Ch11 problem solutions - 202 ' Chapter 'II ALKENES;...

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Unformatted text preview: 202 ' Chapter 'II ALKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY of the most stable alkene (the thermodynamic product). Alcohol dehydrations are susceptible to rearrangement processes. A classic example is encountered in attempted syntheses of terminal alkenes such as l-buterie. The only 100% reliable method is base-promoted E2 elimination of a suitable 1-butyl compound (e.g., 1-bromo- butane, l—butyl tosylate). Any other method will give mixtures: K+_0C CH , CH COH CH3CH2CH2CH2BI ##L CH3CH2CH=CH2 The only elimination product K'*"0C(CH3)3, (CHslac‘OH ____.__—) CH3CH2CHB1‘CH3 CH3CH2CH=CH2 + CH3CH:CHCH3 Major Minor, cis and trans N +‘OCH CH ,CH CH OH CH3CH2CHBICH3 ‘1 2 3 3 2 CHgCHgCH=CH2 + CH3CH:CHCH3 Minor Major. cis and trans C .H S0 ,3. Either l—or2—butanol M» CH3CH2CH=CH2 + CH3CH=CHCH3 Minor Major, cis and trans Additional information pertaining to this process is presented in Chapter 12. Solutions to Problems C1 C1 26’ (a) W/ M I Br (11} HO \ OCH3 (c) H0 _ (d) C1 0:) 1 C 27. (a) cis- or Z—2—Pentene (b) 3-Ethyl—1—pentene (c) trans- or E—6—Chloro—5—hexen—2-ol (d) Z—1-Bromo-2—chloro—2—fluoro-l-iodoethene (Priorities are I > Br on C1 and C1 > F on C2.) (e) Z»2—Ethyl—5,5 ,5—trifluoro-4—methyl-2—penten— 1 ~01 (f ) 1,1-Dichloro— l-butene (g) Z—1,2-Dimethoxypropene (h) Z—Z,3—Dimethyl-3—heptene (i) 1-Ethyl—6—methylcyclohexene. This name is better than 2—ethyl-3-methylcyclohexene, because the first number is smaller. 28. (a) Hsat : 8 + 2 — 1 = 9‘, degrees of unsaturation : (9 - 7);? = 1 1': bond or ring present. The integrated intensities reveal the pieces. 8 = 1.8 (s, 3 H): CH3, attached to an unsaturated functional group 8 I 4.0 (s, 2 H): CH2, most likely attached to Cl 8 = 4.9 and 5.1 (singlets, each 1 H): Two alkene hydrogens Solutions to Problems I 207 33. Lower. because the vibrational frequency varies inversely with the square of the “reduced” mass involving the atoms about the bond. So, bonds involving heavier atoms have lower energies associated with vibrational excitation. Typically, fiche] 2"» 700 cm”, vcflgr % 600 cmil, and ficki 2:, 500 cm-1. 34. 10,000/17 = um. (a) 5.81 pm; (b) 6.06 um; (c) 3.03 urn; (d) 11.24 pm; (e) 9_09 Hm; (f) 442 um. 35. A—(b) (saturated alkarie) B—(d) (alcohol band at 3300 cm") C—~—(a) (alkene band at 1640 cm—1 in addition to alcohol band) D—(c) (alkene band at 1665 cm‘1 but no alcohol band) 36. (i) Both alkene (1660) and alcohol (3350) products have formed. (ii) Only alkene (1670) forms. (iii) Only alcohol (3350) forms. (a) Conclusions: Isomer C is probably a primary bromoalkane. which gives a primary alcohol product (8N2). Isomer B is probably a tertiary bromoalkane, which gives only alkene as product (E2). Isomer A is probably a secondary bromoalkane, which gives a mixture of 3N2 and 132 products_ (b) A possibilities: CH3CHBrCH2CH2CH3, CH3CHBrCH(CH3)2, or CH3CH2CHBrCH2CH3 B possibilities: (CH3)2CBrCH2CH3 (only tertiary isomer) C possibilities: CH3CH2CH2CH2CH2B1‘, (CH3)2CHCH2CH2Br, or CH3CH2CH(CH3)CHQBL but probably not (CH3)3CCH2BI (too hindered to give 8N2 reaction) 37. Process of elimination: we begin by noticing the absence of absorptions in certain regions of the spectrum, and conclude that the corresponding functional groups cannot be in the actual molecule_ 30, no 0—H (no strong broad band around 3300 cm”), no C:O (nothing around 1700 cm” or so), no CEC—H (bands near 2100 and 3300 cmml), and n0 CEC (1530 Gin—i)- All that are left as possibilities are the alkane and the ether. The spectrum shows strong bands between 1000 and 1200 cm“, strongly suggesting the presence of C—0 bonds. The fact that such absorptions do not appear in the sample spectra for alkanes in the chapter (see Figures 11-14 and 15) confirms that the correct answer is the ether, YOY . 