Ch4 problem solutions - 56 0 Chapter 4 CYCLOALKANES numbers...

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Unformatted text preview: 56 0 Chapter 4 CYCLOALKANES numbers for substituent groups, using the same criteria for “lowest numbers” presented earlier. Second, rings have a “top” and a “bottom” face, relatively speaking. Therefore, substituents on different ring carbons may either be on the same face or on opposite faces, necessitating the cis or trans denotation in the name. All other principles of nomenclature follow unchanged. 4-2. Ring Strain and Structure Electron pairs repel each other and try to be as far apart as possible. Rings with only three or four atoms force the electron pairs of the C~—C bonds to be closer together than is normal for carbon atoms in molecules. The repulsion that results is the major cause of the high-energy nature of small ring compounds and is the physi— cal cause of the ring strain referred to in the text. To examine the structural aspects of these molecules, you will find your "set of models to be indispensable. Cyclopropane is the only flat cycloalkane ring. All larger cycloalkanes are nonplanar. Ring distortion away from a planar structure reduces eclipsing interactions between neighboring carbon—hydrogen bonds. 4-3 and 4-4. Cyclohexclnes Before you do anything else, make a model of cyclohexane. Be sure to use the correct atoms. and bonds from your kit. The completed model should not be too floppy and should be easily capable of holding the shape shown in Figure 4—503). This is the chair conformation, with three C—H bonds pointing straight up and three C—H bonds pointing straight down (the axial C—~H bonds). Starting from this point, you should be able to construct the other important cyclohexane conformations by moving an “end” carbon through the plane of the “middle” four carbons of the ring; that is, Move up Cm —> Boat and boatlike conformations (rather floppy, too) Learn to recognize axial and equatorial positions and their cis/trans interrelationships around the ring. Again, use your model in conjunction with the chapter text and illustrations. Note the congestion associated with large groups in axial positions, a result of 1,3-diaxial’ interactions, the main effect that causes differences in energy between the two possible chair conformations of a substituted cyclohexane. This is an example of a trans- armular (literally, “across the ring”) interaction, arising in this case from the ring structure forcing groups to adopt gauche conformational relationships. Be sure to use your models when trying to do the problems at the end of the chapter. 4-5 and 4-6. larger Rings; Polycyclic Molecules The material in these sections is intended only to give a very brief introduction to areas of organic chemistry that are important in current research but are generally beyond the scope of a course at this level. Only a small number of selected molecules are mentioned with relevant points of structure and nomenclature presented where appropriate. Solutions to Problems 17. Start with the largest ring and systematically go through successively smaller rings: 0 ECHs Cyclolientane Methylcyclobutane CH3 CH3 1, l-Dimethylcyclopropane A A 52% CH3 CH3 CH3 CH3 cis-a1, 2-Dimethyl- trans-1, 2-Dimethyl- Ethylcyclopropane cydopmpane CYCIOPI'DP‘me (Did you forget this one? Lots of students miss it.) 7-2-:. , ' 1%?