Organic chemistry section test (3)

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MCAT Subject Tests Dear Future Doctor, The following Subject Test and explanations contains questions not in test format and should be used to practice and to assess your mastery of the foundation content necessary for success on the MCAT Simply memorizing facts is not sufficient to achieve high scores; however, an incomplete understanding of basic science knowledge will limit your ability to think critically Think of building your content knowledge as learning the vocabulary and practicing MCAT-like questions as actually speaking All rights are reserved pursuant to the copyright laws and the contract clause in your enrollment agreement and as printed below Misdemeanor and felony infractions can severely limit your ability to be accepted to a medical program and a conviction can result in the removal of a medical license We offer this material for your practice in your own home as a courtesy and privilege Practice today so that you can perform on test day; this material was designed to give you every advantage on the MCAT and we wish you the best of luck in your preparation Sincerely, Albert Chen Executive Director, Pre-Health Research and Development Kaplan Test Prep © 2003 Kaplan, Inc All rights reserved No part of this book may be reproduced in any form, by Photostat, microfilm, xerography or any other means, or incorporated into any information retrieval system, electronic or mechanical without the written permission of Kaplan, Inc This book may not be duplicated, distributed or resold, pursuant to the terms of your Kaplan Enrollment Agreement _O R G A N I C C H E M I S T R Y S U B J E C T T E S T Organic Chemistry Subject Test Which of the following alkyl halides would be the least reactive substrate in a Williamson ether synthesis? What is the reagent X in the following reaction? CH3 CH=CHCH CH3 A X CH3 CHCHCH2 CH3 CH2 CH2CH2Br B A B C D E CH3CHCH3 Br C HCHO, heat CO2, LiAlH4 CH3OH, H+ CH3COOH, H+ CH2N2, UV light CH3C(CH3)CH3 Cl Which of the following is a pyrimidine? A D CH3CH2CHCH 2CH3 Br E CH2Cl B Which of the following will yield a racemic product mixture? C A CH3 CH3 CCH3 CH2 OH B N N N HBr H+ N N H D 1)CH3MgCl CH3CH2OCH2CH3 2) H2O N C CH3CH2CH CH2 Br2 / CCl4 N E D CH3CH2CH CHCH2CH3 KMnO4 N Heat E HCl CH2 CHCH3 K A P L A N _ O R G A N I C C H E M I S T R Y S U B J E C T T E S T Starting with a molecule of benzyl alcohol ΦCH2OH, we wish to form ΦNH2 (aniline) Which of the following syntheses should be used? [Φ = phenyl group] Addition of Br2 to 2-pentene yields A C2H5 A Φ CH2 OH KMnO4 NH3 PBr3 NaCN H Br H OBr B Φ CH2 OH Br - CH3 H2 , Ni B C Φ CH2 OH PBr3 NH4 + , H+ Cu NH3 C2H5 Br Br H H D Φ CH2 OH LiAlH CH3 E HCN Φ CH2OH H2O NH3 C C2H5 H Br Br H CH3 D C2H5 H Br H Br CH3 E More than one of the above K A P L A N _O R G A N I C C H E M I S T R Y S U B J E C T T E S T What is the major product of the reaction below? (CH3 )3 C COOH Br NaOH KMnO4 (conc.) CH3 CH=CH Predict the product of the following reaction: C ? H heat CH2 CH3 CH3 A (CH3 )3 C A H CH3 H3 C H H C C ? C HO CH2 CH3 CH3 B (CH3)3CC=CH2 C (CH3 )2 C=CCH3 OH OH B COOH H D (CH3)2CCH2CH3 OH CH3CH CH COOH E Either A or B, depending on reaction conditions C COOH COOH COOH D CH3 OH H H3 C C C H OH CH2 CH3 E No reaction will occur K A P L A N _ O R G A N I C C H E M I S T R Y S U B J E C T T E S T 10 Which of the following is an acetal? A C2 H5 KMnO4 SOCl2 HCN HOCCH3 H2 O H+ O B CH3 CCH2 CCH3 A O O C O C CH3 COOH OR C H B OH D O C CH2 OH NH2 O OH OCH3 OH OH E C COOH CH2 OH OH OH O OH OH D CH2 COO- E CN CH OH K A P L A N _O R G A N I C C H E M I S T R Y S U B J E C T T E S T 11 If the compound pictured below undergoes E2 elimination, what is the most likely product? Ha 12 Which of the choices below lists a possible sequence via which the three compounds shown, initially in a chloroform solution, may be separated? NH2 Cl Hc C2 H5 H Hb Hd A C2 H5 Hc Hb Hd B Ha C2 H5 Hd C Ha H N ; ; CH NO NO2 A Extraction with a weakly acidic aqueous solution, followed by extraction with a strongly acidic aqueous solution, followed by distillation B Extraction with a strongly acidic aqueous solution, followed by extraction with a weakly acidic aqueous solution, followed by distillation C Extraction with a strongly acidic aqueous solution, followed by extraction with an even stronger acidic aqueous solution, followed by distillation D Extraction with a strongly basic aqueous solution, followed by extraction with a weakly acidic aqueous solution, followed by distillation E Extraction with a weakly basic aqueous solution, followed by extraction with a strongly basic aqueous solution, followed by distillation C2 H5 Hb Hc D C2 H5 Hc Ha Hd E Ha C2 H5 K A P L A N _ O R G A N I C C H E M I S T R Y S U B J E C T T E S T 13 14 Which monosubstitution