Organic chemistry section test (2)

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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 Questions 3-6 refer to the following choices: Which of the following is the most stable carbocation? A B C D E A H2 C CH+ H2 C CH B CH2 + enantiomers diastereomers structural/constitutional isomers conformational isomers tautomers C CH3CH2CHCH2+ H CH3 C CH3 D CH3 andCH3CH2CH2OH OH + E CHO CH2 CH2 + CHO H C OH H C OH and OH C H H C OH CH2OH CH2OH Which is correctly paired? A O R CH C OCH2CH3 amino este and C C NH2 C C B R C NH R amide O C R H D R CHOH C α− hydroxy acid C2H5 CH3 amine NH2 C OH and C2H5 H C OH CH OH O E All of the above KAPLAN _ 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 hybridization of the carbonyl carbon in N,Ndimethylformamide? A B C D E O sp3 sp2 sp sp3d sp3d2 A H 2N C O O B A compound is separated from an ethereal mixture by extraction with cold, aqueous NaHCO3, then recovered by addition of HCl to the extract Which of the following is a likely identity of the compound? A B C D E 10 Which of the following is an appropriate resonance form of the conjugate base of p-aminobenzoic acid? O O C N O H O H 2N D E 1,3-Cyclopentadiene reacts with sodium metal at low temperatures according to: 196K C H C CH3(CH2)7CH2OH CH3(CH2)7CHO CH3(CH2)7CH2NH2 CH3(CH2)7COOH CH3(CH2)7COONa Na H 2N Na What is the best explanation for this observation ? A The reactant is more unstable at reduced temperatures B The cation formed is stabilized by aromaticity C Sodium metal is highly specific for cycloalkenes D The rehybridization of the saturated carbon atom provides additional stability to the product E The anion formed is stabilized by aromaticity C O H O N C H O 11 Which of the following statements is true of acetone? A It exists as a pair of tautomers, of which the keto form predominates B It exists as a pair of tautomers, of which the enol form predominates C It is useful as a solvent due to its unusually high boiling point D It is useful as a solvent due to its polar, protic nature E It is useful as a solvent due to its lack of reactivity toward nucleophiles 12 The compound pictured below is a member of which compound class? COOH O A B C D E C(CH3)3 carboxymethoxybenzenes methoxybenzoic acids propoxybenzoic acids t-butoxybenzoic acids carboxycumene oxides _ KAPLAN 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 Which of the following could be the formula of an ester? A B C D E C7H12O C7H12O2 C7H14O C7H16O2 C6H12O 14 What is the rate law for the elimination (E2) reaction shown below? t-BuBr + t-BuO–K+ → ← H2C=C(CH3)2 +t-BuOH + KBr A B C D E rate = k[t-BuBr] rate = k[t-BuBr][t-BuO–] rate = k[t-BuBr]2 rate = k[t-BuO–]2 rate = k[t-BuOH][H2C=C(CH3)2] 18 How many distinct organic compounds have the molecular formula C5H12? A B C D E One Two Three Four Five 19 Which of the structures below is most stable? A CH H 3C Cl B CH 15 Cyclohexane has how many fewer hydrogen atoms per molecule than does n-hexane? A B C D E Cl CH D 16 Which of the following has the most exothermic heat of hydrogenation? A B C D E Cycloheptene Cyclohexene Cyclopentene Cyclobutene Cyclopropene 17 What are the bonding types and degree measures of the bond angles normally associated with the hybridizations sp2 and sp3, respectively? A B C D E σ 109.5°, and σ 109.5° σ 109.5°, and π 109.5° σ 120°, and π 120° π 120°, and σ 120° π 120°, and σ 109.5° CH C CH E CH CH 20 Which of the choices below best describes the relationship between Z and E isomers? A B C D E Enantiomers Geometric isomers Conformational isomers Structural isomers Tautomers KAPLAN _ 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 21 What is the approximate C-N-C bond angle in diethylamine? A B C D E 25 How many different stereoisomers does the following compound have? OH OH OH OH 107° 110° 120° 150° 180° HOOC 22 What are the positions occupied by the bromine atom and the tert-butyl group respectively in the most stable conformation of trans-1-bromo-3-tert-butylcyclohexane? C(CH3)3 H A B C D E A B C D E H Bromine: axial, t-butyl: axial Bromine: equatorial, t-butyl: equatorial Bromine: axial, t-butyl: equatorial Bromine: equatorial, t-butyl: