Iodoform reaction

[@NarendraModu-w6k]

thx man helped a lot

[@CVPQ]

In the final step I draw out the Iodoform and the carboxylic acid with the proton present - this is the final product but not the result of the arrows I have drawn which would leave the acid unprotonated (negative charge on the oxygen like in the title slide at the very beginning)

[@oshadayasiru6224]

Best explain, Thank you!

[@aszasq3150]

i'm a bit confused, can someone explained why the OH- ion attack the hidrogen first and then attack the keton? why wouldn't the OH- ion attack the keton right away? the carbon attached to the oxygen should be quite parsially positive no?

[@CVPQ]

It will attack the ketone also though the attack on the ketone will become more likely as there are more halogens added to the adjacent carbon since they make it increasingly more partially positive. Importantly though, the attack on the carbonyl is reversible and the OH will just leave again until there is another leaving group that can leave and the adjacent carbon only becomes a sufficiently good leaving group when it is substituted with three halogens.

[@Abhisekpr]

Ops bro