Consider the compounds I to VII: I. CH3—CH2—CH2—CH2—OH II. CH3—CH2—CH(OH)—CH3 III. (CH3)3C—OH IV. CH3—CH(CH3)—

NCERT Exemplar · Q25

Q. Consider the compounds I to VII: I. CH3—CH2—CH2—CH2—OH II. CH3—CH2—CH(OH)—CH3 III. (CH3)3C—OH IV. CH3—CH(CH3)—CH2—OH V. CH3—CH2—O—CH2—CH3 VI. CH3—O—CH2—CH2—CH3 VII. CH3—O—CH(CH3)2 Identify the pairs of compounds that represents position isomerism.

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Concept understanding — IUPAC Nomenclature

IUPAC Nomenclature (Organic Compounds)

IUPAC nomenclature is a systematic way to name a compound so that its name alone tells you its exact structure, with no ambiguity. Every organic name follows the same underlying recipe, whatever the functional group.

The Recipe

Identify the principal characteristic group. If the molecule has a functional group senior enough to be named as a suffix (carboxylic acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine, and so on down the seniority order), that group decides the suffix and must be included in the parent chain. A halogen is never senior enough to be a suffix — it is always named as a prefix ("halo-"), whatever else is present.
Choose the parent chain. The parent is the longest continuous carbon chain that contains the principal characteristic group (if there is one). Among chains of the same length, the one with the most substituents wins.
Number the chain. Number from whichever end gives the LOWEST LOCANT to the principal characteristic group first. If there is no principal group (e.g. a simple haloalkane, or an alkene with a halogen substituent), lowest locant goes to the site of unsaturation (double/triple bond) first, then to substituents as a set.

Watch out

When two numbering directions give the SAME locant for the principal group/unsaturation (a genuine tie), the tie-break is the lowest locant SET for the substituents as a group — compare the two sets at their first point of difference. Only if the sets are themselves tied does the alphabetically-first substituent get the lower number.
Name and cite the substituents as prefixes, in alphabetical order (ignoring multiplying prefixes like di-/tri- but not ignoring structural prefixes like iso-/cyclo-), each with its own locant.
Assemble the name: locants + substituent prefixes (alphabetical) + parent chain name + suffix (if any).

Worked Example

$CH_3-CH(Cl)-CH(CH_3)-CH_2-CH_3$ : the longest chain is 5 carbons (pentane), no principal characteristic group (just a halogen substituent), so number for the lowest locant set. From the left: Cl at C2, methyl at C3 → set {2,3}. From the right: methyl at C3, Cl at C4 → set {3,4}. {2,3} is lower, so numbering from the left wins: 2-chloro-3-methylpentane. …

Why this formula?

IUPAC Nomenclature: Why the Rules Work the Way They Do

IUPAC nomenclature is not a single formula, but a system of rules designed to give every organic compound a unique, unambiguous name. The "why" behind these rules lies in clarity, consistency, and communication — ensuring that a chemist in Tokyo and one in Toronto draw the same structure from the same name.

1. The Core Principle: The Longest Carbon Chain

Rule: Identify the longest continuous chain of carbon atoms. This becomes the parent chain (e.g., pentane, hexane).

Why?

The longest chain represents the backbone of the molecule.
It gives the most stable, fundamental name — shorter chains would be branches, not the main structure.
Example: In a molecule with 5 carbons in a row and a 2-carbon branch, calling it "pentane" (not "ethane") tells you the core skeleton is 5 carbons long.

Key idea: The parent chain is the maximum continuous path — not necessarily the one that looks "straight" on paper.

2. Numbering: Lowest Locants (The "First Point of Difference" Rule)

Rule: Number the parent chain so that substituents get the smallest possible numbers. When there's a tie, compare the first point of difference.

Why?

This ensures reproducibility — two chemists will always number the same way.
It avoids ambiguity: "2-methylpentane" is unambiguous; "3-methylpentane" would be a different compound.
The first point of difference rule: If you have substituents at positions 2,4 and 3,5, choose 2,4 because 2 < 3 (the first number is smaller).

Example:

For a methyl group on carbon 2 vs. carbon 4 of a 5-carbon chain:
- 2-methylpentane (correct)
- 4-methylpentane (wrong — higher number)

3. Alphabetical Order of Substituents

Rule: List substituents in alphabetical order (ignoring prefixes like di-, tri-, sec-, tert- but not iso-).

Why?

Alphabetical order is a universal sorting convention — no need to remember priority based on size or complexity.
It makes names searchable and predictable.
Example: "3-ethyl-2-methylpentane" (e before m) — not "2-methyl-3-ethylpentane".