axial Bromine: either axial or equatorial, t-butyl: equatorial 23 What is the IUPAC name of the following compound? CH CH 3CH 2CHCH CCH CH CH CH CH 3CH A B C D E CH 5-ethyl-3,3-dimethylnonane 3-ethyl-5,5-dimethyloctane 3-ethyl-5,5-dimethylnonane 7-ethyl-5,5-dimethyloctane 7-ethyl-5, 5-dimethylnonane C C C H H H H H 16 26 Which of the following compounds will exhibit the greatest dipole moment? Br A B C D E C 27 (Z)-1,2-Dichloro-1,2-diphenylethene (E)-1,2-Dichloro-1,2-diphenylethene 1,2-Dichloro-1,2-diphenylethane 1,2-Dichloroethane 1,2-Difluoroethane How many structural isomers of C3H6Br2 are capable of exhibiting optical activity? A B C D E 28 Which of the compounds below would be the best starting material for the synthesis shown? ? A B C D E SOCl CH 3CH 2MgBr CH 3(CH 2) 6CH CH3(CH2)4COOH CH3(CH2)4CH3 CH2=CH(CH2)3CH3 CH3(CH2)4CH2OH CH3(CH2)2CH2OH 24 What is the product of the reaction below? CH3Cl + CH3O- Na+ A B C D E CH3CH2OCH2CH3 CH3OCH2CH3 CH3OCH3 CH2=CH2 CH3CH3 _ KAPLAN 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 elimination reaction below? H 3C H H CH C C C Cl H H KOH CH alcohol H A H 3C CH C C C H H CH CH B H 3C C C C CH H C H 3C C H CH C C Cl H H D H 3C CH C C H H H 3C E H 3C CH C H C C CH H C CH H 30 A mixture of alkyl halides is subjected to elimination (E1) with one mole of aqueous acid and alcohol If the reactant mixture initially contains one mole each of npropyl chloride and isopropyl chloride, what can be predicted about the product mixture? RX + R'X H +/H 2O CH 3CH 2OH products A A single alkene product is formed, and isopropyl chloride is present in excess B A single alkene product is formed, and n-propyl chloride is present in excess C Two distinct alkene products are formed, and npropyl chloride is present in excess D Two distinct alkene products are formed, and isopropyl chloride is present in excess E No reaction occurs because both alkyl halides used are inert to elimination (E2) STOP! END OF TEST KAPLAN _ 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 B B 13 B 19 A 25 D E D 14 B 20 B 26 A C E 15 C 21 A 27 B B 10 C 16 E 22 C 28 D D 11 A 17 E 23 C 29 A A 12 D 18 C 24 C 30 B _ KAPLAN ORGANIC CHEMISTRY SUBJECT TEST Explanations B Cation stability, like that of the free radicals, declines in the order 3°>2°>1°>methyl Choices C, D, and E, as primary cations, can therefore be disposed of first (Each would quickly rearrange via hydride shifts to form tertiary or benzylic cations.) Of the remaining choices, choice A is a vinylic cation which is rather unstable due to the high s character of the sp2 orbital The _ bond and the empty orbital are perpendicular to each other and hence the carbocation cannot be stabilized through charge delocalization Choice B, however, is an allylic cation which can be stabilized via resonance (or equivalently, delocalization of the _ electron cloud): CH2 CH CH2 CH2 CH CH2 E All of the structures listed are correctly paired with their description Choice A is an ester (RCOOR) that has a amine substituent Choice B is an amide (RCONR2), choice C is a simple (in this case primary) amine, and choice D is an acid with a hydroxy (alcohol) substituent on the α-carbon C B D A Both compounds in #3 have the same atoms, but the connectivity among them is different These are structural or constitutional isomers by definition The two compounds listed in #4 have the same two chiral centers on each and differ in the configuration around only one of these; they are thus not mirror images of each other They are therefore diastereomers The two compounds in #5 are chemically identical, differing only in their rotation about a single bond, in this case the central carbon-carbon bond These are referred to as conformational isomers The last pair of compounds are non-superimposable mirror images and are recognizable as such because the two compounds have two of the groups interchanged about a chiral center By definition, they are enantiomers Note that another switch between any two of the four groups will bring the molecule back to its original configuration B A carbonyl carbon is one that is double bonded to an oxygen Since the carbon contains one π bond and three σ bonds (one double bond