Question

Synthesis problem

Answer 1

Answer:

Because the final product is different than the one it's required

Explanation:

The question is incomplete, you are not showing the synthesis required. However, I found a similar to this, so I'm going to answer it based on that. If you have a different one, then, come back and post it on another question.

The synthesis we have is Toluene to form the 2,3-dibromo-benzoic acid, so, the answer is really simple.

Remember that in Aromatic Electrophylic substitutions, we have differents kind of groups. One that activates the aromatic, and other that deactivate. These groups have priority order depending on the strength of the compound.

Halides and COOH are both deactivating groups, however, COOH is deactiviting the molecule and makes that any other electrophyle to react with the molecule, goes to the meta position of the ring, in other words, is meta oriented. Halides like bromine and Chlorine are also deactivating groups but they are ortho - para oriented.

Now, In picture 1, we have the actual synthesis. When we use the Br2/Febr2, we use a bromination reaction. As the methyl group is activating group, is ortho - para director, so, the bromine will go to the para position (It could go to the ortho position too, but it will rather go to the para position, is better cause it do not have steric hindrance with the methyl). When this happens, we make the oxydation with KMnO4 to transform the CH3 into a COOH group, which is deactivating the molecule along with the Bromine. Now the COOH is stronger than Bromine, so, the COOH takes priority in the order position, so, when it reacts with the bromine again, it's reacting to the meta direction of the COOH and not the ortho of the bromine, therefore, it will go to the meta position of the ring from the COOH.

In picture 2, we see why can't we do the oxydation and then two brominations. As it was stated before, the COOH is stronger than bromine, so, when the first bromination comes in, this will wo to the meta position  of the ring, but when the second bromination comes along, instead of going to the para position, it will go to the other meta position of the ring, and the final product is different.

This is why the synthesis should be done as the answer key shows.