Monosaccharide structures can be drawn in an open chain (acyclic) or closed chain form. The Fischer projection is used to represent monosaccharides in their open chain form, while the Haworth structure is used to represent the cyclic form. Glucose is shown here with its Fischer projection and Haworth structure. Let's take a close look at the Fischer projection for the monosaccharide, glucose. Glucose is a monosaccharide with 6 carbons and an aldehyde functional group. It is classified as an aldohexose. The structure of glucose is shown in detail with the six carbon atoms shown and numbered. Next to it is the Fischer projection of glucose with the carbon atoms numbered. In this model, the most highly oxidized carbon, the aldehyde group, is placed at the top and numbered carbon #1, and the intersections of vertical and horizontal lines represent a carbon atom that is usually chiral. Classify the following Fischer projection for a monosaccharide, the Fischer projection for the monosaccharide pictured is fructose. Is it a, b, c or d? The correct answer is (d). There is a ketone functional group at carbon #2, classifying it as a keto, and there are six total carbons, classifying it as a hexose. Putting this all together classifies it as a ketohexose. The Fischer projections of monosaccharides represent the open chain form. However, the most stable form of both pentoses and hexoses are five or six atom rings. These rings can be drawn as Haworth structures. Let's go through converting the Fischer projection of D-glucose to the corresponding Haworth structure. In the Fischer projection of D-glucose, we are going to number the carbons 1-6. The groups attached to carbons 2 through 5, and on the right side of the structure, are shown boxed in green. The groups attached to carbons 2 through 5, and on the left side of the structure, are shown boxed in pink. The oxygen of the hydroxyl group attached to carbon 5 is highlighted pink. The first step in converting a Fischer projection to a Haworth structure is to turn the Fischer projection clockwise by 90°. Step 2 is to fold the horizontal carbon chain into a hexagon and rotate the groups on carbon 5 counterclockwise so that the group containing carbon 6 is on the top of the ring. This will allow the O on carbon 5 to be oriented so that it can bond to carbon 1. The oxygen on carbon 5 bonds to carbon 1. Since carbon can only have 4 bonds, the double bonded oxygen must change. The hydrogen on the oxygen from carbon 5 will bond to the oxygen of the aldehyde functional group and it is converted to a hydroxyl group. When this occurs, a new chiral carbon at carbon 1 is formed. The OH can be below the plane of the ring or the OH can be drawn above the plane, as shown here. These two structures, with the OH group being above or below the ring, are isomers of D-glucose. In the alpha isomer, the OH is below the plane of the ring, and in the beta isomer the OH is above the plane of the ring. Shown is the alpha-isomer of D-galactose, choose the beta-isomer of D-galactose. Is it a, b, c or d? The correct answer is (b). The alpha and beta isomers are designated by the orientation of the OH group around carbon 1. The alpha isomer has the OH below the plane of the ring and the beta isomer has the OH group above the plane of the ring. In an aqueous solution, the Haworth structure of alpha-D-glucose opens to give the open chain of D-glucose. At any given time, there is only a trace amount of the open chain because it closes quickly to form a stable cyclic structure. The open chain form is in equilibrium with the alpha isomer. When the open chain closes again, it can form beta-D-glucose. In this process called mutarotation, each isomer converts to the open chain and back again. As the ring opens and closes, the OH group on carbon 1 can form either the alpha or beta -isomer. An aqueous solution of glucose contains a trace amount of the open chain, and a mixture of 36% alpha-D-glucose and 64% beta-D-glucose. Identify the Haworth structure (beta isomer) of the following monosaccharide, shown is the Fischer projection of D-gulose, with an aldehyde group at carbon 1, and OH groups on the right at carbons 2, 3, and 5, and on the left at carbon 4. Carbon 6 has an OH group and two H's. Is it a, b, c or d? The correct answer is (d). The new chiral carbon formed when the ring closes is at carbon 1 and the OH group is above the plane of the ring for the beta isomer. As we look at the Fischer projection, we see that the OH groups on carbons 2 and 3 are on the right side so they should be below the plane of the ring, and the OH on carbon 4 is on the left side so this should be above the plane of the ring in the Haworth structure.