Q.Write the isomers of the compound having formula C4H9Br.
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🔒 Start your 14-day free trial to unlock the full solution →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.
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 |
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 C4H10, the formula satisfies 4(4)+10(1)=26 valence electrons. But the carbon atoms can be arranged as:
- A straight chain: CH3−CH2−CH2−CH3 (n-butane)
- A branched chain: CH3−CH(CH3)−CH3 (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:
- C4H10: 2 structural isomers
- C5H12: 3 structural isomers
- C6H14: 5 structural isomers
- C10H22: 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 C5H12:
- 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 CnH2n+2O), the position of the -OH group creates isomers:
- CH3CH2CH2OH (propan-1-ol) — OH at end
- CH3CH(OH)CH3 (propan-2-ol) — OH in middle …
The key idea is structural isomerism — specifically, chain and position isomerism for a monosubstituted butane.
Step 1: The formula C4H9Br corresponds to a butane chain (C4H10) with one hydrogen replaced by bromine. The number of isomers equals the number of distinct carbon skeletons and distinct positions for Br.
Step 2: Butane has two carbon skeletons:
- n-butane (straight chain): CH3−CH2−CH2−CH3 — Br can attach at C1 or C2, giving two isomers. …
The key idea is to systematically draw all distinct carbon skeletons for 4 carbons and place the bromine atom at every unique position, avoiding duplicates. The compound C4H9Br has 4 structural isomers.
Why this approach works
Structural isomerism (also called constitutional isomerism) arises when molecules have the same molecular formula but different connectivity — different arrangements of atoms. For C4H9Br, the bromine atom can be attached to different carbon atoms, and the carbon chain itself can be straight or branched. The trick is to first sketch all possible carbon skeletons for 4 carbons, then place the bromine at every chemically distinct carbon position. Two placements that look different on paper but are actually the same molecule (by symmetry or rotation) must be counted only once.
Step-by-step reasoning
-
Draw all carbon skeletons for 4 carbons.
There are two distinct skeletons:
- A straight chain of 4 carbons: C−C−C−C (butane skeleton).
- A branched chain with a methyl group on the middle carbon: C−C(C)−C (isobutane skeleton). No other skeletons exist for 4 carbons.
-
Place bromine on the straight chain skeleton.
Number the carbons 1, 2, 3, 4 from one end.
- Bromine on carbon 1: CH3−CH2−CH2−CH2Br (1-bromobutane).
- Bromine on carbon 2: CH3−CHBr−CH2−CH3 (2-bromobutane).
- Bromine on carbon 3: This is the same as carbon 2 by symmetry (flipping the chain end-to-end gives the same molecule). So no new isomer.
- Bromine on carbon 4: Same as carbon 1. So from the straight chain, we get 2 isomers.
-
Place bromine on the branched skeleton.
The only branched skeleton for C4 is isobutane, CH3−CH(CH3)−CH3: a central tertiary carbon (C2) bonded to one hydrogen and three other carbons — two methyl groups (C1, C3) plus the branch methyl (C4). Counting confirms 4 carbons total (C1, C2, C3, C4). The skeleton is:
C1 - C2 - C3 | C4where C2 is the central carbon.
Now place bromine at each distinct carbon:
- Bromine on C1 (a primary carbon): CH2Br−CH(CH3)−CH3 (1-bromo-2-methylpropane).
- Bromine on C2 (the tertiary carbon): CH3−CBr(CH3)−CH3 (2-bromo-2-methylpropane).
- Bromine on C3: This is equivalent to C1 by symmetry (the two methyl groups on the ends are identical). So no new isomer. …
Method: Carbon Skeleton + Functional Group Placement Method
This method systematically generates all structural isomers by:
- First drawing all possible carbon skeletons (straight and branched chains)
- Then placing the bromine atom at every unique carbon position
Step 1: Draw all possible carbon skeletons for 4 carbons
There are two distinct skeletons:
-
Straight chain (butane):
C−C−C−C
-
Branched chain (isobutane / 2-methylpropane):
C−C−C∣C
Step 2: Place the bromine atom at each unique carbon position
Skeleton 1: Straight chain (n-butane backbone)
Number the carbons:
C1−C2−C3−C4
- Br at C1 → 1-bromobutane
- Br at C2 → 2-bromobutane
- Br at C3 → same as C2 (symmetry)
- Br at C4 → same as C1 (symmetry)
Two isomers from this skeleton.
Skeleton 2: Branched chain (isobutane backbone)
Structure:
C1−C2−C3∣C4
- Br at C1 → 1-bromo-2-methylpropane
- Br at C2 → 2-bromo-2-methylpropane
- Br at C3 → same as C1 (symmetry)
- Br at C4 → same as C1 (symmetry)
Two isomers from this skeleton.
Step 3: Count total distinct isomers
| Skeleton | Isomers |
|----------|---------| …
This is a classic exam favourite. Let's break it down.
🧠 The Core Concept First
The formula C4H9Br has zero degrees of unsaturation — it is fully saturated, just like C4H10 with one H replaced by Br (DoU=22(4)+2−9−1=0). So all isomers will be structural isomers — no rings or double bonds.
