[orgoclear]

see the attached fig.

Q1. It should be in the ring where -NO2 group is attached isnt it as two o-p directing groups are there. But the answer is given in the ring which is para to the nitro-phenyl group

Q2. Im pretty sure iodobenzene is wrong and fluorobenzene is correct. Just want a confirmation

Q3. not sure (thinking 3)

Q4. with conc H2SO4 wont the aniline get protonated to give Ph-NH3+  which becomes a meta directing group? Answer says the product (Y) is 2,4,6-tribromoaniline while I think it should be 3,4-dibromoaniline (due to excess)

[Grundalizer]

I'll try my best to help, although I'm not an organic chemist.

1. EAS would probably not take place at the o-p sites on the same ring as the NO2 because of steric hindrances.  Remember the more functional groups/atoms/electrons in the way of or surrounding the sites where you'd like the reaction to happen leads to instability.  NO2 is a meta-directing group, and at the meta position you also have two huge aromatic rings with decent electron density getting in the way, adding to instability.  So in that case, you want to go as far away from the NO2 group as possible, and the place that is furthest away is on the para ring.  I bet you would also get some EAS on the meta ring, but since it is that little bit closer to the NO2 perhaps it is not as suitable as the para-ring.

2.  Most flourinated compounds aren't very reactive since Flourine has such a high electronegativity...you have to use some powerful reagents to get a Flourine atom to let go.  When in doubt about things like this...look at the periodic table trend.  Iodine is the largest halogen (forget At noone uses it) with regards to atomic radius.  So electron Shielding plays a big roll, and is the reason F>Cl>Br>I in terms of reactivity.  Highschool/Orgo 1 really underplayed and left out how much information you can actually get from looking at periodic table trends, and atomic radius/electron shielding plays a big roll.  Thankfully my inorganic teacher was awesome.

3.  Hmmm...it's either 3 or 6...but I might go with you and guess 3.

4.  NH2 is an ortho-para directing group.  Meta-substitution does happen, but to a much lesser extent. 

See http://en.wikipedia.org/wiki/Electrophilic_aromatic_substitution halfwaydown it's your same reaction.

And http://www.mhhe.com/physsci/chemistry/carey/student/olc/graphics/carey04oc/ref/ch22arylamines.html

Best I can do, good luck Orgo is challenging

[orgoclear]

@grundalizer

thanks for Q1. (forgot steric factor completely)

Q2. Solomons says that fluorobenzene is most reactive as its +R effect is max due to effective orbital overlap

Q3. It was a foolish doubt (Ans: 5)

Q4. Bruice says that the ratio of para product to meta product in sulphonation is almost 1:1. So, I am now expecting the product to be aniline with bromine at positions 2,4,6 and a sulpho- group at 3. along with 2,4,6 tribromo aniline.. (both almost having 50%)

am I correct?

[bromidewind]

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 I₂, 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

[orgopete]

Re: love these questions? Not so much. If I got them right, I might feel differently.

Q1) I saw the nitration differently. If the resonance structure can be use to predict the orientation of the nitration, I would have predicted a different product, see attachment.

I did find nitration of biphenyl gives mainly para substitution (depending on conditions). I couldn't find any nitration results of a nitrobiphenyl. Dinitrobiphenyls are all made by condensation reactions.

Q2) I agree that it is fluorobenzene.

Q4) I had no shot at answering. I didn't expect conc sulfuric acid could sulfonate aniline. Bromine can polybrominate aniline, an anilinosulfonic acid, I don't know. How reactive is it?

[cpncoop]

Re: Benzyllic hydrogens

Benzyllic hydrogens are hydrogens on carbons adjacent to benzene rings.  This is why the answer is 5 (2 each on the methylene carbons, 1 on the tertiary carbon.)

Good explanations of the mechanistic problems for electrophilic aromatic substitution....