HW solutions Ch 4-6

HW solutions Ch 4-6 - CHAPTER 4 4.1 Expected peaks = 2N1+l...

Info iconThis preview shows pages 1–23. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 16
Background image of page 17

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 18
Background image of page 19

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 20
Background image of page 21

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 22
Background image of page 23
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER 4 4.1 Expected peaks = 2N1+l for 13C but some species eg 37Cl 160 do not impart J—ooupling. (a) 13c of CH2C12 1: 1/2 ofH 3 (b) 13c ofCD2C12 I = 1 ofD 5 (c) 13c of CFZCIZ 1: 1/2 oflgF 3 (d) 13c of CHDClZ dt JHC > JCD (e) 13c of CF3C02H CF3 = 4; C02H = 4 (2JFC = 4) 4.2 Max # of 1H NMR lines for ~CH2- groups Ha Hb (a) H3C>\ f3+7Hb :na:ti°t:Pi° H ll dq J: an J: H (b) Br*CH2P(F)C1 enantiotopic CH2 with 2JPCH> 3JFPCH H H Cl Hx (C) M Ha + Hb = diastereotopic ABX or 8 lines Ha Hb on3 Me ((1) CH3 Ha + Hb = diastereotopic .'. i] if Hb P11 Ha 4-3 (a) Instructors note: error in text, revise problemto be DMSO. 0 l l s . . . . . CH3/..\ CH3 two smglets Mes diastereotopic 1n chiral solvent (b) Instructors note: errorin Structure b additional C:C Mes diastereotopic H H (C) JH.Me= 7 JILLNHz = 7 ifNH not exchanging sextet H H . (a) 6‘] =0 mews .. H H HaxBr Br Heq :0 (e) H 3Jeez 3Jea ‘ ' J l Hax (t) EachHisdd H ll H H (g) Instructor note: errorin structure, it should be H>:%O Non—equivalent Mes two singlets H3C 4.4 (a) Figure 4.8 shows pent for —CHD2 via 2NI+ 1, N = 2, I: 1. So, 5 lines expected. J (b) 27—“ Jm. = <2.2><%>=14-3HZ JDH 7D 4.5 Using 1J data (page 136). Ph Br Br ph A (a) 129 A H 15? D: 150 4003 O H N H (b) V M m N200 180 (c) H g 160 130 0 (d) :J h: <0H 0 o E ng : CSC/ 123 ‘ ~50 4.6 (a) 2 HZ (b) 5 Hz (0)4 Hz (d) 12—13 Hz (e) 5 Hz (0 4 Hz (g) 50 Hz (h) 27 Hz 4.7 OH cm WEI JaezJee=3—4 01'16W “2:8 H trans >[\WKOH JeeIJan3-4 : 20+8=28 Jaa =10-12 “ll/2:14 4.8 Note C6 ring Jawax >> Jax_eq w J eq—eq Comments: H1 must be eq H2 Jax.ax = 8 H3 two Jax,ax = 10,8 COZCH3 H4 axial to 3Hs and eq J to one other H . O NOCH] H6 ax1al Jto H73 0 Note stereo assigrmient in literature is incorrect! lit = H OH 4.9 Examine JCH and 8C model compound data for the following assignments which do not agree with the literature. General X5 = C14Br C8 = 119.5 (d), J: 190 C5 = 71.3 (5) C10 2 38.8, J: 163 Next C2 and C4: 8 59 and 8 64 J: 140 Hz C3 and C6 : 5 38.3 and 5 38.8 J= 134 Hz Halogen regiochem Br 38.8 Br 37.1 X e vs. Cl_ 7 9 CI 49.5 All others must be Cl Br 38.8 38.3 119—50 Cl / Cl Cl 59.0 38-8 64.1 Finally, the assignments in the literature given is incorrect. 