March 28, 2024, 02:37:11 PM
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Topic: What makes the cyclic form of glucose be more stable than the open chain form?  (Read 7572 times)

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Offline Anindyagm1

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Is it tautomerism or something else?


Offline Babcock_Hall

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Can you define tautomerism for us and then apply it to the forms of glucose?

Offline pgk

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Can you apply the Hess law and then conclude whether the formation of the cyclic conformer is exothermic or not?
https://en.wikipedia.org/wiki/Hess's_law
http://www.cem.msu.edu/~reusch/OrgPage/bndenrgy.htm



Offline pgk

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Can you also suggest whether glucopyranose (six member ring) is more stable than glucofuranose (five member ring) or not? (or else, less energy consumptive)
https://en.wikipedia.org/wiki/Ring_strain


Offline pgk

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There is still something else.
Enthalpy alone is not a safe criterion for predicting a reaction procedure. Entropy and temperature also play an important role as shown by the Gibb’s equation. However, it is easy to understand that the difference of formation entropy between a flexible linear molecule and the flexible cyclic isomer, is too small and has a little interference at room temperature.
https://en.wikipedia.org/wiki/Gibbs_free_energy

Offline pgk

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And a last thing.
Cyclic forms of glucose are more stable than the linear form and thus, glucopyranose is (almost quantitatively) the main form in solid state. However, hemiacetal formation and hydrolysis are reversible reactions. Therefore there is equilibrium between all isomer forms in aqueous solution, due to the interconversion: glycopyranose ↔ linear form ↔ glucopyranose.
The composition of the isomers mixture is not constant but it depends on a variety of factors, such as concentration, pH, temperature, Hydrogen bonding, etc.
In other words:
When you drink a glass of glucose solution, the composition of the isomers mixture is different in the glass at room temperature, different in the empty stomach at 37oC and pH = 2-4, as well as different in blood at 37oC and pH = 7.35-7.45.

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