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ORGANIC CHEMISTRY TOPICAL: Oxygen-Containing Compounds Test Time: 22 Minutes* Number of Questions: 17 * The timing restrictions for the science topical tests are optional If you are using this test for the sole purpose of content reinforcement, you may want to disregard the time limit MCAT DIRECTIONS: Most of the questions in the following test are organized into groups, with a descriptive passage preceding each group of questions Study the passage, then select the single best answer to each question in the group Some of the questions are not based on a descriptive passage; you must also 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) GO ON TO THE NEXT PAGE as developed by Organic Chemistry Discretes Test Passage I (Questions 1–6) β-keto esters and malonic esters (commonly known as β-dicarbonyl compounds) are compounds that contain two carbonyl groups separated by an intervening carbon The central carbon of these compounds contains highly acidic protons, known as α-hydrogens An α-hydrogen can be abstracted with an appropriate base, forming an enolate ion that can be either acylated or alkylated Two common laboratory syntheses that rely on this principle are the malonic ester synthesis (Reaction 1) and the acetoacetic ester synthesis (Reaction 2) The malonic ester synthesis is one of the most profitable methods of preparing substituted carboxylic acids, while the acetoacetic ester route is often used in the synthesis of methyl ketones O O OEt OEt 1) NaOEt 2) RX 1) H3O R 2) R Which of the following best explains the chemistÕs choice of ethanol as the solvent used in the synthesis described in the passage? A The solvent chosen should have a boiling point similar to that of the starting material B The solvent chosen should have a boiling point similar to that of the desired product C Ethanol solvates the ethoxide anions, rendering them less nucleophilic D Ethanol is a better solvent than polar aprotic solvents such as DMSO COOH +CO 2(g) O O Figure + OEt OEt COOH Reaction Which of the following compounds can most readily be synthesized via the acetoacetic ester pathway shown in Reaction 2? O O O 1) NaOEt 1) H3O+ R 2) 2) RX OEt OEt O R O +CO2(g) Reaction A chemist attempted to synthesize cyclohexanecarboxylic acid (Figure 1) using a slight variation of the malonic ester synthesis Diethyl malonate (1 equiv) was dissolved in a solution of sodium ethoxide in ethanol (2 equiv) The resulting solution was then added dropwise, with constant stirring, to a solution of 1,5dibromopentane (1 equiv) When the reaction was complete, the reaction mixture was hydrolyzed with 6M HCl(aq) and heated to remove any CO2 that formed After the desired product was isolated, the final yield was determined to be 75% A B C D Hexanoic acid 2-Hexanone 3-Hexanone Ethyl hexanoate Which of the following best describes the mechanism of both synthetic pathways shown in the passage? A Alkylation followed by hydrogenation and carboxylation B Deprotonation followed by alkylation and carboxylation C Alkylation followed by protonation and decarboxylation D Deprotonation followed by carboxylation and hydrolysis then then then then GO ON TO THE NEXT PAGE KAPLAN MCAT Which of the following halides would be the most suitable starting material for a malonic ester synthesis of dibenzylacetic acid? A B C D Bromobenzene 2-Chlorophenol 4-Chlorophenol Benzyl bromide Attempts to produce laboratory quantities of asymmetrically substituted carboxylic acids such as 2ethylpentanoic acid will result in lower yields than those usually obtained in the case of symmetric products This difference is primarily attributed to: A reduced solubility of the participating reactants B competitive alkylation resulting in a mixture of products C decreased nucleophilicity of the anionic intermediate D increased difficulty in hydrolyzing the asymmetric ester The two synthetic pathways described in the passage can best be described as: A B C D nucleophilic substitution reactions nucleophilic addition reactions alpha substitution reactions simultaneous alpha substitution and nucleophilic substitution reactions GO ON TO THE NEXT PAGE as developed by Organic Chemistry Discretes Test Passage II (Questions 7–12) Triacylglycerols are fats or oils commonly found in plant and animal systems These molecules are formed by the attachment of three long chain carboxylic acids (fatty acids) to glycerol via an ester linkage The carbon chain of the fatty acids tends to be unbranched and even in number (usually between and 20 carbons) Fatty acids can be either unsaturated or saturated: Unsaturated fats contain double bonds, whereas saturated fats not Since recent research has supported the suggested link between dietary saturated fats and serum cholesterol, the degree of saturation in various animal fats and vegetable oils is of great interest to nutritionists Table contains a number of vegetable oils consisting of various fatty acids that differ in both chain length and degree of saturation Table FATTY ACID CONTENT(%) Palmitic Palmitoleic Steric Oleic Linoleic Linolenic