Question

Draw the major product formed when 1-ethoxyprop-1-ene is heated with one equivalent of HI.

Answer 1

\[\begin{array}{cc}
\phantom{...............................................}\ce{H}\\
\phantom{...............................................}|\\
\ce{\underset{(1-ethoxy-prop-1-ene)}{CH3 - CH = CH} - \underset{\bullet\bullet}{\overset{\bullet\bullet}{O}} - C2H5 ->[H - I][\Delta] CH3 - CH = CH - O - C2H5}
\end{array}\]

\[\ce{\underset{(1-iodo-prop-1-ene)}{CH3 - CH = CH - I} <-[+I^Θ] CH3 - CH = \overset{⊕}{C}H + C2H5OH}\]

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.

\[\ce{CH3 - CH = CH - O - C2H5 ->[HI/\Delta] \underset{(1-iodo-prop-1-ene)}{CH3 - CH = CH - I} + \underset{(ethanol)}{C2H5OH}}\]