Visual walkthrough — Markovnikov vs anti-Markovnikov (peroxide effect, Kharasch)

We work the same molecule throughout so you can compare the two paths side by side:

$\underset{\text{C3}}{\underbrace{C{H}_3}}-\underset{\text{C2}}{\underbrace{CH}}=\underset{\text{C1}}{\underbrace{C{H}_2}}$

Here each label just names one carbon: C1 is the $=C{H}_2$ end (two H's, lonely on substituents), C2 is the middle carbon, C3 is the methyl cap. Keep this numbering in your head — every figure uses it.

See the Hinglish version and the parent Alkenes - Addition Reactions for the overview.

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Step 1 — What a double bond actually is (the loose electrons)

WHAT. A double bond is not "two identical sticks". One stick (the $\sigma$ bond) is a strong, tight handshake holding the two carbons together. The second stick is a $\pi$ bond — a soft cloud of two electrons sitting above and below the carbons, held loosely.

WHY it matters. Loosely-held electrons are greedy: they reach out and grab any hungry, positive thing that comes near. That grab is the first move of the whole reaction. If you don't see this cloud, nothing else makes sense.

PICTURE. In the figure the tight $\sigma$ line sits between the carbons; the pale-yellow $\pi$ cloud floats above and below — those are the electrons that will do the grabbing.

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Step 2 — The ionic path: the $\pi$ hands grab $H^{+}$

WHAT. Bring in $H-Br$. The bond between $H$ and $Br$ is polarised: $Br$ is greedy for electrons, so it pulls the shared pair toward itself, leaving $H$ slightly positive ($H^{\delta +}$). The $\pi$ cloud reaches out and grabs that positive $H$.

WHY this tool — a cation and not something else? Because when the two $\pi$ electrons leave to make a new $C-H$ bond, they came from one side of the double bond. The other carbon is left one electron pair short → it carries a full $+$ charge. That positively-charged carbon is a carbocation. We track it because its stability decides everything downstream.

PICTURE. Watch the curved blue arrow: the $\pi$ cloud swings down onto $H$. A new $C-H$ bond forms; the leftover $+$ sits on the other carbon.

$C{H}_3-\underset{\text{C2}}{\mathrm{CH}}=\underset{\text{C1}}{C{H}_2}+\stackrel{\delta +}{H}-\stackrel{\delta -}{\mathrm{Br}}⟶\text{[carbocation]}+B{r}^{-}$

• $\stackrel{\delta +}{H}$ — the partially positive proton the cloud wants.
• $B{r}^{-}$ — leaves as a negative ion, waiting its turn.

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Step 3 — Two choices, one winner (why Markovnikov)

WHAT. The proton could land on C1 or on C2. Each choice puts the $+$ on the opposite carbon. Only one of these is comfortable.

WHY. A $+$ charge is happier when nearby alkyl groups can share the load (hyperconjugation + inductive donation — see Carbocation Stability and Hyperconjugation). More neighbouring carbons = more sharing = more stable. The order is always $3°>2°>1°$.

PICTURE. Two branches drawn as two hills. The left hill (put $H$ on C1 → $+$ on C2) is a secondary cation: C2 has two carbon neighbours propping up the charge — a low, easy hill. The right hill (put $H$ on C2 → $+$ on C1) is a primary cation, propped by only one carbon — a tall, hard hill. Reaction rolls down the easy hill.

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Step 4 — $B{r}^{-}$ closes in → Markovnikov product

WHAT. The waiting $B{r}^{-}$ (negative) is pulled straight to the $+$ carbon (C2). New $C-Br$ bond forms.

WHY. Opposite charges attract; the nucleophile goes exactly where the positive charge is — no choice here, the earlier step already decided the carbon.

PICTURE. Pink arrow: the lone pair on $B{r}^{-}$ swings onto C2, capping the charge.

$C{H}_3-\stackrel{+}{C}H-C{H}_3+B{r}^{-}⟶C{H}_3-\underset{Br}{\mathrm{CH}}-C{H}_3$

$\boxed{\text{2-bromopropane}(\text{Markovnikov})}$

• $Br$ ends on C2 = the carbon that had fewer H's originally. "The rich (in H) get richer" — C1 kept all its H's and gained one more.

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Step 5 — Add peroxide: a brand-new kind of intermediate is born

WHAT. Now drop organic peroxide $RO-OR$ into the pot and warm it. The $O-O$ bond is feeble (~$150\text{kJ/mol}$; compare a normal $C-H$ near $\sim 410$). Heat snaps it evenly — one electron to each oxygen. That even split is called homolysis, and it makes species with a single unpaired electron: radicals.

WHY this tool — a radical and not a cation? Because a radical carries no charge, just a lonely electron (a "dot"). Its chemistry is driven by pairing that lonely electron, not by charge attraction. This is a totally different game — and it's why the answer will flip. See Free Radical Mechanisms and Bond Dissociation Energies.

PICTURE. The weak $O-O$ spring snaps into two $R{O}^{\bullet }$ pieces; each fragment grabs the $H$ off $H-Br$, spitting out $B{r}^{\bullet }$ — the bromine radical that starts everything.

$RO-OR\stackrel{\Delta }{\to }2R{O}^{\bullet }R{O}^{\bullet }+H-Br⟶RO-H+B{r}^{\bullet }$

• $\bullet$ — one unpaired electron (a single dot, not a charge).
• $B{r}^{\bullet }$ — the initiator that attacks the alkene next.

