Question

An alcohol of formula ${\text{C}}_{\text{9}}{\text{H}}_{\text{12}}\text{O}$ reacts with $\text{N}{\text{a}}_{\text{2}}\text{C}{\text{r}}_{\text{2}}{\text{O}}_{\text{7}}$ to form a compound having a formula ${\text{C}}_{\text{9}}{\text{H}}_{\text{10}}\text{O}$. The original alcohol might be:

A.

B.

C.

D.

Answer 1

Hint: Sodium Dichromate oxidises different organic compounds, under different acidic conditions as it is a strong oxidising agent. Alcohol can oxidise to carbonyl compounds or carboxylic acid.

Complete Step by step solution:
Alcohols are oxidized by oxidising agents like sodium dichromate and potassium permanganate under highly acidic conditions to form aldehydes or ketones. This may be further oxidized to form carboxylic acids. The empirical formula ${\text{C}}_{\text{9}}{\text{H}}_{\text{10}}\text{O}$ has six carbon atoms and 5 hydrogen atoms for the phenyl ring, which leaves us with three carbon atoms, 5 hydrogen atoms, and one oxygen atom. In the above options all the options have the said number of carbon atoms, hydrogen atoms, and oxygen atoms.
Among the given options only option B will be oxidized by the sodium dichromate to form the compound with the empirical formula as given.

Hence, the correct option is B.

Notes: Sodium and potassium dichromate are strong oxidising agents that can oxidise alcohols into aldehydes and ketones. Due to the oxidation, the orange oxidising agent turns green due to the formation of the chromic salt. The type of products depends on the type of the alcohol. Primary alcohols give either aldehydes or ketones depending on the reaction conditions. Secondary alcohols are always converted to form ketones by the action of Sodium and potassium dichromate under acidic conditions. While tertiary alcohols are not converted into either alcohols or ketones due to their inability to react.