Write structures of the compounds whose IUPAC names are as follows: (i) 2-Methylbutan-2-ol (ii) 1-Phenylpropan

Exercises · 7.2

Q. Write structures of the compounds whose IUPAC names are as follows:

(i) 2-Methylbutan-2-ol

(ii) 1-Phenylpropan-2-ol

(iii) 3,5-Dimethylhexane-1,3,5-triol

(iv) 2,3-Diethylphenol

(v) 1-Ethoxypropane

(vi) 2-Ethoxy-3-methylpentane

(vii) Cyclohexylmethanol

(viii) 3-Cyclohexylpentan-3-ol

(ix) Cyclopent-3-en-1-ol

(x) 4-Chloro-3-ethylbutan-1-ol.

Karnataka PUCTextbookSubjective· 3mImportance★★★★★

16% · 21/135 Questions

✓ Free question

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.

Concept: IUPAC Nomenclature — Writing structures from names

Reasoning steps for each:

Identify the parent chain (longest carbon chain containing the principal functional group).
Number the chain to give the functional group the lowest possible locant.
Attach substituents (alkyl, halo, phenyl, cycloalkyl, etc.) at the indicated positions.
Add multiple bonds or rings as specified, and ensure correct stereochemistry (if any) — here all are achiral or racemic.

(i) 2-Methylbutan-2-ol

Parent: butane (4 C). –OH at C2, methyl at C2.

Structure:

    CH3
    |
CH3-C-CH2-CH3
    |
    OH

(ii) 1-Phenylpropan-2-ol

Parent: propane (3 C). –OH at C2, phenyl (–C₆H₅) at C1.

Structure:

C6H5-CH2-CH-CH3
          |
          OH

(iii) 3,5-Dimethylhexane-1,3,5-triol

Parent: hexane (6 C). –OH at C1, C3, C5; methyl at C3 and C5.

Structure:

       CH3   CH3
       |     |
HO-CH2-C-CH2-C-CH3
       |     |
       OH    OH

(iv) 2,3-Diethylphenol

Parent: phenol (benzene ring with –OH at C1). Ethyl groups at C2 and C3.

Structure: a benzene ring with $-\text{OH}$ at C1 and $-\text{C}_2\text{H}_5$ at C2 and C3.

(v) 1-Ethoxypropane

Parent: propane (3 C). –O–CH₂CH₃ (ethoxy) at C1.

Structure:

CH3-CH2-CH2-O-CH2-CH3

(vi) 2-Ethoxy-3-methylpentane

Parent: pentane (5 C). Ethoxy (–OCH₂CH₃) at C2, methyl at C3.

Structure:

CH3-CH(OCH2CH3)-CH(CH3)-CH2-CH3

C2 bears the ethoxy group, C3 bears the methyl group.

(vii) Cyclohexylmethanol

Parent: methanol (1 C). Cyclohexyl ring attached to the carbon bearing –OH.

Structure: a cyclohexane ring with a $-\text{CH}_2\text{OH}$ substituent.

(viii) 3-Cyclohexylpentan-3-ol

Parent: pentane (5 C). –OH at C3, cyclohexyl at C3.

Structure: $CH_3-CH_2-C(C_6H_{11})(OH)-CH_2-CH_3$ — a pentan-3-ol backbone with a cyclohexyl ring also on C3.

(ix) Cyclopent-3-en-1-ol

Parent: cyclopentene (5 C ring, double bond between C3 and C4). –OH at C1.

Structure: a cyclopentene ring with $-\text{OH}$ at C1 and a double bond between C3 and C4.

(x) 4-Chloro-3-ethylbutan-1-ol

Parent: butane (4 C). –OH at C1, chloro at C4, ethyl at C3.

Structure:

Cl-CH2-CH-CH2-CH2-OH
        |
       CH2CH3

✓Final answer

The structures are drawn above for each IUPAC name.

The key to writing structures from IUPAC names is to identify the parent chain (the longest carbon chain containing the principal functional group), number it to give the functional group the lowest locant, then attach all substituents at the correct positions. Each name is decoded step by step below, and the final structures are given in the answer block.

The Concept: Decoding IUPAC Names

IUPAC nomenclature is a systematic way to name organic compounds. To go from a name to a structure, you work backwards:

Identify the parent chain — the longest continuous carbon chain that contains the principal functional group (the one with highest priority, e.g., -ol for alcohols, -oxy for ethers).
Identify the suffix — tells you the functional group (e.g., -ol = alcohol, -oxy = ether, -ene = double bond).
Identify the locants — the numbers before the suffix and before substituent names tell you where things are attached.
Identify substituents — groups like methyl, ethyl, phenyl, chloro, etc., with their locants.
Assemble — draw the parent chain, number it correctly, attach substituents at the given positions, and add the functional group.

Let’s work through each compound.

(i) 2-Methylbutan-2-ol

Parent chain: “butan” means a 4-carbon chain. The suffix “-2-ol” means an alcohol (-OH) at carbon 2.
Substituent: “2-methyl” means a methyl group ( $-CH_3$ ) at carbon 2.
Numbering: The parent chain is numbered so that the -OH gets the lowest number — here it’s already at position 2.
Structure: Butane backbone: C1–C2–C3–C4. At C2, we have both an -OH and a methyl group. That means C2 is a quaternary carbon (bonded to four carbons: C1, C3, the methyl, and the -OH).

So the structure is:

C1: $CH_3-$
C2: $-C(CH_3)(OH)-$
C3: $-CH_2-$
C4: $-CH_3$

In condensed form: $CH_3-C(CH_3)(OH)-CH_2-CH_3$

Tip

The “-2-ol” suffix tells you the -OH is on carbon 2. The “2-methyl” tells you a methyl is also on carbon 2. So carbon 2 has four different groups attached — that’s a tertiary alcohol.

(ii) 1-Phenylpropan-2-ol

Parent chain: “propan” = 3-carbon chain. “-2-ol” = -OH at carbon 2.
Substituent: “1-phenyl” means a phenyl group ( $-C_6H_5$ ) at carbon 1.
Structure: Propane backbone: C1–C2–C3. At C1, attach a phenyl ring. At C2, attach -OH.

So: $C_6H_5-CH_2-CH(OH)-CH_3$

Watch out

A common mistake is to put the phenyl on carbon 2. The locant “1-” explicitly places it on carbon 1. Always check the number before the substituent name.

(iii) 3,5-Dimethylhexane-1,3,5-triol

Parent chain: “hexane” = 6-carbon chain. “-triol” means three -OH groups.
Locants for -OH: “1,3,5-triol” means -OH at carbons 1, 3, and 5.
Substituents: “3,5-dimethyl” means methyl groups at carbons 3 and 5.
Structure: Hexane backbone: C1–C2–C3–C4–C5–C6.
- C1: $-OH$ (so $HO-CH_2-$ )
- C2: $-CH_2-$
- C3: has both an -OH and a methyl group (so $-C(CH_3)(OH)-$ )
- C4: $-CH_2-$
- C5: has both an -OH and a methyl group (so $-C(CH_3)(OH)-$ )
- C6: $-CH_3$

In condensed form: $HO-CH_2-CH_2-C(CH_3)(OH)-CH_2-C(CH_3)(OH)-CH_3$

Note

Carbons 3 and 5 each have two substituents (an -OH and a methyl), making them tertiary alcohol carbons. Carbon 1 is a primary alcohol.

