# How is dioxygen stabilised by coordination to haemoglobin?

I wonder how the binding of $\ce{O2}$ to a molecule of hemoglobin stabilizes it to the extent that I can carried around the body with the risk of it being involved in oxidation reactions minimized. The same applies to hemocyanin.

I am looking for an answer invoking orbital interactions, back-bonding etc.

• Note how I formatted your molecular formula for oxygen using MathJax formatting. This is how we should format chemical equations on this site, and there is a good introduction here. It's really pretty simple to get the hang of it and it makes your posts look much more professional. – airhuff Feb 27 '17 at 20:04
• Are you asking about the bonding of oxygen molecules to the Fe in heamoglobin (and to Cu in hemocyanin) and about the different structures of the protein in different states ? As to the latter see the chapter on this in Stryer 'Biochemistry' and in Kuriyan, Konforti & Wemmer , 'The molecules of Life'. – porphyrin Feb 27 '17 at 20:59
• @porphyrin Thanks! I'm primarily looking at the bonding with the iron-containing haem. It seems like oxygen would be a superoxide, so I wonder if there are any orbital interactions involved that would stop it from reacting with molecules around it. If not, I wonder if the stability is to do with the whole protein (e.g. the oxygen-containing harm being buried deep within its core). – GingerBadger Feb 27 '17 at 21:09
• – Mithoron Feb 27 '17 at 21:36
• Your question incorrectly assumes that oxygen molecule is unstable or reactive at room temperature. No orbitals, sorry – Greg Feb 28 '17 at 5:55

I think you have a misconception. The binding of oxygen to haem and its release are processes that are entirely equilibrium-controlled by the equilibrium shown in $(1)$.

$$\ce{haem + O2 <=> haem{-}O2}\tag{1}$$

Meaning that wherever high partial oxygen pressures are observed, oxygen preferentially binds to haem but likewise it is released where low partial oxygen pressures are observed. Thereby, Le Chatelier’s principle nicely controls where oxygen is released from the red blood cells’ haem groups — it will always be in those areas where oxygen is required. Haem itself has evolutionally acquired the role of binding oxygen rather well but only to make sure that it actually reaches the distant parts of the body.

Oxygen in its $\ce{O2}$ modification is also not as bad as your initial post seems to assume. Yes it can react with quite a few substances and many of its reactions have a low activation barrier — but the concentrations in which it would be released into the blood stream ‘undesiredly’ are concentrations the body has learnt to cope with over millenia of evolution. So while minor damage may occur, most damage is quickly repaired by dedicated enzymes. Most importantly, any liberated oxygen will always be $\ce{O2}$ and never superoxide or peroxide.

Only while being bound to haem do oxygen’s (and iron’s) electronic properties change. On a microscopic level, the oxygen molecule causes a one-electron oxidation of iron converting itself to superoxide. This also causes a spin-flip in the iron centre which goes from high to low spin. The combined $\ce{\overset{\mathrm{+III}}{Fe}-O2}$ system can finally undergo antiferromagnetic coupling to give an overall observed diamagnetic ground state. Details can be found in this answer. However, while it is superoxide the $\ce{O2}$ moiety will never leave the iron centre — or, more precisely described, as soon as $\ce{O2}$ leaves, the entire process is reversed again. Thus, even though the bound state includes a superoxide ion no superoxide can ever be directly released into the surrounding blood stream.

$$\ce{[\overset{\mathrm{+II}}{Fe}(por)] + ^3O2 <=> [\overset{\mathrm{+III}}{Fe}(O2^{.-})(por)]}\tag{2}$$

• Thank you! So while bound to Fe, oxygen (superoxide) isn't in contact with the surrounding solution at all, even with water molecules diffusing to the binding cleft that it can oxidize or surrounding amino acid side chains? – GingerBadger Feb 28 '17 at 10:54
• @substitutedbutadiene Well, it is somewhat close enough to the surrounding solution to be liberated as neutral dioxygen but it does not show the reactivity of free superoxide. – Jan Feb 28 '17 at 12:12