product(s) would be expected from the following reaction? H3C excess O + CO CH2MgBr O C H2O Br2 , FeBr3 H+ A OH C A O BrCH2 CH2 B C B O C O H3C CH2 O C C OH C CH2 CH2 Br C H3C D O C OH C O CH2 CH2 Br D H3C E Br OH C O C CH2 E A mixture of B and D K A P L A N _O R G A N I C C H E M I S T R Y S U B J E C T T E S T 15 Which one of the following compounds can be most easily converted to a Grignard reagent? A 17 A general formula for para-substituted benzoic acids is shown below The Ka of a particular para-substituted benzoic acid will be smallest when X is which of the following? Cl X COOH B COOH Cl A B C D E NH2 OCH NO2 Cl CH3 NH2 18 C Cl O CH3C CH3C OH D Cl OCH3 Cl C O The products of this reaction are A B C D E E O + 2NH3 2CH3CONH2 + H2O CH3COONH2 + HCONH2 + H2 CH3CONH2 + CH3COO–NH4+ CH3COO–NH4+ + CH=CH–O–NH2 CH3CONH2 + CH4 19 In the electrophilic aromatic substitution of phenol, substituents add predominantly in which position(s)? A B C D E ortho to the hydroxyl group para to the hydroxyl group meta to the hydroxyl group A and B A, B, and C C H 16 The base-catalyzed condensation product of two acetaldehydes is A B C D E an α-hydroxyaldehyde an α-ketoaldehyde a β-hydroxyaldehyde a β-keto aldeyhyde an α-hydroxyketone K A P L A N _ O R G A N I C C H E M I S T R Y S U B J E C T T E S T 20 Which of the following compounds will give a single proton NMR signal? I hexane II 1, 2-dichloroethane III tert-butyl chloride IV 2-methyl-2-butene A B C D E III only II and III II and IV I, II, and II None of the above 22 Which of the following compounds would be expected to be the most basic? A NH2 Cl B NH2 21 To prepare a primary alcohol with a Grignard reagent, the Grignard reagent must react only with A B C D E CH3COCH3 CH3CHO HCHO HCOOH CO2 NO2 C NH Cl D NH2 E NH2 OCH3 K A P L A N _O R G A N I C C H E M I S T R Y S U B J E C T T E S T 23 Which of the following compounds would not give a positive result under the iodoform test? A CH3CH2OH B O R1 C C OH H A B C D E CH3CH C O CH3CCH2 CH3 D O O CCH3 heat R1 H C C H + H2O R1 = R2 = H R1 = CH3 and R2 = H R1 = CH3 and R2 = Cl R1 = CH3 and R2 = CH3 R1 = Cl and R2 = Cl 27 FCH2COOH is a stronger acid than BrCH2CH2COOH because 28 The organic acid pictured below will be most acidic when X is which of the following? COOH 24 A compound produces an infrared spectrum with a sharp peak at approximately 2950 and 1700 cm–1, as well as a number of smaller peaks between 900 cm–1 and 1460 cm–1 The substance yields a negative result under Tollens’ test This substance is most likely a(n) A B C D E H R2 H+ A the more electronegative F is closer to the carboxyl group than is the less electronegative Br B Br is more electronegative than F C FCH2COOH has a smaller Ka than does BrCH2CH2COOH D FCH2COO- is a stronger base than BrCH2CH2COO- E F- is a better leaving group than Br- CH E R2 H ketone aldehyde alcohol alkane alkyne X A B C D E CH3 NH2 CH2Cl OH NO2 25 Which of the numbered hydrogens in the molecule pictured below is most acidic? O O CH2CH2CH2CCH2CH A B C D E 26 The dehydration reaction below proceeds most readily when K A P L A N _ O R G A N I C C H E M I S T R Y S U B J E C T T E S T 29 What is the major product of the reaction below? OH (CH3)3C C CH3 H H+ heat ? 30 Which of the following aromatic compounds will undergo electrophilic substitution primarily in the ortho and para positions? A NO2 A (CH3)2C CH CH3 CH3 B OCCH3 B (CH3)2C C(CH3)2 (CH3)3C CH2 CH3 (CH3)3C CH CH2 O C C COOH D E O (CH3)3C C CH3 D CN E CHO STOP! END OF TEST 10 _ K A P L A N _O R G A N I C C H E M I S T R Y S U B J E C T T E S T THE ANSWER KEY AND EXPLANATIONS BEGIN ON THE FOLLOWING PAGE K A P L A N 11 O R G A N I C C H E M I S T R Y S U B J E C T T E S T ORGANIC CHEMISTRY SUBJECT TEST ANSWER KEY C C 13 C 19 D 25 D C E 14 C 20 B 26 D E A 15 D 21 C 27 A D 10 D 16 C 22 E 28 E A 11 C 17 A 23 D 29 B E 12 A 18 C 24 A 30 B 12 _ K A P L A N _O R G A N I C C H E M I S T R Y S U B J E C T T E S T EXPLANATIONS C The Williamson ether synthesis is an SN2 reaction in which an alkoxide ion (OR–) acts as the nucleophile while an alkyl halide serves as the substrate: - RO Na + + NaX ROR' R'X Since this is an SN2 reaction, it follows that those alkyl halides which are most susceptible to nucleophilic attack are most suitable to serve as the substrate in this synthesis The order of reactivity toward SN2 reaction is dictated largely by steric factors and goes 1°>2°>3° The compound in choice C, as a tertiary alkyl halide, is thus the least suitable of the compounds offered in the choices Choices A and E are primary (and benzylic) halides while choices B and D are secondary alkyl halides; these compounds all undergo SN2 reactions more readily than does the tertiary alkyl halide of choice C C In order for a reaction to produce a racemic mixture, the product formed must possess at least