Glucose is the major sugar in mammalian blood. We often see it represented as either the "free aldehyde" or the cyclic hemiacetal forms shown here. Of the two forms of glucose, the cyclic hemiacetal is the preferred form found in blood. Can you suggest two reasons why?

Question

Glucose is the major sugar in mammalian blood. We often see it represented as either the "free aldehyde" or the cyclic hemiacetal forms shown here. Of the two forms of glucose, the cyclic hemiacetal is the preferred form found in blood. Can you suggest two reasons why?

Accepted Answer

All right. Hello everyone. So this question says that galactose is one of the simple sugars also called monosaccharide. Shown below are the free aldehyde and cyclic hem acetal structures of galactose. In aqueous solutions, galactose predominantly exists as a cyclic hemi Aal explain why. So considering the structure shown below here, the two of them, the predominant structure is the one on the right side. So if I scroll down here, I have four different answer choices, galactose predominantly exists as a cyclic hemi acetyl because a the cyclic hem acetyl structure is more strained than the open chain structure. B the cyclic chem ay structure has pi bonds or rather has more pi bonds than the open chain structure. C the cyclic hemi acetyl structure is more stable than the open chain structure or option D the cyclic hem ay structure has fewer oxygens than the open chain structure. So with this in mind, I'm going to go ahead and scroll up one more time to consider the structures provided. In this case, in aqueous solution galactose predominantly exists as a psychic chem Aal. Now recall that the the predominant form of a certain molecule is generally considered to be the more stable and generally more favorable configuration of that compound. So this implies that the cyclic hemi acetyl structure of galactose is more stable and more favorable compared to the open chain or the free aldehyde structure. And that would be true for a couple of different reasons. One of which is the stability that the ring itself provides because notice how the cyclic chem acetyl structure is a six member ring five of which are carbons and one of them oxygen being part of the ring. So this six carbon ring is very stable because the ring itself has no strain associated to it. So thats one factor. Now another factor is that the cyclic he acetyl structure makes galactose itself less reactive and therefore more stable by preventing the formation of any unwanted side products. And to understand the reason why this is the case, I want to draw your attention to the free aldehyde structure. On the left side, specifically, we have the free aldehyde functional group on carbon. One of the chain itself recall that aldehydes and ketones are quite reactive, right. So by having a free aldehyde at the very end of the carbon chain, then this allows for any unwanted reactions to potentially occur. However, by comparing or by converting the free aldehyde structure into the cyclic acetyl structure, then this reactive aldehyde functional group is no longer present. So this means that the cyclic hemi acetyl structure is less reactive and therefore more stable. So with this in mind, our answer is going to be option C in the multiple choice. But I do want to point out here that options A B and D could have been eliminated by examining them. Starting off with option eight. Option A suggests that the cyclic hemal is more strained than the open chain structure. Now, if this were true, then the cyclic chem ay structure would be less stable than the open chain structure which means that it would not be the predominant form of galactose. Option B says that the cyclic ay structure has more pi bonds than the open chain structure. Now, considering the structure provided right, the cyclic Hemel has zero pi bonds because there are no double or triple bonds at present. The free aldehyde however, does have one pi bond considering there is an aldehyde functional group that contains a carbon. So this would be incorrect because it's actually the other way around, right, the open chain structure has more pi bonds than the cyclic hem acetyl. Lastly, we have option D which states that the cyclic chem ay structure of galactose has fewer oxygens than the open chain structure. Now, this is incorrect because both structures have the same amount of oxygen. So just to reiterate here, our answer is going to be option C in the multiple choice galactose predominantly exists as a cyclic hem acetyl because the cyclic hem acetyl structure is more stable than the open chain structure. And with that being said, thank you so very much for watching. And I hope we found this helpful.

Glucose is the major sugar in mammalian blood. We often see it represented as either the "free aldehyde" or the cyclic hemiacetal forms shown here. Of the two forms of glucose, the cyclic hemiacetal is the preferred form found in blood. Can you suggest two reasons why?

Verified video answer for a similar problem:

Key Concepts

Cyclic Hemiacetal Formation

Stability in Aqueous Solutions

Biological Function and Reactivity