Question

Which does not react with ${I_2}\& NaOH$ ?
(A) Ethyl alcohol
(B) Acetaldehyde
(C) Acetone
(D) Propanol

Answer 1

The reaction with iodine and sodium hydroxide ${I_2}\& NaOH$ is known as Haloform reaction and commonly known as Iodoform test. This test is known as the Iodoform test as iodoform is formed as a by-product. The basic condition for the iodoform test is given by the compounds which have methyl ketone.

Complete answer:
First, we will understand the basic haloform reaction. According to haloform reaction, when a methyl ketone (even acetaldehyde) is reacted with halogen and that too in aqueous sodium hydroxide. During the reaction, the ketone gets oxidized to the sodium salt of acid with one carbon atom less than the ketone. At the same time, the haloform $(CH{X_3})$ also gets formed. Therefore, this reaction is also considered as an Iodoform test.
We can use this iodoform test to check which compound does not react with ${I_2}\& NaOH$ . So, when the haloform reaction is carried out with iodine, a yellow precipitate of iodine is formed or obtained. So, the formation of a yellow precipitate of iodoform is used to detect the presence of $- COC{H_3}$ . Now we will use this condition of the iodoform test in the given option. So first, we will study their structures. So, in propanol, there is no methyl ketone even on oxidation. So the group is absent in propanol $C{H_3}C{H_2}C{H_2}OH$ . Acetaldehyde, Acetone, and ethyl alcohol give iodoform tests due to the presence of $- COC{H_3}$ the group. Therefore, propanol will not give an iodoform test and hence does not react with it ${I_2}\& NaOH$ .
Therefore, the correct option is (D).

Note:
The iodoform test is given by methyl ketones, acetaldehydes, ethanol, and compounds such as $C{H_3}CH(OH) -$ which are oxidized to appropriate carbonyl compounds under the conditions used for the reaction.