Write the IUPAC names of the compounds i–iv from their given structures. (i) CH₃CH₂CH(OH)CH₂CH₂CH(CH₃)CH₂CH₃ (

Worked Examples · Example 8.8

Q. Write the IUPAC names of the compounds i-iv from their given structures.

(i) CH₃CH₂CH(OH)CH₂CH₂CH(CH₃)CH₂CH₃

(ii) CH₃CH₂COCH₂COCH₃

(iii) CH₃COCH₂CH₂CH₂COOH

(iv) CH≡C–CH=CH–CH=CH₂

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Concept understanding — Chain Isomerism

Chain Isomerism: The Skeleton of a Molecule

Imagine you have a set of 5 identical Lego bricks. You can snap them together in a straight line — one after the other — to make a long, thin rod. Or, you could take the same 5 bricks and make a "T" shape: a straight chain of 4 bricks, with the 5th brick attached to the middle one. Both structures use exactly the same number and type of bricks, but their shapes are different.

That difference in shape — specifically, in the carbon skeleton — is the heart of chain isomerism.

The Intuition

Carbon atoms love to bond with each other to form chains. But they don't always have to form a single, straight line. They can branch off. When two molecules have the same molecular formula (same number of each atom) but differ in how their carbon atoms are connected — one is a straight chain, another is branched — they are chain isomers.

Think of it like a family tree. The same set of people can be arranged in a direct line (grandparent → parent → child) or with branches (grandparent has two children, each with their own children). The "people" (carbon atoms) are the same, but the "connections" (bonds) form different patterns.

The Precise Statement

Chain isomerism is a type of structural isomerism where two or more compounds have the same molecular formula but differ in the arrangement of the carbon skeleton (the carbon chain). This difference arises because the carbon atoms can be linked in a straight chain or in one or more branched chains.

Important

Chain isomers always have the same molecular formula but different structural formulas due to different carbon skeletons. They are distinct compounds with different physical and chemical properties.

A Concrete Example: Pentane ( $C_5H_{12}$ )

The molecular formula $C_5H_{12}$ can represent three different chain isomers. Let's draw them:

n-Pentane (normal pentane): A straight chain of 5 carbons.

CH_3-CH_2-CH_2-CH_2-CH_3

Isopentane (2-methylbutane): A chain of 4 carbons with a 1-carbon branch (a methyl group, $-CH_3$ ) attached to the second carbon.

CH_3-CH(CH_3)-CH_2-CH_3

Neopentane (2,2-dimethylpropane): A central carbon bonded to four other carbons — a highly branched structure.

C(CH_3)_4

Isomer Name	Common Name	Structure (Condensed)	Boiling Point (°C)
Pentane	n-Pentane	$CH_3(CH_2)_3CH_3$	36.1
2-Methylbutane	Isopentane	$(CH_3)_2CHCH_2CH_3$	27.7
2,2-Dimethylpropane	Neopentane	$C(CH_3)_4$	9.5

Notice how the boiling point drops as the molecule becomes more branched. This is because branching reduces the surface area available for intermolecular forces (London dispersion forces), making the molecule easier to vaporize.

Watch out

A common mistake is to think that any difference in structure is chain isomerism. It is only about the carbon skeleton. If two molecules have the same skeleton but different positions of a double bond or a functional group, that is a different type of isomerism (position isomerism), not chain isomerism.

When Does Chain Isomerism Occur?

Chain isomerism is possible only when the carbon chain is long enough to allow branching. For alkanes:

Methane ( $CH_4$ ), Ethane ( $C_2H_6$ ), Propane ( $C_3H_8$ ): No chain isomers. There is only one way to arrange the carbons.
Butane ( $C_4H_{10}$ ): Two chain isomers (n-butane and isobutane).
Pentane ( $C_5H_{12}$ ): Three chain isomers (as shown above).
The number of possible chain isomers increases rapidly as the number of carbon atoms increases.

Why Does This Matter?

