Question

An ether (A), C₅H₁₂O, when heated with excess of hot HI produce two alkyl halides which on hydrolysis form compound (B) and (C), oxidation of (B) gave and acid (D), whereas oxidation of (C) gave a ketone (E). Deduce the structural formula of (A), (B), (C), (D), and (E).

Answer 1

i. The ether (A) with molecular formula C₅H₁₂O is 

\[\begin{array}{cc}
\ce{H3C - CH2 - O - HC -CH3}\\
\phantom{..............}|\\
\phantom{................}\ce{CH3}\\
\end{array}\]

ii. Reacts with hot HI to produce two alkyl halides as follows:

iii. Oxidation of (B) gives acid

\[\ce{CH3 - CH2 - OH ->[{[O]}] \underset{(D)}{CH3COOH}}\]

iv. Oxidation of (C) gives ketone

\[\begin{array}{cc}
\ce{CH3 - CH - OH ->[{[O]}] CH3 - C = O}\\
\phantom{...}|\phantom{....................}|\\
\phantom{....}\ce{CH3}\phantom{...........}\ce{(E)}\phantom{...}\ce{CH3}\\
\end{array}\]

Hence, structural formulae of compounds (A) to (E) are

(A) (2-Ethoxypropane)

\[\begin{array}{cc}
\ce{CH3 - CH2 - O - CH - CH3}\\
\phantom{.....................}|\phantom{...........}\\ \phantom{.......................}\ce{CH3\phantom{...........}}\\
\end{array}\]

(B) (Ethanol)

CH₃CH₂–OH 

(c) (Propan-2-ol)

\[\begin{array}{cc}
\ce{CH3 - CH - CH3}\\
\phantom{..}|\\ 
\phantom{....}\ce{OH}\\
\end{array}\] 

(D) (Ethanoic acid)

CH₃COOH 

(E) (Propanone)

\[\begin{array}{cc}
\ce{CH3 - C = O}\\
\phantom{...}|\phantom{.}\\
\phantom{....}\ce{CH3}\\
\end{array}\]