«YIWZQHWYJN. L933"? --\-'-'J-'~. ' ‘ 4 Solutions to Problems 0 57 35L (3) Iodocyclopropane (b) trans— 1 -Methyl—3-( 1 -methylethyl)cyclopentane (c) cis-1,2—Dichlorocyclobutane (d) cis-l-Cyclohexyl-5-methylcyclodecane (e) To tell whether this is cis or trans, draw in the hydrogens on the substituted carbons: <— Groups on top of ring ,t ® \ <-—— Groups on bottom of ring One Br on top, one on bottom, trans-1,3—dibromocyclohexane. (1') Similarly, On top ——> 19* CH2CH3 Br CH3CH2 CHECH3 (a) (b) @431 to ‘C1 C1 CH3CH2 CHZCHg C1 CH3 F F CH3 {Cl (11) (c) (1") Br” (:1 Cl \CH3 3‘3- (a) The very low relative radical chlorination reactivity of cyclopropane implies abnormally strong C~—H bonds and an abnormally unstable cyclopropyl radical. (b) Radicals prefer SP2 hybridization, with 120° bond angles. So in the cyclopropyl radical, the bond angle strain at the radical carbon is greater (120° — 60° = 60° bond angle compression) than at a carbon in cyclopropane itself (109.5° — 60° = 49.5° bond angle compression). Forming the radical therefore increases ring strain and is more difficult in cy010propane than in a molecule lacking bond angle distortion to begin with. 23 = In all cases the reference value to begin with is the DH0 for the C—C bond between CH2 (2°) groups, he , DH“ for CH3CHQWCH2CH3, 88 kcal 11101"1 (Table 3-2). (a) Cleavage of a C—C bond in cyclopropane requires a smaller net energy input because ring strain is relieved in the process. Breaking a “norma ” C—C bond would require 88 kcal mol—1 input, but because 28 kcal mol—1 is recovered as a result of strain relief in opening the three—membered 58 0 Chapter 4 CYCLOALKANES ring, the DH” actually required is 88 — 28 = 60 kcal molvl. Note that this is consistent with the Eat of 65 kcal rnol'1 for ring opening (Section 4-2). CH3CH2- + -CH2CH3 28 kcal mol‘1 recovered as result of relief of ring strain 88 kcal mol“ 1/ (88 . d input) vs. fir?“ -CH2CH2CH2- \ a “mum!” C—C bond \\ ‘ \ 60 kcal mol" 1 net input actually required CH3CH2—CH2CH2 C327CH2 CH2 (b) For cyclobutane, our estimated DH° = 88 - 26 a 62 kcai 11101—1. (0) DH“ = 88 — 7 m 81 kcal mol—1 (d) DH“ 2 88 — 0 2 88 kcai mol“1 Thus, the unusual ring-opening reactions of cyclopropane and cyclobutane (relative to other alkanes and cycloalkanes) are thermodynamically reasonable. 22. Here is a drawing of cyclobutane, with axial (a) and equatorial (e) positions labeled. All the carbons are equivalent, and flipping the puckered form exchanges axial and equatorial positions, exactly as does flipping chair conformations in cyclohexane. . mm (a) CH31A HV Transannular (1,3-diaxia1) interaction Equatorial; more stable it CH3 .‘ CH H 3 CH3 (1)) CH3 «~— HIV l H H H a; CH3 CH3 H I, This (the trans-1,2 compound) is CH3 ‘— H more stable because both CH3‘s I can be equatorial at the same (c) H CH3 time. In the cis—l,2 [(b) above] Both CH3’s equatorial; there is always one axial group in more stable either conformation. CH3 CH3 ((1) CH3jArCH3 : HVLH H H Both CH3’s equatorial; more stable ' CH3 H (6) (:14ng,14 : H/JVLCHs H CH3 23. Refer to answers to Problems 18(6) and 18(f) for guidelines. Now it is the cis-1,3 compound that can have both CH3’s equatorial at the same time. It is more stable than the trans (below). Equal in energy: one methyl axial and one‘ equatorial in each conformation. (3) Trans. Not most stable form. Ring flip gives diequatorial conformation: m CH3 (b) Trans! (Surprise!) Note positions of hydrogens. / H OCH Trans $2 (3:25: Trans \\ H Solutions fo Problems ' 59 The two hydrogens are trans, so clearly the NHZ and OCH3 groups must be trans, too. The NH; is cis to the top H, and the OCH3 is cis to the bottom H. Both groups are equatorial, so this is the most stable conformation. H H (c) Cis. WEI : HQ H0 CH(CH3)2 CH(CH3)2 H0 From Table 4-3, we see that CH(CH3)2 prefers an equatorial position more (2.2 kcal mol") than does OH (0.94 kcal mol“). In the structure drawn, CH(CH3)2 is aerial and OH is equatorial. This is not the most stable conformation because the ring can flip to the form on the right, in which CH(CH3)2 is equatorial and OH axial. CH3 (d) Trans- ny Most stable conformation (CH3 equatorial). CH 0/ / \OCH3 60 ' Chapter 4 CYCLOALKANES I CH2CH3 (e) CiS. Most stable form (CH3CH2 equatorial). H H H (f) Trans. WEI Most stable form (both groups equatorial). NH 2 H ‘ H : E :\..§ 2 C (g) Cis. C 3 0 H3 Most stable form (both groups equatorial). H H (h) Cis. F f ? 