one chiral center and its enantiomers must form in a one-to-one ratio The product of the addition reaction in choice C is 1,2-dibromobutane This compound has one chiral center, at carbon number 2, and its R and S enantiomers should be expected to form in equal amounts in the absence of any special effort to produce either in enantiomeric excess H H CH 3CH 2C CH + Br CH 3CH 2C Br Br CH Br + CH 3CH 2C H CH Br As for the other choices, the reaction in choice A should produce a mixture of neopentyl bromide (via substitution of the protonated hydroxyl group by bromide) and 2-bromo-2-methylbutane (via substitution after a 1,2 methyl shift); neither product is chiral Choice B shows a Grignard reagent over the arrow and diethyl ether as the reactant; no reaction will occur since ethers are generally inert to Grignard reagents and are, in fact, often chosen as solvents for Grignard reactions for that very reason The reaction in choice D will produce, upon heating, two equivalents of the achiral compound propionic acid as the permanganate oxidatively cleaves the alkene double bond, while the reaction in choice E will yield the Markovnikov addition product 2-chloropropane which is likewise achiral E The synthesis of cyclopropane via addition to the double bond in an alkene is accomplished by reaction with methylene, H2C:, an example of a class of compounds known as carbenes in which carbon forms two single bonds to make a neutral molecule (Notice the lone pair of electrons on the carbon atom.) This reactive intermediate is most commonly formed by the heat- or light-induced decomposition of diazomethane, as shown in choice E The lone pair of electrons on the carbene intermediate are used together with the _ electrons of the alkene to form the two new single bonds in the cyclopropane product (An alternative route to cyclopropane synthesis involves reacting diiodomethane and a zinc-copper couple to generate a carbene-like species called a carbenoid which will add a methylene group to the double bond.) D The compound in choice D is pyrimidine itself, a heterocyclic analog of benzene containing two nitrogen atoms, positioned as shown Pyrimidine and its derivatives are essential components of nucleic acids such as DNA and RNA K A P L A N 13 O R G A N I C C H E M I S T R Y S U B J E C T T E S T Choice A is almost, but not quite, the general structural formula for a steroid; it differs from the standard steroidal ring structure in that the ring on the far right has six carbons instead of five Choices B and E, known as quinoline and pyridine respectively, each have only one nitrogen atom and thus fail to meet the definition of a pyrimidine Choice C is purine, another essential component of nucleic acids A Since many synthetic pathways might be available to effect any given transformation, it makes most sense to check the choices offered Choice A will yield the desired product: permanganate oxidizes primary or benzylic alcohols to the corresponding carboxylic acids in the first step The benzoic acid formed by this oxidation will, in the presence of aqueous ammonia, form an ammonium carboxylate salt which, upon heating, will produce an amide in the second step Finally, treatment of the benzamide formed in the second step with hypobromite will produce an amine with one less carbon This reaction is known as a Hofmann rearrangement, and it will convert benzamide into aniline, the desired product, in the third step Choice B is incorrect Phosphorus tribromide, step one, will convert an alcohol into an alkyl bromide, in this case benzyl bromide Treatment of this benzyl halide with sodium cyanide, step two, will produce a nitrile via an SN2 pathway Finally, hydrogen over a nickel catalyst will reduce the nitrile to a primary amine; the primary amine formed by this series of steps will have two carbons between the phenyl ring and the amino group, and thus is not aniline Incorrect choice C, like B, produces benzyl bromide after the first step Benzyl bromide should be inert to ammonium and acid in the absence of any nucleophilic species, thus no reaction should be expected in the second step Choices D and E also not lead to the formation of aniline E The electrophilic addition of bromine to an alkene proceeds through a cyclic bromonium ion intermediate Configuration-wise, the addition is anti Regiospecificity also dictates that the geminal dihalide shown in choice B cannot possibly be a product of this reaction Complete scratchwork would show that choices A and D should form with equal probability, and therefore that choice E is correct: H5C2 H C H Br Br H5C2 C + Br C CH3 H C H CH3 H5C2 CH3 C H C Br H Br Br- OR H5C2 + Br C H H5C2 C C H CH3 CH3 Br H C Br H BrA faster way to get to the correct answer here is to notice that both reactants are achiral, and because of that the product(s) as a