\( 0.44 \mathrm{~g} \) of a monohydric alcohol when added to methylmagnesium iodide in ether liberates at
S.T.P., \( 112 \mathrm{~cm}^{3} \) of methane. With PCC the same alcohol forms a carbonyl compound that
answers silver mirror test. The monohydric alcohol is

At STP \(112 \mathrm{~cm}^{3}\) of methane liberated, \(22400 \mathrm{~cm}^{3}=1 \mathrm{~mol}\) at ST
\(1 \mathrm{~cm}^{3}=\frac{1}{22400} \mathrm{~mol}\)
\(112 \mathrm{~cm}^{3}=\frac{1}{22400} \times 112 \mathrm{~mol}\)
\(=0.005\) mol of methane liberated
1 mol of monohydric alcohol react with 1 mol of methyl magnesium iodide to produces \(1 \mathrm{~mol}\) of methane, according to
the reaction,
Monohydric alcohol+ \(\mathrm{CH}_{3} \mathrm{M} \mathrm{gl}-\overline{\text { Either }}\)
\(\Rightarrow 0.005\) mol of monohydric alcohol produces \(0.005\) mol of methane
Now, molar mass of monohydric alcohol
\(=\frac{\text { Given mass of monohydric alcohol }}{\mathrm{Number} \text { of moles of monohydric alcohol }}\)
\(=\frac{0.44}{0.005}=88 \mathrm{~g}\)
Monohydric alcohol from the given option having \(88 \mathrm{~g}\) molar mass is \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}\)
\(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{M} \mathrm{gI}-\mathrm{Either}\)
When react with \(\mathrm{PCC}\), aldehyde is formed which give positive silver mirror test.
( \(\mathrm{CH}_{3}\) ) \(\left._{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}-\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OM} \mathrm{gI}+\mathrm{CH}_{4}\)
( \(\mathrm{CH}_{3}\) ) \(_{3} \mathrm{C}-\mathrm{CHO}\)