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 TEST1 Organic Chemistry Subject Test The least stable free radical is A H Which resonance structure for acrolein would be expected to make the LEAST contribution to the actual hybrid? A H C CH2 CH CH O CH2 CH CH O CH2 CH CH O CH2 CH CH O B B C CH C H 3C CH D C H H 3C C E All of the above would make an equal contribution to the hybrid H D H3 C A CH3CH2OH B CH3CH2CH2OH C E CH2 CH Which of the following would be most soluble in water? CH2 O CH3CCH3 Which of the following is the most stable carbanion? (Φ = phenyl group) A B C D E Φ3C:– Φ2HC:– ΦH2C:– H3C:– (CH3)3C:– D H2C=CHCH3 E CH3OCH2CH3 Which of the following has the lowest heat of hydrogenation per mole of H2 absorbed? A B C D E 1,3-pentadiene 1,4-pentadiene 1,2-pentadiene 1,3,5-hexatriene 1,2,4-hexatriene 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 Sulfanilic acid, illustrated below, can be recrystallized from hot water and has an extremely high melting point When an optically active sample of 2-bromobutane, (288oC) relative to its molecular weight (173.2) depicted below, reacts with I– in an aprotic solvent, second order kinetics is observed Which of the following statements must be true? SO3H CH2CH3 H H H Br H H 2N The best explanation for the physical properties mentioned is that A The reactant has R stereochemistry while the product has S stereochemistry B The reactant has S stereochemistry while the product has R stereochemistry C Both reactant and product have the same R/S designation D The product is a racemic mixture of R and S enantiomers E The product is optically inactive Which of the following is an amide? A the amino group of sulfanilic acid acts as a Lewis acid B sulfanilic acid exists as a pair of enantiomers C sulfanilic acid exists as a pair of diastereomers D sulfanilic acid exists as a zwitterion E the sulfur atom of sulfanilic acid has an unusually low oxidation number Which of the following has the lowest boiling point? A B C D E A NH2 n-butane isobutane cis-2-butene 1-butyne 2-butanol 10 Which of the following has the highest boiling point? CH3 A B C D E B O CH3(CH2)2CH3 (CH3)2CHCH2CH3 CH3(CH2)3CH3 (CH3)2CHCH3 CH3CH2CH3 N(CH3)2 C 11 Which of the following has the highest melting point? N H D N(CH3)3 A B C D E trans-2-butene cis-2-butene n-butane 1-propanol 1-pentanol E CH3 NO2 _ KAPLAN 14 12 Which of the structures pictured below corresponds to 2-methoxy-3-butenal? ORGANIC CHO OH A CH2 CH O O C C H H H OH OH H OH CHO The two molecules pictured above are OCH3 C CHO CHO CH3 B CH3 CH CHEMISTRY SUBJECT TEST1 C A B C D E H O C meso structures enantiomers diastereomers identical racemates OCH3 CH3 CH C C CH3 15 Which of the following will be least reactive towards SN1 substitution? O A D H CH2 CH C H O H C OCH3 H C Br H B E OCH3 CH3 CH CH3 O CH CH C CH3 CH2 C H Br H 13 Which of the following statements is true regarding the relative boiling points of cis- and trans- 1,2dichloroethene? A The boiling point of the cis isomer is higher because it has a net dipole moment B The boiling point of the cis isomer is higher because it is thermodynamically more stable C The boiling point of the trans isomer is higher because the _ electrons are more delocalized D The boiling point of the trans isomer is higher because it is more symmetrical E The boiling point of the two will be equal because cis and trans isomers have identical physical properties C CH3 CH3 C Br CH3 D CH2 CH3 CH3 CH2 C Br H E H CH3 CH2 C Br H 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 16 Which of the following is true regarding SN1 and SN2 reactions? 19 Which of the following statements is true about the A SN1 reactions proceed via a carbocation compounds pictured below? intermediate; SN2 reactions so under certain CH3 CH3 conditions B Both SN1 and SN2 reactions can yield Cl H H Cl rearrangement products C SN1 reactions proceed more readily with a tertiary H Cl Cl H alkyl halide as reactant; SN2 reactions proceed more readily with a primary alkyl halide as reactant CH2CH3 CH2CH3 D The rates of SN1 reactions are largely affected by A They are diastereomers with different melting steric factors; the rates of SN2 reactions are largely points affected by electronic factors B They are diastereomers with identical boiling E Protic solvents favor SN2 over SN1 reactions points C They are enantiomers with different melting points 17 Which of the compounds below has the highest melting D They are enantiomers which rotate plane-polarized point? light in different directions E They are identical meso structures A CH (CH ) COOH B C D E CH3(CH2)4CH3 CH2=CH(CH2)3CH3 CH3(CH2)4CH2OH CH3(CH2)2CH2OH 20 Which of the following statements is true about the compounds pictured below? CH3 18 Which of the structures below corresponds to 3-cyano2-butenoyl chloride? A