ORGANIC CHEMISTRY TOPICAL: Molecular Structure of Organic Compounds Test Time: 23 Minutes* Number of Questions: 18 * 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 Rf (261) 105 Ha (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 Molecular Structure of Organic Compounds Test Passage I (Questions 1–7) Until recently, it had been common practice for pharmaceutical companies to manufacture chiral drugs as racemates, not as single enantiomers Usually, only one enantiomer of a given compound possesses therapeutic value, while the other may have no beneficial pharmacological properties and may even induce serious physiological side effects A typical example of this problem was the use of the drug Thalidomide in 1961 Administered to reduce nausea and vomiting during the early stages of pregnancy, the desired physiological activity lies with the R-isomer The S-isomer of Thalidomide is a teratogen (an agent that produces physical defects in developing embryos) As a result, administration of this drug as the racemate caused congenital malformations in thousands of infants The structure of Thalidomide is shown below O C N L + [Ni] + HCN MeO 6-Methoxy-2-vinylnapthalene CH3 CN H MeO CH3 CO2 H H MeO Naproxen C O N H O O (L + [Ni] = Metal complex catalyst) Figure Today, advances in the synthesis of specific enantiomers should severely limit the number of racemic drugs produced Traditionally, the enantiomer of a chiral drug could be synthesized via resolution (conversion of the racemic mixture into diastereomers, separation, and then reformation of the enantiomers) However, this process is wasteful because of the numerous steps involved in the procedure The emergence of a new technique called enantioselective catalysis may soon be used to avoid this inefficiency Reaction Another example of enantioselective catalysis is that of meso breaking For example, azidotrimethylsilane has been shown to react with cyclohexene oxide in the presence of the enantioselective catalyst titanium isopropoxide: chiral O + R3 SiN OSiR catalyst N3 Reaction The catalysts employed in enantioselective catalysis are usually based on optically active metal complexes and are highly effective in the production of one enantiomer over another The success of this technique is attributed to the production of thousands of chiral products from one molecule of catalyst A good example is the use of a nickel-based catalyst to synthesize the S-isomer of the chiral drug Naproxen (Reaction 1) GO ON TO THE NEXT PAGE KAPLAN MCAT How many chiral centers are present Thalidomide molecule? A B C D in the l What is the most likely reaction mechanism between HCN and 6-methoxy-2-vinylnapthalene? A B C D Electrophilic addition Bimolecular nucleophilic substitution Bimolecular elimination Free radical addition Which of the following statements most accurately describes the properties of Thalidomide? A It is a mixture of enantiomers that have the same chemical and physical properties, with the exception of the direction of optical rotation, but different physiological properties B It is a mixture of diastereomers that have the same physical and chemical properties, but different physiological properties C It is a mixture of enantiomers that have different physiological, chemical, and physical properties D It is a mixture of diastereomers that have different physical, chemical, and physiological properties Which of the following is the R-isomer of Naproxen? A CH3 CO2H C C CH3 H H Reaction proceeds through an S N mechanism Consequently, the attacking nucleophile is: A B C D the alkyl group the azide ion silicon the chiral catalyst The double bond of the vinyl group in 6-methoxy-2vinylnapthalene contains which of the following orbitals? I II III IV A B C D sp orbitals sp2 orbitals sp3 orbitals p orbitals II only III only I and IV only II and IV only MeO B MeO D CH3 CO2H C CH3 CO2H C H MeO CO2H C H MeO The term “meso breaking” can be applied to Reaction because: A the chirality of the catalyst is destroyed B chiral centers in cyclohexene oxide are created when the product is formed C symmetry is retained when cyclohexene oxide is converted to the product D the symmetry of cyclohexene oxide is broken to form an optically active compound GO ON TO THE NEXT PAGE as developed by Molecular Structure of Organic Compounds Test Passage II (Questions 8–13) One of the major problems in using solar energy as a power source is the conversion of the raw material, sunlight, into a usable form of fuel The transformation of energy contained in sunlight to chemical energy via photochemically induced isomerization (photoisomerization) is one method that has been suggested in order to alleviate this problem In this process, a simple organic molecule is