`0.037 g` of an alcohol, `ROH` was added to `CH_(3)MgI` and the gas evolved measured `11.2 cm^(3)` at N.T.P. What is the molar mass of alcohol ? On dehydration, `ROH` gave an alkene which on ozonolysis gave acetone as one of the products. `ROH` on oxidation easily gave an acid containing the same number of carbon atoms. give the structure of `ROH` and of acid with proper reasoning.

Explanation

(i) The molar mass of alcohol can be calculated as follows :

`11.2 cm^(3)` of `CH_(4)` are evolved from alcohol `=0.037 g`
`22400 cm^(3)` of `CH_(4)` are evolved from alcohol `=((0.037 g))/((11.2 cm^(3)))xx(22400 cm^(3))=74.0 g`.
(ii) General formula of alcohol `=C_(n)H_(2n+1)-OH`
`:." "12n+2n+1+17=74` or `14n+18=74`
`:." "14n=74-18=56` or `n=56//14=4`.
The molecular formula of alcohol `= C_(4)H_(9)OH`.
(iii) As the alcohol can be easily oxidised to acid with the same number of carbon atoms, it is primary alcohol. Since on dehydration alcohol gives an alkene which upon ozonolysis gives acetone as one of the products, the alcohol must have `(CH_(3))_(2)CH-` group. Thus, the alcohol is isobutyl alcohol. The reactions involved a+re as follows :
`{:(" "CH_(3)" "CH_(3)" "CH_(3)" "H),(" |"" |"" |"" |"),(CH_(3)-CH-CH_(2)OH underset(-H_(2)O)overset(Conc. H_(2)SO_(4)//"heat")(rarr)CH_(3)-C=CH_(2)underset(Zn//H_(2)O)overset(O_(3))(rarr)CH_(3)-C=O+H-C=O),(" Isobutyl alcohol"" Isobutylene"" Acetone"" Formaldehyde"):}`
`{:(" "CH_(3)" "CH_(3)" "CH_(3)),(" |"" |"" |"),(CH_(3)-CH-CH_(2)OH underset(-H_(2)O)overset((O))(rarr)CH_(3)-CH-CHO overset((O))(rarr)CH_(3)-CH-COOH),(" Isobutyl alcohol"" Isobutyraldehyde"" Isobutyric acid"):}`