Question

An organic compound [A] having the molecular formula C₄H₁₀O forms a compound [B] with the molecular formula C₄H₈O on oxidation. Compound [B] gives a positive iodoform test. The reaction of compound [B] with CH₃MgBr followed by hydrolysis, gives compound [C] with the molecular formula C₅H₁₂O. Identify the compounds [A], [B] and [C]. Write the reaction for the conversion of compound [A] to compound [B].

Answer 1

Compound (B) produces a positive iodoform; hence, the test must be a methyl ketone. Butanone, CH₃COCH₂CH₃ is the only methyl ketone with four carbon atoms.

Butanone will be obtained by oxidising butan-2-ol. -So, chemical (A) is butan-2-ol CH₃CH(OH)CH₂CH₃. Butanone reacts with CH₃MgBr and then hydrolyzes to produce 30 alcohol, so compound (C) is

\[\begin{array}{cc}
\phantom{.......................................................}\ce{CH3\phantom{........................}CH3}\phantom{}\\
\phantom{.....................................................}|\phantom{............................}|\phantom{.}\\
\ce{CH3 - CH - CH2CH3 ->[PCC] CH3 - C - CH2CH3 ->[CH3MgBr] CH3 - C - CH2CH3 ->[H+/H2O]CH3 - C - CH2CH3}\\
\phantom{......}|\phantom{..........................}||\phantom{.............................}|\phantom{............................}|\phantom{.........}\\
\phantom{.}\ce{\underset{(A)}{OH}\phantom{.......................}\underset{(B)}{O} \phantom{............................}OMgBr \phantom{....................}\underset{(C)}{OH}}\phantom{..}\end{array}\]