Q.Write structures of different dihalogen derivatives of propane.
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
Structural isomers can have wildly different properties. Ethanol (C₂H₆O) is a drinkable alcohol; its isomer dimethyl ether is a gas used as a refrigerant. Same atoms, but one is a liquid you can consume, the other is a gas that would kill you. That is why chemists care so much about connectivity — it determines everything.
Quick Check
Question: Are these structural isomers?
Molecule A: CH₃–CH₂–CH₂–CH₃
Molecule B: CH₃–CH(CH₃)–CH₃
Answer: Yes. Both are C₄H₁₀. A is n-butane (straight chain), B is isobutane (branched). Different connectivity → structural isomers.
The molecular formula must be identical. If the formulas differ, they are not isomers at all — just different compounds.
Structural isomerism is introduced in the NCERT/CBSE Class 11 Chemistry chapter on Organic Chemistry: Some Basic Principles and Techniques, and ‘structural isomerism examples class 11’ is a frequently searched important-question topic for board exams, JEE Main and NEET. Correctly distinguishing structural isomers by connectivity, rather than just matching molecular formulas, is a skill tested throughout competitive organic chemistry exams.
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
Why these are distinct: The OH group's position changes the carbon's hybridization environment and the molecule's polarity.
The Ring-Chain Isomerism Reason
For unsaturated formulas like C4H8, the same formula can represent:
- A straight alkene: CH2=CH−CH2−CH3
- A branched alkene: CH3−C(=CH2)−CH3
- A cycloalkane: cyclobutane (ring)
Why rings form: Carbon atoms can bond to form closed loops, reducing the number of hydrogen atoms needed. The formula CnH2n can be either an alkene (one double bond) or a cycloalkane (one ring).
Summary: The Takeaway
| Aspect | Why It Holds |
|---|---|
| Different connectivity | Atoms can bond in multiple sequences while satisfying valency |
| No simple counting formula | The number of possible trees grows combinatorially |
| Branching creates isomers | Carbon chains can have branches at different positions |
| Position matters | Functional groups at different locations change properties |
| Rings vs. chains | Same formula can represent open chains or closed rings |
The key insight: Structural isomerism exists because molecular formula is a constraint, not a blueprint — it tells you the ingredients, not the recipe.
Concept: Structural Isomerism — different arrangements of the same molecular formula, here C3H6X2 (where X is a halogen).
Reasoning:
- Propane has a three-carbon chain. Dihalogen derivatives replace two H atoms with halogen atoms.
- The two halogen atoms can be on the same carbon (geminal) or on different carbons (vicinal or terminal).
- For propane, the possible positions are: both on C-1, one on C-1 and one on C-2, both on C-2, or one on C-1 and one on C-3.
Structures:
- 1,1-Dihalopropane: CH3−CH2−CHX2
- 1,2-Dihalopropane: CH3−CHX−CH2X
- 1,3-Dihalopropane: XCH2−CH2−CH2X
- 2,2-Dihalopropane: CH3−CX2−CH3
The four structural isomers are 1,1-, 1,2-, 1,3-, and 2,2-dihalopropane.
The key idea is to systematically replace two hydrogen atoms on the propane chain with halogen atoms, considering both the carbon skeleton and the positions of the halogens. This gives four distinct structural isomers: 1,1-dichloropropane, 1,2-dichloropropane, 1,3-dichloropropane, and 2,2-dichloropropane.
Why This Approach Works
Structural isomerism arises when molecules have the same molecular formula but different connectivity — how atoms are bonded to each other. For dihalogen derivatives of propane (C₃H₆X₂, where X is a halogen like Cl or Br), the two halogen atoms can be placed on the same carbon or on different carbons. The propane chain itself is unbranched (three carbons in a row), so no chain isomerism is possible here. The only variation comes from the position of the two halogen atoms.
A common mistake is to think that placing both halogens on the middle carbon gives the same compound as placing them on an end carbon — but these are different because the carbon atoms are not equivalent. The two end carbons (C1 and C3) are equivalent by symmetry, so we must be careful not to double-count.
Let’s work through the possibilities step by step.
-
Identify the carbon skeleton of propane.
Propane is CH3−CH2−CH3. Number the carbons: C1 (end), C2 (middle), C3 (other end). C1 and C3 are equivalent due to symmetry.