converted to a product that has far more energy than the starting material This excess energy may be stored in the product as angle, torsional or non-bonded strain The reverse reaction, formation of the starting material, provides a means in which energy can be released and used in a controlled manner The alkane equivalent of norbornadiene, norbornane, is shown in Figure Norbornane Figure The ring strain that arises in small cycloalkanes, such as cyclopropane or cyclobutane, is mainly attributed to: A good illustration of the photoisomerization process is the conversion of norbornadiene to quadricyclane (Reaction 1) Quadricyclane has 62 kcal/mol of excess strain energy relative to norbornadiene and is therefore thermally unfavorable; the reverse reaction, quadricyclane to norbornadiene, is highly favored in the presence of metal catalysts Since it is readily synthesized in high yield, norbornadiene is a convenient starting material in the laboratory However, further research is required before it can be mass-produced for large scale adaptation I compression of the bond angles to less than 109.5° II interaction between eclipsed hydrogens on adjacent carbons III nonbonding interactions between hydrogens on non-adjacent carbons A B C D I only II only I and II only I, II, and III Which of the following will most readily undergo an intramolecular [2+2] ring closure? hv A C CH Ph Norbornadiene Strain energy = 33 kcal/mol Quadricylane Strain energy = 95 kcal/mol C CH B Reaction C Ph D Ph Ph Norbornadiene converts to quadricyclane via a photochemically allowed [2+2] ring closure, analogous to the conversion of 1,3-butadiene to cyclobutene (Reaction 2) hv 1,3-Butadiene Cyclobutene Reaction GO ON TO THE NEXT PAGE KAPLAN MCAT Catalytic hydrogenation of norbornadiene releases approximately 57 kcal of heat per mole of norbornadiene The product, norbornane, is: A more stable, due to the decrease in angle strain accompanying saturation B more stable, due to the increased number of hydrogens in the product C less stable, due to the change in hybridization of the double bonded carbons D less stable, due to steric effects and an increase in angle strain A Diels-Alder process ([4+2] cycloaddition) takes place according to the following scheme: A Diels-Alder reaction between which of the following would result in the formation of norbornadiene? A B C D Propene and 1,3-butadiene Acetylene and 1,3-cyclopentadiene Ethylene and propene Ethylene and 1,3-cyclopentadiene 1 For substituted norbornanes, it has been found that those substituted in the position undergo SN2 reactions with the appropriate nucleophile while those substituted in the position not This occurs because: A it is difficult to form a carbocation intermediate at carbon in the initial step of the reaction B it is easier to form a carbocation at the position, making substitution kinetically favorable C back-side attack at the position by the nucleophile is difficult because it is sterically hindered D the carbon at position is more electrophilic than the carbon at position 1 Which of the following is true of norbornane? A Norbornane is less stable than cyclopentane but more stable than cyclohexane B Norbornane is more stable than both cyclopentane and cyclohexane C Norbornane is less stable than chair cyclohexane but more stable than boat cyclohexane D Norbornane is less stable than both cyclopentane and cyclohexane GO ON TO THE NEXT PAGE as developed by Molecular Structure of Organic Compounds Test Questions 14 through 18 are NOT based on a descriptive passage Which of the following compounds will exhibit the greatest dipole moment? A B C D (Z)-1,2-Dichloro-1,2-diphenylethene (E)-1,2-Dichloro-1,2-diphenylethene 1,2-Dichloro-1,2-diphenylethane 1,2-Difluoroethane Which of the compounds listed below is linear? A B C D Carbon tetrachloride Propyne Acetylene 1,3-Hexadiene What is the order of increasing carbon-carbon bond length in the molecules listed below? How many structural isomers of C3H6Br2 are capable of exhibiting optical activity? A B C D I Acetylene II Benzene III Ethylene A B C D I, II, III I, III, II II, III, I II, I, III C=C, C=O, N=N, and C=N bonds are quite common in organic compounds However, C=S, C=P, C=Si, and other similar bonds are not often found The most probable explanation for this observation is that: A carbon does not combine with elements found below the second row of the periodic table B sulfur, phosphorus, and silicon not form pi bonds due to the lack of occupied p orbitals in their ground state electron configurations C sulfur, phosphorus, and silicon are incapable of orbital hybridization D the comparative sizes of the 2p and 3p atomic orbitals make effective overlap between them less likely than between two 2p