A student argues that hydroboration-oxidation places the hydroxyl group on the less substituted carbon because a primary carbocation is more stable than a tertiary one. What is wrong with this reasoning?
APrimary carbocations are actually less stable than tertiary, so the argument is backwards
BNo carbocation intermediate forms — addition is concerted, and regiochemistry reflects steric preference of boron, not carbocation stability
CThe hydroxyl group actually ends up on the more substituted carbon in hydroboration-oxidation
DThe student is correct — carbocation stability does determine regiochemistry
The critical mechanistic distinction is that hydroboration proceeds through a concerted four-membered transition state — no carbocation ever forms. Boron, as the larger electrophilic atom, bonds to the less sterically hindered (less substituted) carbon to minimize steric strain. The regiochemistry is entirely steric in origin, which is why it is anti-Markovnikov rather than some modified carbocation argument.
Question 2 Multiple Choice
What is the stereochemical consequence of performing hydroboration-oxidation on (Z)-2-butene compared to (E)-2-butene?
ABoth give the same racemic product because the intermediate is planar
BBoth give the same product because stereochemistry is lost during the oxidation step
CThey give different stereoisomeric products because syn addition preserves the relative face of addition from the starting alkene geometry
DThey give enantiomers of each other because the reaction proceeds through a carbocation that can be attacked from either face
Syn addition — both B and H add from the same face — combined with the fixed geometry of (Z) vs (E) alkenes produces different diastereomers. The concerted mechanism locks in the facial selectivity from the starting material, so geometry matters. This contrasts with reactions through carbocation intermediates, where rotation allows nucleophilic attack from either face and stereochemical information is often lost.
Question 3 True / False
Hydroboration-oxidation places the hydroxyl group on the less substituted carbon because boron forms a more stable primary carbocation at that position.
TTrue
FFalse
Answer: False
No carbocation forms in hydroboration-oxidation. The addition is concerted — boron and hydrogen add simultaneously through a four-membered transition state. The anti-Markovnikov outcome results from steric preference: boron, as the bulkier group, bonds to the less hindered carbon. Carbocation stability is irrelevant here, which is precisely why the reaction gives the opposite regiochemistry to acid-catalyzed hydration.
Question 4 True / False
The syn addition observed in hydroboration-oxidation means that boron and hydrogen add to the same face of the alkene double bond.
TTrue
FFalse
Answer: True
The concerted four-membered transition state forces boron and hydrogen to approach the alkene from the same face simultaneously. This syn selectivity is preserved through the oxidation step (which proceeds with retention of configuration at carbon), so the final alcohol retains the stereochemical information about which face of the double bond was attacked.
Question 5 Short Answer
Why does hydroboration-oxidation give anti-Markovnikov regiochemistry, and how does this differ mechanistically from reactions that follow Markovnikov's rule?
Think about your answer, then reveal below.
Model answer: Hydroboration-oxidation is concerted — no carbocation intermediate forms. Boron bonds to the less substituted carbon due to steric preference in the four-membered transition state. Markovnikov reactions (acid-catalyzed hydration, HX addition) proceed through a discrete carbocation intermediate that forms at the more substituted (more stable) carbon, directing the nucleophile there.
The mechanistic distinction is fundamental: carbocation stability controls Markovnikov regiochemistry, while steric effects in a concerted mechanism control anti-Markovnikov regiochemistry. Understanding this lets you predict regiochemistry from mechanism rather than memorizing outcomes — and explains why both reactions can serve as complementary tools to place OH on either carbon of a double bond.