Problem Set Chapter 1 Answer Key

Transcription

1 Problem Set hapter 1 Answer Key 1. a. b. There are eight: N N N N 1

2 2... a) N.... b).. c) N Notice that these Lewis structures have been drawn as line-bond structures, and incorporate bond angles appropriate to the hybridization states of their atoms. It is typical for organic chemists to draw Lewis line-bond structures this way a) 2p 2s ¾½ ¾½ ¾½ ¾½ 1-1s ¾½ b) 2p ¾ ¾ ¾ 2s ¾½ N 1s ¾½ c) 3d 3p 3s 2p 2s ¾½ ¾½ ¾½ ¾½ Al 3 1s ¾½ rganic chemists normally draw electronic configurations using only the core and valence orbitals of atomic species: that s why, of example, 1- above was configured without 3s, 3p, 3d, or higher orbitals. 2

3 4. a) 3d 3p 3s 2p 2s ¾½ ¾½ ¾½ ¾½ Mg 2 1s ¾½ b) 2p ¾ ¾ ¾ 2s ¾½ 1-1s ¾½ c) 2sp 3 ¾½ ¾ 1 1s ¾½ 5. a) 2sp 3 ¾½ ¾½ ¾ ¾ B 3-1s ¾½ b) 2p 2sp 2 ¾½ ¾½ ¾ 1 1s ¾½ alternatively, as the problem does not explicitly ask for the ground state electronic configuration, 2p ¾ 2sp 2 ¾½ ¾ ¾ 1 1s ¾½ 3

4 6. a) E * e 1s 1s e e 2 internuclear distance b) Less stable: e 2 has two electrons in an antibonding orbital, whereas 2 e have none. The rule of thumb is that bonding interactions are better than nonbonding interactions, which in turn are better than antibonding interactions: E σ > E n > E σ* 4

5 7. a) b) ive: sp 3 1s, Nsp 3 1s, sp 3 Nsp 3, Nsp 3 sp 3, sp 3 1s c) < E 8. a) b) Six: σ sp 3 1s, σ Nsp sp, π Nsp sp, σ sp 3 sp, σ sp 3 sp 3, σ sp 3 1s c) < E 5

6 9. a) b) Seven: 2 different sp 3 sp 3, 3 different sp 3 1s, and 2 different sp 3 sp a) Si( 3 ) 4 is a bigger ball (larger surface area for induced dipole-induced dipole effects) than ( 3 ) 4. Therefore, it has the higher boiling point. b) Si bonds are too long to allow for the good Sip p orbital overlap needed to create a π bond. 11. a). N N.... b) 14 c) about 109.5E 6

7 12. a. Na a 2 2- l - - S 2- b. 2p ¾ ¾ 2sp ¾½ ¾½ 1s ¾½ N N : N N hybridization state: sp 3 hybridization state: sp 2 a. learly add lone pair electrons so as to fill the octets of all atoms of caffeine. b. In the spaces provided, provide the hybridization state of the nitrogens indicated. c. The bond that the squiggly arrow is pointed at is at a double bond. This bond has formed from overlap of a sp 2 orbital of carbon and a sp 2 orbital of nitrogen to form a σ bond, and from overlap of a p orbital of nitrogen and a p orbital of carbon to form a π bond. 7

8 14. a. 3 S > > 3 3 b. It s the bond. c. It s the bond. 15. A B D E 8

9 16. This is a combination of two effects. irst, bond polarization is a factor of both electronegativity difference and the length of the bond: the longer the bond between two different atoms, the more polarized that bond is. A l bond is 30% longer than a bond, so the effective polarizing ability of l on is about one-third greater than it would be if the l bond length were the same as a bond length. Put another way, for l bonds, the electronegativity of l that feels really is one-third greater than atomic chlorine s value (4.0 instead of 3.0). hlorine has about the same polarizing effect on carbon as fluorine because of the difference in bond lengths! Second, being in row 3 of the periodic table, chlorine is bigger than fluorine. It occupies a larger volume of space than fluorine, so it has a greater effect on compression of the bond angles p in 3 l compared to 3, effectively increasing the contribution of bond polarization to the dipole moment of 3 l relative to 3. l effect of l about the same as angle compression increases - bond contributions to dipole moment 9

