Chapter 1 Introduction, Formulas and Spectroscopy Overview

Transcription

1 hapter 1 Introduction, Formulas and Spectroscopy verview Problem 1 (p 1) - elpful equations: c = ( )( ) and = (1 / ) so c = ( ) / ( ) c =. x 1 8 m/sec =. x 1 1 cm/sec a. Which photon of electromagnetic radiation below would have the longer wavelength? onvert both values to meters. = 5 cm -1 = (1/ ) = (1/ ) = (1/ cm -1 )(1 m/1 cm) =.9 x 1-6 m = 4 x 1 14 z = (c / ) = (. x 1 8 m/s) / (4 x 1 14 s -1 ) = 7.5 x 1-7 m = longer wavelength = lower energy b. Which photon of electromagnetic radiation below would have the higher frequency? onvert both values to z. = 4 cm -1 = (1/ ) = 1/ [(4 cm -1 )(1 cm / 1 m)] =.5 x 1-5 m = (c/ ) = (. x 1 8 m/s) / (.5 x 1-5 m) v = 1. x 1 1 s -1 = nm = (c/ ) = (. x 1 8 m/s) / (.5 x 1-5 nm)(1m / 1 9 nm) v = (. x 1 8 m/s) / ( nm)(1m / 1 9 nm) v = 1. x x 1 15 s -1 higher frequency c. Which photon of electromagnetic radiation below would have the smaller wavenumber? onvert both values to cm -1. = 1 m = 6 x 1 1 z = (1 / ) = (1 / 1m) = 1 m -1 1 cm -1 = ( / c) 1 m -1 = 1 cm -1 = (6 x 1 1 s -1 /. x 1 8 m/s) = m -1 1 cm -1 smaller wavenumber 1 m -1 =, cm -1 Problem (p 14) - rder the following photons from lowest to highest energy (first convert each value to kjoules/mole). What is the energy of each photon in kcal/mole? a. E = h = hc(1/ ) = hc E = hc(1/ ) E = h E = hc E = (6.6 x 1-4 j-s)(. x 1 8 m/s)(1 5 cm -1 x (1 m -1 /cm -1 ) ) E = x 1-18 j highest

2 E = hc(1/ E = (6.6 x 1-4 j-s)(. x 1 8 m/s)(1 / 1 nm x (1 m / 1 9 nm) ) E = x 1-19 j E = h E = (6.6 x 1-4 j-s)(1 9 / s) E = 6.6 x 1-5 j middle lowest b. E = h = hc(1/ ) = hc E = hc(1/ ) E = h E = hc E = (6.6 x 1-4 j-s)(. x 1 8 m/s)(1 cm -1 x (1 m -1 /cm -1 ) ) E = x 1-4 j E = hc(1/ E = (6.6 x 1-4 j-s)(. x 1 8 m/s)(1 / 1 1 nm x (1 m / 1 9 nm) ) E = x 1-7 j E = h E = (6.6 x 1-4 j-s)(1 15 / s) E = 6.6 x 1-19 j middle lowest highest Problem (p ) - Determine the degrees of unsaturation for each of the following formulas. What combinations of pi bonds and rings are possible in each case? Is it possible for any of the following formulas to have an alkene, alkyne, carboxylic acid, ester, amide, nitrile, aromatic ring, ketone, aldehyde, ether, amine or alcohol? Try and draw some examples of such structures. a. 7 8 b. 5 1 (7) + = 16 degrees unsaturation = (16-8) / = 4 o (5) + = 1 degrees unsaturation = (1-1) / = 1 o + 1 = 15 c. 4 8 d (4) + = 1 (6) + degrees unsaturation = (1-8) / = 1 o degrees unsaturation = (15-11) / = o e (8) = 19 f. 1 9 l o degrees unsaturation = (19-17) / = 1 (1) + + = 4 degrees unsaturation = (4-1) / = 7 o g. 9 9 h. 1 8 l (9) = 1 degrees unsaturation = (1-9) / = 6 o (1) + = 6 degrees unsaturation = (6-1) / = 8 o l l i. 7 5 j. 6 1 S 4 (7) = 16 (6) + = 14 degrees unsaturation = (17-5) / = 6 o degrees unsaturation = (14-1) / = o S

3 Problem 4 (p ) A low resolution MS on compound X was 15 g/mol. Possible formulas considered were 11 18, 1 14, 9 1, 8 6 and What are the possible numbers of pi bonds and rings for each of these formulas? (11) + = 4 degrees unsaturation = (4-18) / = o 9 14 (9) + + = degrees unsaturation = ( - 18) / = o pi 1 pi 1 rings 1 rings (1) + = degrees unsaturation = ( - 14) / = 4 o pi rings (9) + = (8) + = 18 degrees unsaturation = ( - 1) / = 5 o degrees unsaturation = (18-6) / = 6 o pi 4 1 pi rings 1 4 rings

4 Spectroscopy Solutions hapter 5 hapter - I Problem 1 (p 58) 1. Discuss how I could help answer the following questions. Be specific in your analysis, pointing out the values in wave numbers (cm -1 ) for absorption peaks that could resolve each question. a. Which product(s) is(are) obtained in the following Wittig reaction. Is starting material is still present. 1. P Ph Ph Ph. workup ketone = 17 cm -1 aldehyde = 17 cm -1 aldehyde - 8 cm -1 7 cm -1 geminal - bend 89 cm -1 mono alkene - bend 99, 91 cm -1 ketone = 17 cm -1 mono alkene - bend 99, 91 cm -1 aldehyde = 17 cm -1 aldehyde - 8 cm -1 7 cm -1 geminal - bend 89 cm -1 b. Did the nitrile hydrolyze to the amide or the carboxylic acid? Is there any starting material left? / mono aromatic - bend 75, 7 cm -1 nitrile 5 cm -1 mono aromatic - bend 75, 7 cm -1 amide = 166 cm -1 amide - 4, 1 cm -1 mono aromatic - bend 75, 7 cm -1 acid = 17 cm -1 acid cm -1

