8Organic chemistry discretes test w solutions

14 115 0
  • Loading ...
1/14 trang

Thông tin tài liệu

Ngày đăng: 04/05/2017, 10:03

Organic Chemistry Discretes Test Time: 30 Minutes Number of Questions: 30 This test consists of 30 discrete questions—questions that are NOT based on a descriptive passage These discretes comprise 15 of the 77 questions on the Physical Sciences and Biological Sciences sections of the MCAT MCAT ORGANIC CHEMISTRY DISCRETES TEST DIRECTIONS: The following questions are not based on a descriptive passage; you must select the best answer to these questions If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain Indicate your selection by blackening the corresponding circle on your answer sheet A periodic table is provided below for your use with the questions PERIODIC TABLE OF THE ELEMENTS H 1.0 He 4.0 Li 6.9 Be 9.0 B 10.8 C 12.0 N 14.0 O 16.0 F 19.0 10 Ne 20.2 11 Na 23.0 12 Mg 24.3 13 Al 27.0 14 Si 28.1 15 P 31.0 16 S 32.1 17 Cl 35.5 18 Ar 39.9 19 K 39.1 20 Ca 40.1 21 Sc 45.0 22 Ti 47.9 23 V 50.9 24 Cr 52.0 25 Mn 54.9 26 Fe 55.8 27 Co 58.9 28 Ni 58.7 29 Cu 63.5 30 Zn 65.4 31 Ga 69.7 32 Ge 72.6 33 As 74.9 34 Se 79.0 35 Br 79.9 36 Kr 83.8 37 Rb 85.5 38 Sr 87.6 39 Y 88.9 40 Zr 91.2 41 Nb 92.9 42 Mo 95.9 43 Tc (98) 44 Ru 101.1 45 Rh 102.9 46 Pd 106.4 47 Ag 107.9 48 Cd 112.4 49 In 114.8 50 Sn 118.7 51 Sb 121.8 52 Te 127.6 53 I 126.9 54 Xe 131.3 55 Cs 132.9 56 Ba 137.3 57 La * 138.9 72 Hf 178.5 73 Ta 180.9 74 W 183.9 75 Re 186.2 76 Os 190.2 77 Ir 192.2 78 Pt 195.1 79 Au 197.0 80 Hg 200.6 81 Tl 204.4 82 Pb 207.2 83 Bi 209.0 84 Po (209) 85 At (210) 86 Rn (222) 87 Fr (223) 88 Ra 226.0 89 Ac † 227.0 104 Unq (261) 105 Unp (262) 106 Unh (263) 107 Uns (262) 108 Uno (265) 109 Une (267) * 58 Ce 140.1 59 Pr 140.9 60 Nd 144.2 61 Pm (145) 62 Sm 150.4 63 Eu 152.0 64 Gd 157.3 65 Tb 158.9 66 Dy 162.5 67 Ho 164.9 68 Er 167.3 69 Tm 168.9 70 Yb 173.0 71 Lu 175.0 † 90 Th 232.0 91 Pa (231) 92 U 238.0 93 Np (237) 94 Pu (244) 95 Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (251) 99 Es (252) 100 Fm (257) 101 Md (258) 102 No (259) 103 Lr (260) as developed by Organic Chemistry Discretes Test On the graph below, which is the titration curve for an amino acid, what does point A represent? H Which of the following reactions would produce an ester? H2 SO4 A 2CH3OH B CH3CO 2H + SOCl2 C CH3CO 2H + C2 H5 OH D C 6H5 OH + CH3CH 2Br nits of base A The isoelectric point H2 SO4 NaOH H2 O B pH = pKb2 C pH = 14 – pKb2 Which of the following reactions will NOT yield C6H5CH2COCH3? D pH = pKb2 – 14 The diagram below shows a step in which of the following processes? CH2OH O H CH2OH OH OH A C H5CH2 C CH + HgSO4 / H2O B C H5CH2 CH 2CH2 OH + cold, dil KClO C C H5CH2 (CH )C D C H5CH2 (CH )CHOH + cold, dil Na 2Cr2 O7 CH2 + O3 Zn / H2 O O H CH2OH O Which of the following reactions will NOT result in the formation of a carboxylic acid? OH H A B C D Aldehyde formation Hemiketal formation Mutarotation Anomerization A B C D Carbonation of Grignard reagents Hydrolysis of nitrites Reduction of aldehydes Oxidation of primary alcohols The following reaction is an example of: OH O Which of the following products might be formed if benzoyl chloride was treated with excess CH3MgBr? A B C D C6H5COOH C6H5COOCH3 C6H5CHO C6H5C(CH3)2OH and C6H5COCH3 H2 C A B C D esterification tautomerism elimination dehydration GO ON TO THE NEXT PAGE KAPLAN MCAT Which of the following compounds is the most basic? A H2 N The reaction below will primarily yield which of the following products? C H5CH OCH3 A B H2 N CHO C H2 N NO2 CH3 B HNO H2 SO4 CH C CH NO2 NO2 NO D A and B D H2 N Br Which of the following describes the reaction below? Which of the following can be synthesized from an arenediazonium salt? I C6H5Br II C6H5CN III C6H5OH A B C D I only II and III I and III I, II, and III In the following reaction, what is the final product? R2C=O + H2NNHC6H5 → R2COHNHNHC6H5 → R2C=NNHC6H5 A An oxime B A phenaylhydrazone C A semicarbazone D An aromatic nitrile 1 Which of the following processes describes the reaction below? + – R CHCR – N R O H heat → R C = CR + R N:+H O A B C D Alkylation Hoffman elimination Salt formation Conversion into amides H5 C C C CH3 H2 , Pd H H5 C A B C D H CH3 Catalytic hydration Substitution Stereospecific reduction Racemization Which of the following reactions is NOT an electrophilic aromatic substitution reaction? A B C D C6H6 + CH3CH2COCl/AlC13 → C 6H + H2, Rh/C → C6H6 + C6H5CH2Cl/AlCl3 → C6H6 + Br2/FeBr3 → Proteins can be characterized by the fact that