JEE Mains Chemistry · Hydrocarbons

Halogen Addition, Oxidation and Ozonolysis of Alkenes

Bromine adds anti across a C=C through a bromonium ion; cold dilute KMnO₄ turns it into a diol while hot acidic KMnO₄ cuts it; ozonolysis cuts it into two carbonyl compounds whose structures reveal where the double bond was.

All Hydrocarbons notes JEE Mains Chemistry strategy

Why this matters

Twenty-six PYQs, twenty-two of them multiple choice, and four from 2026. Five are about adding a halogen, the stereochemistry of bromine addition, bromine water or substitution at the allylic carbon. Seven use KMnO₄, cold for a diol or hot for cleavage. Fourteen are ozonolysis: predict the carbonyl products, or work back from them to the alkene. Four of the twenty-six ask for a number.

Concept 1 of 3: Adding halogens across C=C, and allylic substitution

Br₂ attacks a C=C to form a three-membered bromonium ion. The second nucleophile must come from the opposite face, so the two new groups end up anti (trans). If water is the solvent, water is that second nucleophile, and it attacks the more substituted carbon because that carbon carries more of the positive charge. In light or at high temperature, with little halogen, the halogen does not add at all: a radical takes the allylic H instead, because the allylic radical is stabilised by resonance.

Definition

$\mathrm{Br_2}$ in $\mathrm{CCl_4}$ : anti addition. trans-But-2-ene gives meso-2,3-dibromobutane; cis-but-2-ene gives the racemic (±) pair. The red-brown colour disappears (test for C=C).
Bromine water: a bromohydrin, OH on the more substituted carbon (Markovnikov), Br on the other.
$\mathrm{Cl_2}$ or $\mathrm{Br_2}$ in light or at high temperature, or NBS: substitution at the allylic (or benzylic) C–H.

Reagent and conditions	Type of reaction	Product from cyclohexene	Stereochemistry
$\mathrm{Br_2}$ in $\mathrm{CCl_4}$ , dark	Electrophilic addition	1,2-Dibromocyclohexane	trans (anti addition)
$\mathrm{Br_2}$ in water	Addition of Br and OH	2-Bromocyclohexan-1-ol	trans (anti addition)
$\mathrm{Cl_2}$ in $\mathrm{CCl_4}$ , dark	Electrophilic addition	1,2-Dichlorocyclohexane	trans (anti addition)
$\mathrm{Cl_2}$ , light or 500 °C (low concentration)	Radical allylic substitution	3-Chlorocyclohexene	C=C kept; racemic at C-3
NBS, light or peroxide	Radical allylic substitution	3-Bromocyclohexene	C=C kept; racemic at C-3

The same halogen adds in the dark and substitutes at the allylic carbon in light: the conditions decide.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 22 · Q42Moderate

Example 1 · Hydrocarbons · Halogen Addition, Oxidation and Ozonolysis of Alkenes

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Treatment of bromine water with propene yields 1-bromopropan-2-ol. Reason (R) : Attack of water on bromonium ion follows Markovnikov rule and results in 1-bromopropan-2-ol. In the light of the above statements, choose the most appropriate answer from the options given below:

Anti addition to trans gives meso

It is easy to swap these. trans-But-2-ene + Br₂ gives the optically inactive meso compound; cis-but-2-ene gives the racemic pair.

Light turns addition into substitution

Cl₂ in CCl₄ in the dark adds across cyclohexene. Cl₂ in light (or at high temperature) replaces an allylic H and keeps the C=C. Read the conditions before choosing.

Drill 4 more on adding halogens across c=c, and allylic substitution

Concept 2 of 3: KMnO₄: cold gives a diol, hot cuts the C=C

Cold, dilute, alkaline KMnO₄ (Baeyer's reagent) adds two OH groups to the same face of the C=C, and its purple colour disappears. Hot acidic KMnO₄ goes further and breaks the C=C. Each doubly bonded carbon becomes a C=O, and any H left on it is oxidised too: a =CH₂ end becomes CO₂, a =CHR end becomes an acid RCOOH, and a =CR₂ end stays as a ketone.

Definition

Cold, dilute, alkaline KMnO₄: syn-diol. Ethene gives ethane-1,2-diol; cyclohexene gives cis-cyclohexane-1,2-diol.
Hot acidic (or hot alkaline, then acid) KMnO₄: cleavage. $\mathrm{=CH_2 \rightarrow CO_2}$ (effervescence), $\mathrm{=CHR \rightarrow RCOOH}$ , $\mathrm{=CR_2 \rightarrow R_2C{=}O}$ .
A ring alkene gives ONE open-chain product with a group at each end: cyclohexene gives hexanedioic (adipic) acid.

Hot KMnO₄ cleavage

\mathrm{R_2C{=}CHR' \xrightarrow{KMnO_4/H^+,\ \Delta} R_2C{=}O + R'COOH},\qquad \mathrm{{=}CH_2 \rightarrow CO_2 + H_2O}

Worked example

What does 2-methylpent-2-ene give with hot acidic KMnO₄?