H3C C N C C A B C D E C Cl B H3C C C N C C H Br Br H H Br Br H CH3 O CH3 CH3 They are enantiomers and are optically active They are diastereomers and are optically active They are diastereomers and are optically inactive They are meso structures and are optically active They are meso structures and are optically inactive O Cl H C H3C C N O C C C O CH2Cl H D H3C C C N C CH2Cl H E N C H H3C C C H H O C Cl _ KAPLAN 21 What is(are) the major organic product(s) of the SN1 reaction below? CHEMISTRY SUBJECT TEST1 22 Which of the compounds below is an acid anhydride? A CH2CH3 Br ORGANIC H+ H H2O O ? CH3 O B A O CH2CH3 OH OH H CH3 C B OH CH2CH3 H OH D CH3 O CH3CH2C C CH2CH3 OH CH2CH3 H and H CH3 OH OCH3 E O CH3CH2C CH3 O O CCH3 D CH2CH3 Br 23 Which of the following compounds will have the lowest boiling point? OH A CH3CH2CH2NH2 B CH3CH2CH2COOH C CH3 E CH2CH3 Br OH and OH CH3 NH2 CH2CH3 Br CH3CHCH3 D CH3CH2NHCH3 E N(CH3)3 CH3 24 Which one of the compounds below is incapable of significant hydrogen bonding? A B C D E ethanol bromoethane propanoic acid propanamide ethanamine 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 28 The IUPAC name for the compound below is 25 Which of the following statements is true of ethene? A Both carbon atoms are sp2 hybridized and the molecule is planar B Both carbon atoms are sp2 hybridized and all bond angles are approximately 109.5° C One carbon atom is sp hybridized while the other is sp2 D Both carbon atoms are sp3 hybridized and all bond angles are approximately 109.5° E Both carbon atoms are sp hybridized and the molecule is planar 26 Which of the following depicts the most stable resonance structure of diazomethane? C N N: H B H C N CH3 CH3 CH2 CHCH2 CH2 CCH3 CH3 A B C D E 2,2-dimethyl-5-ethylheptane 5-ethyl-2,2-dimethylheptane 3-ethyl-6,6-dimethylheptane 6,6-dimethyl-3-ethylheptane Two of the above 29 Which of the following substrates is least reactive towards SN2 reactions? A B C D E A H CH2 CH3 CH3I CH3CH2Br CH3CH2CH2I (CH3)2CHCH2I (CH3)2CHCH2Br N: H C H C N N: H D H C N N: H E They are all equally stable 27 The compounds pictured below are CH3 H Br Br H H Br Br H A B C D E CH3 CH3 CH3 A B enantiomers meso compounds diastereomers conformers configurational isomers _ KAPLAN 30 All of the following represent cis-1,4dimethylcyclohexane EXCEPT ORGANIC CHEMISTRY SUBJECT TEST1 A CH3 H H3 C H B CH3 H CH3 H C H H3 C CH3 H D H3 C H H3 C H E CH3 H3 C 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 D B 13 A 19 D 25.A A D 14 C 20 E 26 B B B 15 A 21 C 27 B A 10 C 16 C 22 E 28 B D 11 E 17 A 23 E 29 E A 12 D 18 A 24 B 30 C _ KAPLAN ORGANIC CHEMISTRY SUBJECT TEST1 Explanations D A trivalent carbon with an unpaired electron does not possess an octet and can hence be considered electron deficient The free radical, then, just like carbocations, is stabilized by the presence of groups which can share electrons inductively (alkyl groups) or defuse the electron deficiency through resonance (conjugated systems) The order of free radical stability goes 3°>2°>1°>methyl Allylic and benzylic radicals are similar to tertiary radicals in terms of stability because of resonance delocalization The free radicals shown in choice A (benzylic), choice B (tertiary), and choice E (allylic) are all relatively stable, and therefore these three choices are incorrect The radicals offered in choices C and D are both relatively unstable, but the methyl radical depicted in choice D is more unstable than the primary radical in choice C Thus choice D, as the most unstable free radical, is the credited choice A Carbanions are stabilized either by electronegative substituent groups (absent in this case) or by resonance with _ electrons Choices D and E not allow for any resonance stabilization and will therefore be less stable than the other three choices The anion in choice A allows the negative charge to be delocalized, through resonance, over nine other carbon atoms, three in each phenyl ring Choices B and C allow for delocalization over six additional carbons and three additional carbons, respectively Choice A thus allows for the greatest amount of charge delocalization of any of the choices offered, and is thus the most stable carbanion B The more energetically favorable a resonance structure is, the more it will contribute to the actual structure of the hybrid In other words, the actual electron density distribution of the hybrid molecule will more likely resemble the more stable resonance structures In looking for the resonance structure expected to make the least contribution, then, we are looking for energetically unfavorable structures It is highly unfavorable for electronegative species such as oxygen to bear a formal positive charge, thus making B the correct choice Choice A is the formal structural formula for acrolein It is the