4.10 Structure B: The key here are the assignments of the isolated Vinyl group proton resonances. Ha “XV” \ Hc 5a=5.84 Jac=17 8b=5.0 :’ Jab:11 50 5.18 chzl-Z 4.11 5.0 Hz/mm C i 1.5 Hz/mm \ f .r“ f A 5.8 W 1.80 min-t, A 3 4 r r—fi *1 ppm Mel MCH/MCHI 2.7 Hz/mm H8 H7 H2 3.25 '320 '3.15 3.10 3.05 3.00 H18 5.8 dd; 3J1,2 = 5.0 H2 8 3.24 pent but 3&1 = 5.0, 35,8 = 3.0, 3ng3 = 2.5 H8 5 3.14 dd Um = 8.1, 3m = 3.0 H7 5 3.06 dd J: 8.1, 3.0 H6 8 3.0 pent J: 3.0 H4/5 5 1.8/ 1.18 muIt Simulated H9 (1 sept 1.3 .111 = 5.0 H3 mull J18 : 3.0 .1213 =25 4.12 Only D : DD, DD, DT, DT 0 DD DT DT Br DD D 4.13 Spin type is AA’XX’. T we sets of chem equivalent nuclei CR3 HA HA. JAX :6 JAXI etc. Note: A v/J= 200/7 z 29 BI’ / . - t (a) W .. maybe non-equiv So expect 1S order but not Ph CN observed HX HX' Br (b) No Js by inspection see 4.1G and 4.1H 4.14 Expect two separate ABX patterns for diastereotopic sets ab and cd with the latter at highest field: va = 3.92 vb = 4.14 vx 4.34 vc = 2.63 vd = 3.07 4.15 Not easily cyclopropyl Jcis : Jtrans ; need nOe data 4.16 (a) Ha Jz 17, 5 6.4 trans (b) T, D, BS, DD implies trans Ha calc: 5 5.27 +1.00 + 0.39 = 6.66 H cis Ha calc15 5.27 +1.10 + 0.06 = 6.33 (Table 3.8) H T——> H H OH 4.17 K and L fit 00 13C resonance at 8 161, but only K has quart, A = 1 at 6 3.6. 4.18 O 14N-H not J coupled because of rapid quadrupole relax CHAN/CH3 and 3JNH_CH3 = 7 is visible. Net result is a non— }:i exchanging NH. However, it is broadened because of quadrupole relaxation effect to 14N. Note NH similar to middle boxed example Fig. 2.14 4.19 13C :> C7H1202 (See Problem 10G and the worked answer). 0.“ O 4.20 C14H18N207 O OCH; H l N N COH HO/% \ \/ 2 H CH3 HO OH 4.21 / I \ Br N Br Br Br / A2B pattern U ‘ 3 With large JHH Note: \N does NOT fit 4.22 AA'BB' pattern Br 0 Br Br 4-23 4.24 verbenone CHAPTER 5 5.1 5.2 5.3 5.4 5.5 5.6 5.7 The pulse sequence is an inversion recovery sequence: (180°)x - t1 - (90°)x - acquire t, is the variable evolution period which produces varying peak signs and intensities. The pulse sequence is a spin echo sequence: (90°)x - r - (180°)X - T — acquire The FID collected at point D would give a negative absorption peak. APT, c, CH2 t , CH, CH3 T 8c (ppm! Solution 1 Solution 2 152 C C 1 06 CH2 CH; 52 CH CH; 41 C C 41 CH CH 27 CH2 CH; 26 CH3 CH3 24 CH2 CH2 22 CH3 CH3 Total C9151l4 c91116 CéHm from 13C and ‘H NMR. Peak at 581 JCH = 250 Hz will show anomalous behaviour in APT but will be OK in DEPT-135. Suspect CEC and C-0 from 13C chemical shifts. Proposed structure: HO : 73 3.26 H : 29 35 1.2 82 2.35 Given that the coupling constant for H—C=C-CH3 is approximately equal to that for CHg'CZC' CH3 then using the N + 1 rule we get: 16 2.06 dq Me COOH H COOH H liq/Ice Me Me . 14 11 1.92 (Slumt 1.76 dq 1.78 quint The AB system is at 5;; 7.59 ppm, giving C1 5c 126 6H 7.57 and C2 5C 124 5;; 7.60. Confirmation that C2 is at 8C 124 ppm is given by the 1JCH = 184.4 Hz. Th6 W is: + ‘i‘ (CH2)3 + (CH3); : C9H14 All structures have an APT MF of CgHM, but structures B and D have a plane of symmetry and will only show 5 peaks in the 13C NMR spectrum, and can therefore be ruled out. J = 10 Hz H gem Jvic : O see Fig 5.11 .