Organic chemistry section test (5)

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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 Organic Chemistry Subject Test What is the major organic product of the reaction sequence shown? CH3CH2I A B C D E What is the major organic product of the synthetic sequence outlined below? [A] H3 C C C CH3 B H3 C C C Br E H3 C H3 C NH3 CH3 Br A B C D E ? independent of concentration first order in alcohol, first order overall first order in acid, first order overall first order in alcohol, second order overall second order in alcohol, second order overall A compound to be analyzed is known to be a six carbon cyclic hydrocarbon The compound is found to be inert to bromine in water and to bromine in dichloromethane, yet it decolorizes bromine in carbon tetrachloride when a small quantity if FeBr3 is added Which of the following could be the identity of the compound? cyclohexane benzene 1,3-cyclohexadiene 1,4-cyclohexadiene cyclohexene Which of the compounds below could be the reagents used to carry out the conversion shown? O H OH O C C X Y HO O H A B C D E C C CH3 H3 C CH2 CH2 CH3 Which of the following factors usually increases the solubility of a compound in a given solvent? I II III IV Higher temperature Similar polarities Higher molecular weight of the compound Lower density of the solvent A B C D E I only I and II only I, II, and III only I, II, and IV only I, II, III, and IV OCH2CH3 OH Br Br H3 C NaNH2 Which of the phrases below best describes the kinetics of the dehydration of 3-methyl-3-hexanol in concentrated (85%) phosphoric acid? A B C D E ? CH3CH2OD CH3CH2MgD CH3CH2D CH3CH2MgI CH3CDO A D D20 Mg ether H3 C C C H C X = ethanol; Y = H2/Pt X = H+/H2O; Y = LiAlH4/ether X = ethanol; Y = LiAlH4/ether X = H+/H2O; Y = NaBH4/H2O X = ethanol; Y = NaBH4/H2O Which of the following would yield more than one major organic product upon reaction with HCl in the absence of peroxides? A B C D E 1,2-dichloroethane 1,1-dichloropropane 3-hexene 2-hexene 2-methyl-1-hexene 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 α-helix is an example of a protein's _ structure A B C D E lewis resonance primary secondary tertiary What reaction type is responsible for the biochemical formation of peptide bonds? A B C D E rearrangement addition resolution elimination condensation 10 What is/are the product(s) if 1-ethyl-2methylcyclohexene is hydrogenated over platinum? A H H CH3 CH2CH3 B H CH2CH3 CH3 H C H CH3CH2 CH3 H D H CH2CH3 and CH3 H H CH3CH2 CH3 H E H H and CH3 CH2CH3 H CH3CH2 H CH3 K A P L A N 11 The compound below is heated in a deuteroacid solution (D+/D2O) and allowed to reach equilibrium Which of the labeled hydrogens will be exchanged for deuterium? H H H H O 2 H H A It exists predominantly in the enol form because its keto tautomer is nonaromatic B It exists predominantly in the enol form because its keto tautomer is antiaromatic C It exists predominantly in the keto form because its enol tautomer is nonaromatic D It exists predominantly in the keto form because its enol tautomer is antiaromatic E It exists predominantly in the keto form because its keto tautomer is aromatic and 1, and 1, and 1, and and 12 What are the major products of the reaction sequence shown below? O CH3 CH2 C OCH3 A B C D E LiAlH4 THF HCl H2 O CH3CH2COOH and CH3OH CH3CH2COOH and HCOOH CH3CH2CH2OH and CH3OH CH3CH2CHO and CH3OH CH3CH2CH2OH and CH2O 13 Which of the following would yield one major organic product upon reaction with HBr in the presence of peroxides? A B C D E OH NH H A B C D E 14 Which of the following statements is true of phenol, pictured below? 1,2-dichloroethane 1,1-dichloropropane 2-hexene 3-hexene 2-methyl-3-hexene 15 Which of the compounds below could be the reagents used to carry out the conversion shown? O O H3C X C Y O Et A B C D E X = H+/H2O , Y = NaOEt/EtOH X = KMnO4/OH-, Y = NaOEt/EtOH X = NaOEt/EtOH, Y = NH3/H2O X = NH3/H2(10 atm), Y = Zn/H+ X = NaOEt/EtOH, Y = H+/heat O O Et 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 16 Which of the following molecules would produce the NMR spectra shown below? 17 How many chiral carbons are there in morphine? HO O N H H CH3 HO H Morphine A B C D E A Br D C CBr3 Br CH3 H C CH3 H CH3 18 The two products formed by the following racemization reaction could best be described as: O B Br C CH3 Br CH3 E C2H5 Br C CH3 H CH3 O C2H5 H3C C Br C CBr3 CH3 H H - R C EtO , EtOH CH3 H S C H3C C2H5 H3C CH3 O R C CH3 H A B C D E enantiomers diastereomers conjugate acid and base conformational isomers same compound K A P L A N 19 Which of the following reactions will lead to the formation of an aldehyde? A 21 The diagram below shows a step in which of the following processes? CH2 OH O O H H20 Cl CH2 OH OH OH O B OH PCC H CH2 OH O H OH C OH D CH3CH2Cl H KMnO4 A B C D E CN2 H+/H2O E + KMnO4/ H heat 20 Which of the following species will form the most stable radical? A Aldehyde formation Hemiketal