Cl2 excess/ OH- will remove _______
two OH’s syn
Cl2 excess/ OH- will remove aldehydes and ketones.
In presence of an excess of chlorine (Cl2) and a base (OH-), the aldehyde or ketone undergoes oxidation to form carboxylic acids. The reaction is called chloroform reaction.
The general equation is:
RCHO + 3Cl2 + 4OH- → RCOOH + 3Cl- + 3H2O
Where R is an alkyl or aryl group.
The reaction starts with the formation of a chlorohydrin intermediate by the addition of one chlorine molecule and one hydroxyl ion to the carbonyl group of the aldehyde or ketone:
RCHO + Cl2 + OH- → RCH(OH)Cl + Cl-
The chlorohydrin intermediate then undergoes further reaction with excess chlorine and base to form a carboxylic acid:
RCH(OH)Cl + Cl2 + 2OH- → RCOOH + 3Cl- + 2H2O
This reaction is used as a qualitative test for the presence of aldehydes and ketones in an organic compound.
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