Alkane Halogenation Mech

8 MCQs3 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 first: NEET questions on alkane halogenation rarely ask you to draw the mechanism. They ask you to predict the product when peroxide is present versus absent — and the trap is applying Markovnikov's rule blindly without checking for peroxide. Worse, the peroxide (anti-Markovnikov/Kharasch) effect works only with HBr, not HCl or HI. Missing that qualifier costs you 4 marks.

The mechanism in brief. Free-radical halogenation of alkanes (e.g., CH₄ + Cl₂ → CH₃Cl + HCl) proceeds through three stages:

  1. Initiation — UV light or heat homolyses the Cl–Cl bond into two chlorine radicals (Cl·).
  2. Propagation — Cl· abstracts an H from the alkane forming HCl and an alkyl radical. The alkyl radical then attacks Cl₂ to form the alkyl halide and regenerate Cl·. These two steps cycle repeatedly.
  3. Termination — Two radicals combine (Cl· + Cl·, R· + Cl·, or R· + R·), ending the chain.

For unsymmetrical alkanes, the hydrogen abstracted determines the product. Tertiary C–H bonds break preferentially because the resulting 3° radical is more stable than 2° or 1°. This is why propane + Cl₂ gives a higher proportion of 2-chloropropane than the statistical ratio predicts.

Bridge to NEET — Markovnikov vs. anti-Markovnikov in HBr addition to alkenes. Without peroxide, HBr adds ionically via a carbocation intermediate — H attaches to the carbon with more hydrogens (Markovnikov). With peroxide (ROOR), the mechanism switches to radical — Br· adds first to the less substituted carbon (anti-Markovnikov, Kharasch effect). This reversal is specific to HBr: HCl's C–Cl bond is too strong for radical propagation, and HI's chain transfer is too fast.

Watch-out: When a NEET stem mentions "in the presence of organic peroxide" or "ROOR" or "benzoyl peroxide," switch your mental model from ionic to radical and reverse the regiochemistry — but only if the hydrogen halide is HBr (NCERT Class 11 Chemistry, Chapter 13).

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

The free-radical halogenation of methane with Cl₂ proceeds through three stages. The correct sequence is:

MCQ 2Easy RecallPractice

In the free-radical chlorination of methane, the propagation step that regenerates the chlorine radical is:

MCQ 3Easy RecallPractice

The anti-Markovnikov addition (Kharasch effect) in the presence of peroxide is observed specifically with:

MCQ 4Direct ApplicationPractice

When propene (CH₃CH=CH₂) reacts with HBr **without** peroxide, the major product is:

MCQ 5Direct ApplicationPractice

When propene reacts with HBr **in the presence of benzoyl peroxide**, the major product is:

MCQ 6Direct ApplicationPractice

In the free-radical chlorination of isobutane (2-methylpropane), the major monochlorinated product is:

MCQ 7Concept TrapPractice

A student claims that adding peroxide to the reaction of HCl with propene will give anti-Markovnikov addition, just as with HBr. This claim is:

MCQ 8CalculationPractice

Propene is treated with HBr under two different conditions:

Quick recall before you leave

Worked Example

  1. 1

    Given

    - Substrate: but-1-ene (CH₃CH₂CH=CH₂) - Reagent: HBr - Condition: in the presence of organic peroxide (ROOR)

  2. 2

    Required

    Identify the major product.

  3. 3

    Concept

    Peroxide triggers the radical (Kharasch) mechanism for HBr addition. The regiochemistry reverses: anti-Markovnikov addition. The Br· radical adds to the less substituted carbon of the double bond because the resulting radical intermediate at the more substituted position is more stable.

  4. 4

    Formula / Rule

    Anti-Markovnikov: Br adds to the carbon with MORE hydrogens (the less substituted end). The H transfers to the more substituted carbon. This is the opposite of ionic Markovnikov addition.

  5. 5

    Substitution

    But-1-ene: CH₃CH₂CH=**CH₂** - C-1 has 2 H atoms (terminal, less substituted). - C-2 has 1 H atom (more substituted, bonded to –CH₂CH₃). Br· attacks C-1 (more H atoms, anti-Markovnikov). The radical forms at C-2 (more stable secondary radical). H then transfers to C-2.

  6. 6

    Calculation

    No arithmetic is needed — this is a regiochemistry-prediction problem. The reasoning is: peroxide → radical mechanism → anti-Markovnikov → Br on C-1.

  7. 7

    Final answer

    **Major product: 1-bromobutane (CH₃CH₂CH₂CH₂Br).**

  8. 8

    Common trap

    Without peroxide, the same reaction gives **2-bromobutane** (Markovnikov, ionic mechanism). The high-frequency NEET trap is applying Markovnikov regiochemistry even when the stem explicitly states "in the presence of peroxide." Also: the Kharasch effect works only with HBr — if the question changed the reagent to HCl + peroxide, the answer would still be Markovnikov (ionic).

  9. 9

    Similar NEET-style question

    "When 2-methylpropene reacts with HBr in the presence of benzoyl peroxide, the major product is ___." (Answer: 1-bromo-2-methylpropane, anti-Markovnikov.) ---

Before solving, remember these

Key Fact

Halogenation of alkanes (mechanism)

Free-radical chain. (1) Initiation: X₂ → 2X• (UV/heat). (2) Propagation: RH + X• → R• + HX; R• + X₂ → RX + X•. (3) Termination: 2R•, 2X•, R•+X•. Reactivity order F > Cl > Br > I.

-- NCERT, p. 10

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)

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