orbitals END OF TEST KAPLAN MCAT ANSWER KEY: B A B D A 8 10 C D C D A 11 12 13 14 15 C D B A A 16 17 18 D C B as developed by Molecular Structure of Organic Compounds Test EXPLANATIONS Passage I B The first three paragraphs talk about the problems associated with manufacturing chiral drugs as their racemates It then goes on to give Thalidomide as a classic example of this problem O * C N C O N H O O A racemic mixture is one that contains equal amounts of two enantiomers, chiral molecules that have opposite configurations from each other at every stereocenter and so are non-superimposable mirror images In organic compounds, a stereocenter or a chiral center is usually defined as a carbon that is attached to four different groups Thalidomide definitely has at least one chiral center since the text talks about its administration as a racemate and the R– and S– isomers Therefore, choice A can be eliminated A chiral center is an atom attached to four different functional groups Starting at the benzene ring, you can see that there are no chiral centers, as all of the carbons are attached to only three substituents The same thing applies to the carbonyl groups Moving on to the carbon in the nitrogen-containing ring, you can see that this is a chiral center It is attached to a hydrogen (not shown explicitly), a nitrogen, a CH group, and a carbonyl group Moving clockwise around the ring, the next group is the carbonyl functionality Again, this center is achiral since it is only attached to three substituents; the same thing applies to the carbonyl group opposite Although the nitrogen in the ring is attached to four different groups (the lone pair of electrons not shown is considered a group), it undergoes rapid inversion of configuration that causes it to not be a chiral center Moving along to the last two carbons in the ring, you can see that neither of these are chiral since they are each attached to two hydrogens atoms So, there is only one chiral center in the Thalidomide molecule: the carbon in the nitrogen containing ring which we discussed earlier A In Reaction 1, you should be able to see that the first step involves the electrophilic addition of hydrogen cyanide to the double bond in the vinyl group Let’s look at the mechanism of this reaction in more detail Since the double bond is electron rich, it is likely to be attacked by positively polarized molecules or electrophiles The hydrogen in HCN acts as an electrophile and, therefore, adds to the least substituted carbon to form the most stable carbocation intermediate This intermediate is formed as electrons are given up from the double bond to form a new bond to the hydrogen, leaving behind a positive charge The formation of a secondary carbocation is favored over a primary carbocation, and so the intermediate –CH+CH3 is formed The cyanide ion (CN–) then adds to the carbon bearing the positive charge to form the nitrile product shown in Reaction Obviously from the mechanism I have just described, there is addition across the double bond This addition is initially electrophilic and so choice B, which describes the reaction as being nucleophilic substitution, is incorrect Elimination, as stated in choice C, is also incorrect Elimination usually involves the removal of fragments to form a multiple bond; the process that occurs in the first step of Reaction is just the opposite of this This leaves choices A and D Markovnikov’s rule states that in the addition of a HX to an alkene, the hydrogen will add to the least substituted carbon in order to form the most stable carbocation intermediate This rule is followed in Reaction 1, since the hydrogen adds to the –CH2 in the vinyl group, not the –CH group so choice A is the correct response Anti–Markovnikov addition can be obeyed when peroxides are added to the reaction mixture This reaction occurs by a radical mechanism and the hydrogen adds to the most substituted carbon This is not the case in this reaction, so choice D is incorrect B Epoxides, such as the one shown in Reaction 2, are highly strained and, because of the electron withdrawing nature of the oxygen, a nucleophile can attack one of the carbons attached to it resulting in a ring opening reaction Simultaneously, the nucleophile attacks one of the epoxide carbons, and since carbon cannot form bonds, the bond between it and the oxygen is broken The electrons from this bond are taken on by the oxygen forming a negatively charged anion In Reaction 2, the R 3SiN molecule forms R3Si+ and N – The latter is called an azide ion and as it is negatively charged; it will attack the electrophilic carbon It does so 180° to the oxygen, which takes on electron density from the carbon–oxygen bond The positively polarized silicon-alkyl group