Question

What are B and C respectively in the following set of reactions? \[ \text{C (1,2-dibromopropane)} \xrightarrow{\text{Zn},\ \Delta} \text{A} \xrightarrow[\text{(ii) NaNH}_2]{\text{(i) alc. KOH}} \text{B} \xrightarrow{\text{Lindlar Catalyst}} \text{C} \]

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Vicinal dihalides on reaction with Zn form alkynes. Lindlar’s catalyst converts alkynes to cis-alkenes (partial hydrogenation).

Answer 1

The Correct Option is A

Step 1:

Reduction of 1,2-dibromopropane with Zn in heat forms an alkyne (A). The Zn eliminates Br atoms from vicinal dihalide:

\[ \text{CH}_2\text{Br}-\text{CHBr}-\text{CH}_3 \xrightarrow{\text{Zn}/\Delta} \text{CH}_3-\text{C} \equiv \text{CH} \ (\text{Propyne}) \]

Step 2:

Propyne reacts with alc. KOH and NaNH₂, leading to dehydrohalogenation and formation of an acetylide intermediate, but eventually alkynes remain (no significant change).

Step 3:

Propyne on Lindlar’s catalyst gets partially hydrogenated to give a cis-alkene (propene).

So:

• Intermediate B = Propyne
• Final compound C = Propene