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Case 1: Both halogens on the same carbon.
- If both are on C1 (or equivalently C3), we get 1,1-dichloropropane: CH3−CH2−CHCl2 (The two Cl atoms are on the terminal carbon.)
- If both are on C2, we get 2,2-dichloropropane: CH3−CCl2−CH3 (Both Cl atoms on the middle carbon.)
These are two distinct isomers because the carbon environment differs.
-
Case 2: Halogens on different carbons.
- Place one Cl on C1 and the other on C2: this gives 1,2-dichloropropane: CH3−CHCl−CH2Cl (Note: the carbon chain remains straight; the Cl atoms are on adjacent carbons.)
- Place one Cl on C1 and the other on C3: this gives 1,3-dichloropropane: CH2Cl−CH2−CH2Cl (The Cl atoms are on the two end carbons, separated by one carbon.)
Could we also have 2,3-dichloropropane? That would be the same as 1,2-dichloropropane because numbering from the other end makes C2 and C3 equivalent to C1 and C2. So no new isomer.
-
Count the total distinct isomers.
We have:
- 1,1-dichloropropane
- 1,2-dichloropropane
- 1,3-dichloropropane
- 2,2-dichloropropane
That’s four structural isomers.
Do not confuse 2,3-dichloropropane with a new isomer — it is identical to 1,2-dichloropropane because the chain is symmetric. Always check for equivalent positions.
A quick way to generate all isomers: list all pairs of carbon numbers (1,1), (1,2), (1,3), (2,2). Since (2,3) is same as (1,2) and (3,3) is same as (1,1), you get exactly four.
The four structural isomers of dihalogen derivatives of propane are 1,1-dichloropropane, 1,2-dichloropropane, 1,3-dichloropropane, and 2,2-dichloropropane.
Method: Systematic Carbon Skeleton + Functional Group Placement
This method works by first fixing the carbon skeleton, then placing the two halogen atoms at all unique positions — avoiding duplicates.
Steps
1. Draw the carbon skeleton of propane
Propane has 3 carbon atoms in a straight chain:
C – C – C
Number the carbons:
C1 – C2 – C3
2. Identify all possible positions for two halogen atoms (X = Cl, Br, etc.)
We place two X atoms on the 3 carbons. The key is to consider all unique combinations of positions.
Possible position pairs (C1, C2, C3):
- (1,1) — both on same carbon
- (1,2) — on adjacent carbons
- (1,3) — on terminal carbons (1 apart)
- (2,2) — both on middle carbon
- (2,3) — same as (1,2) by symmetry
- (3,3) — same as (1,1) by symmetry
So the unique position pairs are: (1,1), (1,2), (1,3), and (2,2).
3. Draw the structures for each unique pair
(a) 1,1-dihalogenopropane — both halogens on C1
X
|
C – C – C
|
X
(Also called geminal dihalide)
(b) 1,2-dihalogenopropane — halogens on C1 and C2
X X
| |
C – C – C
(Also called vicinal dihalide)
(c) 1,3-dihalogenopropane — halogens on C1 and C3
X X
| |
C – C – C
(d) 2,2-dihalogenopropane — both halogens on C2
X
|
C – C – C
|
X
(Another geminal dihalide)
Final Answer
There are 4 structural isomers of dihalogen derivatives of propane:
| S.No. | IUPAC Name | Structure (X = halogen) |
|---|---|---|
| 1 | 1,1-dihalogenopropane | CH3–CH2–CHX2 |
| 2 | 1,2-dihalogenopropane | CH3–CHX–CH2X |
| 3 | 1,3-dihalogenopropane | XCH2–CH2–CH2X |
| 4 | 2,2-dihalogenopropane | CH3–CX2–CH3 |
Key insight: The method avoids duplicates by recognizing molecular symmetry — positions (2,3) and (3,3) are identical to (1,2) and (1,1) respectively.
Here is a breakdown of the common mistakes students make when tackling the question: "Write structures of different dihalogen derivatives of propane," along with the precise reasoning to avoid them.
The Core Concept (The "Why")
Before listing mistakes, understand the goal. Propane is a 3-carbon chain (C3H8). A "dihalogen derivative" means two hydrogen atoms are replaced by two halogen atoms (like Cl, Br, I). The key is to find all unique structural isomers — molecules with the same formula but different connectivity of atoms.