most stable of all the different resonance structures one can draw because there is no separation of charge Choices C and D are reasonable resonance forms which show how the conjugated double bonds delocalize the _ electrons and stabilize the molecule: the electronegative oxygen bears a formal negative charge which is certainly acceptable A Questions on the physical properties of organic compounds are best answered based on general chemistry principles, most notably those related to bond polarity and molecular interactions Hydrogen bonding is the strongest intermolecular force in which water can participate; alcohols also participate in hydrogen bonding, but this capability decreases with the increasing size of the alkyl portion of the alcohol molecule, because the relatively nonpolar alkyl group disrupts the hydrogen bonding network Ethanol, choice A, has a smaller alkyl group than propanol, choice B Ethanol thus has a greater capacity for hydrogen bonding and will be more soluble in water than the propanol will be Choices C and E show a ketone and an ether respectively; these compounds are not as soluble in water since, even though they have polar carbon-oxygen bonds, they have no hydrogens bonded to electronegative atoms that can contribute to hydrogen bonding Choice D, an alkene, is the least soluble in water of the choices offered D While this question is directly asking about the relative heats of hydrogenation of various polyenes, indirectly it is really asking, "Which of the following has the most stable _ system?" Recall that conjugation lends special stability to compounds with more than one _ bond Thus the question could be further translated to "Which of the following has the most extensive conjugated _ system?" Choice D is correct since 1,3,5-hexatriene has three double bonds in complete conjugation; it is thus more stable than choice E, for instance, because the first two double bonds in the compound 1,2,4-hexatriene are cumulated, 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 not conjugated Choice A, 1,3-pentadiene, is conjugated but has only two _ bonds, choice B, 1,4-pentadiene, has two _ bonds which are isolated, not conjugated, and choice C, 1,2-pentadiene, has two cumulated rather than conjugated _ bonds Each of these compounds is thus less stable “per double bond” than the 1,3,5-hexatriene of choice D A The reactant is a secondary alkyl halide; as such, it will react with iodide ions in an aprotic solvent via an SN2 pathway This fact is supported by the statement that the reaction proceeds with second order kinetics, the rate observation expected for an SN2 reaction Since SN2 reactions are accompanied by inversion of configuration, in most cases, it follows that if the reactant is R, then the product will be S, vice versa We must therefore next determine whether the reactant is R or S It would be most helpful to convert the Newman projection shown into a Fischer projection from which to determine the R/S designation The carbon in the background of the Newman projection shown has three hydrogens attached to it, so it is a methyl group Writing a Fischer projection for this compound then requires that we place the methyl group, the ethyl group, the bromine atom, and the hydrogen atom (the four groups attached to the chiral carbon in the foreground of the Newman projection) in the appropriate spatial order H Br C CH2CH3 CH3 It should now be evident that the reactant is the R enantiomer of 2-bromobutane, and therefore the inverted product will be the S enantiomer of 2-iodobutane, and choice A is confirmed as the correct answer B Amides are defined as organic compounds in which an amine functionality is directly bonded to a carbonyl carbon Choices A, C, and D are all simple amines (primary, secondary, and tertiary respectively); Choice E is a nitro compound, which is an organic compound with an —NO2 group Only choice B has an amine group, —N(CH3)2, bonded to a carbonyl carbon D O O O O S S OH H N H H H O- N H A zwitterion is a molecule that has both positive and negative charges Usually amphoteric species, molecules that have both acid and basic functional groups, can form zwitterions (e.g amino acids) Sulfanilic acid has both an acid functional group (–SO3H) and a basic functional group (–NH2), and can therefore have both a positive and negative charge simultaneously (see diagram above of conjugate base and acid) Zwitterions have strong intermolecular attractions due to ionic and hydrogen bonding, making D the correct choice Any correct answer choice has to involve intermolecular attractions, since that is what provides for high boiling points relative to molecular weight No other answer choice addresses intermolecular attractions, so can be eliminated Additionally, all the other answer choices are stated incorrectly