of n in the case of a straight chain aldohexose is four. 24 =16, so there are sixteen possible stereoisomers for an aldohexose. Of these 16 possible stereoisomers for an aldohexose straight chain, eight are D-sugars and eight are L-sugars. This make 8the best answer. Pick B, and do what is best... in terms of answering that is.
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49. Choice A is correct. The addition of HNO3 results in a compound with identical groups on the terminal carbons (with HNO3 they are both carboxylic acid groups). The addition of NaBH4 to the sugar will reduce the aldehyde to a primary alcohol making both terminal carbons identical without affecting the other carbons.
This is analogous to the results when using nitric acid. Nitric acid is cheaper than sodium borohydride, but as a general rule, the MCAT has stayed away from questions about cost effectiveness of chemical reactions. Pick A and show off your correct answer selection skills. Choice B (H2SO4) dehydrates a sugar to carbon, while choice C (Br2(aq)) will oxidize only carbon 1 into a carboxylic acid, not both terminal carbons. Choice D (KMnO^aq)) will oxidize everything on the sugar, so choice D is a bad answer. It is bad to pick bad answers; don't be bad!
50. Choice D is correct. The moststable form of an aldohexose is the pyranose ring structure. This ties up the OH of carbon 5. Because periodic acid (HIO4) cleaves 1,2-diols, the sites are limited to neighboring diols at Ci and C2, C2 and C3, and C3 and C4. The least sterically hindered is the C3 and C4 pair of alcohols. Pick D. This information could be extracted directly from the passage, which is common on the MCAT.
Passage VIII (Questions 51 - 57) Kiliani-Fischer Synthesis
51.
52.
53.
Choice D is correct. Using the Kiliani-Fischer method, a D-aldopentose is used to synthesize a D-aldohexose.
The addition occurs at carbon 1 of the D-aldopentose, turning carbon 1 into carbon 2 in the D-aldohexose products.
This means that the chirality of carbons 2, 3, and 4 in the D-aldopentose match that of carbons 3, 4, and 5 in the D-aldohexose. To determine the D-aldopentose needed to synthesize the D-aldohexose in the question, using retrosynthetic analysis, carbon 1 must be removed and carbon 2 must be turned into a aldehyde group.
H^°
H- H- HO- H-
OH OH H OH
CH2OH
P
Hv^°
H- HO- H-
OH H OH
CH2OH D-Xylose
Upon matching structures with the four D-aldopentoses shown in Figure 2, the best answer is D-xylose, choice D.
Choice D is correct. To be C-2 epimers, the sugars must be diastereomers that vary in their chirality at carbon 2.
Ribose is an aldopentose and altrose is an aldohexose, so they are not even isomers, let alone epimers. Choice A is eliminated. The only way to answer this question, without actually knowing what the sugars are, is to employ the data in Table 1. The products of Kiliani-Fischer synthesis are C-2 epimers. Glucose and talose are not C-2 epimers and neither are mannose and allose. This eliminates choices B and C. Idose and gulose are formed as products when xylose undergoes Kiliani-Fischer synthesis, so idose and gulose are C-2 epimers, making choice D the best answer.
Choice C is correct. Reduction in organic chemistry is most simply viewed as either the gain of bonds to hydrogen or the loss of bonds to oxygen. In step 1, the carbonyl carbon loses one bond to oxygen (in forming a new bond to carbon), so carbon 1 is reduced in Step I. This eliminates choice B. In Step II, reduction occurs, because the cyano group gains a hydrogen on carbon and a hydrogen on nitrogen. The symbol [H] refers to reduction, so Step II is definitely reduction. This eliminates choice A. Step III involves the hydrolysis of an imine into an aldehyde. This is where the general definition needs to be a bit more specific. Both oxygen and nitrogen are more electronegative than carbon, so when carbon goes from being bonded to nitrogen to being bonded to oxygen, it has not been oxidized or reduced. This means that Step III is not oxidation or reduction, so choice C is the best answer. This is a case where oversimplification can hurt. Technically speaking then, oxidation occurs when the number of bonds to more electronegative atoms increases and /or the number of bonds to less electronegative atoms decreases. So, for those of you who considered Step III to be oxidation, you still got lucky in that you didn't figure it to be reduction and choice C was still your choice. But for edification, know the more correct definition of oxidation and reduction for organic chemistry.
