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Organic Chemistry 1 · Radical Chemistry

Radical Addition of HBr to Alkenes

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On this page 7 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Study tools

In 30 seconds

With present, adds to an alkene anti-Markovnikov: bromine attaches to the less substituted carbon, hydrogen to the more substituted one. The mechanism is a radical chain — the peroxide initiates Br•, Br• adds to the less substituted carbon to give the more stable carbon radical, which then abstracts H from HBr to complete the product and regenerate Br•. The works only for HBr: with HCl the H-abstraction step is too endothermic (H–Cl too strong), and with HI the iodine atom adds reversibly and abstracts H poorly.

Why this matters

Anti-Markovnikov hydrobromination installs bromine at the terminal (less substituted) position of an alkene, a ionic addition cannot achieve; terminal bromides undergo clean SN2 substitution (topic 20) and build carbon skeletons (topic 42). The same peroxide/radical logic underlies industrial free-radical polymerization. Safety boundaries: peroxides are shock- and heat-sensitive explosion hazards, and HBr is corrosive; no operational, quantity, PPE, or disposal instructions are given here — all such work must follow approved institutional safety documentation under trained supervision.

The college version

1. The Peroxide Effect and Anti-Markovnikov Addition

Normal ionic hydrohalogenation (topic 28) adds HBr with Markovnikov regiochemistry: H⁺ adds to the less substituted carbon to form the more stable carbocation, then Br⁻ adds to the more substituted carbon. With peroxides (ROOR), the reaction takes the anti-Markovnikov pathway — Br to the less substituted carbon, H to the more substituted carbon. This reversal is the peroxide effect (Kharasch effect). The peroxide is a radical initiator that homolyzes (RO–OR ⇀ 2 RO•) to start a radical chain overriding the ionic pathway.

2. The Radical-Chain Mechanism

  1. Initiation: RO–OR ⇀ 2 RO• (peroxide homolysis, heat/light). The alkoxy radical abstracts H from HBr: RO• + HBr → ROH + Br•. Br• is the chain carrier.
  2. Propagation step 1 (addition): Br• adds to the less substituted carbon. Its unpaired electron (fishhook ⇀) attacks one carbon of the π bond, forming a C–Br bond and leaving the unpaired electron on the other (more substituted) carbon — the more stable carbon radical. Adding Br• to the more substituted carbon would give the less stable primary radical, so radical stability sets this regiochemistry (topic 36).
  3. Propagation step 2 (H abstraction): The carbon radical abstracts a hydrogen from HBr (fishhook on H–Br), forming the C–H bond and regenerating Br•. The product is complete and Br• continues the chain.
  4. Termination: radical combination (Br• + Br• → Br₂; R• + R• → R–R).

3. Why HCl and HI Fail the Peroxide Effect

The peroxide effect is useful only for HBr, for energetic reasons in the propagation steps:

  • HCl: The H–Cl bond (≈ 103 kcal/mol) is much stronger than H–Br (≈ 88 kcal/mol). The step R• + HCl → R–H + Cl• is strongly endothermic (breaking a strong H–Cl to form a weaker C–H), so it is far too slow — the chain stalls.
  • HI: The H–I bond (≈ 71 kcal/mol) is weak, so R• + HI → R–H + I• is thermodynamically allowed, but the resulting I• is too unreactive: iodine atoms add to alkenes reversibly and abstract hydrogen poorly, so the chain does not propagate productively. HI also tends to reduce peroxides.

Net result: only HBr has both propagation steps favorable and fast enough for a useful anti-Markovnikov .

How it works

  1. Peroxide homolyzes to alkoxy radicals; the alkoxy radical abstracts H from HBr to give Br• (the chain carrier).
  2. Br• adds to the less substituted alkene carbon, making the more stable carbon radical.
  3. The carbon radical abstracts H from HBr, completing the product and regenerating Br•; the chain repeats until termination.
  4. Regiochemistry follows radical stability: Br on the less substituted carbon, H on the more substituted carbon.

Common confusions

Do not confuseWithDifference
Anti-Markovnikov HBr (peroxides)Markovnikov HBr (no peroxides)Radical pathway vs ionic (carbocation) pathway
Carbon radical intermediateCarbocation intermediateRadical is neutral (7 e⁻); carbocation positive (6 e⁻)
Peroxide effect (HBr only)General hydrohalogenation (HCl/HI)HCl/HI lack a useful peroxide effect
Alkyl bromide product (saturated)Allylic bromide (topic 39)Radical addition gives a saturated C–Br; allylic bromination keeps the double bond
Peroxide (initiator)Br₂ (bromine source)Peroxide only starts the chain; HBr supplies the bromine
Markovnikov rule (prediction)Mechanism (explanation)The rule predicts; radical stability explains why

Memory aids

"Peroxide Puts Br on the Poor (less substituted) carbon." Chain of events: "ROOR → RO• → Br• → Br adds to the less-substituted carbon → more-stable radical → grab H from HBr." For the "HBr only" rule: "HCl is too Strong, HI is too Sloppy, only HBr is Just Right" — the Goldilocks of radical hydrohalogenation.

Quick review

Topic Recap

In the presence of peroxides, HBr adds to alkenes anti-Markovnikov via a radical chain: peroxide homolysis generates Br•, Br• adds to the less substituted carbon to form the more stable radical, and H abstraction from HBr completes the product and regenerates Br•. Regiochemistry is set by radical stability, not the Markovnikov carbocation rule. The peroxide effect is useful only for HBr — HCl's strong H–Cl bond makes the chain endothermic and HI's iodine atom is too unreactive — so HCl and HI give only the ionic (Markovnikov) product.

