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 Class 11 Chemistry, Ch. 9, p. 303Alkane Halogenation Mechanism
8 MCQs6 revision cards9-step worked example
Source: NCERT HydrocarbonsPYQ coverage: NEET 2025Official key: NTA-verifiedLast updated: 25 Sep 2026
Alkane Halogenation Mechanism, explained for NEET
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:
- Initiation — UV light or heat homolyses the Cl–Cl bond into two chlorine radicals (Cl·).
- 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.
- 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 9).
Can you answer these Alkane Halogenation Mechanism 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:
Show answer and why every option is right or wrong
Answer: C. C is correct. Free-radical halogenation always begins with initiation (homolytic fission of Cl₂ by UV/heat), followed by the self-sustaining propagation chain, and ends when two radicals combine in termination (NCERT Class 11 Chemistry, Chapter 9).
Why A is wrong: A is wrong because propagation cannot begin before radicals are generated in the initiation step.
Why B is wrong: B is wrong because termination ends the chain — it cannot occur before propagation has generated the radical intermediates.
Why D is wrong: D is wrong because termination is the final stage, not the first; chain reactions require radical generation (initiation) to start.
MCQ 2Easy RecallPractice
In the free-radical chlorination of methane, the propagation step that regenerates the chlorine radical is:
Show answer and why every option is right or wrong
Answer: B. B is correct. In this propagation step, the methyl radical attacks Cl₂ to form chloromethane and regenerate a Cl· radical, which re-enters the chain (NCERT Class 11 Chemistry, Chapter 9).
Why A is wrong: A is wrong because Cl₂ homolysis is the initiation step, not a propagation step — it generates the first radicals but does not sustain the chain.
Why C is wrong: C is wrong because two radicals combining is a termination step — it consumes radicals rather than regenerating them.
Why D is wrong: D describes only the first half of propagation (H-abstraction). While it is a propagation step, it does not regenerate Cl· — the question asks specifically for the step that regenerates the chlorine radical.
MCQ 3Easy RecallPractice
The anti-Markovnikov addition (Kharasch effect) in the presence of peroxide is observed specifically with:
Show answer and why every option is right or wrong
Answer: A. A is correct. The peroxide-mediated anti-Markovnikov addition (Kharasch effect) works only with HBr. HCl's H–Cl bond is too strong for the radical chain to propagate, and with HI the iodine free radicals combine to form I₂ instead of adding to the double bond (NCERT Class 11 Chemistry, Chapter 9).
Why B is wrong: B is wrong because the H–Cl bond is too strong for the propagation step of the radical chain to be energetically favorable — the Kharasch effect does not operate with HCl.
Why C is wrong: C is wrong because, although the H–I bond is weak, the iodine free radicals combine to form iodine molecules instead of adding to the double bond (NCERT).
Why D is wrong: D is wrong because HF has the strongest H–X bond in the series, making radical propagation thermodynamically unfavorable.
MCQ 4Direct ApplicationPractice
When propene (CH₃CH=CH₂) reacts with HBr without peroxide, the major product is:
Show answer and why every option is right or wrong
Answer: B. B is correct. Without peroxide, HBr adds via ionic (electrophilic addition) mechanism. The proton adds to the terminal carbon (more H atoms, Markovnikov's rule), generating a more stable secondary carbocation, and Br⁻ attacks at C-2 to give 2-bromopropane (NCERT Class 11 Chemistry, Chapter 9).
Why A is wrong: A is wrong because 1-bromopropane would be the anti-Markovnikov product, requiring peroxide (radical mechanism). Without peroxide, the ionic pathway gives Markovnikov addition — trap: ignoring peroxide condition (mistake: hx no peroxide check).
Why C is wrong: C is wrong because HBr adds only one H and one Br across the double bond in a single addition. Dibromide formation requires Br₂, not HBr.
Why D is wrong: D is wrong because simple addition of HBr to an alkene gives a bromoalkane, not an alkane. Reduction to propane would require H₂/catalyst.
