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Question
Draw the major product formed when 1-ethoxyprop-1-ene is heated with one equivalent of HI.
Solution
\[\begin{array}{cc}
\phantom{...............................................}\ce{H}\\
\phantom{...............................................}|\\
\ce{\underset{(1-ethoxy-prop-1-ene)}{CH3 - CH = CH} - \underset{\bullet\bullet}{\overset{\bullet\bullet}{O}} - C2H5 ->[H - I][\Delta] CH3 - CH = CH - O - C2H5}
\end{array}\]
\[\ce{\underset{(1-iodo-prop-1-ene)}{CH3 - CH = CH - I} <-[+I^Θ] CH3 - CH = \overset{⊕}{C}H + C2H5OH}\]
This reaction follows SN1 mechanism because in this reaction the more stable carbocation is formed that is double bonded carbocation. Therefore, the given molecule reacts with HI to form ethanol and 1-iodo prop-1-ene.
\[\ce{CH3 - CH = CH - O - C2H5 ->[HI/\Delta] \underset{(1-iodo-prop-1-ene)}{CH3 - CH = CH - I} + \underset{(ethanol)}{C2H5OH}}\]
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