\ H The 1H peak at 5 1.47 d J: 9.6 Hz suggests: So, we can rule out G and H as these contain cyclopropyl groups. This leaves: A, C, E, F. See Figure 4.1 la and Figure 5.11 for a similar case. 5.8 Formula is C9H14O. Data from NMR spectra: Atom 5 13c (ppm) 5 1H (2311;) mult lH-BC COSY (LR) 1H-‘H COSY A C=O 213 b, e, f, g B CH 57 2.48 m Mel, M62 e/e’, g C C 41 Me], M62, f D CH 40 2.10 m Me1,Me2 e or f, E CH; 32 2.40 m, 2.25 dd b Mel CH3 25 1.20 5 M62 Mez F CH2 24 1.50 d, 2.40 m Mez CH3 22 0.70 5 Me, G CH; 21 1.90m b,eorf, dor e, M62? Using the COSY and long range data the possible substructures are : Mer'8 k? C=O T T A e 9 . From 5.7 we know that flf must be in 4 membered ring. Therefore, possible structures with the above substructures are: Me Me B:C:D Me\C,Me Me\ ,Me l‘F’l 1 B: :D 0:0:3 G\A/E é F A 2 (I3 F A 5 u ‘E’ \‘o ‘E’ *‘o O Me/‘CZMB Me\ ,Me me‘c’Me lI3‘F’li3 3 B/C‘D D: :B E\ /G PF’i 4 é F A 6 g E‘G’A°o \G’ °0 1 and 3 can be ruled out because they possess a plane of symmetry (see Problem 5.7).Compounds 2, 4, 5, 6 all correspond to structure A, but only 5 conforms to all the 2D NMR data (2 and 4 are ruled out by A —> b) and the 8C ofE is correct next to the C=O, ruling out 6. Hence 5 provides the best fit. 5.9 The instructor should copy the spectrum on the top of page 214 and note the following errors in assignment: revise 46 to 7e, revise 4a to 4e and revise 5a to 621 and renumber the ring as shown below. There are only four, not five, important long range correlations (based on revised numbering and assignments). These are for the following long range correlations: going along the diagonal contour Irom lefit to right: H363 to Hle; Hla to Mel 1; H63. to Me12; H821 to Mel 1 CH3 1 4 H3C CH3 5.10 H19 is at 5.9 ppm (as H1 1/1 1 ' is the two singlets at 4.7 ppm). Using the 1H—“c COSY(J=140 Hz) C19 is at 102 ppm. Similarly, HIS/C15 are at 2.6 ppm (ABq)/34 ppm. H15/15’ are correlated to 3 quaternary carbons in the ‘H-“C cosy (J: 9 Hz) at 182, 154 and 118 ppm. in the same spectrum, H19 is correlated to 182, 162 and 154 ppm and the MeO at 3.9 ppm is correlated to 162 ppm, giving the following situation: 5.11 Atom no 6 13C 5 1H C—1 32 1.3 bin, 1H C-2 45 0.9 m, 1H, 1.9 bd, 1H C—3 71 3.4 dt, 1H C-4 50 1.1 m, 1H C—5 23 0.9m, 1H, 1.55 dt, 111 C-6 35 0.85 In, 1H, 1.6 (it, 1H C-7 22 0.9 d, 3H 08 25 2.2 d sept, 1H C-9 16 0.8 d, 3H C-10 21 0.9 d, 3H 5.12 Look for the peaks in the NOESY that are not present in the COSY, these are the true NOESY peaks. One example is the correlation between the Me at 0.7 ppm and the CH at 2.5 ppm. 5.13 (a) 180 ppm C-2 179 ppm C—6 D-6 8.5 ppm 135 ppm C—4 D—4 7.5 ppm 122 ppm C—3, C-5 D—3, D-5 7.1 ppm (b) There are 2 non~equivalent carbons, so the signal is composed of two singlets and is not a doublet. (c) The 13C NMR was acquired