they: A are composed of a single peptide chain B have a primary structure formed by covalent linkages C retain their conformation above 40°C D always have quaternary structures GO ON TO THE NEXT PAGE as developed by Organic Chemistry Discretes Test Halogenation of alkanes proceeds partly by the mechanism depicted below hv I X   → 2X • II X • + RH → HX + R • III R • + X2 → RX + X • Which of the following statements is correct? I CH2 OH H HO I, II, and III III only II and III I and II II Which of the following compounds is least susceptible to electrophilic aromatic substitution? A C6H5CH3 B p-Br– C6H4– NH 3+ C p-O2N–C6H4–NH2 D p-H3CO–C6H4–OCH3 COOH OH H3 C OH H H3 C OH CH2 OH Which of these steps can be described as chain propagation? A B C D III COOH CH2 OH IV CHO HO H HO HO H H CH2 OH A B C D H OH CH2OH Compounds I and II are diastereomers Compounds I, II, and IV are meso structures Compounds II and IV are optically inactive Compounds III and IV are optically active Which two of the following compounds are geometric isomers? I III II IV How many different stereoisomers can the following compound have? COOH H OH H OH HO H H OH H A B C D A B C D 2 Which of the following is a result of the reaction below? In thin-layer chromatography, which of the following best describes the behavior of the solvent? A It moves downward because of gravity B It moves upward because of capillary action C It moves downward because of capillary action D It does not move I and III I and IV II and III II and IV H C H5 Cl + Br– H A B C D CH Inversion of configuration Retention of optical activity Mutarotation Loss of optical activity GO ON TO THE NEXT PAGE KAPLAN MCAT When placed in an electric field, which of the following compounds will migrate toward the cathode at pH 7.0? A B C D H2NCH2COOH H2NCH2CHNH2COOH HOOCCH2CHNH2COOH HOOCCHNH2CH2–S–S–CH2CHNH2COOH Which of the following compounds will give the greatest number of proton NMR peaks? A CH 3CH3 A B C D vacuum distillation simple distillation sublimation fractional distillation CH2 CHCH3 B At atmospheric pressure, a certain organic liquid has a boiling point of 185°C, and decomposes at 180°C, while its isomer, also a liquid, boils at 215°C and decomposes at 190°C These isomers can be separated by: H2 C C (CH ) C=CH D CH3CHBrCH CH3 A clear liquid is subjected to infrared spectroscopy and produces a spectrum with prominent, sharp peaks at approximately 2950 cm–1 and 1700 cm–1, as well as a number of smaller peaks between 1460 cm–1 and 900 cm–1 This substance is most likely: A B C D a ketone an aldehyde an alcohol an alkane Which of the following compounds form a racemic mixture? Infrared spectroscopy provides a chemist with information about: A B C D functional groups conjugated bonds molecular weights distribution of protons I CHO A B C D CH3(CH2)6CH3 (CH3)2CHCH2(CH2)3CH3 (CH3)2CHCH2CH(CH3)CH2CH3 (CH3)3CC(CH3)3 CHO HO H H OH HO H H OH CHO II A compound with the molecular formula C8Hl8, produces a single NMR signal What is its structural formula? III CHO H OH HO H CHO A B C D CHO IV CHO HO H H OH CHO I and III I and IV II and III II and IV What is the most likely mechanism for the reaction between 1-chloropropane and sodium cyanide? A B C D S N2 E2 SN1 El END OF TEST GO ON TO THE NEXT PAGE as developed by Organic Chemistry Discretes Test THE ANSWER KEY IS ON THE NEXT PAGE KAPLAN MCAT ANSWER KEY: C C D C B 11 12 13 14 15 B D C B B 21 22 23 24 25 C D B A A 10 16 17 18 19 20 C B D B A 26 27 28 29 30 D D A D A C B A D B as developed by Organic Chemistry Discretes Test ORGANIC CHEMISTRY DISCRETES TEST EXPLANATIONS The correct answer for question is choice C This problem deals with the titration of amino acids The titration given begins with a strongly acidic solution As we add base, the pH rises at first But then as the pH approaches the pKa1, protons begin to dissociate from the carboxyl group The value of the pH then levels off because the carboxyl group buffers the solution when the pH is close to its pKa1 Therefore, point A on the graph represents the value of pH equal to the pKa1 That is, point A shows deprotonation of the amino acid carboxyl group But the reverse reaction—protonation of the amino acid, is represented by pKb2 And by the definition of pK, the pH at point A is equal to 14 minus the pKb2 Therefore, choice C is correct Choice A the isoelectric point is in fact point B When we continue to add base after point A, the pH gradually rises until all the molecules of the amino acid are in the neutral