Choice B is correct. A shortcut for evaluating the- 123docz.net

stereocenters is drawn below:

Method 1

H- HO- H-

S for hydroxyls R for hydroxyls

on the left. CH2OH on trie right.

D-Xylose

For sugar is in the standard Fisher projection, hydroxyls on the right have R chirality and hydroxyls on the left have S chirality

•H OH

o r Method 2

H^O

H HO

OH H OH

Solving the first chiral center yields:

CH2OH D-Xylose

Draw the Fisher projection with 3-dimensional accuracy. The Hs are in front in Fisher projections.

CH2OH D-Xylose

Counterclockwise normally indicates an S stereocenter, but because the hydrogen is coming out of the plane,

the stereocenter is reversed to R.

The compound has chirality of 2R, 3S,4R,so choice B is the best answer.

87. Choice Bis correct. From the balanced equation, there are six C02(g) formed for every one glucose thatreacts. To solve this question, you must convert the glucose from grams tomoles, convert moles ofglucose tomoles ofC02(g), and then convert moles of C02(g) into grams of C02(g). This process requires dividing by the molecular mass of glucose (180), multiplying by the ratiofrom the balanced equation for carbon dioxide and glucose (6:1), and then multiplying by the molecular mass of carbon dioxide (44). The long-hand math goes as follows:

10.00gC6H12O6 x Imole x 6C°2 x**gE™ =lOx^x* x44gC02

180 grams 1C6H1206 lmole 180 1

The set up is what is requested in the question, so the best answer is choice B.

88. Choice B is correct. The best explanation for the difference in heats of combustion between the aldohexoses (glucose, galactose, and mannose) is structural in nature. They all have the exact same bonds (thus they are stereoisomers) so according to the simplistic application of Hess's Law, they should yield the same amount of heat. Because the heats are so different, there must be a difference in stability. The more stable the molecule, the less heat that will be given off. Because all of the aldohexoses have the same bonds, choice A is eliminated. Sugars are neither aromatic, nor do they have any stabilizing resonance forms. This eliminates choices C and D. The only choice left is answer B, stating that the sugars have different orientation is space and thus different hydrogen bonding. The sugars also exhibit differences in their steric interactions. The best answer

of the choices is choice B.

89. Choice C is correct. The given in this question is 15 grams ribose which, when divided by the molecular weight of ribose (150 g/mole), gives 0.10 moles of ribose. The AHcombustion for ribose is listed in Table 1 as -2076 kj/mole. This means that the 0.10 moles of ribose generates 207.6 kj (207,600 J) of heat when it is burned (oxidized). Plugging into the equation E = mCAT yields - K, which is approximately

1000 x 4.18 1000 x 4

(roughly 50), making choice C the best choice. The mathematical layout is drawn for you below:

15gC5H10O5 x lmole x 2Q76kT = 2076kJ &E=mCAT /. AT = -E- = 207'600 =2HIA = 200 s 5(rc

150 grams lmole mC 1000x4.18 4.18 4

It is doubtful that you will see a calculation this lengthy on your MCAT, but you should still understand the set up and the theory behind the math.

90. Choice B is correct. The most C02(g) results from the combustion of the compound with the largest mass percent of carbon. This is because for every carbon within a sugar, one C02(g) molecule is formed. Therefore, the more carbon by mass in the compound there is, the more C02(g) that is formed upon complete oxidation. The mass percent of carbon in any monosaccharide is 40%. The mass percent of carbon in sucrose (C12H22O11) is greater than the mass percent of monosaccharide, because the difference in the ratio of C:H:0 between sucrose and the

monosaccharides is due to a loss in H and O from the water. This increases the relative abundance of carbon in

sucrose compared to a monosaccharide giving sucrose a greater mass percent of carbon than a monosaccharide.

The mass percent ofcarbon in sucrose is: 1^4 _ 144 _ _7_2_ an(j _72_ > _72_ =0.40 .-. -Z2_ > 40%. Because the

144 + 22 + 176 342 171 171 180 171

mass percent of carbon in sucrose is greater than 40%, more C02(g) will be produced from the combustion of one gram of sucrose than from the combustion of one gram of any monosaccharide. Choose B and be happy.

91. Choice A is correct. For this, you must use the AH in terms of kj per gram, where the kj/g value is found by dividing the kj/moles by the molecular mass of the sugar. Monosaccharides have masses that are multiples of 30 g/mole, so the numerators should be simple to work with. The following calculations show the energy per gram values for each of the sugars given in the answer choices. The correct answer is the sugar with the highest

numerical value.

