aromatic acidity question

The basicity is relative to the ability of the Nitrogen in donating it's nonbiding electron pair.
The weakest base is 'A'. you gotta phenol group there which is pretty much acidic with a pKa of 10, which will lose its H to turn into O-, converting NH3 to NH4+, which is not basic anymore

The choices in C,D and E all have an electron withdrawing group on their benzene ring [F and CL]. These groups withdraw the lone pari electrons on the nitrogen; therefore, making it harder for nitrogen to donate an electron pair and be basic. F is a stronger withdrawing group, so the next weak base is E. Between C and D, C will be a weaker base because of Chlorine being closer to the NH2 group. So D is a stronger base than C.

Finally, B is the strongest base because CH3 is an electron releaser, contributing to the bacisity of NH2.

Order from weakest to strongest: A < E < C < D< B
I hope this helps.

 

The nitrogen on the NH2 all have a lone pair which is not drawn in. What they are asking by saying the strongest base is.. in which one will this lone pair grab a H most easily. To answer this question you should ask okay which one of these has the lone pair destabilized in the ring making it more basic or more spread out by the aid of the other constituents making it less basic (they can all resonate around the ring but we want to know which has the least to most stable resonance forms) Kinda hard to explain hope that made sense. Now... the negative charge in the ortho position is worst for OH because OH has its own lone pair that it wants to donate to the ring. I would say that makes it most basic. Then B is second more because it also donated via hyper-conjugation and destabilizes the negative charge. Then D its fair nothing special nothing bad. Then C can help stabilize the negative charge because it is more electronegative and same but more so would make E least basic.
Im not 100% sure about this though kinda makes sense I think Was that close to the answer?

 

Electron donor groups add electron density to the benzene ring, making the arylamine more basic than aniline. The -OH group is a better electron donor functional group than -CH3 making compound A the strongest base. Electron-withdrawing groups remove electron density from the benzene ring, making the arylamine less basic. Since F is more electronegative than Cl, compound E would be a weaker base than compound C. Just a guess, but based on resonance effect, C is probably more stable than D, making C less basic than D.

 

Yea i thought the same thing, e- donating groups such as OH- and CH3, OH- being a stronger electron donating group, increases the basicity...while electrong withdrawing groups decrease the basicity of the analine...therefore the right order should be this...

A>B>D>C>E

Is this correct???? thanks

Basically I think of it as this, you have electron withdrawing groups meaning they will take away electrons from the other group (in this case NH2) and cause it to be more acidic...now when you have electron donating groups you have a group donating its electrons to the other group making it more basic...

 

Is anyone taking into consideration that the OH in choice A is a phenolic OH which is so acidic [pka = 10]. I believe that OH will give its acidic H to the basic NH3, converting the OH to O- [a weak base] and NH4+ [not a base anymore! an acid!].
That is why I put A as the weakest base. I might be wrong, but how can we ignore the strong acidity of the phenol in choice A?

 

You need to review your organic chemistry textbook on arylamines.

 

Can you nullify my argument here instead of giving such a condescending advice?

 

Phenol is an (weak) acid as you suggested, however, electrodonating functional (-NH2) groups make phenol (and benzoic acid) less acidic. Since you had difficulty accepting the explanation given, the advice to check the info in a organic chemistry textbook was given since therein you will find a discussion of the subject concerning arylamines. If your source does not have the info you can find it in J.G. Smith, Organic Chemistry p. 912.

 

Both A & B are EDG, but A is a better EDG than B. A works by resonance and B works by the inductive effect;therefore, resonance > inductive.

 

I have the same question as 113zami.

Would "D" be a stronger base because it is less electron withdrawing in the meta position than it is in the ortho position?

 

To add to this question, what if the Cl was para? That way it is still in the favorable location but just a different one.

 

My opinion is this: The Cl's effect is based on its vicinity to the NH3. the closer it is, the weaker the base is gonna be. We are not dealing with electrophilic aromatic substitution in which the ortho and para positions give the carbocation a better stability because of resonance.
This was my 2 cents.

 

Resonance plays a big role in acidity/basicity.

For example, take para-nitrophenol. Once you deprotonate it, the nitro group withdraws the negative charge on the phenoxide oyxgen. As such, the negative chage is delocalized through the pie system.

Now take meta-nitrophenol. There is NO resonance possible, so the once you deprotonate it, the negative charge cannot be delocalized.

Since, the first example is able to delocalize its negative charge after donating its proton, its a much more stable compound. If its more stable, it will be more likely to donate its proton, which makes it more acidic.

Sorry if my explanations are not clear, it would be a lot easier if I could draw these stuff out.

