Why does the cyclic form of glucose exist predominantly in solution and why is the b form more abundant than the a form?
The various cyclic forms and open chain form of glucose are in a dynamic equilibrium in a solution where the;y are free to move. The B-anomeic OH group is equitorial making it the low energy conformer.
In solution, glucose exists as?
In solution, the glucose molecule can exist in an open-chain (acyclic) and ring (cyclic) form (in equilibrium), the latter being the result of a covalent bond between the aldehyde C atom and the C-5 hydroxyl group to form a six-membered cyclic hemiacetal. The correct answer is B
Why does the most stable form of the common sugar glucose contain a six-membered ring in the chair conformation with all the substituents equatorial?
Glucose can exist as a variety of different isomers. It can be linear or cyclic. Of the cyclic forms, it can be a 5 membered ring or a 6 membered ring. Each of the 5 and 6 membered rings can exist as 2 different stereoisomers. Finally, for the 5 and 6 membered rings, there are a variety of different conformations it can potentially exist in. In spite of all of the potential isomers it can exist as, under physiological pH it almost exclusively exists as beta-glucopyranose (see figure below) in the chair conformation with all of the large groups equatorial.First, why is it cyclic instead of linear? This one was the hardest for me to rationalize. This is what I came up with, although i'm open to other answers: Enthalpically, two carbon oxygen single bonds are energetically more favorable than a carbon oxygen double bond. Additionally, the reaction is intramolecular so the effective concentration of the hydroxyl group is high and the entropic penalty is reasonably low. Thus, the equilibrium lies towards cyclization. Second, why a 6 membered ring favored over a 5 membered ring?The ideal bond angles for an sp3 carbon are 109.5 degrees. In a five membered ring, the bond angles are less than 109.5 degrees, which causes torsional strain. In the 6 membered ring, all of the atoms in the ring can form 109.5 degree bond angles so there isn't any torsional strain. Thus, the 6 membered ring is more energetically favored.Third, why is the beta stereoisomer formed over the alpha stereoisomer?If you look at the figure below, you can see that alpha glucopyranose has 1,3 diaxial strain (the red dots) between a hydrogen and a hydroxyl group. If you compare this to the beta-glucopyranose isomer, you'll see that the only groups that are axial are hydrogens, so 1,3 diaxial strain is minimized. Lastly, why is the chair with all the hydroxyl groups equatorial favored?Again, the answer comes from 1,3 diaxial strain. If you look at the figure for beta glucopyranose, you can see that all of the large groups being axial results in a lot of steric strain from 1,3 diaxial interactions.
Glucose can exist in straight-chain or cyclic form. In which combination does glucose link together in a polysaccharide?
a) cyclic forms only
In aqueous solutions there are three forms of glucose;: the a-form (36%), the B-form (64%)and a trace amount of the open-chain form. At equi?
In aqueous solutions there are three forms of glucose;: the a-form (36%), the B-form (64%)and a trace amount of the open-chain form. At equilibrium, the alpha and beta cyclic forms are interconverted by way of the open-chain structure. Explain (a) why the cyclic forms exist predominantly in solutions, and why the B-form is more abundant than the A-form.
What happens when a solution of D-glucose is kept for some time?
If a solution of a pure crystalline sample of glucose is dissolved in water, (like β-D-glucopyranose then structure opens to form the carbonyl group and reclose, the opening and closing repeats continuously, over time some β-D-glucopyranose undergoes mutarotation to become α-D-glucopyranose and the rotation of the solution changes from +18.7 to an equilibrium value of +52.5).The equilibrium mixture is actually about 64% of β-D-glucopyranose and about 36% of α-D-glucopyranose and there are traces of furanoses and open chain form ((less than 0.02%). The solution nevertheless exhibits the characteristic reactions of an aldehyde as there is a shift in the equilibrium to yield more aldehyde as the free aldehyde is used up in a reaction. The α form melts at 146°C and has a specific rotation of +112°, while the β form melts at 150°C and has a specific rotation of +18.7°. The observed rotation of the sample is the weighted sum of the optical rotation of each anomer weighted by the amount of that anomer present.The α - D glucose and β- D glucose are anomers and diastereomers as well.. Anomers are a stable cyclic hemiacetal and have ether linkage between the aldehyde group and the -OH group on the C-5 of glucose. C-1 becomes asymmetricand anomers differ only in the configuration of C-1.The mutarotation of anomers results from the opening and closing of the hemiacetal ring and the equilibration is via the ring opening at the anomeric center of the cyclic sugar with the acyclic form as the intermediate.
