[pacman12]

Hello
I am new to this site and I am a college student currently taking an organic chemistry course. I like this site and how it helps students.

I have 2 questions of my own I am currently having trouble with,

First is why is p-nitro-aniline more acidic than p-bromo-aniline? I know it is because of inductive effect and p-nitro-aniline has more resonance structures. I have attached a link to a picture of the resonance structures I came up with. I can't seem to get the rest of them 

http://img340.imageshack.us/i/imag0419i.jpg/


My second question was how to know whether m-nitro-aniline is a stronger or weaker base than p-nitro-aniline. I know that NO2 is a meta directing group so would that mean it is more acidic?

Any help would be greatly appreciated. Thank you

[nj_bartel]

Using this, try to use your corrected knowledge of the resonance structures to draw the resonance contributors for meta-nitroaniline and para-nitroaniline and see what you come up with as far as basicity

[adianadiadi]

NH₂ group is basic

[Honclbrif]

The reason why something is made more acidic, is because its electron density is being removed to another part of the molecule. Keep in mind the difference between inductive effects (the ability of a group to pull electron density toward it through the sigma-framework), and resonance effects (the ability to draw structures with the pi-electrons redistributed). Both can "pull electron density around a molecule", but in general the resonance effects tend to be stronger.

First, consider the inductive effects only. Generally they're easier to consider because you don't have to redraw anything, you just need to have two groups connected by bonds. Different groups have different abilities to "push" (+I) or "pull" (-I) electrons. You should have a table in your book which ranks the order of various common groups. Check how the inductive power of bromo- and nitro- compare. Does this explain the observed effect?

Secondly, consider resonance effects. If you can draw a resonance structure which removes an electron pair from substituent A, and adds and electron pair to substituent B, you can say that B has removed electron density from A. Is either bromo- or nitro- capable of removing electron density from the aniline nitrogen?

Things can get more complex when the inductive effect of a group is opposite to its resonance effect (e.g: methoxy- is sigma-withdrawing, but pi-donating). When considering these cases, check to see if the two groups you are considering are in conjugation. If they are, the resonance effect almost always wins, if they are not in conjugation, the inductive effect almost always wins.

This brings us to your second question. Again, consider the inductive and resonance effects. In this case, the only difference is the location of the nitro group relative to the aniline. How would location effect inductive effects, and how would it effect resonance effects? How would these various effects stack?

[pacman12]

The reason why something is made more acidic, is because its electron density is being removed to another part of the molecule. Keep in mind the difference between inductive effects (the ability of a group to pull electron density toward it through the sigma-framework), and resonance effects (the ability to draw structures with the pi-electrons redistributed). Both can "pull electron density around a molecule", but in general the resonance effects tend to be stronger.

First, consider the inductive effects only. Generally they're easier to consider because you don't have to redraw anything, you just need to have two groups connected by bonds. Different groups have different abilities to "push" (+I) or "pull" (-I) electrons. You should have a table in your book which ranks the order of various common groups. Check how the inductive power of bromo- and nitro- compare. Does this explain the observed effect?

Secondly, consider resonance effects. If you can draw a resonance structure which removes an electron pair from substituent A, and adds and electron pair to substituent B, you can say that B has removed electron density from A. Is either bromo- or nitro- capable of removing electron density from the aniline nitrogen?

Things can get more complex when the inductive effect of a group is opposite to its resonance effect (e.g: methoxy- is sigma-withdrawing, but pi-donating). When considering these cases, check to see if the two groups you are considering are in conjugation. If they are, the resonance effect almost always wins, if they are not in conjugation, the inductive effect almost always wins.

This brings us to your second question. Again, consider the inductive and resonance effects. In this case, the only difference is the location of the nitro group relative to the aniline. How would location effect inductive effects, and how would it effect resonance effects? How would these various effects stack?

As far as resonance structures go this is what I have so far.

http://img217.imageshack.us/i/imag0421g.jpg/

I'm not sure how to go about the p-bromo-aniline resonance structures but I think it undergoes inductive effect because the bromo group gives it's electrons to the Nitro group making it more acidic? For p-bromo-aniline it is less basic because bromine doesn't pull electrons from the aniline group? Am I on the right track?

[adianadiadi]

for second one, closer the group greater the effect.

[Honclbrif]

"As far as resonance structures go this is what I have so far"

I'll be honest here, you need to go back and re-read the sections of your book about resonance structures and formal charges. If you don't strengthen those areas you are not going to solve these sorts of problems.

The nitrogen in the nitro group should have a positive formal charge. In the third, you've drawn an arrow flowing from a positive charge. Curved arrows represent the flow of an electron pair from the source, to the sink, and therefore originate from lone pairs, or bonds. The arrow ends when you reach an atom you can make a new bond to, or install a new electron pair upon without violating its valence. The overall charge will never change, and atoms will never move. You are pushing around electrons is all. You should instead have the arrow starting on the nitrogen's lone pair, and flowing to the positive charge on the carbon. In the fourth and fifth structures, you are violating the valence of the top and bottom nitrogens respectively.

When drawing resonance structures, what I recommend, is that whenever you can, start from a lone pair (in this case the aniline nitrogen would be the best choice). Then ask yourself, where can the electrons flow from there? Who can accept a bond or a lone pair without violating its valance? Are there any adjacent bonds which I can push electrons from another atom to make a lone pair?

Start with these guidelines, and re-read the relevant sections in your text book and that will get you back on track to answering these questions.