Bromination of a benzene based molecule

Question

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).

Answer 1

You can deduce the end product from considering the change in the molecular formula from start to finish. Aromatic bromination results in the loss of one hydrogen and gain of one bromine, bromination of the double bond results in no loss of hydrogen but gain of two bromines and the Sandmeyer reaction results in loss of two hydrogens. Therefore the intermediate must have fourteen hydrogens which means only one aromatic bromination has taken place.

The aromatic bromination likely produces a major product with the bromine at the ortho position shown. The amine group strongly activates the positions ortho and para to it so substitution will occur at one of these three positions. I suspect the one shown is most likely due to the bulky nature of the other attatched group, which stericically hinders attack at the other two positions.