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Organic Chemistry Part I Sections I-IV Section I Structure, Bonding, and Reactivity Section II Structure Elucidation Section III Stereochemistry Section IV Hydrocarbon Reactions , BERKELEY Specializing in MCAT Preparation ERKELEY R • E • V • I • E • W F.O Box 40140, Berkeley, California 94704-0140 Phone: Internet: (800) 622-8827 MCATprep@berkeleyreview.com (800) M C AT-TBR http://www.berkeleyreview.com The Berkeley Review and The Berkeley Review logo are registered trademarks of The Berkeley Review This publication for The Berkeley Review® was written, edited, and composed on a desktop publishing system using Apple Macintosh® computers and Microsoft® Word Pages were created on the Apple LaserWrite® Pro Line art was created using numerous graphics programs designed tor use on Macintosh computers The majority of the text type and display type was set in Times Roman and Palatino Cover Design by MacGraphics Copyright © 2012, 2010, 2007, 2005, 2004, 2002, 2000, 1995, 1994, 1993, 1992 by The Berkeley Review All rights reserved No part of this publication may be reproduced, stored in a retrieval system, ortransmitted, in any form orby any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission ofthe copyright owner Nomenclature a) IUPAC Nomenclature b) General nomenclature Bonding and Molecular Orbitals a) Section I Lewis Dot Structures b) Bonding Model c) Covalent Bonds d) Molecular Orbitals and Bonds i Structure, Bonding, Reactivity by Todd Bennett H H * / H C Double Bonds iii Triple Bonds e) Molecular Structures Octet Rule (HONC Shortcut) 9) Charged Structures Hybridization and ii Single Bonds ii Hybridization of Atomic Orbitals i sp-Iiybridzation ii sp2-Hybridzation iii sp3-Hybridzation b) Common Shapes a) Bond / Energy a) Bond Dissociation Energy C, Y*H H b) Ionic Bonds Intramolecular Features a) Resonance b) Inductive Effect c) Steric Hindrance d) Aromaticity Fundamental Reactivity a) Proton Transfer Reactions b) Lewis Acid-Base Reactions Acid and Base Strength i Primary Effects ii Secondary Effects iii Values and Terminology d) Electrophiles and Nucleophiles c) Physical Properties a) b) c) d) Hydrogen-Bonding Polarity Van der Waals Forces Solubility and Miscibility Berkeley ,EY Ur-E-V-I^E'W® Specializing in MCAT Preparation Structure, Bonding, and Reactivity Section Goals o» Be able to correlate structures with common and 1UPAC nomenclature It is expected that you can answer questionsabout organicmoleculeswhen given either the molecular structure or the name of the compound You will need to be able to rapidly convert from name to the structure or from the structure to name You are expected to be able to recognize common structural features like substitution and functional groups Youmust be able to name small organic molecules according to IUPAC conventions Be able to predict relative bond lengths, bond strengths, and structural angles There will be questions that require you to comparethe structural features of similar molecules You should know the hybridization-to-bond angle correlation Youshould also know what effect the scharacter of a hybrid orbital has on bond length and strength Youmust know the common molecular geometries and shapes and how they correlate to hybridization Youmust be able to read data tables and explains trends in bonding features Be able to draw resonance structures and determine which is the most stable Some questions on the MCAT require you either to count the resonance structures or determine whichresonance structureis moststable Stable resonance structureshave octetstabilityabout all atoms except hydrogen, haveminimal charges, andwhencharges arepresent, negative charge resides on the more highly electronegative atom and positive charge resides on the less electronegative Know the structure of aromatic compounds and their unusual stability Benzene isthetypical aromatic compound, because itiscompletely planar with a cyclic, conjugated arrangement of ^-electrons that obey Huckel's rule Huckefs rule states that aromatic compounds must have 4n + n electrons in a cyclic array of p orbitals, where n is any integer (or zero) Benzene has aromatic stability due to its sixn electrons in a continuous cyclic array of p orbitals You should be able to recognize aromaticity in structures other than benzene too Know the common organic acids and bases and their reactivity Common organic acids include phenols, thiols, protonated amines, andcarboxylic acids Common organic bases include amines and carboxylates Youshould be able to determine the direction of a proton transferreaction fromthe pKa values You should be ableto giveapproximate ratiosof the conjugate acid and base in a buffered solution You must have a solid understanidng on the relationship between pH and pKa Be able to predict the relative acidity and basicity of organic compounds Acidity is determined by bothprimaryeffects (involving atomsthat aredirectly bonded to the acidic Proton) and secondary effects (involving atoms that are not directly bonded to the acidic proton), rimaryeffects include atomic size, electronegativity, and hybridization Secondary effects include resonance, theinductive effect, aromaticity, andsteric hindrance You mustknow the relative impact of the various effects •> Be familiar with fundamental reactions, energetics, and mechanisms Knowing that reactions in organic chemistry involve the interaction of electron-rich sites with electron-poor sites, be able to identify the reactive sites of organic molecules You must have a fundamental understanding of electrophiles and nucleophiles and how they interact in transition states It is alsoimportant that you be ableto correlate a reaction mechanism to an energy diagram, identifying reactants, transitionstates,intermediates, and products Be able to determine relative boiling and melting points Physical properties like boiling and melting point are the result of intermolecular forces such as hydrogen-bonding, dipole-dipole interactions, and Van der Waals forces You should be able to predictthe effect ofintermolecular forces, molecular mass,and structuraldetails(like branchingand the presence of rc-bonds) on the physical properties of a compound You should be ableto predict relative physical properties from functional groups Organic Chemistry Molecular Structure