Question

Which of the following reagents are used to convert propene to propyne?

• A. SOCl$_2$/Py, alc. KOH, H$^+$/H$_2$O
• B. Br$_2$/CCl$_4$, alc. KOH/$\Delta$
• C. alc. KCN, SOCl$_2$
• D. alc. KOH, B$_2$H$_6$/H$_2$O$_2$

Hint:

Alkene → Alkyne = halogen addition + double elimination.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To convert an alkene (propene) to an alkyne (propyne), we first need to introduce two leaving groups (halogens) and then perform a double elimination reaction (dehydrohalogenation).
Step 3: Detailed Explanation:
1. Halogenation: Propene ($CH_{3}-CH=CH_{2}$) reacts with $Br_{2}$ in $CCl_{4}$ to undergo addition across the double bond.
\[ CH_{3}-CH=CH_{2} + Br_{2} \xrightarrow{CCl_{4}} CH_{3}-CH(Br)-CH_{2}Br \]
The product is 1,2-dibromopropane (a vicinal dihalide).
2. Dehydrohalogenation: The vicinal dihalide is treated with alcoholic $KOH$ and heated. This strong base removes two molecules of $HBr$ in a two-step elimination.
\[ CH_{3}-CH(Br)-CH_{2}Br \xrightarrow{alc. KOH, \Delta} CH_{3}-C \equiv CH \]
The final product is propyne.
Step 4: Final Answer:
The reagents required are $Br_{2}/CCl_{4}$ followed by alcoholic $KOH$ and heating.