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Question
\[\ce{(CH3)2CH - O - CH3 ->[Cold HI] X + Y}\]
\[\ce{X ->[K2Cr2O7][dil. H2SO4] Z}\]
\[\ce{Y ->[NaOH(aq)][\Delta] CH3OH}\]
Identify X, Y and Z.
Options
X - Propan-2-ol
Y - Iodomethane
Z - Propanone
X - Propan-1-ol
Y - Iodomethane
Z - Propanal
X - 2-Iodopropane
Y - Iodomethane
Z - Propanone
X - Propan-2-ol
Y - Methanol
Z - Propanone
Solution
X - Propan-2-ol
Y - Iodomethane
Z - Propanone
Explanation:
In cold, a mixed ether (except the one having a tertiary alkyl group) gives higher alcohol and lower alkyl iodide.
\[\ce{(CH3)2CH - O - CH3 ->[Cold HI] \underset{\underset{(X)}{Propan-2-ol}}{(CH3)2CHOH} + \underset{\underset{(Y)}{lodomethane}}{CH3I}}\]
\[\ce{(CH3)2CHOH ->[K2Cr2O7][dil. H2SO4] \underset{\underset{(Z)}{Propanone}}{CH3COCH3}}\]
\[\ce{CH3I + \underset{(aq.)}{NaOH} ->[\Delta] \underset{Methanol}{CH3OH} + NaI}\]
