Unformatted text preview: ORGANIC CHEMISTRY TOPICAL: Amines Test 1
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
1 H 1.0 3 Li 6.9 11 Na 23.0 19 K 39.1 37 Rb 85.5 55 Cs 132.9 87 Fr (223) 4 Be 9.0 12 Mg 24.3 20 Ca 40.1 38 Sr 87.6 56 Ba 137.3 88 Ra 226.0 21 Sc 45.0 39 Y 88.9 57 La * 138.9 89 Ac 227.0 22 Ti 47.9 40 Zr 91.2 72 Hf 178.5 104 Rf (261) 58 Ce 140.1 90 Th 232.0 23 V 50.9 41 Nb 92.9 73 Ta 180.9 105 Ha (262) 59 Pr 140.9 91 Pa (231) 24 Cr 52.0 42 Mo 95.9 74 W 183.9 106 Unh (263) 60 Nd 144.2 92 U 238.0 25 Mn 54.9 43 Tc (98) 75 Re 186.2 107 Uns (262) 61 Pm (145) 93 Np (237) 26 Fe 55.8 44 Ru 101.1 76 Os 190.2 108 Uno (265) 62 Sm 150.4 94 Pu (244) 27 Co 58.9 45 Rh 102.9 77 Ir 192.2 109 Une (267) 63 Eu 152.0 95 Am (243) 64 Gd 157.3 96 Cm (247) 65 Tb 158.9 97 Bk (247) 66 Dy 162.5 98 Cf (251) 67 Ho 164.9 99 Es (252) 68 Er 167.3 100 Fm (257) 69 Tm 168.9 101 Md (258) 70 Yb 173.0 102 No (259) 71 Lu 175.0 103 Lr (260) 28 Ni 58.7 46 Pd 106.4 78 Pt 195.1 29 Cu 63.5 47 Ag 107.9 79 Au 197.0 30 Zn 65.4 48 Cd 112.4 80 Hg 200.6 5 B 10.8 13 Al 27.0 31 Ga 69.7 49 In 114.8 81 Tl 204.4 6 C 12.0 14 Si 28.1 32 Ge 72.6 50 Sn 118.7 82 Pb 207.2 7 N 14.0 15 P 31.0 33 As 74.9 51 Sb 121.8 83 Bi 209.0 8 O 16.0 16 S 32.1 34 Se 79.0 52 Te 127.6 84 Po (209) 9 F 19.0 17 Cl 35.5 35 Br 79.9 53 I 126.9 85 At (210) 2 He 4.0 10 Ne 20.2 18 Ar 39.9 36 Kr 83.8 54 Xe 131.3 86 Rn (222) * GO ON TO THE NEXT PAGE. 2 as developed by Amines Test 1
Passage I (Questions 18) A student was given a mixture of unknown nitrogen-containing compounds and asked to identify them. After separation by distillation and extraction, the student analyzed the individual compounds. The results are shown in Table 1. Compound Table 1 Molecular Reaction formula with acetyl chloride C4H11N slow reaction C4H11N fast reaction C6H15N no reaction C6H5O2N no reaction Reaction with 5% HCl salt formation salt formation salt formation no reaction 1 . Which of the following is formed when acetyl chloride reacts with an amine? A. B. C. D. A nitrile group A peptide bond A nitro group An amide group 2 . Which of the structures listed below could represent Compound A and Compound B, respectively? A.
NO2 NH2 A B C D and H3C C CH 3 CH 3 B.
H N CH 2CH3 and CH 3CH2CH2CH2NH2 The general reaction of an amine with acetyl chloride is shown in Reaction 1.
O CH3 C Cl + R O CH3 C R N R + HCl R N H CH 2CH3 C.
H3C CH 3 C CH 3 CH 2 NH2 NH2 and H3C C CH 3 CH 3 D.
CH 3CH2CH2CH2NH2 (R = alkyl or H) and H3C N CH 3 CH 2CH3 Reaction 1 The general reaction of an amine with an acid is shown in Reaction 2.