form The correct answer for question is again choice C The 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 phenonenon 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 Again then, choice C is the correct answer Now for question The correct answer here is choice D The reagent here benzoyl chloride is an acyl halide which will form a ketone on reaction with an equimolar quantity of a Grignard reagent like this one in the question With excess of the Grignard reagent however, acyl halides react further to produce primarily tertiary alcohols along with a small amount of ketone Therefore, the product of this reaction will be a mixture of a tertiary alcohol and acetophenone This corresponds to choice D and so this is the correct answer As for the other choices, Choice A benzoic acid would be formed by hydrolysis of benzoyl chloride Choice B methyl benzoate would be formed by reaction of benzoyl chloride with methanol Finally, choice C benzaldehyde would be formed by reduction of benzoyl chloride Again then, the correct answer is choice D The correct answer here is choice 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 is wrong as this is an example 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 Now for question The correct choice is B This question asks you to find the correct answer choice that will NOT lead to the formation of a particular compound, which happens to be benzyl methyl ketone Choice A shows an alkyne being treated with an acidic mercury sulfate solution This leads to Markownikov addition of water—that is, hydrogen adds to the less substituted triple bonded carbon and hydroxyl adds to the other triple bonded carbon This results in the formation of an enol which will then spontaneously rearrange to form the ketone we are looking for So, choice A DOES yield benzyl methyl ketone and therefore it's NOT the answer we're looking for In choice B, a primary alcohol is oxidized with cold, dilute potassium chlorate Although ketones can be synthesized by oxidizing secondary alcohols, oxidation of primary alcohols results in the formation of aldehydes NOT ketones Therefore, B is the correct answer Choice C ozonolysis of an alkene will yield our ketone plus formaldehyde and choice D oxidation of a secondary alcohol also gives our desired ketone, so both of these choices are wrong Again, B is the correct answer choice Choice C is correct This question deals with common methods of preparing carboxylic acids Let's go through the choices given The carbonation of a Grignard reagent leads to the formation of the magnesium salt of a carboxylic acid When treated with mineral acid, the magnesium salt is then converted to a carboxylic acid Therefore choice A CAN be used to prepare carboxylic acids, so this is wrong The acidic or basic hydrolysis of nitriles also yields carboxylic acids, so choice B is also wrong The oxidation of primary alcohols by various strong KAPLAN MCAT oxidizing agents such as potassium permanganate yields carboxylic acids, so choice D can be rejected Choice C is correct because carboxylic acids are formed by the oxidation of aldehydes, not their reduction For question 7, the correct choice is B This diagram represents the tautomeric transformations of a ketone Ketones exist in two spontaneously-interconvertible forms the keto form and the enol form These differ in the placement of a hydrogen atom and a double bond The keto structure is more stable because a carbon-oxygen double bond is more stable thermodynamically than a carbon-carbon double bond with a hydroxyl group attached and the chemical properties of ketones are determined mostly by the keto form None of the other choices describes this process Choice A esterification is a reaction between a carboxylic acid and an alcohol resulting in the formation of an ester Choice C elimination is a reaction characteristic of alkyl halides and leads to the formation of a stable double bond Choice D dehydration is the loss of a hydroxyl group and a proton to yield water and usually a stable double or single bond is left behind So again, the correct choice is B Now for question The correct answer is choice A This question deals with the effect of substituents on the relative basicities of aromatic amines The basicity of aromatic amines or anilines manifests itself in protonation of the amino-group nitrogen atom to make positively charged anilinium ions In general, electron withdrawing substituents decrease the basicity of anilines because the withdrawal of