formation Mutarotation Anomerization Tautomerization 22 Which of the following reactions would produce an ester? A B C D E CH3OH + H2SO4 ∅ CH3COOH + SOCl2 ∅ CH3COOH + C2H5OH + H2SO4 ∅ C6H5OH + CH3CH2Br ∅ CH3CH2Br + CH3CH2O-Na+ ∅ D 23 Which of the following describes the reaction below? H5 C2 C B C C CH3 H2 , Pd E H H5 C2 A B C D E Catalytic hydration Substitution Stereospecific reduction Racemization Grignard reduction H CH3 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 24 Which of the following is a result of the reaction below? H C2 H5 - Cl + Br A B C D E H CH3 A B C D E 28 Reaction of benzoic acid with thionyl chloride followed by treatment with ammonia will yield which of the following compounds? Inversion of configuration Retention of optical activity Mutarotation Double bond formation Loss of optical activity Benzamide p-Aminobenzoic acid p-Chlorobenzamide 3-Chloro-4-aminobenzaldehyde m-Aminobenzoic acid 29 Which of the following bonds is a peptide linkage? A 25 Infrared spectroscopy provides a chemist with information about: A B C D E functional groups conjugated bonds molecular weights distribution of protons ionization patterns CHO CHO HO H H OH HO H H OH CHO II CHO H HO OH H CHO A B C D E H C H E O H III O N O N 26 Which of the following compounds form a racemic mixture? I C B D O C O C H O N C C O CHO IV CHO HO H H OH CHO I and III I and IV II and III II and IV I and II 27 Which of the following compounds is the most susceptible to nucleophilic attack by OH–? A B C D E Propanoic acid Ethanal Benzyl chloride Propionyl chloride Propanal K A P L A N 30 Which of the following pairs of side chain –R groups would tend to associate with each other? A and B (CH2) 2COOH and CH3 C CH3 CH3 D CH3 H E and CH2OH CH3 CH2CH3 and (CH2) 2COOH H N and CH3CH2NH2 STOP! END OF TEST THE ANSWER KEY AND EXPLANATIONS BEGIN ON THE FOLLOWING PAGE 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 D 13 D 19 B 25 A A D 14 A 20 A 26 D B E 15 E 21 C 27 D B 10 E 16 E 22 C 28 A B 11 D 17 C 23 C 29 D C 12 C 18 A 24 E 30 C 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 This is an example of a reaction utilizing a Grignard reagent Whenever you see Mg as a reactant, the first thing that should come to mind is a Grignard synthesis In Grignard reactions, magnesium is used to turn an alkyl halide into a good nucleophile Grignard reagents are utilized in many reactions, because the carbon associated with the magnesium has a partial negative charge, and is therefore a good nucleophile In this reaction, the magnesium nucleophilically attacks the ethylmagnesium iodide (very similar to an SN2 reaction) and forms our Grignard reagent (an organomagesium compound) The Grignard reagent is a strong base (it is proton deficient), and undergoes an acid/base reaction with heavy water (the hydrogen isotope deuterium, D, has a neutron and so is heavier than H) Specifically, the ethyl group abstracts a proton from heavy water, and the magnesiumidodide is displaced as a result CH3CH2I + Mg δ− δ+ CH3CH2 MgI + D OD Grignard reagent CH3CH2 D + OD - + Mg 2+ + I- A This question asks what happens when you react an alkyne with a strong base The alkyne hydrogens are relatively acidic, owing to resonance stabilization of the conjugate base, and so can undergo acid/base reactions NaNH2 (sodium amide) is a strong base, and so will remove the terminal hydrogen The resulting carbanion is a strong nucleophile, and will undergo an SN2 reaction with methylbromide The bromine will be displaced, and the methyl group will add to the terminal carbon, giving us 2-butyne B Most substances tend to become more soluble at higher temperatures and so statement I is correct Statement II is also correct Recalling the phrase “like dissolves like” compounds that are polar or ionic will dissolve in solvents that are also polar, whereas compounds that are non-polar will dissolve in non-polar solvents Therefore, similar polarities will increase the solubility Statement III molecular weight and statement IV density of the solvent not determine solubilities, so I and II are the only true statements, and choice B is the correct response B This question is describing an acid catalyzed dehydration reaction When a 3_ alcohol is treated with strong acid, an E1 reaction occurs The hydroxyl group gets protonated and then leaves in the form of water, giving us a carbocation The released water then abstracts a proton from the carbocation, reforming the acid (that is why it is a catalyst) and converting the carbocation into an alkene Since the reaction proceeds via E1, the reaction rate depends only on the concentration of the substrate, and so is first order with respect to the alcohol only B The question stem informs us that the unknown cyclic hydrocarbon does not react with bromine in dichloromethane, carbon tetrachloride, or water This means that the unknown cannot contain any non-conjugated double or triple bonds, otherwise