Glucose:2538=423 Xylose : 21Q2= 4204 Ribose :2QZfc=il5£

180 30 150 30 150 30

Ribose gets eliminated immediately, because it is equal in mass to xylose but produces less energy. The possible answers are narrowed down to either glucose or xylose. Instead of solving the math, it is easier to reduce the two values to something over a common denominator (in this case 30). Glucose results in the greatest numerator, making choice A correct.

Not Based on a Passage (Questions 92 -100) Carbohydrates

92. Choice D is correct. Addition of an excess of periodic acid to a monosaccharide results in the oxidative cleavage of all of the carbon-carbon bonds in the sugar. Aldehydes are oxidized into formic acid , primary alcohols are oxidized to formaldehyde and the secondary alcohols associated with internal carbons are oxidized by losing both carbon-carbon bonds to form formic acid. Tiie product distribution for the decomposition of glucose is drawn

below:

H^°

H- HO"

H- H-

OH H

OH OH

CH2OH D-glucose

>• 5

H" OH

M H

In the end, the full oxidation of glucose yields 5 formic acids and 1 formaldehyde. Choose D for correctness.

93. Choice C is correct. The molecular formula for xylose is C5H10O5. The molecular weight of xylose (or any other aldopentose or ketopentose) is 60 + 10 + 80 = 150 grams/mole. The mass of the five carbons is 60 grams. This means that the mass percent of carbon in xylose is 60/150 x 100% = 40%. As a point of trivial interest, the mass percent of carbon in any t7ionosaccharide is always 40%. Choose C and feel happily correct.

94. Choice A is correct. In order to be optically inactive, the compound must be either meso or achiral. In the case of sugar derivatives, the diacid (aldaric acid) formed after oxidation of the terminal carbons must be meso to be optically inactive. D-mannose has stereochemistry of 2S, 3S, 4R, 5R, which does not yield a meso aldaric acid when oxidized, so the aldaric acid derivative of mannose is optically active. D-galactose has stereochemistry of 2R, 3S, 4S, 5R, which yields a meso aldaric acid which is optically inactive. D-ribose has stereochemistry of 2R, 3R, 4R and is five carbons long (which you should know from memory), so the third carbon would no longer be a chiral center in the aldaric acid form. The aldaric acid of D-ribose is meso and therefore optically inactive.

D-allose has stereochemistry of 2R, 3R, 4R, 5R, which would form a meso aldaric acid which is optically inactive. The aldaric acid of D-allose is meso and therefore optically inactive. The only sugar to form an optically active aldaric acid derivative of the given choices is D-mannose. The best answer is choice A.

HO. . O

HO- HO- H- H-

•H

•H OH OH

HO^O

Aladirc acid derivative of D-Mannose

(not meso .\ active)

HO. . O

H- HO- HO- H-

•H

•H OH

HO^O

Aladirc acid derivative of D-Galactose

(meso .•. inactive)

HO. . O

H-

—H- H-

•OH OH—

HO^O

Aladirc acid derivative of D-Ribose

(meso /. inactive)

HO. . O

H- H- H- H-

OH OH OH OH

HO^O

Aladirc acid derivative of D-Allose

(meso /. inactive)

CARBOHYDRATES EXPLANATIONS!

95. Choice A is correct. When treated with nitric acid, the two terminal carbons (carbon 1 and carbon 5) of an aldopentose get oxidized into carboxylic acid functionalities. This results in the formation ofa 1,5-dicarboxylic acid (aldaric acid). After the oxidation reaction, only carbons 2 and 4 can be chiral centers, because the third carbon now has two identical substituents attached to it (making carbon 3 achiral.) In a D-aldopentose, carbon 4 has R stereochemistry. This would mean that carbon 2 must also have R stereochemistry in the original sugar, although that changes in the aldaric acid, because the priorities are different. Only choice A has this feature.

The stereochemistry of carbon 3 is irrelevant.