can then add to the negatively charged oxygen to form the product shown As a result, the attacking nucleophile is the azide ion and choice B is the correct response KAPLAN MCAT O– + SiR3+ O N3 OSiR3 N3 N3 – The chiral catalyst is involved in the reaction, but not as a nucleophile A catalyst serves to speed up the reaction and at the end of the reaction, it remains unchanged Therefore, it would not behave as a nucleophile and become incorporated into the product; choice D is wrong Choices A and C are wrong since these molecules constitute the electrophilic portion of the R 3SiN molecule Since they are positively polarized, there is no way they could be nucleophilic D Carbon–carbon double bonds are made up of both sp hybridized orbitals and p orbitals The electron configuration of carbon is 1s22s22p The 2s orbital and two 2p orbitals hybridize to form three sp hybrid orbitals This leaves one free p orbital which can overlap with an adjacent orbital to form a pi bond sp2 hybridized orbitals have a geometry of 120°, and these constitute the carbon-carbon and carbon-hydrogen sigma bonds in the molecule Therefore, the double bond in the vinyl group is formed by the overlap of two sp2 orbitals and two p orbitals A Thalidomide was administered as a racemate which is an equal mixture of two enantiomers You should know that enantiomers have identical chemical and physical properties with one exception: they rotate plane polarized light in opposite directions However, they behave differently in chiral environments, and so they exhibit different behavior in the human body One enantiomer reduces nausea and vomiting while the other is a teratogen which is an agent that causes physical defects in the developing embryo Therefore, Thalidomide is a mixture of enantiomers that have the same chemical and physical properties but completely different physiological effects; choice A is the correct answer Choices B and D are incorrect because the isomers in Thalidomide are enantiomers, not diastereomers If they were diastereomers, then they would possess different chemical and physical properties and would form a mixture of two different compounds, not a racemic mixture as described in the passage Choice C is incorrect because enantiomers have the same chemical and all of the same physical properties except in the direction that plane polarized light is rotated C First, let’s look at Naproxen drawn in Reaction 1, which you are told is the S-isomer Priorities have to be assigned to each substituent directly attached to the stereocenter This is done by atomic number, so the lowest priority goes to the hydrogen The other three atoms attached to the stereocenter are carbons, so the atomic weights of the groups attached to these carbons now have to be considered The substituent that will have the next to last priority is the methyl group, since hydrogens are attached to the methyl carbon In the remaining two substituents, the carbons are attached to two other carbons (in the case of the substituted naphthalene group) and two oxygens (in the case of the carboxyl group) Another rule you need to know, is that in double bonds, the atoms have to be duplicated; a carbonyl group would be classed as a carbon bonded to two oxygen atoms When this is done, the carboxyl group turns out to be of highest priority So to summarize, the order of increasing priority is hydrogen, methyl, naphthalene and then carboxyl In order to assign a configuration, the lowest priority substituent has to be rotated to the back and then arrows are drawn from the highest priority substituent (numbered 1) to the lowest priority substituent (numbered 3) If this is done for Naproxen in Reaction 1, you should see that the arrows are in an anti–clockwise direction so the molecule is an S-isomer In order to qualify as an R–isomer, the answer choice must have an opposite configuration at all chiral centers, making it a non-superimposable mirror image of the S–isomer Choice C is correct because it has the opposite configuration of the Sisomer The hydrogen group is already oriented toward the back, and so in drawing arrows from the carboxyl group through to the methyl group, you can see that the direction is clockwise, or R Choices A and D are wrong because they show the chiral carbon and the methoxy group respectively sticking out from the naphthalene ring These choices are wrong because both groups would be in the plane of the ring In addition, the orientation around the chiral carbon is incorrect in both responses Choice B is incorrect because this is the S-isomer of Naproxen; the configuration around the chiral center is exactly the same as that shown in Reaction D A meso compound is a molecule that contains chiral centers, so you would expect