Alkanes — preparation (Wurtz, Kolbe electrolysis, hydrogenation), properties, free-radical halogenation (Cl₂ - Br₂): Visual walkthrough

This page assumes only what a free radical is — a species with one lonely unpaired electron — and we will re-draw even that from scratch in Step 2. See also the parent: the topic note.

Step 1 — Two calm molecules that ignore each other

WHAT. Put a methane molecule ( $C H_{4}$ ) next to a chlorine molecule ( $C l_{2}$ ) in the dark. Nothing happens. We need to understand why nothing happens before we understand what makes it happen.

WHY. A reaction needs an attacker with a "handle": a lone pair, a $π$ -bond, a positive/negative charge. Methane has none — four identical, strong, non-polar C–H bonds, all electrons paired and buried. Chlorine likewise is a neutral molecule with a shared pair. Two closed-shell molecules just bounce.

PICTURE. Look at Figure s01. Each bond is drawn as a shared pair of dots sitting exactly between the two atoms. Notice: no dot is ever alone. That "no lonely electron" is exactly why the pair is unreactive.

Figure — Alkanes — preparation (Wurtz, Kolbe electrolysis, hydrogenation), properties, free-radical halogenation (Cl₂ - Br₂)

Step 2 — Light snaps the weakest bond in half (Initiation)

WHAT. Shine ultraviolet light. A photon hits a $C l_{2}$ molecule and splits it evenly — one electron of the shared pair goes to each chlorine: $C l : C l h ν C l^{∙} +^{∙} C l$

Let us name each symbol, right where it sits:

$C l : C l$ — the two dots ( $:$ ) are the shared bonding pair.
$h ν$ — the energy of one photon of light ( $h$ is Planck's constant, $ν$ is the light's frequency; together $h ν$ is just "one packet of light energy"). This is the tool that supplies exactly enough energy to break the bond.
$C l^{∙}$ and $^{∙} C l$ — each chlorine walks away with one dot. Two radicals are born.

WHY light, and why this bond? Every bond has a "cost to break" (its bond energy). Among all bonds present, $C l$ – $C l$ is the weakest ( $\approx 242 kJ/mol$ ) — much weaker than C–H ( $\approx 414$ ) or C–C. UV photons carry roughly this much energy each, so a photon can crack $C l_{2}$ but not the sturdy methane. This is why the tool is light and not, say, gentle heating: we need a precise, targeted energy packet aimed at the weak link.

WHY split evenly? An even split (one electron each) is called homolysis ("homo" = same, each side gets the same). The alternative — both electrons to one atom — would make ions, but chlorine's two atoms are identical, so there is no reason for one to hog both. Symmetry forces the even split.

PICTURE. In Figure s02 the yellow burst is the photon; watch the shared pair (two dots) pull apart so each chlorine keeps exactly one dot (red half-arrows show single electrons moving).

Step 3 — A chlorine radical steals a hydrogen (Propagation, half 1)

WHAT. The lonely $C l^{∙}$ collides with methane and grabs one hydrogen atom (the H and one electron), leaving behind a methyl radical: $C l^{∙} + C H_{4} \to C H_{3}^{∙} + H C l$

Term by term:

$C l^{∙}$ — the attacker, carrying its lonely dot.
$C H_{4}$ — the calm methane; one of its C–H bonds is about to be broken.
$C H_{3}^{∙}$ — methyl radical: the carbon now holds the lonely dot (it kept its half of the old C–H pair).
$H C l$ — the escaped hydrogen has paired up with the chlorine's lonely dot to form a brand-new, paired bond.

WHY does the radical steal an H and not attack elsewhere? The radical wants to pair its lonely electron. Forming the very strong H–Cl bond releases energy, which "pays for" breaking the C–H bond. So the radical hunts for an H it can pull off — and in doing so it passes the lonely electron on to carbon. Notice the accounting: one radical went in, one radical came out. Radicals are never destroyed here; they are handed forward.

PICTURE. Figure s03: follow the red half-arrows — one electron of the C–H pair goes to make H–Cl, the other stays on carbon as its new lonely dot. The "radicalness" hops from Cl to C.

Step 4 — The methyl radical grabs a chlorine (Propagation, half 2)

WHAT. The new $C H_{3}^{∙}$ now attacks a fresh $C l_{2}$ molecule, takes one chlorine, and spits the other back out as a radical: $C H_{3}^{∙} + C l_{2} \to C H_{3} C l + C l^{∙}$

$C H_{3}^{∙}$ — the radical from Step 3, still hunting a partner for its lonely dot.
$C l_{2}$ — a fresh, whole chlorine molecule.
$C H_{3} C l$ — the product we wanted: chloromethane, all electrons paired again.
$C l^{∙}$ — a regenerated chlorine radical, identical to the one that started Step 3.

WHY is this the genius step? The chlorine radical produced here is exactly the reactant needed for Step 3. So Step 3 → Step 4 → Step 3 → Step 4 … loops forever from a single original push. That is the chain. This is why one photon (one initiation event) can turn thousands of methane molecules into product.