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The same idea in a real exam question:

JEE Mains · 2022 · 29 June 2022 · Q43Moderate

Example 2 · Hydrocarbons · Halogen Addition, Oxidation and Ozonolysis of Alkenes

Two isomers '

A

' and '

B

' with molecular formula

C_{4}H_{8}

give different products on oxidation with

KMnO_{4}

in acidic medium. Isomer 'A' on reaction with

KMnO_{4}/H^{+}

results in effervescence of a gas and gives ketone. The compound '

A

' is

=CH₂ gives CO₂, not methanal, with hot KMnO₄

Hot KMnO₄ oxidises everything it can. A terminal =CH₂ becomes CO₂ (seen as effervescence), and an aldehyde fragment becomes the acid. Only ozonolysis with Zn/H₂O stops at HCHO and aldehydes.

Baeyer's reagent gives a diol, not cleavage

Cold, dilute, alkaline KMnO₄ keeps both carbons joined and adds two OH groups on the same face. A ring alkene gives a cis-1,2-diol.

Drill 6 more on kmno₄: cold gives a diol, hot cuts the c=c

Concept 3 of 3: Ozonolysis: predicting products and working back to the alkene

Ozonolysis cuts every C=C and puts =O on each of its carbons, keeping any H that was there. So the products are aldehydes and ketones, and nothing else in the molecule changes. To work backwards, take the two carbonyl carbons, delete the two oxygens and join the carbons by a double bond. If the only product is one molecule with two C=O groups, the alkene was a ring.

Definition

Reagents: O₃, then Zn/H₂O (reductive work-up). =CH₂ gives HCHO; =CHR gives RCHO; =CR₂ gives R₂C=O.
With water alone (no Zn), H₂O₂ formed in the reaction oxidises the aldehydes to acids.
One mole of O₃ per C=C. A cycloalkene gives one dicarbonyl chain; a diene gives three fragments (or two if it is cyclic).
Moles of H₂ taken up on hydrogenation = number of C=C (and each C≡C takes two).

Reductive ozonolysis

\mathrm{R_2C{=}CHR' \xrightarrow{(i)\ O_3\quad (ii)\ Zn/H_2O} R_2C{=}O + R'CHO}

Worked example

What does 1-methylcyclopentene give on ozonolysis followed by Zn/H₂O?

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The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q124Moderate

Example 3 · Hydrocarbons · Halogen Addition, Oxidation and Ozonolysis of Alkenes

A hydrocarbon '

X

' with formula

C_{6}H_{8}

uses two moles

H_{2}

on catalytic hydrogenation of its one mole. On ozonolysis, '

X

' yields two moles of methane dicarbaldehyde. The hydrocarbon '

X

' is:

A ring alkene gives one product, not two

The ring atoms stay joined through the rest of the ring, so cleaving a ring C=C opens the ring into one chain with a carbonyl at each end. Do not split it into two molecules.

Zn decides aldehyde or acid

O₃ then Zn/H₂O gives aldehydes. O₃ then H₂O alone (or H₂O₂) gives carboxylic acids from the same carbons. Ketone fragments are the same either way.

cis and trans isomers give the same products

Ozonolysis destroys the C=C, so the geometry is lost. cis- and trans-but-2-ene both give two moles of ethanal.

Drill 13 more on ozonolysis: predicting products and working back to the alkene

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 (2)

KMnO₄: cold gives a diol, hot cuts the C=C
Hot KMnO₄ cleavage
$\mathrm{R_2C{=}CHR' \xrightarrow{KMnO_4/H^+,\ \Delta} R_2C{=}O + R'COOH},\qquad \mathrm{{=}CH_2 \rightarrow CO_2 + H_2O}$
Ozonolysis: predicting products and working back to the alkene
Reductive ozonolysis
$\mathrm{R_2C{=}CHR' \xrightarrow{(i)\ O_3\quad (ii)\ Zn/H_2O} R_2C{=}O + R'CHO}$

Reference tables (1)

Adding halogens across C=C, and allylic substitution5 rows

Reagent and conditions	Type of reaction	Product from cyclohexene	Stereochemistry
$\mathrm{Br_2}$ in $\mathrm{CCl_4}$ , dark	Electrophilic addition	1,2-Dibromocyclohexane	trans (anti addition)
$\mathrm{Br_2}$ in water	Addition of Br and OH	2-Bromocyclohexan-1-ol	trans (anti addition)
$\mathrm{Cl_2}$ in $\mathrm{CCl_4}$ , dark	Electrophilic addition	1,2-Dichlorocyclohexane	trans (anti addition)
$\mathrm{Cl_2}$ , light or 500 °C (low concentration)	Radical allylic substitution	3-Chlorocyclohexene	C=C kept; racemic at C-3
NBS, light or peroxide	Radical allylic substitution	3-Bromocyclohexene	C=C kept; racemic at C-3

The same halogen adds in the dark and substitutes at the allylic carbon in light: the conditions decide.

Watch out for (7)

Anti addition to trans gives meso
→ Adding halogens across C=C, and allylic substitution
Light turns addition into substitution
→ Adding halogens across C=C, and allylic substitution
=CH₂ gives CO₂, not methanal, with hot KMnO₄
→ KMnO₄: cold gives a diol, hot cuts the C=C
Baeyer's reagent gives a diol, not cleavage
→ KMnO₄: cold gives a diol, hot cuts the C=C
A ring alkene gives one product, not two
→ Ozonolysis: predicting products and working back to the alkene
Zn decides aldehyde or acid
→ Ozonolysis: predicting products and working back to the alkene
cis and trans isomers give the same products
→ Ozonolysis: predicting products and working back to the alkene

Test yourself on Hydrocarbons

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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