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The haloform reaction of acetone with sodium hypobromite yields:
(a)- acetic acid
(b)- propionic acid
(c)- acetaldehyde
(d)- isopropyl alcohol

Answer
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Hint: Haloform reaction is used to detect the CH3CO 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 answer:
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 corresponding haloform (CHCl3 (chloroform), CHBr3 (bromoform), and CHI3 (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.
CH3CHO+3Cl2NaOHCCl3CHO+3HCl
Step- 2- Chloral on hydrolysis forms chloroform and sodium formate.
CCl3CHO+NaOHHydrolysisCHCl3+HCOONa
Step- 3- The sodium formate on acidification forms formic acid.
HCOONaDil.HClHCOOH
Therefore, the formic acid formed is having one carbon atom less than acetaldehyde.

So, when acetone with sodium hypobromite yields formic acid.
Step- 1- Acetone reacts with bromine to form a tribromo acetone.
CH3COCH3+3Br2NaOHCBr3COCH3+3HBr
Step- 2- tribromo acetone on hydrolysis forms bromoform and sodium acetate.
CBr3COCH3+NaOHHydrolysisCHBr3+CH3COONa
Step-3- Sodium acetate on acidification forms acetic acid.
CH3COONaDil.HClCH3COOH
So, the correct answer is “Option A”.

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 CH3COgroup.