(iv) 2,3-Diethylphenol

Parent: “phenol” means a benzene ring with an -OH group directly attached. The -OH is always at position 1 by definition.
Substituents: “2,3-diethyl” means two ethyl groups ( $-CH_2CH_3$ ) at positions 2 and 3 on the ring.
Numbering: The ring is numbered starting from the carbon bearing the -OH as position 1, then going around so that the substituents get the lowest numbers — here 2 and 3 are already the smallest possible.

Structure: A benzene ring with -OH at C1, ethyl at C2, and ethyl at C3.

In skeletal form:

    OH
     |
    1
   / \
  2   6
  |   |
  Et  5
  3---4
  |
  Et

(Where Et = $-CH_2CH_3$ )

Tip

In phenols, the -OH is always at position 1, so you never write “1-phenol”. The locants for substituents are given relative to that fixed position.

(v) 1-Ethoxypropane

Parent: This is an ether. The name “1-ethoxypropane” means the parent chain is propane (3 carbons), and there’s an ethoxy group ( $-O-CH_2CH_3$ ) attached at carbon 1.
Structure: Propane backbone: C1–C2–C3. At C1, instead of a hydrogen, we have $-O-CH_2CH_3$ .

So: $CH_3CH_2CH_2-O-CH_2CH_3$

Note

Ethers are named as alkoxyalkanes. The smaller alkyl group (here ethyl) becomes the “alkoxy” part, and the larger chain (propane) is the parent.

(vi) 2-Ethoxy-3-methylpentane

Parent chain: “pentane” = 5-carbon chain.
Substituents:
- “2-ethoxy” means an ethoxy group ( $-O-CH_2CH_3$ ) at carbon 2.
- “3-methyl” means a methyl group at carbon 3.
Structure: Pentane backbone: C1–C2–C3–C4–C5.
- C1: $-CH_3$
- C2: $-CH(OCH_2CH_3)-$ (the ethoxy is attached via oxygen)
- C3: $-CH(CH_3)-$
- C4: $-CH_2-$
- C5: $-CH_3$

In condensed form: $CH_3-CH(OCH_2CH_3)-CH(CH_3)-CH_2-CH_3$

Watch out

The ethoxy group is attached through oxygen, not directly as an ethyl. So at C2, the carbon is bonded to H, the ethoxy oxygen, and the rest of the chain.

(vii) Cyclohexylmethanol

Parent: “methanol” means a one-carbon chain with an -OH group — that’s $CH_3OH$ , but here the methyl is substituted.
Substituent: “cyclohexyl” means a cyclohexane ring attached to the methanol carbon.
Structure: The methanol carbon (C1) has -OH and is bonded to the cyclohexane ring.

So: $C_6H_{11}-CH_2OH$ (where $C_6H_{11}$ is a cyclohexyl group).

Tip

“Cyclohexylmethanol” is the same as “(hydroxymethyl)cyclohexane”. The -OH is on a carbon attached to the ring, not directly on the ring.

(viii) 3-Cyclohexylpentan-3-ol

Parent chain: “pentan” = 5-carbon chain. “-3-ol” = -OH at carbon 3.
Substituent: “3-cyclohexyl” means a cyclohexyl group at carbon 3.
Structure: Pentane backbone: C1–C2–C3–C4–C5. At C3, we have both an -OH and a cyclohexyl group. That makes C3 a tertiary carbon (bonded to C2, C4, the cyclohexyl, and the -OH).

So: $CH_3-CH_2-C(C_6H_{11})(OH)-CH_2-CH_3$

Note

Carbon 3 is a tertiary alcohol carbon because it’s attached to three other carbons (two from the pentane chain and one from the cyclohexyl ring).

(ix) Cyclopent-3-en-1-ol

Parent: “cyclopent” = 5-carbon ring. “-3-en” means a double bond between carbons 3 and 4. “-1-ol” means an -OH at carbon 1.
Numbering: The ring is numbered to give the -OH the lowest number (1), then the double bond gets the lowest possible locant (3).
Structure: A 5-membered ring with a double bond between C3 and C4, and an -OH at C1.

In skeletal form:

    OH
     |
     1
   /   \
  2     5
  |     |
  3=====4

(Where the double bond is between C3 and C4)

Watch out

The double bond locant “3” means the bond is between carbon 3 and carbon 4. The -OH is at carbon 1. Make sure the numbering goes in the direction that gives the -OH the lowest number.

(x) 4-Chloro-3-ethylbutan-1-ol

Parent chain: “butan” = 4-carbon chain. “-1-ol” = -OH at carbon 1.
Substituents:
- “4-chloro” means a chlorine atom at carbon 4.
- “3-ethyl” means an ethyl group at carbon 3.
Structure: Butane backbone: C1–C2–C3–C4.
- C1: $-OH$ (so $HO-CH_2-$ )
- C2: $-CH_2-$
- C3: $-CH(CH_2CH_3)-$ (ethyl group attached)
- C4: $-CH_2Cl$ (chlorine attached)

In condensed form: $HO-CH_2-CH_2-CH(CH_2CH_3)-CH_2Cl$

Tip

The ethyl group at C3 makes that carbon a chiral center (four different groups: H, C2, C4, and the ethyl). This compound can exist as enantiomers, but the name doesn’t specify stereochemistry.

✓Final answer

CH_3-C(CH_3)(OH)-CH_2-CH_3

Method: Retrosynthetic Name-to-Structure Decoding

This method works by systematically breaking down the IUPAC name into its structural components, then assembling the molecule from the parent chain outward.

