Choice Dis correct. The ratio of the integrals for- 123docz.net

Passage VIII Questions 47 - 53) Wolff-Kishner Reduction versus Clemmensen Reduction

75. Choice Dis correct. The ratio of the integrals for each signal in the proton NMR for spectrum in Figure 2is 2:2

3 : 3. This means that the major compound isolated from solution has ten hydrogens. Given that the reactant has eight hydrogens, it can be concluded that some reaction took place, which eliminates choice A. Choice B (ethyl acetate) has eight hydrogens, so choice Bis eliminated. This narrows the question down to either choice C (propyl ethanoate) or choice D (ethyl propanoate). Key features in Spectrum II include the quartet far downfield and the absence of a singlet. Choice C(propyl ethanoate) would show a singlet for the CH3 adjacent to the carbonyl. Because Spectrum II has no singlet, choice C is eliminated. The presence of the quartet downfield implies that the compound must have an ethyl group on the ester oxygen. Pick Dto know what right

feels like. The structures are drawn below:

76.

77.

Methyl Propanoate a: Triplet (3H) b: Quartet (2H) c: Singlet (3H)

a ^^^ b c

H3C OCH2CH3

Ethyl Ethanoate a: Singlet (3H) b: Quartet (2H) c: Triplet (3H)

a y—\ b e d

H3C OCH2CH2CH3

Propyl Ethanoate a: Singlet (3H) b: Triplet (2H) c: Sextet (2H) d: Triplet (3H)

a b

H3CH2C

X OCH2CH3c d

Ethyl Propanoate a: Triplet (3H) b: Quartet (2H) c: Quartet (2H) d: Triplet (3H) The bolded hydrogen will be found the furthest down field.

Choice Bis correct. An ester is more reactive than all carbonyl compounds that have aleaving group worse than the alkoxide leaving group. Leaving groups worse than the alkoxide leaving group would be more basic than the alkoxide anion. Because both halides and carboxylates are good leaving groups (as well as weak bases), acid halides and acid anhydrides are both more reactive than esters. This eliminates choices A, C, and D. The

only choice remaining is amides, choice B. The NR2" leaving group of the amide is a bad leaving group,

therefore amides are not very good electrophiles. Choice B is the best selection.

Choice C is correct. By using an amine nucleophile rather than an alcohol nucleophile, the product that is formed is an amide as opposed to an ester. The amine is a better nucleophile than the alcohol is in Reaction II, so the reaction should be more favorable than transesterification. The equilibrium constant for the reaction is greater than one. Regardless, the question asks for the product of the reaction, which is the amide formed when the amine replaces the alcohol leaving group. The amine nucleophile is N-ethylamine therefore the best answer is N-ethylpropanamide (H3CCH2NHCOCH2CH3), choice C. This can be determined by just adding the ethyl amine in same way as the ethanol is added to the ester in the sample reaction carried out at reflux.

78. Choice Bis correct. The purpose of the acid in the reaction mechanism is to protonate the carbonyl to make the ester more electrophilic. In a later step, the acid protonates the alkoxy leaving group to form an alcohol, which is abetter leaving group than alkoxide. The role of the acid in the reaction is best stated as serving to protonate the ester electrophile and in doing so make itmore reactive. This makes choice Bcorrect and desirable to you.

79. Choice Dis correct. If water were the nucleophile in lieu of the alcohol, the product would be a carboxylic acid.

When carried out under basic conditions, this reaction is known as saponification. Hydrolysis of an ester is common in biological systems. The carboxylic acid formed in the reaction would have the OCH3 group replaced by an OH group. This carboxylic acid would be propanoic acid (CH3CH2CO2H), choice D.

80. Choice C is correct. The deuterated solvent is used so that the solvent (which normally contains hydrogens) does not interfere with proton NMR spectrum of the compound. The deuterium atoms do not register on proton

NMR, so the solvent is invisible when it is deuterated. The best answer is therefore choice C. Deuterium exhibits little to no affect on pH in organic solvents. This eliminates choice D. The solvent should have no effect on the magnetization of the NMR sample nor should thesolvent prevent any reaction as solvents are inert.

Choices A and B can both be eliminated.

Passage XIII (Questions 81 - 87) Malonic Ester Synthesis

81. Choice B is correct. The structure of urea must be determined by comparing the final barbiturate structure (product of the lower pathway) with the substituted malonic ester. Urea must contain two nitrogens bonded to a central carbonyl. The alkoxy groups of the ester are leaving groups when the urea substitutes. The structural deduction is drawn below. Tiie drawing represents retrosynthetic analysis of the final product. Pick choice B.