acid acid acid acid acid acid (C16)* (C16:1) (C18) (C18:1) (C18:2) (C18:3) SOURCE: Corn 10 25 62 Ð Linseed 10 Ð 23 57 Safflower Ð 13 75 Soybean Ð 43 40 Castor Bean Ð Ð 93 *C x:y x = number of carbons y = number of double bonds Ð The fatty acid content of these oils can be easily determined by employing the transesterification reaction (Reaction 1) O O C O R' OH MeO O O C C R' O H+/OHÐ R'' + 3MeOH OH + MeO O O C C R'' O OH MeO R''' Triacylglycerol Methanol Glycerol C R''' Fatty acids Reaction A student attempted to determine the fatty acid composition of a commercially available cooking oil by using Reaction The procedure involved dissolving the oil to be analyzed in toluene; when dissolution appeared to be complete, a solution containing BF3 in methanol was added After heating for 30 minutes, the reaction mixture was cooled to room temperature and extracted with distilled water The organic layer was then retained for analysis What is the purpose of using BF3 in the experiment described? A To absorb excess water formed B To catalyze the transesterification by polarizing the carbon-oxygen double bond C To catalyze the transesterification by acting as the initial nucleophile D To solvate the methanol molecules If the student analyzes the product mixture by proton NMR, which of the following observations would signify that the reaction had reached completion? A The presence of several singlets in the region 3.0 to 45 ppm B The absence of multiplets in the region 3.5 to 4.5 ppm C The presence of peaks in the region 2.0 to 3.0 ppm D The absence of a sharp singlet at 9.5 ppm GO ON TO THE NEXT PAGE KAPLAN MCAT Triacylglycerol is a useful starting material in the production of soaps (salts of long chain carboxylic acids) via a process known as saponification If this process was carried out by boiling the fat in aqueous sodium hydroxide and then extraction with toluene, what products would you expect to see in the aqueous layer? A B C D A mixture of methyl esters and glycerol Glycerol and sodium chloride Both unreacted sodium chloride and triglyceride A mixture of glycerol and sodium salts of fatty acids Partial transesterification of triacylglycerol can result in the formation of ÔdiolsÕ These compoundscould also be described as: A B C D methanol dimers nonalkylated fatty acids monoesters of glycerol diesters of glycerol 1 According to Table 1, which of the following oils has the highest polyunsaturated fat content? A B C D Corn oil Linseed oil Soybean oil Castor bean oil In order to determine the degree of unsaturation in the oil being analyzed, the student treats a portion of the reaction mixture with a solution of bromine in carbon tetrachloride If it is found that 1.50 equivalents of bromine react, what is the most likely source of the oil? A B C D Soybean Safflower Corn Castor bean GO ON TO THE NEXT PAGE as developed by Organic Chemistry Discretes Test Questions 13 through 17 are NOT based on a descriptive passage Which of the following compounds is the strongest acid? A B C D HOOC-CH2Cl HOOC-CH2OCH3 HOOC-CH2F HOOC-CH3 Reaction of benzoic acid with thionyl chloride followed by treatment with ammonia will yield which of the following compounds? A B C D Benzamide p-Aminobenzaldehyde p-Chlorobenzamide 3-Chloro-4-aminobenzaldehyde The lH NMR spectrum below is characteristic of which of the following compounds? Which of the following compounds is the most susceptible to nucleophilic attack by OHÐ? A B C D Propanoic acid Ethanal Benzyl chloride Propionyl chloride The Williamson synthesis of methoxybenzene is outlined below (1) 10 (2) (2) (3) 1) NaOH 2) CH3 Cl ppm (δ) OCH3 OH + NaCl + H2 O A B C D Butanol Butanone Butanoic acid Butanal Which of the following best describes the mechanism of this reaction? A B C D Bimolecular nucleophilic substitution Unimolecular nucleophilic substitution Bimolecular elimination Unimolecular elimination END OF TEST KAPLAN MCAT ANSWER KEY: C B B D B 8 10 D B B D C 11 12 13 14 15 B C C D A 16 17 A D as developed by Organic Chemistry Discretes Test OXYGEN-CONTAINING COMPOUNDS TEST EXPLANATIONS Passage I (Questions 1Ð6) For question number 1, the correct answer is C This question requires an understanding of the malonic ester synthesis described in the passage and the mechanism by which it occurs, as well as the concept of solvation Ethanol is a protic solvent: a solvent whose molecules have a hydrogen attached to an oxygen or a nitrogen Protic solvents solvate anions very effectively by hydrogen bonding to them In this experiment, sodium ethoxide dissociates to produce an ethoxide anion which, depending on the solvent, can act as either a base, abstracting an acidic hydrogen, or as a nucleophile, attacking an electrophile However, as you can see from the reaction, sodium ethoxide abstracts an acidic alpha hydrogen from diethyl malonate, and does not attack the electrophilic carbonyl carbon In using the protic solvent ethanol, the ethoxide anion becomes hydrogen bonded and consequently solvated This "solvation sphere" renders the anion much less nucleophilic and so ensures that alpha deprotonation can proceed