Steps:

Identify the parent chain (the longest carbon chain containing the principal functional group)
Number the parent chain to give the principal functional group the lowest possible locant
Identify and place substituents (alkyl groups, halogens, etc.) at the numbered positions
Add the principal functional group (alcohol -OH, ether -O-, etc.)
Check for stereochemistry or special features (rings, multiple bonds, aromatic rings)

(i) 2-Methylbutan-2-ol

Parent: butane (4-carbon chain)

Functional group: -ol (alcohol) at C-2

Substituent: methyl at C-2

Structure:

IUPAC: 2-methylbutan-2-ol

(ii) 1-Phenylpropan-2-ol

Parent: propane (3-carbon chain)

Functional group: -ol at C-2

Substituent: phenyl (C₆H₅–) at C-1

Structure:

IUPAC: 1-phenylpropan-2-ol

(iii) 3,5-Dimethylhexane-1,3,5-triol

Parent: hexane (6-carbon chain)

Functional group: triol (three -OH groups) at C-1, C-3, C-5

Substituents: methyl at C-3 and C-5

Structure:

IUPAC: 3,5-dimethylhexane-1,3,5-triol

(iv) 2,3-Diethylphenol

Parent: phenol (benzene ring with -OH)

Functional group: -OH at C-1 (by default)

Substituents: ethyl at C-2 and C-3

Structure:

IUPAC: 2,3-diethylphenol

(v) 1-Ethoxypropane

Parent: propane (3-carbon chain)

Functional group: ether (-O-) with ethoxy (C₂H₅O–) at C-1

Structure:

IUPAC: 1-ethoxypropane

(vi) 2-Ethoxy-3-methylpentane

Parent: pentane (5-carbon chain)

Functional group: ethoxy at C-2

Substituent: methyl at C-3

Structure:

IUPAC: 2-ethoxy-3-methylpentane

(vii) Cyclohexylmethanol

Parent: methanol (CH₃OH)

Substituent: cyclohexyl ring attached to the carbon of methanol

Structure:

IUPAC: cyclohexylmethanol

(viii) 3-Cyclohexylpentan-3-ol

Parent: pentane (5-carbon chain)

Functional group: -ol at C-3

Substituent: cyclohexyl at C-3

Structure:

IUPAC: 3-cyclohexylpentan-3-ol

(ix) Cyclopent-3-en-1-ol

Parent: cyclopentene (5-carbon ring with one double bond)

Functional group: -ol at C-1

Double bond: between C-3 and C-4

Structure:

IUPAC: cyclopent-3-en-1-ol

(x) 4-Chloro-3-ethylbutan-1-ol

Parent: butane (4-carbon chain)

Functional group: -ol at C-1

Substituents: chloro at C-4, ethyl at C-3

Structure:

IUPAC: 4-chloro-3-ethylbutan-1-ol

Key Exam Tips:

Always number the parent chain to give the principal functional group the lowest number
Alcohols (-ol) take priority over alkyl substituents and halogens
Ethers are named as alkoxyalkanes (smaller alkyl group as alkoxy)
Phenols retain the name "phenol" with substituent positions relative to -OH
Cyclic compounds use "cyclo" prefix with appropriate numbering

Common Mistakes in IUPAC Nomenclature (Alcohols & Ethers)

Students often lose marks in this topic due to small but critical errors. Here are the most frequent mistakes and how to avoid them.

1. Misidentifying the Parent Chain

Mistake: Choosing the wrong longest carbon chain, especially when multiple chains of equal length exist.

Example from (iii): 3,5-Dimethylhexane-1,3,5-triol

Wrong: Selecting a chain that includes all three –OH groups but is not the longest continuous chain.
Correct: The parent is hexane (6 carbons), not pentane or heptane.

How to avoid:

Always count the longest continuous carbon chain that includes the principal functional group (here, –OH).
For alcohols, the –OH carbon must be in the parent chain.

2. Incorrect Numbering (Lowest Locant Rule)

Mistake: Not giving the principal functional group the lowest possible number.

Example from (i): 2-Methylbutan-2-ol

Wrong: Numbering from the end that gives the methyl group position 1.
Correct: Number from the end that gives the –OH carbon the lowest number (position 2).

How to avoid:

For alcohols, the –OH group gets the lowest locant before any substituents.
Number the chain so that the carbon bearing –OH has the smallest possible number.

3. Forgetting to Show the –OH Group in the Structure

Mistake: Drawing only the carbon skeleton and forgetting to attach the –OH group.

Example from (v): 1-Ethoxypropane

Wrong: Drawing CH₃–CH₂–CH₂–O–CH₂–CH₃ but missing the –OH (this is an ether, not an alcohol).
Correct: For alcohols, explicitly show –OH attached to the correct carbon.

How to avoid:

Always check the suffix: “-ol” means alcohol → must have –OH.
For ethers (“-oxy”), the oxygen is between two alkyl groups.

4. Confusing Ethers with Alcohols

Mistake: Treating an ether as an alcohol or vice versa.

Example from (vi): 2-Ethoxy-3-methylpentane

Wrong: Drawing CH₃–CH₂–CH(CH₃)–CH₂–CH₂–OH (alcohol structure).
Correct: This is an ether: CH₃–CH₂–O–CH(CH₃)–CH₂–CH₂–CH₃.

How to avoid:

“-oxy” means ether (R–O–R’).
“-ol” means alcohol (R–OH).
Read the suffix carefully before drawing.

5. Incorrect Placement of Substituents

Mistake: Placing substituents on wrong carbon atoms.

Example from (iv): 2,3-Diethylphenol

Wrong: Putting ethyl groups on carbons 1 and 2 of the benzene ring.
Correct: The –OH is on carbon 1; ethyl groups are on carbons 2 and 3.

How to avoid:

Number the benzene ring starting from the carbon attached to –OH (carbon 1).
Place substituents at the positions given in the name.

6. Ignoring Stereochemistry (when applicable)

Mistake: Not showing cis/trans or E/Z when the name implies it.

Example from (ix): Cyclopent-3-en-1-ol

The double bond is between C3 and C4.
Common error: Drawing the double bond but not showing the correct geometry (cis/trans).
Correct: For cyclopent-3-en-1-ol, the double bond is cis (both hydrogens on same side) due to ring constraints.

How to avoid:

For cyclic alkenes, the double bond is always cis unless specified otherwise.
If the name includes “(E)” or “(Z)”, show the correct geometry.

7. Wrong Number of Carbons in the Parent Chain

Mistake: Miscounting carbons when drawing branched chains.

Example from (viii): 3-Cyclohexylpentan-3-ol

Wrong: Drawing a pentane chain with only 4 carbons.
Correct: Pentane = 5 carbons; cyclohexyl is a substituent on carbon 3.

How to avoid:

Write the parent chain first (e.g., pentane = 5 carbons).
Then attach substituents at the correct positions.

8. Forgetting to Number the –OH in Cyclic Compounds

Mistake: Not indicating which carbon bears the –OH in cyclic alcohols.

Example from (vii): Cyclohexylmethanol

Wrong: Drawing cyclohexane with –OH directly attached to the ring.
Correct: The –OH is on a side chain (CH₂OH) attached to the ring.

How to avoid:

“Cyclohexylmethanol” means the –OH is on a methyl group attached to cyclohexane.
For cyclic alcohols like cyclopent-3-en-1-ol, the number indicates the –OH position.

9. Incorrect Order of Substituents in the Name

Mistake: Writing substituents in wrong alphabetical order.

Example from (x): 4-Chloro-3-ethylbutan-1-ol

Wrong: Writing “3-ethyl-4-chloro...” (ethyl before chloro).
Correct: Alphabetical order: chloro before ethyl.

How to avoid:

In IUPAC, substituents are listed alphabetically (ignoring prefixes like di-, tri-).
“Chloro” (c) comes before “ethyl” (e).

10. Drawing the Wrong Functional Group for Phenols

Mistake: Treating phenol as a regular alcohol.

