Question:

\((CH_3)_3C-OC_2H_5\) on reaction with HI gives

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When an ether is cleaved by HI, the bond breaks so that the iodide attaches to the alkyl group that can form the more stable carbocation.
Updated On: Jun 16, 2026
  • \((CH_3)_3C-I\) and \(C_2H_5-I\)
  • \((CH_3)_3C-OH\) and \(C_2H_5-I\)
  • \((CH_3)_3C-I\) and \(C_2H_5-OH\)
  • \((CH_3)_3C-OH\) and \(C_2H_5-OH\)
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The Correct Option is C

Solution and Explanation

Concept:
When an ether is cleaved by HI, the bond breaks so that the iodide attaches to the alkyl group that can form the more stable carbocation. For ethers with a tertiary group, the reaction follows an \(SN1\) path through the stable tertiary carbocation.

Step 1:
In \((CH_3)_3C-OC_2H_5\), the tert-butyl group forms a very stable tertiary carbocation \((CH_3)_3C^+\), so it takes the iodide to give \((CH_3)_3C-I\).

Step 2:
The remaining ethyl part leaves as \(C_2H_5OH\) (the oxygen stays with the ethyl group). So products are \((CH_3)_3C-I\) and \(C_2H_5-OH\).

Answer: Option (C) \((CH_3)_3C-I\) and \(C_2H_5-OH\).
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