R R N + HX R (R = alkyl or H) R R N+ R H X- Reaction 2 GO ON TO THE NEXT PAGE. KAPLAN 3 MCAT
3 . Which of the following is the most soluble in water?
A. Ph N H Ph 6 . Compound A reacts with acetyl chloride more slowly than Compound B because: A. B. C. D. it is a more sterically hindered molecule. it is a tertiary amine. it has a higher molecular weight. it possesses a nitrogen with a lone pair of electrons. B. Ph Ph N Ph 7 . Why is an amine salt more soluble in water than the corresponding free amine?
CH2CH3 C. CH3CH2 N+ H CH2CH3 D. NO2 A . It is ionic, and is therefore more soluble than covalent compounds with the same structure. B . It has a higher molecular weight than the corresponding amine. C . The negative charge on the nitrogen atom increases water solubility. D . All amines are insoluble in water. 8 . Compound A and Compound B could be classified as: A. B. C. D. enantiomers. conformational isomers. geometric isomers. structural isomers. 4 . Compound D will NOT react with HCl because: A . the nitrogen atom does not have a lone pair of electrons. B . it is a nonpolar molecule. C . the nitrogen atom carries an acidic proton. D . it is an amide, and is not basic. 5 . Which of the following could NOT be Compound C? A. B. C. D. Triethylamine Ethylmethylpropylamine Dipropylamine Ethylisopropylmethylamine GO ON TO THE NEXT PAGE. 4 as developed by Amines Test 1
Passage II (Questions 914) A number of methods are available to prepare amines. The more common pathways are shown in Figure 1.
O KOH N O Phthalimide H ethanol O Potassium phthalimide RCH 2X, O N R C N LiAlH 4 R NH3 R CH2 X CH2 NH2 1) H + /H2O, 2) OH R' X OH R CH2 O
NH 3 The reaction of an alkyl halide with ammonia to produce a primary amine is limited in use since the reaction does not stop cleanly after monoalkylation: the primary amine can react further with an excess of alkyl halide to produce secondary and tertiary amines and quaternary ammonium salts. In order to investigate amines further, a student reacted ammonia (1 mole) with Compound A, formed by the reaction of propene with hydrogen bromide, to form Compound B. Compound B was then converted to Compound C by reaction with Compound D. When Compound D was reacted with potassium phthalimide and hydrolyzed, n-butylamine was produced. Compound C has the formula C7H17N. These reactions are summarized in Figure 2.
Propene + Hydrogen bromide Compound A O N K+ CH2R R' NH R'' X OH
Potassium phthalimide nButylamine Compound D Compound B R' X R CH2 N+ R''' R'' R''' X R CH2 R' N R''
Compound C C 7H1 7 N Figure 1 The reaction of potassium phthalimide with an alkyl halide (the Gabriel synthesis) and the reaction of ammonia with an alkyl halide are restricted to primary and secondary alkyl halides, since these reactions proceed through an S N 2 mechanism. The most suitable method to produce primary amines are through the Gabriel synthesis and the reduction of nitriles, both shown in Figure 1. Figure 2 9 . According to information given in the passage, which of the following starting materials could be used to synthesize n-butylamine? A. B. C. D. 1-Chloropropane 2-Chlorobutane Butanenitrile Diethylamine GO ON TO THE NEXT PAGE. KAPLAN 5 MCAT
1 0 . What is the most probable structure of Compound A?
A. NH2 B. CH3 CH CH3 C. CH3 CH2 CH2Br Br 1 3 . The reaction of a tertiary alkyl halide with potassium phthalimide to produce an amine would NOT be successful because: A . potassium phthalimide is a poor nucleophile. B . only primary alkyl halides react with potassium phthalimide to produce amines. C . tertiary alkyl halides will not readily undergo nucleophilic substitution reactions. D . potassium phthalimide would act as a base, and elimination products would be formed. D. CH3 CH CH3 NH2 1 4 . Which of the following best represents the transition state in the reaction of ammonia with bromomethane? A. B. C. D. [H3N --CH3 --Br+] [H3N+ --CH3 --Br] H3N+--CH3 + Br CH3NH2 + HBr 1 1 . Which of the following statements is NOT true of amines? A . Primary amines can be produced only from primary alkyl halides. B . The reduction of nitriles yields only primary amines. C . Phthalimide contains an acidic hydrogen atom. D . Quaternary ammonium salts are more soluble in water than their corresponding secondary and tertiary amines. 1 2 . What is the most probable structure of Compound C?