with ‘H and 2H decoupling to remove multiplicity in the 13C spectrum. The COSY was acquired under 1H decoupling to remove the correlations between 1H and 13C, and to let the 2H-“C correlations appear. If 2H decoupling had been used during acquisition of the COSY the OD couplings which we wish to observe would have been suppressed. ‘ MF = + (CI—12);; + (CI—13):; = Cu)ng 5c 72 is a CHOH moiety, thus indicating the formula is CIOHZOO. Unsaturation number = 1, no Sp2 carbons, therefore 1 ring. You might notice a similarity between the 13C spectrum and that in Problem 5.1 1! 5.15 It’s linalool, see Unknown B (structure 10.2C) in Chapter 10 for a worked solution. / \ OH 5.16 (a) Possible structures: 0 OMe 0 OH #3 Me I \ Me I \ O N l OMe HO 1}: o MeO rlq o \ Me Me Me OMe O A B c (b) Given this information the correct answer is A: 55 3.5 0 ob 5.17 (a) Tabulated data for 1. 5.18 5.19 9 10l\\8 H 4 1N o 5 2 NH 13C Peak 5 13C (Rpm) Atom A 165 N-C=O 06 B 150 sp2 C 07 c 148 J: 180 Hz sz CH C—11 D 137 J= 165 Hz sp2 CH 09 E 135J=206Hz q¥CH C6 F 134 SP2 c 01 Gr 126J=155Hz yfiCH C40 H 122J=155Hz ¥¥CH C8 1 116 J=19OHz sp2 CH C—2 I w sp3CHz cs K 27 SP3 CH2 c-4 The ligand forms a complex with Co“: \ I N/ O Cog“? ‘ "N N </ I N H O O O + l O O O OH HO OH O\/|\/O HO 3 OH a+b=5 CHAPTER 6 6.1 Methylcyclohexane, C7H14 MW = 98 (a) NI" 2 m/z 98 (b) Base peak : m/z 83 (c) M" - C3H7 = m/z 98 — 43 = 55 which is a large peak. 62 +0 odd electron m/z 98 m/z = 83 odd electron even electron N0 neutral species are formed in this fragmentation. 6.3. [C6H14]+' A small ion peak is observed at m/z 86 [C5H11]' Not observed as it is uncharged [C3H7]' anions cannot be observed in positive ion MS [C4Hg]+ = 56 is an impossible fragment ion from C6H14, as it would require the loss of CzHe. From the M‘ [CH3]+ = 15 is observed. 6.4 I»? = 128 Hydrocarbons: CgHzo or Clng Oxygenath compounds CsHiéo [0]; C7H1202 [2]; CeHsos [3]; C5H404 [4]; C9H40 [8] “Qt possible 6.5 M)“ = 79 therefore, it must contain nitrogen: C5H5N (4) or C4HNO (5). Other formulas are not reasonable because they will be too unsaturated. 6.6 Both are MW = 108. Use the ratio ofheights M+ : M+ + 2 C5H13C1M*:M++2=100:30 CszBr M:M+2=100:98 6.7 Calculate mass to mmu or 0.0001 amu (Use Table 6.1) C6H120 = 100.0888 csnsoZ = 100.0524 68 Highest m/z 138. Possible formulas: ClOHISa C9H14O, CsHiooz, CsHmNz, C6H10N4, C7H10N20 From 13C NMR (CH3); + (CH2): + (CH)4 = CIOHIQ a CHOH moiety suggested by (SC 72 Assume ClongOH = 156 Therefore m/z 138 is M+ - 18 (dehydration product) “ 6.9 6.10 6.11 e. r11“ ‘(9 W \I/ —_> + CGH.’ f +' +‘ / + : Retro Dlels-Alder \ Scheme 6.1 e +- O —CO , +' RJL R, [ R-R ] even amu _ ‘ -' even amu odd electron Q odd electron ; odd amu R “63 even electron ~NHR2 + ' even amu . [alkene] SP /' odd electron even amu odd electron even electron R/ R" \ dd - ' \ o amu R [IR,N Ruf 6.12 I3C (CH3)1 + (CH2)3 + (CH)1 2 CsHlo as it is a hydrocarbon M+‘ (7) = 98 therefore C7H14 hence symmetiy with two equivalent sets of CHz’s therefore 13C = (CH3)! + (CH2)5 + (CH)! = C7Hl4« Unsaturation number = 1, therefore 1 Ring. 43H; Top / m/z 83 : \ m/z55 'C3H7 Bottom £2145 6.13 0:0 "h —00 [ D 6k [TI/256 or [CsHsO]+ OTIC4H91+ 'CHO m/z 55 This must arise by a complex fragmentation involving cleavage of an odd number of bonds. 