Question

Which is the major product formed when acetone is heated with iodine and potassium hydroxide?
(A)- Iodoacetone
(B)- Acetic acid
(C)- Iodoform
(D)- Acetophenone

Answer 1

Hint: Haloform reaction is used to detect the $C{{H}_{3}}CO-$ group in the molecule. Aldehydes and ketones undergo a haloform reaction to give the corresponding haloform along with sodium salts of carboxylic acids containing one carbon atom less than the parent molecule.

Complete step by step solution:
Methyl ketones or methyl carbonyl on treatment with a solution of sodium hypohalite ($NaOCl$ (sodium hypochlorite), $NaOBr$ (sodium hypobromite), and $NaOI$ (sodium hypoiodite) ) undergo haloform reaction to give the corresponding haloform ($CHC{{l}_{3}}$ (chloroform), $CHB{{r}_{3}}$ (bromoform), and $CH{{I}_{3}}$ (iodoform) ) along with sodium salts of carboxylic acids containing one carbon atom less than their respective parent methyl ketone or carbonyl. The sodium salts of carboxylic acids on acidification give the corresponding acids.
In this reaction, all three H-atoms of the methyl group are first replaced by halogen atoms to form either a trihalo aldehyde or tri haloketone which subsequently reacts with alkali to yield a haloform and the salt of a carboxylic acid-containing one carbon atom less than the starting molecule.
For example, when acetaldehyde reacts with chlorine it forms chloroform and formic acid.
This takes place in 3 steps.
Step- 1- Acetaldehyde reacts with chlorine to form chloral.
$C{{H}_{3}}CHO+3C{{l}_{2}}\xrightarrow{NaOH}CC{{l}_{3}}CHO+3HCl$
Step- 2- Chloral on hydrolysis forms chloroform and sodium formate.
$CC{{l}_{3}}CHO+NaOH\xrightarrow{Hydrolysis}CHC{{l}_{3}}+HCOONa$
Step- 3- The sodium formate on acidification forms formic acid.
$HCOONa\xrightarrow{Dil.HCl}HCOOH$
Therefore, the formic acid formed is having one carbon atom less than acetaldehyde.
So, when acetone with iodine and potassium hydroxide yields Iodoform as a major product.
Step- 1- Acetone reacts with iodine to form a tri iodoacetone.
$C{{H}_{3}}COC{{H}_{3}}+3{{I}_{2}}\xrightarrow{KOH}C{{I}_{3}}COC{{H}_{3}}+3HI$
Step- 2- tri iodoacetone on hydrolysis forms iodoform and sodium acetate.
$C{{I}_{3}}COC{{H}_{3}}+KOH\xrightarrow{Hydrolysis}CH{{I}_{3}}+C{{H}_{3}}COONa$
Step-3- Sodium acetate on acidification forms acetic acid.
$C{{H}_{3}}COONa\xrightarrow{Dil.HCl}C{{H}_{3}}COOH$

Therefore, the correct answer is an option (C)- Iodoform.

Note: When sodium hypoiodite is used, the yellow precipitate of iodoform is produced. Due to the formation of a yellow precipitate of iodoform in the reaction, it is known as iodoform test and used to detect $C{{H}_{3}}CO-$ group.