8. Choice D is correct. When the straight chain form of a sugar is converted into its Haworth projection, the hydroxyl groups originally on the left side of the sugar are found above the ring and hydroxyl groups originally on the right side of the sugar are found below the ring (recall the mnemonic "downright uplefting"). Mannose has the hydroxyls on carbons 2 and 3 on the left. The C-epimer of D-mannose therefore has a straight chain with only the hydroxyl on carbon 2 on the left. In the Haworth projection of the C-3 epimer of 6-D- mannopyranose, the hydroxyl groups on carbons 1 and 2 are above the ring and the hydroxyl groups on carbons 3 and 4 are below the ring. This makes choice D the best answer. Choices A and B should have been eliminated early on, because they have a-orientation at carbon 1. As a point of interest, choice C is fi-D-glucopyranose.
9. Choice B is correct. This question is best answered from straight memorization, although a good guess can be derived from the passage. Lactose is a disaccharide made from glucose and galactose, not a polysaccharide, so choice D is eliminated. Sucrose is a disaccharide made from glucose and fructose and not a polysaccharide, so choice C is eliminated. Glycogen has alpha linkages between saccharide monomers, while cellulose (the polysaccharide we can't digest) has beta linkages between the saccharide monomers. We lack the enzyme to cleave the 6-1,4-glycosidic linkage. Choice B is the best answer.
10. Choice C is correct. When you consider the most common aldopyranose, ribose, it is reasonable to assume that a furanose ring is the most stable cyclic form for aldopentoses in general. Choice A can be eliminated. The penultimate carbon determines the designation of D or L for a sugar. In a five carbon sugar, carbon 4 is the penultimate carbon, so choice B can be eliminated. In the cyclic form, furanose ring, all of the carbons except carbon 5 have chiral centers, so it is true that a cyclic aldopentose has four chiral centers. Choice D is eliminated. In an aldopentose, carbon 1 has two bonds to oxygen, so in its cyclic form, carbon 1 still has two bonds to oxygen, making it the anomeric carbon. Carbon 2 is not the anomeric carbon, so choice C is correct.
11. Choice C is correct. For any trisaccharide, the total weight is the weight of the three monosaccharides minus the weight of the two waters lost in forming the linkages. When three 6-carbon monosaccharides are combined, there are two bridging linkages. Calculating this value gives 3(180) - 2(18) = 540 - 36 = 504 grams/mole.
Choose C to show off those math skills of yours. The fact that the linkages were 1,4-linkages is a moot point when considering the mass of the compound. Either way you look at it, the polysaccharide loses one water per linkage, regardless of the exact linkage.
12. Choice C is correct. This question tests straight memorization, so remember it correctly, then choose C. Should you not recall the fact that polysaccharides form linkages from carbon one to four (when dealing with pyranoses), then the passage also gives you a subtle hint, by giving only one example of a linkage, and the example is a 1,4-linkage.
13. Choice A is correct. An aldopentose is most likely to form a furanose ring when the hydroxyl on the penultimate carbon (carbon 4) attacks the aldehyde carbon. This eliminates choices C and D. Because the original sugar is an aldehyde, the structure is a hemiacetal and not a hemiketal. The best answer is choice A.
14. Choice C is correct. Tiie glycosidic bond forms when a nucleophile displaces the anomeric hydroxyl group of the sugar. The nucleophile is methanol, which attacks using the lone pairs on its oxygen atom. This eliminates choices A and B. This question now focuses on whetherit is the a-anomer or the 6-anomer. The methoxy group is
trans to carbon 6, so it is the a-anomer, making choice C the best answer.
Passage III (Questions 15 - 22) Sugar Conventions
15. Choice A is correct. An aldohexose has one aldehyde functional group, one primary alcohol functional group, and four secondary alcohol functional groups. This makes choice A the best answer. An aldohexose has chiral centers at carbons 2, 3, 4, and 5, so choice B is eliminated. Tiie aldehyde carbon has the most bonds to oxygen of any carbon, so it is most oxidized. This eliminates choice C. For a D-sugar, the penultimate carbon has R- chirality, so for a D-aldohexose, carbon 5 has R-chirality. Choice D is eliminated.