electron density increases the positive charge on the anilinium ion, making it less stable On the other hand, electron-donating substituents decrease the positive charge of the anilinium ion and stabilize it Now look at the choices B, C, and D have electron withdrawing substituents, while A has an electron donating methoxy substituent Since electron donating substituents increase basicity, choice A is the most basic among the compounds shown and therefore this choice is correct Let's now look at question The correct choice is D Arenediazonium salts, which are synthesized from primary aromatic amines, are compounds with an N2+ group attached to the aromatic ring They are useful for synthesizing a wide variety of compounds due to the fact that they can easily be made to undergo replacement reactions in which molecular nitrogen is released and a nucleophilic substituent attaches to the aromatic ring in its place For example, in the presence of cuprous halides, diazonium salts release molecular nitrogen and form halogen-substituted arenes often called the Sandmeyer reaction Similarly, in the presence of cuprous cyanide, nitrogen is released and replaced by the cyanide ion, thus forming aromatic nitriles In the same way, in cold aqueous solutions, the hydroxyl group replaces the nitrogen to form phenol Therefore, compounds I, II, and III can all be obtained by the replacement of nitrogen in diazonium salts, and so the correct choice is D As all three choices are correct, you can discard A, B, and C which state that only one or two of the compounds will be formed So again, D is the right answer 10 For question 10, the correct answer is choice B This question requires you to know the most important reactions between ammonia derivatives and aldehydes In these reactions, the nucleophilic nitrogen of an ammonia derivative attacks the electrophilic carbonyl carbon, while hydrogen released from this nitrogen attacks the nucleophilic carbonyl oxygen This reaction results in the formation of a single carbon-nitrogen bond, as well as a hydroxyl group on the formerly carbonyl carbon Since this system is quite unstable, the addition is always followed by dehydration that is, a water molecule is formed from the carbons hydroxyl and the neighboring nitrogen's hydrogen The release of the water molecule results in the formation of a carbon-nitrogen double bond In this case, the final product of the reaction between the carbonyl compound and phenylhydrazine is phenylhydrazone and therefore choice B is correct Choices A and C are synthesized in reactions with other ammonia derivatives Choice A is wrong because oximes are synthesized by reactions between carbonyl compounds and hydroxylamine Choice C is wrong because semicarbazones are synthesized by reactions between carbonyl compounds and semicarbazides Finally, aromatic nitriles are obtained by reacions between diazonium salts and cuprous cyanide, and so choice D is also wrong Again, the correct choice is B 11 Now look at question 11 The right answer here is choice B The reaction given is an example of Hofman elimination When a basic solution of a quaternary ammonium hydroxide is heated, it undergoes decomposition forming an alkene, a tertiary amine and water This reaction proceeds by an E2 mechanism The base abstracts a hydrogen ion from carbon and the subsequent release of the molecule of tertiary amine results in the formation of the double bond So, choice B is correct Let's now look at the wrong answers Choice A alkylation is the insertion of an alkyl group into a molecule Choice C formation of ammonium salts is possible only in acidic solutions because it requires protonation of of nitrogen Moreover, in this case nitrogen doesn't have unshared electron pairs, therefore protonation and subsequent formation of a salt may occur only upon the breakdown of a carbon-nitrogen 10 as developed by Organic Chemistry Discretes Test bond Finally, amides are formed in reactions between primary and secondary amines with acyl halides Therefore, choice D is also incorrect and choice B is again the correct answer 12 For question 12, the correct answer is choice D The methyl group attached to an aromatic ring has an activating and ortho/para directing effect on the ring that is, it makes these positions very reactive towards electrophilic aromatic substitution Therefore, nitration is most likely to occur as shown in choice A which is ortho and choice B which is para This makes choice D the correct answer The meta