5 Spectroscopy Solutions hapter 6 c. Was the unsaturated ester reduced to an allylic alcohol, the saturated ester or the saturated alcohol? Is there any starting material left? conj. ester = 17 cm -1 mono alkene - bend 99, 91 cm -1 acyl - 15 cm -1 alkoxy - 15 cm LiAl 4. workup alcohol - cm -1 mono alkene - bend 99, 91 cm -1 ester = 174 cm -1 acyl - 15 cm -1 alkoxy - 15 cm -1 alcohol - cm -1 d. Did the alkyne reduce to a cis alkene, trans alkene or an alkane? Did the reaction work? catalyst alkyne cm -1 (weak) cis - bend 69 cm -1 trans - bend 98 cm -1 alkane cm -1 e. Did toluene undergo a successful Friedel-rafts acylation? Was the product ortho, meta or para? + All + + mono aromatic - bend 75, 7 cm -1 anhydride = 181, 176 cm -1 conj. ester = 169 cm -1 para aromatic - bend 81 cm -1 conj. ester = 169 cm -1 meta aromatic - bend 69, 76, 89 cm -1 conj. ester = 169 cm -1 ortho aromatic - bend 75 cm -1 f. Did nitration work in the first reaction (if so, was it ortho, meta or para)? Did the nitro group reduce to the amine in the second reaction? Was the amino group substituted for a nitrile in the third reaction? mono aromatic - bend 75, 7 cm -1 S 4 para aromatic - bend 81 cm -1 nitro = 15, 155 cm -1 (weak) Fe / l para aromatic - bend 81 cm -1 1 o 15, 155 cm -1 (weak) 1. l / a. u para aromatic - bend 81 cm -1 nitrile 5 cm -1 (strong)

6 Spectroscopy Solutions hapter 7 g. Did the alcohol undergo an elimination reaction to an alkene or substitution reaction to an / S 4 ether? alcohol - cm -1 alkoxy - 15 cm -1 alkene - -1 cm -1 cis - bend 69 cm -1 alkoxy cm -1 h. Did the oxidation produce an aldehyde or carboxylic acid? Did the reaction work? r l alkene - -1 cm -1 alcohol - cm -1 mono aromatic - bend 75, 7 cm -1 alkoxy - 15 cm -1 alkene - -1 cm -1 acid cm -1 mono aromatic - bend 75, 7 cm -1 acid = 17 cm -1 alkene - -1 cm -1 aldehyde - 8 cm -1 7 cm -1 mono aromatic - bend 75, 7 cm -1 aldehyde = 17 cm -1 i. Did the esterification reaction work or was the carboxylic acid isolated in the workup? acid cm -1 acyl - 15 cm -1 acid = 17 cm -1 alcohol - cm -1 alkoxy - 15 cm -1 Ts (- ) ester = 174 cm -1 acyl - 15 cm -1 alkoxy - 15 cm -1 j. Did the DIBA reduction of the nitrile produce an aldehyde or an amine or did it work? nitrile 5 cm -1 (strong) 1. Ar DIBAL. workup aldehyde = 17 cm -1 aldehyde - 8 cm -1 7 cm -1 1 o 15, 155 cm -1 (weak) - bend 164 cm -1 (weak)

7 Spectroscopy Solutions hapter 8 k. Did the aldol reaction produce a -hydroxycarbonyl or an, -unsaturated carbonyl or did it work? K /? aldehyde - 8 cm -1 7 cm -1 alkene - -1 cm -1 conj. aldehyde = 169 cm -1 mono aromatic - bend 75, 7 cm -1 ketone = 17 cm -1 alkene - -1 cm -1 mono aromatic - bend 75, 7 cm -1 -hydroxycarbonyl alcohol - cm -1 alkoxy - 15 cm -1 alkene - -1 cm -1 mono aromatic - bend 75, 7 cm -1 -unsaturatedcarbonyl conj. ketone = 168 cm -1 trans - bend 98 cm -1 ketone = 17 cm -1 l. Did the reaction of thionyl chloride with benzoic acid produce an acid chloride? Was this converted into an ester or an amide in the second reaction with -(-methylamino)-1-ethanol? Sl l sp - -1 cm -1 acid cm -1 conj. acid = 17 cm -1 acyl - 15 cm -1 mono aromatic - bend 75, 7 cm -1 sp - -1 cm -1 acid chloride = 18 cm -1 sp - -1 cm -1 amine - cm -1 conj. ester = 17 cm -1 acyl - 15 cm -1 mono aromatic - bend 75, 7 cm -1 sp - -1 cm -1 alcohol - cm -1 conj. amide = 165 cm -1 alkoxy - 15 cm -1 mono aromatic - bend 75, 7 cm -1 m. Did the ketone, the aldehyde, both or neither get protected as acetals in reaction with ethylene glycol? aldehyde - 8 cm -1 7 cm -1 aldehyde = 17 cm -1 ketone = 17 cm -1 Ts (- ) ketone = 17 cm -1 alkoxy - 15 cm -1 aldehyde - 8 cm -1 7 cm -1 aldehyde = 17 cm -1 alkoxy - 15 cm -1 alkoxy - 15 cm -1

8 Spectroscopy Solutions hapter 9 n. Did the alkene get oxidized with mpba to an epoxide ether or a 1,-diol or did the reaction work? l mpba sp - -1 cm -1 mono alkene - bend 99, 91 cm -1 mono aromatic - bend 75, 7 cm -1 sp - -1 cm -1 alkoxy - 15 cm -1 mono aromatic - bend 75, 7 cm -1 alcohol - cm -1 sp - -1 cm -1 alkoxy - 15 cm -1 mono aromatic - bend 75, 7 cm -1 Problem (p 6) Match each of the following compounds with the I spectra below. Generic alkane parts are represented with an symbol. Identify all distinguishing functional group absorption bands to demonstrate the logic of your choices (e.g sp = ; 17 = =, etc.) S ' ' l