MCQ 5Direct ApplicationPractice
When propene reacts with HBr in the presence of benzoyl peroxide, the major product is:
Show answer and why every option is right or wrong
Answer: A. A is correct. Benzoyl peroxide triggers the radical mechanism (Kharasch effect). Br· adds to the less substituted carbon (C-1), generating the more stable secondary radical at C-2, which then abstracts H from HBr. The product is 1-bromopropane — anti-Markovnikov regiochemistry (NCERT Class 11 Chemistry, Chapter 9).
Why B is wrong: B is wrong because 2-bromopropane is the Markovnikov product formed via ionic addition without peroxide. Peroxide switches the mechanism to radical, reversing the regiochemistry — trap: applying Markovnikov regardless of conditions (trap: markovnikov peroxide reversal).
Why C is wrong: C is wrong because allyl bromide would require allylic substitution (e.g., NBS/light), not electrophilic or radical addition of HBr to the double bond.
Why D is wrong: D is wrong because HBr adds one molecule across the double bond. Dibromo products require Br₂ addition, not HBr.
MCQ 6Direct ApplicationPractice
In the free-radical chlorination of isobutane (2-methylpropane), the major monochlorinated product is:
Show answer and why every option is right or wrong
Answer: A. A is correct. Cl· is only mildly selective: per hydrogen, a 3° C–H reacts about 5 times faster than a 1° C–H. Isobutane has nine 1° hydrogens and one 3° hydrogen, so the product ratio is about 9 × 1 : 1 × 5 = 9 : 5, roughly 64% isobutyl chloride to 36% tert-butyl chloride. The statistical advantage outweighs the selectivity. (Br· is far more selective, so bromination gives mainly tert-butyl bromide; that is the case where the 3° product dominates.)
Why B is wrong: B is wrong because free-radical halogenation is selective, not statistical — radical stability (3° > 2° > 1°) makes each 3° hydrogen about 5 times as reactive as a 1° one, so the two products form in about a 64 : 36 ratio, not 50 : 50.
Why C is wrong: C is wrong because the ~5× per-hydrogen preference of Cl· for the 3° C–H does not make up for there being only one such hydrogen against nine 1° hydrogens: 1 × 5 against 9 × 1 leaves tert-butyl chloride the minor product (about 36%). The tertiary product dominates in bromination, not chlorination.
Why D is wrong: D is wrong because n-butyl chloride is a product of chlorinating n-butane, not isobutane (2-methylpropane). The carbon skeleton does not rearrange under radical conditions.
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:
Show answer and why every option is right or wrong
Answer: D. D is correct. The anti-Markovnikov (Kharasch) effect is specific to HBr. For HCl, the H–Cl bond dissociation enthalpy is too high for the radical propagation step to be exothermic, so the radical chain does not sustain. For HI, the iodine free radicals combine to form I₂ instead of adding to the double bond. Therefore, peroxide does not reverse regiochemistry with HCl (NCERT Class 11 Chemistry, Chapter 9).
Why A is wrong: A is wrong because the Kharasch effect does not operate with all hydrogen halides. The radical chain energetics are favorable only for HBr — the H–Cl bond is too strong, and with HI the iodine radicals combine to I₂ instead of adding (trap: markovnikov peroxide reversal).
Why B is wrong: B is wrong because peroxide does have a real mechanistic effect — but only with HBr. Claiming peroxide has no effect on any HX ignores the well-documented Kharasch effect.
Why C is wrong: C is wrong because even though it correctly excludes HI, it incorrectly includes HCl. The Kharasch effect is exclusive to HBr — HCl's radical propagation is endothermic.
MCQ 8CalculationPractice
Propene is treated with HBr under two different conditions:• Condition I: No peroxide• Condition II: In the presence of organic peroxide (ROOR)Which of the following correctly identifies the major product under each condition?
Show answer and why every option is right or wrong
Answer: D. D is correct. Condition I (no peroxide): ionic electrophilic addition — H⁺ to terminal C, Br⁻ to internal C → 2-bromopropane (Markovnikov). Condition II (peroxide): radical addition (Kharasch) — Br· to terminal C, H to internal C → 1-bromopropane (anti-Markovnikov). Two-step reasoning: identify the mechanism switch, then apply the appropriate regiochemistry rule (NCERT Class 11 Chemistry, Chapter 9).