the bromine would have added across the π bonds Since the unknown did react when FeBr3 was added, it must be benzene The FeBr3 acts as a Lewis acid and catalyzes aromatic electrophilic addition reactions Bromine is not a strong enough electrophile to disrupt the conjugated double bonding in the benzene ring, but the addition of the iron(III)bromide catalyst converts the bromine into a strong enough electrophile that it can add to the benzene ring C In this reaction a cyclic anhydride is transformed into an ester, then reduced to an alcohol Cyclic anhydrides are very reactive (they undergo the same reactions as acyl chlorides) because there are two electron withdrawing groups deshielding 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 carbon nucleus; the carbonyl carbon geometry is trigonal planar, so there is little steric hindrance to nucleophilic attack; and there is a good leaving group attached Choices B and D are wrong because if an anhydride is reacted with water it will become a dioic acid (the anhydride was formed by condensing a dioc acid) When the anhydride is reacted with ethanol, the oxygen of the alcohol will add to the carbonyl carbon, and the other side of the molecule acts as a leaving group The product formed will be an ester The carbonyl carbons are then reduced using LiAlH4, a strong reducing agent, to form a diol Choice E is wrong because NaBH4 can only reduce aldehydes and ketones, it is not a strong enough reducing agent to effect an acid (or it's derivatives) If platinum were used, as in choice A, the double bond would be reduced to a single bond The product has the double bond intact, so A is incorrect O *O OCH2CH3 O + HOCHCH OH O * OH LiAlH ether HO OCH2CH3 * OH *O D The addition of acid halides across double bonds has to follow Markovnikov's rule the H adds to the less substituted carbon (the carbon with the most H's) and the halide adds to the other Choices A and B will not react with HCl Answer choice C, 3-hexene, is symmetrical, it does not matter to which side of the double bond the H adds, and only 3chlorohexane can form Answer choice D, 2-hexene, can form more than one product Each carbon that is part of the double bond is equivalent, so the hydrogen can add to either, followed by the chlorine adding to the other The resulting products can be either 2-chlorohexane or 3-chlorohexane The only product that can result from choice E is 2-chloro-2methylhexane D α-helixes and β-pleated sheets are the two common forms the primary amino acid sequence of a protein can assume These secondary structures are held together by hydrogen bonds A protein's primary structure is the sequence of amino acids it contains Tertiary structure is the final 3-dimensional form the functioning protein assumes, and is a result of hydrophilic and hydrophobic interactions between the amino acid R groups E Peptide bonds are the covalent links between two amino acids It is formed by a nucleophilic acyl substitution of the carbonyl carbon of one amino acid by the amine of the other The hydroxyl group of the carboxylic acid leaves in the form of water, and so the reaction is called a condensation 10 E Hydrogenation on a platinum surface leads to syn addition of H atoms Choices B, C, and D can thus be eliminated Choice A may have been tempting, but keep in mind that the hydrogen can add to either side, as long as the two atoms are syn versus H3C H H CH2CH3 H3CH2C H H CH3 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 The two scenarios will lead to the two different products shown in choice E Another way to look at this is that the starting compound, cyclohexene, is achiral Without introducing any chiral agents, we cannot generate optical activity, and so the product overall must be optically inactive The two products in choice E are enantiomers, and so the product is a racemic mixture 11 D This question is asking, "which of the labeled hydrogens is acidic?" The hydrogens labeled and are both α hydrogens they are one carbon away from the carbonyl carbon The hydrogens attached to α carbons are much more acidic than a normal alkyl hydrogen because of the electron withdrawing effect of oxygen and the resonance stabilization of the conjugate base The hydrogens labeled 5, although not acidic, can be replaced by deuterium because the amine group is ← constantly binding and releasing protons: R–NH2 R–NH3+ A hydrogen can be exchanged by deuterium if a D is → ← ← R–NH2 → R–NH2D+ → R–NHD The hydrogens picked up and an H released; labeled and are alkyl hydrogens, and are not acidic 12 C This question requires you to understand the process of reduction The reducing agent, LiAlH4, adds hydrogens (in the form of hydride, H–) to the acyl carbon, reducing it The acid added causes protonation of the methoxide group, and it leaves in the form of methanol Since LiAlH4 is a strong reducing agent, the process does not stop at the aldehyde, but continues until the alcohol