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 Rank the following structures in order of increasing basicity: H Which of the following structures properly represents a zwitterion? A H NH2 N H N H COOH H H B H NH3 I analine II ammonia H COOH H H C H N NH3 N CH3 H CH3 H CH3 III trimethylamine A B C D E I, IV, II, III IV, I, II, III II, III, IV, I III, II, I, IV III, IV, I, II COO D NO IV p-nitro-analine NH2 H COO H E None of the above O CH3OH + CH3COH ? A CH3OH B CH3OCH3 C O CH3OCCH3 D O CH3CCH3 E CH3CH2OH 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 following reaction? NO2 What will be the favored product of the following reaction? O Cl2 CH3 O S AlCl + O A H NO2 Cl CH3 C CH3 OK CH3 A CH3 O Cl C CH3 CH3 B Cl H B NO2 C C O NO2 O S O Cl Cl CH3 O D CH3 Cl C CH3 CH3 NO2 D E NO2 E CH3 O C CH3 CH3 Cl H _ 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 reagents are necessary for the following reaction to occur? For the following reaction, what will be the two major products? N(CH3 )2 CH3CH2MgBr + ? ∅ CH3(CH2)3OH A CH3CH2OH B CO2, HCl C CH3CH HNO3 H2SO4 NO2 O, H2O D A O H2C N(CH 3)2 CH2 , H2O N(CH 3)2 O2N E None of the above NO2 NO2 O2N B N(CH 3)2 N(CH 3)2 O2N NO2 NO2 NO2 C N(CH 3)2 N(CH 3)2 O2N NO2 NO2 N(CH 3)2 D O2N NO2 N(CH 3)2 NO2 NO2 NO2 E None of the above At the isoelectric pH of a certain amino acid solution, the amino acid may exist as a zwitterion This is A B C D a positively charged ion a negatively charged ion either a positively or a negatively charged ion an ion that carries both a positive and a negative charge E an ion without a constant charge 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 ROOR ,∆ A CH3CH=CH2 + HBr → CH3CH2CH2Br B (CH3)2C=CH2 + HCl ∅ (CH3)3CCl OH (CH3) 2CCH2Cl C (CH3)2C=CH2 + Cl2/H2O ∅ H3C Br H Br E Br H Br D + Br2 and H3C Br H Br Which is an example of anti-Markovnikov addition? and H3C Br anti addition H Br H3C H Br Br H E + Br2 syn addition H H Br Br 10 Br2 CCl4 H3C The product(s) of the above reaction is (are) A Br Br H3C CH3 B and H3C Br Br H H3C Br Br H C Br H and Br H3C H H3C 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 12 What is the most effective reducing agent for ethene? 11 OCH3 H2 N NH KOH O C P A B C D E KMnO4 H3O+/H2O H2/Pt Fe/HCl HNO3/H2SO4 OCH3 C4H9 13 Which of the following substituted phenols is the most acidic? The product P is A A OCH3 CH3 O OH B HO C C 4H9 OCH3 H OH B NO2 H C OH O C H C4 H9 Cl D C H H2N E O C OH OH C4H D NO2 OCH C5 H11 OCH E 14 An aldol condensation involves the base-catalyzed reaction between A B C D E OCH3 NH2 OH an ether and an alcohol an alcohol and an alcohol an aldehyde and an alcohol a ketone and an alcohol an aldehyde and an aldehyde NH2 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 When placed in a basic solution, the ketone: CH3 C C O H undergoes racemization, while the ketone of a similar structure: 17 Which of the following compounds would be most reactive towards chlorination of the aromatic ring? A NH2 B NO2 CH3 C C O C2H5 C does not What could account for this observation? A Alkyl groups are electron-releasing B The second compound forms a carbanion in basic solution, while the first one does not C The base catalyzes carbocation rearrangement D There is no a-hydrogen in the second compound E OH– adds to the carbonyl carbon in the second molecule, generating a new chiral center D SO3H E Br 16 In the light-activated chlorination of methane, the chain-initiating step is uv → 2Cl• A Cl2  uv B CH4 + Cl2  → CH3Cl + HCl uv C •CH3 + Cl2  → CH3Cl + Cl• uv D Cl• + CH4  → •CH3 + HCl uv E Cl• + Cl•  → Cl2 18 Which of the following is the most reactive as a hydride donor? A B C D E H2NNH2 CH3MgBr CH3+ LiAlH4 All are equally reactive 19 Which compound below will undergo oxidation without the cleavage of any carbon-carbon bond? A B C D E t-butyl alcohol acetone acetaldehyde ethyl methyl ketone None 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 Questions 20-24 refer to the following choices: A B C D E 25 What is the major organic product of the reaction below? ortho/para-directing with activation ortho/para-directing with deactivation meta-directing with activation meta-directing with deactivation neither directing nor activating C(CH 3) Br hv What is the effect of each of the following substituents on electrophilic aromatic substitution? C(CH 3) 20 —F A 21 —NO3 Br 22 —OCH3 23 C(CH 3) NH C O CH3 B 24 —SO3H Br Br C(CH 3) C Br C(CH 3) D E Br C(CH 3) 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 