Primary alkyl halide C₄H₉Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with

Exercises · 6.21

C_4H_9Br

(a) reacted with alcoholic KOH to give compound (b). Compound

(b) is reacted with HBr to give

(a) is reacted with sodium metal it gives compound (d),

C_8H_{18}

which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of

(a) and write the equations for all the reactions.

CBSENCERTSubjective· 5mImportance★★★★★

27% · 39/147 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

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.

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 …

Start your 14-day free trial to unlock the full solution →

The key is that (a) is a primary alkyl halide with formula $C_4H_9Br$ , but its reaction with sodium gives a $C_8H_{18}$ product different from that of n-butyl bromide — this forces (a) to be isobutyl bromide. The sequence then proceeds through elimination, addition, and Wurtz coupling.

Let’s unpack this step by step. The problem is about structural isomerism and how different isomers behave differently in reactions. The formula $C_4H_9Br$ has four isomers, of which only two are primary — and only one fits the entire puzzle.

1. Identify the possible primary alkyl halides with formula $C_4H_9Br$

A primary alkyl halide has the bromine attached to a terminal carbon. The four isomers of $C_4H_9Br$ are:

n-Butyl bromide: $CH_3CH_2CH_2CH_2Br$
Isobutyl bromide: $(CH_3)_2CHCH_2Br$
sec-Butyl bromide: $CH_3CH_2CHBrCH_3$ — but this is secondary, not primary, so it’s out.
tert-Butyl bromide: $(CH_3)_3CBr$ — tertiary, also out.

So only two primary isomers exist: n-butyl bromide and isobutyl bromide. The problem says (a) is primary, so (a) must be one of these two.

2. Reaction of (a) with alcoholic KOH gives (b) — an elimination

Alcoholic KOH favours elimination (dehydrohalogenation) over substitution. For a primary halide, the major product is the more substituted alkene (Saytzeff’s rule), but with only one possible alkene from each:

n-Butyl bromide gives 1-butene: $CH_3CH_2CH=CH_2$
Isobutyl bromide gives 2-methylpropene: $(CH_3)_2C=CH_2$

So (b) is either 1-butene or 2-methylpropene.

3. Compound (b) reacts with HBr to give (c), an isomer of (a)

Addition of HBr to an alkene follows Markovnikov’s rule — the hydrogen adds to the less substituted carbon, bromine to the more substituted.

If (b) is 1-butene: HBr adds to give 2-bromobutane (secondary), which has formula $C_4H_9Br$ but is not a primary halide — it’s an isomer of (a), but (a) is primary. That’s fine: (c) just needs to be an isomer, not necessarily primary.
If (b) is 2-methylpropene: HBr adds to give tert-butyl bromide (tertiary), also an isomer of (a).

Both possibilities give an isomer of (a). So this step alone doesn’t decide.

4. Reaction of (a) with sodium metal gives (d), $C_8H_{18}$ — the Wurtz reaction

The Wurtz reaction couples two alkyl halides with sodium:

2RBr + 2Na \rightarrow R-R + 2NaBr

For n-butyl bromide, the product is n-octane: $CH_3(CH_2)_6CH_3$

For isobutyl bromide, the product is 2,5-dimethylhexane: $(CH_3)_2CHCH_2CH_2CH(CH_3)_2$

The problem states that (d) is different from the compound formed when n-butyl bromide is reacted with sodium. That means (a) cannot be n-butyl bromide — because if it were, (d) would be n-octane, which is exactly what n-butyl bromide gives. So (a) must be isobutyl bromide.

Watch out

A common mistake is to assume that the Wurtz product from isobutyl bromide is the same as from n-butyl bromide — but they are structural isomers. n-Octane is a straight chain; 2,5-dimethylhexane is branched. They are different compounds.

--- …

Start your 14-day free trial to unlock the full solution →

Method: Retrograde Analysis with Reaction Pathway Mapping

This method works backwards from the given products and constraints to deduce the unknown starting structure.

Step 1 — Identify the key constraint

We are told:

(a) is a primary alkyl halide with formula $\ce{C4H9Br}$ .
(c) is an isomer of (a) — meaning same formula but different structure.
(d) is $\ce{C8H18}$ from Wurtz reaction of (a), and it is different from the product of n-butyl bromide under the same conditions.

