The IUPAC name for the compound shown (a benzene ring bearing –Cl at C-1, –NO2 on the carbon adjacent to –Cl,

NCERT Exemplar · Q3

Q. The IUPAC name for the compound shown (a benzene ring bearing -Cl at C-1, -NO2 on the carbon adjacent to -Cl, and -CH3 para to -Cl) is ______.

(i) 1-Chloro-2-nitro-4-methylbenzene

(ii) 1-Chloro-4-methyl-2-nitrobenzene

(iii) 2-Chloro-1-nitro-5-methylbenzene

(iv) m-Nitro-p-chlorotoluene

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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.

Tip

A locant TIE (both directions give the same first-point-of-difference number) is common on short/symmetric chains — always check both directions explicitly rather than assuming "number from the end nearer the first substituent mentioned in the name" is automatically correct.

Common Mistakes

Picking a chain that is NOT the longest one just because it "looks simpler" — always verify no longer chain exists, including chains that run through what looks like a branch.
Forgetting the alphabetical-order rule for citing substituents (locants are chosen by the lowest-locant rule; the ORDER they're written in the name is alphabetical, not by locant).
Treating a halogen as if it could ever be the principal characteristic group / suffix — it cannot; it is always a prefix, however many are present.

IUPAC nomenclature is a foundational skill taught in the NCERT/CBSE Class 11 Chemistry chapter on Organic Chemistry: Some Basic Principles and Techniques, and ‘IUPAC nomenclature rules and examples’ is one of the most searched important-question topics for board exams, JEE Main and NEET. Naming organic compounds correctly underpins almost every other organic-chemistry question asked in competitive exams.

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

Rule: The principal functional group determines the suffix (e.g., -ol for alcohol, -al for aldehyde). Other groups become prefixes (e.g., chloro-, hydroxy-).

Why?

The suffix tells you the most important chemical feature at a glance.
Prefixes are secondary — they modify the parent name without changing its core identity.
Example: "3-chloropropan-1-ol" — the "-ol" tells you it's an alcohol; "chloro-" is just a substituent.

6. Why "E/Z" and "R/S" Exist

Rule: Use E/Z for alkene geometry (based on Cahn-Ingold-Prelog priority) and R/S for chiral centers.

Why?

Simple cis/trans fails when there are more than two different substituents.
E/Z and R/S are unambiguous — they assign priority based on atomic number, not just "same side" or "opposite side".
This prevents confusion: (E)-3-methylpent-2-ene is a specific isomer; "cis" would be ambiguous here.

Summary: The "Why" in One Table

Rule	Purpose
Longest chain	Defines the core skeleton
Lowest locants	Ensures unique numbering
Alphabetical order	Universal sorting
Functional group priority	Highlights reactivity
E/Z, R/S	Handles stereochemistry

Final thought: IUPAC nomenclature is a language, not a formula. Every rule exists to eliminate ambiguity — so that a name is a perfect blueprint for a molecule.

Number the ring so the alphabetically-first substituent (chloro) is C-1 and the locant set is lowest: Cl at 1, NO $_2$ at 2, CH $_3$ at 4. Citing the substituents alphabetically gives 1-chloro-4-methyl-2-nitrobenzene — option (ii).

For a polysubstituted benzene, IUPAC nomenclature rests on two rules: cite the substituents alphabetically in the name, and number the ring to give the lowest set of locants.

Assign the locants. The ring carries chloro, methyl, and nitro. Fix the geometry from the description: chloro on one carbon, nitro on the carbon adjacent to it, methyl para to the chloro. Placing chloro at C-1, the nitro (adjacent) is at C-2 and the methyl (para) is at C-4:

\text{Cl at C-1}, \quad \text{NO}_2 \text{ at C-2}, \quad \text{CH}_3 \text{ at C-4}

The locant set is $\{1, 2, 4\}$ . Numbering in the opposite direction would put the nitro at C-6, giving $\{1, 4, 6\}$ , which is higher — so $\{1, 2, 4\}$ is correct.

