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MHT CET Chemistry question

ChemistryOrganic Oxidation & Reduction ReactionsMHT CET

0.44 g of a monohydric alcohol when added to methyl magnesium iodide in ether liberates at STP, 112 of methane. With PCC the same alcohol forms a carbonyl compound that answers silver mirror test. The monohydric alcohol is

Answer: (B)

Solution

One mole of alcohol liberates one mole of : 112 cm / 22400 cm = 0.005 mol, so M = 0.44 g / 0.005 mol = 88 g/mol (); the carbonyl compound gives the silver mirror test, so it is an

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Question ID VK-079921