Introduction Molecular Structure The perfect place to start any review of organic chemistry is the basics of molecular structure, which traditionally include bonding, hybridization, and electronic distribution We shall consider a chemical bond to be the result of atomic orbitals overlapping to form molecular orbitals We shall consider all bonds involving carbon to be covalent in nature A covalent bond is thought to involve the sharing of electrons between two adjacent nuclei According to the rules of electrostatics, the region between two nuclei offers a highly favorable environment for electrons, where they can be shared by neighboring atoms However, there are several other factors to consider in bonding If bonding were purely a matter of electrostatics, then all of the electrons would be found between two neighboring nuclei, not just the bonding electrons The sharing of electrons may be either symmetric (when the two atoms of the bond are of equal electronegativity) or asymmetric (when the two atoms of the bond are of unequal electronegativity) Sharing of electrons occurs when the atoms of a bond lack a complete valence electron shell By sharing electrons,each atom moves closer to completing its shell This is the driving force behind the formation of stable covalent bonds Having looked briefly at electron distribution, we can introduce the idea of electronic orbitals, which are three-dimensional probability maps of the location of an electron They represent the region in space where an electron is found 95% of the time We shall consider the orbitals and the overlap of orbitals to describe the electronic distribution within a molecule Once one has established a foundation in bonding, the classification of molecules can be made based on similarities in their bonding of particular atoms, known asfunctional groups Each functional group shall be considered in terms of its unique electron distribution, hybridization, and nomenclature Nomenclature, both that of the International Union of Pure and Applied Chemists (IUPAC) and more general methods describing the substitution of carbon within a functional group, shall be used to describe a particular organic molecule The review of nomenclature is continuous throughout all sections of this book Then, we shall consider the factors that affect the distribution of electron density within a molecule, including resonance, the inductive effect, steric hindrance, aromaticity, and hybridization The distribution of electron density can be used to explain and predict chemical behavior The simplest rule of reactivity in organic chemistry is that regions of high electron density act as nucleophiles by sharing their electron cloud with regions of low electron density, which act as electrophiles If you can correctly label a molecule in terms of the region that carries a partially negative charge (the electron-rich environment) and the region that carries a partially positive charge (the electron-poor environment), you can understand chemical reactions better And so begins your review of organic chemistry Fortunately, much of organic chemistry is taught from the perspective of logic, which makes preparing for organic chemistry on the MCAT easier In organic chemistry courses you are required to process information and reach conclusions based on observations, which is also required on the MCAT Reviewing and relearning this material will help you develop critical thinking skills, which will carry over into your review for other portions of the exam Despite what you may have perceived was a girth of information when you initially studied organic chemistry, you don't need to review that much material to prepare successfully for the MCAT Copyright © by The Berkeley Review Exclusive MCAT Preparation Organic Chemistry Molecular Structure Nomenclature Nomenclature IUPAC Nomenclature (Systematic Proper Naming) IUPAC Nomenclature is an internationally used system for naming molecules Molecular names reflect the structural features (functional groups) and the number of carbons in a molecule In IUPAC nomenclature, the name is based on the carbon chain length and the functional groups The suffix indicates which primary functional group is attached to the carbon chain Table 1-1 lists prefixes for carbon chains between one and twelve carbons in length Table 1-2 lists the suffices for various functional groups Be aware that "R" stands for any generic alkyl group When R is used, it indicates that the carbon chain size is irrelevant to the reaction Table 1-3summarizes the nomenclature process by listing several four-carbon compounds Carbons Prefix Carbons Prefix Carbons Prefix meth- pent- non- eth- hex- 10 dec- prop- hept- 11 undec- but- oct- 12 dodec- Table 1-1 Functionality Compound Name Bonding R-CH3 Alkane C—C & C—H R-O-R Ether C—O—C O Aldehyde R-CO-H II C—C—H R-CH2-OH Alcohol R-CO-R Ketone C—O—H O II C—C—C R-CO-OH Carboxylic acid II C—C—OH Table 1-2 Formula IUPAC Name Structural Class H3CCH2CH2CH3 Butane Alkane H3CCH=CHCH3 Butene Alkene H3CCH2CH2CHO Butanal Aldehyde H3CCH2COCH3 Butanone Ketone H3CCH2CH2CH2OH Butanol Alcohol H3CCH2CH(OH)CH3 2-butanol Alcohol H3CCH2CH2CH2NH2 Butanamine Amine H3CCH2CH2CO2H Butanoic acid Carboxylic acid Table 1-3 Copyright © by The Berkeley Review The Berkeley Review Organic Chemistry Molecular Structure nomenclature Figure 1-1 shows examples of IUPAC nomenclature for four organic compounds with variable functional groups: ® l^.O 4-chloro-5-methyl-3-heptanol Longest chain: carbons Alcohol group 3-methylpentanoic acid Longest chain: carbons Carboxylic acid group Methyl substituent at C-3 Chloro substituent at C-4 Methyl substituent at C-5 © Br Br H 3,3-dibromobutanal 3-ethylcyclopentanone Ring of carbons Ketone group Ethyl substituent at C-3 Longest chain: carbons Aldehyde group Bromo substituents at C-3 Figure 1-1 General Nomenclature (Common Naming Based on Substitution) In addition to the IUPAC naming system, there is a less rigorous method of naming compounds by functional group and carbon type (based on carbon substitution) Carbon type refers to the number of carbon atoms attached to the central carbon atom (carbon atom of interest) A carbon with one other carbon attached is referred to as a primary (1°) carbon A carbonwith two other carbons attached is secondary (2°) A carbon with three other carbons