Identify the product from the following reactionProduct
(a)
(b)
(c)
(d)
Answer
Answer: A ✓ checked by 4AB · confidence high
The source book printed B; on checking, A is correct — see the explanation.
9-BBN hydroboration–oxidation is a syn addition. The new H (on the C–Me carbon) and OH (on the CH carbon) are cis, so Me and OH on adjacent carbons end up trans. Only (a) has trans (diequatorial) Me/OH. In (b) OH is axial on the carbon next to the equatorial Me, which is cis. (c) is cis and (d) is the Markovnikov tertiary alcohol. The printed key was A.
Explanation
9-BBN adds syn and anti-Markovnikov to the trisubstituted alkene. Boron goes to the less substituted CH carbon and H goes to the methyl-bearing carbon, both from the same face. H₂O₂/NaOH replaces B by OH with retention. So the new H and OH are cis, which makes CH₃ and OH on adjacent carbons trans. Only (a) has trans (diequatorial) CH₃ and OH. (b) and (c) are cis isomers, and (d) is the Markovnikov tertiary alcohol. Answer: (a).
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