Markovnikov vs anti-Markovnikov (peroxide effect, Kharasch)

1. The core rule: Markovnikov's Rule

WHY does this happen? Because the addition is ionic and goes through a carbocation.

Worked example Propene + HBr (no peroxide) — derive the product

$CH_3{-}CH{=}CH_2 + HBr \longrightarrow ?$

Step 1 — where does $H^+$ go?

Option A: $H^+$ on C1 (the $=CH_2$ end) → $+$ on C2 → secondary carbocation $CH_3{-}\overset{+}{C}H{-}CH_3$ .
Option B: $H^+$ on C2 → $+$ on C1 → primary carbocation $CH_3{-}CH_2{-}\overset{+}{C}H_2$ . Why pick A? The 2° cation is more stable than 1°. Chemistry takes the easier hill.

Step 2 — $Br^-$ attacks C2: $CH_3{-}CHBr{-}CH_3 \quad (\textbf{2-bromopropane})$ Why this step? The nucleophile $Br^-$ goes to the carbon carrying the $+$ charge.

2. The peroxide effect (Kharasch effect)

WHY does adding peroxide flip the answer? Because the mechanism changes from ionic to free-radical. Now stability of a carbon radical (not a carbocation) decides the orientation. Radical stability also follows $3° > 2° > 1°$ , but the atom that adds first is the bromine radical, not the proton — so the regiochemistry flips.

Formula The radical chain mechanism (derive each step)

Initiation — peroxide breaks the weak $O{-}O$ bond, then abstracts $H$ from HBr: $RO{-}OR \xrightarrow{\Delta}\ 2\,RO^{\bullet} \qquad RO^{\bullet} + H{-}Br \rightarrow RO{-}H + Br^{\bullet}$ Why? $O{-}O$ is a weak bond (~150 kJ/mol), easy to snap → makes radicals.

Propagation step 1 — $Br^{\bullet}$ adds to the alkene to form the more stable radical: $CH_3{-}CH{=}CH_2 + Br^{\bullet} \rightarrow CH_3{-}\overset{\bullet}{C}H{-}CH_2Br \ (2° \text{ radical})$ Why this carbon? $Br$ adds to C1 so the unpaired electron sits on C2 → secondary radical (more stable than primary).

Propagation step 2 — the radical grabs $H$ from another $HBr$ : $CH_3{-}\overset{\bullet}{C}H{-}CH_2Br + HBr \rightarrow CH_3{-}CH_2{-}CH_2Br + Br^{\bullet}$ The regenerated $Br^{\bullet}$ continues the chain. Net product: 1-bromopropane (anti-Markovnikov).

Figure — Markovnikov vs anti-Markovnikov (peroxide effect, Kharasch)

3. Side-by-side summary

	No peroxide	With peroxide (HBr only)
Mechanism	Ionic	Free radical
Intermediate	Carbocation	Carbon radical
Adds first	$H^+$	$Br^{\bullet}$
Product (propene + HBr)	2-bromopropane	1-bromopropane
Rule	Markovnikov	Anti-Markovnikov

Common mistake Steel-man the common errors

Mistake 1: "Peroxide effect works for HCl and HI too." Why it feels right: They're all hydrogen halides, so symmetry suggests they behave alike. The fix: Only $HBr$ has the right bond energetics for BOTH chain steps to be exothermic. Remember: only HBr.

Mistake 2: "In the radical mechanism, H still adds to the more-H carbon." Why it feels right: You memorized "H to richer carbon" as the universal rule. The fix: In radicals, Br adds first — so think about where the radical is most stable, not where $H^+$ goes. The orientation of $Br$ flips.

Mistake 3: "Markovnikov is about more stable product." Why it feels right: Often the major product is more stable, so it seems product-controlled. The fix: It's controlled by the stability of the intermediate (cation/radical), i.e. the transition state leading to it — not the final product.

Recall Feynman: explain to a 12-year-old

Imagine throwing a sticky ball ( $HBr$ ) at a magnet shaped like a double bond. Normally the small piece ( $H$ ) sticks to the crowded side and the big piece ( $Br$ ) sits on the lonely side — because that arrangement is the comfiest (most stable) halfway through. Now add a "troublemaker" chemical (peroxide). It changes the game: now the big bromine piece jumps on first, and it picks the side that makes the leftover comfiest. Because bromine went first to the other side, everything ends up flipped! And this trick only works with bromine — chlorine sticks too hard to its $H$ to play, and iodine is too floppy. So only $HBr$ flips with peroxide.

