375 mg of an alcohol reacts with required amount of methylmagnesium br – InterviewSolution

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1.	375 mg of an alcohol reacts with required amount of methylmagnesium bromide and releases 140 mL of methane gas at STP. The alcohol is
Answer» ethanol n-butanol methanol n-propanol Solution :140 mL of `CH_(4)` is produced from ALCOHOL=375mg `therefore22400`mL of `CH_(4)` is produced from alcohol`=(375)/(1000)xx(22400)/(140)=60` G `mol^(-1)` Thus, alcohol with molecular mass=60g `mol^(-1)` is `CH_(3)CH_(2)CH_(2)OH` (n-propanol).

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