9 Spectroscopy Solutions hapter 1 A. 1 %T 5 41 = overtone 86 7 aldehyde sp - 17 sp - bend rock = = wavenumber = cm aldehyde B. 1 %T 64 = 5 overtone = sp - asym. 15 sym. sp - acyl bend = wavenumber = cm %T alkoxy sp - sp bend acyl 119 = = wavenumber = cm ' D. 1 %T 5 41 = overtone sp - = = wavenumber = cm -1 '

10 Spectroscopy Solutions hapter 11 E. 1-5 acid - 9. %T sp - acyl sp - 86 = bend = wavenumber = cm -1 F. %T o bend sp - sp - bend - bend = wavenumber = cm G. 1 %T 5 1 o 18 amide sp bend sp - amide amide = bend bend = wavenumber = cm %T sp - = rock sp - mono alkene sp - bend - bend = wavenumber = cm

11 Spectroscopy Solutions hapter 1 I. 1 %T sp - alcohol 8 bend bend alcohol alkoxy - sp = wavenumber = cm J. 1 %T 68 = 5 overtone sp - 18 sp - bend = = wavenumber = cm l K. 1 %T 5 sp - bend alkoxy sp = wavenumber = cm L alkyne %T sp - 96 alkene mono = 84 sp - aromatic bend - bend sp = wavenumber = cm

12 Spectroscopy Solutions hapter 1 M. 1 %T 66 thiol 18 S- 7 5 sp rock bend sp = wavenumber = cm S. 1 %T alkyne 185 rock 1 s 147 sp - tretch sp - sp - sp - bend bend = wavenumber = cm %T mono sp sp nitrile 15 aromatic aromatic bend = wavenumber = cm -1 P. 1 %T rock t-butyl 147 doublet sp - sp - bend = wavenumber = cm -1 1.

13 Spectroscopy Solutions hapter 14 Q. 1 %T -5 acid mono sp - 1 aromatic 97 conj. sp - bend acyl - bend = = wavenumber = cm %T sp bend nitro sp - = = wavenumber = cm

14 Spectroscopy Solutions hapter 15 Problem (p 66) Match each of the following aromatic compounds (and occasional alkene) with its I spectra below. Identify all distinguishing absorption bands to demonstrate the logic of your choices. Write each structure next to the spectra. = alkane portion A. 1 %T sp sp mono mono 74 alkene aromatic - - = - - bend 7 bend = wavenumber = cm B. 1 9 %T geminal mono sp sp alkene aromatic - - = - - bend 7 bend = wavenumber = cm %T mono sp 75 aromatic sp - - = - bend = wavenumber = cm -1 5.

15 Spectroscopy Solutions hapter 16 D. 1. %T sp sp 14 para aromatic = - bend = wavenumber = cm -1 E. 1 5 %T 5 sp meta sp aromatic - = - 79 bend = wavenumber = cm F. 1 %T sp 148 sp - = - bend = wavenumber = cm E. 1 %T 5 1 ortho substituted aromatic overtones = wavenumber = cm -1

16 Spectroscopy Solutions hapter 17 Problem 4 (p 68) Match each of the following alkene spectra with the corresponding alkene. Identify distinguishing absorption bands to demonstrate the logic of your choices. Write each structure next to the spectra A. %T sp - sp = geminal - bend cm = wavenumber = cm geminal alkene B. %T sp - sp sp rock 7 cm trans - bend cm = wavenumber = cm trans alkene. 1 %T sp = sp mono alkene - bend , 91 cm = wavenumber = cm monosubstituted alkene

17 Spectroscopy Solutions hapter 18 D. %T 1 symmetrical tetrasubstituted alkene has no sp = very weak symmetrical tetrasubstituted alkene has no sp - bend = wavenumber = cm tetrasubstituted alkene E. %T sp - sp = cis - bend 69 cm = wavenumber = cm cis alkene F. %T 1 5 sp -, too weak sp = tri sp - bend cm = wavenumber = cm -1. trisubstituted alkene

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19 Spectroscopy Solutions hapter hapter Mass Spec Problem 1 (p. 75)- A low-resolution mass spectrum of satosporin (J. rg. Lett., 1, 15, 864-7) showed the molecular weight to be 64. This molecular weight is correct for the molecular formulas 1 5, 6 4 and 4. A high-resolution mass spectrum provided a molecular weight of Which of the possible molecular formulas is the correct one? What is the degree of unsaturation in satosporin? 1 5 (1) x (1.1115) = 5.4 () x (1.797) =.55 (5) x ( ) = total = = () x (1.1115) = (6) x (1.797) = (4) x ( ) = total = = () x (1.1115) = (4) x (1.797) = () x ( ) = total = =.59 experimental molecular mass = appears closest to the second molecular formula Problem (p. 75) Penilactone B, (J. rg. Lett., 1, 15, 891-), a natural product was found by low-resolution mass spectrometry to have a molecular weight of 514. Possible molecular formulas include 8 4 9, 7 1, and igh-resolution mass spectrometry indicated that the precise molecular weight was What is the correct molecular formula of penilactone B? What is the degree of unsaturation? (8) x (1.1115) = 6.1 (4) x (1.797) = 4.71 (9) x ( ) = total = =.18 (7) x (1.1115) = 4.1 () x (1.797) =.91 (1) x ( ) = total = =.845 (6) x (1.1115) = (6) x (1.797) = 6.7 (11) x ( ) = total = =.49 experimental molecular mass = appears closest to the third molecular formula

