Iodoform Test in Organic Chemistry | A unique Video on YouTube | Jee Advanced | Mains | NEET | AIIMS

Hello, Guys. In this Lecture we are going to explain all the important points, concepts and questions from Iodoform test.
This is the most powerful Lecture for jee advanced and AIIMS. Here, We will discuss this topic for Jee Advanced and AIIMS level.
First of all u need to understand the basic mechanism behind this reaction. In this reaction you treat a compound containing methyl carbonyl group with X2 and NaOH. Here NaOH is a strong base which extracts a proton from the alpha position i.e. the methyl group, thus producing an anion. This anionic group attacks the halogen molecule extracting X+. This leaves us with a compound with one alpha hydrogen replaced with a halogen atom. This same process repeates two more times. This produces a compound with all three alpha hydrogen atoms replaced with halide groups.
Now a new molecule of base NaOH attacks the electrophilic carbonyl carbon to replace the tri halide methyl anion. This anion extracts a proton from the carboxylic acid formed to generate a haloform molecule with by product as the carboxylate anion.
This complete reaction is known as haloform reaction.
The rate determining step in this reaction is the formation of a stable alpha carbanion. So stability of this carbanion enhances the forward reaction.
Hope this helps :) . Feel free to ask any further doubts in comment section.
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