attached is tertiary (3°) Figure 1-2 shows some sample structures H CH3 \ ^ Tertiary carbon H,C s dy >dx) asp2-sp2 -_ / OH OH Compound IV Copyright © by The Berkeley Review( and OH Enantiomers 316 HYDROCARBONS EXPLANATIONS 45 Choice A is correct The first paragraph of the passage implies that Compound I is an allylic alcohol If you recall your general nomenclature, then you should know that when a functional group, hi this case the hydroxyl group, is on the carbon bonded to the alkene carbon, it is said to be allylic Choice A fits this description Choice D should have been eliminated early, because it does not contains a JC-bond Choice B is eliminated, because it is a vinylic alcohol (hydroxyl group directly bonded to the alkene carbon) Choice C has the double bond too far from the alcohol group to be allylic, so it is eliminated as well 46 Choice C is correct The option for either 1,2-addition or 1,4-addition occurs when the reactant has conjugated rt-bonds Choices A and D should be eliminated immediately, because when the two numbers describing the kbonds differ by 2, then the 7t-bonds are conjugated Cyclopentadiene has only five carbons, so one Ti-bondmust be between carbons one and two The second 7t-bond must be between carbons three and four, because in a five-carbon ring, no matter how you place two double bonds, for the ring to not be so strained it can't exist, they must be conjugated Only in choice C are the 7t-bondsnot conjugated, so choice C is the best answer Choice A Choice B Choice C ^ \ Choice D CH, 47 Choice B is correct Tlie hydration of Compound II starts with the addition of a proton to the conjugated knetwork The easiest carbon to protonate, because of steric hindrance and resonance stability, is the secondary, terminal carbon of the system This generates the structure in choice C, so choice C is eliminated That structure can undergo resonance to generate thestructure shown in choice D This eliminates choice D If water were to attack the structure shown in choice C, the structure in choice A, a new intermediate, forms This eliminates choice A By default, the best answer is choice B Choice B is not possible, because the structure would have to gain a proton at the more sterically hindered terminal carbon of the 7i-system Choice C Choice A Choice D 48 Choice A is correct At 35°C, the hydration of Compound II using sulfuric acid and water yields about 40% secondary alcohol and 60% tertiary alcohol It is stated in the passage that "the percentage of the secondary alcohol formed increases as the temperature of the hydration reaction increases This is attributed to a shift from kinetic control to thermodynamic control." This means that at 75°C, it is reasonable to suspect that the secondary, allylic alcohol is the major product This eliminates choice C The product is not a vinylic alcohol, so choices B and D are eliminated In choice A, a secondary allylic alcohol is formed, so it is the best answer Taking the information in the third paragraph of the passage and Figure 1, and erasing the cyclopentane ring could also solve this question Diels-Alder Reaction Rate Study Passage VIII (Questions 49 - 54) 49 Choice D is correct It is easiest to start by evaluating which pair represents enantiomers Enantiomers, you recall, are non-superimposable mirror images In this case, it is easier to compare the chiral centers rather than reorient the structures to see if they are mirror images If all of the chiral centers differ, then the two structures are enantiomers In choices A and C, only one of the three chiral centers differs between the pair, so they are diastereomers, not enantiomers The next factor to consider is the alignment of the carbonyl substituents They are cis to begin with (on the alkene reactant), so they should finish cis In choice B, the two carbonyl substituents are trans, sochoice Bis eliminated In choice D, the groups are cis, soit is the best answer Copyright © by The Berkeley Review 317 HYDROCARBONS EXPLANATIONS 50 Choice A is correct In Trial 4, with methyl groups in positions A and B of the conjugated diene, the rate is roughly 0.005 that of 1,3-butadiene The reason for this reduced rate is the steric hindrance associated with the methyl group pointing to the middle of the transition state No matter how the molecule contorts, one of the two methyl groups is always pointing inwards, where the transition state forms If ethyl groups were used in lieu of methyl groups, then the steric hindrance would be even more substantial and the reaction rate would be even slower Only choice A presents a slower reaction rate, so choice A is the best answer 51 Choice A is correct According to the passage, the addition of a Lewis acid to the system increases the rate of the Diels-Alder reaction Choice A, AICI3, is a Lewis acid, because the aluminum lacks a complete octet Choice B, CC14, is a common organic solvent where each atom has a satisfied octet Because CCI4 acts like a solvent and not a Lewis acid, choice B is eliminated Choice C, KH, is a strong base that readily donates electrons Because KH acts like a base and not a Lewis acid, choice C is eliminated Choice D, L1AIH4, is a strong reducing agent where each atom has a satisfied octet Because LiAlHj acts like a reducing agent and not a Lewis acid, choice D is eliminated The best answer is choice A 52 Choice D is correct If the reaction proceeds by a nucleophilic mechanism, then we must determine which molecule is acting as the nucleophile and which is acting as the electrophile The reaction is best when the dienophile has an electron-withdrawing substituent, so let us assume that the diene is the nucleophile and the dienophile is the electrophile Based on this mechanism, the part of the intermediate that originally came from the diene should carry a positive charge (because it donated electrons) and the part of the intermediate that originally came from the dienophile should carry a negative charge (because it accepted electrons) This is observed in each answer choice except choice A, so choice A is eliminated On the bright side, based on the answer choices, we know that our assumption about the diene beingthe nucleophile and dienophile being the electrophile is valid The drawing below shows the formation of the first intermediate after nucleophilic attack CH3 O0 The second intermediate