Question

Although tryptophan contains a heterocyclic amine, it is considered a neutral amino acid.

(a) Explain why the indole nitrogen of tryptophan is more weakly basic than one of the imidazole nitrogens of histidine.

Answer 1

Basically, the conjugate acid of the basic group on tryptophan in the indole functional moiety is so desperate to dump it's proton that it is laterally a factor of less basic than histine's imidazole. In histidine the nitrogen whose lone pair is not donated in the ring is basic enough.

Histidine contains a metal Lynn images all side chain. Because the two nitrogen’s are in an aromatic ring the pi system cannot accommodate all of the loan pairs and the pi bonds.

Therefore though both nitrogens are hybridized sp2 ( one with different strengths (basicity is defined as the ability to form strong bonds)—and the other nitrogen (in the C=N-C arrangement) is sp2 by virtue of its overlap with adjacent p orbitals. It contributes one electron to the ring system as does it’s neighboring carbon to make a pi bond. No more electrons can be in that pi system.

It does not want to form a bond to H+ as this will lessen conjugation and in this cyclic case, destroy aromaticity (a very stable, difficult to disturb electron cloud (set of molecular orbitals). It’s therefore not basic, it doesn’t accept protons well.

The other Nitrogen is not double bonded but something must occupy the p orbital on the N that appears to be sp3. It’s the lone pair. Since it’s a part of the resonance and aromaticity those electrons do not sit there as drawn but move at blinding speed through a cloud of probability density (chi^2).

Therefore, know how to identify hybridization in pyridine (and sum the pi electron count) look at pyrrole (add up the pi electrons). If Hückel’s rule is violated you overcounted!

You will see then if you did the counts for and understand puridine and pyrrole, you must understand Imidazole by extension.

A nitrogen cannot be both double bonded (it lost a p orbital and electron to be double bonded) and also have a lone pair in the same Orbital. That is a violation of Hund’s Rule. You cannot have three electrons in one atomic orbital.

The pi system is perpendicular (orthogonal) to the plane of the benzene ring. The lone pair on that trigonal planar nitrogen can form a bond—creating a formal (+)charge, but NOT disturbing aromaticity which would require much more energy than is available thereby requiring forcing conditions, heat, pressure, strong Lewis acids, etc.

Now consider tryptophan. The lone pair on the 5 ring is on nitrogen, not a double bond. Therefore just like pyrrole, indole has no available electrons to bond to H+. The benzene ring attached doesn’t make a difference.