Question

$0.44\, g$ of a monohydric alcohol when added to methyl magnesium iodide in ether liberates at $S.T.P.,$ $112\, cm^3$ of methane. With $PCC$ the same alcohol forms a carbonyl compound that answers silver mirror test. The monohydric alcohol is

• A. ${(CH_3)_3}C - CH_2OH$
• B. ${(CH_3)_2}CH - CH_2OH$
• C.
• D.

Answer 1

Answer: (A)

The monohydric alcohol is $\left( CH _{3}\right)_{3} C - CH _{2} OH$
0.44g of a monohydric alcohol when added to methylmagnesium iodide in ether liberates $112 cm ^{3}$ of methane at STP.
At STP $112 cm ^{3}$ of methane corresponds to $\frac{112}{22400}=0.005$ moles.
0.005 moles of methane corresponds to $0.005$ moles of alcohol.
$ R - OH + CH _{3}- Mg - I \rightarrow R - O - Mg - I + CH _{4} $
$0.005$ moles of alcohol corresponds to $0.44 g$. Hence, the molar mass of alcohol is $\frac{0.44}{0.005}=88 g / mol$. This corresponds to molecular formula $\left( CH _{3}\right)_{3} C - CH _{2} OH$.
With PCC, the same alcohol forms a carbonyl compound that answers silver mirror test. This indicates that the alcohol is primary alcohol. Note: Silver mirror test (Tollen's reagent, ammoniacal silver nitrate solution) is given by aldehydes but not ketones.