Write the IUPAC name of each of the following two compounds. Compound A is the open-chain diol CH3-CH(CH3)-CH(

NCERT Exemplar · Q18

Q. Write the IUPAC name of each of the following two compounds. Compound A is the open-chain diol CH3-CH(CH3)-CH(OH)-CH(C2H5)-CH(OH)-CH3, i.e. a six-carbon main chain bearing a methyl branch on the second carbon, an ethyl branch on the fourth carbon, and hydroxyl groups on the third and fifth carbons. Compound B is a cyclohexane ring carrying a -NO2 group on one carbon and an -OCH3 (methoxy) group on the carbon that is meta to it (a 1,3 arrangement).

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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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For A, the six-carbon chain is the parent and the two -OH groups are the principal characteristic group (suffix -diol), so numbering gives OH the lowest locants (2 and 4); the branches are an ethyl at C-3 and a methyl at C-5, giving 3-ethyl-5-methylhexane-2,4-diol. For B, both -NO2 (nitro) and -OCH3 (methoxy) are prefix substituents on cyclohexane placed 1,3; alphabetical priority gives methoxy the lower number, so it is 1-methoxy-3-nitrocyclohexane.

Compound A

Longest chain through both hydroxyl-bearing carbons: six carbons -> hexane; two -OH -> hexane-diol.
Number to give the -OH groups the lowest locants. From the right-hand end the OH groups fall on C-2 and C-4 (locant set {2,4}), versus {3,5} from the left; choose {2,4}.
With that numbering the ethyl branch is on C-3 and the methyl branch on C-5.
Assemble alphabetically (ethyl before methyl): 3-ethyl-5-methylhexane-2,4-diol.

Compound B

Parent ring: cyclohexane (saturated six-membered ring). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2026Set eng-2026-04174 marksMCQ
$Q. The IUPAC name of mesityl oxide is (A) 2-Methylpent-2-en-3-one (B) 3-Methylpent-2-en-4-one (C) 4-Methylpent-2-en-3-one (D) 4-Methylpent-3-en-2-one (E) 2-Methylpent-3-en-4-one$
›Reveal solutionSolution
The structure $(CH_3)_2C{=}CH{-}CO{-}CH_3$ names as 4-methylpent-3-en-2-one.

Mesityl oxide has the structure $(CH_3)_2C{=}CH{-}CO{-}CH_3$ .

The longest chain containing the carbonyl is five carbons (pent-). Numbering to give the ketone the lowest locant, start from the methyl next to the C=O:

C1: $CH_3$

C2: $C{=}O$ (ketone → -2-one) …
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KEAM 2026Set eng-2026-04194 marksMCQ
$_3$
›Reveal solutionSolution

The five-carbon skeleton is propane with two methyls on C-2 and Br on C-1: 1-bromo-2,2-dimethylpropane.

Structure. $(CH_3)_3C\text{-}CH_2Br$ = a central carbon bearing three methyls and a $CH_2Br$ . The longest chain is propane (3 C); numbering to give Br the lowest locant puts $CH_2Br$ as C-1, the quaternary carbon as C-2 car …
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KEAM 2026Set pha-2026-0418F4 marksMCQ
$CH_3-CH_2-CH(C_2H_5)-CH_2-CH(CH_3)-CH_2-CH_3$
›Reveal solutionSolution

Longest chain = 7 C (heptane), ethyl at C3, methyl at C5.

The structure $CH_3-CH_2-CH(C_2H_5)-CH_2-CH(CH_3)-CH_2-CH_3$ has a 7-carbon parent chain. Numbering to give the lowest locants (tie {3,5} both ways) gives the lower number to the first-cited substituent alphabetically (ethyl before methyl), so ethyl = 3, methy …
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KEAM 2026Set pha-2026-0420F4 marksMCQ
$Q. The IUPAC name of the following compound is (A) 2-Methylpent-2-en-2-one (B) 3-Methylpent-2-en-2-one (C) 4-Methylpent-2-en-3-one (D) 4-Methylpent-3-en-2-one (E) 1,1-Dimethylbuten-2-one$
›Reveal solutionSolution
Numbering from the carbonyl end (mesityl oxide) gives 4-methylpent-3-en-2-one.

The structure is $\mathrm{CH_3\!-\!CO\!-\!CH\!=\!C(CH_3)\!-\!CH_3}$ (mesityl oxide). Choosing the longest chain containing the carbonyl (the principal group) and numbering to give the ketone the lowest locant:

C1 = $CH_3$ , C2 = $C\!=\!O$ (the 2-one), C3 = $CH$ , C4 = $C$ , C5 = $CH_3$ . …
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KEAM 2025Set eng-2025-04234 marksMCQ
$Q. The IUPAC name of phenyl isopentyl ether is (A) 3-Methtylbutoxybenzene (B) 2-Methylbutoxybenzene (C) 2-Methylphenoxybutane (D) 4-Methylbutoxybenzene (E) 1-Methylbutoxybenzene$
›Reveal solutionSolution
Phenyl isopentyl ether is named 3-methylbutoxybenzene.

Concept and Intuition

Ethers are named as (alkoxy)benzene when one group is phenyl. Isopentyl (isoamyl) is the 3-methylbutyl group, (CH3)2CH-CH2-CH2-. Attaching it via oxygen to benzene gives 3-methylbutoxybenzene.

Step-by-Step Solution

Isopentyl = isoamyl = 3-methylbutyl = (CH3)2CHCH2CH2-.

