A student treats 1-methylcyclohexene with BH₃·THF, then H₂O₂/NaOH. Which product is formed?
A1-methylcyclohexan-1-ol — Markovnikov addition places OH on the more substituted carbon
Bcis-2-methylcyclohexan-1-ol — anti-Markovnikov regiochemistry with syn addition
Ctrans-2-methylcyclohexan-1-ol — anti-Markovnikov regiochemistry with anti addition
DA racemic mixture of 1-methylcyclohexan-1-ol via a planar carbocation intermediate
Hydroboration places boron on the less substituted carbon (C2, next to the methyl-bearing C1) via a concerted syn addition — both B and H add to the same face. Oxidation then replaces B with OH with retention of configuration at that carbon. The net result is anti-Markovnikov regiochemistry (OH on C2, not C1) and syn stereochemistry — hence cis-2-methylcyclohexan-1-ol. Option A is the acid-catalyzed hydration product. Option C would require anti addition, which does not occur here. Option D describes the carbocation pathway that hydroboration specifically avoids.
Question 2 Multiple Choice
Why does boron end up on the less substituted carbon during hydroboration?
AThe less substituted carbon has more hydrogens available to stabilize the developing negative charge on boron
BSteric factors direct the bulkier boron to the less hindered carbon, and the transition state places partial positive charge on the more substituted carbon, which better stabilizes it
CBoron is nucleophilic and attacks the terminal carbon of the double bond in all cases
DMarkovnikov's rule applies to boron just as it does to hydrogen — the less electronegative atom goes to the less substituted carbon
Two factors conspire to place boron on the less substituted carbon. First, steric: boron is a large atom that preferentially occupies the less hindered position. Second, electronic: in the four-centered transition state, there is partial positive charge on the carbon receiving the hydride — the more substituted carbon better stabilizes this partial positive charge through hyperconjugation and inductive effects. Note that option D states 'Markovnikov's rule applies to boron' — this is precisely backwards; hydroboration is the anti-Markovnikov reaction.
Question 3 True / False
The oxidation step (H₂O₂/NaOH) in hydroboration-oxidation inverts the configuration at the carbon that bore the boron substituent.
TTrue
FFalse
Answer: False
This is a common misconception. The 1,2-alkyl migration from boron to oxygen in the oxidation mechanism proceeds with retention of configuration at the migrating carbon. Since boron was placed by syn addition in the hydroboration step, and oxidation retains that configuration, OH ends up on the same face where boron was — no inversion occurs. The overall reaction delivers syn stereochemistry because both new bonds (C–B then C–OH) form on the same face.
Question 4 True / False
Hydroboration-oxidation and acid-catalyzed hydration of an unsymmetrical alkene give the same regiochemical product but differ in stereochemical outcome.
TTrue
FFalse
Answer: False
The two reactions give opposite regiochemistry, not just different stereochemistry. Acid-catalyzed hydration follows Markovnikov's rule — OH ends up on the more substituted carbon via a carbocation intermediate. Hydroboration-oxidation places OH on the less substituted carbon (anti-Markovnikov). For an unsymmetrical alkene like propene, acid hydration gives 2-propanol while hydroboration-oxidation gives 1-propanol. This complementary regiochemistry is precisely why both reactions are taught together.
Question 5 Short Answer
Explain why hydroboration-oxidation is described as giving 'syn' addition overall, even though it involves two chemically distinct reaction steps.
Think about your answer, then reveal below.
Model answer: In the hydroboration step, borane adds in a concerted, four-centered transition state where both the B–C and H–C bonds form simultaneously on the same face of the double bond. This locks in syn stereochemistry — boron and hydrogen are delivered to the same face. In the oxidation step, the carbon that migrates to oxygen does so with retention of configuration. So the oxygen inherits the same facial position that boron occupied. Because boron was on the same face as the hydrogen (from the syn addition), and oxygen replaces boron with retention, the final product has OH and H on the same face — syn addition overall.
The key insight is that syn selectivity is established in the first step (the concerted borane addition) and preserved by the retention mechanism of the second step (oxidation). There is no inversion anywhere in the pathway. This contrasts with reactions that go through carbocations (no stereocontrol) or those where addition occurs from opposite faces (anti addition).