and two single bonds), it must be sp2 hybridized Note that we not need to know the structure of N, N-dimethylformamide to answer this question CH O N CH N,N -Dimethylformamide D The correct answer choice has to be a compound that is hydrophobic (since it started in the ether layer), but when reacted with a weak base becomes a water soluble salt Choice A is an alcohol which is not acidic enough to react with KAPLAN _ 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 sodium bicarbonate, a weak base It would remain dissolved in the ether Choice B is an aldehyde, which can be eliminated for the same reason as A Choices C and E are both bases, so neither will react with another base Choice D, nonanoic acid, will react with sodium bicarbonate and form sodium nonanoate, a water soluble salt Extra Q: How can you tell you have a carboxylic acid by observing the addition of sodium bicarbonate? A: In addition to the formation of a salt, carbonic acid is also formed (H2CO3) which can decompose into H2O and CO2 you'll see bubbles! E When 1,3-cyclopentadiene reacts with sodium, it goes from a nonaromatic to an aromatic compound A planar cyclic compound is aromatic if it satisfies Huckel's rule—that is, it has (4n+2) π electrons, where n is any integer The product of the reaction has π electrons, so is therefore aromatic An aromatic compound is very stable because the π electrons are delocalized throughout the ring In this reaction the products are much more stable, so the reaction will be spontaneous 10 C When aminobenzoic acid loses a proton, the conjugate base assumes a charge of –1 Therefore, any resonance structure must also posses the same net charge Only answer choices B and C satisfy this requirement, and B can easily be eliminated because nitrogen is not part of the benzene ring 11 A Acetone is a compound commonly used as a solvent to promote SN2 and E2 reactions because it is aprotic and polar Since it cannot form strong intermolecular bonds and is of low molecular weight it has a low boiling point Acetone exists in two forms, with the keto form predominating by 99% This is because the carbon-oxygen double bond (the carbonyl bond) is much stronger than the carbon-carbon double bond OH O CH3 12 C CH3 CH3 C CH2 D Benzoic acid is a benzene ring with a carboxylic acid function attached directly This eliminates choices A and E Meta to the acid is a t-butoxy group, leaving only choice D 13 B An ester must have at least two oxygens; both of which are bonded to the carbonyl carbon (one via a double bond and one via a single bond) This leaves only choices B and D The maximum number of hydrogens a seven carbon ester can have is 14, This happens when all the other carbon atoms (except for the carbonyl carbon) are saturated One possibility is given below: O H2 H3C C C H2 H2 C C H2 C O CH3 Again, this is the maximum number of hydrogen atoms So choice D, with 16 H’s, cannot be correct Choice B is the correct response The fact that it has two fewer H’s means that the compound is unsaturated One possibility for the structure of choice B is as follows _ KAPLAN ORGANIC CHEMISTRY SUBJECT TEST O H2 H3C 14 C H C H2 C C H C O CH3 B For SN2 and E2 reactions, the rate law is first order with respect to both the nucleophile and the substrate In this reaction the nucleophile is the t-butoxide anion (t-BuO-) and the substrate is t-butyl bromide (t-BuBr) Thus the rate law must be B 15 C Cyclic hydrocarbons have the general formula of CxH2x, while straight chain hydrocarbons have the formula CxH2x+2, giving a difference of two Cyclohexane has 12 hydrogens, while n-hexane (straight chain) has 14 H C H H C H H H C H C H C C H H H H C H H H H12 cyclohexane,6C 16 H H C C H H H H C H C C H H H H n-hexane, 6CH14 E Heat of hydrogenation (_HH) is the amount of energy released when π bonds are converted to σ bonds by the addition of hydrogens The greater the _HH of a molecule, the greater the potential energy it has The more potential energy a molecule has, the less stable it is By comparing the _HH of different molecules, their relative stabilites can be obtained Cyclopropene, a three-membered ring, has the greatest bond strain, and is the most unstable It will