10 17. a. 2p sp 2 ¾½ ¾½ ¾ ½ b. a nonpolar covalent bond c. carbon or iodine 10

11 Problem Set hapter 2 Answer Key a. N N b. 3 2 ( 3 )(N 2 )()= c. 3. rom top to bottom: g, c, e, h, b, d

12 4. a) N... b) _.. N c) N N.. N.... N.. N N. 5. a) A : arrow indicates resonance structures. A arrow means to move the 2 electrons of the S σ bond and localize them as alone pair of electrons of S. b) S

13 6. a) A.... _. B.. N N N.... _ b)..... _ 7. a) b. No way! Resonance structures differ only by the location of pi and nonbonding electrons. This structure has no pi electrons. Also, the positions of the atoms of the molecule do not change in resoance structures. They sure have here. c. melting point, boiling point, vapor pressure, refractive index, dipole moment d. Tetrahedral geometry allows for only one possible isomer: And no, it does not prove 2 2 is tetrahedral! It proves only that it probably is not square planar. The scientific method relies on the idea that a hypothesis, once stated, can be tested. I the hypothesis is not consistent with the results of a test, it is not a reasonable hypothesis. I a hypothesis is consistant with the results of tests run to check it, it is reasonable until disproven. The facts of science are phenomena that we understand by discovering what they are not, as opposed to what they are.

14 8. a. N b. 2 4 N 2 c. Yes d. In all three resonance isomers, the carbon bonded to the nitrogen is sp 2 -hybridized and the other carbon is sp 3 -hybridized. 9. a) ere are two of the possible structures that can be drawn:.... _.. N N.. b) The circled one has all octets filled and none of the atoms have formal charges. c) 120 E 10. a) two explanations are: (1) there are 10 valence electrons around nitrogen, violating the octet rule, and (2) there is no formal charge indicated for B, but as written, it has a formal charge of -1.

15 11. a) nine b) four c) two (both are on oxygen) :: :: A B D The relative importance of contribution of each resonance isomer to the resonance hybrid, from most important to least important: B D A (most important) (least important) 13. ether ester N nitrile alkene The wedged bond-lines are telling you that the bonds are coming out of the plane of the paper towards the observer.

16 14. a..... N b. π 1s 1s sp 2 Nsp 2 1s 1s sp a. ten b N 2 c. p N = a little less than E 16. a. : l : : l :.... l : l : : l : b. 30 c. yes d. : l : sp 2

17 17. a. : : : : : : N 2 N N b. 29 c. 11 d. 7 e E f. the bond is projecting out of the plane of the paper, towards the viewer.

18 Problem Set hapter 3 Answer Key 1. a) weak b) weak c) greater 2. parts a), b) and c) are: wa wb sb sa d) left e) greater 3. N 2 S (one way to remember this: pk a decreases as you go from left to right and top to bottom in the periodic table) 4. 0r parts a & b: : : : :.. : : : : :.... :.. : : : :.. : -. : : : : : :.. B A A B c. the one with K eq = 1.4 x 10 3.

19 5. 3. S 3 - S 6. a) k WA WB SB SA b) less than one. 7. The rankings are: highest pk a N 2 A B D B 3 lowest pk a A D 8. greater

20 9. a. Equilibrium reaction What are the 2 letters of the reactants acting as acids? What is the letter of the reactant that is the stronger base? Is the equilibrium constant greater than 1, less than 1, or equal to 1? i. B, D less than 1 ii. A, D less than 1 b. N. :N : : Mg.... : : - Mg.. Lewis acid Lewis base

21 Problem Set hapter 4 Answer Key 1. a) 2,3-dimethyl-4-propylheptane b) c) d) 4-tert-butyl-2,3-dimethylnonane or 2,3-dimethyl-4-(1,1-dimethylethyl)nonane 2. octane 2-methylheptane 3-methylheptane 4-methylheptane 2,2-dimethylhexane

22 2,3-dimethylhexane 2,4-dimethylhexane, and so on a) b) induced dipole - induced dipole c) d < c < b < a d) 1, 3, 2, 4 4. a) one b) four c) four 5. a) 5-tert-butyl-3-methyloctane or 3-methyl-5-(1,1-dimethylethyl)octane 6. highest lowest

