Markovnikov vs anti-Markovnikov (peroxide effect, Kharasch): Worked examples

This page is a decision-drill for the parent topic. The parent gave you the two mechanisms; here you will run them on every kind of molecule the exam can throw, including the sneaky degenerate cases (symmetrical alkenes, terminal vs internal double bonds, halides that don't play, energy sign-flips). Nothing here contradicts the parent — we just go one level deeper: before naming a product, we always ask "which intermediate is more stable, and what does that picture look like?"

Prerequisites we lean on: Carbocation Stability and Hyperconjugation, Free Radical Mechanisms, Bond Dissociation Energies, Electrophilic Addition of HX, and Alkenes - Addition Reactions.

The scenario matrix

Before working examples, let's list every case class this topic can produce. Think of it as a checklist — every cell must be hit by at least one worked example below.

Cell	What makes it distinct	Which example
A. Ionic, clear winner	Unsymmetrical alkene, no peroxide, one cation clearly more stable	Ex 1
B. Radical, clear winner	Same alkene, peroxide present, flip the answer	Ex 2
C. Degenerate — symmetrical alkene	Both carbons equal → peroxide changes nothing	Ex 3
D. Wrong halide (HCl / HI)	Peroxide present but chain dies → stays Markovnikov	Ex 4
E. Tie-break by substituent count	Internal alkene where 2° vs 3° decides	Ex 5
F. Zero-substituent edge (ethene)	No "more vs fewer H" distinction at all	Ex 6
G. Real-world word problem	You're told the product, deduce the conditions	Ex 7
H. Exam twist — compare to hydroboration	Same anti-Markovnikov result, different reason	Ex 8

The stability ladder we will use everywhere

Both the carbocation and the carbon radical get comfier the more carbon groups surround them, because those neighbours smear the charge/electron out (this is hyperconjugation). Figure s01 draws that ladder once so we never re-derive it.

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

Example 1 — Cell A (ionic, clear winner)

Figure s02 shows the two possible proton landings and the cation each one leaves behind.

Step 1 — decide mechanism. Why this step? No peroxide → the $π$ electrons act as the base and grab $H^{+}$ first → ionic path.

Step 2 — where does $H^{+}$ go? Why this step? The proton attaches to whichever carbon leaves the more stable $+$ on the other carbon (look at the amber " $+$ " in s02).

$H^{+}$ on C1 → $+$ on C2 = 2° cation ✅ comfier.
$H^{+}$ on C2 → $+$ on C1 = 1° cation ✗ worse.

Step 3 — $B r^{-}$ attacks the cation carbon (C2). Why this step? The nucleophile goes to the positive charge.

$C H_{3} - C H B r - C H_{3} (2-bromopropane)$

Verify: Br sits on the more substituted carbon (fewer H's) → this is Markovnikov. Molecular formula $C_{3} H_{7} B r$ : $3 \times 12 + 7 \times 1 + 80 = 123 g/mol$ . ✔

Example 2 — Cell B (radical, clear winner)

Step 1 — mechanism. Why? Peroxide and HBr → both conditions met → radical chain (see Free Radical Mechanisms).

Step 2 — who adds first? Why? In the radical chain the first thing to touch the alkene is $B r^{∙}$ (not $H^{+}$ ). This is the switch that flips everything.

Step 3 — where does $B r^{∙}$ add? Why? It adds to the carbon that leaves the unpaired electron on the more stable carbon.

$B r^{∙}$ on C1 → $∙$ on C2 = 2° radical ✅
$B r^{∙}$ on C2 → $∙$ on C1 = 1° radical ✗

Step 4 — the 2° radical grabs $H$ from HBr, regenerating $B r^{∙}$ .

$C H_{3} - C H_{2} - C H_{2} B r (1-bromopropane)$

Verify: Same $C_{3} H_{7} B r$ , $M = 123$ g/mol, but Br is now on the less substituted carbon → anti-Markovnikov. Ex 1 and Ex 2 are constitutional isomers — same formula, Br in different spots. ✔

Example 3 — Cell C (degenerate: symmetrical alkene)

Step 1 — inspect symmetry. Why? Both double-bond carbons carry exactly one H and one CH₃. They are identical.

Step 2 — ionic path. Either carbon gives a 2° cation → no preference.

Step 3 — radical path. Either carbon gives a 2° radical → no preference.

$C H_{3} - C H B r - C H_{2} - C H_{3} (2-bromobutane) either way$

Verify: Both mechanisms give the same product because there is no "more vs fewer H" to distinguish. Markovnikov and anti-Markovnikov collapse to the same answer for a symmetrical alkene — the peroxide effect is invisible here. Formula $C_{4} H_{9} B r$ , $M = 137$ g/mol. ✔

Example 4 — Cell D (wrong halide, chain dies)

Step 1 — recall the energetics. Why? A radical chain only runs if both propagation steps are exothermic (see Bond Dissociation Energies).

Step 2 — HCl. The $H - C l$ bond is very strong ( $\approx 431$ kJ/mol). The step "radical grabs H from HCl" would need to break this strong bond → endothermic → chain stalls.

Step 3 — HI. The step " $I^{∙}$ adds to the alkene" is endothermic (weak C–I bond doesn't pay off), and $I^{∙}$ radicals just recombine → chain stalls.