whole cannot be optically active Furthermore, since the groups attached to the two stereocenters are not identical (a methyl group on one and an ethyl group on the other), absence of optical activity cannot be due to the generation of one lone achiral meso compound An enantiomeric pair must therefore be the product of the reaction By this 14 _ K A P L A N _O R G A N I C C H E M I S T R Y S U B J E C T T E S T faster method of reasoning it is not necessary to figure out which product or products will form, but only that more than one will C This reaction is an example of benzylic oxidation Substituted benzene derivatives are easily oxidized by permanganate or dichromate to benzoic acid derivatives if there is at least one hydrogen attached to a benzylic carbon (The first step in the reaction is the abstraction of a benzylic hydrogen to give a benzylic radical, thus making the hydrogen atom necessary.) The reactant shown has three benzylic carbons, one of which, the carboxylic acid carbon at the top of the structure, is already completely oxidized The other two, the benzylic carbon in the alkyl group on the right and that in the alkenyl group on the left, can both be oxidized to carboxylic acid groups, yielding the tricarboxylic acid compound shown in choice C Choices A and D can be dismissed for several reasons, among which is that they suggest reduction of the COOH group in the reactant to a CH3 group in the choices; permanganate is far and away not a reducing agent Choices A and D also show the conversion of the alkene side chain to a diol; this mild oxidation would occur only if the permanganate used were cold, dilute, and dissolved in basic solution One might more quickly eliminate choices A and D by noticing that they differ only by rotation around a carbon-carbon single bond; they are thus equivalent choices Choice B is wrong because it shows the selective oxidation of the alkyl side-chain only, when in fact both side-chains will be oxidized under the conditions indicated in the reaction shown E This question is testing our understanding of fundamental organic reaction mechanisms The reactant shown is a secondary alkyl halide, thus it is susceptible to nucleophilic attack via either an SN1 or SN2 pathway It also has a proton α to the bromine atom (on the methyl group), so it is capable of undergoing elimination to form an alkene Sodium hydroxide is a strong base which can facilitate E2; the hydroxide ion can also behave as a nucleophile and lead to an SN2 reaction under the proper conditions The preference for either E2 or SN2 can be enhanced by the appropriate choice of reaction conditions, justifying choice E as the credited choice A This question is testing our knowledge of organic synthesis, and requires that we work through the steps shown to predict the final product (Partial knowledge of the reagents shown would, however, allow for the elimination of some of the answer choices For instance, noticing that the final step in the synthesis includes acidification allows us to eliminate choice D since this product would be protonated in acidic solution.) Let’s go through the steps: KMnO4 is a strong oxidizing agent which will oxidize most alkylbenzenes, including the ethylbenzene starting material here, to benzoic acid After the first step we thus have benzoic acid, which can now react with thionyl chloride, SOCl2, in the second step Thionyl chloride is most commonly used to convert a carboxylic acid to the corresponding acyl chloride; the product of the second step is thus benzoyl chloride Acyl halides are especially reactive toward nucleophilic acyl substitution, the reaction by which the halide is replaced by any of a large number of suitable nucleophiles while the carbonyl C=O bond remains intact The third step of the synthesis in this question should therefore result in the substitution of the chloride by cyanide from the HCN and the product of this step should be the α-ketonitrile derivative of benzoic acid The cyano group of nitriles is easily hydrolyzed in aqueous acid; the final step of the synthesis should thus be expected to convert the α-ketonitrile to the α-ketoacid shown in choice A The sequence of reactions is as follows: O C2H5 KMnO4 C O OH SOCl2 C O Cl HCN C O N C H+ C COOH = Choice A K A P L A N 15 O R G A N I C C H E M I S T R Y S U B J E C T T E S T 10 D By definition, an acetal must contain a carbon atom bonded to two -OR groups and a hydrogen atom Acetals might thus be thought of as geminal diethers The carbon atom on the right side of the ring in choice D is bonded to the OCH3 group on the far right and to the O atom in the ring which is, in turn, bonded to the rest of the ring; a third substituent would be a hydrogen atom which is not shown by convention Choice D does therefore meet the criteria for classification as an acetal