The Structure of D-Glucose
In the late 1800s, in what still stands as a monument to chemical ingenuity and intellect, Emil Fischer elucidated the structure of D-glucose. The discussion that follows presents the salient features of Fischer's approach. In order to appreciate his logic, it is necessary to be aware of several chemical transformations that Fischer and his colleagues developed.
Oxidation of Sugars: Aldaric Acids
Treatment of aldoses with dilute nitric acid converts them into aldaric acids. The aldehyde function at one end of the molecule and the primary alcohol at the other are both oxidized to carboxylic acids. Equation 1 gives a generalized description of the process. The designation Cn(H,OH)n represents the chiral carbons in the structures without specifying stereochemistry. While all of the aldoses are chiral, not all of the aldaric acids are. Some of them contain internal an symmetry plane.
Exercise 1 How many of the following aldaric acids are not optically active?
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Exercise 2 There are four D-aldopentoses. Draw Fischer projections of each of them. Then draw Fischer projections of the aldaric acids they would yield. How many of those aldaric acids would be optically active?
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Formation of Osazones: Epimers
Like ordinary aldehydes, aldoses undergo nucleophilic addition when treated with phenylhydrazine. When treated with an excess of phenylhydrazine, aldoses form compounds known as osazones. Equation 2 summarizes the overall transformation.
While the mechanism of the reaction is interesting, the important outcome is that the chiral center at C2 is destroyed. Consequently, aldoses that differ in configuration only at C-2 produce identical osazones. Aldoses that differ in configuration only at C-2 are called epimers.
Exercise 3 Select the compounds that would produce the same osazone.
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The Kiliani-Fischer Synthesis: Buildin' 'em up.
Equation 3 summarizes the net transformation achieved by a series of reactions known as the Kiliani-Fischer synthesis.
This reaction sequence involves nucleophilic addition of cyanide ion to the carbonyl group of the aldose, base-promoted hydrolysis of the cyano group of the resulting cyanohydrin, and dissolving metal reduction of a cyclic ester (a lactone) that is formed as an intermediate. Again, the details of the process are secondary to the net result: the Kiliani-Fischer synthesis converts an aldose into two epimeric aldoses which each contain one more carbon.
Exercise 4 Select the aldoses that would be formed by a Kiliani-Fischer synthesis in which D-threose was the starting material:
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The Ruff Degradation: Tearin' 'em down.
The Ruff degradation converts an aldose containing n carbon atoms into a new aldose containing n-1 carbons. The process involves oxidation of the carbonyl group to a carboxylic acid followed by decarboxylation of the acid: C-1 is lost as CO2; C-2 becomes the carbonyl carbon in the new aldose. Equation 4 summarizes the overall process.
Exercise 5 Draw a Fischer projection of L-mannose, then select the aldose that is formed by Ruff degradation of L-mannose.
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The Fischer Proof
At the outset Fischer knew that D-glucose was an aldohexose. As such it contained 4 chiral centers. The question Fischer asked was "What is the relative disposition of the H and OH groups at C-2, C-3, C-4, and C-5?"
The Aldohexose Backbone
Another way of asking this question is "Which of the 16 (24) possible stereoisomers of an aldohexose is D-glucose?" The 16 stereoisomers constitute 8 pairs of enantiomers. Fischer knew that he could not assign absolute stereochemistry to any chiral center, so he arbitrarily designated those aldohexoses with the OH group at C-5 projecting to the right as D sugars. See Figure 2.
The D-Family of Aldohexoses
This reduced his task to determining the relative positions of the H and OH groups at C-2, C-3, and C-4. The 8 possibilites are displayed in Figure 3.
This set the stage for Fischer to interpret the results of a series of experiments involving the transformations outlined in Equations 1-4. A key feature of Fischer's interpretation was the relationship between optical activity and symmetry: molecules that contain a plane of symmetry are not optically active.
Oxidation of D-glucose
When Fischer treated D-glucose with dilute nitric acid, he obtained an aldaric acid that was optically active, i.e. did not contain a plane of symmetry. Figure 4 shows the aldaric acid, 1AA, that would be produced from aldohexose 1. The dashed line in the figure represents an internal symmetry plane that bisects the bond between C-3 and C-4. This aldaric acid would be optically inactive. Therefore aldose 1 cannot be D-glucose.
One of Eight Aldaric Acids
Exercise 6 Which of the remaining aldohexoses in Figure 3 would produce optically active aldaric acids?
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Ruff Degradation of D-glucose
Ruff degradation of D-glucose produced D-arabinose, an aldopentose that Fischer had available to him. This result establishes that the configurations at C2, C3, and C4of D-arabinose are the same as those at C3, C4, and C5 of D-glucose. Notice that Ruff degradation of 1 would give the same aldopentose as obtained from 5. This fact, combined with the oxidation results, allowed Fischer to eliminate aldohexose 5.
Exercise 7 Aldohexose 2 would yield the same aldopentose as aldohexose . Aldohexose 3 would yield the same aldopentose as aldohexose . Aldohexose 4 would yield the same aldopentose as aldohexose
Figure 5 shows the structures of the aldopentoses that are consistent with the data thus far.
Possible Aldopentoses Generated from D-Glucose
Oxidation of D-Arabinose
Oxidation of D-arabinose with dilute nitric acid produced an aldaric acid that was optically active.
Exercise 8 Which of the three alternatives in Figure 5 would yield an optically active aldaric acid?
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At this point Fischer had reduced the possible structures for D-glucose from 8 to 4.
When D-arabinose was subjected to Kiliani-Fischer conditions, D-glucose and D-mannose were formed. This means that D-glucose and D-mannose must have the same configurations at C3, C4, and C5. They differ only at C2. In other words, D-glucose and D-mannose are epimers. Comparing the structures in Figures 5 and 3, it is possible to state that D-glucose and D-mannose must be either 2 and 6 or 4 and 8.
Oxidation of D-Mannose
Treatment of D-mannose with dilute nitric acid yielded an aldaric acid that was optically active.
Exercise 9 Draw Fischer projections of the aldaric acids that would be formed from aldohexoses 2 and 6 and 4 and 8. Select the compounds that would yield optically active aldaric acids.
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Exercise 10 Given the results just described, D-glucose and D-mannose must be 2 and 6 4 and 8.
Now the question comes down to whether D-glucose is aldohexose 2 and D-mannose is 6 or vice versa. To answer this question, consider the transformations shown in Figure 6.
The apparent simplicity of the transformations shown in Figure 6 belies the synthetic challenge they presented to Fischer. Nonetheless, he was equal to the challenge. When he reduced C1 and oxidized C6 of D-glucose, he obtained a new sugar. When he reduced C1 and oxidized C6 of D-mannose, he obtained D-mannose. These results allowed Fischer to conclude that D-glucose was structure 2. Note that if you rotate structure 6' by 180o about an axis perpendicular to the page and bisecting the C3-C4 bond (marked by the red dot) you get structure 6. If you rotate structure 2' about the corresponding axis, you do not get structure 2. In other words, 6 and 6' are the same compounds. If you rotate 2' about the corresponding axis, you do not get 2. These two compounds are isomers. Therefore, D-glucose must be aldohexose 2.