Markovnikov vs anti-Markovnikov (peroxide effect, Kharasch): Question bank

Intuition How to use this bank

These are the questions that separate "I memorized the rule" from "I understand the mechanism." Cover the answer, say your reasoning out loud, then reveal. If your reason is right but you guessed the answer, you still lose. This page is the child of the parent topic; if any word here is unfamiliar, go read the parent first.

Vocabulary anchor (used throughout, no symbol before its meaning):

Carbocation = a carbon atom missing a bonding electron pair, so it carries a $+$ charge. It craves electrons.
Free radical = an atom or group with one unpaired electron (drawn as a dot $^{∙}$ ). Neutral, but hungry to pair up.
Regiochemistry = which carbon each piece lands on (the "address" of $H$ and $B r$ ).
Substituted carbon = a carbon bonded to more other carbons (fewer hydrogens).
Degree of substitution — $1°$ , $2°$ , $3°$ = a shorthand for how many other carbons touch the electron-poor carbon. $1°$ ("primary") = bonded to one other carbon; $2°$ ("secondary") = two; $3°$ ("tertiary") = three. More neighbours = more substituted = more stable cation/radical. Look at the picture below before reading on.
$R$ (the "R-group") = a stand-in for "any carbon chain / organic group." So $R O - O R$ just means "a peroxide with organic tails on each side," and $R O^{∙}$ is the oxygen radical left when that molecule splits.

Before the questions, one shared picture to anchor every " $1°/2°/3°$ " and every "which piece adds first" answer:

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

True or false — justify

TF1. "Markovnikov's rule is a rule about which product is more stable."

False. It is controlled by the stability of the intermediate (carbocation/radical) and the transition state leading to it, not the stability of the final molecule. See Carbocation Stability and Hyperconjugation.

TF2. "Adding peroxide makes the reaction go anti-Markovnikov because it changes the thermodynamics of the product."

False. The product energies barely change; peroxide switches the mechanism from ionic to free-radical, so a different intermediate now decides the address of

B r

TF3. "In the peroxide (Kharasch) mechanism,

H

still adds to the carbon with more hydrogens."

False. In the radical path the bromine radical adds first, not

H

B r

lands on the less-substituted (more-H) carbon so the leftover radical is the more stable one;

H

comes second.

TF4. "Both carbocations and carbon radicals follow the stability order

3° > 2° > 1°

(tertiary > secondary > primary)."

True. Both are electron-deficient at that carbon, so alkyl groups stabilize them (hyperconjugation and

+ I

donation) the same way — more carbon neighbours (

3°

) means more stabilization. See Free Radical Mechanisms.

TF5. "The peroxide effect works for

H C l

H B r

and

H I

False. Only

H B r

has "Goldilocks" bond energetics that make both propagation steps exothermic. See Bond Dissociation Energies.

TF6. "Since radical and ionic paths both favour the

3°/2°

intermediate, both give the same product."

False. They favour the same intermediate stability, but a different piece adds first (

H^{+}

B r^{∙}

). Same logic, opposite

B r

address → opposite products.

TF7. "The peroxide itself ends up inside the final organic product."

False. Peroxide only initiates the chain (it makes

B r^{∙}

). It is consumed in tiny amounts and is not part of the added-across-the-bond product.

TF8. "Ethene (symmetrical alkene) gives different products with and without peroxide."

False. With

C H_{2} = C H_{2}

both carbons are identical, so Markovnikov vs anti-Markovnikov are indistinguishable — you get bromoethane either way. Regiochemistry only matters for unsymmetrical alkenes.

Spot the error

SE1. "Propene + HBr with peroxide → 2-bromopropane."

Error: that's the Markovnikov (no-peroxide) product. With peroxide the radical path gives 1-bromopropane —

B r

on the terminal (more-H) carbon.

SE2. "Initiation:

B r^{∙}

splits the peroxide

O - O

bond."

Error: it's the other way. Heat/light homolyses the weak

O - O

bond into two

R O^{∙}

(where

R

= any organic tail), and that alkoxy radical abstracts

H

from

H B r

to create

B r^{∙}

SE3. "In the ionic mechanism,

B r^{-}

adds first and then

H^{+}

Error:

π

electrons grab

H^{+}

first (making the carbocation); only then does

B r^{-}

attack the positive carbon. See Electrophilic Addition of HX.

SE4. "HCl shows no peroxide effect because

C l^{∙}

won't form."

