Solving Spectroscopy Problems Part 1 Lecture Supplement page 159 + +

Solving Spectroscopy Problems How to deduce structure from spectra? Logical, orderly procedure Compare information between spectra when necessary Conservative analysis - do not discard possibilities until 100% sure Procedure MS gives formula Formula gives DBE Use formula plus DBE to guide IR analysis of functional groups NMR gives skeleton Assemble the pieces Check your work!

Sample Problem #1 Steps 1 and 2: Formula and DBE Mass spectrum m/z = 120 (M; 100%), m/z = 121 (9.8%), and m/z = 122 (0.42%) Formula and DBE Usual procedure reveals only one viable formula: C9H12 DBE = 4 Four rings and/or pi bonds Possible benzene ring

Sample Problem #1 Step 3: Functional Groups from IR 100 Transmittance (%) 4000 3000 2000 1500 1000 Stretching frequency (cm-1) Zone 1 (3700-3200 cm-1) C9H12 DBE = 4 Alcohol O-H: Amide/amine N-H: Terminal alkyne C-H: All absent – no peaks

Sample Problem #1 Step 3: Functional Groups from IR 100 Transmittance (%) 4000 3000 2000 1500 1000 Stretching frequency (cm-1) Zone 2 (3200-2700 cm-1) C9H12 DBE = 4 Aryl/vinyl sp2 C-H: Alkyl sp3 C-H: Aldehyde C-H: Carboxylic acid O-H: Present - peaks > 3000 cm-1 Present - peaks < 3000 cm-1 Absent - no peak ~2700 cm-1; no C=O in zone 4 Absent - not broad enough; no C=O in zone 4

Sample Problem #1 Step 3: Functional Groups from IR 100 Transmittance (%) 4000 3000 2000 1500 1000 Stretching frequency (cm-1) Zone 3 (2300-2000 cm-1) C9H12 DBE = 4 Alkyne CC: Nitrile CN: Both absent – no peaks

Sample Problem #1 Step 3: Functional Groups from IR 100 Transmittance (%) 4000 3000 2000 1500 1000 Stretching frequency (cm-1) Zone 4 (1850-1650 cm-1) C9H12 DBE = 4 C=O: Absent - no strong peak; no oxygen in formula

Sample Problem #1 Step 3: Functional Groups from IR 100 Transmittance (%) 4000 3000 2000 1500 1000 Stretching frequency (cm-1) Zone 5 (1680-1450 cm-1) C9H12 DBE = 4 Benzene ring C=C: Alkene C=C: Present - peaks at ~1610 cm-1 and ~1500 cm-1 Absent - not enough DBE for alkene plus benzene

Sample Problem #1 Step 4: C-H Skeleton from 1H-NMR 1H-NMR: 7.40-7.02 ppm (multiplet; integral = 5), 2.57 ppm (triplet; integral = 2), 1.64 ppm (sextet; integral = 2), and 0.94 ppm (triplet; integral = 3). Step 4a: Copy NMR data to table Shift Splitting Integral # of H Implications 7.40-7.02 ppm multiplet 5 2.57 ppm triplet 2 1.64 ppm sextet 2 0.94 ppm triplet 3

Sample Problem #1 Step 4: C-H Skeleton from 1H-NMR Step 4b: Divide hydrogens according to integrals Shift Splitting Integral # of H Implications 7.40-7.02 ppm multiplet 5 2.57 ppm triplet 2 1.64 ppm sextet 2 0.94 ppm triplet 3 5 H 2 H 3 H Totals 12  12 H

Sample Problem #1 Step 4: C-H Skeleton from 1H-NMR Step 4c: Combine splitting with number of hydrogens to get implications Shift Splitting Integral # of H Implications 7.40-7.02 ppm multiplet 5 2.57 ppm triplet 2 1.64 ppm sextet 2 0.94 ppm triplet 3 5 H 2 H 3 H C6H5 (Ph; monosubstituted benzene ring) CH2 in CH2CH2 CH2 in CHCH2CH 2 x CH in CHCH2 2 x CH in CHCHCH two neighbors CH2 or 2 x CH CH2 in CH2CH5 2 x CH in CH3CHCH2 2 x CH in (CH)2CHCH3 2 x CH in (CH2)2CHCH five neighbors CH2 or 2 x CH CH2 in CH3CH2CH2 CH3 in CH3CH2 3 x CH in CHCH2 3 x CH in CHCHCH two neighbors CH3 or 3 x CH Totals 12  12 H