38. Begin by noting the masses of the most prominent ions in each mass spectrum. Then try to predict how each of the alkanes might be most likely to fragment, using as a guiding principle a preference for forming more rather than less stable carbocations upon bond cleavage. The three compounds are constitu- tional isomers, all with the molecular formula C6H14 and a molecular weight of 86. Spectrum A shows a base peak at m/z : 57 (C4H9) and other significant ions with m/z = 56, 41, 29 (CZHS) and 27. The molecular ion at m/z = 86 is weak. Spectrum B also shows a base peak at m/z = 57 (C4H9). The peak at m/z = 43 (C3H7) is bigger than in spectrum A, while that at m/z = 29 (C2H5) is smaller. The molecular ion is more intense Spectrum C shows a base peak at m/z : 43 (C3l-l7). The molecular ion is weak, but m/z = 71 (C5H11) is prominent in this spectrum. Now consider the three structures and the bonds in the molecular ion most likely to fragment in each: CH3CH2+ (m/z z 29) Hexane: [CH3CH2—CH2—CH2_CH2CH3]+' -e—) CH3CHZCHZ+ (111/2 2 T T T CH3CH2CH2CH2+ (m/z = 57) Cleavage is most favorable between CH2 groups (avoiding formation of methyl cation), but since the best cations that you can get are only primary—"and not very stable—fragmentation as a whole will be less likely. Spectrum B seems the best match because of the prominent molecular ion. 208 ' Chapter II ALKENES,‘ INFRARED SPECTROSCOPY AND MASS SPECTROMETRY CH3 _ . + u 3 3] (m/z u 2 Methylpentane. —) ( l 71) T Cleavage will occur mainly about the CH to give secondary cations. The best match is spectrum C. CH3 l 3-Methylpentane: [CHgCHz—CH—mCHZCH3P“ -—-> T T Fragmentation occurs at the indicated bond to form mainly sec-butyl cation and, to a lesser extent, ethyl cation (m/z = 29). Spectrum A fits best. As is typical in mass spectrometry, the large am0unt of energy imparted to molecules in the process induces rearrangements and other modes of fragmentation as well. Fortunately, ions arising from such processes do not usually dominate the spectrum. [CH3CI-12CHCH3]+ (m/z = 57) 39. Major peaks: m/z 43 (CH3CH2CH2)Jr from M—Br m/z 41 (CHZCH=CH2)+ from M—HBr—H Minor peaks: m/z 109 ((3H2CH2813r)+ _ m/z 107 (CHzcnflar)+ from M CH3 m/z 42 (CH3CH:CH2)+ from Mw-HBr m/z 29 (CH3CHZ)Jr from M—Br——CH2 m/z 28 (CH2=CH2)+ from M—Br—CH3 m/z 27 (CH2=CH)+ from MwBr_CH3—H 40. Compound is saturated (see Section 11-6). Try to use the general guidelines that intense fragment peaks either result from the loss of relatively stable neutral species or are due to relatively stable cations. So, looking at the high intensity m/z 73 peak for isomer C, it corresponds to (M — 15)+, or loss of CH3. This is most likely if the remaining fragment is a very stable cation, for example, CH3 + - ~+~ + CH3CH2 —C|:'—* —* ‘l‘ CH3 ‘ CH3 CH3 m/z 73 cation, stabilized by oxygen lone pair. Looking at the rest of the spectmm, the base peak is at m/z 59, or (M — 29)+, loss of CH3CH2. (EH3 + v + CH3CH2+$—OH —> (CH3)2COH + CH3CH2 ' CH3 m/z 59 This is, all together, good evidence for isomer C being 2-methyl-2—butanol. as shown. Isomer B also has a peak at ml: 73 for loss of CH3. Its base peak (m/z 45) corresponds to loss of 43, 01‘ CH3CH2CH2. These signals are what you might expect for 2—pentanol. sienna Solutions to Problems ' 209 OH l CH3CH2CH29H + CH3' OH + - / m/Z 