Example from (iv): 2,3-Diethylphenol

Wrong: Drawing a hexane ring with –OH and ethyl groups.
Correct: Phenol has a benzene ring with –OH directly attached.

How to avoid:

“Phenol” means –OH is attached to a benzene ring.
The ring is numbered starting from the –OH carbon.

Quick Checklist to Avoid Mistakes

Step	Action
1	Identify the parent chain (longest chain with principal functional group)
2	Number to give the functional group the lowest locant
3	Draw the carbon skeleton first
4	Attach the functional group (–OH for alcohols)
5	Add substituents at correct positions
6	Check alphabetical order in the name
7	For cyclic compounds, verify ring size and substituent positions
8	For ethers, remember R–O–R’ structure

Final tip: Always draw the structure step-by-step — parent chain → functional group → substituents. Double-check the suffix (-ol, -oxy, -phenol) before finalizing.

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

Showing the 12 most recent of 15 on this concept.

COMEDK 2026Set 2026-A1 markMCQ
$\text { Identify the correct structure of o-ethyl anisole. }$
›Reveal solutionSolution
The name “o-ethyl anisole” tells us the substituents are an ethyl group and a methoxy group in the ortho (1,2) position. The correct structure is therefore the one with –OCH₃ and –C₂H₅ on adjacent carbons — option (D).

The key is to decode the name “o-ethyl anisole” systematically. Anisole is the common name for methoxybenzene (C₆H₅–OCH₃). The prefix “o-” (ortho) indicates that the ethyl group is attached to the ring carbon adjacent to the methoxy group. So the molecule is 1-methoxy-2-ethylbenzene.

Let’s check each option:

Option (A) has –OC₂H₅ (ethoxy) and –C₂H₅. That would be o-ethyl phenetole, not anisole.

Option (B) has –COCH₃ (acetyl) and –C₂H₅. That’s an ortho-substituted acetophenone, not anisole.

Option (C) has –OH (hydroxyl) and –C₂H₅. That’s o-ethyl phenol, not anisole.

Option (D) has –OCH₃ (methoxy) and –C₂H₅. That matches exactly: anisole (methoxybenzene) with an ethyl group in the ortho position.

Watch out

A common mistake is confusing “ethoxy” (–OC₂H₅) with “methoxy” (–OCH₃). Anisole specifically means the methoxy group, not any other alkoxy group.

Tip

Remember: “anisole” = methoxybenzene. If you see “phenetole” it’s ethoxybenzene. The prefix tells you the alkoxy chain length.

✓Final answer
$The correct option is (D) . ANSWER: D$
COMEDK 2026Set 2026-M1 markMCQ
$\text { Which one of the following is not the correct IUPAC name of the compound? }$
›Reveal solutionSolution
[!TLDR]

Options A, B and C are each valid IUPAC names for the structures shown, so the name that is NOT correct is option (D).

Concept

An IUPAC name is correct only when the parent chain is the longest chain that carries the principal characteristic group with the lowest set of locants, and every substituent is numbered correctly (CBSE/NCERT nomenclature rules).

Solution

(A) $(\mathrm{CH_3})_2\mathrm{C{=}C(Br){-}CH_2OH}$ : the chain is but-2-en-1-ol (OH at C1, double bond C2=C3) with Br on C2 and a methyl on C3 $\Rightarrow$ 2-bromo-3-methylbut-2-en-1-ol. Correct.

(B) $(\mathrm{CH_3})_2\mathrm{CH{-}CH_2{-}CH(OH){-}CH(CH_2OH){-}CH_3}$ : the longest chain passing through both -OH groups is six carbons, with OH at 1 and 3 and methyls at 2 and 5 $\Rightarrow$ 2,5-dimethylhexane-1,3-diol. Correct.

(C) $\mathrm{C_6H_5{-}O{-}CH_2CH_2CH(CH_3)_2}$ : the alkyl group $-\mathrm{CH_2CH_2CH(CH_3)_2}$ is 3-methylbutyl, so the ether is 3-methylbutoxybenzene. Correct.

Since the names in A, B and C are all correct, the name that is not correct is the one in option (D).

[!ANSWER]

(D) option (D) is the incorrect IUPAC name
KCET 2026Set D31 markMCQ
$CH_3\text{-}C(CH_3)_2\text{-}O\text{-}C_2H_5$
›Reveal solutionSolution

Ethers are named in the IUPAC substitutive system as an alkane parent chain carrying an "alkoxy" substituent, not by the older functional-class "alkyl alkyl ether" style.

Step 1 — Identify the skeleton

The compound is $(CH_3)_3C-O-C_2H_5$ : a central carbon bearing three methyl groups (the tert-butyl-type carbon) connected through an oxygen to an ethyl group.

Step 2 — Choose the parent chain

Stripping away the ethoxy (-O- $C_2H_5$ ) substituent, the remaining carbon skeleton is $CH_3-C(CH_3)_2-CH_3$ , a three-carbon chain (propane) with a methyl branch at C2. This gives the base name 2-methylpropane, and C2 also carries the ethoxy group.

Step 3 — Name the ether as an alkoxy-substituted alkane

The $-O-C_2H_5$ group becomes the substituent prefix ethoxy, attached at C2. Combining the substituents alphabetically on the 2-methylpropane parent gives 2-ethoxy-2-methylpropane.

Step 4 — Why the other options fail

"Tertiary butoxy ethane" and "Ethoxy tertiary butane" use non-IUPAC common-name fragments ("tert-butoxy", "tertiary butane"), and "1,1-Dimethyl-1-ethoxyethane" numbers the chain incorrectly (C1 cannot simultaneously be a chain terminus and carry two methyl substituents in a valid parent-chain sense here).

✓Final answer
$The correct option is (C) — 2-ethoxy-2-methylpropane.$
KCET 2025Set D-41 markMCQ
$\mathrm{CH_3-CH_2-CH(CH_3)-CH_2-CH_3}$
›Reveal solutionSolution

Isomers must share the molecular formula $\mathrm{C_6H_{14}}$ ; the cyclopentane ring in option (C) is $\mathrm{C_5H_{10}}$ , so it is not an isomer of hexane.

Step 1 — What must be true of an isomer of hexane.

Hexane is $\mathrm{C_6H_{14}}$ (a saturated, acyclic alkane, $\mathrm{C_nH_{2n+2}}$ with $n=6$ ). Any isomer of hexane must have exactly the same molecular formula, $\mathrm{C_6H_{14}}$ : six carbons, fourteen hydrogens, zero degrees of unsaturation.

Step 2 — Count the atoms in each option.

(A) $\mathrm{CH_3-CH_2-CH(CH_3)-CH_2-CH_3}$ — 3-methylpentane.

Carbons: $1+1+1+1+1$ in the pentane chain $= 5$ , plus the $\mathrm{CH_3}$ branch $= 6$ . Saturated and acyclic $\Rightarrow \mathrm{C_6H_{14}}$ . Is an isomer.

(B) $\mathrm{CH_3-CH_2-CH_2-CH_2-CH_2-CH_3}$ — n-hexane itself.