position, as shown in choice C, is deactivated by the methyl substituent and so nitration here is negligible in comparison to the other two positions As a result, choice C is wrong and again, D is the correct answer 13 In question 13, the correct answer is choice C This is an example of stereospecific reduction or hydrogenation This reaction, catalyzed by palladium, results in the addtion of two hydrogen atoms on the same side of the molecule refered 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 definately an addition reaction, not a substitution reaction Finally, 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 Again, the correct answer is choice C 14 The correct answer to question 14 is choice B This question deals with different mechanisms of substitution on aromatic rings An aromatic ring is especially susceptible to electrophilic attack, but in some cases, nucleophilic aromatic substitution is also possible Let's go through the choices Choice A is an example of Friedel-Crafts acylation In this reaction, benzene reacts with an acyl chloride in the presence of the Lewis acid aluminum trichloride The Lewis acid removes chloride from the acyl halide thus transforming it into a strongly electrophilic oxycarbonium ion, which immediately attacks the benzene ring Thus choice A is wrong because it IS an example of electrophilic attack Choice C is an example of Friedel-Crafts alkylation, which proceeds similarly to acylation, therefore C is also wrong Choice D can be rejected as it is yet another example of electrophilic attack But in choice B, the benzene ring is being catalytically hydrogenated to form cyclohexane, so this answer choice is NOT an example of electrophilic aromatic substitution making it the correct answer 15 Now look at question 15 The correct choice here is B The primary structure of a protein is the amino acid sequence, which is formed by covalent peptide linkages Choice A is incorrect because many proteins contain more than one peptide chain Choice C is incorrect because proteins are denatured by heating and so they LOSE their conformation at 40°C not retain it Choice D is incorrect because only those proteins containing more than one peptide subunit have quaternary structure 16 The correct answer to question 16 is choice C The reaction equations shown are steps in the free radical halogenation of an alkane Free radical halogenation is one of the few reactions that alkanes will undergo and occurs by the reaction of the alkane with a highly reactive halogen radical A halogen radical is a single, neutrally charged halogen atom, which has an unfilled valence shell containing seven electrons Since it has an odd number of electrons, the halogen radical also has a half filled orbital, which is denoted by Xo Specifically, this might be denoted as Bro, Clo, or another halogen free radical In the halogenation reaction, the step that starts the whole process is the formation of the free radical from a diatomic molecule This process, which is shown in the first reaction equation, is usually catalyzed by ultraviolet light although it can also be catalyzed via attack by another free radical Step I is followed by step II, in which the halogen free radical attacks the alkane, producing an alkyl halide and another highly reactive free radical an alkyl radical In step III, this alkyl radical attacks a halogen molecule, X2, picking up one of the halogen atoms to become another molecule of alkyl halide and at the same time producing another halogen radical This new halogen radical will then start the process again thus causing a chain reaction Anyway, getting back to the question, step I—UV light-induced generation of two free radicals—is called the chain initiating step As a result, you can already rule out answer choices A and D Steps II and III are both chain propagating steps because they serve to push the reaction chain onward by generating a molecule of product plus a molecule of free radical This means that choice C is the correct answer Choice B is wrong as it fails to recognize that step II is a chain propagating step, so again, choice C is correct 17 Now look at question 17 The correct choice here is B The reactivity of an aromatic compound toward electrophilic substitution depends on the nature of the ring's substituents Electron donating substituents enrich the KAPLAN 11 MCAT electron system of a ring and so activate it towards electrophilic attack Electron withdrawing substituents, on the other hand, deactivate the