unhindered Therefore, choice C is the correct response Now for the other answer choices Choices A and B discuss the boiling point of the solvent Of course, ethanol has to be distilled off at the end of the reaction, but its boiling point was not the reason it was chosen Ethanol is used for its protic properties as we discussed, therefore A and B can be discarded Choice D is wrong, since a polar aprotic solvent would be just as good a solvent as ethanol However, aprotic solvents are not suitable for this reaction: they not possess a hydrogen attached to an oxygen or a nitrogen Again, choice C is the correct answer The correct answer to question is choice B This question is testing your understanding of the syntheses described in the passage, along with a bit of nomenclature It's mentioned in the passage that the acetoacetic ester synthesis is commonly used in the production of methyl ketones: ketones of the type CH3COR So, the question is really asking: "which of the following is a methyl ketone?" 2-Hexanone is the only answer choice that fits this definition and so is correct Hexanoic acid choice A and ethyl hexanoate choice D would be more easily synthesized by the malonic ester synthesis, not the acetoacetic ester synthesis Finally choice C, 3-hexanone, has the carbonyl in the middle of the chain and would most likely be synthesized by methods other than those mentioned in the passage such as direct oxidation of 3hexanol Again, choice B is the correct answer The correct answer to question is choice B Just like the previous question, this one tests your understanding of the mechanism of both syntheses According to the reactions shown, both starting materials are reacted with a base (in this case, an alkoxide) which abstracts a proton, forming a carbanion: deprotonation Choices A and C can be eliminated The carbanion, being a nucleophile, attacks the alkyl halide, displacing the halide: alkylation Choice D can be eliminated, leaving choice B as the correct response (By the way, since carbon dioxide is a product in both of the reactions, they also undergo decarboxylation.) The correct answer to question is choice D This question requires you to have a thorough understanding of the malonic ester synthesis probably more so than the other questions It would probably be useful to drawout the target compound, dibenzylacetic acid, and compare it to the cyclohexane carboxylic acid made by the chemist in the passage If you this, you will get a figure showing an acetic acid molecule in which the alpha carbon is attached to two benzyl groups so the overall formula is C6H5CH2 twice and then CHCOOH The CH carbon is the original middle alpha carbon of the malonic ester or more specifically, the one that is deprotonated first by sodium ethoxide So, you can see that two benzyl groups have been substituted onto this carbon, so the alkyl halide that reacts must contain a benzyl group The only answer choice that contains this benzyl group is choice D, so this is the correct response When the alpha carbon of diethylmalonate is deprotonated, a benzyl group adds to this carbon, resulting in loss of bromide from the alkyl halide The same process occurs with a second benzyl bromide so overall, two equivalents of benzyl bromide and sodium ethoxide are needed Finally, the substituted acetic acid is formed by addition of acid and then heating in order to decarboxylate the molecule All the other answer choices are aryl halides and since this reaction does not proceed by nucleophilic aromatic substitution, we could have quickly eliminated them Again, the correct answer is choice D The correct answer to question is choice B This question again tests your understanding of the malonic ester synthesis The named product 2-ethylpentanoic acid could be made from diethyl malonate by first deprotonating the ester with a strong base, then adding one equivalent of propyl halide, deprotonating for a second time and then adding one equivalent of ethyl halide If this were done simultaneously, without removing the first alkylated product, we would expect to wind up with a mixture of 5, 6, and carbon chains attached to the acid groups in the final products, since the propyl and ethyl halides compete for the same nucleophilic site To obtain the desired product, there would have to be a separation step so that the first alkylated product could then be treated independently In either case the yield of desired product will be reduced: by formation of undesired side products or by losses during separation Anyway, back to the question The primary reason the yields are low is that in any case we have competitive alkylation of the anionic intermediate Done in a single step, the nucleophilic carbanion can attack either alkyl halide This competitive alkylation can be avoided by doing the reaction in sequence, resulting in losses from other factors However, the primary reason that the reaction would be carried out in sequence is to avoid competitive alkylation Choice B is the correct response KAPLAN MCAT Solubility problems choice A can be avoided by suitable choice of solvents, and