20 Spectroscopy Solutions hapter 1 Problem (p. 76) alculate the relative intensities (as a percent) of M+, M+1 and M+ for propene ( -= ), ketene ( ==) and diazomethane ( ==). an these three formulas ( 6 vs vs ) be distinguished on the basis of their M+1 and M+ peaks? alculate the exact mass (four decimal places) for all of these formulas. an they be distinguished on the basis of exact mass? elpful data are on page #. propene, (-=) = 6 one = ( ways) =. one = zero 17 = (6 ways) =.7 sum of possibilities = (.) + (.7) + (.) =.7 M+1 peak as a percent of M+ peak = (.7)x(1%) =.7% two 1 = two = zero 18 = x x 1 = (.) ( ways) =.7 6 x 5 x 1 = (5x1-7 )(15 ways) = 7 x 1-6 = too small to consider one 1 and one 1.8 = 11.1 ( ways) x =. = too small to consider sum of possibilities = (.7) + (.) + (.) =.7 (6 ways) M+ peak as a percent of M+ peak = (.7)x(1%) =.7% diazomethane, (==, ) one = (1 ways) =.17 one = one 15 = (ways) = (1 way) =.7 sum of possibilities = (.17) + (.) + (.7) =.181 M+1 peak as a percent of M+ peak = (.181)x(1%) = 1.81% two 1 = not possible two = two 15.7 = 1.7 (1 way) =. sum of possibilities = (.1) =.1 =.1 M+ peak as a percent of M+ peak = (.1)x(1%) =.1% ketene, (==, ) one = ( ways) =.14 one = one 17 = ( ways) =. (1 way) =. two 1 = two = one 18 = = (.11) =.1 = too small to consider (1 way) =. one 1 and one 1.8 = 11.1 = too small to consider ( ways) x ( ways) sum of possibilities = (.14) + (.) + (.4) =. M+1 peak as a percent of M+ peak = (.)x(1%) =.% sum of possibilities = (.1) + (.) + (.) =.1 M+ peak as a percent of M+ peak = (.1)x(1%) =.1%

21 Spectroscopy Solutions hapter Problem 4 (p. 78)- alculate the relative intensities (as a percent) of M+, M+ and M+4 for l and. Use the probabilities from above. M+ peak relative size two 5 l = (.758) (1 way) =.575 (assigned a referenced value of 1%) M+ peak relative size one 5 l and one 7 l = (.758) x (.4) x ( ways) =.67 percent of M+ peak = [(.67)/(.575)] x 1 = 6.8% M+4 peak relative size two 7 l = (.46) (1 way) =.65 percent of M+ peak = [(.65)/(.575)] x 1 = 1.5% 1% 64% 11% M+ M+ M+4 M+ peak relative size two 78 = (.57) (1 way) =.57 (assigned a referenced value of 1%) M+ peak relative size one 79 and one 81 = (.57) x (.49) x ( ways) =.5 percent of M+ peak = [(.5)/(.57)] x 1 = 195% M+4 peak relative size two 81 = (.49) (1 way) =.4 percent of M+ peak = [(.4)/(.57)] x 1 = 95% 195% 1% 95% M+ M+ M+4 Problem 5 (p. 8) Match the MS data with the correct number of l, and/or S. For each molecular weight estimate how many carbon atoms are present. The proton count is provided from the proton M, as is number of pi bonds from the 1 M. Use this information to calculate a chemical formula and degrees of unsaturation and number of rings. Formula 1 has one l = c. This has to be compound c because the M+ peak is % of the M+ peak (= ). Subtract 5 from the mass peak gives the residual mass of 95. Subtract proton mass from 95 leaves 6. Divide the M+1 peak (16.6%) by 1.1 suggests there are 15 carbon atoms (= 18 mass units). Subtract 18 from 6 leaves 8 which is divisible by 16, indicating there are 5 oxygen atoms. Formula 1 = 15 l 5 unsaturation = (15) + = -4 = 8 = 4 degrees unsaturation ( pi bonds, 1 ring)

22 Spectroscopy Solutions hapter Formula has one = e. This has to be compound e because the M+ peak is 97% of the M+ peak (= 87). Subtract 79 from the mass peak gives the residual mass of 8. Subtract 14 proton mass from 8 leaves 194. Divide the M+1 peak (1.%) by 1.1 suggests there are 11 carbon atoms (= 1 mass units). Subtract 1 from 194 leaves 6. The M+ peak is odd so there has to be an odd number of nitrogen atoms. Subtract 14 from 6 leaves 48 which is divisible by 16, indicating there are oxygen atoms. Formula = unsaturation = (11) = 5-15 = 1 = 5 degrees unsaturation ( pi bonds, rings) Formula has one l and one S = d. This has to be compound e because the M+ peak is 6.5% (.% + 4.5%) of the M+ peak (= 45). Subtract 67 (= 5+) from the mass peak gives the residual mass of 78. Subtract 4 proton mass from 78 leaves 54. Subtract.7 (sulfur) from the M+1 peak (18.5%-.7% = 17.8% for carbon) divided by 1.1 suggests there are 16 carbon atoms + 1 sulfur atom (= 4 mass units). Subtract 4 from 54 leaves. The M+ peak is odd so there has to be an odd number of nitrogen atoms. Subtract 14 from leaves 16 which is divisible by 16, indicating there is 1 oxygen atom. Formula = 16 4 ls unsaturation = (16) + +1 = 5-5 = 1 = 5 degrees unsaturation ( pi bonds, rings) Formula 4 has one and one S = b. This has to be compound b because the M+ peak is 11.9% (97.% + 4.5%) of the M+ peak (= 45). Subtract 111 (= 79+) from the mass peak gives the residual mass of 94. Subtract proton mass from 94 leaves 74. Subtract.7 (sulfur) from the M+1 peak (17.%-.7% = 16.6% for carbon) divided by 1.1 suggests there are 15 carbon atoms (= 18 mass units). Subtract 18 from 74 leaves 94. The M+ peak is odd so there has to be an odd number of nitrogen atoms. Subtract 14 from 94 leaves 8 which is divisible by 16, indicating there are 5 oxygen atoms. Formula 4 = 15 S 5 unsaturation = (15) + +1 = -1 = 1 = 6 degrees unsaturation ( pi bonds, rings)

