Markovnikov Peroxide

8 MCQs2 revision cards9-step worked example
Source: NCERT Aldehydes, Ketones and Carboxylic AcidsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

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 9/10, page 14).

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.

Practice 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?

MCQ 2Easy RecallPractice

The peroxide effect (Kharasch effect) in HX addition to alkenes operates with which hydrogen halide?

MCQ 3Easy RecallPractice

What is the mechanism of HBr addition to an alkene in the presence of organic peroxides?

MCQ 4Direct ApplicationPractice

What is the major product when HBr is added to but-1-ene in the presence of benzoyl peroxide?

MCQ 5Direct ApplicationPractice

HCl is added to propene in the presence of organic peroxide. The major product is:

MCQ 6Direct ApplicationPractice

In the radical addition of HBr to propene (with peroxide), which intermediate determines the product orientation?

MCQ 7Concept TrapPractice

A student claims: "Adding peroxide to HI + propene will give 1-iodopropane (anti-Markovnikov product)." This claim is:

MCQ 8CalculationPractice

When 3-methylbut-1-ene reacts with HBr in the presence of peroxide, the major product is:

Quick recall before you leave

Worked Example

Pattern: P.CHE.U15.MARKOVNIKOV_HX_ADDITION (observed 2021, 2022, 2025; frequency 3)

  1. 1

    Given

    - Substrate: propene (CH₃–CH=CH₂), asymmetric alkene - Reagent: HBr - Condition: dibenzoyl peroxide present

  2. 2

    Required

    Major product and its IUPAC name.

  3. 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. 4

    Formula/Rule

    Anti-Markovnikov orientation: Br adds to terminal carbon (forming more stable secondary radical at internal carbon).

  5. 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. 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. 7

    Final answer

    Major product: **1-bromopropane** (CH₃–CH₂–CH₂Br), the anti-Markovnikov product.

  8. 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. 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.) ---

Before solving, remember these

Law

Markovnikov's rule

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, p. 14

Formulas

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.

SymbolQuantitySI Unit
H,Xadded atoms-

Valid when

  • Asymmetric alkene
  • H-X with X = Cl, Br, I
  • Without peroxide for Markovnikov

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Markovnikov reverses with peroxide

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).

Trap

Solvent changes product distribution

Category: Organic Reaction Conditions

Same starting materials give different products depending on solvent. Polar protic (water, alcohols): SN1/E1 favoured. Polar aprotic (DMSO, DMF): SN2 favoured. Affects substitution vs elimination.

When it triggers

Question contrasts product when solvent is changed; or specifies solvent type.

How to avoid

Polar protic stabilises carbocation → SN1/E1 (3° preferred). Polar aprotic doesn't solvate nucleophile → strong SN2 nucleophile (1°/2° preferred). Bulky base (t-BuOK) favours E2 over SN2.

Mistake

Confuses ortho/para directors with meta directors.

Root cause: concept gap

Correction

o,p-directors (activators except halogens): -OH, -OR, -NH₂, -NHR, alkyl. m-directors (deactivators): -NO₂, -CN, -COOH, -CHO. Halogens: o,p-directors but DEACTIVATORS.

Past Year Questions

12 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2022

Given below are two statements Statement I: The acidic strength of monosubstituted nitrophenol is higher than phenol because of electron withdrawing nitro group. Statement II: o-nitrophenol, m-nitrophenol and p-nitrophenol will have same acidic strength as they have one nitro group attached to the phenolic ring. In the light of the above statements, choose the most appropriate answer from the options given below:

1Statement I is incorrect but Statement II is correct.
2Both Statement I and Statement II are correct.
3Both Statement I and Statement II are incorrect.
4Statement I is correct but Statement II is incorrect.
NTA Answer: Option 4(final)
NEET 2022

Given below are two statements : Statement I : The boiling points of aldehydes and ketones are higher than hydrocarbons of comparable molecular masses because of weak molecular association in aldehydes and ketones due to dipole - dipole interactions. Statement II : The boiling points of aldehydes and ketones are lower than the alcohols of similar molecular masses due to the absence of H-bonding. In the light of the above statements, choose the most appropriate answer from the given below

1Statement I is incorrect but Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct but Statement II is incorrect
NTA Answer: Option 2(final)

How NEET usually asks this

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