Relative Strengths of Acids (Organic Chemistry)?

Question

Comparing the relative acidity strengths of aniline and a secondary amine (drawn as a "cyclohexane" where one C has been removed and replaced by an NH). The answer is that aniline is a stronger acid than the secondary amine:
Thanks to resonance, the acidity of aniline is eight powers of ten stronger than the secondary amine.
I do not understand the explanation given. Other than the typical resonance found in the benzene ring, I can't find another resoance strucutres for either!

Thanks for your help in advance!

Answer 1

you are right that aniline is a stronger lewis acid than secondary amine. Let me stress it here that we are not talking about aqueous system, therefore any concept of bronsted acidity such as pKa value or pH value are useless here.

NH2 group of aniline is a very interesting substitute group.
It is an electron withdrawing group by inductive effect and it is an electron donoring group by means of resonance.

Let me explain the latter:
-:NH2 has one lone pair electron which it can share with the adjacent C1 of the pi system. this will give a positive charge on the -NH2 group while the electron that were shared between C1 and C2 are no longer shared in order to obey the octet rule, instead lone electron pair will appear on C2. The lone electron pair of C2 will be shared again with C3 carbon atom of the pi system and will result of C4 atom carbon having a lone electron pair and so on and so forth. This resonance system result in an increase of electron density of the benzene ring hence the name electron donating effect. One thing for sure the increase of electron due to this resonance mechanism will result in DECREASE of LEWIS ACIDITY when compared to unsubstituted benzene. So yes, the increase of aniline lewis acidity IS NOT because of the RESONANCE effect but the outweighing INDUCTIVE effect.

Inductive effect simply telling you that due to different electronegativity of NH2 group and the benzene ring dipole will occurs and since NH2 is more electronegative than benzene, then NH2 will be the negative pole while benzene ring will be the positive one, and you know what happen when positive charges increases, yes the LEWIS ACIDITY of the benzene ring will be higher than the unsubstituted benzene.

so as you said it's all about relative acidity strength. Aniline is after all considered as Lewis base but when compared to a secondary amine, where two unsaturated alkyl groups are attached to amine thus negative charge of nitrogen's lone pair is uniformly distributed as well as no resonance system could be form, Aniline is most definitely stronger in LEWIS acidity.

I can't stress that enough: STRONGER lewis acid and not STRONG lewis acid

I hope that would help

Answer 2

To determine the relative acidity of any given H, you must see how stable the conj base is (this is the H in question missing, with a minus charge instead). The more stable the conj base is, the less reluctant it is to lose said H, thus making said H more "pluckable", or acidic. We use pka's in organic, b/c the pH scale isn't descriptive enough. Weakest (stronger as you go down): Ethanol Methanol Water Ethanoic Acid (aka acetic acid) Methanoic Acid (aka formic acid) Alcohols: The neg charge is localized completely on the O for both...the deal-breaker here is that R groups are electron donating in nature, so the ethanol, having two carbons, will have higher electron density than the methanol (and water, for that matter). Therefore, the ethoxide would be the most unstable and bear the least acidic H of both alcohols and water. Carboxylic Acids: Same idea here...these guys do have resonance stabilization, so that's why they beat the water and alcohols (the electron density is delocalized by resonance, adding to stability of the conj base). As before, the ethanoic acid has the R group that will donate electron density toward the neg charge, making it less stable and therefore less acidic. Hope this helps...

Answer 3

I find it extraordinary that you are comparing the acidity of phenylamine rather than its basicity. Are you sure that you are not comparing the acidity of the phenylammonium ion (C6H5NH3+) with your secondary amine? The desire to regain association of the N lone pair with the benzene ring accounts for the very much larger pKa of the phenylammonium ion.