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0.44 g of a monohydric alcohol when added to methyl magnesium iodide in ether lilberates at STP 112 cm^(3) of methane with PCC the same alcohol form a carbonyl compound that answers silver mirror test. Identify the compound. |
Answer» Solution :`R-OH + CH_(3) MgI to CH_(4)` +  `22400 cm^(3)` at STP is `112 cm^(3)` Mass of alcohol `=(0.44 XX 22400)/112 = 88 G` Alcohol reacts with PCC to give a carboxyl compound which answers silver mirror test. Therefore, the alcohol MUST be a PRIMARY alcohol which on oxidation GIVES aldehyde with mass of 88 g. So the probable primary alcohol is `(CH_(3))_(2)CH-CH_(2)-OH`. |
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