1. There's only one para position on this molecule, which is directly opposite of the nitro group. The other aromatic ring is meta to the nitro group. In a para attack, the intermediate sigma complex will form a tertiary cation opposite the nitro group, which is more preferred to the formation of a cation on the meta position. Intermediate sigma complex of a meta position only results in secondary cations. If you look at an energy profile of meta vs. ortho/para attacks, meta requires much more energy than ortho or para. Since there's no ortho position, para can be the only answer.
2. The fluorination of benzene is a very difficult reaction to control. It can be done with thallium tris(trifluoroacetate) (
Tl(OCOCF3)2), followed by potassium fluoride and boron trifluoride. Since this is a two step process and involves a heavy metal, it is a highly unfavorable reaction. In fact, my textbook has this to say about thallium "Like most heavy metals, thallium is highly toxic and should not be used on breakfast cereal."
You might think that the bromination of benzene would be a better reaction, but it is also a two step process, albeit without the use of heavy metals. Chlorination of benzene requires an aluminum chloride catalyst to proceed. However, the iodination of benzene typically uses nitric acid, which functions as a reagent rather than a catalyst. It also requires only half a mole of I
2, and the iodine cation acts as an electrophile. All of this happens in one step, so it is much faster than the other three halogenation reactions. So the answer was right, it is iodobenzene.
3. There are three on the aromatic ring on the left, and two on each of the benzylic carbons (the ones that neighbor the benzene ring). My answer is seven. I'm not sure why the answer is five.
4. In a reaction with a strong acid, aniline is protonated to form the anilinium ion ([X]). The ammonium ion is a strongly deactivating meta-director. Bromination of this reaction will then give you para substitution. I believe that the ammonium ion stays since the bromine atom will shift the electron density slightly towards itself. So [Y] would be
p-bromoanilinium. I'm not totally sure on this, I'm probably wrong. But that's my best guess. If they made ChemSketch for Mac, I would be able to draw some mechanisms, but sadly it's only for Windows.
PS - I love these questions. Your professor sure makes you guys think a lot. Keep posting them up, I love a good challenge