drawn matches choice D, so choice D is the best answer Choices B and C could have beeneliminated by the incorrect location of the positive charge 53 Choice A is correct The dienophile is the same in Trial and Trial 4, so the difference in reactivity must be attributed to theconjugated diene This eliminates choice D Methyl groups, when bonded to a structure, not act as Lewis acids, because their octets are complete This eliminates choice C The methyl groups are mildly electron-donating, not electron-withdrawing, so choice Bis incorrect The most significant factor in thereaction rate is that the extra methyl group on the diene in Trial causes steric hindrance in the transition state No matter how the diene contorts, one of the two methyl groups interferes with the incoming dienophile The best answer is choice A CH, CH, CH, Significant steric hindrance 54 Choice D is correct It is stated in the passage that a dienophile is enhanced when it has an electronwithdrawing group conjugated to the alkene Choice A is enhanced by the carbonyl groups conjugated to the alkene, so choice A is eliminated Choice B, albeit an alkyne and not an alkene, has electron-withdrawing ester groups conjugated with the 7t-bond, so it is enhancedas a dienophile Choice B is eliminated Choice C is enhanced by the carbonyl groups conjugated to the alkene, so choice C is eliminated In choice D, the amine group is an electron-donating group that lessens the reactivity of the dienophile Choice D is the best answer Copyright © by The Berkeley Review® 318 HYDROCARBONS EXPLANATIONS Passage IX (Questions 55 - 60) Diels-Alder Reaction 55 Choice D is correct If the Y-group is a carbonyl group, then it is the exact same substituent as the other carbonyl group (on the adjacent carbon), making the compound symmetric and thus indistinguishable Both Product A and Product B are the same compound, if the reactant is symmetric, so choice D is the best answer 56 Choice A is correct In the second paragraph of the passage it is stated that when X is an electron-donating group, product A is the major product Because the OCH3 reactant yields more Product A than the CH3 reactant in comparable reactions, it can be concluded that an OCH3 group is more electron donating than a CH3 group The best answer is choice A Choices B and D should have been eliminated, because the major product is product A, not product B 57 Choice A is correct By analogy, OCH2CH3 (ethoxy) is an electron donating group like OCH3 (methoxy) The presence of the electron-donating group makes Product A the more favorable product Product A from the generic reaction of tlie passage is choice A Be careful not to choose B without paying attention to the location of the double bond The double bond in choice B is on the side opposite from where it should be 58 Choice A is correct Two five-carbon species are combined, so the final product can have only ten carbons altogether Choice D is eliminated for having twelve carbons total One of the double bonds is in the wrong location in both choice B and choice C The best answer is therefore choice A The stereochemistry with the new cyclopentyl ring trans to the bridging carbon is what is referred to as the "endo product." The arrowpushing schematic from the reactant to the product is drawn below: 59 Choice D is correct Structural isomers have different bonds (connectivity of atoms) Product A and Product B in the sample reaction in Figure are structural isomers Structural isomers result when both reactants are asymmetric The best answer is choice D 60 Choice C is correct For the reaction as drawn in the question, with two asymmetric reactants, there are two possible structural isomers (corresponding to product A and product Bin the generic reaction) that can form In both structural isomers, there are two new chiral centers formed For a compound with two chiral centers, there are four (22) possible stereoisomers, meaning that there are four possible stereoisomers for each structural isomer The result is that there are eight possible isomers total, so the best answer is choice C In reality, not all eight isomers are observed to any measurable level in a Diels-Alder reaction The major product results from the transition state of least steric hindrance In a typical Diels-Alder reaction such as this, the major products are an enantiomeric pair of one of the two possible structural isomers The less favorable structural isomer may also be formed, resulting in an enantiomeric pair, but it is generally in much lower concentration than the more favorable structural isomer Claisen and Cope Rearrangements Passage X (Questions 61 - 67) 61 Choice D is correct A concerted reaction occurs in one step Given that a sigmatropic rearrangement involves just one molecule, if it occurs in just one step, then only one product can be formed This eliminates choice C, because there are not multiple products, let alone cross products The stereochemistry can be lost at centers that go from sp^-hybridization to s/?2-hybridization and it can be gained at centers that go from sp2-hybridization to sp3-hybridization Carbons that not change hybridization cannot experience a change in stereochemistry This means that there is no set rule about the complete retention or the complete inversion of all stereocenters This eliminates choices A and B The only possible answer is the one that supports no cross products being formed, because the molecule only reacts one way Choice D is the best answer 62 Choice D is correct Step III converts a cyclic ketone into a phenol, so the product has aromaticity that the reactant does not The gain of aromaticity drives the reaction, so choices A and C are eliminated The conversion from a ketone to phenol shifts the71-bond from the carbonyl to the benzene ring, so it is the resultof tautomerization, not reduction The best answer is choice D Copyright © by The Berkeley Review® 319 HYDROCARBONS EXPLANATIONS 63 Choice A is correct The role of heat in any pericyclic reaction is to provide energy for the reactant to realign its orbitals to achieve the transition state The best answer is choice A Choice B should have been eliminated, because exothermic reactions generate heat, so no heat must be added to drive them Heat is released when bonds are formed, sigma or pi, so choices C and D are eliminated 64 Choice B is