Question

0.44 g of a monohydric alcohol, when added to methyl magnesium iodide in ether, liberates at STP 112 cm³ of methane with PCC the same alcohol form a carbonyl compound that answers the silver mirror test. Identify the compound.

Answer 1

0.44g of a monohydric alcohol liberates 112 cm³ of methane.

\[\ce{\underset{(1 mole)}{R - OH} + CH3MgI -> \underset{(1 mole)}{CH4} + MgI(OR)}\]

Mass of monohydric alcohol which gives 22400 cm³ of methane = `(22400 xx 0.44)/112` = 88

C₅H₁₂O molecular formula has mass number 88 and it shows eight possible isomers. But neopentyl alcohol reacts with PCC to form neopentyl aldehyde, which shows positive silver mirror test.

Therefore, the compound is neopentyl alcohol (or) 2, 2-dimethyl propan-1-ol.

\[\begin{array}{cc}
\ce{CH3}\phantom{.}\\
|\phantom{....}\\
\ce{CH3 - C - CH2OH}\\
|\phantom{....}\\
\ce{CH3}\phantom{.}
\end{array}\]