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.

AHSEC Higher Secondary (HS) 1st Year Examination 2024Set ANNUAL1 mark
$Q. Write the IUPAC name of the following compound: main chain H3C-CH2-CH(-CH2-CH3)-C(CH3)(CH3)-CH2-CH2-CH3, where the 3rd carbon of the 7-carbon main chain carries an ethyl (-CH2-CH3) branch, and the 4th carbon carries two methyl (-CH3) branches (one drawn above, one drawn below the chain).$
›Reveal solutionSolution

The compound is a substituted heptane with an ethyl group at C3 and two methyl groups at C4; its IUPAC name is 3-ethyl-4,4-dimethylheptane.

Step 1 — Identify the longest continuous carbon chain: The main chain drawn has 7 carbons (heptane). Checking whether a longer chain exists by routing through the ethyl branch instead shows an alternative 7-carbon chain of exactly the same length, so heptane (7 carbons) is confirmed as the parent chain — the choice between the two equal-length options makes no difference here since both give the same substitution pattern.

Step 2 — Identify substituents: The 3rd carbon of the main chain carries an ethyl group (-CH2CH3). The 4th carbon (a quaternary carbon with no hydrogen) carries two methyl groups (-CH3, -CH3).

Step 3 — Number the chain to give the lowest locants: Numbering from the end nearer the substituted carbons gives locants {3, 4, 4} for the three substituents (ethyl at 3, methyl at 4, methyl at 4). Numbering from the other end would give {4, 4, 5}, which is higher at the first point of difference, so the first numbering is correct.

Step 4 — Assemble the name: List substituents alphabetically (ethyl before methyl), use 'di' for the two identical methyl groups, and cite locants for each:

3-ethyl-4,4-dimethylheptane

✓Final answer
$3-ethyl-4,4-dimethylheptane.$
AHSEC Higher Secondary (HS) 1st Year Examination 2023Set ANNUAL1 mark
$Q. Write the IUPAC name of the following compound:$
›Reveal solutionSolution

The compound CH3–CH=C(CH3)–C≡CH is 3-methylpent-3-en-1-yne.

Step 1 — Longest chain with both multiple bonds: a 5-carbon (pent) chain carrying a terminal C≡C triple bond at one end and an internal C=C double bond, with a methyl substituent on the carbon that bears the double bond.

Step 2 — Numbering: number from the triple-bond end. C1≡C2 is the triple bond; C3 carries both the methyl branch and the C3=C4 double bond; C5 is the terminal CH3. This gives unsaturation locants {1 (yne), 3 (ene)} with methyl at 3 — lower than numbering from the other end, which would give {2 (ene), 4 (yne)}.

Step 3 — Assemble: parent pent-3-en-1-yne with a 3-methyl substituent.

Name: 3-methylpent-3-en-1-yne.

[!ANSWER]

The IUPAC name is 3-methylpent-3-en-1-yne (CH3–CH=C(CH3)–C≡CH).
AHSEC Higher Secondary (HS) Final Examination 2022Set ANNUAL1 mark
$Q. Write the IUPAC name of the following compound: a benzene ring bearing the substituent -CH=CH-CH2-OH (Ph-CH=CH-CH2-OH).$
›Reveal solutionSolution

Number the 3-carbon chain from the -OH carbon; the phenyl group sits on C3 and the C=C is between C2-C3.

The compound is Ph-CH=CH-CH2-OH. The principal characteristic group is -OH (alcohol), so it gets the lowest locant and the suffix '-ol'.

Parent chain: the longest carbon chain containing the -OH carbon and the C=C — here it is a 3-carbon (propene) chain: C1(H2OH)-C2(H)=C3(H)-C6H5.

Numbering from the -OH carbon: C1 = -CH2OH, C2=C3 is the double bond, and C3 also bears the phenyl substituent.

So the parent is prop-2-en-1-ol, with a phenyl substituent at C3.

✓Final answer
$3-Phenylprop-2-en-1-ol (commonly known as cinnamyl alcohol).$
AHSEC Higher Secondary (HS) 1st Year Examination 2022Set ANNUAL1 mark
$Q. Write the IUPAC name of the following compound: OHC-CH2-CH2-COOH$
›Reveal solutionSolution

OHC-CH2-CH2-COOH is named 4-oxobutanoic acid.

Step 1 -- Identify the longest chain containing both functional groups: numbering C1(COOH)-C2(H2)-C3(H2)-C4(HO, the aldehyde carbon). This is a 4-carbon (butane) chain.

Step 2 -- Identify seniority of functional groups: in IUPAC nomenclature, the order of seniority (for choosing the principal characteristic group/suffix) is carboxylic acid > ... > aldehyde. Since both -COOH and -CHO are present, -COOH (senior) is expressed as the suffix '-oic acid', and the aldehyde (-CHO), being junior here, is expressed as the prefix 'oxo-' (because it is not a terminal position after numbering from the acid end).

Step 3 -- Number the chain starting from the carboxylic acid carbon as C1 (senior group gets lowest locant): C1=COOH, C2=CH2, C3=CH2, C4=CHO. The aldehyde (oxo group) is at position 4.

Step 4 -- Assemble the name: 4-oxobutanoic acid.

✓Final answer
$The IUPAC name of OHC-CH2-CH2-COOH is 4-oxobutanoic acid.$
AHSEC Higher Secondary (HS) 1st Year Examination 2020Set ANNUAL1 mark
$Q. Write the IUPAC name of neopentane.$
›Reveal solutionSolution
Neopentane, $(CH_3)_4C$ , has the IUPAC name 2,2-dimethylpropane.

Neopentane has the structure $(CH_3)_4C$ — a single central carbon atom bonded to four methyl ( $CH_3$ ) groups, with molecular formula $C_5H_{12}$ (an isomer of pentane).

To name it by IUPAC rules:

Identify the longest continuous carbon chain: since the central carbon is bonded to 4 separate methyl groups and no two methyls are connected to each other, the longest chain running through the central carbon is only 3 carbons long — propane (C–C–C).

The two 'extra' methyl groups are both substituents on the middle (C-2) carbon of that propane chain.

Number the chain (1,2,3) and name the substituents: two methyl groups on carbon 2 → '2,2-dimethyl' prefix.

Putting it together: 2,2-dimethylpropane.

✓Final answer
$IUPAC name of neopentane is 2,2-dimethylpropane.$
AHSEC Higher Secondary (HS) Final Examination 2018Set ANNUAL1 mark
$Q. Give the structural formula of 2-Methylpropan-2-ol.$
›Reveal solutionSolution
The structural formula of 2-methylpropan-2-ol is (CH3)3C–OH.

Name analysis: the parent chain is propan-2-ol (3 carbons, –OH on C-2); a methyl group is attached at C-2 as well. So C-2 carries the –OH plus a methyl branch, making it a tertiary alcohol.

Structure:

CH3 |

CH3–C–OH

|

CH3

i.e. (CH3)3C–OH, condensed formula C4H10O. This is commonly called tert-butyl alcohol; the carbon bearing –OH is bonded to three other carbons, so it is a 3° (tertiary) alcohol.

✓Final answer
$(CH3)3C-OH — a carbon atom bonded to three methyl groups and one hydroxyl group (tert-butyl alcohol).$

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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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