correct The Cope rearrangement involves a 1,5-diene, so there are six carbons within the molecular orbital of the transition state Choice A is eliminated because it has only four carbons Choice C is eliminated because the orbitals show no 7t-overlap between adjacent carbons Choice D is eliminated because there is no overlap across the complete cycle The best overlap is choice B, where the sigma-bond is present on the left and tlie terminal orbitals are aligned correctly to form a pi-bond 65 Choice B is correct The oxygen is directly bonded to the benzene ring in the reactant, so it is phenylic and not benzylic This eliminates choices A and C The oxygen is also bonded to the carbon alpha to the alkene This makes the carbon allylic, so choice B is the best answer 66 Choice A is correct The Claisen rearrangement converts an ether into a carbonyl, so the spectroscopic evidence must depict either the loss of an ether or gain of a carbonyl group Aldehyde protons show a signal around 9.5 ppm in the ^HNMR, so the formation of an aldehyde would in fact correspond with the appearance of a signal around 9.5 ppm Choice A is the best answer Infrared absorbances around 1700 cm"1 indicate the presence ofa carbonyl group and broadinfrared absorbances around 3400 cm"1 indicate the presence of an alcohol group No hydroxyl group appears in either the reactant or product, so choice B is eliminated The reaction would be supported by the appearance of an absorbance around 1700 cm"1 in infrared spectroscopy, not a disappearance, so choice C is eliminated Signals between and ppm in the -^HNMR correspond to vinylic hydrogens bonded to alkene carbons, which are present before and after the reaction, so choice D is eliminated 67 Choice C is correct The first reaction in the synthesis in Figure 2, Step I, involves the oxygen The Claisen rearrangement involves oxygen as the ether is converted into a ketone This eliminates choices B and D According to the remaining choices, Step II is a Cope rearrangement To determine the best answer, we must decide if the units of unsaturation decrease by one during the Claisen rearrangement or whether they remain constant at five In all compounds in Figure 2, there are four rc-bonds and one ring, so there are always five units of unsaturation This makes choice C the best answer You could also concluded that the units of unsaturation not change by looking at the Claisen rearrangement in Figure 1, where there are two 7t-bonds in both the reactant and product Passage XI (Questions 68- 73) 68 Isoprene Units Choice D is correct Terpenes are composed of isoprene subunits which are made of five carbons To be a terpene, a molecule must have a number of carbons that is divisible by five Stearol has eighteen carbons, so it cannot be a terpene The correct choice is D 69 Choice C is correct If the sesquiterpene were derived from a natural source (such as extraction or distillation from a plant), then any impurities would be naturally occurring impurities If there were two enantiomers present, that would be explained by attack at a planar site from two sides This can occur in nature although enzymes strongly favor synthesis of one enantiomer over another Choices A and B are eliminated, because chiral impurities can occur in nature Tlie dead give-away would be an impurity with sixteen carbons Terpenes have multiples of five for their carbon values Because sixteen carbons is not possible, choice C is the best choice A twenty-carbon impurity is a terpene, thus it is naturally occurring 70 Choice D is correct The carbon that is most susceptible to nucleophilic attack is the carbon with a leaving group attached Carbon four, with the pyrophosphate leaving group, is the most electrophilic Alkene carbons act as electrophiles on occasion, but in this compound, carbon four is more electrophilic than an alkene carbon The best answer is choice D 71 Choice A is correct Combining three acetyl coenzyme A molecules result in six carbons total Isoprene units have only five carbons, so one carbon must be in a side product Carbon dioxide contains only one carbon, so choice A is the best choice Ethanol and acetic acid each contain two carbons and isopropanol contains three carbons Choices B, C, and D are all eliminated Copyright © by The Berkeley Review® 320 HYDROCARBONS EXPLANATIONS 72 Choice C is correct A Diels-Alder reaction forms cyclohexene, so caryophyllene and citronellol cannot have been formed from a Diels-Alder reaction This eliminates choices A and B Both oc-pinene and Vitamin Ai have a cyclohexene moiety, so we must look closer Diels-Alder reactions involve a diene and dienophile, so we can look at the compounds in a retrosynthetic fashion In Vitamin Ai, the retro Diels-Alder reaction does not generate terpene fragments, so choice D is eliminated Choice C is the bestanswer by default 73 Choice B is correct Bond a can be eliminated immediately, because the fragments formed from the break are three carbons andseven carbons Bond ccan be also eliminated immediately, because the fragments formed from the break are nine carbons and one carbon These are not multiples offive, therefore the two fragments cannot be involved in the synthesis This eliminates choices A and C Bond b and Bond d when broken can leave a ten carbon molecule, so neither can be eliminated The trouble with bond d is that the fragment to the right of the break cannot form a 2-methylbutene, because it loses the tertiary carbon Choice D is eliminated Isoprene units must be isopentenyl, not straight chain pentenyl, thus the break is not allowed The two possible retro synthesis pathways are shown below, and only Bond b is involved Choice B is the best answer H3C 'Bondd Bond d CH, CH, or H3C Bond a T Bondc Bondb Bond a T Bond c Bond b Bond b must have been formed None of the labeled bonds to connect the isoprene units formed to connect the isoprene units Passage XII (Questions 74 - 80) were Terpenes 74 Choice A is correct Carvone differs from limin by a carbonyl group To go from limin to carvone, a carbon must lose two bonds to hydrogen and gain a double bond to oxygen This is oxidation, so choice A is the best answer 75 Choice A is correct Ozonolysis is the oxidative cleavage of a double bond between two carbons The resulting products are carbonyl compounds that vary from aldehydes to ketones to carboxylic