26 IR spectroscopy would be most useful in distinguishing between which of the following pairs of compounds? A O O CH 3CH 2CH 2CCH 2CH and 28 Which of the structures below corresponds to paranitrobenzenesulfonic acid? SO 3H A NO CH 3C CH2CH 2CH SO 3H B B CH and CH 3CH CH 2CH 2CH3 NO CH 3CH 2CH 2CH 2CH2CH C NO SH C O 2N and SO 3H NO CHO CHO D CH 3CH 2OCH 2CH 2CH and CH 3CH 2CH 2CH 2OCH E CH 3CH 2CH 2CH 2OCH and CH 3CH 2CH 2CH 2CH2OH D H 2N SO H E O 2N 27 What is the major product of the reaction below? O CH 3CH 2MgBr + H 3O A CH3CH2 C Br D OH B + Br ∅ CH3CH2Br CH CH + Br ∅ CH CH + HBr CH CH + Br ∅ CH Br + CH CH CH + CH CH ∅ CH CH CH CH B CH3CH3 + Br ∅ CH3CH2Br + H OCH 2CH A 29 Which of the following is a productive propagation step in the free radical bromination of ethane? E 3 3 3 3 3 2 CH 2CH Br OCH 2CH C OH D E O _ 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 30 Which of the following is a major organic product of the reaction below? OCH CH 3COCl AlCl Cl OCCH A CH OCH H 3C B C O OAlCl C O C D OCH H 3C OCH E Cl Cl STOP! END OF TEST 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 ANSWER KEY B B 13 E 19 C 25 D C D 14 E 20 B 26 E C A 15 D 21 D 27 B B 10 B 16 A 22 A 28 E B 11 D 17 A 23 A 29 C D 12 C 18 D 24 D 30 B 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 EXPLANATIONS B The basicity of amines is dependent upon the stability of the unshared electron pair Electron-withdrawing groups decrease basicity, while electron-donating groups increase basicity The benzene ring stabilizes the electron pair through resonance, while methyl groups are electron-donating Using ammonia as a reference point, it is clear that the two aniline derivatives (compounds I and IV) must be less basic than ammonia (compound II) which is itself less basic than trimethylamine (compound III) p-Nitroaniline would be even less basic that aniline because the nitro group which is electron-withdrawing would add to the electron-withdrawing characteristics of the benzene ring Thus, the order of increasing basicity is IV, I, II, and III, or choice B C The definition of a zwitterion is a species that has both a positive and a negative charge Choice C, which represents the amino acid glycine at neutral pH, is indeed a zwitterion Choice A is incorrect because it is not even an ion! Choice B is simply a cation, while choice D is an anion C Alcohols and carboxylic acids will condense to form esters, a reaction known, appropriately, as esterification This information in itself is enough to answer the question, since only choice C is an ester The reaction proceeds as follows: the oxygen atom on the alcohol acts as a nucleophile and attacks the carbonyl carbon, which becomes sp3 hybridized with four groups attached The carbonyl double bond is restored as the —OH group originally on the carboxylic acid becomes protonated and leaves as a water molecule B Nitro groups are strongly deactivating meta directors for electrophilic aromatic substitutions The chloro group will attach at the meta position However, halides are also deactivating, so the presence of two deactivating species (nitro and chloro) makes a third substitution unlikely In addition, halides are ortho/para directors, so their influence would be in conflict with that of the nitro group B The p-toluenesulfonate group, or tosylate group, is an excellent leaving group The molecule will therefore be expected to undergo either a nucleophilic substitution or an elimination reaction The other reactant, a tert-butoxide ion, is a strong and bulky base which tends to favor an elimination reaction The extraction of a proton from the cyclohexane and the departure of the leaving group will thus lead to the formation of cyclohexene D The Grignard reagent (ethyl magnesium bromide) is a potent nucleophile Choice A, ethanol, is an unlikely target for a nucleophile, as it does not have a good leaving group in the absence of a proton source Carbon dioxide would be a good target for the Grignard reagent, but would only result in a three carbon species, whereas the target product has four carbons Choice C will give a four carbon alcohol, but it will yield 2-butanol instead of 1-butanol as desired: the ethyl group from the Grignard reagent will add to the carbonyl group However, when the Grignard reagent is added to the reagents in choice D, the nucleophile will add to one of the carbons of the epoxide The bond between that carbon and the oxygen atom would break, thus opening up the ring The other