Question

Identify the correct reagents that would bring about the following transformation.

• A. (i) H₂O/H⁺
  (ii) CrO₃
• B. (i) BH₃
  (ii) \(H_2O_2/O^\ominus H\)
  (iii) PCC
• C. (i) BH₃
  (ii) \(H_2O_2/O^\ominus H\)
  (iii) alk. KMnO₄
• D. (iv)\(H_3O^\oplus\)
  (i) H₂O/H⁺
  (ii) PCC

Answer 1

The Correct Option is B

Step 1: Understanding the Given Transformation 

The transformation likely involves oxidation or hydroboration-oxidation, depending on the functional groups present.

Step 2: Analyze the Reagents

• \( BH_3 \) followed by \( H_2O_2/OH^- \): This is the hydroboration-oxidation reaction, which converts an alkene to an anti-Markovnikov alcohol.
• PCC (Pyridinium Chlorochromate): A mild oxidizing agent that converts primary alcohols to aldehydes and secondary alcohols to ketones.
• Chromic acid (\( CrO_3 \)) in acidic medium: A strong oxidizing agent that converts alcohols to carboxylic acids.
• Acidic water (\( H_2O/H^+ \)): Used for hydration of alkenes to form Markovnikov alcohols.

Step 3: Identify the Correct Set of Reagents

Since the transformation involves hydration followed by oxidation, the correct sequence of reagents is:

1. \( BH_3 \) (Hydroboration step).
2. \( H_2O_2/OH^- \) (Oxidation to form alcohol).
3. \( PCC \) (Mild oxidation to aldehyde).

Step 4: Conclusion

The correct answer is:

Option (2): (i) \( BH_3 \), (ii) \( H_2O_2/OH^- \), (iii) PCC.