acids, depending on the work up step To undergo ozonolysis, the reactantmust containan alkene functional group All of the compounds have an alkene functionality except for camphor This makes choice A, camphor, the best answer 76 Choice B is correct Singlets in the proton NMR are caused by unique hydrogen atoms in an environment where the adjacent atoms have no bonds to hydrogen, and thus there are no neighboring hydrogens with which coupling can take place In camphor, all of the methyl groups are bonded to quaternary carbons, so they all fit this description Because the cyclic structure is incapable of rotation, like an alkene, the two methyl groups bonded to the bridge carbon are not equivalent, causing them to express different NMR signals The result is that each of the methyl groups are represented by a singlet in the proton NMR All of the remaining hydrogens on camphor are on carbons adjacent to neighboring carbons with hydrogens, so there are no other singlets than the ones from the methyl groups This generates three proton NMR singlets, so the best answer is choice B No Hs on neighbor, cannot be rotated to be equivalent with other bridge methyl group (3H singlet) HoC 'CH3 H Quarternary Carbons No Hs on neighbor, cannot be rotated to be equivalent with other bridge methyl group (3H singlet) have no Hs attached No Hs on neighbor, isolated methyl group (3H singlet) 77 Choice C is correct Myrcene contains tencarbon atoms, so the addition of another isoprene unit would result hi a product with fifteen carbons total According to the first paragraph of the passage, terpenes having fifteen carbons are referred to as sesquiterpene, making choice C the best answer Copyright © by The Berkeley Review ® 321 HYDROCARBONS EXPLANATIONS 78 Choice B is correct Camphor has a carbonyl group (water soluble) and a large alkyl ring system (not water soluble) It is hard to decide based on the structure It happens that the compound is water soluble, which you may know first hand from using camphor-containing cleaning agents for skin The question is whether ornot it is highly water soluble Because there is some ambiguity, let's say for now that it likely not highly water soluble, and consider statement I to be invalid Camphor is a liquid at room temperature, as you might have seen if you synthesized it in a lab experiment Being a liquid at standard temperature, itsboiling point is above 298 K Statement II is valid Camphor has two chiral carbons, so it rotates plane-polarized light This makes statement III invalid Choice B is the best answer, but not with one hundred percent certainty 79 Choice D is correct It is stated in the passage that carvone has a strong UV absorbance (e > 10,000) Carvone hasconjugation, which causes its intense UV absorbance Onthe other hand, limin hasno conjugation, so itsUV absorbance is not as intense as that of carvone This means that the UV absorbance for limin has an e less than 10,000 (therefore, loge < log 104 = 4) The best answer is choice D This question required some background information on UV spectroscopy The minimum you should know is that rc-bonds are UV active, and with conjugation, the intensity of the absorbance increases and the energy of the absorbance decreases 80 Choice C is correct A Diels-Alder reaction is a cyclization reaction that involves the addition of a diene to a dienophile (alkene) to form a cyclohexene product Both limin and carvone are cyclohexene compounds, eliminatingchoices A and D, but carvonehas a carbonyl group and isoprene contains only Cs and Hs The best answer is limin, choice C Fatty Acids and Oils Passage XIII (Questions 81 - 86) 81 Choice B is correct It is stated in the passage that corn oil is 63% linoleic acid Looking at table shows that linoleic acid is made up of eighteen carbons and has two rc-bonds Choices C and D are eliminated, because they only have sixteen carbons Choice A is eliminated, because it has three 7t-bonds The best answer is choice B You can try to match the exact location of each 7t-bond from the formula in Table to the drawing in the answer choices, but doing so is not time efficient 82 Choice D is correct Linoleic acid contains two rc-bonds, both with cis geometry When linoleic acid is treated with FADH2, the result is hydrogenation of the diene and the formation of the aliphatic carboxylic acid (of eighteen carbons) stearic acid The gain of four hydrogen atoms increases the molecular mass of the acid (by four), and the loss of unsaturation results in more molecular flexibility, which results in a higher melting point Unsaturated fats, with less flexibility and therefore less ability to engage in intermolecular interactions, have lower melting points than saturated fats of comparable mass This is common organic chemistry knowledge that you should have addressed when comparingvegetable and animal fats The correct answer is choice D 83 Choice C is correct Oleic acid is an eighteen-carbon acid with one rc-bond between the eighth and ninth carbons The rc-bond in oleic acid has cis orientation Treating oleic acid with deuterium (D2) and a catalytic metal like palladium adds two deuterium atoms across the 7t-bond of the alkene molecule The two deuterium atoms add syn to one another at carbons eight and nine The result is the formation of two new chiral centers There are no chiral centers to begin with, so the product has two chiral centers The best answer is choice C 84 Choice D is correct Potassium permanganate reacts with alkenes to form diols by adding two hydroxyl groups is a syn addition fashion to the carbons of the 7i-bond Linolenic acid has three 7t-bonds, located between carbons and 10,carbons 12and 13, and carbons 15and 16when the carboxylic acid carbon is considered to be carbon one (IUPAC convention) Hydroxyl groups form at all sp2-hybridized carbon sites This results in a product with hydroxyl groups at carbons 9,10,12,13,15, and 16, as listed in choice D 85 Choice B is correct Themost bromine per molecule is consumed by the fatty acid with the greatest numberofnbonds present Forevery7t-bond present, onemolecule ofbromine liquid willbe consumed Arachidonic acid has four 7U-bonds Arachidic acid has no 7t-bonds present in its structure, linoleic acid has two7t-bonds presentin its structure, and linolenic acid has three rc-bonds present in its structure Arachidonic acid