73 b I (I CH3CH2CH2 JrCH %CH3 \ C|)H CH3§H + CH3CH2CH2' m/z 45 Both fragmentations give cations stabilized by resonance from an oxygen lone pair. This is; in fact, the correct answer. Isomer A does not lose CH3 or CH3CH2 (no peaks at m/z 73 or 59). That pretty much rules out any tertiary or secondary alcohol structure as a possibility. (Any example that you can write should show those fragmentations.) How about possible primary alcohol structures? Look again to intense fragment peaks for clues. The m/z 70, loss of water, doesn’t really help much, except to rule out (CH3)3CCH20H; which has no B—hydrogens and therefore cannot dehydrate. That leaves three possibilities for A: (EH3 1H3 CH3CH2CH2CH2CH20H CH3CH2CHCH20H CH3CHCH2CHZOH The data that you have are in fact quite consistent with either of the first two (the third is difficult to maneuver into a fragment with m/z 42). If you got this far, you are doing well! (The actual spectrum of isomer A is that of l-pentanol, by the way.) 41- (a) C7Hr4 Hsat : 2(7) + 2 = 16; degrees of unsaturation = (16 — 14)/2 : l (b) C3H5C1 Hsat : 2(3) + 2 — 1 (for the C1) = 7; degrees of unsaturation = (7 — 5)/2 = 1 (C) C7H12 Hsat : 2(7) + 2 = 16; degrees of unsaturatiorl = (16 — 12»? = 2 (d) CsHe Hsalt 2 2(5) + 2 = 12; degrees of unsaturation = (12 — 6)/2 = 3 (e) CéHllN Hsat : 2(6) + 2 + 1 (for the N) 2 15; degrees of unsaturation = (15 W ll)/2 = 2 (f) CSHSO Hsat : 2(5) + 2 = 12; degrees of unsaturation = (12 — 8)/2 = 2 (g) C5H100 Hsat : 2(5) + 2 = 12; degrees of unsaturation = (12 — 10V2 = 1 (h) C10H14 Hsat = 2(10) + 2 = 22; degrees of unsaturation = (22 — 14V? = 4 42. (a) I-Isat = 2(7) + 2 = 16; degrees of unsaturation = ( 16 — 12)/2 = 2 (b) Hsat = 2(8) + 2 + 1 (for the N) = 19; degrees of unsaturation = (19 — 7)/2 : 6 (c) Hsat = 2(6) + 2 — 6 (for the Cl’s) = 8; degrees of unsaturation = (8 — 0)/2 = 4 (d) Hsalt = 2(10) + 2 = 22; degrees of unsaturation = (22 — 22)/2 2 0 (e) Hfiat = 2(6) + 2 = 14; degrees of unsaturation : (l4 — 10)/2 = 2 (f) Hsat : 2(18) + 2 = 38; degrees of unsaturation = (38 — 28)/2 : 5 210 0 Chapter 'I 'I ALKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY 43. Begin by using the molecular mass to determine the molecular formula. Given that the unknown is a hydrocarbon, first determine which combinations of carbon and hydrogen can give an approximate mass of 96. Although 96 is the mass of 8 carbon atoms, that option is not possible (no mass left for hydrogen atoms). Reduce the number of carbon atoms and add the necessary number of hydrogens to arrive at a feasible formula. Thus, 7 carbons + 12 hydrogens also gives a mass of 96, and by using the data in Table 11—5, we quickly find a good match for the exact mass measurement: 7( 12.000) + 120.0078) 2 96.0936. Thus we are working with a formula of C7H12: two degrees of unsaturation. The unknown compound contains a total of two rings or 'rr bonds. The IR is useful: The peaks marked indicate at least one of the degrees of unsatu- ration is from a C=C bond. Also, the sharp band at 888 cm" is diagnostic for a R2C=CH2 group. Can you decide whether the unknown has two double bonds, or one double bond and one ring? Hydrogenation gives you C7Hi4: One degree of unsaturation still remains, suggesting one ring in the original compound. The 1H NMR supports this conclusion: The integration of the alkene H signal (at 8 2 4.8) is 2 H. That limits the structure of one R2C=CH2 group. What else can you learn from the 1H NMR? The alkene signal is a quintet (five lines). On the basis of the N + 1 rule, you can try to make a structure where four equivalent neighbor