HNX NH

O o

Urea O

H2N NH> CH3CH20

O O

OCH2CH3

82. Choice B is correct. Urea has a carbonyl carbon separating the two nitrogen atoms, so if the carbonyl oxygen of urea is replaced by two methyl substituents on the carbon, then the final product will be the same as the product from urea, except that it will have two methyl groups at the formerly carbonyl carbon. It is not different at any site other than the carbon between the two nitrogens (the former carbonyl carbon). The correct choice must have the two methyl groups bonded to the central carbon in lieu of the carbonyl oxygen. The best answer is choice B.

To answer this question requires knowing what urea looks like. If you don't know, then you must predict what would happen if the diamine were added to the diester. Tiie correct choice can be deduced if you know that the product of an amine and ester is an amide.

83. Choice C is correct. Decarboxylation occurs with 6-ketoacids. All four answer choices have carboxylic acid functionalities, but only choices A, B, and D have fi-keto groups on the backbone. The ketone functionality of choice C is on the gamma carbon, therefore the molecule will not undergo decarboxylation. Tiie best answer is

therefore choice C.

A. O

H3C P I

O B. O O

OH HO 1P

OH H3C £ • O

84. Choice A is correct. The strong base used to deprotonate the alpha carbon can also act as a nucleophile by attacking the carbonyl carbon. Sodium ethoxide is chosen as the base so that it matches the substituent on the carbonyl carbon of the ester. This means that if the ethoxide undergoes transesterification, it will generate the same ester (the leaving group of the ester is the same moiety as the nucleophile). Sodium ethoxide is a strong enough base to deprotonate the alpha hydrogen, but it will not change the ester when it undergoes

transesterification. The best answer is choice A.

85. Choice D is correct. Malonic ester has three types of protons, thus there are three signals in its proton NMR spectrum. The structure of malonic ester is shown below with each of the hydrogens labeled:

H3CH2CO'

O O

C\ / C \ b a

C OCH?CHo

H c

a: 6 hydrogens adjacent to 2H thus a 3H triplet b: 4 hydrogens adjacent to 3H thus a 2H quartet c: 2 hydrogens adjacent to OH thus a 1H singlet

Based just on the ethyl groups, resulting in a 2H a quartet and a 3H triplet, the best answer is choice D. The true ratio of hydrogens is 6 : 4 : 2, but because NMR shows only the relative quantities and not the absolute quantities, the NMR shows a ratio of 3 : 2 : 1.

86. Choice B is correct. There are four carbons added to the central carbon of malonic ester, so choices A and C can both be eliminated. The addition of methyl bromide followed by the addition of ethyl bromide to malonic ester only adds three carbons total, not four. This eliminates choice D. A secondary alkyl bromide like 2- bromobutane must be added to form the tertiary R substituent. Choice B is the best choice.

87. Choice Bis correct. The product will be the exact product shown in the lower synthesis pathway in Figure 1, only the generic R group is now a straight chain propyl group. This makes choice B the best answer.

Passage XIV (Questions 88 - 94) Carboxylic Acids

89.

90.

91.

Choice D is correct. To synthesize butanoic acid, one ofseveral methods can be applied. Oxidation ofa primary alcohol will work, so choice A is feasible. Hydrolysis of an ester will form an acid, so the four carbon carbonyl fragment of the ester will become a four carbon carboxylic acid. Choice B is a viable synthetic pathway.

Ozonolysis of an alkene works if the addition of ozone is followed by oxidative workup and the alkene must have vinylic hydrogens (hydrogens on the carbons of the rc-bond). Choice C is thus valid. Treatment of an alkyl bromide with the cyanide nucleophile followed by acidic workup forms a carboxylic acid. The problem with choice D is that the carbon chain increases by one carbon with the cyanide nucleophile so the product will be

pentanoic acid, not butanoic acid. The best answer is choice D.

Choice D is correct. Because the pKa for propanoic acid is 5.0, it is a weak acid, implying that it only

partially dissociates in water. With a pKa value of 5.0, the Ka value is 1.0 x 10"5. This means that for a 1.0 M acid solution, the concentration of both H+ and conjugate base are 10"2-5, which is less than 10"2, or 1%. The dissociation is less than 1%. The Ka value is applied as follows:

. [CH3CH2C02-][H30-] =[CH3CH2CQ2f = ^ ^ = ^

[CH3CH2C02H] [CH3CH2C02H]

This implies that the relative value of [CH3CH2C02"j to [CH3CH2C02H] is VlO x 10"3 to 1 which is roughly

3.1 x 10"- to 1. This value is less than one percent confirming choice D.