therefore release the most energy upon hydrogenation, so will have the greatest _HH 17 E sp2 hybridized carbons are involved in a double bond, which consists of one sigma and one pi bond The bond angle is about 120°: 120° 120° C sp2 hybridized sp3 hybridized carbons are involved in single (sigma) bonds only, and the bond angles are about 109.5° KAPLAN _ 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 109.5° C sp3 hybridized 18 C The molecular formula C5H12 corresponds to that of a saturated alkane, since it is of the form CnH2n+2 Distinct compounds arise because of the different degrees of branching that can occur in the carbon chain: H H C H C H C H C H C H H H H H H H H C H C H H H C H C H H C H H n -pentane 19 isopentane H H C H H H H C H C H C H H H C H H neopentane A The question is asking you to decide between the different positions that substituents of cyclohexane can assume The structure that is most stable will have the least repulsion of the electron clouds of neighboring substituents It is therefore more favorable for substituents to occupy equatorial positions since they are more “out of the way.” Choice A has all the substituents in the equatorial position (designated by a bond that is not directed vertically), while all the other answer choices have substituents in the axial position (designated by a vertical bond) 20 B Z and E isomers (and their cousins cis and trans) describe the relative positions of substituents about a double bond They are by definition geometric isomers Since double bonded carbons are not chiral, they cannot be enantiomers, eliminating choice A Conformational isomers can be interconverted by simple rotation about a single bond (e.g boat/chair, eclipsed/gauche) Since no rotation can occur about a double bond, C is incorrect Structural isomers, choice D, differ about the connections between the atoms in the molecule, but have the same general formula E and Z isomers have the same connections, but differ about their arrangement in space (and are therefore classified as stereoisomers) Tautomers are a specific type of structural isomer 21 A N 3HC2HC CH2CH3 H N,N- Diethylamine The normal bond angle for an sp3 hybridized atom is 109.5o By looking at the drawing of diethylamine, you will notice that nitrogen has a free electron pair Nonbonded (or lone) electrons pairs exert stronger repulsive effects, forcing 10 KAPLAN ORGANIC CHEMISTRY SUBJECT TEST the atoms attached to nitrogen down, and closer together VSEPR predicts that the bond angle must be less than 109.5° leaving only choice A Extra Q: What is the shape of the molecule? A: Trigonal pyramidal 22 C The most stable conformation will have the least steric hindrance The t-butyl is the bulkiest substituent, so it must assume the equatorial position (this eliminates choices A & D) Since the bromine is trans to the t-butyl and is two positions away, it must assume the opposite configuration as the t-butyl group, so is therefore axial This leaves us with choice C If members of the ring are trans, and are an even number of carbons apart they will have the opposite position (equatorial vs axial), and those an odd number apart will have same positions If two members of the ring are cis, and are an even number of carbon apart they will have the same position, and odd number apart will have opposite positions These descriptions need not be memorized all you need to is just visualize the ring, keeping in mind the alternating positions of members on the ring A B Alternating postions : A and B are both above the ring A is axial and B is equatorial CH3 H3C C CH3 H3C CH3 CH3 C H H Br H H Br (If both groups were in the equatorial position, that would of course be more favorable still But then the compound would be cis, not trans.) 