A. CH 3 CH CH 3 B.
2 NH [CH 3 CH2CH 2 CH2]2 NH C. CH 3 CH CH 3 NH (CH 2 ) 3CH3 D. CH 3 CH2CH 2 CH 2NH2 GO ON TO THE NEXT PAGE. 6 as developed by Amines Test 1
Questions 15 through 17 are NOT based on a descriptive passage. 1 5 . Which of the following amines is the most basic in the gas phase? A. B. C. D. H2NCH3 (CH3)3N (CF3)3N NH3 1 6 . What is the absolute configuration of the chiral carbon in the molecule shown below? COOH H H3C CH3 A. B. C. D. S R d trans NH2 CH3 1 7 . Which of the following compounds will have the lowest boiling point? A. B. C. D. Diethylamine n-Butylamine Dimethylethylamine sec-Butylamine END OF TEST KAPLAN 7 MCAT
ANSWER KEY: 1. D 2. B 3. C 4. A 5. C 6. 7. 8. 9. 10. A A D C B 11. 12. 13. 14. 15. A C D B B 16. 17. B C 8 as developed by Amines Test 1 EXPLANATIONS
Passage I D A O=CNR linkage is an amide group. This group is formed in Reaction 1 when the acetyl chloride reacts with the amine. The nitrogen on the amine acts as a nucleophile and attacks the carbonyl carbon. This results in the formation of a tetrahedral intermediate: the amide is generated by the donation of an electron pair from oxygen to the carbon-oxygen bond, resulting in the loss of Cl, which acts as the leaving group. You could have been fooled into thinking that a peptide bond is formed in the reaction since this bond also consists of a O=CNR linkage. However, peptide bonds link together amino acids, and are formed through the reaction of the amine functionality of one amino acid with the carboxyl functionality of another. Neither the acyl chloride nor the amine constitutes an amino acid, and in addition, the product is not a dipeptide, so choice B is wrong. A nitrile group consists of a carbon triple bonded to a nitrogen. A nitro group consists of a nitrogen attached to two oxygens: O C nitrile N N O nitro B From Table 1, you can see that Compounds A and B must be isomers of each other. Choices A and C can be eliminated because one molecule in each pair does not have the empirical formula given. The pairs of compounds shown in choices B and D both conform to the correct empirical formula. Since the two compounds react with acetyl chloride, they must be primary or secondary amines--i.e. they must possess at least one proton that can be plucked off from the nitrogen. Choice D is therefore incorrect since the second molecule in the pair is a tertiary amine which cannot react with acetyl chloride to form the amide. Incidentally, Compound A reacts with acetyl chloride more slowly than Compound B because it is more sterically hindered (secondary versus primary amine), which further supports choice B. 3. C This structure is a quaternary ammonium ion, which forms when a tertiary amine reacts with an acid. The nitrogen in the amine contains a lone pair of electrons that can be used to form a bond to a proton. The nitrogen, therefore, becomes positively charged and the resulting ion becomes soluble in water. 4. 2. N O O 1. A Looking at the molecular formula provided in Table 1 for compound D, one might have noticed the small number of hydrogen atoms it contains. We are therefore led to suspect a high degree of unsaturation. In fact, a very likely candidate for the molecule seems to be nitrobenzene. In order to form a salt, nitrobenzene would have to gain a hydrogen when it reacts with aqueous HCl, forming a positively charged ion. This requires that the nitrogen atom has a lone pair of electrons that would enable it to form a bond with the bare proton. This is not satisfied in the nitro functionality: in fact, if we examine the Lewis structure for it (see explanation to #1 above), we see that the nitrogen already has a formal positive charge. This inability to form a salt is the reason why Compound D will not react with HCl, so choice A is the correct response. Choice C is incorrect because nitrobenzene is not an acidic molecule. Choice B is incorrect since nitrobenzene is a fairly polar molecule; the nitro group withdraws electrons from the ring. (This is why it is a meta-directing deactivator in electrophilic aromatic substitutions!) In addition, the polarity of the molecule does not correlate directly with its acid-base properties. Finally, choice D is incorrect since Compound D is not an amide. Note that even if we cannot say with absolute certainty that the molecule is nitrobenzene, we have been able to arrive at a plausible and consistent response to the question with this supposition. C The clue in answering this question is the reaction of Compound C with acetyl chloride. It can be deduced from the scheme given for reaction 1 that only ammonia, primary amines, and secondary amines will react with acetyl chloride to form an amide, an N-substituted amide, and N,N-disubstituted amide, respectively. This can be seen from the following reaction mechanism: 5. KAPLAN 9 MCAT