Knowledge Check

  1. What product forms when propene reacts with HBr in the presence of peroxides, and what is the regiochemistry called?
  2. In the radical mechanism, which step sets the regiochemistry, and why does Br• add to the less substituted carbon?
  3. What species is the chain carrier in radical HBr addition, and how is it regenerated?
  4. Why does the peroxide effect fail for HCl?
  5. How does the product of radical HBr addition differ from that of ionic HBr addition of the same alkene?

Answers and Rationales

  1. 1-Bromopropane (CH₃CH₂CH₂Br), by anti-Markovnikov addition (Br on the less substituted carbon). Rationale: the peroxide-initiated radical chain reverses the normal regiochemistry.
  2. The addition of Br• to the alkene (propagation step 1). Br• adds to the less substituted carbon so the unpaired electron ends up on the more substituted carbon, giving the more stable radical. Rationale: the pathway through the more stable radical is favored.
  3. Br• is the chain carrier, regenerated when the carbon radical abstracts H from HBr in the final propagation step. Rationale: Br• is consumed in addition and remade in H-abstraction, sustaining the chain.
  4. The H–Cl bond (≈ 103 kcal/mol) is too strong, making R• + HCl → R–H + Cl• strongly endothermic and slow, so the chain cannot propagate. Rationale: propagation energetics gate the reaction.
  5. Radical (peroxide) addition gives the anti-Markovnikov alkyl bromide (Br on the less substituted carbon); ionic addition gives the Markovnikov product (Br on the more substituted carbon). Rationale: radical vs carbocation intermediates select opposite sites.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine hydrogen and bromine arriving at a two-seat park bench (the alkene). Normally hydrogen — the eager electron-grabber — rushes to the seat that best supports a positive charge (the less substituted carbon), leaving bromine the other seat. That is Markovnikov addition.

Add peroxides and the rules flip: a peroxide "magnet" (radical initiator) pulls bromine in first, and bromine — a lone, single-electron traveler — sits at the end that leaves the middle most stable (the more substituted radical). Hydrogen follows to that more substituted side — anti-Markovnikov.

Where this stops being exact: the "bench with friends" is a cartoon of electron density, not a force. The real reason for the flip is which intermediate is more stable — a carbocation (ionic path) versus a carbon radical (radical path) — decided by orbital effects, not "friends." The analogy also hides why the trick works only for HBr: each propagation step's energy must be favorable, and for HCl and HI they are not.

Simple Example

Treating propene (CH₃–CH=CH₂) with HBr in the presence of peroxides gives 1-bromopropane (CH₃CH₂CH₂Br) — bromine on the terminal (less substituted) carbon, the anti-Markovnikov product. Without peroxides, the same HBr gives 2-bromopropane (CH₃CHBrCH₃), the Markovnikov product.

Worked example

Radical addition of HBr to propene, moving electrons before naming the product.

  1. Initiation. The peroxide's weak O–O bond breaks homolytically (two fishhooks ⇀, one electron to each oxygen): RO–OR ⇀ 2 RO•. Then RO• abstracts H from HBr (fishhook): RO• + H–Br → ROH + Br•.
  2. Propagation step 1 — Br• adds (electrons first). Br• uses its unpaired electron to attack one carbon of the π bond; one π electron pairs with Br to make the C–Br bond, the other π electron ends up alone on the second carbon. Br• adds to the terminal (less substituted) carbon, leaving the unpaired electron on the internal (more substituted) carbon — the more stable secondary radical, not an unstable primary radical.
  3. Propagation step 2 — H abstraction. The secondary radical abstracts a hydrogen from HBr (fishhook on H–Br), forming a C–H bond at the more substituted carbon and regenerating Br•.
  4. State the product. Br on the terminal carbon, H on the internal carbon: 1-bromopropane (CH₃CH₂CH₂Br) — the anti-Markovnikov product. Note: this is a saturated alkyl bromide, unlike the allylic bromide of topic 39.
  5. Termination (minor): Br• + Br• → Br₂, or coupling of alkyl radicals.

Key takeaways

  • High yield: With peroxides, HBr adds anti-Markovnikov (Br to the less substituted carbon).
  • High yield: Without peroxides, HBr adds Markovnikov via a carbocation (topic 28).
  • High yield: Regiochemistry is set by radical stability: Br• adds to give the more substituted radical.
  • High yield: The peroxide effect is specific to HBr — it fails for HCl (H-abstraction too endothermic) and HI (I• too unreactive, adds reversibly).
  • High yield: The chain carrier is Br•, regenerated in the final propagation step.
  • High yield: The product is an alkyl bromide (saturated C–Br), not an allylic bromide.
  • Peroxides are radical initiators, not the bromine source — they only start the chain.

Keep learning

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Practice Organic Chemistry 1

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Study tools & related lessonsYou’ll learn to · Key vocabulary · Related

You’ll learn to

  • Write the radical-chain mechanism for the anti-Markovnikov addition of HBr to an alkene in the presence of peroxides (the peroxide effect).
  • Explain the regiochemistry — why bromine ends up on the less substituted carbon — using intermediate radical stability.
  • Explain why HCl and HI lack a useful peroxide effect using propagation-step energies and bond strengths.
  • Compare radical addition with ionic hydrohalogenation and use the conditions to predict the product.

Key vocabulary

Radical addition
Addition across a π bond via radicals
HBr
Hydrogen bromide
Peroxides (ROOR)
Radical initiators that homolyze easily
Peroxide effect
Reversal of HBr regiochemistry with peroxides
Anti-Markovnikov addition
H to more substituted, Br to less substituted carbon
Radical-chain mechanism
Initiation/propagation/termination
Regiochemistry
Which atom ends up on which carbon
Bond-dissociation energy
Energy to break a bond homolytically

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