Hydroboration, Oxidation and Cleavage of Alkenes and Alkynes
A double or triple bond is a handle for putting oxygen where you want it. Different reagents attack the same bond with different regiochemistry and different depth of oxidation, up to cutting the molecule in two.
What a learner can do afterwards
- Predicts the anti-Markovnikov, syn product of hydroboration and oxidation and explains both features from the addition step
- Distinguishes the products of cold dilute and hot concentrated oxidation of the same alkene
- Uses ozonolysis products to work backwards to the position of the original double bond
- Predicts whether an alkyne gives an aldehyde, a ketone or an acid under stated conditions
1 · Read
A carbon double bond is a handle for attaching oxygen. You have two ways to add water across it, and they put the OH on opposite carbons. Direct acid hydration follows Markovnikov: the OH lands on the more substituted carbon, so propene gives 2-propanol. Hydroboration followed by oxidation does the opposite: the OH lands on the less substituted carbon, so propene gives 1-propanol.
Both the reversed position and the geometry come from one step. Boron and hydrogen add together across the same face of the double bond, with boron going to the less crowded carbon. Oxidation then swaps boron for OH while keeping that geometry, so H and OH finish on the same face. Chemists call this an anti-Markovnikov, syn addition.
Stronger oxidants take you further, and temperature decides how far. Cold dilute oxidant stops at a syn diol, with one OH on each carbon of the old double bond. Hot concentrated oxidant cuts the double bond entirely, leaving carbonyl pieces or acids: cyclohexene gives a diol under cold dilute conditions but hexanedioic acid under hot concentrated ones.
Ozonolysis is the cleanest cut: each double bond carbon becomes a carbonyl carbon. To work backwards, split each product at its C=O and rejoin the halves with a double bond: ethanal plus propanone rejoin to 2-methyl-2-butene. Alkynes follow the same choose-your-reagent logic: acid hydration of a terminal alkyne gives the Markovnikov ketone, hydroboration gives the aldehyde instead, and harsh oxidative cleavage gives acids.
Boron sends OH to the less substituted carbon on the same face, mild oxidation stops at the diol, and harsh oxidation cuts the bond into pieces you can rejoin.
2 · Watch
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8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.