# 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

The substrate you have provided (2-butanol) would indeed give a positive test.

The iodoform test is performed in presence of iodine and potassium or sodium hydroxide, which first oxidizes the aforesaid compound to 2-Butanone.

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

Thus 2-Butanone responds to iodoform test successfully.

References:

Ueber Entstehung von Jodoform und Anwendung dieser Reaction in der chemischen Analyse, Lieben, Ann. Spl. Bd. 7, 218 1870