JEE Mains Chemistry · Hydrocarbons

Free-Radical Halogenation of Alkanes

In light or heat, a halogen replaces the hydrogens of an alkane one at a time by a free-radical chain; each set of equivalent hydrogens gives one monohalo product, and bromine strongly prefers a tertiary hydrogen.

All Hydrocarbons notes JEE Mains Chemistry strategy

Why this matters

Eleven PYQs, five of them multiple choice, and three from 2026. Seven count the monohalo or dihalo products of an alkane, sometimes with stereoisomers included; four ask where the halogen goes, how many products excess halogen gives, or the percentage of halogen in the product. Six of the eleven ask for a number.

Concept 1 of 2: Counting monohalogenation products

Replacing any one hydrogen of a set of equivalent hydrogens gives the same compound. So the number of monohalo products is the number of different kinds of hydrogen. To find them, look for the molecule's symmetry: every methyl on the same carbon is one kind, and the two ends of a symmetrical chain are one kind. If the question counts stereoisomers too, check each product for a stereocentre and count a chiral product twice.

Definition

Structural products = number of non-equivalent H sets.
n-Pentane has 3 sets, 2-methylbutane 4, 2,2-dimethylpropane 1.
A product with a carbon carrying four different groups is chiral and exists as two enantiomers.
Read the stem: "excluding stereoisomers" means structural only; "all isomers" or "maximum number" means enantiomers are counted.
Further halogenation of a haloalkane is counted the same way, on the haloalkane's remaining hydrogens.

Counting products

N_{\text{structural}} = \text{number of non-equivalent H sets},\qquad N_{\text{total}} = N_{\text{achiral}} + 2\,N_{\text{chiral}}

Worked example

How many monochloro products does 2-methylpentane give in light, (a) structural only and (b) counting stereoisomers?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 25 July 2022 · Q150Moderate

Example 1 · Hydrocarbons · Free-Radical Halogenation of Alkanes

The total number of monobromo derivatives formed by the alkanes with molecular formula

C_{5}H_{12}

is (excluding stereo isomers)

Do not count equivalent methyls twice

The two methyl groups on C-2 of 2-methylbutane are one set, not two. Chlorinating either gives 1-chloro-2-methylbutane.

Read whether stereoisomers are counted

2-Methylbutane gives 4 structural monochloro products. Two of them, 1-chloro-2-methylbutane and 2-chloro-3-methylbutane, are chiral, so the count with stereoisomers is 6. The wording of the stem decides which number is wanted.

Cyclic isomers do not decolourise KMnO₄

When a question says the isomers of CₙH₂ₙ do not decolourise KMnO₄, it means the cycloalkanes only. Draw every ring size and every position of the side chains.

Drill 6 more on counting monohalogenation products

Concept 2 of 2: Radical selectivity and multiple halogenation

The chain carrier is a halogen atom that pulls off a hydrogen. It pulls off the hydrogen that leaves the most stable radical, and radicals are stabilised by alkyl groups: 3° > 2° > 1° > methyl. A bromine atom is slow and choosy, so it takes a tertiary H almost every time. A chlorine atom is fast and less choosy, so it gives a mixture close to the ratio of hydrogens. With excess halogen the reaction keeps going until every H can be replaced.

Definition

Initiation: $\mathrm{Cl_2 \xrightarrow{h\nu} 2Cl^\bullet}$ .
Propagation: $\mathrm{CH_4 + Cl^\bullet \rightarrow CH_3^\bullet + HCl}$ ; $\mathrm{CH_3^\bullet + Cl_2 \rightarrow CH_3Cl + Cl^\bullet}$ .
Termination: two radicals combine, for example $\mathrm{CH_3^\bullet + CH_3^\bullet \rightarrow C_2H_6}$ .
Reactivity of the halogen: $\mathrm{F_2 > Cl_2 > Br_2 > I_2}$ . Reactivity of the hydrogen: 3° > 2° > 1°.
Each substitution uses one $\mathrm{X_2}$ and releases one HX; the product carries ONE X per substitution.