is formed Carboxylic acid O R C - H OCH3 H HO + 13 - O R C + H OCH3 H Aldehyde O R C H - Alcohol OH H R C H H O R C H - H + CH3OH D This question is very similar to that of question 7, only here the halogenation is in the presence of peroxides Peroxides (or UV) can be used to rapidly propagate halogen substitution in alkanes or halogen addition in alkenes through the formation of free radicals Remember that free radical addition across double bonds follow anti-Markovnikov rules In order to get this question correct, you have to know both reaction types Answer choices A and B involve free radical substitution of alkanes The instability and high reactivity of radicals makes these reactions very hard to control, and produce a mixture of products The reactions are useful when only one type of hydrogen is present 1,2-dichloroethane and 1,1-dichloropropane have different types of hydrogens present and will produce a mixture of products, depending on how many hydrogens are substituted with bromine Choice C, 2-hexene, can form 2-bromohexane and 3-bromohexane Since the carbons and hydrogens associated with the double bond are equivalent, both receive bromine via an anti-Markovnikov addition Choice D, 3-hexene, will only produce 3-bromohexane Choice E, 2-methyl-3-hexene, will also produce a mixture of products due to the possibility of the methyl group switching positions during the free radical intermediate stage of the reaction, in order to stabilize the molecule 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 14 A To get this question correct you have to be familiar with both tautomers and aromatic compounds Aldehydes and ketones exist in an equilibrium between the keto and enol forms (see figure) Most compounds exist predominately in the keto form because the carbon=oxygen double bond is more stable Phenol is one of few aldehyde/ketones that exist predominately in the enol form as part of their keto-enol tautomer equilibrium The conjugated benzene ring system of phenol provides the stability of the enol form When the molecule assumes the keto form, the conjugation is lost, and the molecule becomes less stable In order for a molecule to be aromatic, it must be cyclic, have conjugated double bonds, and posses Hückel's number of π electrons (4n + 2, where n is any integer) An antiaromatic compound is also a conjugated ring system, but it possesses 4n π electrons (and is very unstable) A nonaromatic compound is anything else In looking at the enol form of the molecule, we see that the ring is not conjugated (the * carbon is saturated [sp3 hybridized] and so cannot form any double bonds) The keto form is therefore nonaromatic, leaving only choice A as the answer Enol Keto H R1 O H O R1 R2 H H R3 R2 O O H * H H H R3 Note: =H Aldehyde when3R Ketone when3R =C-R H H * H H H H Phenol Tautomers 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 15 E This question is testing your understanding of the Claisen condensation reaction, which is just a sheepish substitution reaction in wolf's clothing This reaction mechanisms employs concepts that we addressed in question 11, that is understanding the acidity of α hydrogens When strong base is added to an ester, the α hydrogen is removed, and a carbanion is formed (it is stabilized through resonance with the carbonyl carbon) The carbanion is a strong nucleophile, and can attack another ester molecule (remember that carbonyl carbons are susceptible to nucleophilic attack), with alcohol serving as the leaving group At the end of reaction X we have the deprotonated β keto acid, and with the addition of H+/heat, the β keto acid gets protonated and forms our product O H3C X O EtO-Na+ C EtOH O Et H2C C + O Et O H3C C* O Et O Na O H3C C* C Et O C O H Et O Na O H3C C* Et C C O H + Y H+ O H3C C* H C O C O H CH3CH2O- In choice A, reagent X would lead to conversion of the ester to the parent acid, followed by conversion back to the ester with the addition of reagent Y Choice B is also incorrect, as a carboxylic acid and it's derivatives are completely oxidized, so the addition of KMnO4 could not produce the desired product In choice C, we would initially form a β keto acid, but the addition of ammonia would convert it into a β keto amide The reagents used in choice D would form ethanamide The ammonia would convert the ester to an amide, and the Zn/H+ would not anything, as it is not a strong enough reducing agent to effect the amide (it can only reduce ketones and aldehydes) 14 _ K A P L A N Et 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 16 E The NMR spectra shows a distinct absorption split into peaks, the classic absorption of an isopropyl group This absorption is downfield, indicating that it is deshielded From this information alone, we can determine that there is at least one proton with equivalent neighboring protons, that is near some electronegative species Only choice A has neighboring protons that can lead to