Compound (A) `C_(4)H_(10)O` reacts rapidly with metalic sodium, but undergoes almost no reaction with Lucas reagent When (A) is treated with hot concentrated sulphuric acid, a new compound (B) `C_(4)H_(8)` is formed If `C_(4)H_(8)` is hydrated with sulphuric acid a new compound (C) `C_(4)H_(9)OH` is formed, which is almost inert to metalic sodium but reacts rapidly with Lucas reagent what are (A),(B) and (C)?

Explanation

(i) As the compound [C] is almost inert towards sodium metal but reacts rapidly with Lucas reagent, it is a tertiary alcohol.
(ii) The molecular formula `C_(4)H_(10)O` suggests alcohol to be tertiary nutyl alcohol i.e. `(CH_(3))_(3)COH`. Since [C] has been formed by the hydration of compound [B] which is an alkene, the compound [B] is therefore, `(CH_(3))_(2)C=CH_(2)`.
(iii) The compound [B] has been formed from [A], as a result of dehydration with concentrated `H_(2)SO_(4)`. since compound [A] reacts rapidly with metallic sodium but very slowly with Lucas reagent, it is a primary alcohol with the formula `(CH_(3))_(2)CHCH_(2)OH`.
The series of reactions involved are as follows :
`underset(("Isobutyl alcohol "(C_(4)H_(10)O)),([A]))((CH_(3))_(2)CHCH_(2)OH) underset("Heat")overset(H_(2)SO_(4)" (conc.)")(rarr) underset(("Isobutylene "(C_(4)H_(8))),([B]))((CH_(3))_(2)C=CH_(2)) overset(H_(2)O //H^(+))(rarr) underset(("Tert. butyl alcohol "(C_(4)H_(10)O)),([C]))((CH_(3))_(3)COH)`