Question

Which of the following will react with $NaOI$?
A.
B.
C.
D.All

Answer 1

$NaOI$ is known as sodium hypo iodide. It is formed by the reaction of ${I_2}$ and $NaOH$ used in the iodoform reaction. The alcohols, aldehydes and ketones containing $C{H_3}CO$group react with $NaOI$ to form yellow precipitate of $CH{I_3}$ which is called as iodoform. This reaction is used as a test for differentiating between different alcohols, aldehydes and ketones.

Complete answer:
If any molecule reacts with $NaOI$ it should have $C{H_3}CO$ i.e. group and at the end of the reaction it will give $CH{I_3}$ as a yellow precipitate. $NaOI$ takes part in a reaction as $NaOH$ and ${I_2}$.
First let us look at basic reaction where$NaOI$ will react with ( where R is any aromatic or aliphatic group)
The mechanism is as follows:

Now we can clearly see in the above mechanism how the $C{H_3}CO$ group containing molecules is undergoing reaction and in the end $CH{I_3}$ is produced.
So among our options we say that and will react with $NaOI$ to give the product as both the molecules contain $C{H_3}CO$ group in their structure. Now let us look at the aromatic compound.
Here this molecule does not have any $C{H_3}CO$ group in its structure so for knowing whether it will react with $NaOI$ or not we will have to look into its mechanism, which is as follows:

So in the above mechanism base will extract and then $I$ will get attached to it twice, third time there is no $H$left so ${}^ - OH$ will then act as nucleophile
We can see that the compound does not have any $C{H_3}CO$ group but it still reacts with $NaOI$ to give $CH{I_3}$
Therefore all the above molecules will react with $NaOI$.

Hence the correct option is $D.$ All

Note:
Only those molecules will react with $NaOI$ which have$C{H_3}CO$ group in them and yield $CH{I_3}$ as a precipitate. Even if the molecule does not have$C{H_3}CO$, in some cases it may react with $NaOI$ and give the yellow precipitate as in the above case.