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Open-chain and pyranose forms of glucose

Is it tautomerism? Or some other factors involved?

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You get to form a C-O $\mathrm{\sigma}$ bond at the expense of a C-O $\mathrm{\pi}$ bond. The single bond has a higher bond energy, even though it is somewhat de-stabilized by the anomeric effect. It's still an overall win, despite taking the entropic hit to confine your molecule into a ring.

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    $\begingroup$ @Karl That's not correct. The free energy difference between the open-chain and ring forms has a entropic component because the open-chain has more degrees of freedom. $\endgroup$ – Zhe Oct 20 '16 at 0:48
  • $\begingroup$ Irrelevant was the wrong word. I meant to say that taking into account the entropic effect, it becomes temperature-dependent, and no temperature was specified. $\endgroup$ – Karl Oct 20 '16 at 2:14
  • $\begingroup$ Shouldn't the double bond have more energy than the single bond? As far as i remember, the order of bond energy is triple>double>single.. Or am I missing something? $\endgroup$ – Anindya Oct 20 '16 at 6:04
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    $\begingroup$ Yes, you're missing the fact that the double bond is a $\mathrm{\sigma}$ bond and a $\mathrm{\pi}$ bond. The latter has a smaller bond energy, so breaking a $\mathrm{\pi}$ bond for a $\mathrm{\sigma}$ bond if favorable in this case. In other words, two singles bonds > one double bond. $\endgroup$ – Zhe Oct 20 '16 at 13:30
  • $\begingroup$ @Zhe "It's still an overall win, despite taking the entropic hit to confine your molecule into a ring." I am curious about how to measure this entropic difference. $\endgroup$ – Mockingbird Oct 15 '17 at 17:12
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Glucose molecules form rings. The first carbon atom (C1), which is an aldehyde group (-CHO), creates a hemiacetal with the fifth carbon atom (C5) to make a 6-membered-ring (termed a pyranose). The atoms in this cyclic molecule then arrange themselves in space to minimize the amount of strain on each of the covalent bonds.

The carbon atoms in the glucose ring each have four covalent bonds. The best, or optimum angle, between all these bonds is 109.5o, which results in a perfect tetrahedron. If, for any reason, these bonds are forced into greater, or smaller angles then the molecule will be strained or stressed, and be much less stable.

It follows, therefore, that the glucose molecule will be at its most stable when all the carbon atoms can arrange themselves so that their bond angles are all close to 109.5o.

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    $\begingroup$ You should have edited your original answer rather than posting a new one. However, they key problem remains: there is nothing that intrinsically makes carbons with four single bonds any more stable a priori than carbons with multiple bonds. $\endgroup$ – Jan Oct 15 '17 at 17:31
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The carbon atoms in the gulucose ring each have four covalent bonds ......it follows,therefore,that the gulucose molecule will be at its most stable when all carbon atoms can arrange themselves , so that their bond angles are all close to 109.5 degree .

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    $\begingroup$ This is incorrect. There is nothing about four single bonds on carbon that would be inherently more stable than multiple bonds. $\endgroup$ – Jan Oct 15 '17 at 17:17

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