Six carbons in a straight chain $\Rightarrow \mathrm{C_6H_{14}}$ . It is hexane (the identity structure), so it certainly is not the odd one out.

(D) $\mathrm{CH_3-CH(CH_3)-CH_2-CH_2-CH_3}$ — 2-methylpentane.

Five-carbon chain + one methyl branch $= 6$ carbons; saturated, acyclic $\Rightarrow \mathrm{C_6H_{14}}$ . Is an isomer.

(C) the drawn five-membered ring — cyclopentane.

Five carbons, each bearing two hydrogens:
$\mathrm{C_5H_{10}} \quad (\mathrm{C_nH_{2n}},\; n=5)$
This differs from $\mathrm{C_6H_{14}}$ on both counts — one carbon fewer, and it has one degree of unsaturation (the ring), so four hydrogens fewer than a $\mathrm{C_5}$ alkane would need to match. It is not an isomer of hexane.

Step 3 — Sanity check with degrees of unsaturation.
$\text{DoU} = \frac{2(5)+2-10}{2} = \frac{12-10}{2} = 1$
One degree of unsaturation (the ring) — whereas hexane and all its isomers must have DoU $= 0$ . Confirms (C).

✓Final answer
$\mathrm{C_5H_{10}}$

ANSWER: C

KCET 2025Set D-41 markMCQ

Q. The organic compound can be classified as ______________ (A) Allylic halide (B) Benzyl halide (C) Aryl halide (D) Alkyl halide

›Reveal solution

Classify by which kind of carbon bears the halogen: here the C–Cl carbon is $sp^3$ and directly attached to an aromatic ring — the definition of a benzylic halide.

Step 1 — Read the structure.

From the figure: a benzene ring is bonded to a carbon that also carries two $\mathrm{CH_3}$ groups and one $\mathrm{Cl}$ . So the compound is

\mathrm{C_6H_5{-}\overset{\mathrm{CH_3}}{\underset{\mathrm{CH_3}}{C}}{-}Cl} \;\equiv\; \text{2-chloro-2-phenylpropane}

The carbon bearing the Cl is $sp^3$ (four σ bonds: to the ring, to two methyls, to Cl) and it is directly attached to the aromatic ring.

Step 2 — The classification rules for halides.

Halides are classified by the hybridisation and environment of the carbon holding the halogen:

Class	Where the halogen sits
Alkyl halide	on an $sp^3$ carbon of a plain alkyl chain (no ring or double bond adjacent)
Allylic halide	on an $sp^3$ carbon adjacent to a C=C double bond
Benzylic halide	on an $sp^3$ carbon directly attached to an aromatic ring
Aryl halide	on an $sp^2$ carbon of the ring itself (e.g. chlorobenzene)

Step 3 — Apply the rules.

Is it aryl? No — the Cl is not on a ring carbon. In chlorobenzene the C–Cl carbon is one of the six aromatic $sp^2$ carbons. Here the C–Cl carbon is an extra, $sp^3$ carbon hanging off the ring. Rejected.
Is it allylic? No — there is no isolated C=C next to the C–Cl carbon; the unsaturation is an aromatic ring, which makes it benzylic, not allylic. Rejected.
Is it a plain alkyl halide? It is $sp^3$ , but the aryl attachment is the defining structural feature and the more specific class always wins. Rejected.
Is it benzylic? The Cl-bearing $sp^3$ carbon is bonded straight to the benzene ring — yes. Specifically it is a tertiary benzylic halide.

Step 4 — Why the distinction matters chemically.

Benzylic halides ionise readily because the resulting benzylic carbocation is stabilised by resonance delocalisation into the ring, so they are exceptionally reactive by $\mathrm{S_N1}$ . Aryl halides do the exact opposite — the C–Cl bond has partial double-bond character and is inert to ordinary nucleophilic substitution. Getting the class right therefore predicts the reactivity.

✓Final answer

The correct option is (B) — Benzyl halide.

ANSWER: B

KCET 2025Set D-41 markMCQ
$\mathrm{CH_3-\underset{\underset{\textstyle CH_3}{|}}{\overset{\overset{\textstyle CH_3}{|}}{C}}-OCH_3} + HI \longrightarrow A + B$
›Reveal solutionSolution

A tertiary alkyl ether cleaved by HI proceeds via SN1: the bond that breaks is the one that forms the more stable (tertiary) carbocation, sending iodide to the tert-butyl side and leaving methanol as the other fragment.

Step 1 — Protonation.

The ether oxygen is protonated by HI, making it a good leaving group.

Step 2 — C–O bond cleavage (SN1, tertiary substrate).

With a tertiary carbon on one side, the bond breaks heterolytically to form the stable tertiary carbocation $(CH_3)_3C^+$ , releasing methanol ( $CH_3OH$ ) as the neutral leaving fragment.

Step 3 — Nucleophilic attack.

Iodide attacks the tertiary carbocation directly: $(CH_3)_3C-I$ .

✓Final answer
$CH_3OH$
COMEDK 2025Set 2025-A1 markMCQ
$Q. Which of the following is the correct name according to IUPAC rules? (A) (B) (C) (D)$
›Reveal solutionSolution
The key idea is to apply IUPAC nomenclature rules for alkenes, alkynes, alcohols, and ethers. Only option (A) correctly names its structure as 3‑bromoprop‑1‑ene; the others contain errors in numbering, suffix order, or functional group priority.

Concept and Intuition

IUPAC naming follows a strict hierarchy: the principal functional group determines the suffix, the longest carbon chain containing that group is chosen, and numbering gives the lowest locants to the principal group and then to multiple bonds. For compounds with multiple bonds, the suffix for the highest‑priority bond (alkene > alkyne) comes last, and locants are placed before the suffix. For alcohols, the –OH group takes precedence over multiple bonds, and the chain must include the carbon bearing the –OH. For ethers, the smaller alkyl group is named as an alkoxy substituent on the larger alkane. Each option must be checked against these rules.

Step‑by‑Step Analysis

Option (A): Br–CH₂–CH=CH₂ named “3‑bromoprop‑1‑ene”

The longest chain is three carbons (prop‑) with a double bond between C1 and C2.

The bromine is a substituent on C3.

Numbering starts from the end nearer the double bond, so the double bond gets locant 1.

The name “3‑bromoprop‑1‑ene” correctly places the bromine locant before the parent, and the double‑bond locant before “‑ene”.

Verdict: Correct.

Option (B): HC≡C–CH=CH₂ named “3‑butene‑1‑yne”

The longest chain is four carbons (but‑). It contains both a triple bond and a double bond.

IUPAC rule: when both alkene and alkyne are present, the suffix “‑ene” comes before “‑yne” (alphabetical order of suffixes), and numbering gives the lowest locants to the multiple bonds as a set.

Numbering from the left: triple bond at C1, double bond at C3 → locants 1 and 3.

Numbering from the right: double bond at C1, triple bond at C3 → locants 1 and 3 (same).

