A hydrocarbon of molecular mass 72 g mol⁻¹ gives a single monochloro derivative and two dichloro derivatives o

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Concept understanding — Structural Isomerism

Structural Isomerism: The First Meeting

Imagine you have a box of identical Lego bricks — four red, ten blue, and six yellow. You build two different models: a car and a house. Both use exactly the same number of each colour brick, but the structures are completely different. That is the core idea of isomerism: same atoms, different arrangement.

In chemistry, molecules are not just a list of atoms. How those atoms are connected matters enormously. Two molecules can have the exact same molecular formula (same number of each atom) but be connected in different ways. Those are structural isomers (also called constitutional isomers).

The Precise Statement

Structural isomers are compounds that have the same molecular formula but different connectivity of atoms — that is, different structural formulas.

The key word is connectivity. Which atom is bonded to which? If you change that, you get a different substance with different physical and chemical properties.

A Concrete Example: C₄H₁₀

Take butane, C₄H₁₀. There are exactly two ways to connect four carbon atoms and ten hydrogen atoms:

n-Butane — a straight chain: C–C–C–C
Isobutane (2-methylpropane) — a branched chain: a central carbon bonded to three methyl groups

Both have formula C₄H₁₀. But n-butane boils at –0.5 °C, while isobutane boils at –11.7 °C. Same atoms, different connectivity → different substance.

Note

Structural isomers are not the same molecule. They are distinct compounds that happen to share a molecular formula. You cannot rotate or flip one to get the other — you must break and reform bonds.

The Three Main Types

Structural isomerism comes in three flavours:

Type	What changes	Example (C₃H₆O)
Chain isomerism	The carbon skeleton (straight vs. branched)	Butane vs. isobutane
Position isomerism	The location of a functional group or substituent	Propan-1-ol vs. propan-2-ol (OH on carbon 1 vs. carbon 2)
Functional group isomerism	The atoms are rearranged into a different functional group	Propanal (aldehyde) vs. propanone (ketone) — both C₃H₆O

Watch out

Do not confuse structural isomers with stereoisomers. Stereoisomers have the same connectivity but differ in spatial arrangement (like left and right hands). That is a completely different chapter. For now: structural isomers = different bond connections.

Why This Matters …

Why this formula?

Structural Isomerism: Why the Key Ideas Hold

Structural isomerism arises when molecules share the same molecular formula but differ in the connectivity of atoms. There is no single "formula" for structural isomerism — instead, the key is understanding why different arrangements are possible.

The Core Principle: Connectivity ≠ Composition

A molecular formula tells you how many of each atom are present, but not how they are joined. Structural isomers exist because atoms can form bonds in multiple distinct sequences while satisfying valency rules.

Why This Happens: The Valency Constraint

Each atom has a fixed bonding capacity (valency):

Carbon: 4 bonds
Hydrogen: 1 bond
Oxygen: 2 bonds
Nitrogen: 3 bonds

Example: For $C_4H_{10}$ , the formula satisfies $4(4) + 10(1) = 26$ valence electrons. But the carbon atoms can be arranged as:

A straight chain: $CH_3-CH_2-CH_2-CH_3$ (n-butane)
A branched chain: $CH_3-CH(CH_3)-CH_3$ (isobutane)

Both satisfy valency, but the connectivity differs.

The "Formula" for Counting Isomers: Why It's Not Simple

There is no closed-form formula to count structural isomers for a given molecular formula. The number grows rapidly and depends on:

Carbon skeleton branching possibilities
Functional group positions
Ring formation possibilities

Why No Simple Formula Exists

The problem is combinatorial — the number of possible trees (acyclic graphs) with $n$ carbon atoms grows exponentially. For example:

$C_4H_{10}$ : 2 structural isomers
$C_5H_{12}$ : 3 structural isomers
$C_6H_{14}$ : 5 structural isomers
$C_{10}H_{22}$ : 75 structural isomers

The pattern follows Cayley's formula for trees, but even that counts only carbon skeletons — not functional group positions.

Key Reasoning: The Branching Principle

The fundamental reason structural isomers exist is that carbon chains can branch. Consider $C_5H_{12}$ :

Straight chain: $C-C-C-C-C$ (n-pentane)
One branch: $C-C-C(C)-C$ (isopentane) — the branch can be at position 2 or 3, but these are identical due to symmetry
Two branches: $C-C(C)(C)-C$ (neopentane) — a quaternary carbon

Why position matters: The branch location changes the carbon's environment, altering physical and chemical properties.