Choice A is a carboxylic acid, while choice B is a b-diketone Choices C and E can be classified as hemiacetals, but not as acetals Hemiacetals, by definition, contain a carbon atom which is bonded to one -OR group, one -OH group, and a hydrogen atom 11 C E2 elimination occurs in the anti configuration, requiring that the proton abstracted be in a position trans to the leaving group As drawn, only the proton labeled as Hd is in the proper position for anti elimination, i.e., trans to the chlorine atom in the reactant, and thus Hd must be absent from the final product This qualifies choice C as the only possible choice 12 A This question is testing our understanding of organic separation techniques, specifically extraction, as well as our knowledge of the relative acidity of the three compounds shown The first compound, aniline, is the most basic of the three; it has an amino group in which the nitrogen has a lone pair of electrons available for donation to a suitable Lewis acid The amino group nitrogen atom in the second compound, para-nitroaniline, is rendered less basic due to the electronwithdrawing nature of the nitro group; with decreased electron density, it is less eager to donate electrons, i.e., to act as a Lewis base Lastly, the third compound, nitromethane, is not basic at all Applying this relative order of basicity to the extraction, now, requires the understanding that acidic extraction will make organic bases soluble in the aqueous phase via protonation of the base to form a soluble ionic salt Therefore choice A is the best choice Extraction with a sufficiently weak aqueous acid will protonate only the most basic compound, aniline, and allow it to transfer, in protonated form, to the aqueous phase A second extraction with a stronger acid will then protonate the more basic of the two compounds remaining in the chloroform layer, para-nitroaniline, and allow it to enter the aqueous phase The third compound would then remain alone in the organic layer, and the separation could be completed by distillation If a strong enough acid were used first, both aniline and nitroaniline would be protonated and transferred to the aqueous layer, and the separation would not be effective Extraction with a base would not effect a separation at all, since the first two compounds, as bases themselves, are inert to base 13 C The reaction shown is an example of a Grignard reaction One should predict that the Grignard reagent used, benzyl magnesium bromide, will act as a nucleophile and attack the carbonyl carbon in the substrate, phenyl benzoate, forming the tetrahedral intermediate shown below OC O CH2 After formation of the tetrahedral intermediate, a pair of electrons on the negatively charged oxygen can move down to the carbon atom of the original carbonyl group, restoring the carbon-oxygen double bond and displacing the phenoxide leaving group Once the carbonyl group has been restored, it is susceptible to further nucleophilic attack by a second 16 _ K A P L A N _O R G A N I C C H E M I S T R Y S U B J E C T T E S T molecule of the Grignard reagent, resulting, after acid workup, in the formation of dibenzylphenylcarbinol, the compound shown in answer choice C 14 C The reagents shown in this reaction, bromine and ferric bromide, are the reagents of choice to effect the bromination of most benzene derivatives via EAS (electrophilic aromatic substitution) Both the rate and the regioselectivity of EAS reactions are influenced by the presence of electron-donating or withdrawing groups already on the ring Electronwithdrawing groups, such as the acetyl substituent shown here, slow down the substitution by the incoming electrophile while directing it toward the meta position relative to the position of the original substituent Thus the major product of this reaction should be meta bromoacetophenone, the compound shown in choice C Choices B and D, and therefore choice E, would be the results if the original compound had an electron-donating, ortho/para directing group on it in place of the electron-withdrawing acetyl group Choice A is the product of alpha bromination, which should be expected to result only via a radical mechanism Absent an active metal, NBS, hν, peroxides, or extremely high temperatures, radicals are unlikely; choice A should thus be eliminated 15 D This question tests detailed knowledge of the Grignard reaction Usually, such reactions are initiated by the addition of clean magnesium metal to dry ethereal solutions of alkyl halides, though substituted alkyl halides are possible substrates if no other reactive functional group is present Absence of water is a requirement, or else the newly formed alkyl magnesium halide will react with the hydroxy proton in the water, forming an alkane and a halomagnesium hydroxide, thereby killing the reaction Similarly, it