|This is the talk page for discussing improvements to the Histidine article.|
|WikiProject Chemicals / Core||(Rated Start-class, Mid-importance)|
|WikiProject Molecular and Cellular Biology||(Rated Start-class, High-importance)|
The use of the single double-headed arrow between the tautomers of histidine depicted is misleading. It should be a pair of single-headed arrows to denote equilibrium not resonance.
The 3D structure shows the amino acid as an uncharged molecule but it should be rather present as a zwitterion. Also the IUPAC name and formula are not for the endogenous amino acid but the racemate of R and S histidine. Is it on purpose or should I change it? -- Panoramix303 (talk) 19:16, 23 March 2008 (UTC)
I just took a look at the illustration of the histidine molecule http://en.wikipedia.org/wiki/File:L-histidine-zwitterion-from-xtal-1993-3D-balls-B.png
There has to be an error here. Look at the second carbon from the right end. It is attached to a blue nitrogen, but the nitrogen group should be NH2, not the NH3 shown. Am I imagining things or should this be changed?
Shouldn't the sidechain be protonated (rather than the N-terminus) in the formally neutral zwitterion? There should also be a reference for the claim that the two NH groups in the imidazolium ring share the charge "equally" when the sidechain is protonated, because an intramolecular hydrogen bond to the N-terminus, e.g., could make the charge sharing unequal.--JP —Preceding unsigned comment added by 126.96.36.199 (talk) 18:57, 13 December 2010 (UTC)
One of the hydrogens bound to the imidazole carbon appears to be out of plane, which is not correct as the carbons are sp2-hybridized. Experimental structure with hydrogens may be found for example here - Acta Cryst. (1972). B28, 2377-2382. — Preceding unsigned comment added by 188.8.131.52 (talk) 21:56, 8 April 2014 (UTC)
i quote from the second paragraph: One (nitrogen) is bound to hydrogen and donates its lone pair to the aromatic ring and as such is slightly acidic, whereas the other one donates only one electron to the ring so it has a free lone pair and is basic. The last part of this sentence doesn't add up does it? the 'basic' nitrogen in the ring is double bonded to one carbon and single bonded to another, that uses up 3 of its 'outer' electrons, leaving it with 2 left, if it then 'donates one to the ring' its left with one electron; not a pair as is said in the article (nitrogen has only 5 outer e-) i didn't want to change this just in case i have missed something here, but if it is incorrect please change it, or explain my error underneath. ty --JCW 14:02, 11 August 2008 (UTC)
So related to the discussion below: one nitrogen donates a lone pair to the ring, apparently the other lone pair never participates in the ring. Is this because of orbital geometry or what? Qchristensen (talk) 07:44, 20 March 2009 (UTC)
- Yes, whichever nitrogen isn't bonded to hydrogen (this can change rapidly by tautomerism) has a lone pair at 90° (orthogonal) to the ring, so the lone pair cannot participate in the ring's pi system.
- Above: orbitals involved in aromaticity in green, lone pair that is orthogonal to pi system and cannot participate in red
- Below: nitrogen lone pair (far left) and pi molecular orbitals of imidazole
Using the N-1 and N-3 (and N-1-H and N-3-H) nomenclature demands that the article note the numbering discrepancy between IUPAC and conventional biochemical usage. Best would be to couch this entire discussion in terms of tele (TAU) and PI nitrogens. 184.108.40.206 (talk) 13:31, 11 November 2010 (UTC)John Thaden, UAMS, USA.
Can someone please check this statement:
"Histidine can be aromatic. When it is deprotonated, and uncharged, it is not aromatic. It no longer obeys Hückel's rule because 8 electrons are in the ring system (an extra two from the deprotonated nitrogen). Histidine obeys Hückel's rule when it is protonated, so then is aromatic."
I am quite sure this is wrong. First off all, when the imidazole ring is neutral, it has six pi electrons, two from the nitrogen lone pair and four from the double bonds. Thus, it is aromatic. When deprotonated the N-metal bond is orthogonal to the pi-system, so it remains aromatic. I'm going to correct this in a day or two if I don't hear to the contrary. Eugene Kwan (talk) 18:35, 13 February 2009 (UTC)
I wrote it because it is a common question for molecular biologists, but a good answer is hard to find. Now it makes sense to me that the second lone pair doesn't participate in the ring system ever (like pyridine). If anyone could find a reference for this I think this would be a great addition. Quin Christensen (talk) 18:50, 18 March 2009 (UTC)
- Stryer (5th Edn.) p. 47: "Histidine contains an imidazole group, an aromatic ring that can also be positively charged".
- See also Imidazole. Clayden probably has info on the aromaticity of imidazole.
Ok, but "NMR shows that the chemical shift of N-1 drops slightly, while the chemical shift of N-3 drops considerably (about 190 vs. 145 ppm). Because these chemical shifts are relative to nitric acid, a substance which resonantes far downfield, a decrease in chemical shift corresponds to deshielding." So does the deshielding of N-3 suggest an uneven distribution of of the pi electron cloud? Does this affect the "aromaticity" of this form? Quin Christensen (talk) 20:23, 19 March 2009 (UTC)
- Be careful to avoid OR. Here's a good article: J. Mol. Struct. (2003) 655, 397-403 - references within may also prove informative.
- Thanks! It looks like it is very aromatic. It definitely is not considered as such in biochemistry books. Qchristensen (talk) 07:41, 20 March 2009 (UTC)
No. The deshielding indicates the presence of the second-order paramagnetic effect, which arises from a mixing of electrons in the lone pair into the aromatic system. It's kind of complicated, but the point is, yes, it's aromatic. It's always aromatic, no matter what the pH is. —Preceding unsigned comment added by E kwan (talk • contribs) 03:40, 3 April 2009 (UTC)
I think that the histidine ring structure is only aromatic when in the positive state as only then does it obey the Huckel 4n + 2 rule. In the base form, the ring has one nitrogen double bonded and therefore contributing three electrons, the other nitrogen with hydrogen attached only contributing two electrons and the three carbon atoms each contributing one electron each. This makes a total of eight pi electrons in the ring. However, the ring structure may disobey the Huckel rule and actually be aromatic as the rule does not include all aromatic molecules. Geoff e j (talk) 11:41, 22 February 2013 (UTC)
That's incorrect, because you can only count the p-electrons that are orthogonal to the ring. That means that histidine is aromatic at any protonation state. Eugene Kwan (talk) 15:45, 23 February 2013 (UTC)
I deleted some of the content from this section diff as the content made rather extraordinary and sweeping claims that would need good citations to remain. Tim Vickers (talk) 21:50, 2 April 2009 (UTC)
Synthesis and precursors
The article does not mention whether histidine is synthesized in the body from some other precursor substance or is itself a commonly occurring amino acid in food. —Preceding unsigned comment added by Carusus (talk • contribs) 11:19, 16 May 2010 (UTC)
The first paragraph claims that humans begin to synthesize it after "several years." I think this claim needs a citation and a time frame more specific than "several years." Does this mean as a toddler, after puberty, in middle age, or what? —Preceding unsigned comment added by 220.127.116.11 (talk) 00:41, 1 September 2010 (UTC)
It is not synthesized. It is an essential amino acid in humans. The early research didn't find any change in adults in the short term, but longer experiments later confirmed it is essential. I added the reference. Qchristensen (talk) 20:20, 11 November 2010 (UTC)
The imidazole sidechain of histidine has a pKa of approximately 6.0, and, overall, the amino acid has a pKa of 6.5.