The Functional Group Position Rule

For compounds with functional groups (e.g., alcohols $C_nH_{2n+2}O$ ), the position of the -OH group creates isomers:

$CH_3CH_2CH_2OH$ (propan-1-ol) — OH at end
$CH_3CH(OH)CH_3$ (propan-2-ol) — OH in middle …

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The hydrocarbon is 2,2-dimethylpropane (neopentane). Its molecular mass of 72 g mol $^{-1}$ corresponds to C $_5$ H $_{12}$ , and its highly symmetric structure explains why it gives only one monochloro derivative (all 12 H atoms are equivalent) and exactly two dichloro derivatives (one geminal, one 1,3-).

Why This Approach Works

The key to solving this problem lies in structural isomerism and symmetry. When a hydrocarbon undergoes photochlorination, chlorine atoms replace hydrogen atoms. The number of distinct monochloro derivatives tells you how many different types of hydrogen atoms exist in the molecule. The number of dichloro derivatives tells you how many ways two chlorine atoms can be placed on different carbon atoms (or the same carbon) without creating identical compounds.

A molecular mass of 72 g mol $^{-1}$ for a hydrocarbon (only C and H) immediately narrows the possibilities. Let's find the formula first, then use the chlorination data to pinpoint the exact structure.

Step-by-Step Solution

1. Determine the molecular formula

A hydrocarbon has the general formula C $_x$ H $_y$ . Its molecular mass is:

12x + y = 72

Since it's a hydrocarbon, $y$ must be even (for alkanes, $y = 2x + 2$ ; for alkenes, $y = 2x$ ; for alkynes, $y = 2x - 2$ ). Let's test the alkane formula first, as alkanes are most common for photochlorination problems.

For an alkane: $y = 2x + 2$

12x + (2x + 2) = 72

14x = 70

x = 5

So the formula is C $_5$ H $_{12}$ . This is a saturated hydrocarbon (alkane) — no double or triple bonds. There are three structural isomers of pentane:

n-pentane (straight chain)
Isopentane (2-methylbutane)
Neopentane (2,2-dimethylpropane)

Note

If we had tried an alkene ( $y = 2x$ ), we'd get $14x = 72 \Rightarrow x \approx 5.14$ , not an integer. For an alkyne ( $y = 2x - 2$ ), $14x - 2 = 72 \Rightarrow 14x = 74 \Rightarrow x \approx 5.29$ . Only the alkane gives an integer $x$ . So C $_5$ H $_{12}$ is confirmed.

2. Analyze monochlorination — why only one product?

Photochlorination replaces one H atom with Cl. The number of distinct monochloro derivatives equals the number of chemically different hydrogen atoms in the molecule.

Let's examine each isomer:

n-pentane (CH $_3$ -CH $_2$ -CH $_2$ -CH $_2$ -CH $_3$ ): Has three types of H atoms (primary on end carbons, secondary on the two middle carbons, and secondary on the central carbon). This would give 3 monochloro derivatives — too many.
Isopentane ((CH $_3$ ) $_2$ CH-CH $_2$ -CH $_3$ ): Has four types of H atoms (two different primary environments — the pair of equivalent methyls on C2 and the lone methyl at the far end — one secondary on the C3 CH $_2$ , and one tertiary on the C2 CH). This would give 4 monochloro derivatives — also too many.
Neopentane (C(CH $_3$ ) $_4$ ): All 12 hydrogen atoms are identical — they're all on methyl groups attached to the same central carbon. Every H is equivalent. So replacing any one H gives the same product: (CH $_3$ ) $_3$ C-CH $_2$ Cl. This gives exactly 1 monochloro derivative.

Watch out

A common mistake is to think that neopentane has "different" hydrogens because the methyl groups are attached to a quaternary carbon. But all four methyl groups are identical, and within each methyl group, all three H atoms are equivalent by rotation. So all 12 H atoms are chemically equivalent.

3. Analyze dichlorination — why exactly two products?

Now we place two chlorine atoms. For neopentane, the carbon skeleton is: $(CH_3)_4C$

All four methyl carbons are equivalent. When we replace two H atoms with Cl, we get:

Case 1: Both Cl on the same carbon atom (geminal dichloride)

Replace two H atoms on the same methyl group: (CH $_3$ ) $_3$ C-CHCl $_2$ …

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Concept: Structural Isomerism & Symmetry in Free Radical Halogenation

The key idea is that photo-chlorination substitutes chlorine atoms for hydrogen atoms via a free radical mechanism. The number of distinct monochlorinated products tells you how many different types of hydrogen atoms (chemically non-equivalent) are present in the original hydrocarbon. The number of dichlorinated products gives further symmetry clues.