are not related to the symmetry of the final product this answer choice is incorrect Choice C tries to distract you with the nucleophilicity of the anion which is the deprotonated diethyl malonate molecule If sodium ethoxide is present, the nucleophilicity will be exactly the same, whether symmetric or asymmetric products are formed, so choice C is also wrong Meanwhile, choice D is, like some of the choices in earlier questions, looking at the wrong part of the reaction Hydrolysis is basically part of the work-up and so occurs after the alkylation steps are complete The correct answer to question is choice D, simultaneous alpha substitution and nucleophilic substitution reactions The alpha position of both reactants is the position where the alkyl group has been substituted for one of the hydrogens Now, in order for substitution to occur, the alkoxide must first abstract a proton from the alpha position, forming a nucleophilic carbanion This nucleophilic carbanion then attacks the alkyl halide, displacing the halide: a nucleophilic substitution reaction Choice D is the correct response (Even though choices A and C could be considered true, the question asked you to choose the best description of the pathways, and that is choice D.) Choice B nucleophilic addition-is a reaction characteristic of carbonyl compounds that have no leaving groups In these reactions the carbonyl is not attacked, so choice B is incorrect Again, choice D is the correct response Passage II (Questions 7–12) The correct answer to question is choice B This question tests your understanding of Reaction The most likely mechanism for a reaction like the one illustrated is nucleophilic attack by the oxygen of methanol on the carbonyl carbon of the ester followed by the loss of a good leaving group which in this case is an OR group, where R is the glycerol "backbone" This will result in the formation of glycerol and three fatty acids the nature of which will depend on the alkyl chain Neutral alcohols are only slightly nucleophilic and react slowly with the carbonyl carbon Boron trifluoride is a strong Lewis acid and can coordinate to the carbonyl oxygen, drawing even more electron density away from the carbonoxygen double bond thereby making the carbon more susceptible to nucleophilic attack by methanol This is described best by answer choice B As for the wrong choices, A misses the minor detail that water is not a product of the transesterification reaction Choice C wants us to believe that a strong Lewis acid like BF3 can act as a nucleophile, which is impossible by the definition of a Lewis acid as it is an electron pair acceptor, not a donor Choice D solvating the methanol molecule is incorrect as this would decrease its nucleophilicity, not enhance it Boron trifluoride has nothing to with the methanol in the reaction as I said before, it simply polarizes the carbon-oxygen double bond in the triacylglycerol molecule Again, choice B is the correct response The correct answer to question is choice B This question requires a basic understanding of NMR and its relationship to the structural analysis of esters Correctly answering this question also requires a thorough understanding of the experiment described in the passage Well, the product mixture retained for analysis was the organic layer Presumably, this should be a mixture of fatty acid esters and an NMR spectrum should reflect this if the reaction did indeed go to completion If the reaction was not complete then we would expect some water insoluble reactants to be present also Water insoluble reactants would have to be the triacylglycerol, and perhaps some mono- and di- esters of glycerol resulting from the incomplete, though partial, transesterification reaction With that level of understanding, we would be safe to hit the choices Choice A states that 3H singlets are observed This is evidence that methyl groups are present, but does not signify that nothing but methyl esters are present in the product mixture Choice B, on the other hand, indicates that no multiplets-meaning no signals corresponding to the hydrocarbon "backbone" of triacylglycerol are observed If this signal were absent, there would also be no mono- or di-esters produced either This should make for pretty good evidence that starting material and partially reacted starting material are absent, and as such, it should make for a pretty good answer choice As for the rest of the choices, choice C indicates signals in the region where protons alpha to a carbonyl would be observed Since the initial triglyceride and the final fatty acid esters, as well as the partially reacted species, all have CH2 groups next to carbonyl carbons, this observation would not tell us anything about the progress of the reaction, or lack thereof Finally, choice D gives us a singlet at 9.5, or rather its absence to consider This resonance is characteristic of an aldehyde proton, so its absence indicates the absence of aldehydes in the product mixture Since neither reactant nor product is an aldehyde, this observation says nothing at all about the reaction or its progress