Question

Which of the following sequence of reagents convert propene to 1-chloropropane?

• A. (i)\( {(BH}_3)_2 (ii){H}_2{O}_2 / {OH}^- \) ; \( {HCl}, {ZnCl}_2 \)
• B. (i)\( {(BH}_3)_2 (ii){H}_2{O}_2 / {OH}^- \) ; NaCl
• C. (i)\( {dil. H}_2{SO}_4 \) ; \( {HCl}, {ZnCl}_2 \)
• D. (i)\( {dil. H}_2{SO}_4 \) ; Conc. HCl

Hint:

The hydroboration-oxidation reaction provides a way to convert alkenes to alcohols with anti-Markovnikov selectivity. The subsequent treatment with HCl and ZnCl2 leads to the conversion of alcohols to alkyl chlorides.

Answer 1

The Correct Option is A

To convert propene to 1-chloropropane, we need to carry out a two-step reaction process:

1. Step 1: Hydroboration of propene with diborane (\( {(BH}_3)_2 \)) in the presence of hydrogen peroxide (\( {H}_2{O}_2 \)) and a base (\( {OH}^- \)) forms 1-propanol through anti-Markovnikov addition. The boron atom adds to the carbon of the double bond that has the most hydrogen atoms, forming an organoborane intermediate.
2. Step 2: The hydroboration product, 1-propanol, is then treated with hydrochloric acid (\( {HCl} \)) and zinc chloride (\( {ZnCl}_2 \)) in a substitution reaction, where the hydroxyl group is replaced by a chlorine atom to give 1-chloropropane.

Thus, the correct reagents are:

• Diborane (\( {(BH}_3)_2 \))
• Hydrogen peroxide (\( {H}_2{O}_2 \)) and base (\( {OH}^- \))
• Hydrochloric acid (\( {HCl} \)) and zinc chloride (\( {ZnCl}_2 \))

Therefore, the correct answer is (1).