HỘ0 HÔO

HV° H O . . O

H- H- H-

(OH) H-

H-

(OH) i-ony

H-

•(•Grr>

H- OH

-(OH)

•(OH)

CH2OH D-Aldopentose (opticallyactive)

HNQ

HO O HO^O

Aldaric acid derivative Inactive, so it must be meso

of D-Aldopentose OH-4on right .*. OH-2 on right (opticallyinactive) OH-3 is unknown b/c C-3 is achiral

Pằ

H- OH

CHjOH

2R, 3 unknown

96. Choice C is correct. Ribose differs from deoxyribose in that carbon 2 has no hydroxylgroup in deoxyribose. This means that carbon 2 does not have four unique substituents, so it is not a chiral center. This makes choice B a valid statement, which eliminates it. By having one more chiral center, ribose has more possible stereoisomers, making choice A a valid statement, and thus it is eliminated. Ribose, by having a hydroxyl group rather than a hydrogen on carbon 2, is more oxidized than deoxyribose. This makes choice D a valid statement, which eliminates it. Ribose and deoxyribose do not have the same molecular formula, so they cannot be stereoisomers, let alone epimers. This makes choice C a false statement and therefore the best answer.

97. Choice D is correct. When two cyclic monosaccharides form a disaccharide, a glycosidic linkage is formed.

Linkage formation occurs when a hydroxyl on one sugar attacks the anomeric carbon on the other sugar. A water molecule is released as the leaving group, as the hemiacetal with the reactive anomeric carbon gets converted

into an acetal. This makes choices A, B, and C all true. The anomeric carbon starts and finishes with two bonds

to oxygen, one bond to carbon, and one bond to hydrogen. This means it is neither oxidized or reduced, so choice D is not observed, making it the best answer.

HOP^C

OH + HO

HOH,C H2? HOH>C Glycosidic linkage

h\ ori

\ OH

Acetal because

anomeric C is bonded to:

OR, OR', C, and H Hemiacetal because

anomeric C is bonded to:

OR, OH, C, and H

98. Choice A is correct. When copper goes from an oxidation state of +2 to +1, it has been reduced. The color of the solution changes when copper changes oxidation state, so a color change corresponds to an oxidation-reduction reaction involving copper ion and the sugar. Because copper is reduced, the sugar must be oxidized. If an aldose forms a carboxylate (deprotonated form of the carboxylic acid), it has been oxidized. This makes choice A a correct answer. If a sugar forms a compound with only hydroxyl functional groups and no carbonyl group, it has been reduced. This eliminates choice B. If an aldose becomes an acetal, the number of bonds to oxygen and hydrogen has not changed, so it has neither been oxidized or reduced. This eliminates choice C. Ketones do not oxidize into carboxylic acids, so choice D is eliminated.

99. Choice A is correct. A trisaccharide has two glycosidic linkages. For every glycosidic linkage, one water molecule is lost. An aldohexose has a formula of C6H12O6 while a aldopentose has a formula C5H10O5. A trisaccharide from two aldohexoses and one aldopentose has seventeen carbons, so choices C and D are eliminated. Choice B represents the sum of the formulae of all three monosaccharides, but with the loss of two water molecules, the formula is C17H30O15, making choice A the best answer.

100. Choice B is correct. An aldopentose has one 7t-bond and no rings. It can form a cyclic structure, but that has no it- bonds and one ring. Either way, the compound has only one unit of unsaturation. The best answer is choice B.

"How sweet it is to finish sugars!"

Section VII Nitrogen Compounds

by Todd Bennett

Histidine

PKa(R) = 607 (Basic Amino Acid)

Nitrogen Compounds (Non-Biological)

a) Amines

i. Basicity and Acidity ii. Huelcophilicity

iii. Formation of Amines iv. Reactions of Amines v. Hofmann Elimination

b) Imine Chemistry

i. Formation of Imines

ii. Imine-Enamine Isomerization c) Amides

i. Structure

ii. Formation of Amides iii. Reactions of Amides d) Amino Acid Synthesis

i. Hell-Volhard-Zelinskii Synthesis ii. Strecker Synthesis

iii. Reductive Amination of a-Keto Acids

Nitrogen Compounds (Biological)

a) Amino Acids

i. Strcutures and Classification ii. Isoelectric Points

iii. Easy Calculation of pi b) Proteins

i. Structure Features ii. Structural Levels

c) Biological Protein Processes d) Biochemistry Lab Techniques

i. Gel Electrophoresis ii. Affinity Chromatography

iii. Sequencing (Primary Structure) iv. Cutting/Fragmentation

v. Edman's Reagent

Choice B is correct. A shortcut for evaluating the chirality of sugars in the Fischer projection is to realize that all hydroxyl groups on the right side of the backbone are located on an Rstereocenter while all hydroxyl groups

Questions 1-6) Alcohols, Acidity, and Nucleophilicity

Questions 8 -14) Fischer and Haworth Projections