Question

The product formed on reaction of ethyl alcohol with bleaching Powder is
A. \[CHC{l_3}\]
B. \[C{H_3}COC{H_3}\]
C. \[CC{l_3}CHO\]
D. \[C{H_3}CHO\]

Answer 1

Hint:Ethyl alcohol is the only primary alcohol that undergoes haloform reaction, other primary alcohols don’t give haloform test. The reason for this exception is that ethyl alcohol can oxidize to ethanal. Ethanal is a methyl aldehyde, it has a methyl group attached to carbonyl carbon. All the methyl aldehyde and ketone show haloform tests.

Complete step-by-step answer: The bleaching powder is a chloronating agent as well as oxidising agent. It has many industrial uses due to this property. The bleaching powder breaks into calcium hydroxide and chlorine gas which are the main reagents of haloform test.
$CaOC{l_2} + {H_2}O \to Ca{(OH)_2} + C{l_2} \\$

The equation of the ethyl alcohol, and chlorine in the presence of calcium hydroxide is as follows:
1. In the first step ethyl alcohol gets converted into ethanal.
$C{l_2} + C{H_3}C{H_2}OH \to C{H_3}CHO + 2HCl $

2. In the second step, $CC{l_3}COH$ forms by the reaction of ethanal and chlorine.
$C{H_3}CHO + 3C{l_2} \to CC{l_3}COH + 3HCl \\$

3. In the third step, $CC{l_3}COH$ hydrolysed to give chloroform and calcium salts of
carboxylic acids.
$2CC{l_3}COH+ Ca{(OH)_2} \to 2CHCl_3 + 2HCOO^-+Ca^{2+}$

Therefore, the product formed on reaction of ethyl alcohol with bleaching Powder is \[CHC{l_3}\].

Option ‘A’ is correct

Note: Aldehydes and ketones which have at least one methyl group linked to the carbonyl carbon atom are oxidised by bleaching powder to calcium salts of corresponding carboxylic acids having one carbon atom less than that of carbonyl compound. The methyl group is converted to haloform.