Question

An aromatic compound A of the molecular formula ${{C}_{8}}{{H}_{10}}O$ on reaction with iodine and dilute NaOH gives a yellow precipitate. The structure of the compound is expected to be:
A.
B. ${{C}_{6}}{{H}_{5}}-CHOH-C{{H}_{3}}$
C.
D.

Answer 1

The compound has molecular formula ${{C}_{8}}{{H}_{10}}O$, since it has one oxygen, then the compound will be alcohol, an ester, or a ketone. The second information given is that the compound forms a yellow precipitate with iodine and NaOH, this means that the compound gives the Iodoform test.

Complete step by step solution: The compound has molecular formula ${{C}_{8}}{{H}_{10}}O$, since it has one oxygen, then the compound will be alcohol, an ester, or a ketone. The second information given is that the compound forms a yellow precipitate with iodine and NaOH, this means that the compound gives the Iodoform test.
So, the compounds that will positive test for the Iodoform test when it give a yellow precipitate with iodine and dilute NaOH, and there is a condition for the compound to give the Iodoform test is the presence of $-COC{{H}_{3}}$ group.
So, in the given options there are two compounds in which the $-COC{{H}_{3}}$ group is present, i.e., option (a) and option (b). The molecular formula of the compound in option (a) is ${{C}_{8}}{{H}_{8}}O$ and the molecular formula of the compound in option (b) is ${{C}_{8}}{{H}_{10}}O$.
So, the reaction is given below:

So, this compound is secondary alcohol and gives yellow precipitate in the Iodoform test.

Therefore, the correct answer is an option (b).

Note: Iodoform is an important test that is used to distinguish between methyl ketone and non-methyl ketones. A compound that doesn't give the Iodoform test is formaldehyde or methanal.