Q.Arrange primary, secondary and tertiary hydrogens in order of ease of replacement by radical halogenation, and explain why bromination is far more selective than chlorination.
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Concept understanding — Free Radical Mechanism
Free Radical Mechanism – From Intuition to Precision
Imagine you have a long chain of paperclips linked together. Now imagine someone snips one link in the middle. That single cut doesn't just break the chain — it creates two new ends, each hungry to grab onto something. That's the core idea of a free radical mechanism: a reaction that proceeds through species with an unpaired electron — a "hungry" atom or molecule that desperately wants to pair up.
The Intuition: Why Radicals Are Special
Most chemical bonds involve paired electrons — two electrons spinning in opposite directions, like a stable couple. A free radical is the chemical equivalent of a lone wolf: it has one unpaired electron, making it highly reactive. It will do almost anything to find a partner — steal an electron from a neighbour, donate its own, or break another bond to create more radicals.
This creates a chain reaction. One radical reacts, produces another radical, which reacts again, and so on — like a row of dominoes falling one after another. That's why free radical mechanisms are often called chain reactions.
The Precise Statement
A free radical mechanism is a stepwise reaction pathway involving species with unpaired electrons (free radicals). It proceeds through three distinct phases:
Initiation – A stable molecule is broken to produce two free radicals. This usually requires energy — heat (thermolysis) or light (photolysis).
Propagation – Radicals react with stable molecules to produce new radicals. This step repeats many times, forming the chain.
Termination – Two radicals combine to form a stable product, ending the chain.
General pattern:
Initiation: A−Bhν or ΔA⋅+B⋅
Propagation: A⋅+C−D→A−C+D⋅
Termination: A⋅+D⋅→A−D
A Concrete Example: Chlorination of Methane
This is the classic textbook example, and it appears in almost every Indian board exam (Class 11/12, JEE, NEET).
Overall reaction:
CH4+Cl2hνCH3Cl+HCl
Step-by-step mechanism:
Initiation – Chlorine molecule absorbs UV light and splits:
Cl2hν2Cl⋅
Propagation – Two steps that repeat:
Chlorine radical attacks methane:
Cl⋅+CH4→HCl+CH3⋅
Methyl radical attacks another chlorine molecule:
CH3⋅+Cl2→CH3Cl+Cl⋅
Notice: the Cl⋅ consumed in step 1 is regenerated in step 2. This is the chain — one radical keeps producing another.
Termination – Any two radicals meet:
Cl⋅+Cl⋅→Cl2
CH3⋅+CH3⋅→C2H6
CH3⋅+Cl⋅→CH3Cl
Watch out
A common mistake: students think termination only happens when the same radicals combine. In reality, any two radicals can terminate — including cross-combination (like CH3⋅+Cl⋅). Also, termination steps are rare because radical concentrations are very low.
Key Characteristics to Remember
Free radicals are neutral — they have no charge, only an unpaired electron. Don't confuse them with ions.
They are highly reactive — lifetimes are typically microseconds or less.
The mechanism requires initiation energy — heat or light. Without it, the reaction won't start.
Chain length can be thousands of propagation cycles before termination.
Product mixtures are common — because termination can produce multiple products (e.g., C2H6 in the example above).
Where You'll See This
Free radical mechanisms appear in:
Halogenation of alkanes (Class 11/12 organic chemistry)
Polymerisation (addition polymers like polyethylene)
To spot a free radical mechanism in an exam: look for light (hν) or high temperature (Δ) as reaction conditions, and halogens (Cl2,Br2) as reactants. The products often include HCl or HBr alongside the substituted product.
The Bottom Line
Free radical mechanism = a chain reaction where reactive species with unpaired electrons (radicals) are generated, propagate through repeated steps, and eventually combine to terminate. The chain is sustained because each propagation step regenerates a radical.
This topic is commonly searched as "Free Radical Mechanism 11 chemistry important questions" or "Free Radical Mechanism formula and examples", and it maps cleanly onto the Class 11 Chemistry portion of the NCERT/CBSE syllabus. Because free radical mechanism shows up repeatedly in JEE Main, NEET and state CET Chemistry papers, mastering the underlying idea (not just the formula) is genuinely worth the extra time.
Ease of radical formation is 3 degree greater than 2 degree greater than 1 degree; bromine radical is less reactive and so more selective for the most stable radical than chlorine radical.
