In which of the following, functional group isomerism is not possible? (i) Alcohols (ii) Aldehydes (iii) Alkyl

NCERT Exemplar · Q5

Q. In which of the following, functional group isomerism is not possible?

(i) Alcohols

(ii) Aldehydes

(iii) Alkyl halides

(iv) Cyanides

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Concept understanding — IUPAC Nomenclature

IUPAC Nomenclature (Organic Compounds)

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

The Recipe

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

Watch out

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

Worked Example

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

Why this formula?

IUPAC Nomenclature: Why the Rules Work the Way They Do

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

1. The Core Principle: The Longest Carbon Chain

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

Why?

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

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

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

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

Why?

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

Example:

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

3. Alphabetical Order of Substituents

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

Why?

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

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

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

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

Why?

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

Priority order:

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

5. The "Suffix" and "Prefix" System …

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Functional group isomerism requires compounds to have the same molecular formula but different functional groups. Among the given options, alkyl halides cannot exhibit functional group isomerism because there is no other class of organic compounds with a different functional group that shares their general molecular formula. The correct option is (iii).

Isomerism refers to the phenomenon where two or more compounds have the same molecular formula but differ in the arrangement of atoms. Functional group isomerism is a specific type of structural isomerism where the compounds share the same molecular formula but possess different functional groups. This means they belong to different classes of organic compounds and, consequently, exhibit distinct chemical and physical properties. To determine if functional group isomerism is possible for a given class of compounds, we look for another class of compounds that can have the exact same general molecular formula but a different characteristic functional group.

Let's analyze each option:

Alcohols:
- Alcohols contain the hydroxyl functional group ( $-\text{OH}$ ). The general molecular formula for saturated acyclic alcohols is $C_nH_{2n+2}O$ .
- Ethers contain the ether functional group ( $-\text{O}-$ ). The general molecular formula for saturated acyclic ethers is also $C_nH_{2n+2}O$ .
- Example: Ethanol ( $CH_3CH_2OH$ ) and Dimethyl ether ( $CH_3OCH_3$ ) both have the molecular formula $C_2H_6O$ . Ethanol is an alcohol, and dimethyl ether is an ether. They are functional group isomers.
- Therefore, functional group isomerism is possible for alcohols.
Aldehydes:
- Aldehydes contain the aldehyde functional group ( $-\text{CHO}$ ). The general molecular formula for saturated acyclic aldehydes is $C_nH_{2n}O$ .
- Ketones contain the ketone functional group ( $-\text{CO}-$ ). The general molecular formula for saturated acyclic ketones is also $C_nH_{2n}O$ .
- Example: Propanal ( $CH_3CH_2CHO$ ) and Propanone ( $CH_3COCH_3$ ) both have the molecular formula $C_3H_6O$ . Propanal is an aldehyde, and propanone is a ketone. They are functional group isomers.
- Therefore, functional group isomerism is possible for aldehydes.
Alkyl Halides:
- Alkyl halides contain a halogen atom ( $-\text{X}$ , where $\text{X} = \text{F, Cl, Br, I}$ ) attached to an alkyl group. The general molecular formula for saturated acyclic alkyl halides is $C_nH_{2n+1}X$ .
- For functional group isomerism to occur, there must be another class of compounds with a different functional group that shares this same general molecular formula. …

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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 96 on this concept.

CBSE 2026Set A1 markMCQ
$Q. The IUPAC name of CH3CH2COCH2CH3 is$
(a) Diethyl ketone
(b) 3-Pentanone
(c) 2-Pentanone
(d) Methyl propyl ketone
›Reveal solutionSolution

Number the five-carbon chain to give the C=O the lowest locant: the ketone carbon is C-3, so the IUPAC name is pentan-3-one (3-pentanone).

Structure: CH3-CH2-CO-CH2-CH3. The longest chain containing the carbonyl has 5 carbons -> pentanone. Numbering from either end places the carbonyl carbon at position 3. Hence the IUPAC na …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. IUPAC name of [a benzene ring with an -OC2H5 (ethoxy) group attached] is$
(a) Benzyl ethoxide
(b) Benzene ethoxide
(c) Ethoxybenzene
(d) Ethoxybenzoyl
›Reveal solutionSolution

Simple ethers with one aromatic and one alkyl group are IUPAC-named as 'alkoxy' + 'benzene', treating the -OR group as a substituent on the ring.