carbon will retain the oxygen, which will become a primary alcohol when water is added 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 B The —N(CH3)2 group is a strongly activating ortho/para director for electrophilic additions While it is true that nitro groups are deactivating and meta directing, the effect of the dimethylamino group overpowers that of the nitro group This will lead to nitration in the ortho/para positions relative to the —N(CH3)2 substituent, which are shown in choice B The second product shown in choice C is also an ortho-substituted product, but compared to the para-substituted product in choice B, it will not be as favored because of steric effects D Zwitterions are neutral molecules that carry both a positive and a negative charge Many amino acids are zwitterions in neutral aqueous solutions because they have a positively charged group (—NH3+) and a negatively charged group (— COO–) At other pHs, one of the groups may change their protonation state, thus losing the charge it carries, resulting in the amino acid acquiring a net positive or negative charge Choices A and B are incorrect because they are cations and anions, respectively Choice C is actually the definition of an ion, while choice E describes multivalent ions, such as iron (Fe2+, Fe3+) A It is perhaps best to first understand what Markovnikov addition is before addressing the case of anti-Markovnikov addition In the addition of a hydrogen halide (or any nonsymmetrical reagent) to a double bond, the reaction often proceeds through an ionic mechanism: the bond between the two parts of the molecule breaks heterolytically as the positive part acts as an electrophile and adds to one of the carbon atoms, leading to a carbocation intermediate The other, negative portion of the molecule then add to the other carbon atom In the case where the substituent groups on the two carbon atoms are different, Markovnikov’s rule predicts which portion of the molecule will add to which carbon: the addition will proceed such that the most stable (most highly substituted) carbocation is formed as the intermediate For a hydrogen halide, then, the hydrogen (as a proton) will add to the less substituted carbon atom, so that the resulting positive charge will reside on the more heavily substituted one The halide ion will then add to this more highly substituted carbon With HBr (but not any other kind of hydrogen halide!) in the presence of peroxides (the conditions in choice A), however, it has been observed that the addition product is different from what is predicted by Markovnikov’s rule: the bromine ends up on the less substituted carbon This is known as anti-Markovnikov addition This phenomenon arises because in the presence of peroxides, the addition proceed through a radical mechanism rather than the usual ionic one: the peroxide ROOR cleaves upon heating to generate two alkoxyl radicals which extract the hydrogen atom from HBr The neutral bromine atom (not bromide ion!) then attacks the less substituted (and hence less sterically hindered) carbon atom to produce the more stable (more highly substituted) radical intermediate Addition is complete with the extraction of hydrogen atom by the radical intermediate Choice B is incorrect because it depicts a normal Markovnikov addition: the proton adds to the terminal carbon, leading to a stable tertiary carbocation intermediate, to which the chloride ion then adds Choice C represents halohydrin formation: the first step is identical to halogen addition, i.e the formation of a cyclic carbocation The next step, however, has water as the attacking nucleophile that opens up the ring Subsequent deprotonation then yields the halohydrin The water molecule attacks the more highly substituted carbon atom preferentially Choices D and E are incorrect because in the case of a symmetrical addition reagent (molecular bromine here) the Markovnikov vs anti-Markovnikov designation is irrelevant 10 B The bromination of alkene proceeds through a cyclic bromonium intermediate, forming a three-membered ring consisting of bromine and the two carbon atoms involved in the double bond The remaining bromide ion, acting as a nucleophile, will add in an orientation anti to the first one as it opens up the ring Two possible products (enantiomers of each other) will form (See diagram on the next page.) 