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I am trying to imagine the cyclization of the Fischer projection of glucose as it forms an $\unicode[Times]{x3B1}$-D-glucofuranose and an $\unicode[Times]{x3B1}$-D-glucopyranose (right now I am trying to understand the Haworth projection, and as my 3D visualization gets better, I will extend to the more thermodynamically stable chair representations of these monosaccharides). Now, if I understand correctly, in the pyranose form, the alcohol on $\mathrm{C5}$ attacks the carbonyl carbon on $\mathrm{C1}$. In the furanose form, the alcohol on $\mathrm{C4}$ attacks the carbonyl carbon on $\mathrm{C1}$. These results in 6-membered and 5-membered rings, respectively.

I understand in nature, these are the most stable rings. However, what thermodynamic incentive does the furanose form have? That is, why would $\mathrm{C4}$'s alcohol attack and form a 5-membered ring at the expense of the $\ce{CH2OH}$ group and the $\ce{OH}$ group being near each other as a side chain on the 5-membered ring? Doesn't this introduce considerable strain? My idea is that the 6-memebered ring is more thermodynamically stable and it doesn't have the problem of a hydroxymethyl and a hydroxyl group on a single carbon of the 5-membered ring.

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If you look up the statistical distribution of glucose isomers in aquaeous solution, you will find $36\,\%$ $\unicode[Times]{x3B1}$-pyranose and $64\,\%$ $\unicode[Times]{x3B2}$-pyranose. All the remaining forms lumped together only account for less than $1\,\%$ of glucose in solution; that includes both furanoses, the open-chain form and the (possible, but unlikely) seven- and four-membered rings.

Neither five-membered rings nor six-membered rings contain any relevant amount of angle strain. You can easily verify this buy getting yourself a molecular model kit and attempting to bild these rings from four-bond sp³ carbons: both can be built smoothly, like a charm. So angular strain is not a reason why the pyranose forms are preferred.

Considering reaction mechanisms, the formation of five-membered rings is usually kinetically preferred due to easier reachable transition states. So this also cannot be the reason for preferring six-membered rings.

The actual reason is not so much angular strain, but rather dihedral angle strain: the same strain that prevents ecliptic conformations in alkanes. The pyranose form has the perfect $60^\circ$ dihedral angle between two non-ring atoms which minimises this type of strain. Furanoses must choose between envelope or twist conformations to at least relieve some of this strain.

All this is, however, happening at an equilibrium, which is semi-rapid in adjusting itself. So there will always be a certain percentage of furanose form in solution being formed and breaking apart. If any kind of reaction selectively only happens with furanose anomers, then this will cause the formation of new furanose which again will be captured away. As I wrote before, the attack which creates a furanose is preferred over that which creates a pyranose. (But the breaking apart of pyranose is slower than that of furanose.)

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  • $\begingroup$ The notion of a statistical distribution's area being <1% sold me quite well. I hadn't thought about that. Also, thank you for eloquently explaining the difference between preference of attack and preference of intermediates. $\endgroup$ – Alvin Nunez Oct 3 '15 at 18:47
  • $\begingroup$ Then what is the reason that Furanose is more stable in case of Fructose ? $\endgroup$ – Aditya Shrivastav Sep 25 '17 at 6:10
  • $\begingroup$ @AdityaShrivastav According to the Wikipedia article, pyranose is the predominant configuration ($70~\%$) of fructose in solution. Please state your source. $\endgroup$ – Jan Sep 25 '17 at 8:24
  • $\begingroup$ Thank you for the fact ,I never came across it. I always seen only furanose structure of the fructose in many books. Then why Sucrose consist of furanose form of Fructose ? en.m.wikipedia.org/wiki/Sucrose#/media/File%3ASaccharose2.svg $\endgroup$ – Aditya Shrivastav Sep 25 '17 at 9:23
  • $\begingroup$ @AdityaShrivastav In sucrose, you are no longer dealing with a hemiketal function in the fructose moiety that can react back and forth rapidly. Instead, you have a kinetically rather inert full ketal that is not subject to equilibration like free fructose is. $\endgroup$ – Jan Sep 25 '17 at 10:40

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