The compound is a benzene ring bearing an ethoxy substituent (-O-C2H5). Following IUPAC substitutive nomenclature for ethers, the smaller/simpler group's oxygen chain is named as an 'alkoxy' prefix attached to the parent (here benzene):

-OC2H5 = ethoxy group

So the compound is Ethoxybenzene (common name: phenetole).

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. IUPAC name of CH3COOH is:$
(a) Ethanol
(b) ethanoic acid
(c) Ethanal
(d) ethane
›Reveal solutionSolution

CH₃COOH is acetic acid, IUPAC name ethanoic acid.

CH₃COOH is a two-carbon carboxylic acid, commonly called acetic acid. In IUPAC nomenclature, carboxylic acids are named by replacing the "-e" of the corresponding alkane name with "-oic …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The IUPAC name of Formic acid is :$
(a) Methanoic acid
(b) Ethanoic acid
(c) Ethandioic acid
(d) Methandioic acid
›Reveal solutionSolution
Formic acid, HCOOH, is the one-carbon carboxylic acid, named methanoic acid by IUPAC rules.

Formic acid has the structure H–COOH (a single carbon bearing the carboxyl group). To name a carboxylic acid by IUPAC nomenclature:

Select the longest carbon chain including the –COOH carbon.

Replace the terminal '-e' of the corresponding alkane name with '-oic acid'.

Here the chain has just one carbon (methane), so the name becomes methanoic acid.

…
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CBSE 2026Set ANNUAL1 markMCQ
$Q. The IUPAC name of isobutyl chloride is$
(a) 2-chlorobutane
(b) 1-chloro-2-methyl propane
(c) 2-chloro-2-methyl propane
(d) 1-chlorobutane
›Reveal solutionSolution

Isobutyl chloride has the structure (CH3)2CH-CH2-Cl. Number the longest chain (propane, 3 carbons) so the substituent (Cl) gets the lowest locant, then name the methyl branch.

Structure of isobutyl chloride: (CH3)2CH-CH2-Cl

This can be redrawn as: Cl-CH2-CH(CH3)-CH3

Step 1: Identify the longest carbon chain containing the point of attachment of Cl. That's a 3-carbon (propane) chain: C1(CH2Cl)-C2(CH, bearing a methyl branch)-C3(CH3).

…

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CBSE 2026Set ANNUAL1 mark
$Q. Write IUPAC name of the following compound:$
›Reveal solutionSolution
Numbering the 6-carbon chain from the -COOH carbon (C1) puts the ketone at C4 and the terminal bromine at C6, giving 6-bromo-4-oxohexanoic acid.

The drawn structure is a straight 6-carbon chain: BrCH2-CH2-CO-CH2-CH2-COOH, i.e. a terminal -CH2Br at one end and -COOH at the other, with a ketone in between.

The carboxylic acid (-COOH) is the senior group and is numbered C1.

Numbering from that end: C1 (COOH), C2 (CH2), C3 (CH2), C4 (the ketone C=O), C5 (CH2), C6 (the terminal CH2Br). …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. The IUPAC name of CH3COOH is$
(a) Acetic acid
(b) Formic acid
(c) Ethanoic acid
(d) Methanoic acid
›Reveal solutionSolution

CH3COOH = ethanoic acid (IUPAC).

CH3COOH has two carbon atoms and a −COOH group. In IUPAC nomenclature, a two-carbon carboxylic acid is 'ethan' (two carbons) + 'oic acid' = ethanoic acid. Its common (trivial) …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The IUPAC name of CH3-CH(OH)-CH3 is$
(a) Propan-1-ol
(b) Propan-2-ol
(c) n-propyl alcohol
(d) Isopropyl alcohol
›Reveal solutionSolution

CH3−CH(OH)−CH3 = propan-2-ol.

The chain has three carbons (propane). The −OH group is attached to the central (second) carbon. Numbering to give the OH the lowest locant places it at position 2, so the …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The structure of a valuable organic compound used as solvent in many chemical industries is shown below (a central carbon bearing a CH₃ group above, a CH₃ group to the left, an OH group below, and a -CH₂-OCH₃ group to the right). The IUPAC name of the organic compound is$
(a) 1-methoxy-2-methylpropan-2-ol
(b) 3-methoxy-2-methylpropan-2-ol
(c) 1-methoxy-3-methylpropan-3-ol
(d) 2-methoxy-2-methylpropan-2-ol
›Reveal solutionSolution
The structure is (CH₃)₂C(OH)CH₂OCH₃; the parent is propan-2-ol with a 2-methyl and a 1-methoxy substituent, giving 1-methoxy-2-methylpropan-2-ol — option (A).

