A compound of the formula $\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}$ reacts with sodium and undergoes…

A compound of the formula $\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}$ reacts with sodium and undergoes oxidation to give a carbonyl compound which does not reduce Tollen's reagent, the original compound is
  1. Diethyl ether
  2. $n$ -Butyl alcohol
  3. Isobutyl alcohol
  4. sec-Butyl alcohol

Solution

Since the compound $\left(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}ight)$ 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 $\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}$ should be a secondary alcohol, i.e. sec-butyl alcohol; other two given alcohols are $1^{\circ}$. ^

Asked in: JEE-TOPICTESTS-CHEMISTRY

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