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When HX or water adds to an unsymmetrical alkene, the hydrogen goes where it leaves the more stable, more substituted carbocation, and the other group ends up on that carbon; Markovnikov's rule describes the result, and a radical mechanism reverses it.
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You will name the major products of HBr addition and hydration, say where the hydrogen goes and why, classify the carbocation formed, and explain how peroxides reverse the outcome.
You know addition to an alkene goes through a carbocation. When the alkene is unsymmetrical, the carbocation can form in two places, and they are not equally stable.
Regiochemistry is which atom a new group ends up on. Markovnikov's rule: in adding HX to an alkene, H goes to the carbon that already has more hydrogens. The more substituted carbon has more carbon groups attached.
Add HBr to propene, CH2=CH–CH3. H+ can go to either carbon:
Secondary cations are far more stable than primary ones, so almost all the reaction goes the first way, and bromide adds to the middle carbon: 2-bromopropane.
Markovnikov's rule describes the result — H went to the carbon that already had more hydrogens — but the reason is the carbocation. Reasoning from the intermediate works even when counting hydrogens is awkward, and it explains the exceptions.
Another way: steps
To predict regiochemistry:
Acid-catalysed hydration follows the same path: H+ adds to give the more stable cation, water attacks it, and the OH ends up on the more substituted carbon — propene gives propan-2-ol. And when the mechanism changes, the outcome can flip: with peroxides present, HBr adds by radicals, a bromine atom adds first, and it goes where it leaves the more stable radical, so propene gives 1-bromopropane. The rule was never about hydrogens; it was about the intermediate.
Regiochemistry is chosen by placing every new group on the same carbon. H and X go on different carbons.
Markovnikov's rule is a law. It describes the usual result of a carbocation path.
Both products form equally. The more stable cation's product dominates.
Peroxides make primary cations more stable. They change the mechanism.
H+ to the CH2 end leaves a tertiary cation, (CH3)3C+.
Three carbons on the positive carbon.
H+ to the other carbon would leave a primary cation.
Far less stable.
So bromide adds to the tertiary carbon: 2-bromo-2-methylpropane.
More stable cation, major product.
Which carbocation does the favoured route make?
A secondary cation on carbon 2.
Where does the OH go?
On carbon 2: butan-2-ol.
HBr adds to but-1-ene, CH2=CHCH2CH3. Name the major product.
The major product is p.
HBr adds to propene, CH2=CH–CH3. To which carbon of the double bond does the hydrogen add?
When H+ adds to 2-methylbut-2-ene, (CH3)2C=CHCH3 by the favoured route, how many carbons are bonded to the carbon that carries the positive charge?
Answer:
Match each addition to its major product.
| 2-bromopropane | 2-bromo-2-methylpropane | propan-2-ol | 2-methylpropan-2-ol | |
|---|---|---|---|---|
| propene + HBr | ||||
| 2-methylpropene + HBr | ||||
| propene + H2O, acid | ||||
| 2-methylpropene + H2O, acid |
HBr adds to propene in the presence of peroxides, which start a radical chain. The product is 1-bromopropane. Why is the outcome reversed?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Water adds to 2-methylpropene with an acid catalyst. Name the major alcohol.
The major alcohol is p.
You can predict which carbon a group ends up on. Tell someone why propene gives 2-bromopropane. Next: why some carbocations are more stable than others.
8. Your turn: but-1-ene with water and acid, step 3