The correct name is but‑1‑en‑3‑yne (or 1‑buten‑3‑yne), not “3‑butene‑1‑yne”. The locant for the double bond should be before “‑ene”, and the triple bond locant before “‑yne”. The given name reverses the order and misplaces locants.

Verdict: Incorrect.

Option (C): H₃C–CH(OH)–CH₃ named “propan‑1‑ol”

The structure is propan‑2‑ol (the –OH is on the middle carbon).

The longest chain is three carbons, and the –OH group must get the lowest locant. Numbering from either end gives the –OH at C2.

The correct name is propan‑2‑ol (or isopropyl alcohol). “Propan‑1‑ol” would have the –OH on a terminal carbon, which is not the case.

Verdict: Incorrect.

Option (D): H₃C–O–CH₂–CH₃ named “ethoxymethane”

This is an ether. IUPAC names ethers as alkoxyalkanes: the smaller alkyl group (methyl) becomes the alkoxy substituent (methoxy‑), and the larger alkyl group (ethyl) is the parent alkane (ethane).

The correct name is methoxyethane, not “ethoxymethane”. (The older common name “ethyl methyl ether” is also acceptable, but IUPAC prefers the alkoxyalkane form with the smaller group as the prefix.)

Verdict: Incorrect.

Watch out

A common mistake is to think that “ethoxymethane” is acceptable because it swaps the order of the alkyl groups. IUPAC rules specify that the smaller (or less complex) alkyl group is named as the alkoxy prefix, so “methoxyethane” is correct.

Tip

For option (B), remember the mnemonic: when both double and triple bonds are present, the suffix order is “‑ene” then “‑yne” (alphabetical), and the numbering is chosen to give the lowest locants to the multiple bonds as a whole, not just to the highest‑priority bond.

✓Final answer
$The correct option is (A). ANSWER: A$
COMEDK 2025Set 2025-E1 markMCQ
$\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{Q}$
›Reveal solutionSolution
Name each drawn structure yourself using IUPAC rules — identify the parent benzene, the substituents, their positions (ortho/meta/para) and the lowest locants — then compare with the given names P, Q, R, S. The matching is A→R, B→P, C→S, D→Q, which is option (A).

Concept & Intuition

This is a "match the structure to the name" problem. The reliable method is to name each structure yourself using IUPAC rules and then compare with the given names, rather than guessing from the drawings.

Benzene is the parent ring.

Substituents are named as prefixes (bromo, bromomethyl, alkyl groups).

Number the ring for the lowest set of locants; bromo comes first alphabetically here, so it takes position 1.

Know the common alkyl-group names: isopropyl = 1-methylethyl, isobutyl = 2-methylpropyl, sec-butyl = 1-methylpropyl, neopentyl = 2,2-dimethylpropyl.

Step-by-step naming

Structure A

Two substituents: bromine (Br) and a side chain –CH(CH₃)CH₂CH₃, which is a 1-methylpropyl group (sec-butyl).

They sit on adjacent carbons (ortho), so bromo is at position 1 and the side chain at position 2.

IUPAC name: 1-Bromo-2-(1-methylpropyl)benzene → matches R.

Structure B

Two substituents: –CH₂Br (bromomethyl) and –CH₂–C(CH₃)₃, a CH₂ attached to a carbon carrying three methyls, i.e. 2,2-dimethylpropyl (neopentyl).

The only name in Column II built from a bromomethyl group together with a 2,2-dimethylpropyl group is P (1-Bromomethyl-3-(2,2-dimethylpropyl)benzene).

So B matches P.

Structure C

A –CH₂–CH(CH₃)₂ side chain (2-methylpropyl, isobutyl) para to Br.

IUPAC name: 1-Bromo-4-(2-methylpropyl)benzene → matches S.

Structure D

An isopropyl group, –CH(CH₃)₂ = 1-methylethyl, para to Br.

IUPAC name: 1-Bromo-4-(1-methylethyl)benzene → matches Q.

Matching summary

Structure IUPAC name (derived) Column II label
A 1-Bromo-2-(1-methylpropyl)benzene R
B 1-Bromomethyl-3-(2,2-dimethylpropyl)benzene P
C 1-Bromo-4-(2-methylpropyl)benzene S
D 1-Bromo-4-(1-methylethyl)benzene Q

Thus A→R, B→P, C→S, D→Q.

Watch out

A common mistake is to misidentify the alkyl groups: "1-methylpropyl" is sec-butyl (a four-carbon chain branched at the first carbon), while "2-methylpropyl" is isobutyl (a three-carbon chain with a methyl on the second carbon), and "2,2-dimethylpropyl" is neopentyl. Draw the group out if you are unsure.

Tip

First identify the functional groups (bromo, bromomethyl), then the alkyl substituent, then the positions (ortho = 1,2; meta = 1,3; para = 1,4). This quickly narrows the options.

✓Final answer
$The correct option is (A) . ANSWER: A$
COMEDK 2025Set 2025-M1 markMCQ
$\mathrm{A}=\mathrm{Q} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{R}$
›Reveal solutionSolution
The key is to systematically match each drawn structure to its IUPAC name by identifying the parent chain, functional groups, and substituents. The correct matches are: A → R, B → S, C → Q, D → P, which corresponds to option (B).

Let’s walk through each structure step by step, naming them from scratch so the matching becomes clear.

Concept & Intuition

IUPAC nomenclature is like a recipe: first find the longest carbon chain that includes the principal functional group (the one with highest priority, e.g., carboxylic acid > aldehyde > alcohol > alkane). Number the chain to give the functional group the lowest possible number, then name substituents (like bromo, phenyl, alkyl) with their positions. For rings, treat the ring as a substituent if the chain is longer, or as the parent if the ring is the main feature. Here, we have four distinct cases: a substituted benzene (A), a polyfunctional acid (B), a dialdehyde with bromines (C), and a phenyl-substituted alkane with bromines (D).

1. Structure A – The Ibuprofen-like molecule

What we see: A benzene ring with a para-substituted pattern. On one side (top) is a carbon that has a methyl group and a carboxylic acid group – that’s a -CH(CH₃)COOH group. On the opposite side (bottom) is a -CH₂-CH(CH₃)₂ group (isobutyl).

Parent chain: The longest chain attached to the ring is the isobutyl part? No – the carboxylic acid is the highest priority functional group. The ring is a substituent here because the acid-containing chain is only 2 carbons long (propanoic acid). So the parent is propanoic acid.

Numbering: The carbon of the acid is C1. The carbon next to it (C2) bears the phenyl ring. So the name is 2-(4-isobutylphenyl)propanoic acid.

Match: This is name R in Column II.

Tip

The “isobutyl” group is a common name; in IUPAC it’s (2-methylpropyl). But here the given name uses “isobutyl”, so we match directly.

2. Structure B – Citric acid

What we see: A central carbon with an –OH and a –COOH. From that central carbon, two -CH₂- arms each end in a –COOH. That’s three carboxylic acid groups total.