splitting into (N+1) seven peaks The downfield shift is caused by the bromine bonded to the same carbon as the single hydrogen these equivalent neighboring protons split the single protons peak into peaklets Br H 3C C CH H this proton's absorption is shifted downfield (to higher PPM) Choice A is incorrect because there are no neighboring protons, and so there will be no splitting Choice B is incorrect because there are no neighboring protons and so, splitting into seven peaklets will not occur Choice C is incorrect because the single proton has only neighboring protons and will only be split into peaklets (a quadruplet) Note that the absorption for the single proton in this molecule will be shifted more downfield because there are many electronegative species around it 17 C Any carbon bonded to four different substituents will be chiral Morphine has five such carbons as shown below: HO O * * * H HO H * * H N CH3 Morphine Note that the nitrogen atom is also a chiral center If nitrogen's lone pair of electrons is considered a fourth substituent bonded to nitrogen, it can be seen that nitrogen can also act as a chiral center Some amines can invert configuration (and thus are not chiral) The tertiary amine of morphine cannot invert – it is locked into its configuration by the ring structure However, the question asks for the number of chiral carbons, not the number of chiral centers 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 18 A In the reaction shown, the two products formed differ only about their arrangement around one chiral carbon Since it is given that one product is S and the other R, they must be enantiomers The racemization reaction occurs as a result of the strong base in solution The base abstracts an α hydrogen, forming an achiral carbanion The carbanion can then remove a hydrogen from ethanol, reforming the parent compound Since the intermediate is achiral (similar to SN1 and E1 reactions) the resulting products will be racematic Choice B is incorrect because diastereomers are not mirror images Diastereomers may have optical rotations of different magnitudes Thus, a solution with equal concentrations of two diastereomers may still rotate plane polarized light In contrast, a racemic mixture does not rotate plane polarized light Choice C is incorrect because the compounds are not conjugate acids and bases Choice D is incorrect because meso compounds are compounds that have chiral centers but lack optical activity due to an internal plane of symmetry Choice E is incorrect because the compounds are not the same (the only difference being in which direction they rotate plane polarized light) 19 B Of all of the reactions shown, choice B is the only reaction that will result in the formation of an aldehyde PCC (Pyridinium chlorochromate) will convert a primary alcohol into an aldehyde, without further oxidation to the carboxylic acid Choice A is incorrect because the addition of water to an acyl halide (alkanoyl halide) will result in carboxylic acid formation Choice C is incorrect because the oxidation of a secondary alcohol forms a ketone Choice D results in carboxylic acid formation The CN– displaces the Cl in via a SN2 reaction The cyanide group is then hydrolyzed into a carboxylic acid group Choice E is incorrect because the addition of a strong oxidizing agent to an alkene causes oxidative cleavage of the double bond, and the formation of molecules of carboxylic acid 20 A The stability of free radicals decreases from tertiary>secondary>primary>methyl Choice A is the only choice that can form a tertiary radical Choices B, C, D, and E will form secondary radicals 21 C This question calls upon your understanding of the phenomenon of mutarotation Cyclic hemiacetal forms of monosaccharides with different configurations around the first carbon that is, anomers are easily broken in aqueous solutions In such reactions either the alpha or beta anomer becomes an open chain In an aqueous solution, especially if it is slightly acidic, this open chain is easily recyclized, forming a mixture containing both anomers in their equilibrium concentrations Thus the initial opening and the subsequent closing of the chain results in a mixture of anomers This phenomenon is called mutarotation and so choice C is correct Choice A, aldehyde formation , is wrong because the open chain form has a carbonyl group and therefore an aldehyde is already formed Choice B hemiketals are formed as a result of the nucleophilic addition of a hydroxyl donated by an alcohol to a ketone carbonyl group Since in this question you have an aldehyde carbonyl group, this choice is also wrong Although this reaction is an example of anomerization, choice D is wrong since mutarotation is a more accurate description of this process Choice C is incorrect, as tautomerization is the interconversion of an aldehyde from it's keto and enol forms 22 C Esterification is a process in which a