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 Choice A is incorrect: where did the extra methyl group come from? Choices C and E are both incorrect because in the absence of radical-generating agents, the bromine will add only to the unsaturated carbons involved in the double bond Choice D is incorrect because the two bromine atoms will add to adjacent carbons rather than the same carbon Br C Br H3C Br + Br C C H3C C H3C Br H Br+ Br Br OR C H C H3C H3C Br Br- 11 D This is an example of the Wolff-Kishner reduction of a ketone to an alkane (conversion of —C=O groups into —CH2 groups), often seen following a Friedel-Crafts acylation reaction This is a synthetically useful pathway because a straightforward Friedel-Crafts alkylation may lead to the rearrangement of the alkyl group one is trying to add Another way to reduce the acyl group to an alkyl group is by the Clemmensen reduction which, unlike the Wolff-Kishner reduction, takes place under acidic conditions Choice A is a partially reduced compound that would result if we had used hydride donors like LiAlH4 or NaBH4 instead With the conditions and reagents in this question, however, the oxygen is completely removed Choices B and C are incorrect because the reaction conditions are insufficient to break the relatively stable Ar–O bonds (requires HI or HBr at high temperatures) Choice E is incorrect for the same reasons as B and C, and also because hydrazine is not a reagent that is used to donate amine groups to a benzene ring 12 C Hydrogenation of alkenes in the presence of a metal catalyst (Pt, Ni, or Pd) is a standard reaction that every student should be well familiar with Choice A is incorrect because KMnO4 is a strong oxidizing agent that will cleave a double bond by oxidation or produce a geminal diol, depending upon reaction conditions Choice B lists the conditions for acidcatalyzed hydration: it will lead to the addition of a hydrogen atom on one side but a hydroxyl group on the other Neither choice D nor choice E is capable of addition of hydrogen atoms: in particular, you should recognize the reagents in choice E as those for the nitration of aromatic compounds in electrophilic aromatic substitution reactions 13 E To stabilize the negative charge of the phenolate ion (the resultant anion when a phenol acts as an acid and deprotonates), an electron-withdrawing group is required Both methoxy and amine groups are electron-donating, so choices A and D can be eliminated Chlorine is electron-withdrawing, but only weakly so compared to the nitro group, so choice C can also be eliminated This leaves either the m-nitrophenol or o-nitrophenol (choices B and E respectively) Nitro-substituted aromatic rings are meta directing This means that the electron density will be higher in the m position 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 than in the o/p position Since we are looking for low electron density (to stabilize the negative charge) at the hydroxyl group position, choice E would be more acidic than choice B 14 E Aldol condensations occur between aldehydes, one acting as a carbanion nucleophile and the other as an electrophile The end product is a b-hydroxyaldehyde, i.e it has both an aldehyde and an alcohol functionality, hence the name aldol condensation The base catalyzes the reaction by increasing the rate of tautomerization of the aldehyde into the enolate ion The enolate ion is a nucleophile and can attach to the carbonyl carbon of the other aldehyde molecule The reaction is illustrated below for the case of two formaldehyde molecules: O- O CH2 CH CH2 O -:CH2 CH CH H H3C -:OH CH O OH O O CH3CHCH 2CH CH3CHCH 2CH : OH OH Note that the β-hydroxyaldehyde product is rather unstable and can undergo dehydration to give an unsaturated compound in which a carbon-carbon double bond is conjugated with the carbonyl bond 15 D The hydrogen atom attached to the carbon atom next to the carbonyl carbon (the α-hydrogen, the carbon to which it is attached being the α-carbon) is relatively acidic because the resulting anion upon deprotonation can be stabilized via resonance by the carbonyl oxygen The resulting carbanion can then be reprotonated from above or below the plane of the ion, giving a racemic mixture Since there is no α-hydrogen in the second compound, it will not racemize in a basic solution While the statement in choice A is correct, it has nothing to with the phenomenon we are trying to account for Choice B is false: none of the hydrogens in the second compound is expected to be acidic enough to be plucked off, whereas the first molecule definitely yields a carbanion