is the most unsaturated of the choices The correct answer is thus choice B 86 Choice B is correct Palmitoleic acid has sixteen carbons and one 7t-bond When palmitoleic acid is fully hydrogenated, it forms the aliphatic acid of sixteen carbons (listed in Table as palmitic acid) The best answer choice is B Copyright ©by The Berkeley Review® 322 HYDROCARBONS EXPLANATIONS Passage XIV (Questions 87 - 92) 87 Occidentalol Synthesis Choice C is correct Figure shows occidentalol with three specified chiral centers, so choices A and Bcan be eliminated All of the carbons in the ring structure either have s/?2-hybridization or they are methylene (CH2) groups The methyl group and tertiary alcohol carbon not have stereocenters, so there are no other chiral centers on the structure The best answer is choice C 88 Choice D is correct The first paragraph states that occidentalol is a sesquiterpenoid Occidentalol has fifteen carbons total, so it is a reasonable conclusion that sesquiterpenes have fifteen carbons Choice D is the best answer Monoterpenes have ten carbons, diterpenes have twenty carbons, and triterpenes have thirty carbons 89 Choice A is correct The maximum wavelength of absorbance, Xmax, increases as the conjugation of the nnetwork increases All four choices are conjugated dienes, but Compound 1also has a carbonyl in the conjugated networks As such, the compound with the longest ^^ is Compound 1,choice A 90 Choice B is correct Step involves a Diels-Alder reaction followed by decarboxylation The intermediate compound is the Diels-Alder product There is no nitrogen present in either reactant, so thecompound cannot be a lactam (cyclic amide) This eliminates choice D The product of a Diels-Alder reaction involving a diene and an alkene (dienophile) is cyclohexene, so choice B is the best answer The diene in Compound is regenerated from cyclohexene after decarboxylation, so choice A is eliminated The reaction is shown below: 150°C Cyclohexene Lactone (not lactam) CO Me Compound CO Me Carbonyl is NOT conjugated to alkene Compound 1.5( intermediate compound in Step 1) The compound is a cyclohexene with a lactone that is not conjugated to the alkene 91 Choice B is correct Occidentalol has one more methyl group than Compound 6, so the role of methyl lithium, MeLi, must be to add a methyl group to Compound This eliminates choices C and D Methyl lithium has an anionic carbon, so it acts as a nucleophile rather than an electrophile This eliminates choice A and makes choice B the best answer The methyl anion attacks the carbonyl carbon in the same fashion as an alkyl magnesium bromide anion attacks a carbonyl in a Grignard reaction 92 Choice A is correct A conjugated diene can be protonated at either terminal carbon of the 7t-network, because the carbocation that results is resonance stabilized This eliminates choices B and C Carbon a is a secondary carbon while carbon d is a tertiary carbon It is easier to protonate the less hindered site, so carbon a is the site that is most likely to gain H+ The best answer is choice A Not Based on a Descriptive Passage Questions 93-100 93 Choice D is correct Leukotriene A4, LTA4, has four alkene rc-bonds, one carbonyl Tt-bond, and an epoxide ring This results in six units of unsaturation Choice A is a valid statement, which eliminates it LTA4 has three alkene rc-bonds in conjugation, resulting hi six 7t-electrons hi a conjugated system Choice B is a valid statement, which eliminates it Because of the extensive conjugation, there are several sites at which a nucleophile and electrophile may add For instance, if the epoxide oxygen were protonated, a nucleophile could attack the ring or the left carbon of any 7t-bond to add across the system This means that 1,2-addition, 1,4-addition, 1,6addition, and 1,8-addition are all possible Choice C is a valid statement, which eliminates it In all likelihood, this answer choice earned the coveted "huh?", meaning you can't eliminate it, because you're just not sure On a multiple choice exam, this is not a problem You just need to look at choice D and use your testing logic LTA4 has four alkene 7t-bonds and one carbonyl rc-bond, so there are nine sp2-hybridized carbons There are twenty carbons total in LTA4, so eleven ofthem are s/^-hybridized There are more s/?3-hybridized carbons than sp2-hybridized carbons, so choice D is an invalid statement, which makes it the best answer Copyright © by The Berkeley Review® 323 HYDROCARBONS EXPLANATIONS 94 Choice D is correct Bromine, in the presence of light, adds to the mostsubstituted carbon of an alkane by way of a free radical mechanism The most substituted carbon is tertiary, so the choice with bromine added to the tertiary carbon (most substituted) is the best answer This makes choice D the correct choice The product shown, as well as its enantiomer, are both formed 95 Choice D is correct Free radical propagation reactions keep the free radical reaction going, so to be a propagation reaction, there mustbe the same number offree radicals on each side of the equation In choice A, there is one free radical on the reactant side and three free radicals on the product side, so it is not a propagation step Choice A is eliminated In choice B, there are two free radicals on the reactant side and no free radicals on the product side, so it is a termination step and not a propagationstep Choice B is eliminated In choice C, there is one free radical on the reactant side and two free radicals on the product side, so it is not a propagation step Choice C is eliminated In choice D, there is one free radical on the reactant side and one free radical on the product side, so it is a propagation step Choice D is the best answer You may not recognize the reaction from the overall mechanism, but it converts a less stable free radical into a more stable free radical, which ultimately impacts the product distribution 96 Choice A is correct Bergamontene contains fifteen carbons, so it is likely made from three 5-carbon isoprene units We can't be sure without analyzing the structure to find the isoprene fragments, but that is not time efficient Choice A is the best answer so far, and shall remain our choice until a better one comes Bergamontene is a hydrocarbon with no heteroatoms, so it is a terpene and not a terpenoid This eliminates choice B Bergamontene has two 7t-bonds