The 10 KAPLAN ORGANIC CHEMISTRY SUBJECT TEST1 nitrogen of the amine group acts as a Lewis base (it has a free electron pair to donate), not a Lewis acid Sulfanilic acid has no chiral centers, so answer choices B and C can be eliminated The sulfur atom is attached to three oxygens, so it must have a very high oxidation state B Boiling requires that the molecules of a compound in the liquid phase overcome the intermolecular attractive forces and escape into the gas phase: the stronger these interactions are, the more thermal energy one will have to put in to pull the molecules apart, and hence the higher the boiling point Choice E is obviously incorrect, as the alcohol will exhibit hydrogen bonding, and will have a very high boiling point compared with similar compounds with no hydrogen bonding Choices C and D are incorrect because unsaturated compounds have an inherent (albeit small) dipole moment that raises their boiling point compared with their saturated counterparts When evaluating choices A and B, the rule to follow is that compounds have a lower boiling point with increased branching, since branching disrupts the spatial packing of molecules in the condensed phase and hence compromises intermolecular attractions Choice B, the most branched of the saturated compounds without hydrogen bonding, is the correct choice 10 C All of the listed compounds are unsubstituted alkanes Thus, the highest boiling point will belong to the longest, least branched compound: the longer the alkane, the more it weighs and the more difficult it will be to “push” the molecule into the gas phase Also, longer alkanes experience stronger van der Waals forces The two longest compounds are choices B and C, each a five-carbon molecule The highest boiling point belongs to the less branched alkane, in this case the straight chain pentane, or choice C 11 E Melting point is by and large dictated by the same characteristics that determine boiling point: intermolecular attraction, weight, and geometry In examining all of these compounds, only two have hydrogen bonding Hydrogen bonding greatly increases the interaction between molecules, giving them much higher melting points than similar compounds of similar size without hydrogen bonding Thus, choices D and E are likely candidates Of the two, choice E is larger than choice D (a fivecarbon chain versus a three-carbon chain) So, choice E has the highest melting point 12 D 2-methoxy-3-butenal is a five carbon (but + meth) unsaturated (buten) aldehyde (butenal), or an alkene with a terminal carbonyl group This eliminates choices A and C, which are not aldehydes at all but ketones Choice E is a six-carbon molecule (pentenal with a methoxy group) and also is incorrect In numbering the carbon atoms, the carbonyl carbon is labeled carbon This means that the methoxy group at carbon must be next to the aldehyde carbon, with the double bond starting at the next carbon (#3) This is the compound shown in choice D Choice B is incorrect because it is 2-methoxy-2butenal 13 A Cis and trans isomers are diastereomers with generally different physical properties, and so choice E can be eliminated The question comes down to which of the two isomers has the stronger intermolecular attractions For the cis isomer, the dipole moments of the two C-Cl bonds not point directly opposite each other, and therefore the molecule will have a permanent dipole moment when compared to the trans counterpart, in which the two dipole moments cancel each other vectorially Choice A is thus correct because this asymmetry in the cis isomer leads to a net dipole moment, which in turn gives the stronger intermolecular forces and higher boiling point 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 14 C Diastereomers are stereoisomers that are not mirror images of each other Choice A is incorrect because only the compound listed on the right is meso Choice B is incorrect because enantiomers are mirror image isomers, which the two compounds are not This also eliminates choice E, because racemates are enantiomeric pairs Choice D is obviously incorrect because the two molecules, while similar, are not identical 15 A SN1 reactions proceed through a carbocation intermediate, and are therefore dependent upon the quality of the leaving group and the stability of the carbocation In the compounds listed in choices A–E, bromine is the leaving group This leaves only the stability of the carbocation intermediate to determine which compound is least likely to undergo an SN1 reaction Carbocation stability follows degree of substitution, i.e 3°>2°>1°>methyl Since choice A, methyl bromide, will form the least stable carbocation, it will be the least likely to undergo an SN1 reaction 16 C The correct choice is C Choice A is incorrect because while it is true that SN1 