O C H3 C Cl HNRR' H3C H O C N R Cl R' H3 C O C H N R R' + Cl H+ O C H3 C NRR' The first step in the reaction is an addition: the amine, with its lone pair of electrons on the nitrogen, acts as a nucleophile and attacks the carbonyl carbon which has a partial positive charge due to the electronegative oxygen atom. This step results in the loss of the carbonyl functionality as a tetrahedral intermediate is formed. The nitrogen also gains a formal positive charge as its previously non-bonding pair of electrons now has to be shared with the carbon atom. In the next step, the carbon-oxygen double bond is reformed as Cl acts as a leaving group (in the form of chloride ion). The last step involves the deprotonation of the nitrogen: a proton needs to be plucked off so that nitrogen can regain a lone pair of electrons. It is this step that distinguishes a tertiary amine from others: a tertiary amine does not have a hydrogen bonded to the nitrogen that can be lost in the last step. In other words, the nitrogen is "stuck" with the positive charge at the end, resulting in an unstable acylammonium chloride salt, which can be easily hydrolyzed to a carboxylic acid in the presence of water. A tertiary amine, however, would still be able to form a salt with HCl since it does have a lone pair of electrons that enables it to act as a Lewis base. The information given in Table 1 is consistent with compound C being a tertiary amine. So we are looking among the answer choices for the compound that is NOT a tertiary amine.
N N NH N triethylamine ethylmethylpropylamine dipropylamine (N, N-diethylethanamine) (N-ethyl-N-methyl-1-propanamine) (N-propyl-1-propanamine) ethylisopropylmethylamine (N-ethyl-N-methyl-2-propanamine) The structures for the answer choices are shown above. (The IUPAC names are in parentheses.) Answer choice C; dipropylamine, is a secondary amine: there are 2 propyl groups and 1 hydrogen attached to the nitrogen. Therefore, this molecule would react with acetyl chloride to form an N,N-disubstituted amide. All of the other answer choices are tertiary amines. 6. A If the reacting functionality of a compound is sterically hindered by adjacent groups, it will react at a slower rate. Compounds A and B are amines, but Compound A must be more sterically hindered since it reacts at a slower rate. Both compounds possess a nitrogen with a lone pair of electrons, so choice D is wrong. As we discussed in question 5, tertiary amines will not react with acyl chlorides to form an amide, so choice B is incorrect. Choice C is incorrect because both compounds have exactly the same molecular weight. 7. A As we discussed in question 3, amine salts are soluble in water because the nitrogen has a positive charge. Therefore, this molecule is ionic and will be more soluble than the corresponding neutral amine. Choice B is true in that the amine salt has a higher molecular weight than the amine, but it is not the reason why it is more soluble. Choice C is incorrect because a nitrogen donates its lone pair of electrons when forming a salt. If a lone pair of electrons is donated, a positive charge will be left behind on the nitrogen, not a negative charge. Finally, choice D is wrong since some amines are soluble in water. Amines follow much the same solubility pattern as carboxylic acids; up to 6 carbons, they are pretty soluble. 10 as developed by Amines Test 1