23 Spectroscopy Solutions hapter 4 Formula 5 has two l = a. This has to be compound a because the M+ peak is 6.9% (.% +.%) of the M+ peak (= 61). Subtract 7 from the mass peak gives the residual mass of 91. Subtract 1 proton mass from 91 leaves 7. The M+1 peak (17.7% carbon) divided by 1.1 suggests there are 16 carbon atoms (= 196 mass units). Subtract 196 from 7 leaves 74. The M+ peak is odd so there has to be an odd number of nitrogen atoms. Subtract 14 from 74 leaves 6 which is divisible by 16, indicating there are 5 oxygen atoms. Formula 4 = 16 1 l 5 unsaturation = (16) + +1 = 5 - = 1 = 6 degrees unsaturation ( pi bonds, rings) Formula 6 has two = f. This has to be compound f because the M+ peak is 194.6% (97.% + 97.%) of the M+ peak (= 445). Subtract 158 (= x) from the mass peak gives the residual mass of 87. Subtract 1 proton mass from 87 leaves 66. The M+1 peak (17.7% carbon) divided by 1.1 suggests there are 16 carbon atoms (= 19 mass units). Subtract 19 from 66 leaves 74. The M+ peak is odd so there has to be an odd number of nitrogen atoms. Subtract 14 from 74 leaves 6 which is divisible by 16, indicating there are 5 oxygen atoms. Formula 4 = unsaturation = (16) + +1 = 5 - = 1 = 6 degrees unsaturation ( pi bonds, rings) Problem 6 (p. 114) +, + and 5 + have fragment masses of approximately 9, yet + has a M+1 peak of 1.1% and M+ peak of.%, + has a M+1 peak of 1.51% and M+ peak of.1% and 5 + has a M+1 peak of.4% and M+ peak of.1%. igh resolution mass spec shows +, + and 5 + all have different masses. Explain these observations and show all of your work. elpful data follow. one 1 = 1.8 (1 way) = two 1 = not possible one = (1 way) =.1 two = not possible one 17.4 = (1 way) =.4 one = 1.4 (1 way) =. sum of possibilities = (.17) + (.1) + (.4) =.11 M+1 peak as a percent of M+ peak = (.11)x(1%) = 1.1% sum of possibilities = (.) M+ peak as a percent of M+ peak = (.)x(1%) =.%

24 Spectroscopy Solutions hapter 5 one = (1 way) =.17 three = one 15 = ( ways) = (1 way) =.7 sum of possibilities = (.17) + (.4) + (.7) =.181 M+1 peak as a percent of M+ peak = (.148)x(1%) = 1.48% 5 one = ( ways) =.14 one = (5 ways) =.6 sum of possibilities = (.14) + (.6) =. M+1 peak as a percent of M+ peak = (.)x(1%) =.% two 1 = not possible two = two 15.7 = 1.7 (1 way) =.1 sum of possibilities = (.1) =.1 ( ways) =. M+ peak as a percent of M+ peak = (.1)x(1%) =.1% two 1 = two = one 1 and one 1.8 = 11.1 = too small to consider sum of possibilities = (.1) =.1 = (.11) =.1 = too small to consider ( ways) x ( ways) M+ peak as a percent of M+ peak = (.1)x(1%) =.1% Average uclide Element Atomic Mass (elative Abundance) Mass (1) 1.78 (.1) (1) 1. 1 (1.8) (1) (.7) (1) (.4) (.) Problem 7(p. 114) adical cations of the following molecules (e- + M M+ + e-) will fragment to yield the indicated masses as major peaks. The molecular ion peak is given under each structure. The base peak is listed as 1%. ther values listed represent some relatively stable possibilities (hence higher relative abundance), or common fragmentations (expected), even if in low amount. Explain logical peaks (alkyl fragments, pi bond, functional group and heteroatom fragments). This may be as easy as drawing a line between two atoms of a bond, or it may require drawing curved arrows to show how electrons move (e.g. McLafferty). This may also involve drawing resonance structures or indicating special substitution patterns ( o + > o + > 1 o + > + ). If a fragment has an even mass and there is a pi bond, think McLafferty (unless an odd number of nitrogen atoms are present). Even masses can also be formed by elimination of a small molecule such as loss of water from an alcohol or loss of an alcohol from an ether or a retro-diels-alder reaction, etc. Make sure you show this. Some fragments may be an extension of a logical fragment by units of 14 amu ( ).

25 Spectroscopy Solutions hapter 6 a. M + = 86 The base peak is not logical from the starting skeleton. m/e % base < mass = 9 (alkene) (M-75 = 1%) b M+ = octane - all expected alkane fragments are observed, except 99 (M-15) M+ = (6%) (%) (6%) (%) (4%) (1%) (7%) (1%) not observed mass = 7 (alkene) (M-44 = %) MW = 114 MW = 114 MW = 114 M- (ethane) M-44 (propane) M-58 (butane) mass = 84 (alkene) (M- = 7%) mass = 7 (alkene) (M-44 = 1%) mass = 56 (alkene) (M-58 = 18%) lost ethane ( = 1%) lost propane (44 = %) lost butane (58 = %) Possible elimination reactions forming alkenes. All are observed in the actual MS of octane. Further alkene fragments (see next functional group) allylic fragmentation 41 = 44% 55 = 11% 69 = % Alkene & McLafferty fragmentation 8 = 4% 4 = 15% 56 = 18% 7 = 1% 84 = 7% mass % base M+ mass = 9 (alkene) (M-75 = 1%) mass = 5 (alkene) (M-61 = %) c. M + = 84 m/e % base base o easy explanation for 56, but if ring opens and forms alkene, McLafferty might work ? or 4 4 or 56 8 mass = 87 (alkene) (M-44 = %)

