Question:medium

Identify the product ' Z ' in the following series of reactions. $\text{Ethanol} \xrightarrow[\Delta]{\text{SOCl}_2} \text{X} \xrightarrow[\text{Dry ether}]{\text{Mg}} \text{Y} \xrightarrow{\text{NH}_3} \text{Z}$

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Grignard reagents are "proton hungry." If they see $\text{NH}_3$, $\text{H}_2\text{O}$, or alcohols, they immediately grab an H and turn into a simple alkane.
Updated On: May 14, 2026
  • Ethyl chloride
  • Ethyl magnesium chloride
  • Ethyl amine
  • Ethane
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
This synthetic sequence involves three classical organic transformations: conversion of a primary alcohol to an alkyl halide, generation of a Grignard reagent, and finally, the acid-base reaction of the Grignard reagent with a compound containing an acidic hydrogen.
Step 2: Key Formula or Approach:
Approach: Systematically trace the functional group transformations step-by-step through the given reagents to determine structures X, Y, and Z.
Step 3: Detailed Explanation:
Let's break down the reaction sequence step-by-step: 1. Formation of X: Ethanol ($\text{CH}_3\text{CH}_2\text{OH}$) reacts with thionyl chloride ($\text{SOCl}_2$) upon heating. This is a highly efficient method (Darzens procedure) for preparing alkyl chlorides because the byproducts are gases. \[ \text{CH}_3\text{CH}_2\text{OH} + \text{SOCl}_2 \xrightarrow{\Delta} \text{CH}_3\text{CH}_2\text{Cl} \, (\text{X}) + \text{SO}_2 \uparrow + \text{HCl} \uparrow \] Intermediate X is ethyl chloride. 2. Formation of Y: Ethyl chloride (X) is treated with magnesium metal ($\text{Mg}$) in an anhydrous solvent (dry ether) to form a Grignard reagent. \[ \text{CH}_3\text{CH}_2\text{Cl} + \text{Mg} \xrightarrow{\text{Dry ether}} \text{CH}_3\text{CH}_2\text{MgCl} \, (\text{Y}) \] Intermediate Y is ethyl magnesium chloride. 3. Formation of Z: Grignard reagents behave as strong bases and strong nucleophiles. When exposed to compounds containing an active (acidic) hydrogen—such as water, alcohols, or ammonia ($\text{NH}_3$)—they rapidly abstract a proton to form the corresponding alkane. Here, ammonia provides the acidic proton. \[ \text{CH}_3\text{CH}_2\text{MgCl} + \text{H}-\text{NH}_2 \longrightarrow \text{CH}_3\text{CH}_3 \, (\text{Z}) + \text{Mg(NH}_2\text{)Cl} \] The final product Z is ethane ($\text{CH}_3\text{CH}_3$).
Step 4: Final Answer:
The final product Z in the sequence is Ethane.
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