Error:

C l^{∙}

forms fine. The block is at propagation step 2: the very strong

H - C l

bond makes the radical's H-abstraction endothermic, so the chain dies.

SE5. "HI shows no peroxide effect for the same reason as HCl (bond too strong)."

Error: opposite cause. For HI the problem is the

C - I

/ addition step:

I^{∙}

addition to the alkene is endothermic and

I^{∙}

radicals just recombine. Different failure point, same "chain won't run" result.

SE6. "The

1°

carbocation is chosen in propene + HBr because it's less crowded."

Error: less crowding is irrelevant here; the $2°$ carbocation is more stable (more hyperconjugation) and that's the one the reaction picks.

SE7. "Hydroboration–oxidation gives anti-Markovnikov product by the same radical chain as Kharasch."

Error: Hydroboration-Oxidation is anti-Markovnikov for a different reason — boron (not a radical) adds to the less-hindered carbon in a concerted step. Same outcome, unrelated mechanism.

Why questions

WHY1. "Why does the reaction 'choose' the more stable intermediate?"

A more stable intermediate sits at lower energy, and the transition state leading to it is also lower — so that pathway has the smaller energy barrier and runs faster. Chemistry takes the easier hill.

WHY2. "Why does

B r

(not

H

) adding first flip the regiochemistry?"

The species that adds first decides which carbon becomes the intermediate.

H^{+}

first → cation logic puts

B r

on the substituted carbon;

B r^{∙}

first → radical logic puts

B r

on the unsubstituted carbon. The "first mover" determines

B r

's address.

WHY3. "Why must BOTH propagation steps be exothermic, not just one?"

A chain only sustains itself if every link is downhill; one endothermic step becomes a bottleneck that stalls the cycle and the radical chain fizzles out.

WHY4. "Why is the

O - O

bond the one that breaks in initiation?"

It is one of the weakest bonds around (~150 kJ/mol), so mild heat/light supplies enough energy to snap it homolytically before stronger bonds break. See Bond Dissociation Energies.

WHY5. "Why do more alkyl groups stabilize both the cation and the radical?"

Adjacent

C - H

C - C

bonds donate electron density into the electron-poor centre (hyperconjugation + inductive

+ I

), spreading the deficiency out. More alkyl neighbours = more spreading = more stable.

WHY6. "Why doesn't peroxide affect the addition of

H B r

to a symmetrical alkene's rate-controlling choice?"

Because there is no choice to make — both carbons are equivalent, so no matter which mechanism runs, only one product is possible.

Edge cases

EC1. "What if the alkene is

2

-methylbut-

2

-ene type, where BOTH ionic and radical routes could give a

3°

(tertiary) centre — do the products differ?"

They can still differ: the ionic route builds the

3°

cation (so

B r

ends on that carbon), while the radical route makes the

3°

radical by putting

B r

on the other carbon. Same stability class, opposite

B r

address.

EC2. "What happens with

H B r

+ peroxide but the alkene is fully symmetrical like but-2-ene?"

Both carbons carry one H and one methyl — they are equivalent, so anti-Markovnikov and Markovnikov coincide. You cannot tell the mechanisms apart from the product.

EC3. "If you use

H B r

with NO peroxide and NO other initiator, what do you get?"

The ionic path dominates → Markovnikov product. Radicals need an initiator (peroxide/UV) to start; without one, the ionic route wins.

EC4. "You run

H B r

+ peroxide, but the peroxide is stale/decomposed. Prediction?"

No usable

R O^{∙}

→ no

B r^{∙}

initiation → the chain never starts, so the reaction defaults to the ionic (Markovnikov) product.

EC5. "Could adding a radical inhibitor to an

H B r

+ peroxide reaction switch the product back to Markovnikov?"

Yes. An inhibitor mops up radicals, killing the chain, so the ionic pathway takes over and you recover the Markovnikov product — direct evidence the peroxide effect is radical-based.

EC6. "Terminal alkyne / very hindered alkene aside — what if the two possible radicals are equally stable (e.g. a symmetric internal position)?"

With no stability preference,

B r

shows no regioselectivity and you get a mixture; the whole "flip" only matters when one intermediate is clearly more stable than the other.

Connections

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

Recall One-line self-test

If someone says "peroxide makes it anti-Markovnikov because the product is more stable," what's the single-word category of their mistake? Answer ::: Mechanism vs thermodynamics confusion — it's intermediate-controlled (kinetic), not product-controlled.