Chemistry 240Summer 2001

Monosaccharides -- Structure of Glucose
Mono- Di- and Polysaccharides"Oses" and D-SugarsWhich D-aldohexose is Glucose?Cyclic Structures - Anomers
Last time we learnedhow a chiral compound"s absolute configuration can be described by theR/Snaming system. We also considered the situations which can arise when acompound has two (or more) stereogenic carbons. Our examples for that werein fact sugars; monosaccharide aldotetroses. We"ll begin by making somestructural sense of those terms.Sugars are small molecules which belong to the class of carbohydrates.As the name implies, a carbohydrate is a molecule whose molecular formulacan be expressed in terms of just carbon and water. For example, glucosehas the formula C6(H2O)6 and sucrose (tablesugar) has the formula C6(H2O)11. Morecomplex carbohydrates such as starch and cellulose are polymers of glucose.Their formulas can be be expressed as Cn(H2O)n-1.We"ll look at them in more detail next time.The difference between a monosaccharide and a disaccharide can be seenin the following example:A quick glance tells us that a monosaccharide has just one ring, a disaccharidehas two, and a polysaccharide has many. Beyond that, though, there"s anotherimportant structural feature. Look at the disaccharide and focus on theoxygen which links the two rings together. The atom above it is connectedto two oxygens, both of which are in ether-type situations. The carbonand these oxygens are in an acetal linkage.(The bonds are heavier and in blue.)If we look at the corresponding location in the monosaccharide and askwhat the functional group might be, we see that it is a hemiacetal.(Here the bonds are heavier and in red.) So, another way to describe thesituation is that a monosaccharide has a single ring with a hemiacetalin it, a disaccharide has two rings linked by an acetal functional group,and a polysaccharide has many rings linked by many acetal functional groups.( Check this last statement against the polysaccharide structure above).How about the "sugars" we saw last time with just 4 carbons. Why arethey monosaccharides when there is no ring? If we consider that the OHgroup on the bottom carbon could form a hemiacetal with the aldehyde function,then we get a ring, and that structure fits our description of a monosaccharide.