16. Choice B is correct. A ketohexose is a six-carbon sugar with a ketone functionality (likely on carbon 2) in its linear structure. If it were in the cyclic form, the term becomes pyranose or furanose depending on the ring size.
Since fructose is the only ketose listed in the choices given, it is best for you to choose B.
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17. Choice A is correct. When the hydroxyl group on carbon 5 of an aldohexose is inverted, thesugar becomes the opposite type of sugar. This means that theC-5 epimer ofL-glucose mustbe a D-sugar, eliminating choice B. In order to form D-glucose from L-glucose, all of the chiral centers must differ, because the two structures are enantiomers. Only one chiral center differs, so choice D is eliminated. The C-5 epimer of L-glucose has hydroxyl groups on the right on carbons 3 and 5, and hydroxyl groups 2 and 4 on the left, which according to
Figure 1 is D-idose. The best answer is choice A.
18. Choice D is correct. D-Gulose has hydroxyl groups onthe right for all carbons except carbon 4. The C-2 epimer of D-gulose has hydroxyl groups on left for carbons 2 and 4. The C-3 epimer of glucose has all of the hydroxyl groups on the right, so choice A is eliminated. The C-3 epimer ofD-idose has the hydroxyl group on the leftfor carbons 2,3, and 4, so choice Bis eliminated. The C-4 epimer of D-allose has thehydroxyl group on the left on carbon 4 only, so choice C is eliminated. The C-4 epimer ofD-altrose has hydroxyl groups on the leftfor carbons
2 and 4, so choice D is the best answer.
19. Choice Bis correct. Ketals cannot be oxidized byTollen's reagent, because they are stable under basic conditions and they cannot be converted into aldehydes (which can be oxidized). The key feature in a ketal preventing oxidation is the absence of a C-H bond on the ketal carbon (which is the feature that prevents a ketone from being oxidized) . The C-H bond is necessary for oxidation to occur. Oxidation from the organic chemistry perspective involves the loss ofhydrogen as well as the gain ofoxygen. For this question, choose Bfor greatest
satisfaction.
20. Choice C is correct. Fructose is a ketose, so its cyclic structure cannot be an acetal or hemiacetal. This eliminates choices A and B. Ketones, when added to alcohols in basic medium, go on to form hemiketals. The hemiketal will be in equilibrium with ketone. The question does not state whether the medium is basic or acidic, so you
must answer this from experience and memory. Monosaccharides for hemiketals (or hemiacetals if the structure
starts as an aldose). This means C is the best choice.
21. Choice D is correct. Epimers differ by one chiral center in their backbones. Talose differs from galactose at carbon 2, differs from idose at carbon 3, and differs from mannose at carbon 4. This means talose is an epimer of galactose, idose, and mannose, eliminating choices A, B, and C. It differs from glucose at carbons 2 and 4,
meaning it is not an epimer of glucose. This makes choice D the best answer.
22.
*s^°
HO- HO- HO- H-
•H
•H H OH
CHjOH D-(+)-Talose
*s^°
H- HO- HO- H-
OH
•H
•H OH
CBjOH
D-(+)-Galactose
H^*°
HO- H- HO- H-
•H OH
•H OH
CH2OH D-(-)-Idose
Hx^°
HO- HO- H- H-
•H
•H OH OH
CH2OH D-(+)-Mannose
H^°
H- HO' H- H-
OH
•H OH OH
CH2OH D-(+)-Glucose
Choice Cis correct. Treatment of an aldehyde with areducing agent such as NaBHj reduces the carbonyl group (C=0) into a primary alcohol. Following that same reactivity, treatment of an aldohexose (or ketohexose) will convert the sugar into a hexa-ol (a six carbon chain containing six hydroxyls, one on each carbon). Treatment of an aldopentose will convert it into a penta-ol. The results from an analytical reasoning perspective are similar to treatment of the sugar with nitric acid, because the product now has the possibility ofbeing meso (indicated by a lack of optical activity). Fructose is eliminated, because the new hydroxyl would result in the formation of
a new chiral center, thus there would be the formation of two diastereomers. At least one of the two
diastereomers has to be optically active. Glucose and mannose are eliminated, because they have no mirror plane between carbons three and four. This makes ribose the best choice. Ribose will generate a penta-ol with a mirror plane slicing through carbon three. Pick choice C for the tingly sensation of correctivity. Galactose would also yield an optically inactive product although it isnot given as an answer selection.