reactions proceed through a carbocation intermediate, SN2 reactions never do: it is a single-step reaction that goes through a transition state instead of an intermediate Choice B is false because only the cation intermediate in SN1 reactions can undergo rearrangement to form a more stable (more highly substituted) carbocation; SN2 reactions occur in one step and not allow time for rearrangement to take place Choice C is a true statement because SN1 reactions favor substituted alkyl halides for the stable carbocations that they form, whereas SN2 reactions favor unsubstituted reactants for their minimal steric hindrance Choice D is incorrect from what we just said: the exact opposite is true Choice E is incorrect because SN2 reactions are favored by polar aprotic solvents which cannot stabilize the (usually anionic) nucleophile, making it more reactive 17 A The student should be able to eliminate Choices B and C immediately, as they have absolutely no hydrogen bonding Choices D and E are both terminal straight chain alcohols; thus the heavier compound (Choice D) has the higher melting point of the two This leaves choices A and D, compounds with similar chain length but different functional groups Choice A is correct because in addition to the hydroxyl group, it also has a polar carbonyl bond which serves to increase intermolecular attractions 18 A The correct choice is A Choices C and D can be eliminated straight away because neither is an acid chloride Numbering of the four-carbon chain is simple The first carbon is the carbonyl carbon Proceeding down the carbon chain, the double bond begins at the second carbon (2-buten), and the cyano group is on the third carbon (3-cyano) This is compound A 19 D A quick examination of the two compounds will reveal that they are non-superimposable mirror images, and therefore enantiomeric This eliminates choices A and B because diastereomers are stereoisomers that are not mirror images Of choices C and D, C is incorrect because the physical properties of enantiomers are identical Choice E is incorrect because meso structures possess a plane of symmetry and are hence achiral 20 12 E The two compounds are actually identical compounds, and correspond to a meso structure since it has two chiral centers but nonetheless possesses a plane of symmetry Meso compounds are all optically inactive, and therefore choice D cannot be true KAPLAN 21 ORGANIC CHEMISTRY SUBJECT TEST1 C This answer is easily obtainable if two criteria for this reaction can be remembered First, SN1 reactions proceed through a planar intermediate in which the central carbon atom is sp2 hybridized This means that the nucleophile can attack from above or below the plane of the cation, yielding a racemic mixture of compounds Since we started with an optically active compound, the correct choice will have two compounds listed that differ in the configuration around one carbon atom This eliminates choices A, B, and D Secondly, the bromide ion is the obvious leaving group (stable in water, or will form HBr in an organic solvent in the presence of the H+ ion), so the compounds listed in the correct choice will not include a bromide ion This eliminates choice E Therefore, choice C is the correct answer 22 E Acid anhydrides are two acid molecules that are linked with an ester-like bond with the loss of water They are very reactive, and will regenerate the acids used to form them when water is added This relationship for the acid anhydride in question can easily be visualized as follows: O CH3CH2C + OH HO O -H2O CCH3 +H2O O CH3CH2C O O CCH3 Choice A is a diketone Choice B is an a-keto alcohol Choice C is phenol Choice D is an ester 23 E Since we are looking for the lowest boiling point, we should look for compounds in which there is little or no hydrogen bonding All the choices except for choice E possess hydrogens bonded to an electronegative atom (O or N in this case): choice B is a carboxylic acid which, in addition to the capacity for hydrogen-bonding, also possesses a polar carbonyl bond which increases its boiling point Choices A and C are primary amines, each with two hydrogen atoms directly bonded to nitrogen Choice D is a secondary amine with one hydrogen capable of hydrogen bonding Compound E, however, is a tertiary amine, which means that the nitrogen atom does not directly bond to a hydrogen atom This lack of possibility for hydrogen bonding translates into the lowest boiling point of the five 24 B Hydrogen bonding is only found in compounds where a hydrogen atom is bonded to a very electronegative atom, such as oxygen, fluorine, or nitrogen It is readily apparent that the only compound where a hydrogen atom is not bound to an electronegative atom is choice B, where all of the hydrogens are bound to carbon atoms 25 A The two carbon atoms in