Question

Given below are two statements:
Statement I: Propene on treatment with diborane gives an addition product with the formula \( \left( (CH_3)_2 CH \right)_3 B \)
Statement II: Oxidation of \( \left( (CH_3)_2 CH \right)_3 B \) with Hydrogen peroxide in the presence of NaOH gives propan-2-ol
In the light of the above statements, choose the most appropriate answer from the options given below:

• A. Statement I is correct but Statement II is incorrect
• B. Statement I is incorrect but Statement II is correct
• C. Both Statement I and Statement II are correct
• D. Both Statement I and Statement II are incorrect

Hint:

Key points to remember:
- Hydroboration gives anti-Markovnikov addition product
- Overall process converts alkene to alcohol with opposite regioselectivity to acid-catalyzed hydration

Answer 1

The Correct Option is C

Step 1: Analyze Statement I
The hydroboration of propene with diborane $(BH_3)_2$ follows the reaction: \[ 3 CH_3-CH=CH_2 + (BH_3)_2 \rightarrow \left( (CH_3)_2 CH \right)_3 B \] This forms triisopropylborane, confirming Statement I is correct.
Step 2: Analyze Statement II
The oxidation of triisopropylborane with $H_2O_2/NaOH$ follows: \[ \left( (CH_3)_2 CH \right)_3 B + 3 H_2O_2 + 3 NaOH \rightarrow 3 (CH_3)_2 CHOH + Na_3BO_3 \] This yields propan-2-ol, confirming Statement II is correct.
Step 3: Verify both statements
- Hydroboration-oxidation converts alkenes to alcohols via anti-Markovnikov addition - The mechanism involves:

1. Formation of alkylborane (Statement I)
2. Oxidation to alcohol (Statement II)

Step 4: Conclusion
Both statements accurately describe the hydroboration-oxidation process of propene.