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Passage IV (Questions 23 - 29) Monosaccharides versus Disaccharides
23. Choice C is correct. Galactose is the C-4 epimer of glucose. The C-2 epimer of galactose varies from glucose at both carbon 2 and carbon 4, so choices A and D are eliminated. The C-2 epimer of mannose is glucose, so choice B is also eliminated. The only remaining choice, and the correct answer, is choice C. Talose (which is the C-2 epimer of galactose as shown in Figure 1) is the C-4 epimer of mannose. Pick choice C.
H^°
OH
H C-4
H- HO- H- H-
QH epimers
OH
CH2OH
D-Glucose
H-^°
H- HO- HO- H-
OH H
rj epimers
OH
CH2OH
D-Galactose
C-2
H\^°
HO- HO"
HO- H-
H H
„ epimers
OH
CH2OH
D-Talose
C-4
H-^°
HO- HO- H- H-
H H OH OH
CH2OH
D-Mannose
24. Choice B is correct. The glycosidic linkage involves carbon 1 of the glucose on the left and carbon 2 of the fructose on the right. This eliminates choice D. The glucopyranose ring has a-orientation, so choice C is eliminated. In the fructose structure, the anomeric oxygen (involved in the linkage) is cis with carbon 6, so it has 6-orientation.
This makes choice B the best answer.
25. Choice C is correct. Glycosidic linkages involve an anomeric carbon with two OR groups, so the functionality cannot contain the prefix "hemi". This eliminates choices A and B. Because maltose is formed from glucose, an aldose, the functional group is an acetal. This makes choice C the best answer. Aldoses go on to form hemiacetals as monosaccharides and acetals as polysaccharides. Ketoses go on to form hemiketals as monosaccharides and ketals as polysaccharides.
26. Choice D is correct. In fi-D-glucopyranose, all of the substituents on the pyranose ring have equatorial orientation. This is a piece of information you should have committed to memory. The structure in question differs from 6-D-glucopyranose at carbon 3, where the hydroxyl group has axial orientation. This makes the structure a C-3 epimer of 6-D-glucopyranose, making choice D the best answer. An anomer would vary in chirality at the anomeric carbon. This eliminates choice A. A conformer is the identical molecule rotated or contorted. If only one substituent changes from equatorial to axial while keeping the ring in the same orientation, then the structures are not conformers. This eliminates choice B. Enantiomers differ at every chiral center, so all of the centers would need to be axial for it to be an enantiomer. This eliminates choice C.
27. Choice A is correct. A trisaccharide has two glycosidic linkages. This eliminates choices B and D. When the trisaccharide is made from three unique aldohexoses, then the linkages will be acetal linkages, rather than ketal linkages. The linkages involve carbons with two OR groups, so they are acetal linkages, rather than hemiacetal groups. This makes choice A the best answer.
28. Choice D is correct. Hydrolysis under acidic conditions hydrolyzes acetals and hemiacetals, but does not affect ethers. When a disaccharide is exhaustively methylated, all of the hydroxyl groups become methoxy groups.
All of the methoxy groups are ethers except for the anomeric carbon of the glycoside (sugar on the right), which goes from a hemiacetal to an acetal. This means that the anomeric carbon of the glycoside and the anomeric carbon of the glycosyl group making the linkage are both acetals. When hydrolyzed, they will generate hydroxyl groups. All of the other oxygen atoms will be part of methoxy groups. Not all of the OCH3 groups return to being hydroxyl groups, so choice A is eliminated. The glycoside loses the methoxy group on its anomeric carbon, but not any others. The glycosyl group does in fact not lose any methoxy groups. Choice B is not a good answer, but it cannot be eliminated. The disaccharide only racemizes at the anomeric carbons, so choice C is eliminated. Following hydrolysis, the two carbons involved in the glycosidic linkage gain hydroxyl groups
when water is added. The best answer is choice D.
