Write a test to detect the presence of double bond in a molecule.

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Q. Write a test to detect the presence of double bond in a molecule.

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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 key idea is that a double bond adds unsaturation to a molecule, making it reactive toward addition reactions. The classic test is the bromine water test — if the reddish-brown colour of bromine water is decolourised when added to the compound, a double bond is present. The final result: decolourisation of bromine water confirms a double bond.

Why this approach works

A double bond consists of one sigma ( $\sigma$ ) bond and one pi ( $\pi$ ) bond. The pi bond is weaker and more exposed than a sigma bond, making it susceptible to electrophilic attack. In the bromine water test, bromine ( $\text{Br}_2$ ) acts as an electrophile — it adds across the double bond in a characteristic addition reaction. This consumes the bromine, removing its colour.

The reaction is:

\text{R–CH=CH–R} + \text{Br}_2 \rightarrow \text{R–CHBr–CHBr–R}

The reddish-brown colour of bromine water comes from molecular bromine. When bromine adds to the double bond, it is no longer present as free $\text{Br}_2$ , so the colour disappears. This visual change is the basis of the test.

Watch out

A common mistake is to assume that any decolourisation means a double bond. But alkynes (triple bonds) and phenols (aromatic compounds with –OH groups) also decolourise bromine water. So the test is not exclusive to double bonds — it indicates unsaturation in general. For a double bond specifically, you must also rule out a triple bond (e.g., by a separate test like ammoniacal silver nitrate).

Step-by-step procedure

Prepare the sample — Take a small amount of the organic compound (about 0.5 mL or a few crystals) in a clean, dry test tube. If the compound is a solid, dissolve it in a few drops of a non-reactive solvent like carbon tetrachloride ( $\text{CCl}_4$ ) or chloroform. Do not use water if the compound is insoluble in water — the test works best in a homogeneous solution.
Add bromine water — Add 2–3 drops of bromine water (a dilute, reddish-brown solution of $\text{Br}_2$ in water) to the test tube. Shake gently to mix.
Observe the colour change — Look for the disappearance of the reddish-brown colour. If the colour fades or vanishes completely within a few seconds to a minute, a double bond is likely present. If the colour persists, the compound is saturated (no double bond).
Confirm with a control — Always run a blank test with the solvent alone (no compound) to ensure the bromine water itself is not decolourised by impurities or light. Also test a known saturated compound (e.g., hexane) — it should not decolourise bromine water.

Tip

For a quicker and more sensitive test, use bromine in carbon tetrachloride instead of bromine water. The colour change is sharper because the reaction medium is non-polar, and side reactions (like substitution) are minimised. This is the standard method in many lab manuals.

Why the colour disappears — the chemistry

Bromine water is reddish-brown because of the presence of $\text{Br}_2$ molecules. When a double bond is present, the following addition reaction occurs: …

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Common Mistakes: Testing for a Double Bond (Unsaturation)

Students often lose marks on this seemingly simple question. Here are the most frequent errors and how to avoid each.

Mistake 1: Naming the Wrong Reagent

The error: Students write "bromine water test" but then describe using chlorine water, potassium permanganate, or even iodine solution.

Why it happens: Confusion between halogen tests (Br₂, Cl₂, I₂) and oxidation tests (KMnO₄).

How to avoid: Memorise the two standard tests for unsaturation:

Bromine water test — decolourisation of reddish-brown Br₂
Baeyer's test — decolourisation of purple alkaline KMnO₄

Key fact: Only bromine water is the specific test for a double bond. Baeyer's test is for unsaturation generally (double or triple bonds).

Mistake 2: Describing the Wrong Colour Change

The error: "Bromine water turns from brown to colourless" — but bromine water is reddish-brown, not brown.

The error: "KMnO₄ turns from purple to colourless" — but in Baeyer's test, KMnO₄ turns from purple to brown (MnO₂ precipitate), not colourless.

How to avoid: Learn the exact colours:

Test	Initial Colour	Final Colour
Br₂ water	Reddish-brown	Colourless
Baeyer's (KMnO₄)	Purple	Brown precipitate

Mistake 3: Forgetting to Mention the Observation

The error: Students write "Add bromine water" and stop — no observation, no conclusion.

How to avoid: Always write the full sequence:

Procedure: Add a few drops of bromine water to the compound.
Observation: The reddish-brown colour of bromine water is discharged (decolourised).
Conclusion: This indicates the presence of a double bond (addition reaction occurs).

Mistake 4: Confusing Addition with Substitution

The error: "Bromine water decolourises because bromine substitutes hydrogen atoms."

Why it happens: Students mix up alkene addition (C=C + Br₂ → dibromo compound) with alkane substitution (C-H + Br₂ → C-Br + HBr).

How to avoid: Remember the mechanism:

Double bond: Addition — Br₂ adds across the π-bond → decolourisation is immediate
Saturated compound: Substitution — requires UV light or heat → decolourisation is slow or absent

Exam tip: If a question asks "test for unsaturation", the answer is always about addition, not substitution.

Mistake 5: Writing an Incomplete Chemical Equation

The error: Writing only R-CH=CH-R + Br₂ → without showing the product.

How to avoid: Write the complete addition reaction: …

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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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Q.Write a test to detect the presence of double bond in a molecule.

Concept understanding — Structural Isomerism

Structural Isomerism: The First Meeting

The Precise Statement

A Concrete Example: C₄H₁₀

The Three Main Types

Why This Matters …

Why this formula?

Structural Isomerism: Why the Key Ideas Hold

The Core Principle: Connectivity ≠ Composition

Why This Happens: The Valency Constraint

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

Why No Simple Formula Exists

Key Reasoning: The Branching Principle

The Functional Group Position Rule

Why this approach works

Step-by-step procedure

Why the colour disappears — the chemistry

Concept: Unsaturation Test (Bromine Water / Baeyer’s Test)

Method: Bromine Water Test

Principle

Steps

Common Mistakes: Testing for a Double Bond (Unsaturation)

Mistake 1: Naming the Wrong Reagent

Mistake 2: Describing the Wrong Colour Change

Mistake 3: Forgetting to Mention the Observation

Mistake 4: Confusing Addition with Substitution

Mistake 5: Writing an Incomplete Chemical Equation

Q.Write a test to detect the presence of double bond in a molecule.

Why this approach works

Step-by-step procedure

Why the colour disappears — the chemistry

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