Method: Symmetry & Product Count Analysis

Step 1: Determine the molecular formula from the molar mass

Hydrocarbon means only C and H.
General formula: $C_nH_{2n+2}$ for alkanes (saturated).
Molar mass: $12n + (2n+2) = 14n + 2 = 72$
Solve: $14n = 70 \implies n = 5$
Molecular formula: $\mathbf{C_5H_{12}}$ (pentane isomers).

Step 2: List possible isomers of pentane

Isomer	Structure
n-Pentane	$CH_3-CH_2-CH_2-CH_2-CH_3$
Isopentane (2-methylbutane)	$CH_3-CH(CH_3)-CH_2-CH_3$
Neopentane (2,2-dimethylpropane)	$C(CH_3)_4$

Step 3: Count distinct monochlorination products for each

n-Pentane: 3 types of H → 3 monochloro derivatives ✗
Isopentane: 4 types of H → 4 monochloro derivatives ✗
Neopentane: All 12 H atoms are equivalent (all methyl groups identical) → 1 monochloro derivative ✓ …

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Let’s break this down step-by-step. The question tests your understanding of structural isomerism, symmetry in molecules, and free radical chlorination selectivity.

Step 1: Identify the hydrocarbon from molecular mass

A hydrocarbon contains only carbon and hydrogen.

General formula for alkanes: $C_n H_{2n+2}$ .

Mass:

$12n + (2n+2) = 14n + 2 = 72$

$14n = 70 \implies n = 5$

So the hydrocarbon is $C_5H_{12}$ — a pentane isomer.

Step 2: What the question tells us

Single monochloro derivative → all hydrogen atoms are equivalent in terms of substitution (high symmetry).
Two dichloro derivatives → only two distinct ways to place two chlorine atoms.

This points to neopentane (2,2-dimethylpropane).

Structure: $(CH_3)_4C$

All 12 hydrogens are identical (methyl groups are equivalent).

Monochlorination gives only one product: (CH₃)₃C–CH₂Cl.

Dichlorination gives exactly two products:

Cl on two different methyls (one Cl each): (CH3)2C(CH2Cl)2 -- two separate CH2Cl groups
Both Cl on the same methyl: (CH3)3C-CHCl2 -- one single CHCl2 group

Common Mistakes & How to Avoid Them

✗ Mistake 1: Assuming the hydrocarbon is n-pentane or isopentane

Why it’s wrong:

n-Pentane gives 3 monochloro derivatives (primary, secondary, secondary).
Isopentane gives 4 monochloro derivatives. Both violate the “single monochloro derivative” condition.

How to avoid:

Always check symmetry — count distinct types of hydrogen atoms. If the molecule has only one type of H, it must be highly symmetric (like neopentane or cyclohexane).

✗ Mistake 2: Forgetting to check dichloro derivative count

Why it’s wrong:

Even if monochlorination gives one product, dichlorination might give more than two — which would disqualify the structure.

How to avoid:

Systematically place two Cl atoms on the carbon skeleton.

For neopentane:
- Both Cl on the same CH3 (one CHCl2 group) -> 1 product: (CH3)3C-CHCl2
- One Cl on each of two different CH3 groups (two CH2Cl groups) -> 1 product: (CH3)2C(CH2Cl)2
- On the central carbon? Not possible (no H there to replace). Total = 2. If you get 3 or more, reject the structure.

✗ Mistake 3: Ignoring stereoisomerism in dichloro derivatives

Why it’s wrong:

Sometimes students count optical or geometrical isomers as separate — but the question likely means structural (constitutional) isomers unless specified.

How to avoid:

In photo chlorination problems, “dichloro derivatives” usually means structural isomers (different connectivity). Only consider stereoisomers if the question explicitly mentions optical activity or cis-trans.

--- …

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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 17 on this concept.

CBSE 2026Set DZ1 markMCQ
$Q. Which of the following is an aldehyde?$
(a) $CH_3-\overset{O}{\overset{||}{C}}-H$
(b) $CH_3-CH_2-\overset{O}{\overset{||}{C}}-CH_3$
(c) $CH_3-CH_2-\overset{O}{\overset{||}{C}}-CH_2-CH_3$
(d) $CH_3-\overset{O}{\overset{||}{C}}-CH_3$
›Reveal solutionSolution
An aldehyde has the $-CHO$ group (a carbonyl carbon bonded to at least one H). Only option (a) has this; the rest are ketones.

The functional-group test: in an aldehyde the carbonyl carbon $(>\!C=O)$ carries at least one hydrogen ( $R-CHO$ ). In a ketone the carbonyl carbon is bonded to two carbon atoms ( $R-CO-R'$ ).