✓Final answer
Reactivity order: tertiary > secondary > primary. Bromination is more selective than chlorination because the less reactive bromine radical is choosier and strongly favours forming the more stable tertiary radical, whereas the highly reactive chlorine radical abstracts almost any hydrogen with little preference.
In the propagation step, a halogen radical abstracts a hydrogen to form a carbon radical; the more alkyl groups attached to that radical carbon, the more the radical is stabilised by electron donation (+I effect) and hyperconjugation from those alkyl groups, so radical stability -- and hence the ease and rate of forming it -- runs tertiary > secondary > primary. Chlorine radicals are highly reactive and their hydrogen-abstraction step is only mildly exothermic, so the transition state resembles the starting alkane more than the product radical (an early transition state, by Hammond's postulate) -- the reaction barely 'notices' how stable the eventual radical will be, so chlorination shows only mild selectivity (roughly 5:3.8:1 for 3°:2°:1° per hydrogen). Bromine radicals are far less reactive, so their abstraction step is only weakly exothermic (or even endothermic), giving a late transition state that resembles the product radical closely -- the reaction is therefore very sensitive to which radical is more stable, so bromination is highly selective (roughly 1600:82:1) and strongly favours the tertiary hydrogen even when it is vastly outnumbered by primary hydrogens.
✓Final answer
Ease of H-abstraction: tertiary > secondary > primary. Bromination is more selective than chlorination because the less reactive bromine radical reacts through a later, more product-like transition state that is far more sensitive to radical stability.
Rank hydrogens by the stability of the radical each would form on abstraction, then relate each halogen's overall reactivity to how much that reactivity 'cares about' radical stability (Hammond's postulate: a less reactive/more endothermic step has a later, more product-like, more selectivity-sensitive transition state).
Assuming chlorination and bromination are equally selective because they both follow 3° > 2° > 1°; forgetting that the STATISTICAL number of each type of hydrogen (e.g. nine primary H's versus one tertiary H in isobutane) must be weighed against the per-hydrogen reactivity to predict the actual major product ratio.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2023Set annual1 mark
Q.Which of the following radical is more stable? (CH3)3C-radical, CH3CHCH3-radical, CH3CH2-radical
›Reveal solutionSolution
Alkyl substitution stabilizes a free radical, so (CH3)3C-radical (tertiary) is more stable than CH3CHCH3-radical (secondary), which is more stable than CH3CH2-radical (primary).
A free radical is an electron-deficient (odd-electron) species. Its stability is governed mainly by two effects of the alkyl groups attached to the radical carbon:
Inductive (+I) effect: Alkyl groups are electron-releasing relative to hydrogen, so more alkyl groups around the radical centre push electron density towards it, partially compensating for the electron deficiency.
Hyperconjugation: Adjacent C-H sigma bonds can donate electron density into the singly-occupied p orbital on the radical carbon. A tertiary radical has more adjacent C-H bonds (9, from three methyl groups) available for hyperconjugation than a secondary radical (6) or a primary radical (3), so more hyperconjugative structures stabilize it.
Both effects increase with the number of alkyl groups attached, giving the stability order:
Among the three given, (CH3)3C-radical is tertiary, CH3CHCH3-radical is secondary, and CH3CH2-radical is primary.
✓Final answer
(CH3)3C-radical (the tertiary radical) is the most stable, because hyperconjugation and the +I effect of its three methyl groups best disperse the unpaired electron's charge (stability order: tertiary > secondary > primary).
CBSE 2020Set ANNUAL1 mark
Q.Match the Column-A item 'Homolysis' with the correct entry from Column B:
(a) Anion
(b) Free radical
(c) Cation
(d) 7
(e) delta-G = 0
(f) g cm^-3
›Reveal solutionSolution
In homolysis, a covalent bond breaks symmetrically so that each atom retains one electron of the shared pair, generating two free radicals — matching Column B entry (b).
A covalent bond A-B can break in two ways:
Homolytic fission (homolysis): the bond breaks symmetrically, and each fragment takes one electron from the shared pair, giving two neutral species each with an unpaired electron — these are called free radicals (A. and B.). This typically requires energy input such as heat or UV light and is common in reactions like the halogenation of alkanes.
Heterolytic fission (heterolysis): the bond breaks asymmetrically, with both electrons going to one fragment, producing a cation and an anion.
Since the Column A term is specifically 'homolysis', its direct product is a free radical, matching option (b).
✓Final answer
Homolysis matches with (b) Free radical — homolytic bond fission produces free radicals.