In addition of HX to unsymmetrical alkene, H goes to carbon with more H atoms; X to carbon with fewer H atoms. Driven by carbocation stability. Anti-Markovnikov in presence of peroxide (Kharasch effect): radical mechanism.
-- NCERT Class 11 Chemistry, Ch. 9, p. 311Markovnikov Peroxide
8 MCQs9 revision cards9-step worked example
Source: NCERT HydrocarbonsPYQ coverage: NEET 2021Official key: NTA-verifiedLast updated: 24 Sep 2026
Markovnikov Peroxide, explained for NEET
The trap: You see "HBr + propene + peroxide" and write the Markovnikov product. You just lost 4 marks — one for the wrong answer, three more from negative marking confidence. The peroxide flips everything, but only for HBr.
Markovnikov's rule states that when HX adds to an asymmetric alkene, the hydrogen attaches to the carbon already bearing more hydrogens, while the halide goes to the carbon with fewer hydrogens. The mechanistic basis is carbocation stability: the ionic pathway generates the more substituted (more stable) carbocation as the intermediate.
This is the default pathway — no peroxide, no radical initiator, straightforward electrophilic addition (NCERT Class 11 Chemistry, Chapter 9, page 312).
The peroxide effect (Kharasch effect) reverses the orientation — but exclusively for HBr. With organic peroxides (ROOR) or UV light, the mechanism switches from ionic to radical. The bromine radical adds first, forming the more stable (more substituted) carbon radical. The result: anti-Markovnikov product, where Br ends up on the terminal carbon.
Why only HBr?
- HCl: the H–Cl bond is too strong for homolytic cleavage under peroxide conditions — the radical chain doesn't propagate.
- HI: the H–I bond is too weak — the radical chain terminates prematurely (iodine radical recombines).
- HBr sits in the thermodynamic sweet spot for sustained radical propagation.
Watch-out for NEET: If the question mentions peroxide/ROOR/light with HBr → anti-Markovnikov. If it mentions peroxide with HCl or HI → still Markovnikov (peroxide effect doesn't operate). A common mistake is applying Markovnikov regardless of stated conditions.
Can you answer these Markovnikov Peroxide MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
According to Markovnikov's rule, when HBr adds to propene in the absence of peroxide, the bromine atom attaches to which carbon?
Show answer and why every option is right or wrong
Answer: D. Markovnikov's rule states H goes to the carbon with more hydrogens, so Br goes to carbon-2 (the carbon with fewer H atoms), forming 2-bromopropane. This follows from carbocation stability in the ionic mechanism (NCERT Class 11 Chemistry, Chapter 9, page 312).
Why A is wrong: A is wrong because carbon-1 has MORE hydrogens — Markovnikov's rule places H (not Br) on the carbon with more H atoms. This confusion reverses the rule's assignment.
Why B is wrong: B is wrong because HBr readily undergoes electrophilic addition to alkenes under normal conditions. The reaction proceeds via an ionic mechanism without any special catalyst.
Why C is wrong: C is wrong because Markovnikov addition is regioselective, not random. The more stable secondary carbocation intermediate ensures Br adds preferentially to carbon-2.
MCQ 2Easy RecallPractice
The peroxide effect (Kharasch effect) in HX addition to alkenes operates with which hydrogen halide?
Show answer and why every option is right or wrong
Answer: D. The peroxide effect operates exclusively with HBr. HCl's bond is too strong for effective homolytic cleavage, and HI's bond is too weak for sustained radical chain propagation (NCERT Class 11 Chemistry, Chapter 9/10).
Why A is wrong: A is wrong because the H–Cl bond dissociation energy is too high for peroxide-initiated homolytic cleavage to sustain a radical chain. Peroxide has no practical effect on HCl addition orientation.
Why B is wrong: B is wrong because the peroxide effect is specific to HBr due to its intermediate bond strength being thermodynamically suited for radical chain propagation. Neither HCl nor HI participates.
Why C is wrong: C is wrong because the H–I bond is too weak — iodine radicals recombine before the chain can propagate. HI addition remains Markovnikov regardless of peroxide.
MCQ 3Easy RecallPractice
What is the mechanism of HBr addition to an alkene in the presence of organic peroxides?
Show answer and why every option is right or wrong
Answer: B. In the presence of peroxides, HBr adds via a free radical mechanism: initiation (peroxide homolysis generates RO·), propagation (Br· adds to alkene forming carbon radical, then H abstraction from HBr), and termination (NCERT Class 11 Chemistry, Chapter 9/10).
Why A is wrong: A is wrong because the ionic (electrophilic) mechanism operates WITHOUT peroxide. Peroxide specifically triggers homolytic cleavage of HBr, switching to the radical pathway.
Why C is wrong: C is wrong because nucleophilic addition applies to carbonyl compounds (C=O), not to alkene addition of HBr. Alkenes are electron-rich and undergo electrophilic or radical attack.
Why D is wrong: D is wrong because no elimination step occurs. The addition is a single concerted radical chain process — Br· adds directly to the double bond carbon.