H’s split the alkene H’s: — CH2 \ C = CH2 7 CH2 / This structure fits the signal pattern, and the J of 3 Hz is right for allylic coupling, too (Table 11—2). The two CH2 groups fit the 1H NMR signal for 4 H at 8 = 2.2 as well. This takes care of four carbons and six hydrogens, leaving C3H6 unaccounted for. Try the simplest way to make a ring: add three CH2 groups to get C>=CH2 Does this structure fit the rest of the spectrum? The structure contains two more equivalent CH2 groups, and the unique CH2 at the opposite side of the ring from the double bond, fitting the 1H NMR, and the 13C NMR reflects the symmetry with five peaks. This is indeed the answer. 44. A saturated 60—carbon alkane has the formula C60H122. Therefore, “bucky~ball” possesses 122/2 = 61 degrees of unsaturation. In its hydrogenation product, C60H35, there are (122 -— 36)/2 = 43 degrees of unsaturation. Therefore, there are at least 61 — 43 = 18 n bonds in C60. (As you will see later, there are actually 30 1T bonds, but not all undergo hydrogenation.) 45. The two orders are opposite each other because very stable alkenes are low in energy to start with and give off less heat upon hydrogenation than do less stable alkenes. The stability orders are given below. (a) CH2=CH2 < (CH3)2C=CH2 < (CH3)2C=C(CH3)2 Increasing substitution b H H H H ( ) \ / \ / C=C < C=C / \ / \ (CH3)2CH CH(CH3)2 CH3 CH(CH3)2 (Two large groups cis) (A small and a large group cis) H CH CH < \C_C/ ( 3)2 / _ \ CH3 H (Trans) . Tu- '='IJ 0‘ P Solutions to Problems 0 2'! I (c) £13 < (p < 03 Similar to (a) CH2 CH3 CH3 (d) Cmfi < CHké < 313% Similar to (a) and (c) (e) A < II} < O Ring strain ‘ 46. Generally the alkenes are written left—to-right in decreasing order of stability. <a>>=/>>—/> >—/ (trisubstituted) (disubstitnted) (monosubstituted) (b) >=<>>fi< (tetrasubstituted) (disubstituted) (c) 66 (only possibility) (d) > z (trisubstituted) (disubstituted) H H CH \ / H\ / 3 47. CH3CH2CH= CH2 /C 2 C\ /C = C\ CH3 CH3 CH3 H Smallest amount Major product 48. A haloalkane of general structure R—CHg—CHX—R’ will have two conformations with H anti to X; one gives the cis and the other gives the trans alkene as product. For example, consider 2-bromobutane: H H CH3 H CH3 H CH3 H H CH3 Br Br l l H CH H\ / H\ / 3 C: C C: C / / \ CH3 CH3 CH3 H 212 - Chapter 11 ALKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY A haloalkane of general structure RR’CH—CHX—R” will have only one conformation with H anti to X. Therefore, only a single alkene stereoisomer can form. Its stereochernistry will be determined by the stereochemistry of the two chiral carbons in the haloalkane. 49. The more hindered (CH3)3C0_ K+ will favor removal of protons toward the less crowded “ends” of molecules, thus giving less stable products. Ethoxide eliminations will favor more stable products. Product with NaOCHZCHg in CH3€H20H Structure of starting material Product with (CH3)3COK in (CH3)3C0H (a) CH3OCH2CH3 (b) CH3CH2CH2CH2CH20CH2CH3 + lupentene (c) trans- and cis-Z—pentenes CH3 (d) 6 H CH3 (r) CH3 H O / \ CH3CH2 CH2CH2CH3 (E) (s) CH3 H / \ CH3CH2CH2 CHZCHB (Z) (h) H CH3 \C/ II C / \ CH3CH2 CH2CH2CH3 (Z) CH3C1 CH3CH2CH2CH2CHQBI' CH3CH2CH2CHBICH3 C1 CH3CH2 CH2CH2CH3 CH3 i H H CH3 (:1 H = H CH2CH3 CH2CH2CH3 c1 CH3CH2CH2 CHZCHa CH3 : H H CH3 C] H _ CH3CH2 H CHZCH2CH3 Cl CH3th2 CHZCHZCHg H CH3 H CH3 H C1 _ H CH3CH3 CH2CH2CH3 CH3OC(CH3)3 l-Pentene 1 -Pentene CH2 CH2CH2CH3 (R) CH= CH2 CH3CH2+H CH2CH2CH3 (S) CH2CH2CH3 (R) Solutions to Problems ' 213 50. The energy of the E2 transition state for 1-bromopropane is similar to that for 2-bromopropane, because the products are the same (propane). So the E2 rate constants for these two compounds will be similar. However, l-bromopropane, being primary, will have a faster competing 8N2 reaction than will the secondary 2-bromopropane. Overall, it is reasonable to estimate that the actual E2 rate for l-bromopropane will be intermediate between the rates for bromoethane and 2—bromopropane. 