Exception: Prefixes like iso- and neo- are considered part of the name (e.g., isopropyl comes before methyl because "i" < "m").

4. Multiple Bonds: The "Lowest Locant" Rule for Alkenes/Alkynes

Rule: Number the chain so that the double or triple bond gets the lowest possible number, even if it means giving a substituent a higher number.

Why?

The functional group (alkene/alkyne) is more important than alkyl substituents.
The bond position defines the compound's reactivity and geometry.
Example: In pent-2-ene (not pent-3-ene), the double bond is between carbons 2 and 3 — the lower number (2) is used.

Priority order:

Principal functional group (e.g., -OH, -COOH, C=C)
Multiple bonds
Substituents (alkyl, halo, etc.)

5. The "Suffix" and "Prefix" System …

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Position isomerism occurs when compounds have the same functional group and carbon skeleton but differ in the location of the functional group on the chain. For the given set, the pairs showing position isomerism are I & II (alcohols with –OH at C1 vs C2) and V & VI (ethers with the oxygen atom at different positions along the chain).

The key to identifying position isomers is to first ignore everything except the functional group and the carbon skeleton. If two compounds share the same molecular formula, the same functional group, and the same carbon backbone, but the functional group is attached at a different carbon, they are position isomers.

Let’s examine each compound systematically.

Identify the functional group and molecular formula for each compound.
- I: Primary alcohol, $C_4H_{10}O$ (butan-1-ol).
- II: Secondary alcohol, $C_4H_{10}O$ (butan-2-ol).
- III: Tertiary alcohol, $C_4H_{10}O$ (2-methylpropan-2-ol).
- IV: Primary alcohol, $C_4H_{10}O$ (2-methylpropan-1-ol).
- V: Ether, $C_4H_{10}O$ (diethyl ether).
- VI: Ether, $C_4H_{10}O$ (methyl propyl ether).
- VII: Ether, $C_4H_{10}O$ (methyl isopropyl ether).
All seven compounds have the same molecular formula $C_4H_{10}O$ , so they are all functional group isomers of one another (alcohols vs ethers). But within each functional group class, we look for position isomerism.
Focus on the alcohols (I, II, III, IV).
- I and II share a straight chain of four carbons. The –OH group is on carbon-1 in I and on carbon-2 in II. Same skeleton, same functional group, different position → position isomers.
- III has a branched skeleton (a tertiary carbon with three methyl groups). Its carbon skeleton is different from I and II (it’s a 3-carbon chain with a branch, not a straight 4-carbon chain). So III is a chain isomer of I and II, not a position isomer.
- IV also has a branched skeleton (a 3-carbon chain with a methyl branch at carbon-2). Again, different carbon backbone from I and II, so it’s a chain isomer.
- Between III and IV: both have branched skeletons but different branching patterns — they are chain isomers of each other, not position isomers.
Watch out

A common mistake is to think that moving the –OH group anywhere on a $C_4$ chain gives position isomers. But if the carbon skeleton itself changes (e.g., from straight to branched), the relationship becomes chain isomerism, not position isomerism. Position isomerism requires the same carbon skeleton.
Now examine the ethers (V, VI, VII).
- V: $CH_3CH_2–O–CH_2CH_3$ — the oxygen is between two ethyl groups. The carbon chain is effectively $C–C–O–C–C$ .
- VI: $CH_3–O–CH_2CH_2CH_3$ — the oxygen is between a methyl and a propyl group. The chain is $C–O–C–C–C$ .
- VII: $CH_3–O–CH(CH_3)_2$ — the oxygen is between a methyl and an isopropyl group. The chain is $C–O–C(C)C$ .
For ethers, the “functional group” is the $C–O–C$ linkage. Position isomerism in ethers means the oxygen atom is placed at a different location along the same carbon skeleton. …

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Q.Consider the compounds I to VII: I. CH3—CH2—CH2—CH2—OH II. CH3—CH2—CH(OH)—CH3 III. (CH3)3C—OH IV. CH3—CH(CH3)—CH2—OH V. CH3—CH2—O—CH2—CH3 VI. CH3—O—CH2—CH2—CH3 VII. CH3—O—CH(CH3)2 Identify the pairs of compounds that represents position isomerism.

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Q.Consider the compounds I to VII:
I. CH3—CH2—CH2—CH2—OH
II. CH3—CH2—CH(OH)—CH3
III. (CH3)3C—OH
IV. CH3—CH(CH3)—CH2—OH
V. CH3—CH2—O—CH2—CH3
VI. CH3—O—CH2—CH2—CH3
VII. CH3—O—CH(CH3)2
Identify the pairs of compounds that represents position isomerism.

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