# Why does CH3CH(OH)CH2CH3 give iodoform test?

The iodoform test is given by compounds containing $$\ce{-(C=O)CH3}$$ group or compounds which can be oxidized to such. $$\ce{ CH3CH(OH)CH2CH3}$$

Is a 2 degree alchohol and even though it were to be oxidized it would not contain a $$\ce{-(C=O)CH3}$$ group but according to a question asked by JEE in 1997 it as a matter of fact does give the test.

HOW?

• This alcohol can be oxidised to a ketone as $\ce{I2/NaOH}$ is also an oxidising agent (as it produces $\ce{NaOI}$ in which iodine is at higher oxidation number than general). – Soumik Das Nov 12 '18 at 16:14

Mechanism (taken from this Wikipedia page) which shows $$\ce{BrO-}$$ instead of $$\ce{IO-}$$, but would be executed in a similar fashion: