$0.44 g$ of a monohydric alcohol when added to methylmagnesium iodide in ether liberates $112 {c m}^{3}$ of methane at STP. With PCC, the same alcohol forms a carbonyl compound that undergoes a silver mirror test. The monohydric alcohol is :

{(C H_{3})}_{3} C - C H_{2} O H

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{(C H_{3})}_{2} C H - C H_{2} O H

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\begin{matrix} C H_{3} - & C H & - C H_{2} - C H_{3} | O H \end{matrix}

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\begin{matrix} C H_{3} & - C H & - C H_{2} - C H_{2} - C H_{3} | O H \end{matrix}

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Solution

The correct option is A ${(C H_{3})}_{3} C - C H_{2} O H$
The monohydric alcohol is ${(C H_{3})}_{3} C - C H_{2} O H$
0.44g of a monohydric alcohol when added to methylmagnesium iodide in ether liberates 112cm $^{3}$ of methane at STP.
At STP 112cm $^{3}$ of methane corresponds to $\frac{112}{22400} = 0.005$ moles.
0.005 moles of methane corresponds to 0.005 moles of alcohol.
$R - O H + C H_{3} - M g - I \to R - O - M g - I + C H_{4}$
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 ${(C H_{3})}_{3} C - C H_{2} O H$ .
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.

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