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0.037 g of an alcohol, ROH was added to ...

0.037 g of an alcohol, ROH was added to `CH_3MgI` and the gas evolved measured `11.2 cm^3` at S.T.P. What is the molecular weight of ROH? On dehydration, ROH gives an alkene which on ozonolysis gives acetone as one of the products. ROH on oxidation easily gives an acid containing the same number of carbon atoms. Give structures of ROH and the acid with proper reasoning.

Text Solution

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(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"):}`
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