Write the products of the following reaction :
\[
(CH_3)_3C-O-C_2H_5
\xrightarrow{HI}
?
\]
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Solution and Explanation
Step 1: Identify the ether The given ether is: \[ (CH_3)_3C-O-C_2H_5 \] This compound is tert-butyl ethyl ether. One side contains a tertiary alkyl group while the other side contains a primary ethyl group.
Step 2: Protonation of oxygen The oxygen atom first accepts a proton from HI. \[ (CH_3)_3C-O-C_2H_5 + H^+ \rightarrow (CH_3)_3C-OH^+-C_2H_5 \] This weakens the carbon--oxygen bond.
Step 3: Formation of tertiary carbocation The bond breaks preferentially at the tertiary carbon because a tertiary carbocation is highly stable. \[ (CH_3)_3C^+ + C_2H_5OH \]
Step 4: Attack by iodide ion The tertiary carbocation combines with iodide ion. \[ (CH_3)_3C^+ + I^- \rightarrow (CH_3)_3CI \]
Step 5: Final products Thus the products formed are: \[ (CH_3)_3CI \] and \[ C_2H_5OH \] \[ \boxed{ (CH_3)_3C-O-C_2H_5 + HI \rightarrow (CH_3)_3CI + C_2H_5OH } \]
Final Answer \[ \boxed{ (CH_3)_3CI + C_2H_5OH } \]
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