MCQ 4Direct ApplicationPractice
What is the major product when HBr is added to but-1-ene in the presence of benzoyl peroxide?
Show answer and why every option is right or wrong
Answer: C. Peroxide triggers radical mechanism (anti-Markovnikov). Br· adds to terminal carbon-1 (forming more stable secondary radical at carbon-2), then H from HBr completes at carbon-2. Product: 1-bromobutane (NCERT Class 11 Chemistry, Chapter 9/10).
Why A is wrong: A is wrong because 2-bromobutane is the Markovnikov product (ionic mechanism, no peroxide). With peroxide present, the radical pathway reverses regioselectivity — Br goes to carbon-1, not carbon-2 (trap: ignoring peroxide condition).
Why B is wrong: B is wrong because only ONE molecule of HBr adds across the double bond. Dibromo products require Br₂ addition, not HBr. The question specifies HBr — single addition only.
Why D is wrong: D is wrong because 2,3-dibromobutane would require Br₂ addition to but-2-ene. Neither the reagent (HBr, not Br₂) nor the substrate (but-1-ene, not but-2-ene) matches this product.
MCQ 5Direct ApplicationPractice
HCl is added to propene in the presence of organic peroxide. The major product is:
Show answer and why every option is right or wrong
Answer: A. The peroxide effect does NOT operate with HCl (H–Cl bond too strong for radical chain propagation). Therefore, normal ionic addition occurs following Markovnikov's rule: Cl goes to carbon-2, giving 2-chloropropane (NCERT Class 11 Chemistry, Chapter 9/10).
Why B is wrong: B is wrong because anti-Markovnikov addition requires the radical mechanism, which operates ONLY with HBr. HCl's bond is too strong for peroxide-initiated homolysis — the peroxide has no effect on the reaction orientation (trap: assuming peroxide reverses all HX additions).
Why C is wrong: C is wrong because HCl does add to alkenes — it proceeds via the normal ionic (electrophilic addition) mechanism regardless of peroxide presence. The reaction occurs; only the mechanism remains unchanged.
Why D is wrong: D is wrong because electrophilic addition is regioselective via Markovnikov's rule (carbocation stability). Equal mixtures would imply no selectivity, which contradicts the established ionic mechanism.
MCQ 6Direct ApplicationPractice
In the radical addition of HBr to propene (with peroxide), which intermediate determines the product orientation?
Show answer and why every option is right or wrong
Answer: B. In radical addition, Br· adds to the terminal carbon, generating a secondary carbon radical at carbon-2 (more stable than primary). This radical stability controls regioselectivity in the anti-Markovnikov pathway (NCERT Class 11 Chemistry, Chapter 9/10).
Why A is wrong: A is wrong because carbocations are intermediates of the IONIC mechanism (without peroxide). In the radical mechanism, no carbocation forms — the key intermediate is a carbon radical.
Why C is wrong: C is wrong because the mechanism favours the MORE stable radical, not the less stable one. Br· adds to carbon-1 specifically because the resulting radical at carbon-2 is secondary (more stable than a primary radical at carbon-1).
Why D is wrong: D is wrong because bromonium ions are intermediates in Br₂ addition to alkenes, not in HBr radical addition. The peroxide-initiated pathway involves open carbon radicals, not cyclic intermediates.
MCQ 7Concept TrapPractice
A student claims: "Adding peroxide to HI + propene will give 1-iodopropane (anti-Markovnikov product)." This claim is:
Show answer and why every option is right or wrong
Answer: A. The peroxide effect is exclusive to HBr. HI's bond is too weak — iodine radicals recombine before sustaining a chain. HI + propene gives 2-iodopropane (Markovnikov) regardless of peroxide (NCERT Class 11 Chemistry, Chapter 9/10).
Why B is wrong: B is wrong because the peroxide effect is NOT universal across hydrogen halides. It operates exclusively with HBr due to the specific bond dissociation energy of H–Br being suitable for radical chain propagation.
Why C is wrong: C is wrong because HI does add to alkenes — it undergoes normal electrophilic (ionic) addition following Markovnikov's rule. HI is actually quite reactive toward alkenes; only its radical pathway fails.
Why D is wrong: D is wrong because temperature does not enable the peroxide effect for HI. The fundamental issue is thermodynamic — the I· radical recombines too readily regardless of temperature conditions.
MCQ 8CalculationPractice
When 3-methylbut-1-ene reacts with HBr in the presence of peroxide, the major product is:
Show answer and why every option is right or wrong
Answer: C. Step 1: Peroxide → radical mechanism (anti-Markovnikov). Step 2: Br· adds to carbon-1 (terminal). The resulting radical at carbon-2 is secondary (stabilised by adjacent methyl and the 3-methyl group). Step 3: H· from HBr completes at carbon-2. Product: 1-bromo-3-methylbutane (CH₂Br–CH₂–CH(CH₃)–CH₃). Numbering from the Br-bearing end confirms 1-bromo-3-methylbutane.
