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Hydroboration–oxidation gives anti-Markovnikov alcohol; using BD\(_3\) introduces D at the more substituted carbon.
Updated On: Apr 23, 2026
- A
- B
- C
- D
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
NaBD\(_4\) in presence of H\(_2\)O\(_2\)/OH\(^-\) behaves like hydroboration–oxidation using BD\(_3\), leading to anti-Markovnikov addition.
Step 2: Detailed Explanation:
In hydroboration, boron attaches to the less substituted carbon and deuterium (D) adds to the more substituted carbon.
After oxidation, OH replaces boron at the less substituted carbon.
Thus, product formed is CH\(_3\)CH(D)CH\(_2\)OH.
Step 3: Final Answer:
Hence, the correct option is (B).
NaBD\(_4\) in presence of H\(_2\)O\(_2\)/OH\(^-\) behaves like hydroboration–oxidation using BD\(_3\), leading to anti-Markovnikov addition.
Step 2: Detailed Explanation:
In hydroboration, boron attaches to the less substituted carbon and deuterium (D) adds to the more substituted carbon.
After oxidation, OH replaces boron at the less substituted carbon.
Thus, product formed is CH\(_3\)CH(D)CH\(_2\)OH.
Step 3: Final Answer:
Hence, the correct option is (B).
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