(a) $CH_3-CHO$ → carbonyl C bonded to one H → aldehyde (ethanal) ✓ …
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CBSE 2026Set A1 markMCQ
$Q. Which of the following is Isopropyl amine ?$
(a) CH3-CH2-CH2-NH2
(b) CH3-NH-C2H5
(c) CH3-CH(NH2)-CH3
(d) CH3-CH(CH3)-CH2-NH2
›Reveal solutionSolution

Isopropyl amine = propan-2-amine, CH3-CH(NH2)-CH3.

The isopropyl group is (CH3)2CH-, so isopropylamine has the amino group attached to the central carbon of a propane chain: CH3-CH(NH2)-CH3 (propan-2-amine). For reference: …

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CBSE 2026Set ANNUAL1 mark
$Q. How many structural isomers of C5H11Br are possible?$
›Reveal solutionSolution
$C_5H_{11}Br$ has 8 possible structural isomers, arising from bromine substitution at different positions on the three possible pentane carbon skeletons.

$C_5H_{11}Br$ is derived from pentane ( $C_5H_{12}$ ) by replacing one H with Br. Pentane itself has 3 carbon-skeleton isomers, and Br can go on different, non-equivalent carbon positions of each:

From n-pentane skeleton ( $CH_3CH_2CH_2CH_2CH_3$ ):

1-bromopentane

2-bromopentane

3-bromopentane (C4 and C5 positions are equivalent to C2 and C1 by the molecule's symmetry)

From isopentane / 2-methylbutane skeleton ( $(CH_3)_2CHCH_2CH_3$ ):

4. 1-bromo-2-methylbutane

5. 2-bromo-2-methylbutane …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Ethylidine dichloride is a:$
(a) vic-dihalide
(b) gem-dihalide
(c) allylic dihalide
(d) vinylic halide
›Reveal solutionSolution

Ethylidene dichloride, CH₃CHCl₂, has both chlorine atoms on the same carbon — a gem-dihalide.

Ethylidene dichloride has the structure CH₃–CHCl₂. Both chlorine atoms are attached to the same carbon atom (the second carbon). Dihalides in which both halogens sit on one carbon are called geminal (gem) dihalides; when they are on adjacent …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The correct IUPAC name of the organic compound CH₃—CH(C₂H₅)—CH₂Br is-$
(a)
(i) 1-Bromo-2-ethyl-2 methyl ethane
(b)
(ii) 1-Bromo-2-ethyl propane
(c)
(iii) 1-Bromo-2-methyl butane
(d)
(iv) 2-Methyl-1-bromo butane
›Reveal solutionSolution
$\text{CH}_3\text{–CH}(\text{C}_2\text{H}_5)\text{–CH}_2\text{Br}$ names as 1-bromo-2-methylbutane. Correct option: (iii).

Concept. IUPAC naming of a haloalkane: (1) pick the longest carbon chain that contains the carbon bearing the halogen; (2) number so the substituents get the lowest set of locants; (3) cite substituents alphabetically as prefixes.

Steps.

The structure is $\text{BrCH}_2\text{–CH(CH}_3\text{)–CH}_2\text{–CH}_3$ .

Longest chain through the C–Br carbon = 4 carbons (butane): $\text{C1(CH}_2\text{Br)} - \text{C2(CH)} - \text{C3(CH}_2\text{)} - \text{C4(CH}_3\text{)}$ .

Number from the Br end: bromo at C-1, methyl branch at C-2.

Name: 1-bromo-2-methylbutane. …
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CBSE 2025Set JZ1 markMCQ
$CH_2=CHCH_2NHCH_3$
(a) Allylmethylamine
(b) 1-amine-4-pentene
(c) 4-aminopent-1-ene
(d) N-methylprop-2-ene-1-amine
›Reveal solutionSolution
Naming $CH_2=CHCH_2NHCH_3$ as a substituted amine gives N-methylprop-2-en-1-amine — option (d).

Concept. For a secondary amine, choose the longest carbon chain attached to nitrogen as the parent amine; the smaller alkyl group on N is named as an N-substituent.

Working:

Parent chain: $CH_2=CH-CH_2-$ is a 3-carbon chain with a double bond → prop-2-ene; the amino group is on C-1 → prop-2-en-1-amine. …
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CBSE 2025Set D1 markMCQ
$Q. The number of isomeric alcohols of molecular formula C4H10O is$
(a) 2
(b) 4
(c) 7
(d) 8
›Reveal solutionSolution
The four alcohols of formula C4H10O are 1-butanol, 2-methyl-1-propanol, 2-butanol and 2-methyl-2-propanol.