8. D Structural isomers are defined as molecules with the same molecular formula but different atomic connectivities. In question 2, it was shown that compounds A and B could have been diethylamine and butylamine. These are only two of a number of possible isomers. Because they possess the same molecular formula, and react with acetyl chloride at different rates, indicates that they are the same type of compound. They must, therefore, be structurally different. Conformational isomers are the isomers that result from rotation about single bonds of the same molecule--like the eclipsed and staggered conformations of ethane. Conformational isomers are generally indistinguishable because of the rapid rotations about single bonds. Conformational isomerism is therefore unable to account for the difference in reactivities of the compounds. Choice A is incorrect because enantiomers have identical chemical and physical properties (apart from their ability to rotate plane polarized light). As a result, they would react in the same way, so you would not see a difference in their reactivities toward acetyl chloride. Finally, choice C is incorrect because geometric isomers differ in the arrangement of substituents around a double bond. If Compounds A or B possessed double bonds, the molecular formula would have fewer hydrogens. Passage II C From Figure 1, you can see that upon reaction with lithium aluminum hydride, nitriles are reduced to amines. Since the nitrile functionality is at the end of the alkyl chain, reduction results in the formation of a primary amine. Butanenitrile, choice C, consists of an alkyl chain 4 carbons long where one of the end carbons is triple-bonded to a nitrogen. Upon reduction, the CN group is converted to a CH2NH2 group; the molecule formed isn-butylamine. Reacting 1chloropropane, choice A, with potassium phthalimide, followed by hydrolysis, would result in the formation of npropylamine. By the same token, reacting 1-chloropropane with ammonia would also result in the formation of npropylamine, so choice A is wrong. If 2-chlorobutane, choice B, was reacted with potassium phthalimide and hydrolyzed, a structural isomer of n-butylamine would be produced--namely sec-butylamine. This molecule would also be produced if 2chlorobutane was reacted with ammonia. Choice D is wrong since this is a secondary amine. If this was reacted with another alkyl halide molecule, a tertiary amine would be produced, not a primary amine. 10. B You are told in the passage that Compound A is produced by the reaction of propene with HBr. HBr will add to propene according to Markovnikov's rule. Markovnikov's rule states that the hydrogen in HBr will add to the least substituted carbon. The reasoning behind this rule is that in the ionic addition mechanism, the first step in the reaction is the addition of the proton from HBr to one of the carbons in the double bond. The site of the addition is determined by the stability of the resultant carbocation intermediate.
H3C H3 C C H H Br C H H3C H C H H C H H C H H H C H primary carbocation: not stable 9. secondary carbocation: more substituted, hence more stable In the initial step, the carbocation CH3CH+CH3 is formed. Br can then add to this carbocation to produce the secondary alkyl halide CH3CHBrCH3--this structure is shown in choice B. Choices A and D are incorrect because they contain an amine group. You are told that Compound A is formed by the reaction of propene with hydrogen bromide--two reactants that do not contain nitrogen. Choice C would be formed if anti-Markovnikov addition was obeyed. This type of addition occurs through a radical mechanism and needs UV light or peroxide to initiate it. There is, however, no mention of UV light or peroxide in the passage, so choice C is incorrect. A The statement in choice A is incorrect because primary amines can be produced from a number of molecules, not just primary alkyl halides. Methods not mentioned in the passage include the reduction of nitro compounds, the reaction of an aldehyde or ketone with ammonia, and the degradation of a primary amide. In terms of the passage, choice A is incorrect because the second paragraph states that ammonia can react with primary or secondary alkyl halides: if you react 2chloropropane with ammonia, you get isopropylamine, which is a primary amine. 11. KAPLAN 11 MCAT