Choice Bis correct. A carboxylic acid will turn blue litmus paper red when added to it. From the nature of the passage, it should be implied that the litmus paper is used to detect the acid. In choice A, a primary alcohol is treated with chromic oxide and sulfuric acid, which oxidizes the primary alcohol into a carboxylic acid.

Choice A results in a product that will convert blue litmus paper to red, so choice A is valid. In choice B, the alkene is tetrasubstituted, therefore the carbonyls that are formed must be ketones. They cannot be oxidized (or reduced) into a carboxylic acid. This means that the product in choice Bwill not convert blue litmus paper to red, so choice B is the best answer. In choice C, the anhydride is readily hydrolyzed into two equivalents of acetic acid. This eliminates choice C. In choice D, an ester is hydrolyzed into an alcohol and a carboxylic acid.

This eliminates choice D. The best answer is choice B.

Choice D is correct. The sequence of reagents used will add a carboxylic acid group to the alcohol carbon (carbon 2). It required background information to know that PBr3 converts an alcohol into an alkyl bromide. If you didn't know that before, then it's officially background knowledge now. This final product is composed of a carbon with an ethyl group, methyl group, and a carboxylic acid group attached. This makes choice D the best answer. Choice C should have been eliminated, because it contains six (rather than five) carbons.

92.

93.

Choice Bis correct. As a general rule, hydrolysis reactions are reversible, because each step is an equilibrium

step. The Grignard reaction is not reversible however, therefore the best answer is choice B. Choices A and D should have been eliminated, because a lactone is an ester (a cyclic ester), thus they are the same answer.

Choice D is correct. As a general rule, the pKa value of an acid is roughly 5.0. This makes choice D not true.

Acids do result inaqueous pH values less than 7.0, because they are acidic (choice A). The broad IR peak results from the hydrogen bonding and is found between 2500 cm"1 and 3000 cm"1 (choice B), less than an alcohol stretch because the bond is weaker (and thus easier to stretch). Acids will exchange protons with a protic solvent, so the acidic hydrogen can readily be deuterated. This results in the disappearance of a peak in the proton NMR

(choice C).

94. Choice D is correct. As listed in the passage, the addition of thionyl chloride (SOCl2) followed by an amine to a carboxylic acid will result in the formation of an amide. This eliminates choices A and C. An aldehyde is oxidized into a carboxylic acid by the addition of H2S04 and Cr03 as shown in the passage. The best answer is

therefore choice D.

Not Based on a Passage (Questions 95 -100) Carbonyls and Alcohols

95. Choice A is correct. Under basic conditions, the alkoxy nucleophile group attacks the s/?2-hybridized carbonyl carbon of the ester (see below) to form an s/?3-hybridized tetrahedral intermediate with the negative charge

shifted to the carbonyl oxygen. The best answer is choice A.

R'O R"0"

R OR"

Intermediate

R OR'

96. Choice B is correct. The Claisen condensation reaction involves the deprotonation of an alpha hydrogen on an ester, resulting in a highly nucleophilic carbanion, which attacks another ester in solution. The ultimate product of a Claisen condensation reaction is a 6-ketoester. The Claisen condensation reaction definitely uses an ester reactant, so choice A is eliminated. A Grignard reaction involves the addition of a nucleophilic alkyl magnesium bromide species to a carbonyl compound. An ester certainly qualifies as a carbonyl compound. Tiie Grignard reagent can add twice to an ester, resulting in the synthesis of a tertiary alcohol with at least two identical alkyl groups. The Grignard reaction can use an ester reactant, so choice C is eliminated.

Transesterification, as the name implies, involves the changing of an ester by changing the alkoxy group. This is done by substituting one alkoxy group for another on the carbonyl carbon. A transesterification reaction definitely uses an ester reactant, so choice D is eliminated. Friedel-Kraft alkylation is not a reaction you are required to know, but that is irrelevant in this question. The term alkylation implies that an alkyl group is being added, and this does not lend itself to the use of an ester. An ester is likely to add an acyl group when it reacts. This means that a Friedel-Kraft alkylation does not require an ester reactant, so choice B is the best

a n s w e r .