is necessary that no other acidic protons be present in the solvent system nor in the substrate, or else these would react with the newly formed alkyl magnesium halide It is this last fact which accounts for the elimination of the four wrong answer choices: choices A, B, and C can be eliminated because each contains a proton bonded to an electronegative element (oxygen in A and C, nitrogen in B), while choice E contains an acetylenic proton Elimination of choice E requires that we recall the relatively high acidity of an acetylenic proton; the proton attached to the sp hybridized carbon is sufficiently acidic to react with strong bases such as hydride or amide anions or, as relevant here, to protonate a Grignard reagent Choice D thus remains as the only viable choice 16 C The base-catalyzed condensation of acetaldehyde is an example of an aldol condensation and is outlined below: K A P L A N 17 O R G A N I C C H E M I S T R Y S U B J E C T T E S T O- O CH2 CH CH2 O -:CH2 CH CH H H3C -:OH CH O OH O O CH3CHCH 2CH CH3CHCH 2CH : OH OH The newly formed hydroxyl group thus winds up on the second carbon from the aldehyde carbonyl; the product is therefore classified as a β-hydroxyaldehyde, choice C It should be noted that this product can easily undergo further dehydration (elimination of water to form double bond) to yield a compound where a carbon-carbon double bond is in conjugation with the carbonyl bond 17 A This question requires us to translate acid strength into Ka Recall that a smaller Ka corresponds to a weaker acid; not confuse Ka with pKa! We can thus translate the question into: "Which substituted benzoic acid will be the weakest acid?" Recall next that electron-withdrawing groups increase acid strength since they help to stabilize the anion formed upon dissociation, so a further translation is, "Which of the following is the least electron-withdrawing as a substituent?" Of the choices offered, B, C, and D are electron-withdrawing groups, while choices A and E are electron-donating Of the electron-donating groups listed, NH2 is more so than is CH3 due to the lone pair of electrons on the nitrogen atom (In electrophilic aromatic substitutions, the amine group is a strongly activating ortho/para director because of its electrondonating ability.) Choice A is thus the most electron-donating and has the most destabilizing effect on the conjugate base (the anion), thereby disfavoring dissociation of the proton The compound para-aminobenzoic acid is therefore the least acidic, and has the smallest Ka 18 C The reaction is that of an acid anhydride, acetic anhydride to be precise, and a nucleophilic base, ammonia Nucleophilic attack on one of the anhydride carbonyl carbon atoms by the ammonia nitrogen should be expected, followed by the loss of an acetate leaving group concurrent with the restoration of the carbon-oxygen double bond O :O- O O + :NH3 O +NH3 O O O + +NH3 -O The protonated amide thus formed can then release a proton (to the acetate ion or to another molecule of ammonia) producing one molecule of acetamide Choices B and D can be eliminated at this time due to their lack of acetamide The remaining species in the product mixture (acetate, ammonia, and proton) would then combine to form ammonium acetate, 18 _ K A P L A N _O R G A N I C C H E M I S T R Y S U B J E C T T E S T the second product shown in choice C Note that in the presence of ammonia, carboxylic acid cannot remain protonated and exists as a salt 19 D Phenol, or hydroxybenzene, has an -OH group on the ring The hydroxy oxygen has two nonbonded pairs of electrons, either of which can be donated (via resonance) to the aromatic ring after addition of an electrophile Recall that electron-donating groups stabilize most the cations formed upon addition to the ortho and para positions, thus they serve as ortho/para directing activators The -OH group is not particularly bulky, and thus there will be no steric hindrance that prevents substitution at the ortho position Both ortho and para substituents will result 20 B A single proton NMR signal means that all the hydrogen atoms in the molecule find themselves in identical chemical environment with respect to the other atoms in the molecule Hexane has three different types of protons: the terminal methyl ones, the methylene ones on the adjacent carbon atoms, and the methylene protons on the innermost carbon atoms Because they will have approximately similar chemical shifts, the observed spectrum may be quite complex Compound II, 1, 2-dichloroethane, possesses four equivalent protons and will give one signal: Cl Cl C H C H H H Compound III, tert-butyl chloride, has nine equivalent protons and will also give a single proton NMR signal H3C H3C H3C C Cl Choice B is therefore correct The last compound, 2-methyl-2-butene, has four distinct types of protons: the three methyl groups attached to the double bond are in different chemical environments, plus the vinylic proton 21 C This question is about the reaction of a Grignard reagent, RMgX, with various carbonyl compounds to form alcohols Choices D and E can be eliminated first since a carboxylic acid, choice D, will neutralize a Grignard reagent (RMgX + R'COOH ∅ RH + R'COOMgX) while reaction with carbon dioxide, choice E, results in the formation of a carboxylate anion (RMgX + CO2 ∅ RCOOMgX) The aldehydes and ketone in the remaining choices yield alcohols as the major product However, only choice C, formaldehyde, will result in the formation of a primary alcohol Choice A, acetone, would react with the Grignard reagent to produce a tertiary alcohol: the two methyl groups in the original acetone molecule will wind up on the same carbon as that attached to the hydroxyl group in the product Acetaldehyde, choice B, would react with RMgX to form a secondary alcohol, i.e., the methyl group of the acetaldehyde would be attached to the central carbon atom Only in choice C, formaldehyde, are there two hydrogen atoms already present; these two hydrogens are necessary to form the -CH2OH moiety which would define the end product as a primary alcohol 22 E The less stable an unshared pair of electrons, the likelier it is to extract a proton from other molecules around, and thus the more basic the compound This question about the order of basicity can therefore best be approached by examining the expected stability of the lone electron pair on the nitrogen atom in the amino group of the molecules shown The stability of K A P L A N 19 O R G A N I C C H E M I S T R Y S U B J E C T T E S T the electron pair will be affected by the nature of the other substituent group present, specifically in terms of whether it is electron-withdrawing or electron-donating (either inductively or by resonance) and how effective its influence is based on its position relative to the amino group The first thing to notice is that aniline itself (choice D) is a rather weak base because the electron pair is stabilized via delocalization through the _ cloud of the phenyl group (i.e by resonance): (+)NH : NH2 (+)NH (+)NH 2 (-) (-) (-) One would expect that electron-withdrawing groups would lead to a further stabilization (and therefore weakening) of the base Choices A and C, which contain the electronegative chlorine, would make the compound less basic, and are therefore not correct Choice B, which has the nitro group at the meta position, would also stabilize the electron pair inductively, even though not as effectively as it would if it were at the o- or p- position, where it could also stabilize through resonance Choice E, however, with the methoxyl group at the para position, is destabilizing because of the highly unfavorable resonance contributions: : NH2 : NH2 (-) : OCH3 etc (+)OCH The proximity of the electron pair of nitrogen and the electron density released by the methoxyl group causes the compound to be more basic than aniline itself, overwhelming the inductive effect of the electronegative oxygen atom Choice E is the most basic, and hence the correct answer 23 D The iodoform test is useful in identifying the presence of the following groups: O C OH CH3 , C CH3 Compounds containing these groups react with iodine in NaOH to give a yellow precipitate Choices C and E are methyl ketones and will give positive results Choice B, acetaldehyde, also possesses the —COMe group that leads to the formation of the precipitate Choice A, ethanol, possesses the group on the right above and will first be oxidized to acetaldehyde, yielding a positive result Choice D, benzaldehyde, is the only compound shown that will not react in the solution 24 A Students are expected to know the characteristic frequencies of certain functional groups in the infrared The strong peak at ~1700 cm-1 indicates the presence of a carbon-oxygen double bond, i.e a carbonyl group Choices C, D, and E can thus be eliminated At this point, one may use either of two pieces of information to distinguish between the two remaining alternatives: (1) Tollens’ test is also known as the silver mirror test, and gets the name from the fact that silver is deposited on the walls of a clean test tube or flask as a mirror if an aldehyde is present The negative result of the test indicates that the compound is a ketone, choice A (2) The C-H stretch of the aldehyde group would give two weaker bands in the infrared spectrum at ~2700 cm–1 and ~2850 cm–1, both of which are absent This latter approach, however, probably requires more knowledge about infrared spectroscopy than the student is expected to know 20 _ K A P L A N _O R G A N I C C H E M I S T R Y S U B J E C T T E S T 25 D The most acidic hydrogen is the hydrogen that, upon leaving the molecule, will