The formula for a dihalogen derivative of propane is C3H6X2 (where X is the halogen).
Mistake #1: Forgetting the Terminal Carbon Positions are Different
The Mistake: Students often think that putting both halogens on the two end carbons (C1 and C3) is the same as putting them on C1 and C2. They treat the chain as symmetrical in a way that ignores the distance between the halogens.
Why it’s wrong: Propane is symmetrical, but the positions are numbered. C1 and C3 are equivalent (both are terminal). However, 1,2-dihalopropane (halogens on adjacent carbons) is a different molecule from 1,3-dihalopropane (halogens on carbons separated by one carbon). They have different physical properties (e.g., boiling points, reactivity).
How to Avoid:
- Number the carbon chain (1-2-3).
- Systematically place the two halogens on every possible pair of carbon numbers: (1,1), (1,2), (1,3), (2,2), (2,3), (3,3).
- Eliminate duplicates due to symmetry (e.g., (2,3) is the same as (3,2) when read from the other end). The unique pairs are: (1,1), (1,2), (1,3), (2,2).
Mistake #2: Missing the "Geminal" vs. "Vicinal" Distinction
The Mistake: Students write only one structure for "both halogens on the same carbon" (geminal) and one for "halogens on adjacent carbons" (vicinal), but they forget that the same carbon can be at the end (C1) or in the middle (C2).
Why it’s wrong: A geminal dihalide on C1 (1,1-dihalopropane) is structurally different from a geminal dihalide on C2 (2,2-dihalopropane). The central carbon (C2) is attached to two other carbons, while C1 is attached to only one. This changes the molecule's shape and stability.
How to Avoid:
- Draw both geminal possibilities explicitly:
- 1,1-dihalopropane: CH3−CH2−CHX2
- 2,2-dihalopropane: CH3−CX2−CH3
- Draw both vicinal possibilities explicitly:
- 1,2-dihalopropane: CH2X−CHX−CH3
- 1,3-dihalopropane: CH2X−CH2−CH2X
Mistake #3: Confusing "Dihalogen" with "Dihalide" or "Alkyl Halide" Naming
The Mistake: Students write the correct structure but label it with the wrong IUPAC name (e.g., calling 1,2-dichloropropane as "propylene dichloride" without realizing that name is ambiguous).
Why it’s wrong: In exams, the question asks for "structures," but if you write a name, it must be precise. Common names like "propylene dichloride" can refer to either 1,2-dichloropropane or 1,3-dichloropropane. This shows a lack of clarity.
How to Avoid:
- Always use IUPAC numbering in your mind when drawing.
- Write the condensed structural formula clearly, e.g., CH3−CHCl−CH2Cl (for 1,2-dichloropropane) and CH2Cl−CH2−CH2Cl (for 1,3-dichloropropane).
- Do not rely on common names unless the question explicitly asks for them.
Mistake #4: Forgetting Optical Isomerism (Advanced but Common)
The Mistake: Students stop at 4 structural isomers (1,1; 1,2; 1,3; 2,2). They forget that 1,2-dihalopropane has a chiral carbon (the middle carbon, C2, is attached to four different groups: H, X, CH3, and CH2X). This means it exists as a pair of enantiomers (optical isomers).
Why it’s wrong: The question asks for "different dihalogen derivatives." Enantiomers are different molecules (they rotate plane-polarized light in opposite directions). If the exam expects you to list all isomers, missing optical isomers is a loss of marks.
How to Avoid:
- Check each carbon for chirality: A carbon with four different substituents is chiral.
- For 1,2-dihalopropane (CH2X−CHX−CH3), C2 has: H, X, CH3, and CH2X. All four are different. Therefore, it has two optical isomers (R and S).
- Draw both as wedge-dash structures (or write "d/l pair" or "racemic mixture").
- Note: 1,3-dihalopropane and the geminal ones have no chiral carbons.
The Complete, Correct Answer (Summary)
For propane (C3H8), the dihalogen derivatives (C3H6X2) are:
- 1,1-dihalopropane: CH3−CH2−CHX2
- 2,2-dihalopropane: CH3−CX2−CH3
- 1,2-dihalopropane: CH2X−CHX−CH3 (exists as two optical isomers)
- 1,3-dihalopropane: CH2X−CH2−CH2X
Total distinct structural isomers (including stereoisomers): 5 (4 structural + 1 optical pair counted as 2).