23 7. a b c

24 8. a b c

25 9. a. N 2 b. N intramolecular -bonding! 10. a A E b B D

26 c. going counterclockwise from conformation A: A A E B D

27 11. a. trans-1-sec-butyl-4-methylcycloheptane or trans-1-methyl-4-(1-methylpropyl)cycloheptane b. c. 3-cyclobutyl-2,4-dimethylhexane 12. d < a < b < c 13. a) 1-isobutyl-3,3-dimethylcyclopentane or 1,1-dimethyl-3-(2-methylpropyl)cyclopentane b) one example is: c) d) 2-cyclobutyl-3,4-dimethylpentane

28 14. more less 1 axial substituent 1 axial substituent 1 equatorial substituent 1 equatorial substitutent 2 1,3-diaxial interactions 2 1,3-dianxial interactions a. A. Ï B. Ð. Ò D. Ô Õ b. onformation 3-3 gauche 3-1,3-diaxial

29 c. 5 is a chair conformation 6 is a boat conformation a d. a e e 17. The all axial conformation actually has fewer steric interactions: specificially, the isopropylisopropyl 1,3-diaxial isopropyl- gauche interactions of the all axial conformation are less substantial than the - isopropyl-isopropyl gauche interactions of the all equatorial conformation. It s helpful to make a model and see that this is so! 18. a. b. The tert-butyl group on 2 is anti to 6 of the cyclohexane ring. The ethyl group on 1 is anti to the 3 group of 2. c. ive. d. It is on the opposite side.

30 19. a. b. c. All substituents are equatorial. 20. They are conformational isomers. 21. A B The eclipsing interactions of the methyl and sec-butyl in B are worse than the methyl-methyl 1,3-diaxial interaction indicated in A. Therefore, A will have the lower potential energy. 22. a. trans-1-(1-chloro-2,2-dimethylpropyl)-2-ethylcyclohexane b. three

31 23. N 2 N a. b. constitutional isomers c. cis-1,4-diethylcyclobutane has significant angle strain due to its cyclic p = 90 E that trans-1,4-dimethylcyclohexane does not. Its ethyl substituents also are pseudoeclipsing. 25. a. 3-ethyl-4-(fluoromethyl)-2,3,4-trimethylheptane b. staggered c d. 60 E e. zero

32 Problem Set hapter 5 Answer Key 1. a) in order, from top to bottom: enantiomers enantiomers constitutional isomers identical enantiomers b) in order, from top to bottom: (R) (R) (S) (R) (S) (S) 2. a) (R) 3. Yes. It s mirror image is not superimposable on itself. (You ll find it useful to build models to help answer this question.) 4. in order, from top to bottom: a) identical b) conformational isomers c) diastereomers d) enantiomers e) meso 5. B. A is a meso isomer (that is, it has a mirror plane of symmetry bisecting the 3 4 bond.)

33 a) conformational isomers b) (S) (S) identical (R) (R) enantiomers diastereomers

34 8. a) b) bromo (draw the product s structure in its chair conformation) 9. a) (S) (R) (R) b,c) yes: (3S,4R) in eclipsed conformation the isher projection to draw isher projection clearly shows it to be meso from top to bottom: diastereomers, diastereomers, enantiomers

35 12. a. constitutional isomers b. conformational isomers c. diastereomers 13. a. l ( 3 ) b. It s trans. It has a nonsuperimposable mirror image. The cis-isomer is meso.

36 meso meso a pair of enantiomers

37 Problem Set hapter 6 Answer Key 1. a) 3. S 3 - S b) Energy E a ) rxn Reaction c) There is no intermediate formed. The reaction is a one-step reaction and interediates only form in multistep reactions. d) 3 * * - S

38 e) and f) 3 3 S - 3 S 3 2) initiation step: propagation steps: ) termination steps (any one of them is K):

39 a. elimination b. substitution 4. l Na I- B 3 Na I- (Bl 3 )- I Na (Bl 3 )-

40 5. a) from left to right: 2, 3, 1. b) There are 4 possibilities, and all of them would be favored as they generate more stable cations: : : : :.... irst case: a 2E to 3E rearrangement, and the 3E carbocation is resonance stabilized: : Second case: a 2E to 2E and allylic rearrangement; again, carbocation is resonance stabilized: : : Third case: a 2E to 3E rearrangement: : : ourth case: rearrangement to an all octets filled bromonium ion: :....