6.14 The fragment at m/z 67 is due to W - 15. We might assume that this is NF - Me, but this fragmentation is not possible by a simple one bond cleavage process, but must involve a three bond cleavage. A retro Diels-Alder (Scheme 6.1 e) accounts for the fragment at m/z 54 (M * C2PL). 6.15 CI CH2+ W2 9 Should be ‘clean’ M/M + 2 pattern at m/z 126/128 and also peak at m/z 91. Left -H- h» c. CH; Get two superimposed clusters at [W + 2]+' at m/z 126/128 and [(M/M+2) — H]+ at m/z 125/127. a cleavage -CH3 Ark Note that intensities are 73 >> 101 > 115, which is in ace C4H9' >> CH3~ owing to the fact that the largest radical is lost preferentially if all are primary. +. O ——> OH REC“ -c2H4 Rr 0 O +- 6.17 m/2128 m/z 100 6.18 OH O O a.) McLafferty with Ha (-C3H6) \ WM, 0 Ha m/z= 100 for di 160 MW= 142 for di 160 m/z= 104 for di 180 MW=144fordi180 OH \ b.) McLafferty with Hb (— CH2=C=O) m/z= 100 for di ‘6 m/z = 102 for dl “3 Only m/z 102 observed, therefore pathway b.) is operating. 6.19 (a) Peaks at m/z 154, 139, 122 occur via the following fragmentations: +0 O: \> OH m/z154 m/2112 -X C +0 0 O . 0+0 m/z139 m/z111 This is consistent with the following: (a) _ D bO/v M+'-Me at m/z142 D — O a i D \ ' D D m/z157 OH D m/z115 (13) CD3 8/, [Tl/2115 22/, m/z112/113(OH OI'OD) b L b 0 m/z 142 0 171/2 139/142 C C \ CD3 \ m/z 157 "1/1139 W215;- m/z 142 6.20 C4H5N should Show M' = 67, but the apparent m/z 53 would have to be due to M‘ — 14 which is impossible. Therefore we suspect M' = 70 and that m/z 69 is [M - HT 6.21 (a) NH3 (b) H20 (0) HCN C2H4 (e) C2H4 0r N2 (2%) C02 6.22 (a) 43 = C3H7 [0]; 57 = C4H9 [0]; 71 = €an [0] (b) 39 = C3H3 [2]; 67 = C5H7 [2] (C) 55 = C4H7 [1]; 69 = €ng [1]; 99 = €7H15 [1] 6.23 Measure the ratio of 99 : 96 to follow the reaction. Other peaks to look at in the product include mowatmk96—15(merbhm6J4)mflbrm&96—28UmwmeDkkdUdmywermkm 6.90. H\ IO® T . OH -CH3 m/z 99 \ +. -H 0 m/z 114 2 W2 95 6.24 (a) MW = 104 C —C H 3 7 W2 61 -CH§ *—————"——’ Iné89 m/z 104 m/z 104 (b) MW = 138 m/Z123 . .CSH9 o (0) MW = 260 Note to instructors: correct error in problem text CBHMCI4 to CBHacl4O —Cl CBHBCI4O n CgHscrao m/2225, 227, 229, 231 CeHeBrz 'Br CSH6Br m/z181,183 (d) Mw=101 Nx -NMe° -CH'/ -C1-f /\ff 2 W257 /\K\ 3 or 2 5 0 W2 86/mz 72 MeO’ 6.25. Two major ions at m/z 92 and 91. CH3 CH2 M W 'mp. 97 + T ’> ' CH2 92 6.26 Look for m/z = 69 & 43 a b H30 CH2 H30 J 9CH2 a = CZHaO = 43 o H o / b = 0ng = 69 +gc H C 3 CH3 CH3 6.27 Different fragmentation routes as