it to be optically active However, meso compounds also contain an internal plane of symmetry, so the molecule is optically inactive A good example of this is 10 as developed by Molecular Structure of Organic Compounds Test cyclohexene oxide, shown in Reaction The two carbons attached to the oxygen are in fact chiral centers; each one is attached to four different groups; oxygen, hydrogen, CH 2, and CH However, intersecting the bond perpendicular to these two chiral carbons is a plane of symmetry Therefore, this molecule is superimposable on its mirror image and is optically inactive The addition of azidotrimethylsilane, R 3SiN to cyclohexene oxide results in the loss of this plane of symmetry The two chiral carbon centers still remain, but the molecule is now optically active since it is nonsuperimposable on its mirror image Therefore, the symmetry of cyclohexene oxide is broken and an optically active compound is formed: choice D is the correct answer Choice A is incorrect since the chirality of the catalyst is retained in the reaction Remember: although a catalyst may take part in a reaction, it comes out the same at the end of the reaction Choice B is incorrect because the chiral centers are not generated in the reaction: they have always been present Choice C is wrong because the plane of symmetry is lost in the conversion of the meso compound to the optically active product Passage II C You should be aware that in saturated ring systems (ring systems consisting of all single bonds), the carbons are sp3 hybridized sp hybridized carbons have preferred bond angles of 109.5° However, in cyclopropane, the bond angles are compressed to 60°, whereas in cyclobutane, the bond angles are about 90° (cyclobutane takes on a ‘folded’ or ‘bent’ conformation) In both cases, the sp3 orbitals cannot assume the ideal bond angle of 109.5°, and so there is a great deal of angle strain in the molecule This makes statement I correct and answer choice B wrong Because of the rigidity of the cyclopropane ring, all six hydrogens are eclipsed This results in a great deal of repulsive interaction, so the molecule suffers from torsional strain The diagram below illustrates one such pair of eclipsed hydrogens: eclipsing interactions H H H C H H C C H Cyclobutane also suffers from torsional strain, but to a lesser degree If the bond angles were 90°, the cyclobutane molecule would be planar and all eight hydrogens would be eclipsed To alleviate this, the molecule folds and results is a bond angle of 88° At the expense of slightly more angle strain, the torsional strain is reduced Both molecules possess eclipsed hydrogens though and suffer from torsional strain; statement II is also correct and choice A is out Non–bonding or steric interactions occur when atoms bonded to non-adjacent atoms compete for the same position in space The classic example is the repulsion between substituents on opposite ends of cyclohexane in the unfavorable boat conformation There are no such interactions on cyclopropane or cyclobutane, so statement III is false D The passage gives you an example of a [2+2] ring closure by the conversion of 1,3–butadiene to cyclobutene It is pretty evident from this reaction that ring closure involves electrons in relatively close proximity to each other Ring closure involves the movement of these electrons; in the case of a [2+2] reaction, electrons are used up in forming a new sigma bond, while the other two simply move between carbons Right away, you can rule out choices A and B Benzene as shown in choice A, contains electrons and is very stable A [2+2] ring closure in this molecule is highly unlikely since it will involve breaking the aromaticity of the ring Choice B, norbornene, won’t undergo a [2+2] ring closure since it only contains electrons This leaves choices C and D You may think that ring closure in these molecules is equally likely since both contain electrons However, in choice C, the double bonds are not in close proximity to each other In order for cyclization to occur, the double bonds must be on the same side, in other words, the molecule has to be in the cis configuration To achieve this, the phenyl groups would end up on the same side of the molecule This would result in a great deal of steric interaction, and, as a result, the trans isomer is favored and ring closure is extremely difficult On the other hand, the double bonds in choice D are in closer proximity to each other, making cyclization somewhat easier 10 A Catalytic hydrogenation of norbornadiene involves the addition of two molecules of hydrogen to the double bonds, the result being the formation of the saturated ring system, norbornane You should be aware that double bonded carbons are sp2 hybridized so a p orbital