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Step 6 — $B{r}^{\bullet }$ adds FIRST — and picks the stable radical

WHAT. Here is the twist. In the ionic path $H^{+}$ went first. In the radical path the bromine radical goes first. It adds to one alkene carbon, and the other carbon keeps the lonely electron — it becomes a carbon radical.

WHY that carbon? Same logic as cations but for dots: a radical is more stable with more carbon neighbours ($3°>2°>1°$). $B{r}^{\bullet }$ therefore attaches to C1 so the unpaired electron lands on C2 → a secondary radical, the comfy one.

PICTURE. Blue dot-arrow: $B{r}^{\bullet }$ locks onto C1; the leftover dot appears on C2, propped by two carbons.

$C{H}_3-\underset{\text{C2}}{\mathrm{CH}}=\underset{\text{C1}}{C{H}_2}+B{r}^{\bullet }⟶C{H}_3-\stackrel{\bullet }{\underset{\text{C2}}{C}}H-\underset{\text{C1}}{C{H}_{2}Br}$

• $Br$ is now on C1 (the more-H carbon) — the opposite side from the ionic case!
• $\stackrel{\bullet }{C}$ — the secondary radical, chosen for stability.

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Step 7 — The radical grabs $H$ → anti-Markovnikov product + chain restarts

WHAT. The carbon radical (dot on C2) snatches an $H$ from a fresh $H-Br$. That completes the molecule and — crucially — regenerates a new $B{r}^{\bullet }$, which runs off to attack the next alkene. This self-feeding loop is a chain.

WHY it self-continues. One $B{r}^{\bullet }$ started it; step 7 spits out another $B{r}^{\bullet }$. So a tiny amount of peroxide can convert huge amounts of alkene.

PICTURE. The dot on C2 reaches to $H-Br$, plucks the $H$ (yellow arrow); $B{r}^{\bullet }$ flies out to begin again.

$C{H}_3-\stackrel{\bullet }{C}H-C{H}_{2}Br+H-Br⟶C{H}_3-C{H}_2-C{H}_{2}Br+B{r}^{\bullet }$

$\boxed{\text{1-bromopropane}(\text{anti-Markovnikov})}$

• $Br$ on C1 — opposite to 2-bromopropane. Same reagents, flipped product, all because the first-moving species changed.

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Step 8 — The edge cases: why ONLY $HBr$ (and the degenerate no-flip)

WHAT. For the chain to run, both propagation steps (Step 6: radical adds to alkene; Step 7: radical grabs H) must be downhill (exothermic). Check each halide.

WHY. A chain is only as fast as its slowest, most uphill step. If either step is uphill, the chain stalls and the molecule quietly reverts to the ordinary ionic (Markovnikov) path.

PICTURE. Three energy tracks. $HBr$'s two steps both slope down (green, chain runs). $HCl$'s H-grab step slopes up because $H-Cl$ is too strong to break ($431\text{kJ/mol}$). $HI$'s addition step slopes up because $I^{\bullet }$ barely sticks to the alkene (and $I^{\bullet }$ just recombines). Both broken chains → Markovnikov survives.

Degenerate case — no peroxide at all, or a symmetric alkene: with no peroxide there is no $B{r}^{\bullet }$ initiator, so we simply stay on Steps 2–4 (ionic, Markovnikov). And if the alkene were symmetric like $C{H}_2=C{H}_2$, both carbons are identical — both paths give the same product, so "Markovnikov vs anti" becomes meaningless. Orientation only matters for an unsymmetrical alkene.

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The one-picture summary

One diagram, both roads. The fork is a single question: is a $B{r}^{\bullet }$ around to move first? No peroxide → $H^{+}$ first → $+$ on the comfy (2°) carbon → $Br$ on C2 → 2-bromopropane. Peroxide → $B{r}^{\bullet }$ first → dot on the comfy (2°) carbon → $Br$ on C1 → 1-bromopropane. Compare with the different-reason anti-Markovnikov route in Hydroboration-Oxidation.

Recall Feynman: tell the whole walk in plain words

A double bond is two carbons holding hands twice; the second handshake is a loose cloud of electrons sticking out. Throw $H-Br$ at it. Normally the loose cloud grabs the little $H$ first; whichever carbon is left holding a "positive hot potato" wants the most friends around it to share the heat — that's the middle carbon — so $Br$ finishes on the middle carbon. That's 2-bromopropane, the "rich get richer" answer.

Now add a troublemaker (peroxide). Warm it and it falls apart into hungry loners called radicals, and one of them makes a bromine loner. This bromine loner is impatient — it jumps on first, onto the end carbon. That leaves the lonely electron on the middle carbon (its comfiest spot). The middle carbon then steals an $H$ from another $HBr$, finishing the molecule and burping out a fresh bromine loner to repeat forever. Because bromine jumped first onto the other side, the product is flipped: 1-bromopropane.

Why only bromine? Chlorine holds its $H$ too tightly to ever get stolen (the chain stalls); iodine is so floppy it won't even stick to the double bond (the chain never starts). Bromine is just right, so only $HBr$ does the flip.

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Connections

• Electrophilic Addition of HX
• Carbocation Stability and Hyperconjugation
• Free Radical Mechanisms
• Bond Dissociation Energies
• Alkenes - Addition Reactions
• Hydroboration-Oxidation