Write the name. List the substituents in alphabetical order (chloro, methyl, nitro), each with its locant, and append the parent "benzene": 1-chloro-4-methyl-2-nitrobenzene.

Watch out

Do not order the substituents by their position numbers (that would give "1-chloro-2-nitro-4-methyl"). IUPAC orders substituents alphabetically by name, independent of their locants.

Checking the options:

(i) 1-Chloro-2-nitro-4-methylbenzene — correct locants but the substituents are not in alphabetical order.
(ii) 1-Chloro-4-methyl-2-nitrobenzene — correct locants and alphabetical order. ✓
(iii) 2-Chloro-1-nitro-5-methylbenzene — numbers from the wrong atom; chloro (alphabetically first) should take the lowest locant.
(iv) m-Nitro-p-chlorotoluene — uses common/relative names, not a systematic IUPAC name.

✓Final answer

The correct IUPAC name is 1-chloro-4-methyl-2-nitrobenzene, option (ii) .

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)

C3=C4: the double bond (pent-3-en)

C4 also bears a methyl substituent (4-methyl)

C5: terminal $CH_3$

This gives 4-methylpent-3-en-2-one.

✓Final answer
$The correct option is (D) .$
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 carrying two methyl substituents.

Name: 1-bromo-2,2-dimethylpropane (common name neopentyl bromide).

✓Final answer
$The correct option is (C) .$
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, methyl = 5.

Name: 3-ethyl-5-methylheptane.

✓Final answer
$The correct option is (D) . Heptane chain with 3-ethyl and 5-methyl substituents.$
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 = $CH_3$ .

The double bond is between C3 and C4 → pent-3-en.

A methyl substituent sits on C4 → 4-methyl.

Combining: 4-methylpent-3-en-2-one.

✓Final answer
$The correct option is (D) .$
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.

Common Mistakes

Numbering the methyl at position 2 instead of 3 (start numbering from the point of attachment, the CH2-O end).

✓Final answer
$The correct option is (A) — 3-Methylbutoxybenzene.$

ANSWER: A
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:

C1 = $CH_3$ , C2 = $C=O$ (ketone), ..., C7 = $CH_2OH$ .

The ketone is at C-2 (suffix 'heptan-2-one') and the hydroxyl at C-7 (prefix '7-hydroxy'). Name: 7-hydroxyheptan-2-one.

✓Final answer
$The correct option is (A) .$
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 bond 'en' at 2, and the amine at 1, giving the IUPAC name prop-2-en-1-amine.

✓Final answer
$The correct option is (C) .$
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.

Naming: The IUPAC name for the 20-carbon straight-chain alkane is eicosane (from the Greek eikosi = twenty). The other names offered (didodecane, didecane, etc.) are not valid IUPAC alkane names.

✓Final answer
$The correct option is (E) .$
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 ethers, not the aromatic ethyl phenyl ether named phenetole.

✓Final answer
$The correct option is (A) .$
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:

From the CH $_3$ end: C1 = CH $_3$ , C2 = C=O, ..., C7 = CH $_2$ OH → ketone at 2, OH at 7.

From the OH end: ketone would be at 6 — higher.

Lowest locant for the principal group wins, so the ketone is at position 2 and OH at position 7:
$\textbf{7-hydroxyheptan-2-one}$
(Names like "2-oxo-heptan-7-ol" are wrong because they treat the lower-priority alcohol as the principal group.)

[!ANSWER]

(B) 7-hydroxyheptan-2-one
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.

Common Mistakes

Confusing resorcinol with catechol (1,2) or hydroquinone (1,4).

✓Final answer
$The correct option is (A) — Benzene-1,3-diol.$

ANSWER: A
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.

Common Mistakes

Selecting a branched C5 aldehyde (isovaleraldehyde) instead of the straight-chain pentanal.

✓Final answer
$The correct option is (A) — CH3CH2CH2CH2CHO.$

ANSWER: A

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Q.The IUPAC name for the compound shown (a benzene ring bearing –Cl at C-1, –NO2 on the carbon adjacent to –Cl, and –CH3 para to –Cl) is ______.

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