41 6. a) from most stable to least stable: 2, 5, 3, 1, 4 b) rom most stable to least stable: D,, A, B 7. δ- N 3 δ- 3

42 Problem Set hapter 7 Answer Key 1. a) I b) N 3 c) NR d) N

43 2) a) A chemical species capable of acting as an electron donor; a Lewis base that undergoes a bond-forming reaction. b) 1-Iodo-2,3,3-trimethylbutane. It has less branching on the carbon adjacent to the I bond. c) As mesylate is a better leaving group than bromide, it s the compound on the right. 3) The first reaction (the one using 1-methyl-1-cyclobutanethiol as the starting material) is slower. Because the rate limiting step of the S N 1 reaction is formation of a carbocation intermediate, inspection of those intermediates formed in either reaction clearly shows that, due to angle strain, is a higher energy carbocation than It is easier to see this using perspective drawings: 2 2 p. 90E 2 2 p. 108E

44 4. a) This is an S N 2 reaction. In the first step, the leaving group is protonated by the solvent in oder to form a better leaving group. This resonance stabilized leaving group cleaerly now has partial positive charge on the oxygen bonded to the electrophilic carbon that undergoes backside attack by the nucleophile in the second step S Na S Na -

45 b) This is an S N 1 reaction. As in part a), the leaving group is protonated by the solvent to generate a better leaving group. Water is a weak nucleophile: the electrophilic carbon of the intermediate cation that has formed is not sufficiently positively charged to attract this nucleophile in a bond forming substitution reaction. So, alternatively, the leaving group leaves first, to make a carbocation. Now the fully positively charged carbocation is attractive enough to the weak nulceophile, and bond formation occurs. inal deprotonation of the resultant oxonium ion by the conjugate base of the solvent provides the observed product

46 5. 3 fastest slowest 6. a. l 1. Na I -, acetone 2. Na N 3-, DM N 3 b. N K N - acetone, 23 E

47 Problem Set hapter 8 Answer Key 1. a) (1E, 4Z)-1-chloro-1-iodo-4-methyl-1,4-hexadiene b) the three remaining isomers are I l l l I I (1E, 4E) (1Z, 4E) (1Z, 4Z) 2. a) I b) Yes, the double bond at 5 6. c) I I I (1E, 5E) (1Z, 5E) (1Z, 5Z) 3. a. This is because the E2 mechanism requires the and the be anti.

48 b. c. d. e. f I

49 4. a. The major product is the alkene. The minor product could be minimized by (1) lowering the temperature of the reaction or (2) using an RI or RTs instead of a Rl. b. The first one on the left. c. It s lowest energy (and therefore most populated, highest relative concentration) conformation is the reactive conformation with the on 2 and on 1 anti to each other and both axial. 5. a) (most stable) B > D > A > (least stable) 6. a. 3 (There are other conformations that could be drawn) b. 3 This is the reactive conformation, with the 3 and 4 anti. In this conformation, the 3 and 4 ethyl groups are anti, resulting in the observed trans relationship in the product alkene. Likewise, the 3 alkenyl substituent and the 4 methyl group are anti, resulting in the observed trans relationship in the product alkene. c.there s no on 2, so you can t make a = bond by elimination of between 2 and 3. d. Yes. 7. l anti-elimination requires that double bond formation occur between carbons 2 and 3 of the ring: the on carbon 1 is gauche to the chlorine on carbon 2.

50 8. a) S 2 2 -S 2 2 -S 2 A or 2 -S 2 B b & c) They are both tetrasubstituted alkenes, so should be of similar stability. A has no conformation without considerable eclipsing interactions about the double bond, however. So in fact B is more stable, and is the major product that forms.

51 9. RLS N E ac G rxn

52 10. a. trans-4-(2-chloroethyl)-3-vinyl-1-cyclohexene or trans-4-(2-chloroethyl)-3-ethenyl-1-cyclohexene or, if named as a single enantiomer, (3S,4S)-4-(2-chloroethyl)-3-vinyl-1-cyclohexene or (3S,4S)-4-(2-chloroethyl)-3-ethenyl-1-cyclohexene b. 11. Z N E

53 12. Z Z Z N S N

54 Problem Set hapter 9 Answer Key 1. a. b. NR l c. d. l (plus its enantiomer) l e. f. 2. a) is faster (the double bond is less sterically hindered) b) 3 3 c) from left to right: 2, 3, 1 d) from left to right: 2, 4, 1, 3 e)

55 3. 3 ( 2 ) 7 ( 2 ) 12 3 (Note that (E) or (Z) stereochemistry cannot be determined given the information of the problem) 4. S 2 -S 2 -S 2 2 S 4 5. a. S 2 -S S 2 2 S 4