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29. Choice C is correct. The glycoside is typically the sugar on the right, recognized because it does not use its anomeric carbon for the linkage. fi-D-Glucopyranose has hydroxyl groups that alternate up-to-down-to-up-to- down in the Haworth project. The sugar on the right follows the same pattern as fi-D-glucopyranose, except for carbon 2, which has its hydroxyl group up instead of down. This means that the glycoside is the C-2 epimer of
6-D-glucopyranose, making choice C, D-mannose, the best answer.
Passage V (Questions 30 - 36) Amylose, Amylopectin, and Cellulose
30. Choice Dis correct. Maltose is adisaccharide made from two a-D-glucopyranose structures. The linkage must be alpha, so choices A and Care eliminated. Choice Bis eliminated, because the hydroxyl group on the glycoside
(sugar on the right) has beta orientation. The best answer is choice D.
Choice A is correct. Amylase breaks down the a-glycosidic linkage. However, it is not stable under highly acidic conditions, where it is readily hydrolyzed. Gastric fluids are highly acidic, so amylase is destroyed in the stomach. It must be released again by the pancreas. This makes choice A the best answer. Both pancreatic and salivary amylase cleave a-glycosidic linkages, so choice Dis absolutely wrong. The two forms of amylase
are identical enzymes, so they are equivalent in strength. This eliminates choices B and C.
Choice D is correct. The blood can only absorb monosaccharides. Fructose, galactose, and glucose are monosaccharides, so choices A, B, and Care eliminated. Lactose is a disaccharide of galactose and glucose, so it
cannot be absorbed into the blood. Choice D is the best answer. In lactose intolerance, we don't have the
disaccharase lactase, therefore you cannot break lactose into galactose and glucose. Since we cannot absorb the disaccharide, the lactose does not get absorbed and instead goes to feed our bacteria (which metabolize it to gas and toxic metabolites). This results in pain and an increase in the number of molecules in the gut and lumen,
which draws water in and results in diarrhea.
Choice C is correct. It is stated in the passage that amylose has no branching, so choices A, B, and D are eliminated. It is also stated in the passage that amylopectin has a-l,6-branching at about one out of every
thirty glucose residues. This confirms that choice C is the best answer.
Choice Bis correct. Glycogen is astarch made from exclusively a-D-glucopyranose with a significant amount of 1,6-branching. This eliminates choice A. The standard glycosidic linkage in the polysaccharide is a-1,4, so glycogen contains both a-l,4-linkages and a-l,6-linkages. This eliminates choice Dand makes choice Bthe best answer. Because glucose is an aldehyde, all of the linkages involve anomeric carbons that are part of an acetal
functional group. This eliminates choice C.
Choice Dis correct. Amylopectin is a polysaccharide of a-D-glucopyranose that is held together by a-1,4- glycosidic linkages with an occasional 1,6-glycosidic linkage for branching. This means that the oxygen atoms on carbons 1, 4, and 6are involved in glycosidic linkages. Hydroxyl I is on carbon 1, so the correct choice must contain I. This does not help, because every answer selection contains I. Hydroxyl II is on carbon 2, so the correct choice must not contain II. This eliminates choices Aand C. Hydroxyl III is on carbon 4, so the correct choice must contain III. This does not help, because the remaining answer selections (choices Band D) both contains III.
Hydroxyl IV is on carbon 6, so the correct choice must contain IV. Only choice Dcontains IV, so itis correct.
Choice C is correct. Cellulose is another polymer of glucose, similar to amylose, but with 6-1,4-glycosidic Imkages. This eliminates choice A. Lactose is adisaccharide made from glucose and galactose, so lactose cannot be formed from the break down of amylose. Choice Bis eliminated. Sucrose is adisaccharide made from glucose and fructose, so sucrose cannot be formed from the break down of amylose. Choice Dis eliminated. Amylose is broken down into glucose, which is absorbed by the blood and transported. Some of the glucose undergoes
glycolysis (that which isn't stored), resulting in the formation of pyruvate. Choice Cis the best answer.
31.
32.
33.
34.
35.
36.