0H Not most stable form. Ring flip makes it diequatorial. H H Br (1) C15 COOH Not most stable form. Ring flip makes H ‘i 111 HO—C group equatorial, which is preferable (Table 4-3). (j) Trans. Most stable form [compare (b), above]. 24. The sign for AG" will be negative if the conformation shown in the problem is the less stable one and will be positive if the conformation shown is the more stable one. (a) —(1.70 + 0.52) = —2.22 kcal mol” (b) 1.4 + 0.75 = 2.2 kcal mol’1 (c) —(2.20 — 0.94) = - 1.26 kcal molul (d) 1.70 — 1.29 = 0.41 kcal mol—1 (e) 1.75 - 0.46 = 1.29 kcal mol‘1 (f) 1.4 + 0.55 = 2.0 keal molfl1 (g) 1.70 + 0.75 = 2.45 kcal {1101—1 (h) —(0.94 + 0.25) = —1.19 kcal mol_] (1) —(1.29 — 0.55) = —0.74 keal mol’1 (j) 2.20 + 0.52 = 2.72 kcal 11101—1 25. Most stable Least stable conformation conformation «023% % OH OH How 0’) M CH3 CH CH CH (C) 3% ( 3)2 (EH3 Solutions to Problems I 61 CHgO CH30 CH2CH3 Cl C CH (e) 0% ( 3)3 ‘ C<CH3)3 26. From Table 4—3 Ratios, using AG" = eRT ln Keg (a) 0.94 kcal mol‘1 (less stable conformation Keq 2 4.8; 4 4'8 = 0.83; is higher in energy) . , '8.+ 1 ' - - 83/17 ratio (in favor of more stable conformation) (b)17—094=081maimr1 K =3s- 3'8 —079- ‘ ' ' 3‘4 ‘ ’ 3.8 + 1 _ ‘ ’ 79/21 ratio (c) 2.2 + 1.7 = 3.9 kcal moi“1 Kml 2: 103; 999/01 ratio ((1) 1.75 — 0.75 = 1.00 kcal 1nol‘I Kecl = 5.3; 84/16 ratio (e) 5 + 0.52 = 5.5 kcal mol“ Keq z 104; >> 99.9/0.1 ratio In each case the more stable conformation is the one in which the group with the largest AG° value from Table 4-3 is equatorial. 27. The basic idea is that the two extremes of the diagram, the two chair conformations, will no longer be equal in energy: One has the methyl group equatorial, but the other has it axial. You do not have sufficient information to estimate the energies of the twist—boat and boat conformations in the middle of the diagram, except to assume that they will probably be equal to or (more likely) higher in energy than the corresponding conformations of cyclohexane itself, relative to the more stable chair conformations. Reaction coordinate to conformational interconversion —.— 62 G Chapter 4 CYCLOALKANES 28. Both rings can flip, so there are four possible combinations, two of which are identical: H H W H These two are identical H H ‘ H Most stable: each ring 15 fiflaChEd by an Least stable: each ring equatonal bond to the other is attached by an H axial bond to the other 29. Notice how some positions around a boat conformation of cyclohexane are axial-like (“pseudoaxial”) and some are equatorial-like (“pseudoequatorial”): a a e e a = pseudoaxial e e e = pseudoequatorial a a If you draw conformations placing the methyl group in each different type of position and examine each conformation for strain, you will see the following: Methyl is pseudoequatorial CHE/H U CH3 H Best conformation L Eclipsed J Methyl is pseudoaxial CH3 H CH3 H Worst conformation: Diaxial transannular interactions interaction Of the two with the methyl in an equatorial—like position, the one on the left has the bond to the CH3 staggered with respect to neighboring C—H bonds. The other possibility is higher in energy as a result of the eclipsing interaction shown. The two conformations with pseudoaxial CH3 are both quite high in energy. One actually has three diaxial interactions involving the methyl group (only one is shown; make a model to see the rest!), and the worst of them all has a serious transannular interaction due to the close approach between the CH3 and the H shown. 