Since the compound $$\left( {{C_4}{H_{10}}O} \right)$$   react with sodium, it must be alcohol ( option B, C, or D ). As it is oxidised to carbonyl compound which does not reduce Tollen’s reagent, the carbonyl compound should be a ketone and thus $${{C_4}{H_{10}}O}$$   should be a secondary alcohol, i.e. $$sec$$  - butyl alcohol; other two given alcohols are $${1^ \circ }.$$

\[\begin{align} & C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}\xrightarrow[\text{Anti -Markownikoffs rule}]{\frac{HBr}{{{H}_{2}}{{O}_{2}}}} \\ & C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br\to \\ & \text{Bromo-butanane (}{{\text{1}}^{\circ }}\text{product)} \\ & C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br\xrightarrow[\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\,\,\,{{S}_{N}}2\,\text{reaction} \\ \text{(Williamson }\!\!'\!\!\text{ s synthesis)} \end{smallmatrix}]{{{C}_{2}}{{H}_{5}}ONa} \\ & \underset{\text{Ethoxy-butane}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}O{{C}_{2}}{{H}_{5}}}}\, \\ \end{align}\]

Electron releasing inductive effect of $$ - OC{H_3}$$   group facilitates the protonation of alcohol involved in dehydration mechanism.

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Pyridiminum chloro-chromate $$\left( {PCC} \right)$$  is specific for the conversion.