ring making it less susceptible to electrophilic attack In choice B, the ring is severely deactivated by the combined electron withdrawing effects of the bromine and ammonia substituents Choice B is therefore almost totally unreactive towards electrophilic aromatic substitution and so it is the correct answer As for the other choices, choice A contains an electron-donating, activating methyl substituent, so this compound is very reactive towards electrophilic attack The ring in choice D is even more highly activated because it contains two electron-donating methoxy substituents Finally, in choice C, the strong electron withdrawing effect of the nitro group is compensated by the strong electron donating effect of the amino group, so this compound is moderately reactive towards electrophiles Again, the correct answer is choice B 18 For question 18, the correct answer is choice D The number of different stereoisomers a molecule can have depends on the number of chiral centers it has If the number of chiral centers in n, then the number of different stereoisomers is 2n, where a chiral center is defined as a carbon bonded to different substituent groups For convenience, let's number all the carbons in this compound, beginning with the carboxyl carbon You can see that carbons and are not chiral centers, because carbon is attached to only three substituents while carbon has two identical substituents namely hydrogen On the other hand, carbons 2, 3, and are chiral centers, since each of them is bonded to different groups Therefore, the molecule has chiral centers and the number of different stereoisomers is to the power of that is So, the correct choice is D 19 Let's now take a look at question 19 The correct answer here is choice B This question tests your understanding of how thin-layer chromatography works A mixture is spotted onto the thin layer of adsorbant that covers a plastic or glass sheet The sheet is then placed upright in a developing chamber that contains a solvent, called the eluant Capillary action forces the eluant up the plate, and with it, portions of the mixture that have been dissolved by the eluant Since the different compounds move at different rates based on their size and how well they dissolve in the eluant, the mixture gradually separates When the plate is dried, the spots of different compounds can be visualized, or made visible, by UV light Okay, I said that the mixture moves by capillary action, so choices A and D are wrong Also, the capillary action forces the mixture up, not down, so choice B is correct and C is wrong Incidentally, the downward movement of solvent under the influence of gravity is utilized in chromatography, but in column chromatography, not thin-layer chromatography Again then, the correct choice is B 20 For question 20, the correct choice is A This question describes the terminology used in describing stereochemistry It can best be answered by going through the answer choices Choice A states that compounds I and II are diastereoisomers These are structural isomers that are not mirror images Looking at compounds I and II, you should be able to see that that these compounds are indeed structural isomers, but not mirror images Therefore, choice A is correct Choice B states that compounds I, II, and IV are meso compounds Meso compounds have chiral carbons, but also have planes of symmetry that is they are their own mirror images and are thus optically inactive You can see from the structures that compound II is definately meso but compounds I and IV are not Therefore, choice B is incorrect Choice C states that compounds II and IV are optically inactive As we just said, compound II is meso so its optically inactive, but compound IV has chiral carbons but no plane of symmetry and so IS optically active Therefore, choice C is incorrect Compound III is meso and so it is optically inactive making choice D wrong To repeat then, the correct choice is A 21 Now for question 21 The correct choice here is C Geometric isomers are isomers that differ in the orientation of their substituents around a carbon-carbon double bond In cis isomers, the substituents are on the same side of the double bond, while in trans isomers, the substituents are on opposite sides Compounds I and IV have no double bonds, so they can't possibly be geometric isomers Therefore, if you knew the definition of geometric isomers, you could immediately have eliminated every choice except C You can easily see that compound II is the trans