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.


  • $\begingroup$ 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). $\endgroup$
    – 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:

Mechanism of secondary alcohol to ketone by the hypohalite ion

Thus 2-Butanone responds to iodoform test successfully.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.