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प्रश्न
Draw the major product formed when 1-ethoxyprop-1-ene is heated with one equivalent of HI.
उत्तर
\[\begin{array}{cc}
\phantom{...............................................}\ce{H}\\
\phantom{...............................................}|\\
\ce{\underset{(1-ethoxy-prop-1-ene)}{CH3 - CH = CH} - \underset{\bullet\bullet}{\overset{\bullet\bullet}{O}} - C2H5 ->[H - I][\Delta] CH3 - CH = CH - O - C2H5}
\end{array}\]
\[\ce{\underset{(1-iodo-prop-1-ene)}{CH3 - CH = CH - I} <-[+I^Θ] CH3 - CH = \overset{⊕}{C}H + C2H5OH}\]
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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