Key inference: If n-butyl bromide gives n-octane ( $\ce{CH3(CH2)6CH3}$ ), then (d) must be a branched $\ce{C8H18}$ isomer. This means ** (a)** must be a branched primary alkyl bromide.

Step 2 — List possible primary $\ce{C4H9Br}$ isomers

Name	Structure
n-Butyl bromide	$\ce{CH3CH2CH2CH2Br}$
Isobutyl bromide	$\ce{(CH3)2CHCH2Br}$

Only isobutyl bromide is primary and branched.

Step 3 — Verify with the reaction sequence

Reaction 1: Dehydrohalogenation with alcoholic KOH

\ce{(CH3)2CHCH2Br + alc. KOH -> (CH3)2C=CH2 + KBr + H2O}

(b) is isobutylene (2-methylpropene).

Reaction 2: Addition of HBr to (b)

\ce{(CH3)2C=CH2 + HBr -> (CH3)3CBr}

(c) is tert-butyl bromide, an isomer of ** (a)** (both $\ce{C4H9Br}$ ). …

Start your 14-day free trial to unlock the full solution →

Here are the common mistakes students make on this exact problem, along with how to avoid each one.

Mistake 1: Assuming the primary alkyl halide is n-butyl bromide

Why it’s wrong:

The problem explicitly says that when (a) reacts with sodium (Wurtz reaction), the product $C_8H_{18}$ is different from the product formed when n-butyl bromide reacts with sodium. If (a) were n-butyl bromide, the Wurtz product would be n-octane — the same as the reference. So (a) cannot be n-butyl bromide.

How to avoid:

Read the “different from” condition carefully. It forces (a) to be a branched primary alkyl halide. The only primary $C_4H_9Br$ that is not n-butyl bromide is isobutyl bromide:

n-butyl bromide: $CH_3CH_2CH_2CH_2Br$
Isobutyl bromide: $(CH_3)_2CHCH_2Br$

So (a) must be isobutyl bromide.

Mistake 2: Forgetting that alcoholic KOH causes elimination, not substitution

Why it’s wrong:

Many students write substitution products (like an alcohol) when they see “KOH”. But alcoholic KOH favours elimination (dehydrohalogenation), especially with a primary halide and a strong base.

How to avoid:

Remember the rule:

Aqueous KOH → substitution (alcohol)
Alcoholic KOH → elimination (alkene)

Here, (b) is an alkene. For isobutyl bromide, elimination gives isobutylene (2-methylpropene):

(CH_3)_2CHCH_2Br + KOH \ (\text{alc.}) \xrightarrow{\Delta} (CH_3)_2C=CH_2 + KBr + H_2O

Mistake 3: Adding HBr to the alkene and getting the wrong isomer

Why it’s wrong:

When HBr adds to an unsymmetrical alkene, Markovnikov’s rule applies — the hydrogen goes to the carbon with more hydrogens. For isobutylene:

(CH_3)_2C=CH_2 + HBr \rightarrow (CH_3)_3CBr

This gives tert-butyl bromide, which is an isomer of (a) (both are $C_4H_9Br$ ). Some students incorrectly add HBr the other way and get back isobutyl bromide — but that would mean (c) = (a), which contradicts “isomer of (a)”.

How to avoid:

Always apply Markovnikov’s rule for HX addition. Check that the product is different from (a) — here, (c) is a tertiary halide, while (a) is primary.

Mistake 4: Writing the Wurtz reaction incorrectly

Why it’s wrong:

The Wurtz reaction couples two alkyl halides with sodium:

2RBr + 2Na \rightarrow R-R + 2NaBr

For isobutyl bromide: …

Start your 14-day free trial to unlock the full solution →

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) ✓ …
Start your 14-day free trial to see the full solution →
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: …

Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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°) …
Start your 14-day free trial to see the full solution →
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}.

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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).

…

Start your 14-day free trial to see the full solution →
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), $(CH_3)_3CCl$ . …
Start your 14-day free trial to see the full solution →
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).

…

Start your 14-day free trial to see the full solution →

←Previous question6.20 — The treatment of alkyl chlorides with aqueous KOH leads to the formati…→Next question6.22 — What happens when (i) n-butyl chloride is treated with alcoholic KOH, …

Q.Primary alkyl halide C4H9BrC_4H_9BrC4​H9​Br

🎓Unlock everything free for 14 days

Q.Primary alkyl halide $C_4H_9Br$

Unlock everything free for 14 days