logo

Fischer Projections of Monosaccharides

Pearson
15 views
Was this helpful ?
0
The sweet taste of honey is due to the monosaccharides glucose and fructose. These monosaccharides are two of the simplest carbohydrates and cannot be split, or hydrolyzed, into smaller carbohydrates. Monosaccharides can be classified according to the number of carbons and the carbonyl functional group present. Glucose contains an aldehyde group and six carbons. Thus, it is classified as an aldohexose. Fructose contains a ketone functional group and six carbons. Thus, it is classified as a ketohexose. The following monosaccharide is classified as a, b, c or d? The correct answer is (a). The functional group on carbon 1 is an aldehyde and there are five carbons in the structure so it is an aldopentose. Shown are the wedge-dash structures of the simplest monosaccharide, glyceraldehyde. There is one chiral carbon in the structure of glyceraldehyde, a carbon with four different groups bonded to it. Because of this chiral carbon, there are two enantiomers, or nonsuperimposable mirror images, of glyceraldehyde. For a simple molecule like glyceraldehyde, with only one chiral carbon, using wedges and dashes might not be a problem. But imagine showing those chiral carbons with those dashes and wedges for a more complex molecule like glucose. Emil Fischer received the Nobel Prize in Chemistry in 1902 for his contributions to carbohydrate and protein chemistry. It was Fischer who devised a simplified system for drawing isomers that shows the arrangements of the atoms around the chiral carbons. Now we use his model, called a Fischer projection, to represent a three-dimensional structure of a carbohydrate. In a Fischer projection, vertical lines represent bonds that project backward from a carbon atom (the dashes) and horizontal lines represent bonds that project forward (the wedges). In this model, the most highly oxidized carbon, the aldehyde group, is placed at the top, and the intersections of vertical and horizontal lines represent a carbon atom that is usually chiral. Now let's take a look at that glucose molecule with all of those chiral carbons. The Fischer projection of glucose has simplified the representation of the three dimensional structure of those chiral carbons. The structure of glucose is shown with the six carbons numbered from 1 to 6. Here is the Fischer projection of glucose with the carbon atoms numbered. The most oxidized carbon is drawn at the top. There are four chiral carbons at positions 2, 3, 4, and 5. The dashes have now been replaced with vertical lines and the wedges have been replaced with horizontal lines, simplifying the structure. How many chiral carbons does this monosaccharide contain? Is it a, b, c or d? The correct answer is (a). In this structure, the chiral carbons are found at carbons 3, 4, and 5. They are the carbons with four different groups attached. The enantiomers, or nonsuperimposable mirror images, of a monosaccharide are designated as D or L isomers. The D or L designation is assigned according to the position of the OH group on the chiral carbon farthest from the carbonyl carbon. Here are the enantiomers of erythrose. There are two chiral carbons in each enantiomer of erythrose. The one farthest from the carbonyl carbon is circled in each enantiomer. The letter L is assigned to the structure with the OH on the left. This is the L-isomer and thus named L-erythrose. The letter D is assigned to the structure with the OH on the right. This is the D-isomer and is thus named D-erythrose. Sorbose is a monosaccharide often used in the commercial production of vitamin C. Identify the compound as D or L-sorbose. Step 1 is to number the carbon chain starting at the top of the Fischer projection. We start numbering the carbon chain nearest the ketone functional group. There are six carbons in this structure. Step 2 is to locate the chiral carbon farthest from the top of the Fischer projection. Carbons 3, 4, and 5 are chiral carbons. Carbon 5 is farthest from the top of the Fischer projection or, farthest from the most oxidized carbon, which is the keto group. Step 3 is to Identify the position of the OH group as D or L. Looking at carbon number 5, the OH group is on the right, which is the D enantiomer. This then is the structure of D-sorbose. Identify this compound as the D or L enantiomer. Is it a, b, c or d? The correct answer is (c). The chiral carbon farthest from the top of the Fischer projection is carbon number 5. The OH group is drawn on the right, which makes this a D sugar.