Again, choice B is our answer For question 9, the correct answer is choice D Saponification involves the basic hydrolysis of triacylglycerol to yield glycerol and sodium carboxylates which are in fact soaps When triacylglycerol is boiled in aqueous sodium hydroxide, the hydroxide group attacks and adds to the carbonyl carbon of the ester, forming a tetrahedral intermediate, which in turn decomposes to give a carboxylic acid and an alkoxide ion The alkoxide then deprotonates the acid yielding the sodium salt of the acid and glycerol Choice D is the correct response Because the question states that triacylglycerol is hydrolyzed in aqueous sodium hydroxide and then extracted with toluene, choices B and C can be eliminated since there is no mention of chloride ions As far as choice C is concerned, if there were any unreacted triacylglycerol it would be found in the organic layer, not the aqueous layer Finally choice A is wrong because methyl esters would not be formed in the reaction This answer choice is trying to confuse saponification 10 as developed by Organic Chemistry Discretes Test with transesterification Methyl esters would be formed if say, methanol were present, but again, only sodium hydroxide and toluene are mentioned, so this answer choice is wrong and again the correct response is choice D 10 The correct answer to question 10 is choice C The diol formed would be the remains of the triglyceride after two fatty acid residues have been removed The term "diol" would refer to glycerol with one remaining fatty acid attached to it and so could be described as a monoester of glycerol The other choices are not diols Choice A methanol dimers-would just be composed of two molecules of methanol bonded to each other Choice B nonalkylated fatty acids are acids not diols, and finally, choice D a diester of glycerol will have only one free OH group Again, choice C is the correct response Questions 11 and 12 both require an understanding of the table provided in the passage The table lists several common oils, along with a breakdown of their fatty acid composition by percentage 11 The correct answer to question 11 is choice B This question requires us to determine which is highest in polyunsaturated fat Remember that an unsaturated compound is one in which double bonds are present, so a polyunsaturated compound will have two or more double bonds This corresponds to the last two columns in the table, and if you add the percentages up, you can see that linseed oil has the greatest percentage of polyunsaturated fatty acids out of all of the answer choices at 63% Of the choices given, corn oil has 62% polyunsaturated fatty acids, soybean oil is 44%, and castor bean is 4% So choices A, C, and D can be eliminated and again, choice B is the correct answer 12 The correct answer here is choice C This question requires you to a bit more than just table reading; here we need to consider the reaction of bromine with the fatty acids isolated in the experiment Bromine in carbon tetrachloride is a standard test solution for double bonds One equivalent of bromine would indicate one equivalent of double bonds and in this reaction we are told 1.5 equivalents of bromine react so there should be 1.5 double bonds per molecule, or rather, an average of 1.5 double bonds per molecule We might translate this mathematically as for every one hundred molecules there needs to be 150 double bonds This is a convenient number since the table gives us percentages, so if we just multiply the given percentage by the number of double bonds in that component we will come up with the number of double bonds which that component contributes to each one hundred molecules of the sample and hence the equivalents of bromine So let's try the answer choices starting with choice C corn oil From the table, corn oil has 1% of fatty acids with 16 carbons containing one double bond, 25% C-18 with one double bond for a total of 26% with one double bond It also has 62% times two double bonds, or 124 double bonds per one hundred molecules Adding it all up, we have 26 + 124 = 150 which is the correct answer Trying the others, safflower oil is 13% mono-, times 75% diunsaturated and times 3% triunsaturated making a total of 172, so this is more than 150 making choice B incorrect Soybean oil has 43% mono- plus 80 for the di- and times or 12 for the triunsaturated for a total of 135, so choice A is wrong as well Finally choice D, castor bean is 93% mono- and times 4% diunsaturated so it totals up to 101 double bonds per one hundred molecules or 1.01 equivalents Again then, choice C is the correct answer Discrete Questions 13 The correct answer here is choice C Remember that acid strength is increased by the inductive effect of electron withdrawing groups on the neighboring carbon atoms You should know that: (1) the more electronegative the substituents, the greater the inductive effect; (2) the closer the electronegative group is to the carbonyl carbon, the greater the effect; (3) the greater number of electronegative substituents there are, the greater the inductive effect With this in mind let's