26 Spectroscopy Solutions hapter 7 d. M + = 84 You might have to move the = around to get 56. m/e % base base ? or 4 4 or e. f. M + = 68 m/e % base base or or 84 Don't remove the sp -, there is a better spot to lose an atom (resonance) m/e % base M + = 1 base Use a bridging ring to make 15 and 91 is easy g. M + = is an even mass, but not McLafferty. A small molecule might help explain it. m/e % base <1 base

27 Spectroscopy Solutions hapter 8 h. M + = 74 Peak 1 is harder to explain, but common. m/e % base <1 base 8 (- ) i. M + = 88 8 and 4 are even, but not McLafferty. Think like "g", but "organic" water. 1 requires some drastic rearrangements. j. k. M + = 7 Peak dominates. Think of a small molecule elimination for peak 56. m/e % base m/e % base M + = can be different than m/e base % base l. m/e % base M = can be different than is an even mass, so... base base (- ) or 58 4 or 44 8

28 Spectroscopy Solutions hapter 9 m. m/e % base base M = is an even mass or 74 8 n. M + = 88 An even mass strikes again at is not common, but expected here. m/e % base base or 6 8 o. m/e % base M = ormally 59 would be e 7 19 ven, but there is nitrogen 87 present. base or 59 8 p. M + = 69 ormally 41 would be even, but there is nitrogen present (M-8). m/e % base <1 base or 41 8 q. M + = 16 Two famous names goes with 68. m/e % base base M + =

29 Spectroscopy Solutions hapter Problem 8 (p. 117) n the following pages are 11 compounds (these are lettered A-K) from the 11 functional groups numbered below. Try to match each spectrum (A-K) to the class of functional group numbered 1-11, and then try to solve the exact structure of each compound. These are simple monofunctional group compounds. Use the I data to help identify the functional groups. Explain the major peaks that helped decide on your structure. Why are these peaks formed in preference to others (what is the reason for their special stability)? This problem would be a lot easier if Ms were available. lasses of compounds (I and MS clues) A brominated alkane B thiol aromatic D ketone S 4-methyl-1-bromopentane MW = 164 E -ethylhexanal MW = 1 p-thiomethyltoluene MW = 18 pentylamine MW = 87 propylbenzene MW = 1 aldehyde F amine G alcohol hexan-1-ol MW = 1 octan--one MW = 1 carboxylic acid hexanoic acid MW = 144 I ester J amide K chlorinated alkane l 1-chloropentane MW = 16 propyl pentanoate MW = 144 heptanamide MW = 14

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31 Spectroscopy Solutions hapter 4 hapter 4 M Problem 1 (p. 146): ow many different protons and carbons are in each of the 5 1 isomers and 5 11 isomers? There are three 5 1 isomers 1 # = # = # = 4 # = 4 # = 1 # = There are eight 5 11 isomers 4 # = 5 # = # = 7 # = 5 # = 4 # = # = 4 # = # = 5 # = 5 # = # = 4 # = 7 # = 5 # = # = Problem (p. 149) Use the table to estimate the chemical shifts of each type of sp - in the structures below. a.experimental values (ppm) a b c calculated sp - values (ppm) = nly sp - calculations shown here. a b c 1..8 ( ).1 ( ) ( ).1 ( ).4.9. ( ) alkene. ( ) ether "" 1.

32 Spectroscopy Solutions hapter 4 b. a experimental values (ppm) b c d calculated sp - values (ppm) = nly sp - calculations shown here. a b c d.9 1. ( ). ( ) ( ).1 ( ) ( ). ( ) ( ) 1.1 ( ). ketone acid c. 1 4 diastereotopic next to chiral center calculated sp - values (ppm) = nly sp - calculations shown here. 1 a b ( ). ( ).8( ). ( ).1 ( ).4 ( ) ( ).4 ( ) ( ) 1.1 ( ).4 nitro aldehyde d. 1 Ph 4 Ph = phenyl = aromatic diastereotopic next to chiral center calculated sp - values (ppm) = nly sp - calculations shown here. 1 a b.9.4 ( ). ( ) ( ).1 ( ) ( ). ( ) ( ). ( ) ( ). ( ).6 aromatic alcohol e. 1 4 l calculated sp - values (ppm) = nly sp - calculations shown here ( ). ( ) ( ). ( ) ( ).4 ( ) ( ) 1.8 ( ).4 ketone acid chloride f. 1 diastereotopic next to chiral center nly sp - calculations shown here. 1 a b 4a 4b calculated sp. ( ).8 ( ). ( ). ( ).1 ( ).1 ( ). ( ). ( ).1 ( ).1 ( ).4 ( ).4 ( ) - values (ppm) = ( ) 1.5 ( ) ( ) 1.5 ( ).7 alkene epoxide

33 Spectroscopy Solutions hapter 4 4 Additional Problems - ompare your calculated shifts to the chemical shifts values from hemdraw g. h i. j. i..45*.8*.4* * diasterotopic protons * * diasterotopic protons.45* *.5* S * diasterotopic protons * diasterotopic protons k l * * * diasterotopic protons * diasterotopic protons l ot all diastereotopic protons next to a chiral center have different chemical shifts. In example 'h' both protons at 1.6 have the same chemical shift, while on the other side of the chiral center there are different chemical shifts. Problem (p. 15) alculate chemical shifts for the following alkene and aromatic sp protons (and compare to the values given). a f k b g h c l m n d i j e