Step 4 — so both stay ionic: $HCl \Rightarrow C H_{3} - C H C l - C H_{3} (2-chloropropane, Markovnikov)$ $HI \Rightarrow C H_{3} - C H I - C H_{3} (2-iodopropane, Markovnikov)$

Figure s03 shows the "Goldilocks" energy window: only HBr sits where both arrows point downhill.

Verify: Only HBr has both steps exothermic → the peroxide effect is HBr-exclusive. Sanity check via bond strengths: $431 (HCl) > 366 (HBr) > 298 (HI)$ kJ/mol; HBr is the middle "just right." ✔

Example 5 — Cell E (tie-break: 2° vs 3°)

Step 1 — mechanism. No peroxide → ionic → $H^{+}$ first.

Step 2 — count both cation options. Why? We compare 3° vs 2°, not 2° vs 1°.

$H^{+}$ on the $C H$ carbon → $+$ on the $C (C H_{3})_{2}$ carbon = 3° cation ✅ (three carbons attached).
$H^{+}$ on the $C (C H_{3})_{2}$ carbon → $+$ on the $C H$ carbon = 2° cation ✗.

Step 3 — $B r^{-}$ attacks the 3° carbon.

$(C H_{3})_{2} C B r - C H_{2} - C H_{3} (2-bromo-2-methylbutane)$

Verify: Br on the carbon with zero hydrogens (most substituted) — textbook Markovnikov, and the 3° cation is the most stable choice on our s01 ladder. Formula $C_{5} H_{11} B r$ , $M = 5 \times 12 + 11 + 80 = 151$ g/mol. ✔

Example 6 — Cell F (zero-substituent edge: ethene)

Step 1 — look for asymmetry. Why? Both carbons are $C H_{2}$ — perfectly identical, like Ex 3 but even simpler.

Step 2 — both paths. Ionic gives $+$ on a 1° carbon; radical gives $∙$ on a 1° carbon — but since both ends are identical, "where Br lands" is meaningless.

$C H_{3} - C H_{2} B r (bromoethane) regardless of conditions$

Verify: With no substituent difference, Markovnikov's rule has nothing to decide — this is the degenerate floor of the whole topic. Formula $C_{2} H_{5} B r$ , $M = 109$ g/mol. ✔

Example 7 — Cell G (real-world / reverse-engineering word problem)

Step 1 — identify the product's regiochemistry. Why? 1-bromopropane has Br on the terminal (more-H) carbon → anti-Markovnikov.

Step 2 — map back to mechanism. Why? Anti-Markovnikov HBr addition can only come from the radical chain.

Step 3 — state the required conditions. A radical initiator (organic peroxide, or UV light) must be present, and the halide must be HBr (from Ex 4 we know HCl/HI can't do it).

$Conditions: HBr + peroxide (or h ν), free-radical mechanism$

Verify: Cross-check with Ex 2 — that forward reaction (propene + HBr + peroxide) gave 1-bromopropane. The reverse deduction is consistent. ✔

Example 8 — Cell H (exam twist: same result, different reason)

Step 1 — the reagent. Why? Hydroboration-Oxidation: $B H_{3}$ then $H_{2} O_{2} / O H^{-}$ puts the $O H$ on the less substituted carbon.

Step 2 — the anti-Markovnikov reason. Why? Here boron (electron-poor) adds to the less hindered / terminal carbon in a concerted, non-radical step — the regiochemistry is set by sterics + boron's electronics, not by radical stability.

Step 3 — contrast. The peroxide effect (Ex 2) is anti-Markovnikov because a $B r^{∙}$ radical picks the more stable radical. Hydroboration is anti-Markovnikov for a completely different reason. Same outcome, different cause.

$propene 1) B H_{3} 2) H_{2} O_{2} / O H^{-} C H_{3} C H_{2} C H_{2} O H (1-propanol)$

Verify: Product is the anti-Markovnikov alcohol (OH on terminal C). Formula $C_{3} H_{8} O$ , $M = 60$ g/mol. The reasoning differs from the Kharasch radical route — exactly the point the examiner tests. ✔

Recall drills

Recall Which cell does each phrase belong to?

"Symmetrical alkene, peroxide irrelevant" ::: Cell C (Ex 3). "Peroxide present but nothing flips because it's HCl" ::: Cell D (Ex 4). "Anti-Markovnikov but NOT radical" ::: Cell H (Ex 8, hydroboration). "3° vs 2° cation tie-break" ::: Cell E (Ex 5).

Recall Answers to lock in

Propene + HBr, no peroxide ::: 2-bromopropane. Propene + HBr, peroxide ::: 1-bromopropane. But-2-ene + HBr, ± peroxide ::: 2-bromobutane (both). Ethene + HBr, ± peroxide ::: bromoethane (both). 2-methylbut-2-ene + HBr ::: 2-bromo-2-methylbutane.

Connections

Markovnikov vs anti-Markovnikov (peroxide effect, Kharasch) (Hinglish) — parent (Hinglish)
Electrophilic Addition of HX · Carbocation Stability and Hyperconjugation · Free Radical Mechanisms · Bond Dissociation Energies · Alkenes - Addition Reactions · Hydroboration-Oxidation