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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?

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  • $\begingroup$ 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
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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.


References:

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

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