Question

Which of the following gives yellow precipitate with NaOI?
(A)
(B)

Answer 1

This test is called an iodoform test. The final product of the substance has a medicinal smell like that of antiseptic solutions

Complete step by step solution:
As per the question asked, the yellow precipitate formed is due to the formation of iodoform or called triiodomethane. This reaction is a test for molecules having the groups like$R-CO-C{{H}_{3}}$ $R-CO-C{{H}_{3}}$,$R-CH(OH)-C{{H}_{3}}$
-When ${{I}_{2}}$ reacts with NaOH it produces NaOI, and this on reaction with the compounds containing the groups like $R-CO-C{{H}_{3}}$, $R-CH(OH)-C{{H}_{3}}$ will give yellow precipitate as a result of iodoform formation.
And this is used as a test for ketones and aldehydes, secondary alcohol groups which can be converted to carbonyl group by oxidation also shows positive results towards this test.
The secondary alcohol must have at least a methyl group in the alpha position to show this test.
Description of Iodoform test- To carry out this test, first iodine and NaOH is warmed, which produces NaOI and this is allowed to pass through aldehydes and ketones containing $R-CO-C{{H}_{3}}$ and through secondary alcohol which has at least one methyl group in the alpha position . And we will observe a yellow precipitate with an antiseptic smell.
$R-CO-C{{H}_{3}}\xrightarrow[NaOH]{{{I}_{2}}}RCO{{O}^{-}}N{{a}^{+}}+CH{{I}_{3}}$
$R-CH\left( OH \right)-C{{H}_{3}}\xrightarrow[NaOH]{{{I}_{2}}}RCO{{O}^{-}}N{{a}^{+}}+CH{{I}_{3}}$
If we are testing for an aldehyde and the test is positive then it will be acetaldehyde as it is the only group having $C{{H}_{3}}CO$ group.
In the options above the first, option (A) is a secondary alcohol which has $R-CH(OH)-C{{H}_{3}}$ group and gets oxidized to carbonyl group during the reaction to give the final yellow precipitate.

So the correct answer for this question is option (a).

Note: In the mechanism of the iodoform reaction, first an acidic hydrogen is removed from the aldehyde, ketone or secondary alcohol. So there must be an acidic hydrogen available other than the hydrogen available in the methyl group which is attached to the carbonyl group that is why the second option given in the question gives a negative test result. Compounds which show positive results towards the iodoform test are – dimethyl ketone, acetaldehyde, secondary alcohols with methyl group in alpha position, ethanol etc.