is free to overlap, forming a double bond The bond angle of an sp2 hybridized carbon is 120° Looking at norbornadiene, you can see that the sp2 bond angles are much less than 120°, and as a result, the molecule suffers from angle strain When catalytic hydrogenation occurs, the double bonded carbons become sp hybridized, and the preferred bond angle lowers to 109.5° The bond angles in norbornane are much closer to this bond angle so saturation of norbornadiene is KAPLAN 11 MCAT accompanied by a decrease in angle strain In other words, catalytic hydrogenation results in the formation of a more stable molecule This is best described by choice A Further evidence that the product is more stable is provided from the question stem: it states that 57 kilocalories of heat per mole are released upon hydrogenation; in other words, mole of product is more stable than mole of reactant by 57 kilocalories Choice B is incorrect because the increase in the number of hydrogens is not the cause of increased stability of the product Choice C is correct in that the change in stability is due to the change in hybridization However, this answer choice also states that the product is less stable If the product was less stable than the reactant, 57 kcal would be taken in upon formation of norbornane For this reason, choice D is also incorrect 11 C You really need to be familiar with SN1 and SN2 reaction mechanisms in order to answer this question The rate of an SN1 reaction is dependent only on the concentration of the substrate The rate-limiting step is the formation of a carbocation by the loss of a good leaving group The nucleophile then adds to the carbon bearing this positive charge As a carbocation is formed, the order of reactivity of an alkyl halide, for example, will be tertiary, then secondary, then primary (in fact, primary alkyl halides not react via an SN1 mechanism) On the other hand, SN2 reactions depend on the concentration of both the substrate and the incoming nucleophile In a concerted mechanism, the incoming nucleophile attacks the electrophilic carbon 180° to the leaving group No carbocation intermediate is formed, and any factor that effects the incoming nucleophile and the substrate will affect the rate of reaction One of these factors is steric hindrance: if the electrophilic carbon is highly substituted, for instance, if it is a tertiary carbon, then it will be difficult for the nucleophile to get to this carbon On the other hand, if the carbon is primary, nucleophilic attack will be easier since there is less steric hindrance For this reason, reactivity of alkyl halides toward SN2 decreases in the order primary, then secondary, then tertiary (tertiary won’t react) Looking at norbornane, you can see that substitution with a halide at the position would result in the formation of a tertiary alkyl halide, whereas substitution at the position would result in the formation of a secondary alkyl halide Tertiary alkyl halides won’t react by an SN2 mechanism due to steric hindrance, so at position 1, the incoming nucleophile would be sterically hindered and would not be able to attack 180° to the leaving group SN2 would be significantly easier at position since the incoming nucleophile would be less sterically hindered This makes choice C the correct response Now for the wrong answers It would be very difficult to form a carbocation at position as stated in choice A Remember: carbocations prefer a planar geometry, with a bond angle of 120° There is no way this geometry could be achieved at position However, this is relevant only in an SN1 reaction, which is not what we have here Therefore, choice A is incorrect Choice B also talks about carbocation formation and so is incorrect Finally, choice D is wrong because the carbons at positions and would be more or less equally electrophilic if substituted by the same group 12 D Breaking norbornane down into its constituent rings, you can see that it is actually composed of a six-membered ring (numbers through 6) and two cyclopentane rings (the first ring is made up of carbons 4, 5, 6, and 7, and the second ring is made up of carbons 1, 2, 3, and 7) The six-membered ring is not in its preferred chair conformation and the bond angles are highly compressed Therefore, the molecule will be less stable than cyclohexane, eliminating answer choices A and B Both the boat and the chair conformations of cyclohexane are more stable than the norbornane molecule, so choice C is wrong as well Cyclopentane is almost as stable as cyclohexane Just like cyclobutane, cyclopentane assumes as slightly bent conformation to alleviate torsional strain at the expense of slightly more angle strain The cyclopentane rings in norbornane are far more strained than