Flashcards

What does Markovnikov's rule say about H addition to an unsymmetrical alkene?

H adds to the carbon already bearing more hydrogens; X adds to the carbon with fewer H (more substituted).

What is the REAL mechanistic reason for Markovnikov addition?

The ionic mechanism forms the more stable carbocation (

3°>2°>1°

);

Br^-

then attacks that carbon.

What is the peroxide (Kharasch) effect?

In presence of organic peroxides, HBr adds anti-Markovnikov via a free-radical mechanism (Br on the carbon with more H).

Which hydrogen halides show the peroxide effect?

Only HBr (not HCl or HI).

Why doesn't HCl show the peroxide effect?

The H–Cl bond is too strong, making the H-abstraction propagation step endothermic, so the chain breaks.

Why doesn't HI show the peroxide effect?

Addition of I• to the alkene is endothermic and I• radicals recombine; the chain cannot propagate.

In the radical mechanism, which species adds to the alkene first?

The bromine radical Br•.

Product of propene + HBr WITHOUT peroxide?

2-bromopropane (Markovnikov).

Product of propene + HBr WITH peroxide?

1-bromopropane (anti-Markovnikov).

What initiates the radical chain in the Kharasch reaction?

Homolysis of the weak O–O bond of the peroxide, generating RO• which then abstracts H from HBr to give Br•.

What controls the regiochemistry — intermediate stability or product stability?

Stability of the intermediate (carbocation or radical), via the lower-energy transition state.

Connections

Alkenes - Addition Reactions
Carbocation Stability and Hyperconjugation
Free Radical Mechanisms
Bond Dissociation Energies
Electrophilic Addition of HX
Hydroboration-Oxidation (another anti-Markovnikov route, different reason)

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho yaar, jab HBr jaisa unsymmetrical reagent kisi alkene ke double bond par add hota hai, to sawaal yeh hai ki $H$ kis carbon par jaayega aur $Br$ kis par. Markovnikov ka rule kehta hai: "ameer aur ameer banta hai" — yaani jis carbon par pehle se zyada hydrogen hain, usi par naya $H$ jaata hai, aur $Br$ kam-hydrogen wale (more substituted) carbon par. Lekin yeh rattne ki cheez nahi hai — iska asli reason hai carbocation stability. Reaction ionic mechanism se chalti hai, aur woh wahi raasta chunti hai jisme zyada stable carbocation bane ( $3° > 2° > 1°$ ). Isiliye propene + HBr se 2-bromopropane banta hai.

Ab twist aata hai — agar tum reaction mein peroxide daal do (Kharasch effect / peroxide effect), to mechanism poora badal jaata hai. Ab ionic ki jagah free radical mechanism chalta hai. Peroxide ka weak $O{-}O$ bond toot ke radicals banata hai, aur phir $Br^{\bullet}$ pehle add hota hai (na ki $H^+$ ). $Br$ us carbon par jaata hai jisse zyada stable radical bane. Result: orientation ulta ho jaata hai — propene + HBr + peroxide se ab 1-bromopropane (anti-Markovnikov) banta hai.

Sabse important baat jo exam mein poochi jaati hai: yeh peroxide effect sirf HBr ke saath hota hai, HCl aur HI ke saath nahi. Kyun? Kyunki radical chain ke dono propagation steps exothermic hone chahiye. HCl ka $H{-}Cl$ bond bahut strong hai (H-abstraction step endothermic ho jaata hai), aur HI mein $I^{\bullet}$ ka alkene par add hona endothermic hai aur radicals aapas mein recombine kar jaate hain. Sirf HBr "Goldilocks" hai — bilkul sahi bond energies. Isiliye sirf HBr flip karta hai.

Yaad rakhne ka trick: regiochemistry hamesha intermediate ki stability se decide hoti hai — cation ho ya radical, dono mein $3°>2°>1°$ . Bas farak yeh hai ki ionic mein $H^+$ pehle jaata hai, aur radical mein $Br^{\bullet}$ pehle jaata hai — isliye answer flip ho jaata hai.