PICTURE. Figure s04 shows the two propagation steps drawn as a closed loop: $C l^{∙}$ enters at top, leaves at bottom, and re-enters — a self-feeding cycle.

Step 5 — How the chain finally dies (Termination)

WHAT. The loop keeps running until, by chance, two radicals meet each other instead of a molecule. Two lonely dots pair up, and no new radical is made: $C l^{∙} + C l^{∙} \to C l_{2}$ $C H_{3}^{∙} + C H_{3}^{∙} \to C_{2} H_{6}$ $C H_{3}^{∙} + C l^{∙} \to C H_{3} C l$

WHY is this rare (and why does the chain last so long)? Radicals are present in tiny amounts — most collisions are radical-with-molecule (which continues the chain). Only occasionally do a radical and a radical bump. When they do, both lonely dots vanish into one shared pair, and that branch of the chain stops. The appearance of ethane $C_{2} H_{6}$ (two methyls joined!) is the fingerprint that termination happened.

PICTURE. Figure s05: three little "hand-shake" pictures — two lonely dots merging into one shared pair. Above each, note "no radical out" — that is what ends it.

Step 6 — Edge case: bromine is pickier than chlorine

WHAT. Repeat the whole story with $B r_{2}$ instead of $C l_{2}$ . The mechanism is identical (initiation → propagation → termination), but bromine is slower and choosier: given a molecule with different kinds of H, $B r^{∙}$ strongly prefers the H that leaves behind the most stable radical.

WHY? The H-stealing step (Step 3) for bromine is only slightly favourable — it is a "reluctant" reaction. A reluctant reaction has a late, product-like transition state, so the stability of the radical being formed matters a lot. Bromine therefore waits for the "best" H: $radical stability: 3^{\circ} > 2^{\circ} > 1^{\circ} > C H_{3}^{∙}$ More neighbouring alkyl groups spread out (share) the lonely electron by hyperconjugation and inductive donation, stabilising the radical. Chlorine, being violently reactive, is not fussy and grabs almost any H.

PICTURE. Figure s06: a molecule with a 1° H and a 3° H. Chlorine's arrows point at both (unselective); bromine's single arrow points only at the 3° H (selective), with a small "energy hill" cartoon showing bromine's later, radical-sensitive transition state.

Step 7 — Degenerate case: no light, or a radical trap

WHAT. Two "nothing happens" scenarios that confirm we understand the chain:

No UV, no heat → no initiation → no radicals → no reaction (Step 1 forever).
Add a radical scavenger (a molecule that eats radicals) → the chain is starved and stops even if it started.

WHY this matters. It proves the reaction lives or dies by radicals. Kill the initiation, or kill the carriers, and the whole edifice collapses. This is the acid test of a chain mechanism.

PICTURE. Figure s07: on the left, dark flask, whole molecules, a big "0 radicals"; on the right, a scavenger (mint blob) mopping up a $C l^{∙}$ so the loop can't close.

The one-picture summary

Figure s08 compresses the entire derivation: the one photon at top (initiation) feeds a spinning propagation loop (Steps 3–4) that pumps out product molecule after molecule, until two radicals shake hands and a branch of the loop snaps (termination). One push, many products, messy mixture.

Recall Feynman retelling — the whole walkthrough in plain words

Methane and chlorine sitting in the dark are like two people with their hands full — no free hand to grab anything, so they ignore each other (Step 1). Now flash a light: the light is exactly strong enough to snap the flimsiest handshake in the room — the Cl–Cl one — into two people each with one free hand (Step 2). A free-handed chlorine grabs a hydrogen off methane, and in the swap the methane is left with the free hand now (Step 3). That free-handed carbon grabs a chlorine off a fresh Cl₂, becomes the finished product, and hands a free hand back to a chlorine (Step 4). Same free hand, over and over — one push, a thousand products (the chain). It only stops when, by dumb luck, two free-handed people finally shake each other's hands and both are done (Step 5). Bromine is the slow, fussy version that only bothers with the easiest, best hydrogen (Step 6), and if you never turn on the light — or send in someone who grabs every free hand — the whole dance never starts (Step 7).

Recall

Why does one photon make many product molecules? ::: Each propagation step regenerates the radical it consumed, so the cycle self-feeds from a single initiation event. Why is $C l_{2}$ (not C–H) broken in initiation? ::: Cl–Cl is the weakest bond present, and a UV photon carries about its bond energy. Why is the product a mixture? ::: $C H_{3} C l$ still has hydrogens, so it can be chlorinated again → $C H_{2} C l_{2}, C H C l_{3}, C C l_{4}$ . Why is bromine more selective than chlorine? ::: Its slow, late transition state is sensitive to radical stability, so it prefers the H giving the most stable ( $3^{\circ}$ ) radical. What confirms termination occurred in the methane case? ::: Traces of ethane $C_{2} H_{6}$ (two methyl radicals joined).