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Write the products of the following reaction : \[ (CH_3)_3C-O-C_2H_5 \xrightarrow{HI} ? \]

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In ether cleavage by HI, if one alkyl group is tertiary, cleavage occurs at the tertiary carbon through the \(S_N1\) pathway because tertiary carbocations are highly stable.
Updated On: Jun 29, 2026
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Solution and Explanation

Step 1: Selectivity rule for ether cleavage by HI.
When an ether has one tertiary alkyl group and one primary group, HI cleaves at the tertiary C-O bond via $S_N1$, because a tertiary carbocation is far more stable than a primary one.
Step 2: Protonation and bond cleavage.
$H^+$ from HI protonates the ether oxygen, then the C-O bond at the tertiary side breaks: $(CH_3)_3C-O-C_2H_5 + H^+ \rightarrow (CH_3)_3C^+ + C_2H_5OH$.
Step 3: Iodide attacks the carbocation.
$I^-$ combines with $(CH_3)_3C^+$ to give tert-butyl iodide; the two products are:
\[ \boxed{(CH_3)_3CI + C_2H_5OH} \]
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