 

You are right about your example, and I know how those work. In this case, chlorine is acting as a deactivator [which decreases the basicity]. However its deactivation has nothing to do with resonance. It's just its inductive effect. In your example, the nitro group is an ecelctron withdrawer too, however it does its job by resonance which is possible only in ortho an para positions
Give this some thought, and make your argument. My overall point: Cl has a negative effect on the basicity of aniline, no matter it is in o,m, or para position [im sure everyone agrees with this]. How does Cl exert its negative effect? by resonance or inductive effect? the answer is by inductive effect. How is the inductive effect increased? By getting closer to the target

 

The position of Cl does matter. It is deactivating in the O/P postion, and not the meta. Hence why D is more basic than C.

Resonance is preferred over inductive because the electrons are able to delocalize through the pie bonds. In the inductive effect, delocalization is through sigma bonds. Hence resonance > inductive.

 

Cl's pi electron contribution will enhance the electron density on the ring. I don't see how it could be deactivating. Can you please explain?

 

HUH? Did you just not type that Cl is deactivating above? Now you are saying it cant be deactivating?

Anyhow, Halogens are EWG groups, and EWG groups reduces the electron densities, not enhance. Unlike other EWGs, Halogens has O/P selectivity, not meta.

 

My friend, I said "Cl's pi electron contribution" will enhance the electron density on the ring. I assume that you know Cl's engagement in resonance with the ring is the part that acts as an electron releaser [like OH] and it's inductive effect is the part that acts as an electron withdrawer. However, the withdrawing inductive effect of the halogens overcomes the releasing resonance effect of them, because the pi orbitals of Cl, Br, and I is much bigger than the pi orbitals of Carbon which makes the overlap not so efficient (mentioned by my Fryhle text book).
Don't tell me that you think Cl's involvement in the resonance stucture of the ring withdraws electrons from the ring.
Any halogen on a benzene ring is deactivating. Halogens are very electronegative, but why are they weak deactivators? The reason is that most part of their strong inductive effect [withdrawing] is neutralized by their involvement in the resonance through sharing a lone pari e- with the ring. The sharing is not efficient based on the reason I went through in the prev paragraph.
Now I'm saying, any Cl on the ring in an position will weaken the basicity because of its deactivating character. WHere does the deactivating character come from? from inductive effect or resonance? From inductive effect as I explained. How can we increase the inductive effect? By getting the Cl closer to the NH3.
Please read this without any previous bias. I'm just presenting my perspective, and I'm ready to debate each part of this with you. I did not find anything contradictory in your statements except the fact that the resonance effect in ortho and para is gonna help the basicity. Cl is not good for helping the basicity. It is weakening the basicity. But is it through resonance? no. It is through inductive effect. We have to focus on the inductive effect.

 

I got this in email from my ochem teacher. Check it out.
It says "

Since inductive effects drop off with distance, and since F​

is a stronger EWG than Cl, the order B>D>C>E makes sense."

There is no talk about resonane while comparing the ortho and metha positions. Read my previous arguments, as I was correct about this.

 

Ok everyone, I hope this will clarify this question. I finally got the response back from my Ochem professor.
Here is my email to him, first, and then you can see his reply. In comparing the ortho, meta, and para positions, the inductive effect is the most important factor. Check it out.

From me to professor--->

[FONT=verdana, helvetica, sans-serif]Dr. Byrd,

[FONT=verdana, helvetica, sans-serif]I'm so grateful for the explanation, and also the descriptive attachment. I don't want to keep on asking questions audaciously, but I can't resist this..
I just want to make sure I'm right. while comparing ortho, meta, and para chloroanilines, we take care of the inductive effect only, by looking at the distance of the Cl from NH2. Therefore, we should say Para is the most, and ortho is the least basic.
Chlorine could have the same effect of OH on the basicity, only if the overlap of Cl's pi orbitals with the ring's pi system was efficient. However, we know that Cl's pi orbitals are bigger than carbon's, so we somehow ignore the positive effect that Cl could exert on the basicity of aniline in the ortho and para positions by favorable ion-dipole interactions. Am I thinking right?

[FONT=verdana, helvetica, sans-serif]Thank you again.
[FONT=verdana, helvetica, sans-serif]Harry.



[FONT=verdana, helvetica, sans-serif]Harry, .

You have summarized the effect of Cl precisely; its effect is strictly inductive, thus dropping off with distance from C bearing N. Electron donation via resonance when Cl is -o or -p to amino is essentially zero for the reasons you stated.

Good luck,

Dr. Byrd
.

 

i did not read all of these threads, but i did read most and i didnt see anybody mentioning the fact that an o,p director in a meta position substantially increases the acidity of the base in question and can be exemplified by resonance structures.

 

Yes, the resonance effect of Cl at the ortho position was in the debate, while I kept saying that resonance is not playing a role here. Read my and TUNDERCATZ debates again if you are interested.