54. Choice D is correct. In the Kiliani-Fischer synthesis as shown in Figure 1, and aldehyde is first converted into a hydroxy nitrile compound. The nitrile is reduced into an amine. The amine is hydrolyzed into the aldehyde.
Of the choices, only choice D, an imine, is an intermediate in the synthesis. Choice D is the best answer.
Copyright © by The Berkeley Review® 147 CARBOHYDRATES EXPLANATIONS!
55. Choice C is correct. The two products formed from Kiliani-Fischer synthesis, according Table 1, are C-2 epimers
of one another. Epimers are diastereomers, which eliminates choices A, B, and D. The two diastereomers are formed in unequal concentration, because the presence of a chiral center in the reactant makes one side of the aldehyde easier to attack than the other. Choice C is the best answer.
56.
57.
Choice C is correct. Talose differs in chirality from glucose at carbons 2 and 4. In the Haworth projection offi-D- glucopyranose, the hydroxyl groups are up (above the pyranose ring) on carbons 1 and 3, and down in carbons 2 and 4. This means that talose has the hydroxyl groups on carbons 2 and 4, up, so all of the hydroxyl groups are up in 13-D-talopyranose. For talopyranose, the hydroxyl groups oncarbons 2, 3, and 4 must be up, sochoice C is
the best answer.
Choice D is correct. Kiliani-Fischer synthesis adds one carbon to the sugar structure. Choice A is eliminated,
because a carbon is lost, not gained. Choice Bis isomerization, where the number of carbons does not change.
This eliminates choice B. In choice C, the number of carbons in the product is not listed, so the choice can neither be confirmed nor refuted. In choice D, the number ofcarbons has increased by one. This makes choice D the best
a n s w e r .
Passage IX (Questions 58 - 64) Osazone Derivative Test
58.
59.
60.
61.
62.
Choice C is correct. The unknown disaccharide in the passage is comprised oftwo cyclic monosaccharides formed
from aldohexoses (aldopyranose structures). Acyclic aldohexose molecule has five stereocenters (one at every
carbon except the terminal carbon, carbon 6). Given that each monosaccharide has five chiral carbons, a
disaccharide must have ten stereocenters total. Pick choice C.
Choice Bis correct. Choice Ais eliminated, because the structure on the left is glucose. Neither sugar can be glucose, because the specific rotations of the osazones of both sugars do not match the specific rotation of the osazone of glucose. Choice C is eliminated, because the structure on the left is mannose, the C-2 epimer of glucose. Neither sugar can be mannose, because the specific rotations of the osazones of both sugars do not match the specific rotation of the osazone of glucose (which would also be the osazone of mannose). Choice D is eliminated, because the two structures are C-2 epimers of one another. Because the two sugars form osazones with different specific rotations, the two sugars cannot be C-2 epimers ofone another. In choice B, neither of the
sugars will lead to an osazone that matches either glucose or mannose, neither sugar leads to an optically
inactive aldaric acid, and the two sugars are not epimers of oneanother.
Choice D is correct. When an aldopentose isoxidized with nitric acid to form an aldaric acid, the two carbons to observe for optical activity are carbons 2and 4. If the hydroxyl groups on carbons 2 and 4 are on the same side of the chain in the Fischer projection, then the compound is meso and thus optically inactive. It is only in choice D
that the two hydroxyls oncarbons 2 and 4are on opposite sides. Choice D is the best answer.