upon deprotonation Choice C is incorrect because no carbocation is formed; besides, rearrangement would change the constitutional structure of the molecule, instead of its chiral properties Choice E is incorrect for two reasons First, both compounds could conceivably be attacked by OH- groups Second, attack of the carbonyl by OH- would result in a gem diol, which would not be chiral, as is stated 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 16 A There are three steps in free radical reactions The first is the initiation, where a bond is cleaved homolytically to produce free radicals This is depicted in choice A: the UV radiation breaks the Cl-Cl bond to give two neutral chlorine atoms, which are radicals because they each possess an unpaired electron Chain propagation steps are steps in which one free radical is used to generate another This is shown in choices C and D Finally, chain termination steps describe reactions where two free radicals combine to form a molecule, as in choice E These steps “soak up” free radicals, and eventually will stop the chain reaction Choice B does not involve any radicals, and can be thought of as just a double displacement reaction 17 A Amine groups are electron-donating, and thus will activate electrophilic substitutions to the benzene ring Choices B, D, and E are all incorrect because the substituents to the benzene rings shown are all electron-withdrawing, and thus deactivating Choice C is incorrect because electron-donating species increase the rate of electrophilic aromatic substitution over and above that of the straight benzene ring 18 D This question may be very easily answered if one remembers the name of LiAlH4 (or LAH) to be lithium aluminum hydride It is a powerful hydride donor and is commonly used as a reducing agent Choice A, hydrazine, although a reducing agent, is not a hydride donor Choice B is a Grignard reagent and can be considered a carbanion donor, but not a hydride donor per se Choice C is incorrect because a cation is much more likely to grab hydrides than to donate them 19 C This question can be answered without knowledge of any specific organic reactions, as long as one is able to identify the structures of the molecules and knows what oxidation involves In the context of organic chemistry, oxidation is most conveniently thought of as increasing the number of bonds to oxygen The carbon atoms in molecules A–D all have four bonds already (either four single bonds or two single and one double bond) To oxidize them further would require the formation of another carbon-oxygen bond (either forming a new C-O single bond or turning a C-O bond to a double, i.e carbonyl bond), which can only be formed at the expense of cleaving another (non C-O) bond In choices A, B, and D, all the non-C-O bonds are carbon-carbon bonds, and so oxidation would involve the cleavage of one such bond Aldehydes, however, possess a C-H bond which can be cleaved as a hydroxyl group attaches to the carbonyl carbon, yielding a carboxylic acid 20 B 21 D 22 A 23 A 24 D In electrophilic aromatic substitution reactions, the rate-determining step is the formation of the positively-charged arenium ion intermediate Substituents that are electron-donating would stabilize this intermediate, thereby increasing the rate of reaction (activating the aromatic substrate) Resonance structures also show that in such cases, further substitution at the o/p positions (relative to the electron-donating group) are favored in general Examples of such groups include the methoxy group (#22), the amide group (#23), the hydroxyl group, and, to a more modest extent, alkyl groups On the other hand, electron-withdrawing groups destabilize the intermediate, deactivating the aromatic substrate Substitutions that are forced to occur by imposing more rigorous reaction conditions would take place preferentially in the meta position, because it is there that their unfavorable effects are least felt in general Examples of such groups include the nitro group (#21), the sulfonic acid group (#24), the cyano group, and the carboxyl group Halides (such as fluorine in #20) are weakly electronwithdrawing and are ortho/para-directing As such, they are the exception to the general rule which states that all electronwithdrawing species are deactivating and meta-directing The reason for this is that their high electronegativity exerts an inductively electron-withdrawing effect (hence deactivating the ring), yet they can stabilize