ethene are bonded to each other via a double bond They are thus both sp2 hybridized and the three attached groups each has will be arranged in a planar configuration roughly 120_ apart, since that will minimize the electron-pair repulsion Choice B is wrong because of the bond angle The other choices are wrong because both carbons are sp2 hybridized; sp hybridization is found on carbon atoms with two double bonds (allenes) or, more commonly, a triple bond and a single bond (alkynes), while sp3 hybridization is found on saturated carbon atoms KAPLAN _ 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 26 B Choice B is the only resonance structure shown in which both nitrogen atoms and the carbon atom have an octet of valence electrons Choice A is incorrect because it shows the carbon atom with 10 valence electrons One may argue that it does not even constitute a valid resonance structure The carbon atom in the structure shown in choice C only has valence electrons, and will therefore not be as stable as B Choice D is also not as stable because the nitrogen atom on the end only has valence electrons It also involves more charge separation than choice B 27 B The two structures shown actually represent the same compound It is a meso structure because it has a plane of symmetry while possessing two chiral centers Choice A is incorrect because enantiomers are mirror images that are not superimposable upon each other; the two compounds are mirror images of each other but are superimposable (identical) Choice C is incorrect because diastereomers are stereoisomers (isomers that differ only in the way their atoms are arranged in space) that are not mirror images and likely to exhibit different physical and chemical properties Choices D and E are incorrect because conformers or conformational isomers are compounds that differ only in rotation around one or more single bonds, and are essentially different rotated or bent shapes of the same molecule 28 B Choice E is incorrect because any given organic compound has only one IUPAC name (with the exception of certain common names that are recognized by the IUPAC) The first rule is to find the longest carbon chain, which is 7, making this a heptane (no multiple bonds or non-alkyl side groups) Then, the side chains (two methyl on the same carbon and one ethyl on another) are named, and their positions numbered to yield the lowest set of numbers is given for the side chains (2 and 5, respectively) Add prefixes to show how many side chains there are (ethyl and dimethyl), and alphabetize (not counting the prefix) The compound is 5-ethyl-2,2-dimethylheptane 29 E SN2 reactions are inhibited by bulky, branched side chains and favored by good leaving groups Choice A, iodomethane (methyl iodide), is the most unhindered and equipped with a good leaving group: the iodide ion It will therefore react readily via the SN2 mechanism Straight chain alkyl halides such as choices B and C are also likely to undergo SN2 type reactions because they are primary, straight chain (unhindered) alkyl hydrocarbons Of choices D and E, neither is a very good candidate for an SN2 reaction because even though they are still primary halides, they are branched and therefore slightly more sterically hindered relative to the other choices The only difference between these two choices is the nature of the halogen which will act as the leaving group in its ionic form Therefore, the choice for the worst SN2 reactivity would be choice E, which has the worse leaving group: bromide rather than iodide ion 30 C Cis compounds are compounds in which the substituents (methyl groups in this case) are on the same side of the molecule in question It should be obvious that D and E are incorrect because the methyl groups are on the same side of the cyclohexane ring in both species In choices A and B, the structures drawn depict one methyl group in the axial and one methyl group in the equatorial position; they are both on the same side of the molecule relative to the other substituents (the hydrogen atoms) Only in the compound shown in choice C are both methyl groups equatorial, and on opposite sides of the molecule (trans) Note that this molecule would still be trans if both methyl groups had been portrayed in the axial position 14 KAPLAN ... ORGANIC CHEMISTRY SUBJECT TEST1 Organic Chemistry Subject Test The least stable free radical is A H Which resonance structure... mentioned is that A The reactant has R stereochemistry while the product has S stereochemistry B The reactant has S stereochemistry while the product has R stereochemistry C Both reactant and product... 21 What is(are) the major organic product(s) of the SN1 reaction below? CHEMISTRY SUBJECT TEST1 22 Which of the compounds below is an acid anhydride? A CH2CH3 Br ORGANIC H+ H H2O O ? CH3 O B