As an -O- substituent it becomes 3-methylbutoxy.

On benzene → 3-methylbutoxybenzene. …
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KEAM 2025Set eng-2025-04254 marksMCQ
$HOCH_2(CH_2)_3CH_2COCH_3$
›Reveal solutionSolution

The molecule is a seven-carbon chain bearing a ketone and an alcohol. The ketone (higher priority) gets the suffix '-one' with the lowest locant, and $-OH$ becomes the 'hydroxy' prefix: 7-hydroxyheptan-2-one.

Expanding $HOCH_2(CH_2)_3CH_2COCH_3$ gives a continuous chain of 7 carbons:
$HO-CH_2-CH_2-CH_2-CH_2-CH_2-CO-CH_3$
Priority: the ketone (C=O) outranks the alcohol, so it defines the suffix and gets the lower locant. Numbering from the methyl-ketone end: …
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KEAM 2025Set eng-2025-04264 marksMCQ
$Q. The IUPAC name of allylamine is (A) But-2-en-1-amine (B) But-1-en-2-amine (C) Prop-2-en-1-amine (D) Prop-1-en-2-amine (E) 2-Amino 1-propene$
›Reveal solutionSolution

Allylamine is a 3-carbon chain with a C=C at position 2 and $–NH_2$ at C1: prop-2-en-1-amine.

Allylamine is $CH_2=CH-CH_2-NH_2$ . Numbering to give the amine the lowest locant: C1 bears the $–NH_2$ , and the double bond starts at C2. The three-carbon parent is 'prop', the double …
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KEAM 2025Set pha-2025-0424F4 marksMCQ
$C_{20}H_{42}$
›Reveal solutionSolution

$C_{20}H_{42}$ fits the alkane formula $C_nH_{2n+2}$ with $n=20$ ; the straight-chain $C_{20}$ alkane is named eicosane.

Derivation: Alkanes obey $C_nH_{2n+2}$ . Setting $2n+2 = 42$ gives $n = 20$ , so the molecule is a 20-carbon alkane. …
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KEAM 2025Set pha-2025-0424A4 marksMCQ
$Q. Phenetole is (A) Ethoxybenzene (B) Methoxyethane (C) Methoxybenzene (D) 1-Methoxypropane (E) 2-Methoxypropane$
›Reveal solutionSolution

Phenetole $=$ ethyl phenyl ether $= C_6H_5OC_2H_5 =$ ethoxybenzene.

By analogy, anisole is methoxybenzene ( $C_6H_5OCH_3$ ); phenetole is its ethyl homologue, ethoxybenzene. Methoxyethane and 1-/2-methoxypropane are aliphatic eth …
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KEAM 2024Set eng-2024-06074 marksMCQ
$_2$
›Reveal solutionSolution
[!TLDR]

The compound is a 7-carbon ketone with a terminal OH; naming the ketone as the senior group gives 7-hydroxyheptan-2-one.

Concept

When a molecule contains more than one functional group, the principal characteristic group (chosen by IUPAC seniority) takes the suffix and the lowest locant; others become prefixes. Ketones rank above alcohols in this order — a standard NCERT/CBSE nomenclature rule.

Solution

Expand the structure:
$\text{HO-CH}_2\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-CO-CH}_3$
Counting carbons gives a chain of 7 (heptane skeleton). The functional groups are a ketone (C=O) and a hydroxyl (-OH).

Since a ketone is senior to an alcohol, the suffix is -one and the OH becomes a hydroxy prefix. Number the chain to give the ketone the lowest locant: …
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KEAM 2022Set eng-2022-P1-A14 marksMCQ
$Q. Resorcinol is (A) Benzene-1, 3-diol (B) Benzene-1, 4-diol (C) Benzene-1, 2-diol (D) 3-Methylphenol (E) 4-Methylphenol$
›Reveal solutionSolution
Resorcinol is benzene-1,3-diol.

Concept and Intuition

Resorcinol is a common dihydroxybenzene isomer; the three isomers are catechol (1,2), resorcinol (1,3) and hydroquinone (1,4).

Step-by-Step Solution

Resorcinol has two -OH groups on a benzene ring.

They occupy the meta (1,3) positions.

Therefore resorcinol = benzene-1,3-diol. …
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KEAM 2022Set eng-2022-P1-A14 marksMCQ
$CH_3CH_2CH_2CH_2CHO$
›Reveal solutionSolution
Valeraldehyde is pentanal, CH3CH2CH2CH2CHO.

Concept and Intuition

The common name valeraldehyde denotes the straight-chain five-carbon aldehyde, pentanal.

Step-by-Step Solution

Valer- corresponds to a five-carbon (valeric acid, pentanoic acid) chain.

The -aldehyde suffix places -CHO at the chain end.

Straight-chain pentanal = CH3CH2CH2CH2CHO. …
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Concept understanding — IUPAC Nomenclature

IUPAC Nomenclature (Organic Compounds)

The Recipe

Worked Example

Why this formula?

IUPAC Nomenclature: Why the Rules Work the Way They Do

1. The Core Principle: The Longest Carbon Chain

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

3. Alphabetical Order of Substituents

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

5. The "Suffix" and "Prefix" System …

Compound A

Compound B

Method: Principal-Group-First Nomenclature Method for Polyols and Multi-Substituted Rings

Core Concept

Steps

Concept and Intuition

Step-by-Step Solution

Concept

Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Compound A

Compound B

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