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The chlorination of methane is the first reaction mechanism most students meet, and it shows up in JEE Main and NEET as a "which step is which" question or as a trap about by-products. Learning the three steps by heart is not enough; examiners also ask why light is needed, why ethane appears, and why iodine hardly reacts. This post walks through the mechanism with the energy of each step, worked out from NCERT's own bond enthalpies, so the answers follow from the numbers.
Where is this in NCERT and the syllabus?
The current NCERT Class 11 Chemistry Part II chapter Hydrocarbons (2026-27 reprint) describes halogenation of alkanes as a free radical chain mechanism with initiation, propagation and termination. It gives the reactivity order F₂ > Cl₂ > Br₂ > I₂, calls fluorination too violent to control and iodination very slow and reversible, and explains ethane as a by-product. The JEE (Main) 2026 syllabus lists "mechanism of halogenation of alkanes" and "stability of carbocations and free radicals"; the NEET (UG) 2026 syllabus has the same lines. Check your exam year's syllabus.
If you want to count how many monochloro products a larger alkane gives, see our post on counting monochlorination products. This post stays with methane and the mechanism itself.
What happens in each step?
Initiation. Light or heat splits a chlorine molecule evenly, one electron to each atom (homolysis):
Cl–Cl → 2Cl·
Propagation. Two steps repeat in a loop:
(a) Cl· + CH₄ → CH₃· + HCl
(b) CH₃· + Cl₂ → CH₃Cl + Cl·
Step (b) hands back a chlorine radical, which starts step (a) again. One initiation can therefore make many molecules of product. NCERT also shows how CH₃Cl can react on in the same way to give CH₂Cl₂, and further.
Termination. Two radicals meet and pair up, removing radicals from the chain:
Cl· + Cl· → Cl₂; CH₃· + CH₃· → CH₃–CH₃; CH₃· + Cl· → CH₃Cl
The second of these is why some ethane appears in the product mixture.
Why does the chain start with Cl₂?
NCERT says the Cl–Cl bond is weaker than the C–C and C–H bonds and so is easiest to break. Its table of mean bond enthalpies in the Thermodynamics chapter puts Cl–Cl at 243 kJ/mol, C–C at 347 and C–H at 414.
You can turn that into the light needed. Energy per molecule = 243,000 ÷ (6.022 × 10²³) = 4.04 × 10⁻¹⁹ J. Using λ = hc/E, λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) ÷ (4.04 × 10⁻¹⁹) ≈ 492 nm. Light of wavelength about 492 nm or shorter, which includes blue and violet light, carries enough energy per photon. Breaking a C–H bond would need ultraviolet light shorter than about 289 nm.
Worked example: is each propagation step downhill?
Estimate ΔH for each step as (bonds broken) − (bonds formed). NCERT's Thermodynamics chapter gives 427 kJ/mol for removing the first H from CH₄; for the others use the mean values H–Cl 431, Cl–Cl 243 and C–Cl 330 kJ/mol.
Step (a): break C–H (427), form H–Cl (431). ΔH = 427 − 431 = −4 kJ/mol.
Step (b): break Cl–Cl (243), form C–Cl (330). ΔH = 243 − 330 = −87 kJ/mol.
Sum: −91 kJ/mol, the estimate for CH₄ + Cl₂ → CH₃Cl + HCl.
Both steps release energy, so once started the chain runs easily.
Why does fluorine explode and iodine barely react?
Repeat the calculation with NCERT's values for each halogen (X–X, H–X, C–X in kJ/mol: F 155, 569, 439; Br 192, 368, 276; I 151, 297, 238):
Fluorine: step (a) −142, step (b) −284, total −426. Enormously exothermic, which fits NCERT's "too violent to be controlled".
Chlorine: −4, −87, total −91.
Bromine: step (a) +59, step (b) −84, total −25. The first step is uphill, so the chain is slower.
Iodine: step (a) +130, step (b) −87, total +43. The overall reaction is uphill, which is why NCERT calls iodination very slow and reversible and says it needs an oxidising agent such as HIO₃ or HNO₃ to remove the HI.
These are estimates from average bond enthalpies, so treat them as trends, not exact values. The trend matches NCERT's order F₂ > Cl₂ > Br₂ > I₂.
Which mistakes cost marks?
Calling Cl· + Cl· → Cl₂ initiation. It removes radicals, so it is termination.
Putting H· in a propagation step. Cl· takes an H atom to make HCl; free H atoms are not part of the main chain.
Forgetting ethane. It comes from two methyl radicals in termination.
Assuming one product. CH₃Cl can be chlorinated again. NCERT's mechanism shows CH₂Cl₂ forming the same way. In practice, a large excess of methane keeps CH₃Cl as the main product.
Using heterolysis. Radical reactions use homolysis, with fishhook arrows for single electrons.
When a mechanism question mixes all of this, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
A practice routine
Write the full mechanism from memory, then label every step and count radicals on each side.
Redo the energy table for bromine with the numbers above, and explain why its first step is the slow one.
Find the longest wavelength that can split Br₂ (192 kJ/mol). Check: about 623 nm.
Write the termination step that gives ethane, and one more that gives CH₃Cl.
Today, draw the two propagation steps as a loop and explain to someone why one Cl· can make many CH₃Cl molecules.
