# Bromination of a benzene based molecule

Using two equivalents of $\ce{Br2}$ and catalytic $\ce{FeBr3}$, compound A can be converted into B. In the same reaction vessel but in multiple steps, the conversion of B into C ($\ce{C11H12Br3I}$) can be obtained using $\ce{H2SO4, NaNO2}$ and $\ce{NaI}$.

The molecule A is below:

My Attempt

What I am struggling with is determining what double bonds the $\ce{Br2}$ is added on. I am pretty sure that the double that is not on the benzene ring will get brominated, but I am not sure what other double bond will also get brominated.

I know the mechanism for the conversion for B into C, which is basically the sandmeyer reaction followed by substitution of iodine. However what further confuses me is that the end product only has 3 bromine atoms, not 4 bromine which is what I expected (since bromination occurs twice).

• No other double bond gets brominated. Electrophilic substitution in the ring, that's what happens. – Ivan Neretin Feb 12 '16 at 12:26
• @IvanNeretin Thanks. How do I know which carbon in the ring gets substituted with the bromine atom? – Nanoputian Feb 12 '16 at 12:32
• The one in para position to the hydrocarbon tail, I'd say. – Ivan Neretin Feb 12 '16 at 12:43
• I think you are likely to get a mix, but ortho to NH2/para to alkyl seems likely dominant. – Lighthart Feb 12 '16 at 22:55
• – Jan May 1 '16 at 16:41

• It might probably be worth mentioning that, while in this particular case the electrophilic addition does take place on the pi bond (as justified by the change in molecular formula), BUT in general, with the reagent $\ce{Br2 / FeBr3}$, only EAS can take place. – Gaurang Tandon Mar 2 '18 at 13:11