# Acidities of amines and carboxylic acids

I have been told that the first of the two acids is the stronger acid. According to this person,deprotonation for both occur at the nitrogen.The resulting conjugate bases include,respectively a neutral amine group,and a negatively charged amine group.Because negatively charged nitrogens are highly unstable from an acid/base standpoint (consider lithium diisopropyl amide)... the first must be the stronger acid (because of its more stable conjugate base).

The problem I have with this line of reasoning is that it's based on a faulty premise - that the nitrogen is the site of deprotonation.I see a much better site of deprotonation for both - and that is the carboxylic acid site. A negative charge here will not only be resonance stabilized but also inductively stabilized by oxygen.Plus, the Ka of ammonium ion is 5.6*10^-10, while the Ka of acetic acid is 1.8*10^-5.In other words - orders of magnitude of difference. I can't see why something that is structurally analogous to ammonium ion (much less ammonia) would be deprotonated preferentially over a carboxylic acid.

• This question is nonsensical. The acidity of compound two cannot be determined because it does not exist in most environments. It actually exists as a zwitterion, that is, $$H_3^+N+CH(CH_3)COO^-$$. Compare that to compound 1. – Lighthart Sep 12 '14 at 19:05
• @Aditya It's not necessary to add to the situation. – jonsca Sep 13 '14 at 8:45

The first compound is more acidic. The ammonium functionality is inductively electron withdrawing, stabilizing the conjugate base (carboxylic acid).

This answer is assuming that we are talking about the drawn structures exactly. User137 is correct that it's not possible to have the second compound, because the amine is a strong enough base to deprotonate the carboxylic acid, and it exists as the zwitterion. However, assuming that we could hold the amine as neutral, it would be less inductively electron withdrawing than the cationic ammonium.

• The extent to which the zwitterion exists is very low. – Martin - マーチン Sep 12 '14 at 17:23

Compound 1's conjugate base IS compound 2. The positive NH3 loses a proton to become the neutral NH2, but deprotonation of the NH2 group is extremely unlikely. Compound 2 will lose a proton at the carboxylic acid group first, so your friend is wrong to say deprotonation will occur at the nitrogen in both.

This will likely form a zwitterion, where the nitrogen is NH3+ and the carboxylic acid is COO-. This compound will have 2 pKas, one for the carboxylic acid, 1 for the amine. The carboxylic acid will be the lower of the 2 pKas. So if we start at very low pH and add base, the compound will start like compound 1. As we approach the pKa for the carboxylic acid, it will deprotonate to form the zwitterion. Then, as we approach the pKa of the NH3+, it will also deprotonate, forming a compound with an NH2 and COO-.

• Which one is the stronger acid? From what you're saying you are saying that compound 2 is the stronger right? @user137 – Dissenter Sep 9 '14 at 17:25
• I believe compound 2 is stronger, since compound 1 exists at low pH. Compound 2 would be hard to make because the NH2 only exists at pH well above where COOH converts to COO-. So if you had compound 2 in a very high pH solution, it would immediately give up that proton. – user137 Sep 9 '14 at 17:29
• Are you really saying that compound 2 $\ce{H2N(CHMe)COOH}$ is more likely to release a proton than compound 1 $\ce{H3+N(CHMe)COOH}$? Also the amount of zwitterion in compound 2 is very low, but non negligible. – Martin - マーチン Sep 12 '14 at 17:21
• @Martin My reasoning is that compound 1 can be easily made by putting it in low pH solution, using the pKa's for alanine provided by Aditya's answer, the pH must be below 2.33. Since increasing the pH will cause the COOH to deprotonate first, and NH3 holds on to its proton until about pH 9.71, NH3 must have higher affinity for its proton. If you could get compound 2 in a high pH solution, the COOH would most likely lose the proton very quickly. And these are all equilibriums, with small amounts of each protonation state, but compound 2 would be present in very small amounts. – user137 Sep 12 '14 at 17:42

These are two possible forms of Alanine(Ala||A), an amino acid with hydrophobic side chain. Here are the exact $pK_A$ values: $$-COOH:2.33\\-NH_3^+:9.71$$ Compund two is stronger as you anticipated.Cheers!