Why A is wrong: A is wrong because it reverses the assignments — without peroxide the ionic pathway gives Markovnikov (2-bromopropane), not anti-Markovnikov. This is the classic trap of confusing which condition gives which product (trap: markovnikov peroxide reversal).
Why B is wrong: B is wrong because it gives anti-Markovnikov product under both conditions, which would require a radical mechanism in both cases. Without peroxide, the ionic mechanism dominates and gives the Markovnikov product.
Why C is wrong: C is wrong because it gives Markovnikov product under both conditions, ignoring the peroxide-triggered mechanism switch. With peroxide, the radical mechanism gives anti-Markovnikov 1-bromopropane (mistake: mistake: hx no peroxide check).
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Alkane Halogenation Mechanism: quick recall before you leave
How do you solve a Alkane Halogenation Mechanism question? A worked example
Pattern: free-radical substitution in alkanes (NCERT Class 11 Chemistry, Chapter 9, page 303).
- 1
Given
• Substrate: 2-methylpropane, (CH₃)₃CH• Reagent: Cl₂• Condition: diffused sunlight (hν), one chlorine substituted per molecule
- 2
Required
(a) The number of structurally different monochloro products. (b) Which kind of C–H bond is replaced most readily, per hydrogen. (c) The steps of the mechanism.
- 3
Concept
Light splits Cl₂ into chlorine radicals, which pull a hydrogen atom off the alkane to leave an alkyl radical. Which product forms depends on WHICH hydrogen is removed, so the question is how many different kinds of hydrogen the molecule has. The ease of removal per hydrogen follows the stability of the radical left behind: 3° > 2° > 1° (NCERT Class 11 Chemistry, Chapter 9, page 303).
- 4
Formula / Rule
• Initiation: Cl₂ → 2Cl· (hν)• Propagation: Cl· + R–H → R· + HCl, then R· + Cl₂ → R–Cl + Cl·• Termination: any two radicals combine (Cl· + Cl·, R· + Cl·, R· + R·)
- 5
Substitution
2-Methylpropane has 10 hydrogens of two kinds:• 9 primary hydrogens, on the three equivalent CH₃ groups• 1 tertiary hydrogen, on the central carbon
Replacing a primary H gives 1-chloro-2-methylpropane, (CH₃)₂CHCH₂Cl. Replacing the tertiary H gives 2-chloro-2-methylpropane, (CH₃)₃CCl. - 6
Calculation
Distinct hydrogen environments = 2, so there are 2 monochloro products. Per hydrogen, the tertiary C–H is replaced most readily, because it leaves the most stable (tertiary) radical.
- 7
Final answer
(a) Two monochloro products: 1-chloro-2-methylpropane and 2-chloro-2-methylpropane. (b) The tertiary C–H bond, per hydrogen. (c) Initiation, two propagation steps, termination, as in Step 4.
- 8
Common trap
Counting hydrogen ATOMS instead of hydrogen ENVIRONMENTS. There are ten hydrogens but only two kinds, and equivalent hydrogens give the same product. A second trap is assuming the tertiary chloride must be the main product because the tertiary C–H is most reactive. Reactivity per hydrogen is only half the story: there are nine primary hydrogens against one tertiary, and the product mixture reflects both the number of each kind and its reactivity.
- 9
Similar NEET-style question
"How many structurally different monochloro products are formed when n-pentane is chlorinated in sunlight?" (Answer: 3 — 1-chloropentane, 2-chloropentane and 3-chloropentane, from the three kinds of hydrogen on C-1/C-5, C-2/C-4 and C-3.)
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What to remember before solving Alkane Halogenation Mechanism questions
More in Hydrocarbons: 4 exam traps and mistakes · 1 formula · 2 question patterns from its other lessons.
Alkane Halogenation Mechanism questions from past NEET papers
1 question from NEET 2025. Answers verified against NTA official keys.
NEET 2025
16
22
33
45
NTA Answer: Option 1(final)
Sources
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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