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We"ll take a more detailed look at the cyclic and non-cyclic structuresof sugars shortly.Now let"s see what aldotetrose means. Taking thename apart from right to left, the ending "ose" means a sugar, which maybe a monosaccharide, a disaccharide or an oligosaccharide (a "short" polysaccharide).The middle part "tetr" means that our sugar has four carbons linked ina straight unbranched chain. Terms like "pent" for five carbons and "hex"for six carbons are also in common use. The beginning "aldo" means thatthere is an aldehyde in the compound. The alternative would be a ketonegroup, for which we would use the prefix "keto."Glucose, the most common monosaccharide, is an aldohexose. We understandthat to mean that it is a sugar having six carbons in a straight unbranchedchain which ends in an aldehyde group. We can represent that structurein this fashion:This structure includes four stereogenic carbon atoms (marked with anasterisk *). There are a total of sixteen stereoisomers possible. Eightof these are enantiomers of the other eight. The rest of the relationshipsare diastereoisomeric. Since the groups at the top and the bottom of thechain are not the same, there are no meso isomers. Eight of theisomers are shown here. The other eight are mirror images of these andmay be readily drawn.The question is "which of the sixteen stereochemical representations(Fischer projections, remember that each stereoisomer shown also has anenantiomer which is not shown) describes the absolute configuration ofglucose? When Emil Fischer took up this problem about 100 years ago, herealized that there was no way to determine if glucose was one of the eightstructures above or one of the unshown enantiomers. He made the assumptionthat it was one of the ones above so that he could work on the diastereoisomericpart of the problem, hoping that later work would resolve the questionof which enantiomer best represented glucose.Fischer also developed the D/L system for specifyingthe structures of sugars. If the OH group on the stereogenic carbon farthestfrom the aldehyde group is to the right in the Fischer projection, thenthe compound is a D-sugar. All of the sugars in thefigure above are D-sugars. If the OH group on thestereogenic carbon farthest from the aldehyde group is to the left in theFischer projection, then the compound is an L sugar.The enantiomers of all the sugars in the figure above are Lsugars. Fischer"s assumption amounts to saying that glucose is a D-sugar.Later work resolved this issue, and Fischer was right.How did Fischer determine which of the eight structuresabove was glucose? He had available samples of glucose and mannose, bothaldohexoses, and arabinose, an aldopentose. He also learned how to reducethe aldehyde functional group to a primary alcohol. (We"ll illustrate thiswith NaBH4 to avoid learning a new reaction, but he used anotherreagent.) He developed a method for extending the carbon chain of an aldose(called the Kiliani-Fischer chain extension). He also had a polarimeterso he could determine whether a sample was optically active or not. Perhapsmost importantly, he had a group of talented and dedicated students.Now, some data.Experimental result: When the aldehyde group of arabinose was reduced toa primary alcohol group, the product was optically active.Conclusion: Arabinose has either structure 2 or 4 in the scheme below.This is because if arabinose were either 1 or 3, the product would havea plane of symmetry (mirror plane) and would be optically inactive.Experimental result: When the aldehyde group of glucose was reduced toa primary alcohol group, the product was optically active. The same resultwas obtained for mannose.Conclusion: The structures "X"d" out below do not represent either glucoseor mannose since the products from these structures would be meso compounds.Experimental result: Kiliani-Fischer chain extension applied to arabinoseproduces glucose and mannose.Conclusion: The bottom three stereogenic carbon atoms of glucose and mannoseare have identical configurations to the three stereogenic carbon atomsof arabinose. This means that glucose and mannose differ only in the configurationof the stereogenic carbon atom nearest the aldehyde functional group. Wecan further conclude that if one member of a pair of aldohexoses (pairedbecause their bottom three stereogenic carbons are identical) is ruledout, so is the other.Now let"s see what we have left. There are four structures remaining ascandidates. They are on the right below. If we go back to the possibilitiesfor arabinose, we find that the two on the top come from structure 2 forarabinose, which was a possibility, while the two on the bottom come fromstructure 3, which was ruled out earlier. The conclusion is that arabinoseis represented by structure 2, and glucose and mannose are the two structuresto its right.But which is glucose and which is mannose? Fischer noticed that if reactionscould be developed which changed the aldehyde group into a primary alcoholand the primary alcohol into an aldehyde (switch ends) one of these structureswould give itself, and the other would give back a new Lsugar. The reactions are complex and we will not look at them, but whenthe chemistry was applied to the sample called mannose, the product wasmannose. When the chemistry was applied to the sample called glucose, anew sugar was formed.There was a great deal more to be done to confirm this conclusion andto synthesize the other six aldohexoses, but Fischer"s exercise in logicand dedicated experimentation led to the conclusion that the eight D-aldohexosesare:Notice that the new sugar which was produced from glucose by the "exchangeends" experiment is L-gulose. The names of the hexosestell us which diastereoisomer we have; the D or Ldesignation gives us which enantiomer we have.The corresponding names for the aldopentoses are:To finish today, we"ll see what happens when ahemiacetal is formed between the aldehyde carbon and one of the OH groupson the chain. We"ll look at two examples, ribose, which is a key componentof RNA, and glucose because of its abundance. (You may wish to review themechanismfor hemiacetal formation.)Since there are four OH groups in ribose, we could anticipate four differentring sizes. In three atom rings and four atom rings the bond angles arefar from 109.5o, so these rings are strained, have higher energiesand are hard to form. (See page 68 in Brown.) Remember that there is anequilibrium between a hemiacetal and the aldehyde/alcohol it comes from,and that high energy materials don"t persist at equilibrium. We are leftwith rings which have either five or six atoms in them. In the case ofribose the important ring (found in RNA) is the five membered ring.

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Notice that the carbon in the newly formed hemiacetal group is stereogenic.This means that there are two possible diastereoisomers for the cyclicstructure. Usually both are formed, and they have a special name -- theyare anomers of each other. The carbon between the two oxygens in the hemiacetalgroup is called the anomeric carbon. If the OH group is down (in a drawingwith the ring oxygen to the rear center or right), the designation forthat anomer is alpha. If the OH group is up, the designation is beta. Sincethe alpha and beta anomers are diastereoisomers, they have different properties;in particular, different optical activities. The term for a five atom sugarring is "furanose."Glucose usually makes hemiacetal cyclic structures with six atom rings,although five membered rings can also be formed when the six membered ringsare precluded. Such six membered rings are named by the term "pyranose."The ring forms look like this, keeping in mind that alpha and beta anomersare also involved here:Again, we have an equilibrium between the open chain form and the twodiastereoisomeric anomers.There is one further point to be made about these glucopyranoses. Thestructure we have drawn for the ring is flat. The bond angles would be120o, quite far from the normal tetrahedral value of 109.5o.The atoms in the ring can have bond angles of about 109.5o ifthe ring puckers as shown here:Of course, molecules adopt these puckered shapes (called chair conformationsfrom their resemblance to a rather wide lounge chair) automatically. Youcan review the material on cyclohexane on pp 69-70 of Brown fora more detailed analysis of this material. It has been established thatin these chair conformations, the molecules have a lower energy if thelarger substituents on the carbons are roughly in the plane of the ringitself. These positions are called "equatorial" to distinguish them fromthe other positions (roughly perpendicular to the ring, called "axial").A compound which can have all of its larger substituents (everything islarger than hydrogen) in an equatorial position is more stable than onewhich cannot. beta-D-glucose has all of its substituentsin equatorial positions, and is thus the most stable hexopyranose. It isalso the most abundant.Back to the Course Outline