positive charges at the o/p positions by resonance The resonance structure involves it forming a double bond to the phenyl ring and bearing a formal positive charge It is therefore not a significant resonance structure and is unable to overcome the deactivating effect caused by induction 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 25 D This reaction is an example of free radical halogenation In the initiation step, UV light is used to homolytically split the bromine into two bromine free radicals During the propagation steps a bromine radical will attack the alkane, specifically it will remove a hydrogen from the carbon atom that can form the most stable fee radical (the most substituted carbon atom) The alkyl radical will then attack another Br2 molecule, resulting in the addition of Br to the radical carbon and the formation of another bromine radical The carbon attached to the ring, at the center of the t-butyl group, has no hydrogens to give, so the bromine cannot add there The carbon atom to which the t-butyl group is attached is the next most substituted (it is 3o), so that is the site of the bromine addition Br Br H3C H3C 26 Br Br CH3 C C H H3C H3C CH3 C C Br Br H3C H3C CH3 C C Br E IR spectroscopy is useful in identifying the functional groups a molecule possesses In order for IR to be useful in distinguishing between molecules, they must have different functional groups Only choice E, an ether and an alcohol, can be differentiated (the C–O of an ether peaks at 1050-1150 cm-1, the O–H of the alcohol has a broad peak at 3100-3500 cm-1 Choices A and B could be differentiated using H-NMR, since the have different amounts of hydrogens Choices C and D could also be differentiated using H-NMR, because the different locations of the electron withdrawing groups would cause different downfield shifting of the associated hydrogens 27 B This reaction is an example of nucleophilic acyl addition, with a Grignard reagent acting as the nucleophile The carbonyl carbon of cyclobutone (or any carbonyl carbon) is susceptible to nucleophilic attack for two reasons: (1) it has a partial positive charge because of the electron withdrawing effects of oxygen (2) it is trigonal planar, so there is no steric hindrance The alkyl group of the Grignard reagent has a partial negative charge (because it is more electronegative than magnesium, a metal), and so adds to the carbonyl carbon You can also think of this reaction as electrophilic addition across a double bond, with Mg as the electrophile The intermediate of the reaction, bromomagnesiumcylcobutoxide, removes a proton from H3O+ and becomes an alcohol 28 E Two substituents on a benzene to are para when they are three carbons away from each other The groups in choice A are ortho (next to one another), and the groups in choice B are meta (2 carbons away) Choices C and D are wrong because they are not nitrobenzenesulfonic acid Choice C is para-nitrothiolphenol and D is para-aminobenzenesulfonic acid 29 C A productive propagation step is when a free radical attacks a non-radical compound, producing another free radical that continue the reaction to form the desired product Choices A and D are termination steps: two free radicals are combining, decreasing the amount of free radicals that can continue the reaction Choice B is incorrect; when bromine radical attacks ethane, it removes a proton and becomes HBr, and the ethane becomes a radical Choice D is a nonproductive propagation: the desired product is ethyl bromine, not methyl bromine 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 30 B This reaction is an example of Friedel-Crafts acylation The AlCl3 serves as a Lewis acid catalyst, and makes the ethanoyl chloride a stronger electrophile (a strong electrophile is necessary to disrupt the stable conjugated benzene ring) The ethanone adds to the benzene ring para to the methoxide group, because –OR groups are ortho/para directing The other products of the reaction are HCl and the AlCl3 catalyst 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 Organic Chemistry Subject Test Rank the following structures in order of increasing basicity: H Which of the... 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 E 19 C 25 D C D 14 E 20 B 26 E C A 15 D 21 D 27 B B 10 B 16... knowledge of any specific organic reactions, as long as one is able to identify the structures of the molecules and knows what oxidation involves In the context of organic chemistry, oxidation is