The Question is from neet-chemistry alcohols-phenols-ethers ethers, and it was created by the author Mr. S M HUSSAIN

Question

Which of the following ethers, upon treatment with excess HI and heat, will yield a mixture of three different alkyl iodides?

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
 **Primary Concept**: Cleavage of ethers with excess HI
 **Related Topics**: Nucleophilic substitution, alkyl halides
 **Prerequisites**: Understanding of ether structure, reaction mechanisms of HI with ethers.
## SOLUTION PATHWAY
**Given Information**:
  The reaction involves treatment of ethers with excess HI and heat.
  We need to identify the ether that produces three different alkyl iodides.
**Method & Steps**:
 1. Understand the reaction mechanism: Ethers react with excess HI under heat to undergo cleavage, forming alkyl halides (iodoalkanes in this case).
  Reasoning: The reaction proceeds via protonation of the ether oxygen followed by nucleophilic attack by iodide ion, leading to the formation of two alkyl iodides.  If the ether is unsymmetrical, two different alkyl iodides will be formed. However, to get three alkyl iodides, one of the alkyl groups must also undergo further cleavage. This only happens if one of the alkyl groups has a branching that allows for the formation of a carbocation that reacts with another iodide.
  Formula/Rule:  $R-O-R' + 2HI \xrightarrow{heat} RI + R'I + H_2O$
 2. Analyze each option:
  Working:
		* **A) Diethyl ether ($CH_3CH_2OCH_2CH_3$)**:  This will yield only one type of alkyl iodide: $2CH_3CH_2I$.
		* **B) Diisopropyl ether ($(CH_3)_2CH-O-CH(CH_3)_2$)**: This will yield only one type of alkyl iodide: $2(CH_3)_2CHI$.
		* **C) Ethyl isopropyl ether ($CH_3CH_2-O-CH(CH_3)_2$)**: This will yield two alkyl iodides initially: $CH_3CH_2I$ and $(CH_3)_2CHI$. However, the isopropyl iodide can undergo further reaction with HI to form a secondary carbocation and eventually produce a tertiary alkyl iodide.  This leads to three different alkyl iodides.
		* **D) Methyl tert-butyl ether ($CH_3-O-C(CH_3)_3$)**: This will yield methyl iodide ($CH_3I$) and tert-butyl iodide ($ (CH_3)_3CI$). Although the tert-butyl group is branched, it doesn't lead to the formation of a new alkyl iodide after the initial cleavage.
 3. Final step:
  Result: Ethyl isopropyl ether produces three different alkyl iodides.
  Verification: The reaction mechanism and the analysis of each option confirm the result.
 ## OPTION ANALYSIS
A :x:  Only one type of alkyl iodide is formed.
B :x: Only one type of alkyl iodide is formed.
C ✓: Two alkyl iodides are initially formed, but further reaction of one of them (isopropyl iodide) with HI produces a third different alkyl iodide.
D :x: Only two alkyl iodides are formed.
 ## LEARNING AIDS
 **Common Mistakes**:
  1. [Error 1]: Incorrectly predicting the products without considering the possibility of further reaction with excess HI. [How to avoid]: Carefully analyze the structure of each alkyl group and consider carbocation stability and subsequent reactions.
  2. [Error 2]: Overlooking the effect of excess HI. [How to avoid]: Remember that excess HI is crucial for complete cleavage and potential further reactions.
 **Quick Tips**:
  1. [Helpful shortcut]: Check for unsymmetrical ethers, and analyze if any alkyl group can lead to a stable carbocation that undergoes further reaction with HI.
  2. [Memory aid]:  Remember that the reaction with HI is not just a simple cleavage; it can involve secondary reactions that influence the number and type of products formed.

Mr. S M HUSSAIN · Answer

Diethyl ether

Mr. S M HUSSAIN · Answer

Diisopropyl ether

Mr. S M HUSSAIN · Answer

The correct Option is "Ethyl isopropyl ether"

Mr. S M HUSSAIN · Answer

Methyl tert-butyl ether

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