For the molecular formula C4H10O with an -OH group, the possible alcohols are:

n-Butanol (butan-1-ol): CH3CH2CH2CH2OH (1°)

Isobutyl alcohol (2-methylpropan-1-ol): (CH3)2CHCH2OH (1°)

sec-Butyl alcohol (butan-2-ol): CH3CH2CH(OH)CH3 (2°) …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. The number of isomers in C2BrClFI is$
(a) 3
(b) 4
(c) 5
(d) 6
›Reveal solutionSolution

C2BrClFI, an ethylene bearing all four halogens (Br, Cl, F, I) with no hydrogens, has 6 possible isomers.

Since the formula has 2 carbons and exactly 4 substituents (Br, Cl, F, I) with no hydrogens, this corresponds to a fully-substituted ethylene, i.e. an alkene of type (X)(Y)C=C(Z)(W), where each carbon bears 2 of the 4 different halogens.

Step 1 — constitutional (positional) isomers: choose which 2 of the 4 halogens sit on one carbon (the other 2 automatically go on the other carbon). The number of distinct ways to split 4 different halogens into two unordered pairs is 3: {Br,Cl}|{F,I}, {Br,F}|{Cl,I}, {Br,I}|{Cl,F}.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. An isomer of ethanol is$
(a) Methanol
(b) Dimethyl ether
(c) Diethyl ether
(d) Ethylene glycol
›Reveal solutionSolution

Dimethyl ether (CH3–O–CH3) is the classic functional isomer of ethanol (C2H6O).

Ethanol (CH3CH2OH) has molecular formula C2H6O. Dimethyl ether (CH3–O–CH3) has the identical molecular formula C2H6O but a completely different functional group (ether linkage instead of an –OH group) — this is an example of functional group isomerism. Methanol (CH4O) and diethyl ether (C4H1 …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The total number of isomers for the compounds having molecular formula C4H10O is$
(a) 7
(b) 6
(c) 3
(d) 4
›Reveal solutionSolution

C4H10O has 7 total structural isomers: 4 alcohols + 3 ethers.

Alcohols (C4H9OH, 4 isomers): butan-1-ol, butan-2-ol, 2-methylpropan-1-ol (isobutanol), 2-methylpropan-2-ol (tert-butanol).

…

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CBSE 2023Set ANNUAL1 mark
$Q. How many isomeric monochloro derivatives will be formed when 2-methylpropane is subjected to photochlorination ?$
›Reveal solutionSolution
2-Methylpropane has only two chemically distinct kinds of hydrogen (9 equivalent primary H's and 1 tertiary H), so free-radical photochlorination gives exactly two monochloro isomers.

2-Methylpropane (isobutane), $(CH_3)_3CH$ , has two types of hydrogen atoms:

9 primary hydrogens (three equivalent $CH_3$ groups), all chemically equivalent by symmetry — substitution at any of these gives the same product: 1-chloro-2-methylpropane, $(CH_3)_2CHCH_2Cl$ .

1 tertiary hydrogen (on the central carbon) — substitution here gives: 2-chloro-2-methylpropane (tert-butyl chloride), $(CH_3)_3CCl$ . …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. CH2=CH-CH2-CH3 and CH3-CH=CH-CH3 are:$
(a) Chain isomers
(b) Position isomers
(c) Functional isomers
(d) Metamers
›Reveal solutionSolution

CH2=CH-CH2-CH3 (1-butene) and CH3-CH=CH-CH3 (2-butene) share the same carbon skeleton and functional group but differ only in where the double bond sits — that is positional isomerism.

Both molecules have molecular formula $C_4H_8$ and the same unbranched 4-carbon chain, and both are alkenes (same functional group, so not functional isomers; same chain, so not chain isomers).

…

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←Previous questionQ59 — Write down the structure and IUPAC name for neo-pentylbromide.→Next questionQ61 — Name the alkene which will yield 1-chloro-1-methylcyclohexane by its r…

Q.A hydrocarbon of molecular mass 72 g mol−1^{-1}−1 gives a single monochloro derivative and two dichloro derivatives on photo chlorination. Give the structure of the hydrocarbon.

Why This Approach Works

Step-by-Step Solution

1. Determine the molecular formula

2. Analyze monochlorination — why only one product?

3. Analyze dichlorination — why exactly two products?

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Q.A hydrocarbon of molecular mass 72 g mol $^{-1}$ gives a single monochloro derivative and two dichloro derivatives on photo chlorination. Give the structure of the hydrocarbon.

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