Question

Compound (A) $C_4{H}_{10}O$ reacts rapidly with metallic 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 metallic sodium but reacts rapidly with Lucas reagent. What are (A), (B) and (C)?

Answer 1

Alcohols react with metallic sodium but with Lucas' reagent, no visible reaction at room temperature and thus, $\left(A\right)$ would be primary alcohol, $1-$Butanol.

Dehydration reaction of butanol$takesplacewhentreatedwith$H_2SO_4$ and an alkene is formed i.e. Butene.

When alkene again hydrated it forms alcohol. which is almost inert to metallic sodium but reacts rapidly with Lucas reagent.
Hence, $C$ should be tertiary alcohol.

From the above estimations,

$\underset{\left(A\right)}{({CH}_3{)}_2{CHCH}_{2}OH};\underset{\left(B\right)}{({CH}_3{)}_{2}C={CH}_2};\underset{\left(C\right)}{({CH}_3{)}_{3}COH}$