Chain isomers are different compounds. They have different:

Physical properties: Boiling point, melting point, density.
Chemical reactivity: The branching affects how easily the molecule reacts in certain reactions (e.g., combustion, halogenation).
Biological activity: In organic chemistry and biochemistry, the shape of a molecule determines how it interacts with enzymes and receptors. A branched isomer might be a drug, while its straight-chain isomer might be inactive or even toxic.

So, when you see a molecular formula like $C_6H_{14}$ , don't just think "hexane." Ask yourself: How many different ways can I arrange these six carbon atoms? That question is the beginning of understanding chain isomerism.

If you've searched "Chain Isomerism class 11 chemistry notes" or "Chain Isomerism NCERT solutions", this page covers exactly that ground — the concept is a standard part of the Class 11 Chemistry NCERT/CBSE syllabus. It also carries real weight in JEE Main, NEET and state CET Chemistry papers, where questions on chain isomerism test both conceptual understanding and calculation speed.

The key idea for naming these compounds is applying IUPAC Nomenclature principles, which involve identifying the principal functional group, determining the longest parent carbon chain, numbering the chain to give the lowest possible locants to functional groups and substituents, and then assembling the name using appropriate prefixes, suffixes, and locants.

(i) CH₃CH₂CH(OH)CH₂CH₂CH(CH₃)CH₂CH₃

The principal functional group is the alcohol ( $\text{-OH}$ ). The longest carbon chain containing this group is 8 carbons (octane).
Numbering the chain from the left gives the $\text{-OH}$ group at $\text{C3}$ and the methyl group at $\text{C6}$ , providing the lowest locant for the principal functional group.
The parent name is octan-3-ol, with a methyl substituent at $\text{C6}$ .

✓Final answer

The IUPAC name for compound (i) is 6-Methyloctan-3-ol .

(ii) CH₃CH₂COCH₂COCH₃

The compound contains two ketone ( $\text{-CO-}$ ) functional groups. The longest carbon chain containing both is 6 carbons (hexane).
Numbering the chain from the right gives the ketone groups at $\text{C2}$ and $\text{C4}$ , which provides the lowest set of locants ( $\text{2,4}$ ) compared to ( $\text{3,5}$ ) if numbered from the left.
The parent name is hexane, with two ketone groups.

✓Final answer

The IUPAC name for compound (ii) is Hexane-2,4-dione .

(iii) CH₃COCH₂CH₂CH₂COOH

The principal functional group is the carboxylic acid ( $\text{-COOH}$ ), which has higher priority than the ketone ( $\text{-CO-}$ ). The longest carbon chain containing the carboxylic acid is 6 carbons.
The carbon of the $\text{-COOH}$ group is designated as $\text{C1}$ . This places the ketone group at $\text{C5}$ .
The parent name is hexanoic acid, with an "oxo" (ketone) substituent at $\text{C5}$ .

✓Final answer

The IUPAC name for compound (iii) is 5-Oxohexanoic acid .

(iv) CH≡C–CH=CH–CH=CH₂

The compound contains both double ( $\text{=}$ ) and triple ( $\text{≡}$ ) bonds. The longest carbon chain containing all multiple bonds is 6 carbons.
Numbering from the right gives double bonds at $\text{C1}$ and $\text{C3}$ , and a triple bond at $\text{C5}$ . When the set of locants for multiple bonds is the same from both ends ( $\text{1,3,5}$ ), the double bond is given the lower number.
The parent name is hexa, with two double bonds and one triple bond.

✓Final answer

The IUPAC name for compound (iv) is Hexa-1,3-dien-5-yne .

The IUPAC names are determined by identifying the longest carbon chain containing the principal functional group, numbering it for lowest locants, and then naming substituents. The names are: (i) 6-methyloctan-3-ol, (ii) Hexane-2,4-dione, (iii) 5-oxohexanoic acid, and (iv) Hexa-1,3-dien-5-yne.