carboxylic acid reacts with an alcohol in the presence of an acidic or basic catalyst to form an ester plus water Therefore choice B formation of an acyl halide can be rejected at once Choice A illustrates the formation of dimethyl ether from two molecules of methanol, therefore this choice is also wrong Choice D and E are wrong as both are examples of the Williamson Ether synthesis which involves reaction of a phenoxide ion with an alkyl halide to produce an ether So the only choice that results in ester formation is choice C 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 23 C This is an example of stereospecific reduction or hydrogenation This reaction, catalyzed by palladium, results in the addition of two hydrogen atoms on the same side of the molecule referred to as syn-addition This puts the methyl and ethyl substituents on the same side of the double bond and so the cis isomer is formed Therefore, choice C is correct Metallic sodium in ammonia solution would cause a reaction with the opposite stereospecificity, yielding the trans isomer This is called anti-addition Let's go through the remaining answer choices Choice A is wrong because a hydration reaction is one in which water is added to a molecule; don't confuse this with hydrogenation in which hydrogen is added Choice B is also incorrect because this is definitely an addition reaction, not a substitution reaction Choice D is wrong because racemization means loss of optical activity Since the reactant is achiral and therefore optically inactive, it is incapable of being racemized Finally, choice E, Grignard reduction, doesn't even exist (besides, Grignard reagents use Mg not Pd) 24 E This question tests your familiarity with the stereochemistry of substitution reactions In this case, we're clearly dealing with an SN1 reaction, because tertiary alkyl halides can't undergo SN2 reactions because of steric hindrance The central carbon of the alkyl halide reactant is bonded to different substituents, so this compound is chiral It's also asymmetrical, so it will be optically active In the first step of the reaction that's given here, the alkyl chloride will dissociate to form a stable tertiary carbocation This will result in the loss of optical activity, which always happens in SN1 reactions In any carbocation, the positively charged carbon is always achiral because it has only three substituents This means that all carbocations lie pretty much within a plane Therefore, in the second step, the bromide anion can attack the carbocation from either side of the plane with equal probability As a result, the reaction will yield equal amounts of two chiral products, which are enantiomers non-superimposable mirror images of each other These enantiomers rotate the plane of polarized light to the same extent but in opposite directions, so the product will be an optically inactive racemic mixture Getting back to the question, since this reaction results in the loss of optical activity, choice E is correct and choice B is clearly wrong Choice A is also wrong because inversion of absolute configuration only occurs in SN2 reactions, where a nucleophile is attached from the back in a one-step reaction Since this is an SN1 reaction, that won't happen Some of the individual molecules will wind up inverted, but overall, the absolute configuration of the sample will be lost Finally, choice C will be irrelevant to this question, because mutarotation concerns the equilibrium between openchain forms and cyclic hemiacetal forms of monosaccharides in aqueous solutions 25 A Information about functional groups is provided by IR spectroscopy Choice B is wrong because information about conjugated double bonds is obtained by UV spectroscopy, not IR Choice C is also wrong because information about molecular weight is obtained by mass spectrometry Information about protons is provided by nuclear magnetic resonance spectroscopy, therefore choice D can also be rejected Choice E is also wrong, as ionization patterns are the means by which scientists measure the masses of particles using a mass spectrometer 26 D This question deals with optical activity in stereoisomers A racemic mixture contains equal quantities of two enantiomers that is, isomers that are non-superimposable mirror images of one another Let’s go through the choices that we're given Compounds I and III are mirror images, but if you rotate one of them by 180 degrees, you will see that you can superimpose it on the other That's because each compound has a plane of symmetry So even though these molecules contain chiral carbons, they are optically inactive, and would be called meso compounds, not enantiomers In fact, compounds I and III are the same compound so choice A is wrong It is somewhat easier to eliminate choice B, because isomeric compounds I and IV are clearly not mirror images at all In fact, they are diastereoisomers The same is true of compounds II and III in