isomer of compound III, which is the cis isomer The proper term for compounds I and IV is structural isomers, because they have the same molecular formula but their atoms are connected differently Compounds II and IV have different molecular formulas as compounds I and III , so these pairs are not isomers at all Again, the correct answer is choice C 22 The correct choice for this one is D 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 hinderance 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 12 as developed by Organic Chemistry Discretes Test will result in the loss of optical activity, which always happens in SN1 reactions Let's see why 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 amion 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 D is correct and choice B is pretty 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 open-chain forms and cyclic hemiacetal forms of monosaccharides in aqueous solutions Again, the correct answer is choice D 23 For question 23, the correct answer is choice B This question deals with the amphoteric properties of amino acids At neutral or alkaline pH values, the COOH group of an amino acid dissociates to become a negatively-charged carboxylate At a neutral or acidic pH, the NH2 group is protonated and becomes positively charged Thus at neutral pH an amino acid is a dipolar ion The overall charge of an amino acid depends on the ratio of positively charged amino groups to negatively charged carboxylates in the molecule In order for a molecule to migrate to the cathode, its overall charge must be positive Of the choices given, choice B is the only one in which the positively charged amino groups would outnumber the negatively charged carboxylate groups Therefore, only choice B would be positively charged at pH Again then, the correct answer is B 24 Now let's look at question 24 The correct answer here is choice A To figure out the best way of separating two compounds, you have to take into effect their chemical and physical properties What you're told about these compounds is that at atmospheric pressure, they both decompose before they reach their boiling point-the temperatures of which are given Okay, three of the answer choices here are types of distillation In simple distillation, a mixture is heated until each component boils off in turn, and can be collected separately Since both of these compounds decompose before they reach their boiling point at one atmosphere of pressure, you can't separate them by simple distillation, since they'll just decompose before they are separated So choice B is wrong Instead, you should use vacuum distillation, choice A, which is actually distillation at very low pressure This makes compounds boil at lower temperatures, which will make it possible for these compounds here to to reach their boiling point before they decompose Choice D, fractional distillation, is a procedure used to separate compounds that have boiling points very close together It's generally done at atmospheric pressure, so as with simple distillation, these compounds would decompose instead of distilling, so D is also wrong Finally, choice C, sublimation, is the direct conversion of a solid into a gas This is likely to require lower pressures and/or higher temperatures than vacuum distillation, meaning it would be more difficult to carry out and would also carry more risk of the compounds decomposing, so choice C is wrong too Again then, the correct answer is choice A 25 The correct answer to question 25 is choice 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 Again, choiec A is the correct answer 26 Now for question 26 The correct answer to this one is choice D This question requires you to understand the main principals of NMR spectroscopy This technique provides a chemist with information about the proton enviroment Equivalent hydrogens that is, hydrogens whose positions are identical with respect to the surrounding atoms give a single NMR signal, whereas nonequivalent hydrogens give separate NMR signals Since the compound given shows just one NMR signal, you can assume that all its hydrogens are equivalent So let's now go through the choices given Choice A is wrong because it will give three different signals, one from the two terminal methyl hydrogens, one from the hydrogens on the CH2 groups next to them and one for the hydrogens on the two innermost CH2 groups Choice B gives different signals: one from the two methyls on the far left end of the molecule