When naming organic compounds according to IUPAC rules, the core idea is to systematically describe the molecule's structure so that anyone can draw it unambiguously from its name. This involves identifying the main carbon skeleton, the primary functional group, and any substituents, then assigning locants (numbers) to indicate their positions. The "why" behind the rules is to ensure consistency and clarity across all chemical nomenclature.

Here's how we approach each compound:

(i) $\text{CH}_3\text{CH}_2\text{CH(OH)}\text{CH}_2\text{CH}_2\text{CH(CH}_3)\text{CH}_2\text{CH}_3$

Identify the principal functional group: The presence of the $-\text{OH}$ group indicates an alcohol. The suffix will be '-ol'.
Determine the longest continuous carbon chain containing the principal functional group:
- Tracing the main chain from left to right: $\text{CH}_3-\text{CH}_2-\text{CH(OH)}-\text{CH}_2-\text{CH}_2-\text{CH(CH}_3)-\text{CH}_2-\text{CH}_3$ . This chain has 8 carbons.
- This is an octane derivative.
Number the carbon chain: We must number the chain such that the principal functional group (the hydroxyl group) receives the lowest possible locant.
- If numbered from left to right: The $-\text{OH}$ group is at $\text{C}3$ . The $-\text{CH}_3$ substituent is at $\text{C}6$ . Locants: $(3, 6)$ .
- If numbered from right to left: The $-\text{OH}$ group is at $\text{C}6$ . The $-\text{CH}_3$ substituent is at $\text{C}3$ . Locants: $(6, 3)$ .
- Comparing the first point of difference, $3 < 6$ , so numbering from left to right is correct.
Identify and name substituents: There is one methyl group ( $-\text{CH}_3$ ) at $\text{C}6$ .
Assemble the name: The substituent is named first, followed by the parent chain, and then the functional group suffix with its locant.
- Substituent: 6-methyl
- Parent chain: octan-
- Functional group: 3-ol
- Combining these gives 6-methyloctan-3-ol.

(ii) $\text{CH}_3\text{CH}_2\text{COCH}_2\text{COCH}_3$

Identify the principal functional group(s): There are two ketone ( $\text{-CO-}$ ) groups. The suffix will be '-dione'.
Determine the longest continuous carbon chain containing both principal functional groups:
- The chain is $\text{CH}_3-\text{CH}_2-\text{CO}-\text{CH}_2-\text{CO}-\text{CH}_3$ . This chain has 6 carbons.
- This is a hexane derivative.
Number the carbon chain: We must number the chain such that the ketone groups receive the lowest possible locants.
- If numbered from left to right: The ketone groups are at $\text{C}3$ and $\text{C}5$ . Locants: $(3, 5)$ .
- If numbered from right to left: The ketone groups are at $\text{C}2$ and $\text{C}4$ . Locants: $(2, 4)$ .
- Comparing the first point of difference, $2 < 3$ , so numbering from right to left is correct.
Identify and name substituents: There are no substituents on the main chain.
Assemble the name: The parent chain name is followed by the locants for the ketone groups and the '-dione' suffix.
- Parent chain: hexane-
- Functional groups: 2,4-dione
- Combining these gives Hexane-2,4-dione.

(iii) $\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_2\text{COOH}$

Identify the principal functional group(s): There is a carboxylic acid ( $-\text{COOH}$ $- COOH$ ) group and a ketone ( $-\text{CO-}$ $- CO-$ ) group.
- According to IUPAC priority rules, carboxylic acid has higher priority than ketone. Therefore, $-\text{COOH}$ is the principal functional group, and the suffix will be '-oic acid'.
- The ketone group will be treated as a substituent, using the prefix 'oxo-'.
Determine the longest continuous carbon chain containing the principal functional group:
- The chain is $\text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{COOH}$ . This chain has 6 carbons.
- This is a hexanoic acid derivative.
Number the carbon chain: The carbon of the carboxylic acid group is always assigned $\text{C}1$ $C 1$ .
- $\text{C}1$ : $-\text{COOH}$
- $\text{C}2$ : $\text{CH}_2$
- $\text{C}3$ : $\text{CH}_2$
- $\text{C}4$ : $\text{CH}_2$
- $\text{C}5$ : $-\text{CO}-$ (ketone)
- $\text{C}6$ : $\text{CH}_3$
Identify and name substituents: There is an oxo group (from the ketone) at $\text{C}5$ .
Assemble the name: The substituent prefix is named first, followed by the parent chain and the functional group suffix.
- Substituent: 5-oxo
- Parent chain: hexanoic acid
- Combining these gives 5-oxohexanoic acid.