Choice Dis correct. The osazone of Compound Ihas specific rotation +42.6°, the osazone of Compound II has specific rotation +37.2°, and glucose yields anosazone with a specific rotation of +54.6°. The osazone of mannose will also show an optical rotation of +54.6°, because mannose is the C-2 epimer of glucose. This means that the unknown sugars cannot be either glucose or mannose (selection I). It is possible for the unknown sugars to be either talose or galactose according to the osazone test. Because the aldaric acids of both Compound Iand Compound II are optically active, they are not meso. Galactose leads to an optically inactive (meso) aldaric acid. This means that the unknown sugars cannot be galactose (selection III). Compounds I and II cannot be mannose or
galactose, making choice D the best answer. For those of you who chose B, because you forgot it was a
"CANNOT" question, try to limit your scream to a few seconds... "AAAAHHHHHHHHHHHHHHHHH!!!"
Choice C is correct. Choice A will yield an optically inactive aldaric acid when treated with nitric acid, because it will have a mirror plane slicing the carbon-carbon bond between carbons three and four. Neither
Compound I nor Compound II yields a meso aldaric acid, so choice Acannot be Compound I or Compound II.
Choice Bis mannose, the C-2 epimer of glucose. Choice Dis fructose, the ketose isomer of glucose. Both choices B and Dare eliminated, because they will both yield the same osazone as glucose and Compounds Iand II generate a different osazone than glucose. Choice C will not lead to an optically inactive aldaric acid, because it will not
bemeso after oxidation. The best option for you Is choice C.
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63. Choice D is correct. Identical osazones will form from sugars with matching chiral centers from carbon three down. This means that C-3 and C-4 epimers will form different osazones, so choices A and B are eliminated.
Choice C is a pair of enantiomers, which would generated enantiomeric osazones, so choice C eliminated.
Fructose is the ketose formed when mannose isomerizes from an aldose into a ketose. Mannose and fructose are identical from carbon 3 on, so they form identical osazones. The best choice is thus answer D.
H^°
HO- HO"
H- H-
H 0
•H 3eq N~ N
H h H
OH OH
CH2OH
D-mannose
H^ .N—NH3
CH2OH Osazone product
CH2OH
D-fructose
64. Choice D is correct. Because the final osazone product after treatment of the sugar with three equivalents of phenyl hydrazine contains two phenyl hydrazine moieties, it is safe to assume that two of the three equivalents are substituting onto the sugar. The third phenyl hydrazine gains two hydrogens to form ammonia (NH3) and aniline (C6H5NH2). The gain of hydrogen is reduction, thus one of the three phenylhydrazine
molecules is reduced. Tiie best answer is thus choice D.
Passage X (Questions 65 - 71) Glycolysis Reactions
65. Choice D is correct. Glyceraldehyde-3-phosphate and dihydroxyacetonephosphate are linear sugar structures with the same formula, so they are isomers of some sort. Dihydroxyacetonephosphate has no chiral center, while glyceraldehyde-3-phosphate has one chiral center, so the two structures cannot be configuration isomers (isomers that have the same bonds, but different spatial arrangement). Epimers and diastereomers are configurational isomers, so if the two structures are not configurational isomers, they cannot be diastereomers or epimers. This eliminates choices A, B, and C. An aldose and ketose of the same carbon length are structural isomers. In this example, they each have one phosphate group, so choice D is the best answer.
ketose with no chiral centers
CHnOPOo
CH7OH
C3H506P2-
(same formula but different bonds)
H^°
H- OH alcbse with one chiral center
CH20P032-
Dihydroxyacetonephosphate Glyceraldehyde-3-phosphate
66. Choice D is correct. Carbon 4 of glucose becomes the aldehyde carbon in glyceraldehyde-3-phosphate. It loses its original chirality when it takes on sp2-hybridization (associated with the aldehyde carbon), thus it has no chirality in the end. Galactose (the C-4 epimer of glucose) should therefore yield the exact same products as glucose if the enzymes were able to recognize galactose. The remaining chiral centers are identical between D- glucose and D-galactose, so the chirality of the products is the same. The best answer is choice D.