Biology: Why does both glucose ring structure and linear structure have exact same functions despite the ...?
::Ring formation:: In 3-dimensional space, a glucose chain can easily curl up, such that the oxygen attached to carbon 5 can be juxtaposed next to carbon 1. A 6-membered ring forms preferentially in water, by attack of the hydroxyl of carbon-5 (C5) on the carbonyl double bond at C1. One bond of the carbonyl double bond opens up and forms a new bond between carbon-1 with the O of C5. The H leaves C5 and a new OH group is formed on carbon 1. So a 6-membered ring is formed, with O as one of its members (one of the vertices). One carbon (C6) is left sticking out away from the ring. Unlike most biochemical reactions, which require a catalyst to help them take place at a reasonable rate (more on this in a week or so), - *****this intramolecular cyclization reaction takes place all on its own, as soon as a sugar is put into a water (aqueous) solution. The ring structure can also open up, re-forming the straight chain. *****The 2 forms are in a dynamic equilibrium, but because the ring form is more stable, this species predominates in water.
Why is the natural form of glucose (dextrose) called d-glucose?
Glucose has five chirality centers in the cyclic form and four in the acyclic form. If a compound has one chirality center, it can exist as mirror images. If it has two centers, then then four isomers can be formed, two of which are mirror images and two of which are diastereomers. Diastereomers are different compounds. Adding more chirality centers leads to more diastereomers, that is more compounds. (GET RAXL is an acronym for all C3-C5 sugars, glyceraldehyde, erythrose, threose, ribose, arabinose, xylose, and lyxose.) So, the other (stereo)isomers of glucose are all different compounds and have different names. However, every form of glucose (by a different name) has a mirror image. By degradation, all sugars have the same penultimate chirality center, its configuration is common to all. It would not make sense to refer to the configuration of C-1 of ribose as it is not chiral yet leads to two different isomers in glucose/mannose. The configuration of D-glucose, D-ribose, D-threose, and D-glyceraldehyde are all the same at C5, C4, C3, and C2, respectively. The name is consistent for all sugars as their absolute configuration are related to D-glyceraldehyde. In nature, only the D-isomer is produced most commonly. So, glucose is referred to as D-glucose as that isomer with a specific chirality at C-5.
How do glucose molecules join to form other molecules?
Glucose (Glc), a monosaccharide (or simple sugar), is an important carbohydrate in biology. The living cell uses it as a source of energy and metabolic intermediate. Glucose is one of the main products of photosynthesis and starts cellular respiration in both prokaryotes and eukaryotes. Glucose (C6H12O6) contains six carbon atoms one of which is part of an aldehyde group and is therefore referred to as an aldohexose. The glucose molecule can exist in an open-chain (acyclic) and ring (cyclic) form (in equilibrium), the latter being the result of an covalent bond between the aldehyde C atom and the C-5 hydroxyl group to form a six-membered cyclic hemiacetal. In water solution both forms are in equilibrium, and at pH 7 the cyclic form is predominant. As the ring contains five carbon atoms and one oxygen atom, which resembles the structure of pyran, the cyclic form of glucose is also referred to as glucopyranose. In this ring, each carbon is linked to a hydroxyl side group with the exception of the fifth atom, which links to a sixth carbon atom outside the ring, forming a CH2OH group
Why is the β-D-glucopyranose form of glucose more abundant than the α-D-glucopyranose form in aqueous solutio?
This has to do with sterics. In organic chemistry, we learn that substituents on six-membered rings are most stable in equatorial positions than in axial positions because this minimizes 1,3-diaxial interactions and other unfavorable repulsive interactions in the molecule. In the beta form of glucose, all of the substituents on the ring are equatorial. But in the alpha form, one of the hydroxyl groups will necessarily be axial, making the molecule higher in energy. Consequently, the beta form will be favored at equilibrium.