Why A is wrong: A is wrong because 2-bromo-3-methylbutane would be the Markovnikov product (Br on carbon-2 via ionic mechanism without peroxide). With peroxide, the radical mechanism places Br on carbon-1 (terminal), not carbon-2 (trap: ignoring peroxide reversal).
Why B is wrong: B is wrong because this product implies Br at carbon-3 of the parent chain, which would require addition at an internal position not adjacent to the double bond. The double bond is between C1 and C2; radical Br adds only to carbons of the original π bond.
Why D is wrong: D is wrong because 1-bromo-2-methylbutane implies a different parent chain numbering. In 3-methylbut-1-ene, the methyl branch is at carbon-3. Anti-Markovnikov addition puts Br at carbon-1 of the original chain, giving 1-bromo-3-methylbutane.
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Markovnikov Peroxide: quick recall before you leave
How do you solve a Markovnikov Peroxide question? A worked example
Pattern: P.CHE.U15.MARKOVNIKOV_HX_ADDITION (observed 2021, 2022, 2025; frequency 3)
- 1
Given
• Substrate: propene (CH₃–CH=CH₂), asymmetric alkene• Reagent: HBr• Condition: dibenzoyl peroxide present
- 2
Required
Major product and its IUPAC name.
- 3
Concept
Peroxide initiates radical chain mechanism. In radical addition of HBr, Br· adds first (not H⁺). Radical stability (not carbocation stability) governs regioselectivity.
- 4
Formula/Rule
Anti-Markovnikov orientation: Br adds to terminal carbon (forming more stable secondary radical at internal carbon).
- 5
Substitution/Application
• Br· adds to carbon-1 of propene (terminal, less substituted).• This generates a secondary radical at carbon-2 (CH₃–ĊH–CH₂Br).• Secondary radical is more stable than the primary radical alternative (at carbon-1 if Br· had added to carbon-2).• H· from HBr then adds to carbon-2.
- 6
Calculation
No arithmetic needed. The regiochemistry is determined by radical stability comparison:• Path A: Br on C-1 → radical on C-2 (secondary) ✓ More stable• Path B: Br on C-2 → radical on C-1 (primary) ✗ Less stable
- 7
Final answer
Major product: 1-bromopropane (CH₃–CH₂–CH₂Br), the anti-Markovnikov product.
- 8
Common trap
Forgetting the peroxide condition and writing 2-bromopropane (Markovnikov). Also: assuming peroxide works for HCl or HI — it does not. If the question had said "HCl + peroxide," the answer would still be 2-chloropropane (Markovnikov).
- 9
Similar NEET-style question
"What is the major product when HBr adds to 2-methylpropene in the presence of organic peroxide?"
(Answer: 1-bromo-2-methylpropane — Br on terminal carbon, radical at tertiary carbon.)
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What to remember before solving Markovnikov Peroxide questions
Which Markovnikov Peroxide formulas do you need for NEET?
Markovnikov's rule (and anti-Markovnikov)
Without peroxide: ionic mechanism — H goes to carbon with MORE hydrogens (carbocation stability rule). With peroxide (HBr only, Kharasch): radical mechanism — anti-Markovnikov.
| Symbol | Quantity | SI Unit |
|---|---|---|
| H,X | added atoms | - |
Valid when
- Asymmetric alkene
- H-X with X = Cl, Br, I
- Without peroxide for Markovnikov
Where do students lose marks on Markovnikov Peroxide?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Organic Reaction Conditions
HBr addition to alkene: WITHOUT peroxide → Markovnikov (H to C with more H). WITH peroxide (Kharasch effect) → anti-Markovnikov (radical mechanism). Specific to HBr only — not HCl, HI.
When it triggers
Question gives HX addition to alkene with explicit peroxide condition or hints (e.g. ROOR, light).
How to avoid
Without peroxide: ionic mechanism, carbocation stability → Markovnikov. With peroxide: radical mechanism, radical stability → anti-Markovnikov. Effect ONLY for HBr (HCl too strong, HI too weak).
Root cause: concept gap
Correction
Without peroxide: Markovnikov (carbocation). With peroxide: anti-Markovnikov (radical) — only with HBr.
More in Hydrocarbons: 2 exam traps and mistakes · 1 question pattern from its other lessons.
Markovnikov Peroxide questions from past NEET papers
1 question from NEET 2021. Answers verified against NTA official keys.
NEET 2021
The major product of the following chemical reaction is : (CH₃)₂CH–CH=CH₂ + HBr —[(C₆H₅CO)₂O₂]→ ?
1(CH₃)₂CBr–CH₂–CH₃
2(CH₃)₂CH–CH₂–CH₂–Br
3(CH₃)₂CH–CH₂–CH₂–O–COC₆H₅
4(CH₃)₂CH–CH(Br)–CH₃
NTA Answer: Option 2(final)
How does NEET ask about Markovnikov Peroxide?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Predict major product of HX addition to alkene. Markovnikov rule (without peroxide); anti-Markovnikov with peroxide.
InterpretationEasy
Common distractors
ignores peroxide effect
Same product regardless of conditions
Sources
NCERT refs: Class 11 Chemistry Chapter 9, p.312
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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