Passage VI (Questions 37 - 43) Unknown L-Aldopentose Elucidation
37. Choice C is correct. An aldopentose has three stereocenters, located at carbons 2,3, and 4. The maximum number ofstereoisomers possible is found by2n where n is the number ofstereocenters. The reason the word "maximum"
is chosen is that if one of the stereoisomers happens to be ameso compound, then the total number of possible stereoisomers will decrease by one. This won't be a problem with the sugars, because sugars are not meso.
Because there are three stereocenters, there will be eight possible stereoisomers. Choose C.
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38. Choice B is correct. Because the cyano nucleophile can attack the carbonyl carbon from either side (not necessarily equally from either side however), carbon one in the reactant monosaccharide (which becomes carbon 2 in the product) generates a mixture of two chiral centers on carbon two of the product. No other chiral centers will change, so only the chiral center at carbon two differs between the two product molecules. This makes the two compounds diastereomers (non-superimposable and not mirror images). More specific than diastereomers when dealing with sugars is the term epimer, given to sugar diastereomers that vary at only one
carbon in the backbone. This makes choice B the best answer.
39. Choice C is correct. Compound A is said to be an L-aldopentose. You should know from your information base that glucose is a six carbon sugar (aldohexose). This means that the unknown sugar cannot be glucose. The correct answer is choice C. Ribose you should know is an aldopentose, so choice A definitely should have been eliminated. Without knowing exactly what xylose and lyxose are, they cannot be eliminated. Xylose is the C-3 epimer of ribose and lyxose is the C-2 epimer of xylose.
40. Choice D is correct. Compound A is an aldopentose with a molecularmass of 150 grams per mole. When oxidized at both terminal carbons, the mass increases. This eliminates choices A and B. The formula for the aldaric acid formed is C5H8O7 so the molecular mass is equal to 60 + 8 + 112 = 180 grams per mole. Choice D is therefore the correct answer. You could also have solved this knowing that when it is oxidized, it gains two oxygen atoms and loses two hydrogen atoms, resulting in a net gain in mass of 30grams/mole.
41. Choice A is correct. Compound Bhas increased hydrogen bonding compared to Compound A,because carboxylic acids form stronger hydrogen bonds than alcohols. The increase in hydrogen bonding will manifest itself as an increase in melting point. The best answer is therefore choice A. Choices C and D should exclude one another,
because if choice C were true, then choice D would have to be true.
42. Choice B is correct. Aldohexoses in their most stable form are actually aldohexapyranoses, the six-membered ring form ofa monosaccharide. The ring forms when the hydroxyl group oncarbon 5 attacks the carbonyl carbon.
This means that there is no carbonyl absorbance observed in the IRspectrum of the most stable form of the sugar.
The carbonyl peak in the IR isjust above 1700 cm-1, so no peak is observed there. Choice Bis the best answer.
43. Choice B is correct. Because Compound Bis optically inactive, it must be a meso compound. In order for the compound to be meso, there must be a mirror plane through the molecule. The molecule contains five carbons so the mirror plane must slice through carbon three reflecting carbon 2 onto carbon 4. Compound Bis drawn with carbon 2 having accurate stereochemistry as determined by the optical inactivity of the diacid.
O
On the left because it is an L-suga
H^°
H- H- HO-
"(OH) -(OH)
H
5 CH-.OH
HO,S ^
HO
HNQ
HO
On the left because
of the mirror plane
Compound A Compound B
Unknown L-aldopentose optically inactive diacid
If the hydroxyl group on carbon 2is on the left in the Fischer projection of compound B, then it must be on the left in the Fischer projection of compound A (the original sugar). Hydroxyl groups on the left in the Fischer
projection are assigned S stereochemistry. Pick B.
Passage VII (Questions 44 - 50) Unknown D-Aldohexose Elucidation
44. Choice C is correct. Treatment of Compound X, an aldohexose, with NaBH4 will reduce the aldehyde functionality of carbon 1 to a primary alcohol. Because carbon 6 is also a primary alcohol, the two groups are identical. Generating matching groups at the terminal ends of the sugar issimilar to what occurs when a sugar is oxidized with nitric acids (in which case both terminal carbons become carboxylic acid groups). Because the diacid of Compound X is optically active (it has an optical rotation associated with it), the reduced form (a
hexa-alcohol) is also optically active. The best answer is choice C.
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