[collin_det]

Hi.

I have a question regarding the mechanism of the iodination of benzocaine. There is added Ag2SO4 and elemental iodine in ethanol @ rt. The iodine should be incorporated next to the amine. I have no clue really about the mechanism. I can see that it is halogenated. Normally on activated rings, you should just be able to add bromine or similar and then it would be brominated ortho or/and para, because it is highly activated by an amine, alcohol or methoxy group. But in this case Ag2SO4 is added. I am wondering if it is because the Ag acts as a Lewis acid, activating the amine, which creates a carbocat ion ortho, which the elemental iodine can attack, yielding the desired product. I just think it sounds weird. Otherwise is the Ag2SO4, just to help to split up the iodine and then it simply attacks the ring ortho to the amine?

And how do I know that the lewis acid Ag do not activate the ester and I then get it incorporated ortho to the ester instead?

I have attached a PDF with my crazy mechanism if anyone dares to see it :-)

Thanks

[NTUstudent]

hey your attached mechanism is really nice looking, but i don't think it's correct!

i think that Ag⁺ is just polarize I₂, so ring can attack iodine, and second iodine will leave more rapidly because of silver ion!
ortho position because amines activate ortho positions toward this kind of reactions

[collin_det]

Thank you very much.

I think you are right. So the Ag+ activates one of the halogen bonds of iodine, so it can break, and the other I+ is attacking the ring ortho/para, but ortho because it is substituted by the ester? But I will get HI acid from the halogenation then?

But I dont understand how it makes the I ion leave easier? Could you explain it maybe? Thanks.

[NTUstudent]

ok
positive charged silver polarize iodine, so that one iodine will have partial minus charge and second one will be partially plus charged. So I-I bond is weakens.
don't worry about H+, you will have I- as well.

[jake.n]

Thank you very much.

I think you are right. So the Ag+ activates one of the halogen bonds of iodine, so it can break, and the other I+ is attacking the ring ortho/para, but ortho because it is substituted by the ester? But I will get HI acid from the halogenation then?

But I dont understand how it makes the I ion leave easier? Could you explain it maybe? Thanks.

As the Ag⁺-I-I species is attacked by the nucleophillic ring, AgI will fall out of solution.  Either the sulfate counter-ion will deprotonate the positively charged ring, or it would be deprotonated upon base workup.  So, you will likely get AgI as a solid and some sulfuric acid species.

The addition of the silver ion is a good indication that halogen activation is going on.  The precipitation of the silver halide salt drives the equilbrium ala Le Chatlier's Principle.