34 Spectroscopy Solutions hapter 4 5 a b c d e f g.6 h i j 7.7 k l m n o Problem 4 (p. 178)- Predict approximate M s for the following compounds. Draw a sketch of your predicted M. Include calculations of 1 chemical shifts, estimates of coupling constants and multiplicities as well as the number of hydrogen atoms to be integrated at each chemical shift (by writing the appropriate number of hydrogen atoms above the multiplet). Assume 4 J (or higher) = z. Actual chemical shifts are written by each type of hydrogen. a 1.71,s ,d J= 4.6 1,d J= 1.96,q 1.7,sex 1.7,sex PPM 1

35 Spectroscopy Solutions hapter 4 6 b.9,s 1.1 6,d ,d ,non PPM 1 c ,dd J=17, ,dd J=1, ,dd J=17,.96,t 1.66,qnt 1.,sex,t PPM 1 d.9,s ,d ; ,dd J=1,7.64 1,dd J=1,7.61 1,sex. 1,brd s 4 PPM 1 e.5,t.,td, 1.8 1,t J= ,m ,brd s PPM 1

36 Spectroscopy Solutions hapter 4 7 f 1. 6,t.4 4,q ,t.51,t.4.7 PPM 1 g.7,s ,s 8.1 1,s ,s 5. 1,brd s PPM h 1.11,t 1.,t 4.1,q.76,t.41,q.4,t PPM 1 i ,d J= ,dt J=17, ,dd J=17,6 1.96,q 1.7,sex,t PPM 4 1

37 Spectroscopy Solutions hapter 4 8 j 7.89,d J= ,t J=9 7.4,t J=9.8,t.4,t ,brd s PPM k.7,s.47,s.5,s S ,d J= ,d J= ,s PPM 1 l 4.41,t 1.91,qnt,t ,sex PPM 1 m. 7.75,d J=9 6.54,d J=9 4.9,q 4. 1,brd s.6,t 1.5,qnt 1.,sex 1.,t,t PPM 1

38 Spectroscopy Solutions hapter 4 9 n. 7.1,t J= ,t J=9 6.46,d J= ,s 4.19,q 4. 1,brd s 1.71,s 1.,t PPM 1 o ,d J= ,d J= 4.5,t.64,t.4 4,q 1.9,s 1. 6,t real J values J allyl = 1. z (not included) PPM 1 J gem = 1.6 z p ,dt J=17, ,dt J=17,7 4.75,d.5,t 1.96,q 1.7,sex 1.7,sex,t,t PPM 1 Problem 5 (p. 18) ( 1 M) A large number of fairly straight forward predict the product M problems are listed below. Some are really easy and some are a bit more challenging. Use the molecular formulas along with the chemical shifts, multiplicities and integrations to determine a reasonable structure for the following molecules. Approximate coupling constants are indicated. The multiplicities are s = singlet, d = doublet, t = triplet, q = quartet, qnt = quintet, sex = sextet, se = septet, o = octet, n = nonet, m = multiplet. The degree of unsaturation can help determine the possible number of rings and/or pi bonds. Match each type of hydrogen in the problems with the hydrogen atoms in your structures. Several M problems follow. Propose reasonable structures.

39 Spectroscopy Solutions hapter ; ; ; ,s,s , 5.1, l 4.5 l l l. l.67 l l ; l

40 Spectroscopy Solutions hapter ; l l l l l 1.8 l l l l l l l ; ; ; ;

41 Spectroscopy Solutions hapter ; ; ; ;

42 Spectroscopy Solutions hapter S S S S S S ;

43 Spectroscopy Solutions hapter

44 Spectroscopy Solutions hapter 5 45 hapter 5 - ommon Proton M Problems and Possible Solutions, SY D Interpretation Problem 1 (p. ) Answers to SY problems. a. Formula = 11 l, Possibly helpful I bands: 17, 1, 18 cm -1. (assume that sp peaks are in all I spectra) l l Exact Mass: ; M+ = 98. (77.4%), M+1 = 99.4 (9.4%), M+ =. (1.%), M+ = 1. (11.9%), M+4 =. (4.%), b. Formula = 1 1 l, Possibly helpful I bands:, 85, 78, 17, 169, 166, 98 cm -1 (assume that sp peaks are in all I spectra) l Exact Mass: ;.9.6; l m/e: 6.1 (1.%), 61.1 (14.4%), 6.11 (.%), 6.1 (4.6%), c. Formula = 1 l, Possibly helpful I bands:,, 164, 115, cm -1 (assume that sp peaks are in all I spectra). int: There is one carbon with any protons. 1 l Exact Mass: 4.14 l ; M+ = 4.14 (1.%), M+1 = 5.14 (1.%), M+ = 6.14 (.4%), M+ = 7.14 (4.%) d. Formula = , Possibly helpful I bands: 174, 17, 15, 1, 11, 17, 14 cm -1 (assume that sp peaks are in all I spectra). int: Ether linkage across.4 and ; Exact Mass: M+ = 66.1 (1.%), M+1 = (16.7%), M+ = 68.1 (97.%), M+ = (16.5%),.1

45 Spectroscopy Solutions hapter 5 46 e. Formula = , Possibly helpful I bands:, 85, 78, 17, 169, 166, 155, 14, 98 cm -1 (assume that sp peaks are in all I spectra) ; Exact Mass: 55.4 M+ = 55.4 (1.%), M+1 = 56.5 (16.6%), M+ = 57.4 (97.%), M+ = 58.4 (16.%) Additional possible structures for SY problems f ; Exact Mass: 1.17 M+ = 1.17 (1.%), M+1 = (1.%), M+ = (1.%) g l l Exact Mass: M+ = (1.%), M+1 = 5.15 (1.4%), M+ = (.4%), M+ = 5.15 (4.%) h l Exact Mass: l M+ =.1 (1.%), M+1 =.14 (19.8%), 7.84 M+ = 4.1 (.%), M+ = 5.1 (6.%), 5.7 i ; j k l Exact Mass: l M+ = 8.14 (1.%), 1.1 M+1 = 8.14 (17.6%), 7.78 M+ = (.4%),.15; M+ = (5.6%),.91; ; l Exact Mass: M+ = (1.%), M+1 = (1.%), M+ = (1.%) l Exact Mass: 16.9 M+ = 16.9 (1.%), M+1 = 17.1 (1.%), M+ = 18.9 (.%), M+ = 19.9 (.8%), 1.59

