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SolutionsHydroboration

1. Three ways to hydrate

Predict the result of the following reactions:

1-Methylcyclobutene, 1-methylcyclopentene and 1-methylcyclohexene with dilute H₂SO₄, with the mercury procedure and with 1) BH₃; 2) H₂O₂, HO⁻
Dilute acid and the mercury procedure give the tertiary alcohol (1-methylcyclobutanol, 1-methylcyclopentanol and 1-methylcyclohexanol); hydroboration-oxidation gives the racemic trans-2-methylcycloalkanols

Hydroboration is a syn, anti-Markovnikov addition: boron bonds to the less substituted carbon and H to the other from the same face; OH then replaces boron with retention, so OH and the methyl group end up trans.

2. Design the synthesis

How would you prepare 3-methylbutan-1-ol from an alkene?

By hydroboration-oxidation of 3-methylbut-1-ene: BH3 in THF then H2O2, HO−. The OH goes onto the terminal carbon.

3. Why no rearrangements?

3,3-Dimethylbut-1-ene gives 3,3-dimethylbutan-1-ol by hydroboration-oxidation, with no rearrangement products. Why?

Because hydroboration is concerted: boron and hydrogen bond at the same time through a four-centre transition state, and no carbocation that could rearrange is ever formed.

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