Question:
What are X and Y in the following reaction sequence? $Ethanal \rightarrow C_4H_8 \xrightarrow{Br_2/CCl_4} X \xrightarrow{alc. KOH, NaNH_2} Y$
What are X and Y in the following reaction sequence? $Ethanal \rightarrow C_4H_8 \xrightarrow{Br_2/CCl_4} X \xrightarrow{alc. KOH, NaNH_2} Y$
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$NaNH_2$ is a strong base used to achieve double dehydrohalogenation for terminal alkynes.
Updated On: Jun 10, 2026
- $X=$ alcoholic KOH, $NaNH_2$; $Y=CH_3CH_2C \equiv CH$
- $X=$ alcoholic KOH, $NaNH_2$; $Y=CH_3C \equiv CCH_3$
- $X=$ alcoholic KOH; $Y=CH_3CH_2C \equiv CH$
- $X=$ aq KOH; $Y=CH_3C \equiv CCH_3$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Organic transformations: conversion of aldehydes to alkenes and then to alkynes.
Step 2: Analysis
This involves terminal alkyne formation from an alkene. $X$ must be the dehydrohalogenation agent (alc. KOH + $NaNH_2$) to form the alkyne $Y$.
Step 3: Conclusion
Option A represents the standard sequence for this conversion.
Final Answer: (A)
Organic transformations: conversion of aldehydes to alkenes and then to alkynes.
Step 2: Analysis
This involves terminal alkyne formation from an alkene. $X$ must be the dehydrohalogenation agent (alc. KOH + $NaNH_2$) to form the alkyne $Y$.
Step 3: Conclusion
Option A represents the standard sequence for this conversion.
Final Answer: (A)
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