The key is to systematically vary the carbon skeleton and the position of Br.
✓ The Correct Isomers (for reference)
There are 4 structural isomers:
-
1-Bromobutane (straight chain, Br at end)
CH3CH2CH2CH2Br
-
2-Bromobutane (straight chain, Br on C2)
CH3CH2CHBrCH3
-
1-Bromo-2-methylpropane (branched chain, Br at end)
(CH3)2CHCH2Br
-
2-Bromo-2-methylpropane (branched chain, Br on tertiary carbon)
(CH3)3CBr
✗ Common Mistakes & How to Avoid Them
1. Counting the same isomer twice (due to symmetry)
- Mistake: Writing both CH3CH2CHBrCH3 and CH3CHBrCH2CH3 as different isomers.
- Why it's wrong: The carbon chain is symmetric — numbering from either end gives the same molecule.
- How to avoid: Always number the carbon chain from the end nearest the Br (or functional group). If the Br is on carbon 2 from both ends, it's the same compound.
2. Forgetting branched-chain isomers
- Mistake: Only writing straight-chain isomers (1-bromobutane and 2-bromobutane).
- Why it's wrong: C4 can form a branched skeleton (isobutane).
- How to avoid: Always draw all possible carbon skeletons first — for C4, there are two: n-butane and isobutane (2-methylpropane). Then place Br on each unique carbon.
3. Creating a "new" isomer by rotating the molecule
- Mistake: Writing CH3CHBrCH2CH3 and CH3CH2CHBrCH3 as different.
- Why it's wrong: These are the same molecule (2-bromobutane) — just drawn backwards.
- How to avoid: If you can rotate the molecule in space to match another structure, it's not a new isomer. Use systematic naming to check.
4. Adding a branch that creates a longer chain
- Mistake: Writing CH3CH(Br)CH2CH3 as a "branched" isomer.
- Why it's wrong: That's just 2-bromobutane — the longest chain is still 4 carbons.
- How to avoid: Always identify the longest continuous carbon chain first. If it's 4 carbons, it's a butane derivative, not a branched one.
5. Missing the tertiary bromide isomer
- Mistake: Writing only 1-bromo-2-methylpropane but forgetting 2-bromo-2-methylpropane. …
Showing the 12 most recent of 17 on this concept.
- CBSE 2026Set DZ1 markMCQQ.Which of the following is an aldehyde?(a) CH3−C∣∣O−H(b) CH3−CH2−C∣∣O−CH3(c) CH3−CH2−C∣∣O−CH2−CH3(d) CH3−C∣∣O−CH3
›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) CH3−CHO → carbonyl C bonded to one H → aldehyde (ethanal) ✓ …
- CBSE 2026Set A1 markMCQQ.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: …
- CBSE 2026Set ANNUAL1 markQ.How many structural isomers of C5H11Br are possible?
›Reveal solutionSolution
C5H11Br has 8 possible structural isomers, arising from bromine substitution at different positions on the three possible pentane carbon skeletons.
C5H11Br is derived from pentane (C5H12) 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 (CH3CH2CH2CH2CH3):
- 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 ((CH3)2CHCH2CH3):
4. 1-bromo-2-methylbutane
5. 2-bromo-2-methylbutane …
- CBSE 2026Set ANNUAL1 markMCQQ.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 …
- CBSE 2026Set ANNUAL1 markMCQQ.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
CH3–CH(C2H5)–CH2Br 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 BrCH2–CH(CH3)–CH2–CH3.
- Longest chain through the C–Br carbon = 4 carbons (butane): C1(CH2Br)−C2(CH)−C3(CH2)−C4(CH3).
- Number from the Br end: bromo at C-1, methyl branch at C-2.
- Name: 1-bromo-2-methylbutane. …
- CBSE 2025Set JZ1 markMCQQ.The correct IUPAC name for CH2=CHCH2NHCH3 is :(a) Allylmethylamine(b) 1-amine-4-pentene(c) 4-aminopent-1-ene(d) N-methylprop-2-ene-1-amine
›Reveal solutionSolution
Naming CH2=CHCH2NHCH3 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: CH2=CH−CH2− is a 3-carbon chain with a double bond → prop-2-ene; the amino group is on C-1 → prop-2-en-1-amine. …
- CBSE 2025Set D1 markMCQQ.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°) …
- CBSE 2025Set ANNUAL1 markMCQQ.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}.
…
- CBSE 2024Set ANNUAL1 markMCQQ.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 …
- CBSE 2024Set ANNUAL1 markMCQQ.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).
…
- CBSE 2023Set ANNUAL1 markQ.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), (CH3)3CH, has two types of hydrogen atoms:
- 9 primary hydrogens (three equivalent CH3 groups), all chemically equivalent by symmetry — substitution at any of these gives the same product: 1-chloro-2-methylpropane, (CH3)2CHCH2Cl.
- 1 tertiary hydrogen (on the central carbon) — substitution here gives: 2-chloro-2-methylpropane (tert-butyl chloride), (CH3)3CCl. …
- CBSE 2023Set ANNUAL1 markMCQQ.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 C4H8 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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