Choice B can be eliminated because this statement is true--if you look at Figure 1, you can see that reduction of a nitrile does result in the formation of a primary amine. Choice C is also wrong--in forming potassium phthalimide, the nitrogen in phthalimide loses a proton. Therefore, this proton is acidic, and choice C is true. Finally, choice D can be eliminated because quaternary ammonium salts possess a nitrogen with a full positive charge, and, as we discussed in question 3, this makes them more soluble in water than their corresponding amines. 12. C You should have already established from question 10 that Compound A is 2-bromopropane. This compound is then reacted with ammonia to form Compound B. This reaction is a straightforward nucleophilic substitution reaction where an NH2 group replaces a halide; therefore, Compound B is isopropylamine. (The general scheme for this reaction is included in Figure 1.) Compound B was then reacted with Compound D to form Compound C. Having established what Compound B is, you need to identify Compound D. The passage states that when Compound D was reacted with potassium phthalimide and then hydrolyzed, n-butylamine was produced. Working backwards, you should be able to see that the compound that would have reacted to form n-butylamine would have been an n-butyl halide, or indeed, any n-butyl chain to which a good leaving group is attached. Therefore, isopropylamine (Compound B) reacts with CH3CH2CH2CH2X, where X could be any good leaving group, to produce Compound C. This reaction also proceeds through a nucleophilic substitution mechanism: the lone pair on the nitrogen attacks the positively polarized carbon attached to the leaving group. As a result, the nitrogen becomes substituted with the n-butyl group, still remaining attached to the isopropyl group. Therefore, Compound C is nbutylisopropylamine, choice C. In addition, this molecule has the correct molecular formula--C7H17N. Choice D is a primary amine. We have already established that the reaction of a primary amine with a primary alkyl halide yields a secondary amine, so this answer choice is incorrect. Choices A and B are secondary amines, but choice A has the molecular formula C6H15N and choice B has the molecular formula C8H19N, making them both wrong. The reaction scheme of Figure 2 is shown below:
CH 3CH=CH2 + propene NH3 CH 3CHCH3 (1) potassium phthalimide CH 3CH2CH2CH2X CH 3CH2CH2CH2NH2 (2) hydrolysis (Compound D) CH 3 H C CH 3 (Compound C) N H CH 2CH2CH2CH3 NH2 (Compound B) HBr CH 3CHCH3 Br (Compound A) 13. D The reaction of potassium phthalimide with alkyl halides is limited to primary and secondary alkyl halides since the reaction proceeds through an SN2 mechanism. Tertiary alkyl halides are useless in this synthesis because they will not undergo SN2 reactions--the incoming nucleophile is sterically hindered by the alkyl groups. Potassium phthalimide is not only a good nucleophile, but a strong base as well. As just mentioned, tertiary halides will not undergo substitution because of steric hindrance. In the presence of a strong base, such as potassium phthalimide, bimolecular elimination will occur instead. As a result, an alkene will form. Therefore, choice D is the correct response. Choice A is incorrect because potassium phthalimide is a good nucleophile. Choice B is incorrect because secondary amines can also react with potassium phthalimide to produce amines. This statement is redundant anyway, since it has nothing to do with the lack of success of the reaction of tertiary alkyl halides with potassium phthalimide. Finally, choice C is wrong because tertiary alkyl halides will readily undergo nucleophilic substitution reactions. Although they will not undergo SN2 reactions, they will readily undergo SN1 reactions given the right conditions. 12 as developed by Amines Test 1
14. B Bromomethane is a primary alkyl halide; therefore, it will undergo SN2 reactions. You should know the following about SN2 reactions: (1) the rate of the reaction depends on the concentration of both the nucleophile and the substrate; (2) the nucleophile attacks the substrate at 180 to the leaving group; (3) as the bond between the nucleophile and the substrate is forming, the bond between the substrate and the leaving group is breaking, this is known as a concerted mechanism; (4) in the transition state, the nucleophile acquires a partial positive charge, because it is transferring electrons to the substrate, and the leaving group acquires a partial negative charge, because it is taking electrons. Therefore, the correct representation is choice B. Choice A is wrong because it shows the attacking ammonia with a partial negative charge and the leaving bromine with a slight positive charge. Choice C is wrong because this is the intermediate of the SN2 reaction. The free amine is recovered by reaction with aqueous base. Finally, choice D is wrong because this is the product of the substitution reaction.