The sweet taste of honey is due to the monosaccharides glucose and fructose. These monosaccharides are two of the simplest carbohydrates and cannot be split, or hydrolyzed, into smaller carbohydrates. Monosaccharides can be classified according to the number of carbons and the carbonyl functional group present. Glucose contains an aldehyde group and six carbons. Thus, it is classified as an aldohexose. Fructose contains a ketone functional group and six carbons. Thus, it is classified as a ketohexose. The following monosaccharide is classified as a, b, c or d? The correct answer is (a). The functional group on carbon 1 is an aldehyde and there are five carbons in the structure so it is an aldopentose. Shown are the wedge-dash structures of the simplest monosaccharide, glyceraldehyde. There is one chiral carbon in the structure of glyceraldehyde, a carbon with four different groups bonded to it. Because of this chiral carbon, there are two enantiomers, or nonsuperimposable mirror images, of glyceraldehyde. For a simple molecule like glyceraldehyde, with only one chiral carbon, using wedges and dashes might not be a problem. But imagine showing those chiral carbons with those dashes and wedges for a more complex molecule like glucose. Emil Fischer received the Nobel Prize in Chemistry in 1902 for his contributions to carbohydrate and protein chemistry. It was Fischer who devised a simplified system for drawing isomers that shows the arrangements of the atoms around the chiral carbons. Now we use his model, called a Fischer projection, to represent a three-dimensional structure of a carbohydrate. In a Fischer projection, vertical lines represent bonds that project backward from a carbon atom (the dashes) and horizontal lines represent bonds that project forward (the wedges). In this model, the most highly oxidized carbon, the aldehyde group, is placed at the top, and the intersections of vertical and horizontal lines represent a carbon atom that is usually chiral. Now let's take a look at that glucose molecule with all of those chiral carbons. The Fischer projection of glucose has simplified the representation of the three dimensional structure of those chiral carbons. The structure of glucose is shown with the six carbons numbered from 1 to 6. Here is the Fischer projection of glucose with the carbon atoms numbered. The most oxidized carbon is drawn at the top. There are four chiral carbons at positions 2, 3, 4, and 5. The dashes have now been replaced with vertical lines and the wedges have been replaced with horizontal lines, simplifying the structure. How many chiral carbons does this monosaccharide contain? Is it a, b, c or d? The correct answer is (a). In this structure, the chiral carbons are found at carbons 3, 4, and 5. They are the carbons with four different groups attached. The enantiomers, or nonsuperimposable mirror images, of a monosaccharide are designated as D or L isomers. The D or L designation is assigned according to the position of the OH group on the chiral carbon farthest from the carbonyl carbon. Here are the enantiomers of erythrose. There are two chiral carbons in each enantiomer of erythrose. The one farthest from the carbonyl carbon is circled in each enantiomer. The letter L is assigned to the structure with the OH on the left. This is the L-isomer and thus named L-erythrose. The letter D is assigned to the structure with the OH on the right. This is the D-isomer and is thus named D-erythrose. Sorbose is a monosaccharide often used in the commercial production of vitamin C. Identify the compound as D or L-sorbose. Step 1 is to number the carbon chain starting at the top of the Fischer projection. We start numbering the carbon chain nearest the ketone functional group. There are six carbons in this structure. Step 2 is to locate the chiral carbon farthest from the top of the Fischer projection. Carbons 3, 4, and 5 are chiral carbons. Carbon 5 is farthest from the top of the Fischer projection or, farthest from the most oxidized carbon, which is the keto group. Step 3 is to Identify the position of the OH group as D or L. Looking at carbon number 5, the OH group is on the right, which is the D enantiomer. This then is the structure of D-sorbose. Identify this compound as the D or L enantiomer. Is it a, b, c or d? The correct answer is (c). The chiral carbon farthest from the top of the Fischer projection is carbon number 5. The OH group is drawn on the right, which makes this a D sugar.