1.

Why we get tribromoaniline in bromation with polar sovlent?

Answer»

That's true for benzene, but the methoxy-group in anisole isn't just o/p-directing. It is ACTIVATING as well. Actually, it's that much activating that anisole will react with bromine even without the presence of a Lewis acid.

So, why glacial acetic acid? It's a polar solvent and THUS nice to stabilize the Wheland intermediates. Moreover, it's easily removed in aqueous workup.

Edit 1 Does the acetic acid react with the (MOSTLY undissociated) acetic acid? At least a polarization of bromine (Brδ+−Brδ−⋅⋅⋅HOAcBrXδ+−BrXδ−⋅⋅⋅HOAc) is conceivable and I remember that BrOAcBrOAc has been postulated as the electrophilic species.

Edit 2 I remembered some brominations on dimethoxy-substituted hydroxymethylnaphthalines I did back in the days. The REACTION was performed at 60 °C in CHCl3CHClX3 in the presence of about 3 equivalents of anhydrous sodium acetate. Excess of bromine was reduced by addition of aqueous sodium sulfite. The yield was in the 85% range and I don't find any notes on dibrominated products.



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