H N H H + H H H C Br H H H H H N
+ H C Br transition state H H2NCH3 H
+ H C H + Br N H H H Independent Questions 15. B Amines can act as bases (Brnsted-Lowry and Lewis) because the lone pair of electrons on nitrogen can be used to form a coordinate covalent bond to a proton: N + HA N H + A Given a particular acid HA, the more the equilibrium shown above lies to the right, the more basic the amine is: an equilibrium that lies to the right means that the forward acid-base reaction as written is favorable, i.e. it is favorable for the amine to pluck off a proton from the acid. What would cause the equilibrium to lie to the right? If the ammonium salt formed as a product is stabilized, for example. Electron-donating groups attached to the nitrogen atom in the amine will make the amine more basic because by donating electron density, they will stabilize the positive ion formed. (An equivalent way to look at this is to say that electron-donating groups DEstabilize the lone pair of electrons on the original amine, making it more reactive.) Conversely, substituents that are electron-withdrawing will make the amine less basic by the same reasoning as above. We need to know that alkyl groups, compared to hydrogen atoms, are electron-donating. We would thus expect basicity to decrease in the order trimethylamine>methylamine>ammonia in the gas phase. (The gas phase is specified because in aqueous solutions, hydrogen bonding also plays a role in stabilizing the salt, and thus may disrupt the ordering.) In addition, electronegative atoms like the fluorines in choice C, (CF3)3N, will strongly reduce the basicity of the amine because the inductive electron-withdrawing effect of the F atoms make it very unfavorable for the nitrogen to acquire a positive charge as it forms a salt. Choice C is virtually non-basic. 16. B The carbon atom attached to the carboxyl group, the amine group, the hydrogen atom and the tert-butyl group is a chiral center. The first step in determining the configuration is to assign priorities to the substituent groups based on atomic number: in this case the priority decreases in the order amine, carboxyl, t-butyl, and then hydrogen. In the convention of the KAPLAN 13 MCAT
Fischer projections, the hydrogen atom and the amine group, attached along the horizontal positions, are coming out of the plane of the paper. To correctly determine the configuration, however, we want to reorient the structure so that the hydrogen points towards the back. One way to do this is to switch the positions of the hydrogen and the carboxyl group: the configuration of the resultant molecule would be the opposite of that of the original one. After switching the positions of the two, we can see that the amine, carboxyl, and t-butyl groups are arranged in a counterclockwise fashion. The configuration is therefore S. But the configuration of the molecule given would be the opposite, which is R. COOH H NH2 C(CH3)3 mirror images (2) HOOC H (1) NH2 C(CH3)3 (3) S Choice C is wrong since lower case d stands for dextrorotatory, a term used to describe the clockwise rotation of planepolarized light by an optically active compound. You should be aware that the direction of optical rotation of a molecule has nothing to do with its absolute configuration. This molecule may well have a clockwise rotation, but experimental data would be needed to determine this. Finally, choice D is wrong since the term trans only applies to the relationship between substituents attached to a ring or a double bond. 17. C Boiling points of compounds are determined by two general factors: molecular weight and intermolecular interactions. The higher the molecular weight, the harder it will be to "push" it into the gas phase, and hence the higher the boiling point. Similarly, the stronger the intermolecular interactions, the more energy will be required to disrupt them and separate the molecules in the gas phase, hence the higher the boiling point. In this question, all four of the answer choices have the same empirical formula C4H11N; they are structural isomers of one another.
N H NH2 N NH2 diethylamine (N-ethylethanamine) n-butylamine (1-butanamine) dimethylethylamine (N,N-dimethyl-1-ethanamine) sec-butylamine (2-butanamine) Molecular weight thus does not play a part here. In terms of intermolecular interactions, the strongest of these between neutral molecules is hydrogen bonding. Tertiary amines do not possess any hydrogen atoms that are directly bonded to nitrogen, and hence cannot participate in hydrogen bonding. Choice C, dimethylethylamine, is a tertiary amine and will therefore be expected to have the lowest boiling point (weakest intermolecular interactions). 14 as developed by ...
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- Amine, a. b. c., b. c. d., Potassium phthalimide