Substrate and conditions	What happens	Product	Reason
2-Methylpropane, $\mathrm{Br_2}$ , light	Bromine takes the tertiary H	2-Bromo-2-methylpropane (major)	Br· is highly selective for the most stable radical
Propane, $\mathrm{Cl_2}$ , light	Chlorine attacks both kinds of H	1-Chloropropane and 2-chloropropane in similar amounts	Cl· is fast and less selective
Methane, excess $\mathrm{Cl_2}$ , light	Substitution continues	$\mathrm{CH_3Cl,\ CH_2Cl_2,\ CHCl_3,\ CCl_4}$	Each product still has H to replace
Ethane, excess $\mathrm{Br_2}$ , light	Every degree of substitution forms	9 bromoethanes, from $\mathrm{C_2H_5Br}$ to $\mathrm{C_2Br_6}$	Counts per formula: 1, 2, 2, 2, 1, 1
Cyclopropane, $\mathrm{Br_2}$ , light	One Br replaces one H when the data show one Br per molecule	Bromocyclopropane, $\mathrm{C_3H_5Br}$	One $\mathrm{Br_2}$ used, the second Br leaves as HBr

Use the product's C : X ratio to tell substitution (one X per

\mathrm{X_2}

used) from addition (two X).

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q46Moderate

Example 2 · Hydrocarbons · Free-Radical Halogenation of Alkanes

The cycloalkane (X) on bromination consumes one mole of bromine per mole of (X) and gives the product (Y) in which

C:Br

ratio is

3:1

. The percentage of bromine in the product (Y) is

\_\_\_\_

%. (Nearest integer) (Given : Molar mass in

g{mol}^{- 1}H:1,C:12

O:16,Br:80

)

Substitution puts one halogen in the product

One mole of

\mathrm{X_2}

per mole of alkane in substitution gives RX + HX, not RX₂. An addition product would carry both halogen atoms, so the product's C : X ratio tells the two apart.

Chlorination does not pick only the tertiary H

Because Cl· is fast, primary hydrogens, which are often more numerous, give a large share of the product. "Tertiary product only" is true for bromination, not chlorination.

Drill 3 more on radical selectivity and multiple halogenation

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Counting monohalogenation products
Counting products
$N_{\text{structural}} = \text{number of non-equivalent H sets},\qquad N_{\text{total}} = N_{\text{achiral}} + 2\,N_{\text{chiral}}$

Reference tables (1)

Radical selectivity and multiple halogenation5 rows

Substrate and conditions	What happens	Product	Reason
2-Methylpropane, $\mathrm{Br_2}$ , light	Bromine takes the tertiary H	2-Bromo-2-methylpropane (major)	Br· is highly selective for the most stable radical
Propane, $\mathrm{Cl_2}$ , light	Chlorine attacks both kinds of H	1-Chloropropane and 2-chloropropane in similar amounts	Cl· is fast and less selective
Methane, excess $\mathrm{Cl_2}$ , light	Substitution continues	$\mathrm{CH_3Cl,\ CH_2Cl_2,\ CHCl_3,\ CCl_4}$	Each product still has H to replace
Ethane, excess $\mathrm{Br_2}$ , light	Every degree of substitution forms	9 bromoethanes, from $\mathrm{C_2H_5Br}$ to $\mathrm{C_2Br_6}$	Counts per formula: 1, 2, 2, 2, 1, 1
Cyclopropane, $\mathrm{Br_2}$ , light	One Br replaces one H when the data show one Br per molecule	Bromocyclopropane, $\mathrm{C_3H_5Br}$	One $\mathrm{Br_2}$ used, the second Br leaves as HBr

Use the product's C : X ratio to tell substitution (one X per

\mathrm{X_2}

used) from addition (two X).

Watch out for (5)

Do not count equivalent methyls twice
→ Counting monohalogenation products
Read whether stereoisomers are counted
→ Counting monohalogenation products
Cyclic isomers do not decolourise KMnO₄
→ Counting monohalogenation products
Substitution puts one halogen in the product
→ Radical selectivity and multiple halogenation
Chlorination does not pick only the tertiary H
→ Radical selectivity and multiple halogenation

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Related notes

Halogen Addition, Oxidation and Ozonolysis of Alkenes — allylic substitution →