choice C, and compounds I and II in choice 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 D The correct answer is choice D, because, although compounds II and IV are mirror images, they can't be superimposed even if you rotate one of them by 180 degrees 27 D This question deals with the susceptibility of different compounds to nucleophilic attack All of the answer choices will undergo nucleophilic attack, but the compound that will undergo attack the easiest is propionyl chloride Choice A-propanoic acid has a carbonyl carbon that is slightly susceptible to nucleophilic attack, since the double bonded oxygen has an electron-withdrawing effect However, if you compare propanoic acid and its functional derivative propionyl chloride, you can see that this molecule not only has withdrawing effects from the oxygen, but also the chloride, making it even more reactive You should be aware that out of all of the carboxylic acid derivatives, acyl halides are the most reactive towards nucleophiles due to the electron withdrawing effects of oxygen and a halide Choices B, C, and E are susceptible to nucleophilic attack since they both have electronegative substituents, but not to the same extent as propionyl chloride Choice D is the correct answer 28 A The first step in this reaction involves substitution of the -OH group in benzoic acid by the -Cl group from thionyl chloride (it has the formula SOCl2) The resulting compound is benzoyl chloride a highly reactive acyl halide which is also susceptible to nucleophilic substitution Treating this molecule with ammonia will result in substitution of the Cl group by an NH2 group, resulting in formation of an amide The molecule is named benzamide, which corresponds to choice A Avoiding the other choices here requires a bit of knowledge regarding reagents As benzoyl chloride is so susceptible to nucleophilic substitution, there is no way that a chlorine will be found in the final product if another nucleophile, such as ammonia, is around Therefore, you can discard answer choices C and D Choices B and E suggest that the ammonia can add to the benzene ring This is not possible as ammonia is not a strong enough nucleophile, and the acid functional group deactivates the ring Choices B and E also ignore the fact that the thionyl chloride will react with the benzoic acid 29 D The –CO–NH– linkage that arises between individual amino acids are known as amide linkages or more commonly, peptide linkages This linkage forms when the carboxyl group of one amino acid reacts with the amino group of another This process is characteristic of a condensation reaction and so results in the loss of water as well as peptide bond formation These bonds can link an amino acid sequence to form a dipeptide, tripeptide or a polypeptide Choice A is incorrect because this is an ester bond Choice C is also wrong because this is an ether bond Finally, choice B is incorrect because this bond is a hydrogen bond While this bond may be important in determining the secondary and tertiary structures of proteins, it does not hold amino acids together in the chain 30 C The association rule in proteins is similar to the solubility rule you learned in general chemistry Groups of similar polarity will tend to group together and influence the secondary and tertiary structure of a protein Choice C is correct since both groups are hydrocarbons and non polar in nature If these were side chains in two different amino acids, we may expect them to interact on the interior of the protein, thus contributing to its tertiary structure Choice A is incorrect since the first side chain is hydrophobic and the second is polar Remember we can recognize polar groups by the presence of electronegative atoms such as oxygen and nitrogen Anyway, this polar–non polar interaction will not occur making choice A incorrect In choice B, the first side chain is polar whereas the second is nonpolar; these two also will not interact Choice D is the same as choice A in that the first side chain is hydrophobic and apolar whereas the second is polar In choice E the first molecule a non-polar (the amine groups polarity is not enough to overcome the benzene rings non-polar nature) and the second molecule is polar These side chains will not interact and again, choice C is the correct response 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 K A P L A N 19 .. .Organic Chemistry Subject Test What is the major organic product of the reaction sequence shown? CH3CH2I A B C D E What is the major organic product of the synthetic... END OF TEST THE ANSWER KEY AND EXPLANATIONS BEGIN ON THE FOLLOWING PAGE 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... doesn't even exist (besides, Grignard reagents use Mg not Pd) 24 E This question tests your familiarity with the stereochemistry of substitution reactions In this case, we're clearly dealing with
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