97. Choice A is correct. Treatment of a fi-diacid with heat will drive off water (dehydrate the 6-diacid) to form the corresponding acid anhydride. Choice A is an acid anhydride, so it is the best answer. The reaction is reversible, so acids can be formed from anhydrides upon the addition of water.

98. Choice C is correct. In order to be reduced, the compound must have a carbon with at least two bonds to oxygen (both a CT-bond and a 7t-bond from carbon to oxygen). Only choice C (a ketone) has two bonds to oxygen, so it must be the best answer. Phenols, tertiary alcohols, and ethers all involve carbons that only have one bond to oxygen, so they are all eliminated.

99. Choice C is correct. Carbon number three, the third carbon, of the alcohol is involved in no 7t-bonds and it is

bonded to four other atoms, so it must have sp3-hybridization. Pick C for optimal results.

100. Choice C is correct. Secondaryalcohols, when oxidized, form ketones. The product following the oxidation of 2-butanol using a chromium VI species is 2-butanone. The product is drawn below along with its proton NMR peaks. Pick choice C.

O|| a: 3 hydrogens next to a carbonwith no hydrogens: Singlet (3H) C b ^*CH3 b: 2 hydrogens next toa carbon with3 hydrogens: Quartet (2H) H-C CH2 c: 3 hydrogens next to a carbonwith 2 hydrogens: Triplet (3H)

C4J%0 f

/ C

i I

1 \ \ I 1 1

2 ppm 1 ppm1 Oppm1

:y,\-/r}-4??'K'^l ^?^W'^>]Ay --"• iT> '%'>'* >

"Better living through chemistry!"

Section VI

Carbohydrates

by Todd Bennett D-Glucose Hj>0

O^OH

H- HO- H- Hô

OH H OH OH

CBjOH

OH HOH2C

} 33 2H 3 OH

2 H OH

fi-D-glucopyranose

HOH2C HOH2C

4-0-(6-D-glucop3n:anosyl)-6-D-glucopyranoside

Monosaccharides

a) Straight Chain Form i. Aldoses and Ketoses ii. Pentoses and Hexoses

iii. Common D-Sugars b) Stereoisomerism

Epimers

ii. Anomers

iii.Haworth Projections

iv. Mutarotation v. Isomerization

Oligo- and Polysaccharides a) Linkages (Acetal Connectivity) b) Disaccharides

c) Trisaccharides d) Polysaccharides

Linkages

ii. Cellulose and Starches

iii. Branching

Chemical Reactions and Tests a) nitric Acid Sugar Oxidation b) Osazone Test

c) Benedict's Test d) Tollen's Test

e) Periodate Oxidation f) Kiliani-Fisher Synthesis g) Ruff Degradation

h) Wohl Degradation

Biological Applications

a) Blood Types b) Glycolysis

Berkeley

JLJr-E-V-KE'W®

Specializing in MCAT Preparation

•>

Carbohydrates

Section Goals

Be able to recognize common monosaccharides.

There are certain monosaccharides that recur in both organic chemistry and bicKhemistrythat you should recognize andbeable todraw. These include glucose, ribose, fructose, mannose, andgalactose.

It may be easiest to recall how the other sugars relate to glucose. For instance, mannose is the C-2 epimer of glucose. If you know the structure of glucose, then you know the structure of mannose.

Be able to inter-convert between Fischer and Haworth Projections.

Some passages and questions will assume that you can translate between structures. For instance, the Fischer projection may be given in the passage, but the question may center around the structure

in a Haworth projection. In translating structures, there arethree separate points to observe. The chirality ofthe anomeric carbon is dependent onthe direction of the attack. The chirality ofthe penultimate carbon (carbon five in aldohexoses) is constant according to theD or Lstatus, and the backbone hydroxyl groups on the left in the Fischer projection are up in the Haworth projection.

Be able to solve the chirality of an unknown sugar from reaction data.

Fromorganicchemistryclass,you may recall solvingthe identity of a sugar by evaluating the reaction data presented. This makes for an ideal passage, although the MCAT writers rarely use this information in a passage, just single questions. \ou mayrecall evaluating whether a sugar would oxidize into an optically active or inactive diacid.

Be able to recognize the chirality of cyclic and linear sugars.

It is rather simple to solve for the chirality of selected carbons in both the Fischer and Haworth projections, after you have done this once. The rules are the same, so determine a quick way to identify the R and S configuration of eachhydroxyl group of tine sugar.

Be able to identify common disaccharides.