(iv) $\text{CH}\equiv\text{C–CH=CH–CH=CH}_2$

Identify the principal functional group(s): There are both triple bonds (alkyne) and double bonds (alkene).
- When both are present, the compound is named as an 'enyne'. The chain must include both types of multiple bonds.
Determine the longest continuous carbon chain containing all multiple bonds:
- The chain is $\text{CH}\equiv\text{C–CH=CH–CH=CH}_2$ . This chain has 6 carbons.
- This is a hexene-yne derivative.
Number the carbon chain: The chain is numbered to give the multiple bonds the lowest possible set of locants. If there is a choice (i.e., the locant set is identical from both ends), the double bond is given priority for the lower number.
- If numbered from left to right: Triple bond at $\text{C}1$ , double bonds at $\text{C}3$ and $\text{C}5$ . Locants: $(1, 3, 5)$ .
- If numbered from right to left: Double bond at $\text{C}1$ , double bond at $\text{C}3$ , triple bond at $\text{C}5$ . Locants: $(1, 3, 5)$ .
- Since both numbering directions give the same set of locants $(1, 3, 5)$ , we apply the tie-breaking rule: the double bond gets the lower number. Therefore, numbering from right to left is correct.
Identify and name substituents: There are no substituents.
Assemble the name: The parent chain name is followed by the locants for the double bonds, then 'diene', then the locant for the triple bond, and finally 'yne'. The 'a' in 'hexa' is retained before 'diene'.
- Parent chain: hexa-
- Double bonds: 1,3-diene
- Triple bond: 5-yne
- Combining these gives Hexa-1,3-dien-5-yne.

✓Final answer

The IUPAC names for the given compounds are: (i) 6-methyloctan-3-ol, (ii) Hexane-2,4-dione, (iii) 5-oxohexanoic acid, and (iv) Hexa-1,3-dien-5-yne .

←Previous section · 8.5.3Nomenclature of Organic Compounds having Functional Group(s)→Next question · 8.5.3Example 8.9 — Derive the structure of (i) 2-Chlorohexane, (ii) Pent-4-en-2-ol, (iii)…

Q.Write the IUPAC names of the compounds i–iv from their given structures.

(i) CH3CH2CH(OH)CH2CH2CH(CH3)CH2CH3\text{CH}_3\text{CH}_2\text{CH(OH)}\text{CH}_2\text{CH}_2\text{CH(CH}_3)\text{CH}_2\text{CH}_3CH3​CH2​CH(OH)CH2​CH2​CH(CH3​)CH2​CH3​

(ii) CH3CH2COCH2COCH3\text{CH}_3\text{CH}_2\text{COCH}_2\text{COCH}_3CH3​CH2​COCH2​COCH3​

(iii) CH3COCH2CH2CH2COOH\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_2\text{COOH}CH3​COCH2​CH2​CH2​COOH

(iv) CH≡C–CH=CH–CH=CH2\text{CH}\equiv\text{C–CH=CH–CH=CH}_2CH≡C–CH=CH–CH=CH2​

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(i) $\text{CH}_3\text{CH}_2\text{CH(OH)}\text{CH}_2\text{CH}_2\text{CH(CH}_3)\text{CH}_2\text{CH}_3$

(ii) $\text{CH}_3\text{CH}_2\text{COCH}_2\text{COCH}_3$

(iii) $\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_2\text{COOH}$

(iv) $\text{CH}\equiv\text{C–CH=CH–CH=CH}_2$

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