67. Choice C is correct. The lower three carbons (carbons 4, 5, and 6) of D-glucose form the glyceraldehyde-3- phosphate molecule, so carbon 5 finishes as the middle carbon (C-2) of glyceraldehyde-3-phosphate. Choice C is the best answer. Because of isomerization, the label is also found at C-2 of DHAP, but that is not a choice.
HN^°
H- HO-
H -
OH -H
OH
Step I
HV^°
H- HO-
H -
OH -H
OH H-*C-OH
1
SteP g HO-
H-
CH2OH
O
H Step HI HQ.
OH H- OH
CH2OP032"
_H Step IV
OH
CH20P032-
(=°
CH2OH
H - * C - O H
1
H-*C-OH
1
CH2OH CH2OP032-
H-*C
1
CH2OP032-
H - * C - O H
1
CH2OP032- Copyright © by The Berkeley Review® 149
CH2OP032- CARBOHYDRATES EXPLANATIONS!
68. Choice C is correct. This is a question of Hess's Law as applied in thermodynamics. The overall free energy change is the sum of the individual free energy changes for Step I through Step IV. The math to determine the sum of the four steps is carried out as follows: -4.0 + (0.4) + (-3.4) + (5.7) = -7.4 + 6.1 = -1.3, which is choice C.
69. Choice B is correct. The two terminal carbons in dihydroxyacetone phosphate (DHAP) have two equivalent hydrogens and the middle carbon has sp2-hybridization, so none of the carbons are chiral. The first carbon in
glyceraldehyde-3-phosphate (G-3-P) has s/?2-hybridization and the third carbon has two equivalent
hydrogens. Only the second carbon of G-3-P has four different substituents attached to it. The total number of chiral centers between the two compounds is therefore just one. The best answer is choice B.
70. Choice A is correct. Fructose is a ketohexose with a
furanose ring structure. The phosphate group is Fructose, a-D-fructofuranose, fructose-6-phosphate,
CH2OH
ketone on carbon 2, so carbon 2 is the anomeric carbon in the located on the last hydroxyl group (carbon 6) of fructose, and a-D-fructofuranose-6-phosphate are shown below.
CH2OH
= o
— H O
H
CH2OH CH2OH
HO- H - H -
OH OH
6CH2OH
Fructose
P
OH OH
a-D Fruc tofuranse
Choice C should be eliminated, because carbon 3 because the phosphate is on the wrong carbon.
HO- H - H -
OH OH
6CH2OP03^2-
Fructose-6-phosphate a-D-Fructofuranse-6-phosphab has no hydroxyl group. Choices B and D are eliminated,
^>
6CH2OP032"
•O
OH
CH2OH
OH
71. Choice A is correct. This question tests your knowledge of glycolysis. Glucose is a six-carbon sugar while pyruvate is a three-carbon structure, so two pyruvates form from one glucose molecule. This eliminates choices C and D. Early in glycolysis, two ATPs are invested, but two ATPs are formed per pyruvate, so the net result is that two ATPs are formed in glycolysis. This makes choice A the best answer.
Passage XI (Questions 72 - 78) Blood Types
72. Choice D is correct. The linkage can be determined directly by looking at the two sugars of interest out of the three sugars in the type-O blood antigen. N-acetylglucosamine is the first sugar linked to the protein (the N- acetyl should give this away), thus the galactose must be the second sugar from the protein in the antigenic determinant trisaccharide. The linkage is from carbon 1 of the galactose to carbon 4 of N-acetylglucosamine by way of the 6-anomer ofgalactose. This means that the linkage is a 6-1,4- glycosidic linkage, so choice D is best.
The structure is drawn below.
HO 6I CH2OH
hoVvt %°^T^i O 1 3 NF
IHO
HO
oW 9
protein CH,