As with the monosaccharides, there are certain disaccharides thatrecur in organic chemistry and biochemistry. The disaccharides you must recognize include sucrose, lactose, and maltose. Know the two monosaccharides that comprise the disaccharide and the glycosidic linkage bindingthem.

Be able to recognize the linkage associated with a given disaccharide.

The linkage of a disaccharide is definedby the chiralityof the anomericcarbon and the carbon of the sugar containing oxygen ofthe linkage. The disaccharide that presents thegreatest difficulty is sucrose, which is composed of two anomeric carbons (one that is a-1 and the other that is fi-2).

Be able to distinguish epimers and anomers.

You mustbe able to identify the difference in chirality between diastereomers, whichin the sugar questions are commonly presented as either alpha or beta in regards to an anomer or epimers.

Know the common reactions involving sugars and sugar derivatives.

This booklet contains a largenumberofsugarreactions, perhapsmorethan you need. Keyreactions include theKiliani-Fischer synthesis, Ruff degradation, osazone formation, and nitric acid oxidation.

The passage usually provides plenty of reactions, so don't memorize them, understand them.

Know the common polysaccharides and their biological significance.

You must recognize the difference between glycogen and cellulose. Thestructuraldifference simply involves the linkage,but the difference in reactivity is significant.

Organic Chemistry Carbohydrates

Carbohydrates

Carbohydrates are organic compounds that contain carbon, oxygen, and

hydrogen in a 1 : 1 : 2 ratio. The term is derived from carbon that has been

hydrated in a ratio of one carbon to one water, which explains the C : H : O ratio of 1 : 2 : 1. When we consider carbohydrates, we typically think of them as sugars. For this section, we will focus exclusively on sugars and their organic chemistry and biochemistry applications and examples. While we take a decidedly organic chemistry perspective on this material, you should pay extra attention to any subjects that bridge organic chemistry and biochemistry. Some of this material will overlap with topics in the metabolic components and metabolic pathways sections of the biology books, but it will be presented in a manner that aims at chemistry concepts and short cuts.

We shall consider rules and definitions for sugars first. There is generic nomenclature that describe sugars by their functionality and carbon count. In addition to knowing generic nomenclature, you must be able to identify specific examples of aldohexoses, aldopentoses, and ketohexoses. Common sugars that you should recognize in all traditional structural representations (Fischer, Haworth, and chair) are glucose, ribose, fructose, galactose, and mannose. We will present mnemonic devices to help with the recall of these structures.

Stereochemistry is a significant part of sugar nomenclature, so it is necessary to know the stereochemistry associated with the prefixes of D and L.

Monosaccharides of five carbons or more are typically found in cyclic structures rather than straight chain structures. To form the cyclic structure, a hydroxyl group attacks a carbonyl carbon to form either a hemiacetal (if the sugar is an aldose) or a hemiketal (if the sugar is a ketose). Converting between linear and cyclic representations involves knowing how to draw the hydroxyl groups at each carbon. We shall also consider the relationship of sugars. In the cyclic form, you must be able to determine what anomer is represented. In both the straight chain and cyclic forms, epimers are possible. We shall emphasize stereochemistry terminology.

Considering much of sugar chemistry involves multiple sugars, we will address glycosidic linkages and how to recognize the type of linkage. We will consider many examples from disaccharides, to trisaccharides, to oligosaccharides, and finally polysaccharides. Alpha and beta linkages impact the reactivity and structural nature. To cleave linkages, specific enzymes are required. Of most notoriety is the enzyme to cleave alpha linkages in glycogen, amylose, and amylopectin, which allows us to break down starches. We lack the enzyme to cleave beta-linkages, so we are unable to breakdown cellulose.

We will also consider chemical reactions of sugars. There are three types of chemical reactions we shall address. One type is used to identify the chirality at specific stereocenters, such as treatment with HNO3 and osazone formation.

Another type increases the length of the carbon chain of the sugar. The last type breaks down the sugar, one carbon at a time. The last feature we will consider is the biological reactivity. We will look at glycolysis from a practical perspective, rather than the detail seen in biochemistry. We will also consider blood types and the impact of stereochemistry on the recognition of the glycoproteins that determine the blood type.

Introduction

Choice Dis correct. The ratio of the integrals for each signal in the proton NMR for spectrum in Figure 2is 2:2

Questions 1-6) Alcohols, Acidity, and Nucleophilicity

Questions 8 -14) Fischer and Haworth Projections