Question:
What are B and C respectively in the following set of reactions?
\[
\text{C (1,2-dibromopropane)} \xrightarrow{\text{Zn},\ \Delta} \text{A} \xrightarrow[\text{(ii) NaNH}_2]{\text{(i) alc. KOH}} \text{B} \xrightarrow{\text{Lindlar Catalyst}} \text{C}
\]
What are B and C respectively in the following set of reactions?
\[
\text{C (1,2-dibromopropane)} \xrightarrow{\text{Zn},\ \Delta} \text{A} \xrightarrow[\text{(ii) NaNH}_2]{\text{(i) alc. KOH}} \text{B} \xrightarrow{\text{Lindlar Catalyst}} \text{C}
\]
Show Hint
Vicinal dihalides on reaction with Zn form alkynes. Lindlar’s catalyst converts alkynes to cis-alkenes (partial hydrogenation).
Updated On: Mar 18, 2026
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The Correct Option is A
Solution and Explanation
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
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