look at the answer choices Choice A, chloroacetic acid, has a pKa of 2.86 Choice B is methoxyacetic acid and it would be hard to predict how it's acidity compares with that of chloroacetic acid without experimental evidence However, choice C, fluoroacetic acid, would definitely be more acidic than both of them, since it is the most electronegative Choice D, acetic acid, would be the least acidic since it has no electronegative substituents at all choice C is the correct response 14 For question 14, the correct answer is choice D This question deals with the susceptibility of different compounds to nucleophilic attack All of the answer choices will undergo nucleophilic attack, but the compound that will undergo attack the easiest is propionyl chloride Choice A propanoic acid has a carbonyl carbon that is slightly susceptible to nucleophilic attack, since the double bonded oxygen has an electron-withdrawing effect However, if you compare propanoic acid and its functional derivative propionyl chloride, you can see that this molecule not only has withdrawing effects from the oxygen, but also the chloride You should be aware that out of all of the carboxylic acid derivatives, acyl halides are the most reactive towards nucleophiles due to the electron withdrawing effects of oxygen and a halide Choices B and C are susceptible to nucleophilic attack since they both have electronegative substituents, but not to the same extent as propionyl chloride Choice D is the correct answer 15 The correct answer to question 15 is choice A This question asks you about the mechanism of the Williamson ether synthesis If you look at the reaction, and even draw out the mechanism, you can see that this is an SN2 reaction, with a deprotonated alcohol acting as the nucleophile on an alkyl halide Sodium hydroxide reacts with phenol to form sodium phenoxide The phenoxide ion, being a good nucleophile, attacks the alkyl halide, displacing the chlorine The KAPLAN 11 MCAT fact that the alkyl halide is primary should point you towards bimolecular nucleophilic substitution SN1 choice B and E1-choice D definitely won't occur because dissociation of the alkyl halide would leave a pretty unstable carbocation Always remember that primary alkyl halides only undergo SN2 and E2 reactions Choice C is wrong because there are no elimination products the acidic proton of phenol has been replaced by a methyl group via the nucleophilic attack of the phenoxide ion on the alkyl halide, so it must be substitution Again, choice A is the correct response 16 The correct answer to question 16 is choice A The first step in this reaction involves substitution of the -OH group in benzoic acid by the -Cl group from thionyl chloride which incidentally has the formula SOCl2 The resulting compound is benzoyl chloride a highly reactive acyl halide which is also susceptible to nucleophilic substitution Treating this molecule with ammonia will result in substitution of the Cl group by an NH2 group, resulting in formation of an amide The molecule is named benzamide, which corresponds to choice A Avoiding the other choices here requires a bit of knowledge regarding reagents As benzoyl chloride is so susceptible to nucleophilic substitution, there is no way that a chlorine will be found in the final product if another nucleophile, such as ammonia, is around Therefore, you can discard answer choices C and D Choice B, p-aminobenzaldehyde, is wrong as well p-Aminobenzaldehyde is a benzene ring with an NH2 substituent para- to an aldehyde functionality This may be confuse you a bit if you're not sure of the reactions of carboxylic acids and their derivatives Again, choice A is the correct response 17 The correct answer here is choice D One thing about chemical shifts before I answer the question: the chemical shift of the aldehyde proton is 9-10 ppm Remember it! Of all the chemical shifts it is probably the most important one to have memorized for the MCAT All right, looking at the spectrum, what you see? Low and behold, a peak at 9.8 ppm an aldehyde peak! The only aldehyde in the answer choices is D, making it the correct response All of the other signals are due to alkyl protons and you don't need to worry about these too much as long as you can identify the aldehyde proton Since none of the other answer choices has an aldehyde functionality, none would have a proton that appeared at 9.8ppm, so they can be eliminated from the answer choices 12 as developed by ... elimination END OF TEST KAPLAN MCAT ANSWER KEY: C B B D B 8 10 D B B D C 11 12 13 14 15 B C C D A 16 17 A D as developed by Organic Chemistry Discretes Test OXYGEN-CONTAINING COMPOUNDS TEST EXPLANATIONS... developed by Organic Chemistry Discretes Test Passage I (Questions 1–6) β-keto esters and malonic esters (commonly known as β-dicarbonyl compounds) are compounds that contain two carbonyl groups... the following compounds? A B C D Benzamide p-Aminobenzaldehyde p-Chlorobenzamide 3-Chloro-4-aminobenzaldehyde The lH NMR spectrum below is characteristic of which of the following compounds?
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