51 . H H H H CH3 H H CH3 CH3 H H CH3 CH3 CHZCH3 CH3CH2 CH3 CH3CH2 CH3 CH3 CH2CH3 Br Br Br “ Br 2R,3R 28,35 213,35 28,3R These are probably not the most stable conformations because all contain three gauche interactions between alkyl groups or bromine. For the RR isomer, a better conformation would probably be H BI‘ CH3 CH3 CH2CH3 The others would be similar. 52. Referring to problem numbers in Chapter 7: (22b) CH3CH=C(CH3)2 > CH2=C(CH3)CH2CH3 (22c) Comparable amounts These are the only elimination product mixtures in these problems. 53. Major products are labeled for (a) through (f). Generally, more highly substituted alkene isomers form in highest yields. CH3 C(CH3)3 _ CH3 CH (g) 2 > J6 ~ QB > bf z J5 a 6 (h) was 214 ' Chapter I 'l ALKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY M__, ——> / 54. (a) L \ (b) J H H wiQCHZCHs (cangcézd ._ /\ H+ .+. —> —-—> (0 1H (carbocation rearrangement) + U H (most stable alkene) (again!) ' (ternary carbocation—finally I) 55. A would be expected to give the tetrasubstituted alkene BIC. CH3 \\ H0” B gives the (trisubstituted) alkene shown in the problem in the text. C is a special situation. 56. l-Methylcyclohexene contains a trisubstituted double bond and is therefore more stable than methyl-enecyclohexane, whose double bond is only disubstituted. The stability order in the three-membered rings is affected by angle strain because the geometry compresses bond angles to about 60°: This compression is worse for SP2 carbons (which prefer 120° angles) than for sp3 carbons (109°). l-Methylcyclopropene, with two 5p2 carbons, will be much more strained than methylenecyclopropane, which has only one Sp2 carbon in the ring. 57. Rewrite each compound in a Newman projection with the Br and the [3-H in an anti conformation. (a) BI‘\ H C6H5 H ‘ C6H5 B? H H C6H5 C6H5 I, E E H CH3 H CH3 CH3 C6H5 C6H5 BI‘ CH C H 3\ / 6 5 E2 C —"*’ H /C\ H C6H5 (Z) Solutions to Problems O 215 (b) Br H C6H5 H “ C6H5 131‘ H H C6H5 C6H5 I, E E H3C H CH3 H CeHs CH3 C6H5 Br C H CH 6 5\ / 3 E2 C “we ll /C\ H C6H5 ‘ (E) Elimination of (b) should be faster than that of (a) because the necessary conformation for reaction in (b) has the large C6H5 groups anti to one another, whereas the reactive conformation for (a) requires the C6H5 groups to be gauche. The latter requires additional energy input, raising the E,1 and 10Wering the rate. 58. When E2 eliminations occur in cyclohexanes, the leaving group and the [3—H to be removed must have a 1,2-trans diaxial relationship. So, first, draw the two possible chair conformations for each starting compound, and analyze the one in which the C1 leaving group is axial. (CH3)2CH or 4; Axial CH —\ (osmium 3 H H Equatorial CH3 This is the only H anti to Cl. It is therefore the only one that can be removed in an E2 process. NaOC2H5 —“~+ (CH3)2CH-— -- CH3 C2H5OH Only possible E2 product (CH3)2CH C145— Axial —.<: CH3 Cl (CH3)2CH H t. , CH3 Equatorial I T4 Now there are two H‘s anti to Cl and available for E2 elimination. (CH3)2CH Q CH3 and (CH3)2CH-—<;>~--CH3 \ NaOCgl-ls ( * C2H50H ...
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Ch11 problem solutions - 202 ' Chapter 'II ALKENES;...

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