Examples