Quick Checklist to Avoid Mistakes
- Number the chain (1-2-3).
- List all unique carbon pairs: (1,1), (1,2), (1,3), (2,2).
- Draw each structure with correct connectivity.
- Check for chirality in any carbon with four different groups.
- Use IUPAC names in your mind to verify uniqueness.
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′).
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(a) CH3−CHO → carbonyl C bonded to one H → aldehyde (ethanal) ✓
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(b) CH3CH2−CO−CH3 → butan-2-one → ketone
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(c) CH3CH2−CO−CH2CH3 → pentan-3-one → ketone
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(d) CH3−CO−CH3 → propanone (acetone) → ketone
✓Final answer(a) CH3CHO (acetaldehyde/ethanal).
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- 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:
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(a) CH3-CH2-CH2-NH2 is n-propylamine (propan-1-amine)
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(b) CH3-NH-C2H5 is N-methylethylamine (a secondary amine)
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(d) is isobutylamine
✓Final answer(c) CH3-CH(NH2)-CH3.
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- 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
6. 3-bromo-2-methylbutane
7. 1-bromo-3-methylbutane (isoamyl bromide)
From neopentane / 2,2-dimethylpropane skeleton ((CH3)3CCH3, i.e. (CH3)4C):
8. Neopentyl bromide (1-bromo-2,2-dimethylpropane)
Of these 8: 4 are primary bromides, 3 are secondary, and 1 is tertiary.
✓Final answer8 structural isomers.
- 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 carbons they are called vicinal (vic) dihalides. Since both Cl atoms here are on one carbon, this is a gem-dihalide.
✓Final answer(b) gem-dihalide.
- 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.
Why other options are wrong. (i) and (ii) use "ethane/propane" as the parent — they fail to select the longest chain (butane). (iv) "2-methyl-1-bromobutane" mis-orders the substituents (alphabetical order requires bromo before methyl).
✓Final answer(iii) 1-Bromo-2-methylbutane. This IUPAC nomenclature is common to the NCERT/CBSE-aligned UBSE Class-12 Chemistry course.
- 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.
- The nitrogen also carries a −CH3, written as the prefix N-methyl.
Combining: N-methylprop-2-en-1-amine (the common name is allylmethylamine, but the IUPAC name is required).
✓Final answer(d) N-methylprop-2-ene-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°)
- tert-Butyl alcohol (2-methylpropan-2-ol): (CH3)3COH (3°)
That gives 4 isomeric alcohols. (The remaining C4H10O isomers such as diethyl ether are ethers, not alcohols.)
✓Final answer(B) 4 isomeric alcohols of formula C4H10O.
- 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}.
Step 2 — geometric (cis/trans) isomers: for each of these 3 constitutional arrangements, since each carbon carries 2 different substituents, the molecule can exist as 2 geometric isomers (cis and trans, i.e. Z and E).
Total isomers = 3 (constitutional arrangements) x 2 (geometric isomers each) = 6.
✓Final answer(D) 6.
- 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 (C4H10O) have different molecular formulas, so they are not isomers of ethanol; ethylene glycol (C2H6O2) also has a different formula.
✓Final answer(B) Dimethyl ether.
- 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).
Ethers (C4H10O, 3 isomers): diethyl ether (C2H5–O–C2H5), methyl n-propyl ether (CH3–O–CH2CH2CH3), methyl isopropyl ether (CH3–O–CH(CH3)2).
Total = 4 + 3 = 7 constitutional isomers.
✓Final answer(A) 7.
- 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.
Since photochlorination is a free-radical substitution that can occur at any C–H bond, but chemically equivalent hydrogens always give the identical product, the number of distinct isomeric monochloro products depends only on the number of structurally different types of hydrogen, not the total count of hydrogens.
✓Final answer2 isomeric monochloro derivatives: 1-chloro-2-methylpropane (from the primary H's) and 2-chloro-2-methylpropane / tert-butyl chloride (from the tertiary H).
- 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).
The only difference is the location of the C=C double bond: between C1–C2 in the first compound (1-butene) versus between C2–C3 in the second (2-butene). This difference in position of the double bond (or functional group) along an otherwise identical skeleton defines position isomers.
✓Final answerPosition isomers (option b).
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