30. Only a boat-related conformation permits both bulky groups to avoid axial positions. This molecule will adopt a shape in which both groups are “pseudoequatorial.” It will be based on the twist-boat of _-i11 Solutions to Problems ' 63 cyclohexane in order to minimize eclipsing interactions of the true boat conformation (Section 4-3). (Make a model!) H C(CH3)3 (CH3)3C H I 31 . Models may be helpful here. You should be able to construct structures similar to those pictured below. H H H trans-Hexahydroindane cis-Hexahydroindane Notice how the ring—fusion hydrogens are trans to each other in trans-hexahydroindane and cis to each other in cis-hexahydroindane. In the drawings, the cyclohexane rings are chairs and the cyclopentane rings are envelopes. A slight twist around the cyclopentane bond away from the envelope “flap” would give rise to the cyclopentane half-chair conformation, which is similar in energy but harder to draw. 32. In trans-decalin each of the carbon—carbon bonds attached to the ring fusion occupy an equatorial po- sition with respect to the ring on which they are attached. In the illustration below we have labeled the rings as A and B, and numbered the four relevant bonds 1—4: Bonds 1 and 2 (which are part of ring B) are equatorial substituents with respect to ring A. Bonds 3 and 4 (which are part of ring A) are equatorial substituents with respect to ring B. Both hydrogen atoms on the ring fusion carbons are axial with respect to both rings. In cis-decalin the situation is different. Look at the picture: Bond 1 (in ring B) is now axial with respect to ring A (rotate the page clockwise by 60° to see this more clearly). Bond 2 is still equatorial. Also, bond 3 (in ring A) is now axial with respect to ring B (bond 4 is still equatorial). So two of these four bonds are axial with respect to the ring on which they are substituents and, as a result, give rise to 1.3-diaxial interactions that raise the enthalpy of the compound. Notice that the ring-fusion hydrogens are now equatorial with respect to one ring and axial with respect to the other. If we were to assign an energy of about 1.75 kcal mol‘I to each axial bond to a carbon in cis-decalin, we would conclude that the cis isomer is 3.5 Real mol‘l higher in energy (less stable) than the trans. This turns out to be an overestimate, in part because carbon atoms in rings cannot rotate freely and therefore do not generate as much steric interference as do simple alkyl groups, which can rotate a full 360°. An alterna- tive way to estimate the energy difference is to search for butane structural fragments that possess gauche conformations. The cis isomer has three more than does the trans; assuming each gauche butane raises the energy content by about 0.9 Real moi“, one arrives at an energy difference of 2.7 kcal mol”. 64 I Chapter 4 CYCLOALKANES 33. You really do need to go to models to address this question. Start by examining a simpler compound, one in which a cyclopropane ring is fuscd to a single cyclohexane ring (below, left). H Is the cyclohexane in a chair conformation at all? Actually, it is not. The geometry of the three-membered ring forces the cyclohexane ring bonds associated with it and four of the cyclohexane ring carbons to all be coplanar. Only the two cyclohexane ring carbons farthest from the ring fusion are movable. In tn'cyclo[5.4.01’3.01’7]undecane (above, right), the ring labeled ‘A’ is similarly constrained. Only two of its carbons (the one at the bottom and the one at the lower left) are capable of significant motion. This isn’t much more than a wiggling motion, one up and one down, relative to the plane of the remaining four carbons in that six-membered ring. As for ring ‘B’, it is capable of holding one reasonable approximation of a chair conformation, with the appearance shown below. H B However, if you attempt a chair-chair flip in ring ‘B’, you will encounter much more resistance from the model set. The flip can be forced, but the conformation that results is a distorted twist-boatlike shape. This is because the two carbons at the ring fusion cannot undergo the corresponding rotation that would finish the chair-chair interconversion. The rigidity of ring ‘A’ prevents that. 34. (a) The [3 form is more stable because all of its substituent groups are equatorial. One of the —OH groups (the one on the ring carbon at the right of the picture) in the or form is axial. (b) K6,,l = 64/36 = 1.78. Using the equation AG" = —l.36 loglo Keg (Section 2—1) for an equilibrium constant at room temperature (25°C), we get AG" = —0.81 kcal mol”. In Table 4-3 the axial/ equatorial free energy difference for an —OH substituent on cyclohexane is #094 kcal morl. The difference is small but real, and it derives largely from the fact that the ring in glucose is not a cyclohexane but an oxacyclohexane—a cyclic ether. Replacing a ring CH2 group with an oxygen has several effects, including removal of the steric interactions associated with the hydrogens on the carbon, and introduction of two polar C—O bonds whose dipoles can give rise to either attractive or repulsive interactions with the C—0 bonds of nearby substituted ring carbons. 35. Count carbons in the molecule. If there are 10, the molecule is a monoterpene; if 15, a sesquiterpene; if 20, a diterpene. (a) 10 carbons, monoterpene (b), (c), and (d) 15 carbons, sesquiterpene (e) 11 carbons, but only 10 in the contiguous molecular “skeleton”; monoterpene (f) 15, sesquiterpene (g) 10, monoterpene (h) 20, diterpene 36. H3C (a) Alkenes ch Soluiions to Problems 0 65 37. Each isoprene unit can be set off by dashed lines: CH3 (a) CH3 : 66 o Chapter 4 CYCLOALKANES 38. Cortisone provides a good example for this exercise: Ketone Carboxylic acid Benzene (In cholic acid) (In estradiol) Alkyne Ether (In norethynodrel and mestranol) (In mestranol) C 3CH3 39. u—Pinene is a monoterpene (10 carbons): Africanone is a sesquiterpene: i 40. If pure p orbitals were used for the Cw—C bonds of cyclobutane, then each carbon could be Sp hybridized: WK sp orbitals used for C—H bonds A p orbitals 1| Solutions to Problems 0 67 The H—C—H bond angle would be 180° and cyclobutane would look like this: H H H H H H In reality, cyclobutane uses “bent” bonds just like those of cyclopropane, and all four bonds to each carbon involve hybridized orbitals. In addition, cyclobutane is not flat at all, and the H‘—C—MH bond angle is not much different from the normal tetrahedral value of 109° (see Figure 4-3). . 41. Notice that making “all-chair” cyclodecane is essentially the same as removing the ring-fusion bond from trans-decalin and replacing it with two hydrogens-u—one on each former ring-fusion carbon: H H H I [idd Remove .' l two H- l H H H bond The resulting molecule has these two new hydrogens pointing into the center of the ring, in prohibitively close contact with the carbon and hydrogens on the opposite side of the ring. The steric strain of this transannular interaction makes this a very high energy conformation. 42. (a) From left to right: chair cyclohexane, boat (or twist—boat) cyclohexane, chair cyclohexane, envelope cyclopentane. (b) All are trans. (c) 0t means below and [3 means above. Therefore, Zia—OH, 4a-CH3, Set-CH3, IOB-CH3, lla-OH, 14B-CH3, 16B—OCCH3. ll 0 (d) The boatlike cyclohexane ring is the most unusual feature; most steroids have only chair cyclohexanes. The boat shape is a result of the unusual cis relationship of the groups at positions 9 and 10, and also at positions 5 and 8. Note also the unusual number and location of methyl groups: at positions 4, 8, 10, and 14, instead of the more common pair of methyl groups at 10 and 13. on 43,©+o&>©+no-fl+© For (a), AHn = +985 — 102.5 = —4.0 kcal mol_1; for (b), AH" = -96 kcal mol—l. Overall, AH° = ~—100 kcal mol‘l. 44. (a) Propagation: C1- + RH ——> HCl + R' Can happen initially R- + <;>—Ic12 —> RC1 + Qi—Cl Main steps of chain reaction QI—Cl + RH —> HC1+ R. + Q1 68 0 Chapter 4 CYCLOALKANES Termination (one possibility): R- + Q—mia ——> RC] + Gr (1)) There are four tertiary hydrogens, but the one that is [3 (up) is too hindered to be chlorinated, because of its 1,3—diaxial interactions with the two B methyl groups. So the sites of chlorination are the three tertiary (1 (down) hydrogens: Too hindered CH3 Major sites of chlorination 45. Addition of C12 in each case generates a substituted