Q.0.44 g of a monohydric alcohol, when added to methylmagnesium iodide in ether, liberates 112 cm3 of methane at STP. With PCC, the same alcohol forms a carbonyl compound that answers the silver mirror test. Identify the compound.

Step 1. Moles of CH4 liberated = 112 cm3 / 22400 cm3 mol-1 (STP) = 0.005 mol.

Step 2. R-OH + CH3MgI gives R-OMgI + CH4 is a 1:1 reaction (the classic Zerewitinoff active-hydrogen determination), so moles of alcohol = moles of CH4 = 0.005 mol.

Step 3. Molar mass = 0.44 g / 0.005 mol = 88 g/mol.

Step 4. A silver-mirror (Tollens') positive result after PCC oxidation means the PCC product is an ALDEHYDE, which only forms from a PRIMARY monohydric alcohol (a secondary alcohol would give a ketone, which is Tollens'-negative). …

Convert the gas volume to moles of active hydrogen (=moles of alcohol) via the 1:1 Grignard/methane stoichiometry, get the molar mass, then use the silver-m …

• Forgetting the Zerewitinoff reaction's 1:1 alcohol-to-methane stoichiometry and mis-relating moles of gas to moles of alcohol. …