46 Spectroscopy Solutions hapter 6 47 hapter 6-1 M, DEPT Experiments and ther D M Methods (ET/SQ and MB) Problem 1 (p. 9) Estimate the sp carbon chemical shifts of the following molecules. If real values are known, they are listed next to the appropriate carbon atom (experimental), otherwise estimated values are from hemdraw (estimated). ompare your calculated values to those below. a 16.6 b c 18.1 d e l 11. experimental experimental estimated experimental estimated f.5 g h experimental experimental estimated S i estimated Problem (p. 4) alculate the alkene carbon chemical shifts in the following structures. If you don't find an exact match for a particular substituent, look for a close analog to use in place of the actual substituent. The chemical shifts provided are experimental or come from hemdraw estimates. a b c d e f g h 18.7 i j k l m n

47 Spectroscopy Solutions hapter 6 48 Problem (p. 4) alculate the aromatic proton chemical shifts in the following structures. If you don't find and exact match for a particular substituent, look for a close analog to use in place of the actual substituent. The chemical shifts provided are experimental or come from hemdraw estimates. a b c d e f g h i j k l m n o p

48 Spectroscopy Solutions hapter 6 49 Problem 4 (p. 4) In the following molecules, calculate a chemical shift of any carbon atoms for which a formula is available. If we don t have a formula to calculate a chemical shift, then estimate a range of possibilities from the generic table of shifts. The experimental values are provided for comparison to the estimated values. a b c d e f g h i j Problem 5 (p.54 ) 6 a. b. c d Problem 6 (p. 55) 6 14 a. b c. d e

49 Spectroscopy Solutions hapter 6 5 Problem 7 (p. 56) 5 11 a b. 5.1 c. d e. f g. h Problem 8 (p. 57) Problem 9 (p. 57) Provide an acceptable structure for each DEPT spectra below. Formula: 4 1 (bottom = 1 spectrum, middle = DEPT-9, top = DEPT-15) a. b c. 4.7 d e. f g

50 Spectroscopy Solutions hapter 6 51 Problem 1 (p. 58) Provide an acceptable structure for each DEPT spectra below. Formula: 5 1 (bottom = 1 spectrum, middle = DEPT-9, top = DEPT-15) a b c d e f g h i j k l m 5.8 n Problem 11 (p. 6) Provide an acceptable structure for each DEPT spectra below. Formula: 4 8 (bottom = 1 spectrum, middle = DEPT-9, top = DEPT-15) a b c d e f g h i j k l m n o p q r Problem 1 (p. 61) Provide an acceptable structure for each DEPT spectra below. Formula: 4 8 (bottom = 1 spectrum, middle = DEPT-9, top = DEPT-15) a b c d e

51 Spectroscopy Solutions hapter 6 5 f g h i 19. j k l m n o p q r s t u v w x y z aa bb cc dd ee Problem 1 (p. 64) Provide an acceptable structure for each DEPT spectra below. Formula: 1 14 (bottom = 1 spectrum, middle = DEPT-9, top = DEPT-15) a. b. c d e f. g h

52 Spectroscopy Solutions hapter 6 5 i. 1.4 j. k. l m

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54 Spectroscopy Solutions hapter 7 55 hapter 7 - Medium Difficulty Structure Problems Problem 1 (p. 79) Exact Mass: M+ = 14.1 (1.%), M+1 = 14.1 (8.7%) I peaks (cm -1 ): - 9 sp - sp = 1695 alkoxy - 15 = 16 trans alkene - bend 98 sp - bend 147, 18 Problem (p. 8) Exact Mass: 14.1 M+ = 14.1 (1.%), M+1 = 14.1 (8.7%) x I peaks (cm -1 ): sp - aldehyde - 8 and 75 sp alkoxy - 17, 1 = 16 = 1698 sp - bend 147, 18 trans alkene - bend Problem (p. 81) Exact Mass: Mol. Wt.: m/e: 14.1 (1.%), 14.1 (8.7%) , 67.57;, 9.9;, I peaks (cm -1 ): sp - = 16 alkoxy - 15 sp sp - bend 147, 18 geminal alkene - bend 89 conj. ester = 17 acyl - 1 sp rock Problem 4 (p. 8) I peaks (cm -1 ): sp - sp conj. ester = Exact Mass: 6.14 Mol. Wt.: 6.1, 71.16;, 8.5;,.1 = 16 sp - bend 147, 18 acyl alkoxy - 15 meta aromatic - bend 88, 76, x 1.

55 Spectroscopy Solutions hapter 7 56 Problem 5 (p. 8) I peaks (cm -1 ): acid sp - sp Exact Mass: 6.14 M+ = 6.14 (1.%), M+1 = 7.14 (15.1%), M+ = 8.15 (1.7%) conj. acid = 169 = 16 sp - bend 147, acyl - 11 alkoxy - 117, Problem 6 (p. 84) ; ; I peaks (cm -1 ): alcohol - sp - sp Exact Mass: 9.8 M+ = 9.8 (1.%), M+1 = 4.9 (17.6%), M+ = 41.8 (97.%), M+ = 4.8 (17.%), M+4 = 4.9 (1.8%) conj. ketone = 169 = 16 sp - bend 147, alkoxy - 17, Problem 7 (p. 86) ; ; ;.4.; Exact Mass:.8 M+ =.8 (1.%), M+1 =.9 (1.%), M+ = 4.9 (1.5%) I peaks (cm -1 ):