OICIZ unit. Then, in the presence of light, these become Oi" C1 groups, which can chlorinate nearby C—H bonds according to the propagation steps shown in the answer to Problem 44(a). The selectivity comes from the fact that in the reaction for (a) the Cl group can reach the H at position 9 most easily, whereas in (b) it most easily reaches the H at position 14. Your models should look something like these: (a) CH3 (b) CH3 CH3 CH3 vs. 14 46. (a) Without making a model, this problem may initially make no sense to you at all. Begin by finding the better chair conformation for A. Of the two possibilities, the one with both alkyl groups equatorial is preferable? H3C Diaxial Diequatorial CH3 If we digress a moment for a closer look at these structures, we find that the conformations about the external bond connecting the ring to the large substituent at C4 are not optimal. As discussed briefly in the solution to Problem 33 of this chapter, the free rotation available to a simple alkyl P Solutions to Problems 0 69 group (such as CH3) causes it to be more sterically demanding than is a comparable molecular fragment contained in a ring. Furthermore, the ring angles in cycloprOpanes are only 60°, further shrinking the volume in space taken up by this substituent. As a result, a better conformation for the diequatorial structure is one in which a 120° rotation about the aforementioned bond takes the two methyl groups as far apart as possible and puts the compressed cyclopropane ring on the side of the bond closest to the methyl group at C3. Similar rotation in the diaxial moves the CH3 group away from the axial hydrogens on the same face of the cyclohexane ring: ,CHs , 0 D. . w_ CH3 lama! CH3 0 CH3 Diequatorial (b) Compound B results from opening of the cyclopropane ring in A. This process generates a 47. (a) 48. (d) 49. (d) 1,1-dimethylethyl (tert-butyl) group in place of the 1-methylcyclopropyl substituent. For purposes of evaluating steric interactions, this new tert—butyl group is effectively much larger, for two reasons. First of all, the rigid —CI-I2—CH2— fragment of the cycloprOpane ring is replaced by two freely spinning CH3 groups. Secondly, the original 60° angle between the bonds to these carbon atoms widens to the full tetrahedral value of 109.5“ upon ring-opening. The increase in steric size of the substituent is considerable and not at all conununicated adequately by line drawings. Now that we know what to look for, let’s examine conformations of B comparable to those of A drawn above. The diaxial form is just about impossible, there being no available conformation lacking prohibitive steric compression between one of the CH3’s of the terr—butyl group and one or both axial H‘s on the same ring face. However, as indicated by the intersecting arcs, the diequatorial conformer suffers from a similar interaction between methyl groups, which is unavoidably present in all staggered conformations of the substituent. Basically, hydrogen atoms on these two CH3’s are being forced to try to occupy the same volume of space: #1343 CH3 H CH3 Diaxial 0 CH3 CH3 CH3 CH3 Diequatorial 0 CH3 The molecule escapes this dilemma by adopting a shape based upon the boat shown below, but twisted to relieve eclipsing interactions. This unorthodox conformation places the methyl group at C3 in a pseudoaxial position, out of the immediate proximity of the tert-butyl group. (The situation is not terribly different from that described in Problem 30.) 0 CH3 H CH3 CH3 CH3 H 50. (a) Smallest A Eomb = most stable (diequatorial) ...
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This note was uploaded on 09/30/2009 for the course CHEM 3A taught by Professor Fretchet during the Spring '08 term at University of California, Berkeley.

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Ch4 problem solutions - 56 0 Chapter 4 CYCLOALKANES numbers...

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