leave behind the most stable anion Alpha-carbons, i.e carbon atoms adjacent to carbonyl carbons, have protons that are acidic compared to protons attached to normal carbon atoms because the carbonyl group can stabilize by resonance the conjugate base that results from deprotonation: O C C C O O C C C H :OHChoices C and D are both α-hydrogens and will thus be expected to be more acidic than the other choices In addition, choice D will be more acidic than choice C because the anion will be stabilized by two carbonyl groups rather than one 26 D Dehydration of an alcohol to yield an alkene under acidic conditions proceeds via the E1 mechanism The hydroxyl group is protonated, and departs as a water molecule, leaving behind a carbocation which then loses a proton to the solvent while a double bond forms between the two carbon atoms: CH3 CH3 R1 C R2 OH H+ R1 C R2 CH3 + OH H -H2O R1 C+ R2 CH2 -H+ R1 C R2 Formation of the carbocation is the rate-determining step in this reaction Therefore, the more easily it is formed (i.e the more stable it is), the faster the reaction will go The carbocation intermediate will be stabilized by the presence of electron-donating substituent groups Alkyl groups are electron-donating and thus will increase the rate of reaction by facilitating the formation of the intermediate The more numerous they are, the more stable the cation will be, and the faster the reaction Choices C and E are incorrect because the electronegative chlorine will destabilize the carbocation intermediate instead 27 A Because F is the most electronegative atom, it helps to stabilize the anionic conjugate base, thus increasing the strength of the corresponding acid, especially when, as in this case, it is close to the acid functional group Choice B is a false statement, as F is more electronegative than Br Choice C is also a false statement, as Ka is a direct measure of acid strength: the higher the Ka, the stronger the acid (The lower the pKa, the stronger the acid: keep in mind the difference between the two.) Choice D is false as well: the stronger the acid, the weaker its conjugate base As for choice E, it is a false statement that is irrelevant at best Only very indirectly can one rationalize acid strength from how good a leaving group it contains: high electronegativity is one factor that makes a species stable as an anionic leaving group and hence one may expect to see a correlation between the ability of the species to leave and the acid strength of a compound that contains this species, but in this case Br- is actually the better leaving group, although Br is the less electronegative atom 28 E The group that will stabilize the anionic conjugate base will yield the most acidic compound An electron-withdrawing group stabilizes the negative charge either by induction or by delocalization Choices A–D are all electron-donating, and K A P L A N 21 O R G A N I C C H E M I S T R Y S U B J E C T T E S T would tend to lower the acidity of the molecule (compared with X = H) Only E, —NO2, is electron-withdrawing, thereby increasing the acidity of the molecule 29 B Upon heating under acidic conditions, alcohols will undergo dehydration to yield alkenes This is an E1 reaction that proceeds via the formation of a carbocation intermediate Choices A, C, and E are all incorrect and easily dismissed because they are not alkenes Of the two choices B and D, B is correct because it is the most substituted alkene Choice D is indeed a product of the reaction as written, but a minor one because it is relatively unsubstituted The secondary carbocation formed upon the departure of the protonated hydroxyl group will rearrange (via migration of a methyl group) to generate the more stable tertiary carbocation which in turn will lead to the more substituted alkene: CH3 CH3 H3C C + C CH3 H 30 CH3 H3C C+ C CH3 -H+ (CH3)2C C(CH 3)2 CH3 H B Electron-donating species will activate aromatic rings for ortho/para electrophilic substitution Only choice B, which has an ester substituent, is electron-donating All of the other choices have electron-withdrawing groups, which are deactivating and meta-directing (The exceptions to this are the halogen atoms, which are deactivating but ortho/paradirecting.) 22 _ K A P L A N ... _O R G A N I C C H E M I S T R Y S U B J E C T T E S T Organic Chemistry Subject Test Which of the following alkyl halides would be the least reactive substrate in... G A N I C C H E M I S T R Y S U B J E C T T E S T ORGANIC CHEMISTRY SUBJECT TEST ANSWER KEY C C 13 C 19 D 25 D C E 14 C 20 B 26 D E A 15 D 21 C 27 A D 10 D 16... remaining alternatives: (1) Tollens’ test is also known as the silver mirror test, and gets the name from the fact that silver is deposited on the walls of a clean test tube or flask as a mirror if
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