isolated cyclopentane rings, so the molecule is more unstable than both cyclopentane and cyclohexane making choice D the correct response 13 B The reaction given here involves three pairs of electrons One pair is pushed to a new position in the product molecule, while the other two pairs are involved in the formation of new sigma bonds that link the two reactants together Choice C is incorrect since there are only a total of pairs of electrons in the reactants and so a Diels-Alder reaction cannot occur Among the other choices, only one, choice B, leads to the correct product: 12 as developed by Molecular Structure of Organic Compounds Test A 1, 3-butadiene propene 1, 3-cyclopentadiene acetylene 1, 3-cyclopentadiene ethylene B D In choice B, 1,3–cyclopentadiene is the diene in the reaction and acetylene is the dienophile Don’t be put off by the fact that acetylene contains a triple bond; it still donates only two electrons in the cyclization reaction The overall process results in cyclization to form norbornadiene Two double bonds remain; one from the original diene and one from acetylene (as a triple bond was involved in the reaction, not a double bond) Cyclization involves the formation of a six–membered ring and the formation of a bridgehead carbon, originating from carbon of 1,3–cyclopentadiene Choice D is different from choice B in that ethylene is used instead of acetylene This molecule contains a double bond, not a triple bond, so when this molecule acts as a dienophile, it will give up its electrons to leave behind a single bond, not a double bond Therefore, the product will have one double bond less and so norbornene is produced instead of norbornadiene Choice A will certainly undergo a [4+2] cycloaddition reaction, but the product will be 4–methylcyclohexene, not norbornadiene Therefore, choice A is incorrect Independent Questions 14 A The dipole moment of a molecule is determined by magnitudes of the individual bond dipoles and the spatial arrangement of these individual bonds As for magnitudes, bond dipoles are greatest when there is a large difference in the electronegativities of the atoms involved in the bond Therefore, carbon–carbon bonds have little or no dipole moment, while carbon–halogen bonds have relatively large dipole moments Looking at the answer choices starting with A, (Z)–1,2–dichloro–1,2–diphenylethene has two phenyl rings attached on one side of the double bond and two chlorine groups attached on the other side of the double bond Remember, the (Z)–designation is assigned when the two highest priority substituents attached to a double bond are on the same side (priority is assigned according to molecular weight, so the two chlorines take highest priority) The electronegative chlorines pull electron density toward themselves, setting up a large net dipole moment in that direction This makes choice A the correct answer The (E)-isomer (choice B) differs in that the two highest priority substituents lie across the double bond As the chlorines pull electron density toward themselves, both dipoles cancel each other out to give a net dipole moment of zero Therefore, choice B can be eliminated Choice C contains the same substituents, but there is a single bond in this molecule, not a double bond The key difference in this molecule is that the single bond rotates, so any effective dipole moment that is set up will be canceled out by an opposite dipole as the molecule rotates So, just like in choice B, there will be no net dipole moment in this molecule Choice D is also incorrect for the same reason as choice C The carbon–fluorine bonds are highly polar, but rotation about the carbon–carbon single bond will destroy any net dipole that is set up 15 A Structural isomers have the same molecular formula, but a different atomic connectivity With a molecular formula of C3H6Br2, the carbon skeleton obviously consists of a propane chain Four structural isomers can exist; both bromines can attach to carbon of the propane chain forming 1,1–dibromopropane, both bromines can attach to carbon forming 2,2–dibromopropane, a bromine can attach to carbons and forming 1,2–dibromopropane or finally, a bromine can attach to carbons and forming 1,3–dibromopropane All of these are structural isomers since they have the same molecular formula, KAPLAN 13 MCAT but different atomic connectivity So, which of these will exhibit optical activity? Recall that in order to possess optical activity, a molecule must be chiral Most chiral molecules are identified as possessing atoms attached to four different substituents Of the four isomers, only the 1,2–substituted propane has a chiral center The second carbon is attached to a bromine, a hydrogen, a CH2Br group and a CH group Therefore, this