At STP 112cm3 of methane liberated, 22400cm3=1mol at ST 1cm3=
1
22400
mol 112cm3=
1
22400
×112mol = 0.005 mol of methane liberated 1 mol of monohydric alcohol react with 1 mol of methyl magnesium iodide to produces 1 mol of methane, according to the reaction, Monohydric alcohol+CH3Mgl
Either
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CH4 ⇒ 0.005 mol of monohydric alcohol produces 0.005 mol of methane Now, molar mass of monohydric alcohol
=
Given mass of monohydric alcohol
Number of moles of monohydric alcohol
=
0.44
0.005
=88g Monohydric alcohol from the given option having 88 g molar mass is (CH3)3C−CH2OH
(CH3)3C−CH2OH+CH3MgI
Either
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(CH3)3C−CH2OMgI+CH4
When react with PCC, aldehyde is formed which give positive silver mirror test.