Questions about Insight '11 Trial

[Unseen]

Just want to clear up a few questions:

MC 12 - The suggested solutions say that lysine is more basic than histidine, whereas I thought the opposite. I was under the impression that the nitrogens in the ring structure on the side group of histidine were able to accept a single proton, considering that the nitrogens in the side group already have three covalent bonds (with a lone pair still free). Am I correct in thinking that these nitrogens will be able to accept protons? If not, how would I know this is the case as opposed to being basic?

SA 6b - The question is: "Would you expect a solution of the dipeptide you drew in part a dissolved in water to be acidic, basic or neutral?" The dipeptide of concern contains two NH2 groups and one COOH group. I considered this in light of the dipeptide being converted into a zwitterion the instant it is dissolved into water. Therefore there are two NH3+ groups and one COO- group. If you recall conjugate acid-base pairs (NH2 base / NH3+ acid; COO- base / COOH acid) the presence of the extra NH3+ group would make it acidic, would it not? This was my line of thought, but the suggested solutions claimed that the solution would be basic.

If anyone could clear these questions up, it would be very much appreciated.

[Mao]

MC12 - you are correct in saying that histidine is basic. However, the side group is aromatic, it is quite similar to benzene. You can think of the lone pairs to be delocalized over the ring, and thus has a reduced basicity.

SA 6b - You are correct in noting that the amino acid, when dissolved in water, would contain two NH3+ groups and one COO- group. If we look at what has changed in the solution, two molecules of OH- and one molecule of H3O+ are produced. Therefore the solution is basic, and the amino acid has become the 'conjugate acid' in a sense.