follows: n-propylbenzene: CH3 CH+ V : / 2 ©/\/ . —C2H5 (29) m/Z= 91 o-nitropropyl benzene: CH3 C H 3 C H 3 0 0+ +/ M/ \H N? o 4- $11 CH3 m/z = 148 + N\ \O 6.28 Important note error in problem 4th line should read fragmentation ion patterns: 6-heptene-2—one, M+- (43%), M+-'58 at m/z 54 (5%), M+--64 at m/z 48 (100%); 7-octene-2-one... Examples of McLafferty rearrangement at Ha, but not clear Why the m/z ion at 48 is reduced in intensity in case (b). (a) McLafferty rearrangement produces large m/z ion at 58. 0 __._CH2 Jr OH 3* McLafier’ty H C CH2 3 Ha J + H Ha rearrangement 3 2 C7H120 = 112 at Ha m/z = 54 m/z = 48 [5%] [100%] (b) McLafi‘erty rearrangement produces no m/z ion at 48 O + CH H C CH T OH . 2 2 2 H30 / _> W ] + A Ha Ha H30 CH3 1 m/z = 68 m/z = 48 [1 00%] [0 %] C2Hs+ m/z = 29 6.29 There are four isomeric alcohols of C4H100, MW = 74. The (it—cleavage and -H20 (m/z = 56) are major fragmentation pathways. The m/z = 45 is —C2H5 for B and the m/z = 31 is -C3H7 for A. /\/\ CH3 H30 OH H3C/\[/ A B OH CH3 HO CH3 OH >4 H80 H3C CH3 C D 6.30 The right specmnn is aniline (MW 93) and the left spectmm is 3-methylpyn'dine (MW 93). The fragment at m/z 92 from a tropylium type ion is especially diagnostic. light spectmm NH2 ‘1‘ -HCN ————> (-27) m/z = 66 lefi spectrum CH3 / N m/z = 92 6.31 (a) EIMS on-methyl butanol m/z 73, 55, 45, 43 + H C OH -CH3' 0+ 3 />< WW» H3C/>/ \CH3 H30 CH3 H30 C5H120 = 80 . 3H7. m/z = 73 \ -CZHSO- -? (CH3)20+ = 43 C4H7+ or C3H30+ = 55 (b) EIMS of 2-methyl butanol TMS m/z 160 145, 117 + ésiMe4 -CH3' 0+ Hep/X “fl Hap/V \SiMe4 H3C CH3 H30 m/z = 160 _03H7' m/z = 145 m/z = 117 structure not obvious 6.32 left spectrum: 3-methyl-2—pentanone; right spectrum: 4-methyI-2-pentanone lefi spectrum + . + H O . o ~02H4 (McLaf.) —————> H30 H3C CH3 CH3 m/z = 72 right spectrum + + H ' - O ~03H6 (McLaf.) 0 k H3C CH3 H30 m/z = 58 CH3 6.33 MW = 218 & important m/z = 176, 161 (15%), 147 (65%). O 0 Of 0‘.“ “5% CH3 -C2H20 H506 -29 H506)J\/\CH3 ———> ———> CH3 CH3 CH3 CH3 ‘0 Y 176 147 C14H1802= 06)/ o o 71 H506 ' CH3 H 161 6.34 Explain El fragment ions at m/z 128, 127, 126, 69 O oc~cleavage M» CF; = 69 CmHMONF3 = 197 ~CF3' 0+ J! -H J WNH _____> H OWN “30 3 127 128 41/ /‘0+ HSCWN/I 126 Note: possible McLafferty Rearrangement not observed ‘3'OH C4H9 H A of HN CF3 NH CF3 C2H20F3 = 115 6.35 The odd fragments indicate the even molecular ion at 162 (C3H1803) is not being observed. Loss of 87 at m/z = 75 and loss of 89 at m/z = 73 agrees best with the siructuure and fragment ion masses shown below H30 CH3 \0 0/ H30 H C Cl) V 73 CH3 75 ...
View Full Document

This note was uploaded on 08/06/2008 for the course CHEM 124 taught by Professor Amagata during the Winter '08 term at UCSB.

Page1 / 23

HW solutions Ch 4-6 - CHAPTER 4 4.1 Expected peaks = 2N1+l...

This preview shows document pages 1 - 23. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online