The central carbon bears two $CH_3$ groups, an $OH$ , and a $-CH_2-OCH_3$ group, so the molecule is

$(CH_3)_2C(OH)-CH_2-OCH_3.$

Naming:

The longest carbon chain containing the –OH is three carbons (propane): $C1 = CH_2(OCH_3)$ , $C2 = C(OH)(CH_3)$ , $C3 = CH_3$ . …
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CBSE 2025Set 56/5/11 markMCQ
$CH_3 - CH_2 - N(CH_3) - CH_2 - CH_2 - CH_3$
›Reveal solutionSolution
The compound is a secondary amine with an ethyl and a methyl group on the nitrogen, and a three-carbon chain as the parent. The correct IUPAC name is N-ethyl-N-methylpropan-1-amine, which corresponds to option (B).

The key to naming amines under IUPAC rules is to identify the longest continuous carbon chain attached to the nitrogen — that becomes the parent alkane name, with the suffix "-amine". The other groups on the nitrogen are treated as substituents, prefixed with "N-" to show they are attached to the nitrogen atom, not to the carbon chain.

Let's break this down step by step.

Draw the structure from the given condensed formula.

The formula is $CH_3 - CH_2 - N(CH_3) - CH_2 - CH_2 - CH_3$ .

The nitrogen has three bonds: one to an ethyl group ( $-CH_2CH_3$ ), one to a methyl group ( $-CH_3$ ), and one to a propyl group ( $-CH_2CH_2CH_3$ ). So the molecule is:

CH3 | CH3-CH2-N-CH2-CH2-CH3

The nitrogen is connected to three different alkyl groups: ethyl, methyl, and propyl.

Identify the parent chain.

The IUPAC rule for amines: the parent chain is the longest continuous carbon chain attached directly to the nitrogen. Here, the chain on the right is $CH_2-CH_2-CH_3$ — that's three carbons, a propyl group. The chain on the left is $CH_2-CH_3$ — two carbons, an ethyl group. The methyl group ( $CH_3$ ) is just one carbon.

So the longest chain is the three-carbon chain (propyl). The parent name becomes propan-1-amine (since the nitrogen is at the end of the chain, carbon 1).

Name the substituents on the nitrogen.

The other two groups attached to the nitrogen are an ethyl group ( $-CH_2CH_3$ ) and a methyl group ( $-CH_3$ ). Since they are on the nitrogen, not on the carbon chain, they are prefixed with "N-".

Alphabetically, ethyl comes before methyl. So we write N-ethyl-N-methyl.

Combine the parts.

The full name is: N-ethyl-N-methylpropan-1-amine. …
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CBSE 2025Set X11 markMCQ
$\text{H}_3\text{C}-\underset{\underset{\text{Br}}{|}}{\text{CH}}-\text{CH}_2-\underset{\overset{\text{O}}{||}}{\text{C}}-\text{H}$
(a) 3–bromobutyraldehyde
(b) 2–bromopropanaldehyde
(c) 3–bromobutanal
(d) 2–bromobutanal
›Reveal solutionSolution
The compound is $\text{CH}_3\text{-CH(Br)-CH}_2\text{-CHO}$ , a 4-carbon aldehyde with bromine on C-3 → 3-bromobutanal.

The structure is $\text{H}_3\text{C-CH(Br)-CH}_2\text{-CHO}$ .

The principal functional group is the aldehyde (–CHO), so it gets the lowest locant. Numbering starts at the carbonyl carbon: C1 = CHO, C2 = $\text{CH}_2$ , C3 = CH(Br), C4 = $\text{CH}_3$ .

Longest chain = 4 carbons → butanal. …
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CBSE 2025Set D1 markMCQ
$Q. The IUPAC name of CH3COOC2H5 is$
(a) Methyl propanoate
(b) Ethyl ethanoate
(c) Acetoethane
(d) Ethoxyethane
›Reveal solutionSolution
CH3COOC2H5 = ethyl ethanoate (common name ethyl acetate).

The compound is an ester derived from ethanoic acid (CH3COOH) and ethanol (C2H5OH). An ester R-COO-R' is named as 'alkyl alkanoate':

The alkyl group from the alcohol part (C2H5) -> ethyl …
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Q.In which of the following, functional group isomerism is not possible?

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