0.037 g of an alcohol was added to methyl magnesium iodide and the gas evolved at STP, occupies the volume 11.2 cm^(3). On dehydration, the alcohol gives an alkene. The alkene upon ozonolysis produces acetone as one of its products. The alcohol gives the carboxylic acid upon oxidation with same number of carbon atom. The alcohol is ...

0.037 g of an alcohol was added to methyl magnesium iodide and the gas

1.	0.037 g of an alcohol was added to methyl magnesium iodide and the gas evolved at STP, occupies the volume 11.2 cm^(3). On dehydration, the alcohol gives an alkene. The alkene upon ozonolysis produces acetone as one of its products. The alcohol gives the carboxylic acid upon oxidation with same number of carbon atom. The alcohol is ...
Answer» n-butyl ALCOHOL Isopropyl alcohol Isobutyl alcohol Sec. Butyl alcohol Solution :The molecular mass of `R- OH= (22400 xx 0.037)/(11.2)= 74 g` `:. C_(n)H_(2n+2)O= 74 IMPLIES 12n +(2n+2)+16= 74` `implies n=4` `to` The possible alcohols are n-butyl alcohol, isobutyl alcohol, sec. Butyl alcohol and tertiary butyl alcohol. As the alcohol gives the carboxylic acid with same number of carbon atoms UPON oxidation, the option (d) is ruled out. Also, option (b) is ruled out as the alcohol is of 4-carbon atoms. The one of the products of ozonolysis of Alkene is acetone. Thus, alkene has = `C(CH_(3))_(2)` GROUP. Thus, alkene must be isobutylene, i.e., `(CH_(3))_(2) = CH_(2)`. So, the given alcohol is isobutyl alcohol.