Parent chain: The longest chain that includes all three carboxyls is a three-carbon chain (propane) with the central carbon bearing the –OH. The systematic name for citric acid is 2-hydroxy-1,2,3-propanetricarboxylic acid.

Match: This is name S.

Watch out

A common mistake is to think the central carbon is part of a longer chain, but the three carboxyls are on a three-carbon backbone. The numbering starts from the end nearest the first carboxyl, giving the –OH at position 2.

3. Structure C – Dialdehyde with bromines

What we see: A six-carbon straight chain with an aldehyde (–CHO) at both ends. Two bromine atoms are on adjacent carbons near the right end.

Parent chain: The longest chain includes both aldehyde carbons, so it’s a hexane chain with two aldehyde groups. The suffix for two aldehydes is “dial”.

Numbering: Aldehyde carbons are always C1 and C6 (the ends). The bromines are on C2 and C3 (from the right end, they are the second and third carbons from the right-hand CHO). So the name is 2,3-dibromohexanedial.

Match: This is name Q.

Tip

When both ends are the same functional group (here aldehyde), numbering can start from either end; the bromines get the lowest set of locants, which is 2,3.

4. Structure D – Phenyl-substituted dibromopentane

What we see: A benzene ring attached to a five-carbon chain. The chain: phenyl–CH₂–CHBr–CHBr–CH₂–CH₃. So it’s a pentane chain with a phenyl on C1 and bromines on C2 and C3.

Parent chain: The longest carbon chain is pentane (5 carbons). The benzene ring is a substituent (phenyl).

Numbering: Start from the end nearest the phenyl (to give it the lowest number). So C1 is the carbon attached to the ring. Then bromines are on C2 and C3. The name is 2,3-dibromo-1-phenylpentane.

Match: This is name P.

Watch out

Do not number from the other end – that would give the phenyl at C5 and bromines at C3 and C4, which is a higher set of locants. Always give the first substituent the lowest number.

Final Matching

A → R

B → S

C → Q

D → P

This corresponds to option (B).

✓Final answer
$The correct option is (B) . ANSWER: B$
COMEDK 2024Set 2024-A1 markMCQ
$Q. Names of some organic compounds are given. Which one is not in IUPAC system? (A) (B) (C) (D)$
›Reveal solutionSolution
The key idea is to check each name against IUPAC rules for numbering, substituent order, and functional group suffixes. Option (A) violates the rule that the carboxylic acid carbon must be number 1, making its numbering incorrect; the correct option is (A).

The question asks which name is not in the IUPAC system. That means three names follow IUPAC rules, and one does not. To find the odd one out, we must recall the core principles of IUPAC nomenclature:

The principal functional group (here, the highest priority group) determines the suffix and gets the lowest possible locant.

For carboxylic acids, the carbon of the –COOH group is always number 1.

Double bonds and other substituents are numbered to give the lowest set of locants, but the acid carbon’s position is fixed.

Let’s examine each option step by step.

Option (A): “4-oxo-2,3-dimethylpent-2-en-1-oic acid”

The structure is: H₃C–C(=O)–C(CH₃)=C(CH₃)–C(=O)OH. This is a five-carbon chain with a ketone (oxo) on carbon 4, two methyl groups on carbons 2 and 3, a double bond between carbons 2 and 3, and a carboxylic acid at the end.

In IUPAC, the carboxylic acid carbon must be carbon 1. So the chain is numbered from the –COOH end: C1 is the acid carbon, C2 has a methyl and a double bond, C3 has a methyl and a double bond, C4 has the ketone, and C5 is the terminal methyl.

The correct name should be: 5-oxo-2,3-dimethylpent-3-enoic acid (because the double bond is between C3 and C4, not C2 and C3, when numbering from the acid). The given name says “pent-2-en-1-oic acid,” which incorrectly places the double bond at position 2 and uses “-1-oic” (redundant, since the acid carbon is always 1). This is a clear violation.

Conclusion: Option (A) is not in the IUPAC system.

Option (B): “1,3,3-trimethylcyclohex-1-ene”

Structure: a cyclohexene ring with a double bond between C1 and C2, a methyl on C1, and two methyls on C3 (gem-dimethyl).

Numbering: The double bond gets priority, so carbons 1 and 2 are the double-bonded atoms. The methyl groups are then at positions 1 and 3 (with two at 3). The name “1,3,3-trimethylcyclohex-1-ene” follows IUPAC: the locant for the double bond is given as “-1-ene” (the lower-numbered carbon of the double bond), and substituents are listed alphabetically (trimethyl is fine). This is correct.

Option (C): “but-3-enoic acid”

Structure: CH₂=CH–CH₂–COOH. This is a four-carbon chain with a double bond between C3 and C4 (if numbered from the acid carbon, C1 is –COOH, C2 is CH₂, C3 is CH, C4 is CH₂). The double bond is between C3 and C4, so the locant is 3. The name “but-3-enoic acid” is perfectly IUPAC: the suffix “-enoic” indicates the double bond, and the locant 3 is correct. (Note: The older name “vinylacetic acid” is not IUPAC, but the given name is IUPAC.)

Option (D): “2-methyl-3-phenylpentane”

Structure: H₃C–CH(CH₃)–CH(C₆H₅)–CH₂–CH₃. This is a pentane chain with a methyl on C2 and a phenyl on C3. Numbering from the end nearest the first substituent gives the lowest locants: 2-methyl and 3-phenyl. The name “2-methyl-3-phenylpentane” follows IUPAC rules (substituents in alphabetical order: methyl before phenyl). This is correct.

Watch out

A common mistake is to think that “-1-oic” is acceptable for carboxylic acids. In IUPAC, the “1” is implied and never written; writing “pent-2-en-1-oic acid” is incorrect because the acid carbon is always position 1, so the locant is redundant and the double bond numbering is wrong.

Tip

For carboxylic acids, always number from the –COOH carbon as 1. Then assign locants to double bonds and other groups. If the name includes “-1-oic,” it’s a red flag — it’s either redundant or indicates a numbering error.

✓Final answer
$The correct option is (A). ANSWER: A$
COMEDK 2024Set 2024-M1 markMCQ
$Q. Which one of the following is the correct IUPAC name of the given compound? (A) 5-Bromo-2, 3-dimethylheptanoyl chloride (B) 5,6-Dimethyl-3-bromohexanoyl chloride (C) 1-Chloro-5-bromo-2, 3-dimethyl-1-oxoheptane (D) 3-Bromo-5, 6-dimethylhexanoyl chloride$
›Reveal solutionSolution
The compound is a seven-carbon acyl chloride with methyl groups on C2 and C3 and a bromine on C5; the correct IUPAC name is 5-Bromo-2,3-dimethylheptanoyl chloride, which corresponds to option (A).

The key to naming this compound correctly is to identify the principal functional group and the longest carbon chain that includes it. Here, the functional group is an acyl chloride (–COCl), which takes priority over all other substituents. The chain must be numbered starting from the carbonyl carbon of the acyl chloride, giving it the lowest possible locant (C1). Once the chain is numbered, we name the substituents (methyl groups and bromine) with their positions, and then assemble the name in alphabetical order.

