IIT JEE , AIEEE Forum - Chemistry , Organic Chemistry

hit_ur_heart

I think it need more explanation , the above lines arent enough :-

First important think to note about that it is the carbonyl compound , so either it will proceed through their acidic hydrogen or necleophilic centre cabonyl carbon, in the present case , NaOH acts as a base to extract the acidic hydrogen of ------CH3
and this will continue until no hydrogen is left and these hydrogen will be replaced by halogen , as soon as all hydrogen are consumed. Now -OH will attack carbonyl carbon as it is now more electron deficient due to -I effect of halogens , after nucleophilic attack , instead of additon , -CX3 grop will leave , leadin to final product.

Few special points:-

1) Knowing importance of leaving group is very important , it is this ability that decide the fate of the reaction. I suppose u must have notice here a carbanion act as a leaving group!!!!!!, one of the few rare case where carbanion act as a leaving group( obviuosuly becuase of three halogens)
2) second , rate of the reaction is decide by this step , we can have various exception cases where haloform reaction did not occur , u shd be able to able figure that using this step.
3) One important fact also, ethyl alchol also gives Iodoform test , because they are oxidised to aldehyde in NaOH and X2.

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