23 C In following the IUPAC naming procedure outlined in the review notes, we start by identifying the longest carbon chain, here being This molecule is a hydrocarbon with no double bonds, so we call it nonane Next, the carbons must be numbered such that the substituents receive the lowest numbers possible, so this molecule gets numbered left to right Finally, attached to carbon is an ethyl group, and to carbon is two methyl groups Putting this together we get answer choice C 24 C In order to identify the product, we start by recognizing and classifying the reactants Methyl chloride is a primary alkyl halide, and sodium ethoxide is a strong nucleophile What reaction mechanism does this scream out? SN2 : the ethoxide attacks while the chloride leaves, forming dimethyl ether, choice C (The sodium is just a spectator ion here) This reaction type goes by the name of the Williamson ether synthesis KAPLAN 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 H C CH 3O Cl H3C O CH3 + Cl - H H 25 D The number of stereoisomers can be calculated from the formula 2n, where n = the number of chiral carbons (since each center gives two isomers: S and R) This number is a maximum because the presence of meso compounds will reduce the number of stereoisomers (This rule holds as long as we not have double bonds, which may lead to geometric isomers.) In this case, there is no possibility of meso compound, since no matter how the groups are arranged, there will not be an internal plane of symmetry in the molecule This molecule has chiral carbons, and thus stereoisomers (Note that if the carbon on the right end had been connected to a carboxylic group (COOH) rather than a second hydrogen, there would have been chiral centers, but there would also be the possibility of having meso compounds.) 26 A (Z)-1,2-Dichloro-1,2-diphenylethene φ (E)-1,2-Dichloro-1,2-diphenylethene φ C φ C Cl Cl C Cl C φ Cl net dipole moment no net dipole moment As seen above, the Z isomer has a net dipole moment, where in the E isomer the dipole moments cancel one another out Dicholoroethane and difluorethane also have no net dipole moment 27 B Four structural isomers exist for the formula C3H6Br2: H3C C H2 H C Br Br 1, 1-dibromopropane H3C Br C Br CH3 2, 2-dibromopropane H3C H C Br H C H Br 1, 2-dibromopropane H H C Br H C H H C H Br 1, 3-dibromopropane Among these, only the third structure, 1, 2-dibromopropane, contains a chiral carbon The middle carbon is attached to four different groups: an H, a Br, a methyl group, and a –CH2Br group It can therefore be optically active None of the other choices contains a chiral carbon The compounds are all achiral and optically inactive 28 D This is a SN2 reaction made to look difficult SOCl2 reacts with primary and secondary alcohols to generate something with a very good leaving group, ClSO2– (OH– is a poor leaving group.) 12 KAPLAN ORGANIC CHEMISTRY SUBJECT TEST O O R OH + Cl S Cl R O S Cl + HC good leaving group The compound formed is reactive in nucleophilic substitution reactions The Grignard reagent is a good nucleophile To get the question correct, you have to identify R Since our product has carbons, and came from the ethyl magnesium bromide, our starting compound must have carbons attached to a leaving group; answer D 29 A KOH dissociates in to K+ and OH– ions, the latter of which is a strong base The elimination will therefore proceed via the E2 mechanism Since the hydroxide ion is not a bulky base, and the substrate is a secondary alkyl halide (rather than a more sterically hindered tertiary alkyl halide), the resulting alkene will be the more thermodynamically stable, more substituted one depicted in choice A I.e the double bond forms between the second and third carbon atoms 30 B When either of the alkyl chlorides undergoes elimination, the same alkene will be formed: propene Since this is an E1 reaction, the substrate that can form the more stable carbocation will be most reactive Isopropyl chloride, a 2o alkyl halide, can form a more stable carbocation than n-propyl chloride, a 1o alkyl halide Therefore isopropyl chloride will react the most and n-propyl chloride will be present in excess KAPLAN 13 ... (E2) STOP! END OF TEST KAPLAN _ 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 B... 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 Questions 3-6 refer to the following choices: Which of the following is the most... 11 A 17 E 23 C 29 A A 12 D 18 C 24 C 30 B _ KAPLAN ORGANIC CHEMISTRY SUBJECT TEST Explanations B Cation stability, like that of the free radicals, declines in the
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