as it's shown, one from the CH group next to that one, from the four CH2 groups in the chain, and from the methyl group on the right end Thus choice B is also wrong Using the same principle, you can easily see that choice C will give signals However, in choice D, all the methyl groups are equivalent and so this compound will give one NMR signal It is interesting to note that in choices B and C, the signals probably won't be as simple KAPLAN 13 MCAT as we have described Coupling effects would mean that these NMR patterns would be highly complex due to overlapping and split signals Again, the correct answer is choice D 27 Now for question 27 The correct answer here is choice D Let's go through the answer choices All the hydrogens in choice A are equivalent, so this compound would produce one NMR proton peak In choice B, the methyl group hydrogens are identical so they would produce one NMR signal The two hydrogens of the CH2 group are also identical and so would produce another NMR peak Therefore, choice B would give two peaks By the same principle, choice C produces three NMR peaks one from the two CH2 groups, one from the CH group and one from the CH3 group Finally, choice D has a chiral carbon, the one with the bromine group attached to it The three hydrogen atoms in each methyl group are equivalent, but the two methyl groups are not, so each methyl group will produce a peak The two hydrogens of the CH2 peak will produce another peak and the single hydrogen bonded to the brominated carbon will produce a fourth Thus, choice D gives different NMR signals most of any of the choices and therefore it's the correct answer 28 The correct answer is choice A This question tests your knowledge of IR spectroscopy and what sorts of spectra different kinds of compounds produce In the spectrum that's described, the various peaks between 1600 and 900 cm–1, which aren't described in detail, are in the fingerprint region The tall peak at 1700 cm–1 indicates the carbon-oxygen double bond of a carbonyl group Well, knowing that, we can eliminate choices C and D since neither of them have a carbonyl group in them That leaves either a ketone or an aldehyde Well, the peak at 2950 cm–1, which lies in the range that describes the hydrogen stretching patterns of a molecule, doesn't show the characteristic double bond absorption at about 2800 cm–1 of the carbonyl hydrogen in aldehydes The absorption is more like the normal absorptions of hydrocarbon hydrogens That means that since the characteristic hydrogen pattern of aldehydes is missing but we know we have a carbonyl group, the compound must be a ketone, which is choice A 29 For question 29, the correct choice is 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 Lets 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, which is also wrong 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 Again then, the correct choice is D 30 For question 30, the correct choice is A To answer this question, you need to know under what conditions SN2 reactions will occur Well, you can see that the reactant is a primary alkyl halide and chloride is also a good leaving group You should also know that CN– is an extremely strong nucleophile, so you should be able to narrow the choices down to at least A and B, since SN1 and E1 mechanisms are highly unlikely in primary alkyl halides So now, how we know that SN2 will occur over E2? Well, remember we said that CN– is an extremely strong nucleophile Therefore, substitution will occur exclusively over elimination Only when strong bulky bases are used is E2 favored over SN2 So again, the correct answer is choice A 14 as developed by ... C B D B A 26 27 28 29 30 D D A D A C B A D B as developed by Organic Chemistry Discretes Test ORGANIC CHEMISTRY DISCRETES TEST EXPLANATIONS The correct answer for question is choice C This problem... 1-chloropropane and sodium cyanide? A B C D S N2 E2 SN1 El END OF TEST GO ON TO THE NEXT PAGE as developed by Organic Chemistry Discretes Test THE ANSWER KEY IS ON THE NEXT PAGE KAPLAN MCAT ANSWER KEY:... 40°C D always have quaternary structures GO ON TO THE NEXT PAGE as developed by Organic Chemistry Discretes Test Halogenation of alkanes proceeds partly by the mechanism depicted below hv I X 
- Xem thêm -

Xem thêm: 8Organic chemistry discretes test w solutions , 8Organic chemistry discretes test w solutions , 8Organic chemistry discretes test w solutions

Gợi ý tài liệu liên quan cho bạn

Nhận lời giải ngay chưa đến 10 phút Đăng bài tập ngay