and a cyclohexane ring, so at first look choice C is tempting But the molecule is bicyclic, meaning it has a second ring, the four-membered ring connected to the cyclohexane ring Bergamontene has four units of unsaturation, not three This eliminates choice C To verify this, bergamontene has 24 hydrogens and therefore a formula of C15H24 The units of unsaturation are (2(15) + -24)/2 = 4, so choice C is eliminated There are three chiral centers on bergamontene, so (2^) is the maximum number of stereoisomers, not 16 This eliminates choice D and secures choice A as the best answer 97 Choice D is correct Both the Claisen rearrangement and the Cope rearrangement require dienes, but they need not be conjugated The two rc-bonds must be separated by three sigma bonds, so choicesA and C are eliminated Clemmensen reduction converts a carbonyl into an alkane, so no diene of any kind is required Choice B is eliminated A Diels-Alder reaction involves the cyclization of a conjugated diene and a dienophile, so it must have a conjugated diene This makes choice D the best answer 98 Choice C is correct Foran E2 reaction, thebase mustbestrong enough to remove a protonfrom carbon and bulky enough to not undergo substitution Thismakes choice A a valid statement and thereby eliminatesit For an E2 reaction, the leaving group and proton being lost from carbon must be positioned anti to one another, so the geometry of the product is dependent upon the alignment of the reactant Cis versus trans results from the orientation and stereochemistry, so choice B is a valid statement and thereby eliminated For an Ei reaction, a leaving group first leaves, resulting in a carbocation With carbocations, rearrangement can be observed, so it is with Ei reactions that we see rearrangement, not E2 reactions Because E2 reactions are concerted, there is no rearrangement, so choice C is an invalid statement and thereby the correct answer Heat is required to drive both Ei and E2 reactions, so choice D is a valid statement It is eliminated, leaving choice C as our choice 99 Choice D is correct Terpenes are hydrocarbons of 10,15,20, etc carbons, so they are somewhatmassive lipids Because they are hydrocarbons, they are lipid soluble, so choice A is a valid statement Choice A is consequently eliminated Terpenes have molecular masses of about 140 g/mole, about 210 g/mole, about 280 g/mole, etc., so they have somewhat high boiling points High is a relative term, so we can't be certain in eliminating choice B However, choices B and C are the same concept, so they mutually exclude one another from consideration It is important that you use all of your test taking skills The specific rotation of a compound is dictated by its chiral centers, which a terpene may or may not have Given that there is no general rule about the chirality of terpenes, we cannot conclude that they have high specific rotations Choice D is the best answer 100 Choice C is correct To have a dipole not equal to zerois to have a dipole To have a dipole is to be polar Cis compoundsare alwayspolar so Compound IIIis polar Ethylene is perfectly symmetric, so choice Compound II is nonpolar The question comes down to: "Is Compound I polar?" Compound I is not polar, becausethe methyl groups on the alkene cancel one another and sum to a resultant vector of Choose C for best results Copyright © by The Berkeley Review® 324 HYDROCARBONS EXPLANATIONS Organic Chemistry Section Answers Sections I - IV Organic Chemistry Bubble Sheet (Make four copies, one for each section of the book.) 1-1 CD CD CD CD CD CD CD CD 37 CD CD CD CD 38 CD CD CD CD 73 CD CD CD CD 74 CD CD CD CD CD CD CD 39 CD CD CD CD 75 CD CD CD CD CD CD CD CD 40 CD CD CD CD 76 CD CD CD CD CD CD CD CD 41 CD CD CD CD 77 CD CD CD CD CD CD CD CD 42 CD CD CD CD 78 CD CD CD CD CD 43 CD CD CD CD 79 CD CD CD CD CD CD CD CD 44 CD CD CD CD 80 CD CD CD CD CE> CD CD CD 45 CD CD CD CD CD CD CD CD 81 CD CD CD CD 10 CD CD CD CD 46 CD CD CD CD 82 CD CD CD CD 11 CD CD CD CD 47 CD CD CD CD 83 CD CD CD CD 12 CD CD CD CD 48 CD CD CD CD 84 CD CD CD CD 13 CD CD CD CD 49 CD CD CD CD 85 CD CD CD CD 14 CD CD CD CD 50 CD CD CD CD 86 15 CD CD CD CD 51 CD CD CD 16 CD CD 17 CD CD CD CD CD CD 18 CD CD CD CD 54 87 CD CD CD CD CD CD 88 CD CD CD CD CD CD CD 89 CD CD CD CD 52 CD CD 53 CD CD CD CD CD CD CD CD CD CD CD CD CD 90 CD CD CD CD 19 CD CD CD CD 55 CD 91 CD CD CD CD 20 CD CD CD CD 56 CD CD CD CD 92 CD 21 CD CD CD CD 57 CD CD CD CD 93 CD CD CD CD CD CD CD CD CD CD CD 22 CD CD (D CD 58 CD CD CD CD 94 23 CD CD CD CD 59 CD 95 CD CD CD CD CD 96 CD CD CD CD CD CD CD 24 CD CD CD CD CD CD CD 60 CD CD CD 25 CD CD CD CD 61 CD CD CD CD 97 CD 26 CD CD CD CD 62 CD CD CD CD 98 CD CD CD CD 63 CD CD 99 CD CD CD CD 27 CD CD CD CD CD CD CD CD CD 64 CD CD CD CD 29 CD CD CD CD 65 CD CD CD CD 30 CD 66 CD CD CD CD 28 CD CD CD CD 100 CD CD CD CD 67 CD CD CD CD Chemistry CD CD CD 68 CD CD CD CD Section CD CD CD CD 69 CD CD CD CD 34 CD CD CD CD 70 CD CD CD CD Raw Score 35 CD CD CD CD 71 CD CD CD CD 36 CD CD CD CD 72 CD CD CD CD 31 CD CD CD CD 32 CD 33 Copyright © The Berkeley Review 328 Estimated Scaled Score The Berkeley Review Specializing in MCAT Preparation 'ife DERKELEY IJR'E'V'i'e'W" PERIODIC TABLE OP THE ELEMENTS H He 1.0 4.0 10 Li Be B C N O F Ne 6.9 9.0 10.8 12.0 14.0 16.0 19.0 20.2 11 12 13 14 15 16 17 18 Na Mg AI Si P S ci Ar 23.0 24.3 27.0 28.1 31.0 32.1 35.5 39.9 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85.5 87.6 88.9 91.2 92.9 95.9 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 80 81 82 83 84 85 86 Hg TI Pb Bi Po At Rn (209) (210) (222) 56 57 Cs Ba La+ 132.9 137.3 55 87 88 Fr Ra 138.9 89 K Ac§ (223) 226.0 227.0 f 72 74 75 76 78 79 W Re Pt Au Hf Ta Os Ir 178.5 180.9 183.9 186.2 190.2 192.2 195.1 104 105 106 107 108 109 110 197.0 200.6 HI 204.4 207.2 209.0 126.9 131.3 112 Uub Uuu Mt Uun Hs Bh Sr (277) (272) (269) (266) (262) (265) (261) (262) (263) Rf Db 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Sm Eu Gd Tb Dy Ho Er Tin Yb Lu 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0 96 97 98 99 100 101 102 103 Cm Bk Cf Es Fm Md No Ce Pr Nd Pm 140.1 140.9 144.2 (145) 90 73 77 Th 91 Pa 92 U 93 Np 232.0 (231) 238.0 (237) Specializing in MCAT Preparation 94 95 Pu Am (244) (243) (247) (247) (251) (252) (257) (258) (259) Lr (260) Organic Chemistry
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