56 b) No. c) No rearrangements are observed when alkenes are submitted to oxymercuration demercuration 6. a. b l l l l 2 3 highest lowest 7. a. S S S 2-2 S 4

57 b. Markovnikov addition. The more stable carbocationic intermediate leads to the observed product. 8. N N 2 Let = R- : N R N R N.. N N N N.. R N N N rotate around N-N bond N R N N N R N.. N N N N R 9. Error 1. l is a Lewis acid: the is * and the l is *-. The B bond is electrophilic or *- and can act as a Lewis base. *- doesn t attack *-! Error 2. Argh! An intermediate with a pentavalent carbon has been generated!

58 orrect mechanism: l : :l: : l :

59 10. (a) l l I 4 (b) 1. B T , - (c) l l l dark, 0 E l (d) 1. g( 2 3 ) 2, 2 2. NaB 4, - (e) l 2 Pt, acetic acid l

60 carene Y 12. a. NR b. peroxides 2 l 4 KI acetone I

61 13. N.. : : - N.. : : - N.. : : - starting material carbocation leading carbocation leading to M to L lower E: less repulsion higher E: more repulsion between and between and electropositive N electropositive N 14. a. A b. D c. B 15. a. : : b. δ : : δ... c. yes d. yes

62 Problem Set hapter 10 Answer Key 1. a) d) b) e) l l c) f) :

63 2. This question just points out that there are chemical tests with observable outcomes that let you distinguish between functional group classes in the laboratory: a) 1-Pentyne reacts with NaN 2 (N 3 evolution is observed); 1-pentene does not (N 3 does not evolve). b) 2-hexyne does not react with NaN 2 (N 3 does not evolve); isopropanol does (N 3 evolution is observed). 3. c) 2-Pentyne 2 mol 2 = a colorless tetrahaloalkane, but (Z)-2-pentene 2 mol 2 = a colorless dihaloalkane excess, red 2. l l- l l l l l l- or l l l- l l (postulation of an intermediate chloronium ion is K here) or l- l l l l l- l

64 4. a. Na N 3 (l) b Zn, 2 5. a. : l l- l b. * *- because, and not l must act as the electrophile in this Markovnikov-like addition. c. No (there are no carbocation intermediates)

65 6. a. 3 2 l b. from top to bottom: no, no, yes 7. a. b (S)-3-(2-butynyl)-1,1-dimethylcyclopentane

66 Problem Set hapter 11 Answer Key 1. a) starting material name: (S)-1,1,2-trimethyl-2-(1-methylethyl)cyclobutane or (S)-1-isopropyl-1,2,2-trimethylcyclobutane product: This product exists as a single enantiomer. b) starting material: l product: l This product forms as a mixture of diastereomers. c) starting material: (3R,4S)-3,4-dimethyl-1-cyclohexene product: This product forms as a mixture of diastereomers. 2. a) b) NR

67 3. a. It s this step:.. b. E = Σ(energies of disassociation of bonds broken) - Σ(energies of disassociation of bonds formed) so E = ( ) - ( ) = = 17.5 kcal/mol

68 4. initiation: N ) N.. propagation:. N. N N.... N one example of a termination: N.... N

69 5. a. l l l l l 6. initiation 2. propgation terminations......

70 7. a.. (mixture of enantiomers) (mixture of enantiomers) b. nly the ones indicated c. None of the tertiary bromide-containing constitutional isomers above form as the energy required to form a planar sp 2 -hybridized radical precursor for them is too great.. too much angle strain in molecule for radical to form

71 Problem Set hapter 12 Answer Key 1 a) one reasonable way: NaN 2 Na 2 Lindlar's catalyst light b) one reasonable way: 2 2 NaN 2 2 l 2 0 E 2 Lindlar's catalyst c) one reasonable way: NaN 2 Na NaN 2 Na Na N 3 (l) 2.

72 2. ne way: 2 Lindlar Pd ) NaN 2 Na 2 Lindlar Pd ) Na 1. B T , - 3. ne way is: light 2 xs NaN 2 Na 2 Lindlar Pd

73 4. a. ne way is NaN 2 Na 2 Lindlar Pd s 4 t- 4 9 racemic b. ne way is 2 Lindlar Pd NaN 2 Na NaN 2 Na 1. B T , -