Question

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

• A. $(C{H}_3{)}_{3}C-C{H}_{2}OH$
• B. $(C{H}_3{)}_{2}CH-C{H}_{2}OH$
• C.
• D.

Answer 1

Answer: (A)

The monohydric alcohol is ${\left(C{H}_3\right)}_{3}C-C{H}_{2}OH$
0.44g of a monohydric alcohol when added to methylmagnesium iodide in ether liberates $112c{m}^3$ of methane at STP.
At STP $112c{m}^3$ of methane corresponds to $\frac{112}{22400}=0.005$ moles.
0.005 moles of methane corresponds to $0.005$ moles of alcohol.
$R-OH+C{H}_3-Mg-I\to R-O-Mg-I+C{H}_4$
$0.005$ moles of alcohol corresponds to $0.44g$. Hence, the molar mass of alcohol is $\frac{0.44}{0.005}=88g/mol$. This corresponds to molecular formula ${\left(C{H}_3\right)}_{3}C-C{H}_{2}OH$.
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.