Let’s work through it step by step.

Identify the principal functional group and the parent chain.

The structure shows a carbonyl (C=O) with a chlorine atom attached to the same carbon – that’s an acyl chloride group (–COCl). This group is the highest priority functional group, so the parent chain must include it. The longest continuous carbon chain that includes the carbonyl carbon has seven carbons (heptane). The acyl chloride suffix replaces the “-e” of heptane with “-oyl chloride”, giving the parent name heptanoyl chloride.

Number the chain from the carbonyl carbon.

In IUPAC nomenclature, the carbonyl carbon of an acyl halide is always assigned position 1. So we number the chain as:

C1 = carbonyl carbon (part of –COCl),

C2 = next carbon (has a methyl group up),

C3 = next carbon (has a methyl group down),

C4 = CH₂,

C5 = carbon with Br (down),

C6 and C7 = ethyl end.

Locate and name the substituents.

At C2: a methyl group → 2-methyl

At C3: a methyl group → 3-methyl

At C5: a bromine atom → 5-bromo Since there are two methyl groups, we combine them as 2,3-dimethyl.

Assemble the name in alphabetical order.

Substituents are listed alphabetically (ignoring prefixes like di-). “Bromo” comes before “methyl”, so the order is:

5-bromo-2,3-dimethylheptanoyl chloride.

Check the other options for common mistakes.

Option (B): “5,6-Dimethyl-3-bromohexanoyl chloride” – wrong chain length (hexanoyl instead of heptanoyl) and wrong numbering.

Option (C): “1-Chloro-5-bromo-2,3-dimethyl-1-oxoheptane” – this treats the acyl chloride as a ketone and a chloro substituent, which is incorrect; the acyl chloride is the functional group, not a separate chloro and oxo.

Option (D): “3-Bromo-5,6-dimethylhexanoyl chloride” – again, wrong chain length (hexanoyl) and incorrect numbering.

Watch out

A common pitfall is to number the chain from the end farthest from the carbonyl, or to treat the acyl chloride as a chloro ketone. Always remember: the carbonyl carbon of an acyl halide is C1, and the suffix is “-oyl chloride”, not “-one” or “-oxo”.

Tip

To quickly verify, count the carbons in the longest chain including the carbonyl: here it’s seven, so the parent must be heptanoyl. Any option with “hexanoyl” is automatically wrong.

✓Final answer
$The correct option is (A) . ANSWER: A$
KCET 2023Set D-21 markMCQ
$Q. IUPAC name of the compound is (A) 2, 3-dimethylbut-2-ene (B) 2, 3-dimethyl butyne (C) 1, 1, 2, 2-tetra methylethene (D) 2, 3-dimethyl butene$
›Reveal solutionSolution
The compound is a symmetrical alkene with four methyl groups on a two-carbon double bond; the longest chain is but-2-ene, and the correct IUPAC name is 2,3-dimethylbut-2-ene.

The question asks for the IUPAC name of a compound, but the compound itself is not drawn in the text. From the options, it is clear we are dealing with a six-carbon alkene — specifically, the one where the double bond is between C2 and C3 of a butane chain, and each of those two carbons carries two methyl groups. That structure is $(CH_3)_2C=C(CH_3)_2$ , commonly called tetramethylethene.

The key to IUPAC naming is to identify the longest continuous carbon chain that includes the double bond. Here, the longest chain that contains the double bond has four carbons — that is the but-2-ene backbone. The two extra carbons (the methyl groups) are then treated as substituents.

Let’s work through the naming step by step.

Identify the parent chain. The longest chain containing the double bond is a four-carbon chain: C1–C2=C3–C4. The double bond is between C2 and C3, so the parent name is but-2-ene.

Number the chain to give the double bond the lowest possible locant. Since the double bond is between C2 and C3, numbering from either end gives the same locant (2). So the parent is but-2-ene.

Identify and locate substituents. On C2, there are two methyl groups; on C3, there are also two methyl groups. So we have four methyl substituents — two at position 2 and two at position 3.

Assemble the name. The substituents are listed alphabetically (ignoring multiplying prefixes like di-, tri-). Here, all substituents are methyl, so we combine: 2,3-dimethyl (for the two methyls on C2 and C3) — but wait, there are four methyls total. The correct prefix is tetramethyl, with locants 2,2,3,3. So the full name is 2,2,3,3-tetramethylbut-2-ene? That would be wrong — let’s check.

Watch out

A common mistake is to write "2,2,3,3-tetramethylbut-2-ene". But that name implies the parent chain is butane with four methyl substituents, which is correct in terms of substituent count, but the IUPAC convention for alkenes requires the double bond to have the lowest possible locant, and here the double bond is already at position 2. However, the name "2,2,3,3-tetramethylbut-2-ene" is actually acceptable but not the simplest. The simpler name uses the fact that the two methyls on C2 and the two on C3 can be described as "2,3-dimethyl" if we consider that each of C2 and C3 already has one methyl from the parent chain? No — the parent chain is but-2-ene, which has no methyl groups on C2 or C3 in the parent. So each methyl is a substituent.

Let’s re-evaluate: The structure $(CH_3)_2C=C(CH_3)_2$ has a four-carbon backbone: C1 (CH3), C2 (C with two CH3), C3 (C with two CH3), C4 (CH3). So the substituents are: at C2, two methyls; at C3, two methyls. That gives 2,2,3,3-tetramethylbut-2-ene. But option (A) is 2,3-dimethylbut-2-ene — that would correspond to only two methyl substituents, i.e., $(CH_3)_2C=CHCH_3$ ? No, that’s 2-methylbut-2-ene? Let’s check: 2,3-dimethylbut-2-ene means the parent is but-2-ene with a methyl at C2 and a methyl at C3 — that structure is $CH_3C(CH_3)=C(CH_3)CH_3$ , which is exactly $(CH_3)_2C=C(CH_3)_2$ ! Because the parent chain already includes the four carbons: C1 (CH3), C2 (C with one CH3 substituent), C3 (C with one CH3 substituent), C4 (CH3). So the two methyl substituents are the ones attached to C2 and C3, giving the same molecule.

Tip

In IUPAC naming, the parent chain is the longest chain that includes the double bond. For $(CH_3)_2C=C(CH_3)_2$ , the longest chain is indeed four carbons: CH3–C(=C)–C–CH3. The two "extra" methyls are the ones on C2 and C3. So the name is 2,3-dimethylbut-2-ene, not 2,2,3,3-tetramethylbut-2-ene. The latter would imply a butane parent (no double bond) with four methyls, which is a different compound (2,2,3,3-tetramethylbutane). Always ensure the parent chain includes the double bond.

Thus, the correct IUPAC name is 2,3-dimethylbut-2-ene.

✓Final answer
$The correct option is (A) 2,3-dimethylbut-2-ene .$