Question:
Which from following is obtained by hydroboration - oxidation of propene?
Which from following is obtained by hydroboration - oxidation of propene?
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Anti-Markovnikov addition of water gives the primary alcohol.
Updated On: Oct 1, 2026
- Propanal
- Propanone
- Propan-1-ol
- Propan-2-ol
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Hydroboration-oxidation adds water to an alkene with anti-Markovnikov orientation. Boron adds to the less substituted carbon and is later replaced by \(-\text{OH}\).
Step 2: Key Formula or Approach:
Reagents: (i) \(\text{B}_2\text{H}_6\) (diborane) in ether, (ii) \(\text{H}_2\text{O}_2/\text{OH}^-\).
Step 3: Detailed Explanation:
\(3\text{CH}_3\text{CH=CH}_2 + \text{BH}_3 \to (\text{CH}_3\text{CH}_2\text{CH}_2)_3\text{B}\).
Oxidation then gives \(3\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\) (propan-1-ol).
Propan-2-ol would be the Markovnikov product obtained by acid catalysed hydration. Propanal and propanone are oxidation products, not formed here.
Final Answer:
The product is propan-1-ol, option (C). \[ \boxed{\text{Propan-1-ol}} \]
Hydroboration-oxidation adds water to an alkene with anti-Markovnikov orientation. Boron adds to the less substituted carbon and is later replaced by \(-\text{OH}\).
Step 2: Key Formula or Approach:
Reagents: (i) \(\text{B}_2\text{H}_6\) (diborane) in ether, (ii) \(\text{H}_2\text{O}_2/\text{OH}^-\).
Step 3: Detailed Explanation:
\(3\text{CH}_3\text{CH=CH}_2 + \text{BH}_3 \to (\text{CH}_3\text{CH}_2\text{CH}_2)_3\text{B}\).
Oxidation then gives \(3\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\) (propan-1-ol).
Propan-2-ol would be the Markovnikov product obtained by acid catalysed hydration. Propanal and propanone are oxidation products, not formed here.
Final Answer:
The product is propan-1-ol, option (C). \[ \boxed{\text{Propan-1-ol}} \]
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