isomer may exhibit optical activity All of the other isomers don’t exhibit optical activity because they don’t possess any chiral centers, so A is the correct response H H H H C * C C Br Br H H 16 D In the case of carbon–carbon double bonds, they are made up by the overlap of sp orbitals and p orbitals on adjacent carbon atoms The sp2 orbitals overlap head on to form a sigma bond, whereas the p orbitals overlap to form a bond Sigma bond lengths and strengths are largely determined by the size and shape of the atomic orbitals involved in formation of the bond and the ability of these orbitals to overlap effectively Sigma bonds are stronger than pi bonds because head to head overlap is more efficient than sideways overlap Sigma bonds formed from two 2s orbitals are shorter than those formed from two 2p orbitals or two 3s orbitals The difference between the first four bonds named in the question stem and final three, is the position of their respective atoms in the periodic table Carbon, oxygen and nitrogen can be found in the second period, while sulfur, phosphorus and silicon are found in the third period Therefore, sulfur, phosphorus and silicon would use 3p orbitals in the formation of pi bonds, not 2p orbitals in the case of carbon, nitrogen and oxygen The key problem, however, is that 3p orbitals are much larger than 2p orbitals and so overlap of the 2p orbital of carbon and the 3p orbital of silicon, phosphorus or sulfur is unlikely to occur This makes choice D the correct response Choice A is incorrect since carbon does combine with elements below the second row of the periodic table For example, carbon can form bonds with the elements in the halogen group Choice B is also wrong In their ground state electron configurations, silicon, sulfur and phosphorus all have partially occupied p orbitals which can form bonds Finally, choice C is incorrect since sulfur, phosphorus and silicon can undergo hybridization These elements can not only mix s and p orbitals, but d orbitals as well 17 C In order for a molecule to be linear, the bond angles must be 180° The orbital that is associated with this bond angle is an sp hybridized orbital In sp hybridized carbons, the 2s orbital combines with a 2p orbital You should know that a triple bond contains sp hybridized carbons Acetylene, choice C, having a triple bond, contains two carbons that are sp hybridized Therefore, choice C is the correct answer Now for the wrong answers Choice B, like choice C is an alkyne However, there are three carbons in propyne; two are sp hybridized, while the remaining carbon is sp3 hybridized sp3 hybridization results from the combination of the 2s and all of the 2p orbitals Four sp3 carbons form and the preferred bond angle becomes 109.5° In other words, sp bonds are tetrahedral in their geometry Therefore, propyne won’t be linear since the carbon not involved in the triple bond will have a tetrahedral arrangement of atoms around it Choice A is wrong since the carbon in this molecule is sp3 hybridized as well The carbon is attached to four chlorine substituents, so in order to achieve maximum separation between the bonds, a tetrahedral geometry is sought, hence carbon undergoes sp hybridization Finally, choice D is wrong because 1,3–hexadiene contains sp hybridized orbitals (when the double bonds occur) and sp orbitals where the single bonds occur Bond angles of 120° are associated with sp2 orbitals and as we have already discussed, bond angles of 109.5° are associated with sp orbitals Therefore, there is no way this molecule can be linear and so choice D is incorrect 18 B If we compare bonds between the same atoms, the longer the bond, the weaker it is In this case we are comparing all carbon-carbon bonds, and so essentially we are asked to arrange the bonds in decreasing bond strength The carbon-carbon bond in acetylene (ethyne) is a triple bond, and so is the strongest In ethylene, the carbon-carbon bond is a double bond and so is the next in strength The carbon-carbon bonds in benzene are intermediate between single and double bonds, and so are the weakest (and longest) among the three 14 as developed by ... developed by Molecular Structure of Organic Compounds Test Questions 14 through 18 are NOT based on a descriptive passage Which of the following compounds will exhibit the greatest dipole moment?... However, meso compounds also contain an internal plane of symmetry, so the molecule is optically inactive A good example of this is 10 as developed by Molecular Structure of Organic Compounds Test cyclohexene... symmetry of cyclohexene oxide is broken to form an optically active compound GO ON TO THE NEXT PAGE as developed by Molecular Structure of Organic Compounds Test Passage II (Questions 8–13) One of