Sample Problem #1 Step 4: C-H Skeleton from 1H-NMR Step 4d: Select most likely implications, then total the atoms Most likely implication = least number of atoms Shift Splitting Integral # of H Implications 7.40-7.02 ppm multiplet 5 2.57 ppm triplet 2 1.64 ppm sextet 2 0.94 ppm triplet 3 5 H 2 H 3 H C6H5 (Ph; monosubstituted benzene ring) CH2 in CH2CH2 CH2 in CHCH2CH 2 x CH in CHCH2 2 x CH in CHCHCH two neighbors CH2 or 2 x CH CH2 in CH2CH5 CH2 in CH3CH2CH2 2 x CH in CH3CHCH2 2 x CH in (CH)2CHCH3 2 x CH in (CH2)2CHCH five neighbors CH2 or 2 x CH CH3 in CH3CH2 3 x CH in CHCH2 3 x CH in CHCHCH two neighbors CH3 or 3 x CH Totals 12  12 H C6H5 + CH2 + CH2 + CH3 = C9H12

Sample Problem #1 Step 5: Checks Step 5: Check to see that all atoms and DBE are used Atom Check Formula - atoms used = atoms left over C9H12 - C9H12 (from 1H-NMR) = all atoms used DBE Check DBE from formula - DBE used = DBE left over 4 - 4 (benzene ring) = all DBE used

Sample Problem #1 Step 6: Assembly Step 6: Now to put it all together... Ph CH2 of CH2CH2 CH2 of CH2CH2CH3 CH3 of CH3CH2 Pieces: CH2CH2CH3 Step 7: Check your work! Formula Functional groups Number of signals Splitting How to assemble? Pay attention to splitting patterns All pieces form one molecule Trial and error Pentavalent carbons XXXXXXXXXXXXXXX Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice

Solving Spectroscopy Problems Part 2 Lecture Supplement page 166 + +

Implications, Blind Men, and Elephants An NMR Fable + + + Ph- + -CH2CH2- + CH3CH2CH2- + -CH3 http://en.wikipedia.org/wiki/Blind_men_and_an_elephant

Sample Problem #2 Steps 1 and 2: Formula and DBE Mass spectrum m/z = 118 (M; 100%) Molecular mass (lowest mass isotopes) = 118 Even number of nitrogen atoms 5.7% / 1.1% = 5.2 m/z = 119 (5.7%) Five carbon atoms m/z = 120 (0.63%) No sulfur, chlorine, or bromine Formula 118 - (5 x 12) = 58 amu for oxygen, nitrogen, and hydrogen Usual procedure gives three possible formulas: C5H10O3 DBE = 1 One ring or one pi bond C5H14N2O Rejected: 14H does not fit NMR integration (sum of integrals = 5.0) C5H2N4 Rejected: More than two signals in NMR; no oxygen for C=O (IR zone 4)

Sample Problem #2 Step 3: Functional Groups from IR 100 Transmittance (%) 4000 3000 2000 1500 1000 Stretching frequency (cm-1) Zone 1 (3700-3200 cm-1) C5H10O3 DBE = 1 Alcohol O-H: Amide/amine N-H: Terminal alkyne C-H: Absent - no strong peak Absent - no peak; no nitrogen in formula Absent - no peak; no CC peak in zone 3

Sample Problem #2 Step 3: Functional Groups from IR 100 Transmittance (%) 4000 3000 2000 1500 1000 Stretching frequency (cm-1) Zone 2 (3200-2700 cm-1) C5H10O3 DBE = 1 Aryl/vinyl sp2 C-H: Alkyl sp3 C-H: Aldehyde C-H: Carboxylic acid O-H: Absent - no peaks > 3000 cm-1 Present - peaks < 3000 cm-1 Absent - no peak ~2700 cm-1 Absent - not broad enough

Sample Problem #2 Step 3: Functional Groups from IR 100 Transmittance (%) 4000 3000 2000 1500 1000 Stretching frequency (cm-1) Zone 3 (2300-2000 cm-1) C5H10O3 DBE = 1 Alkyne CC: Nitrile CN: Absent - no peak; not enough DBE Absent - no peak; no nitrogen in formula

Sample Problem #2 Step 3: Functional Groups from IR 100 Transmittance (%) 1741 cm-1 4000 3000 2000 1500 1000 Stretching frequency (cm-1) Zone 4 (1850-1650 cm-1) C5H10O3 DBE = 1 C=O: Present 1741 cm-1 = ester (1750-1735 cm-1) ketone (1750-1705 cm-1) aldehyde (1740-1720 cm-1)

Sample Problem #2 Step 3: Functional Groups from IR 100 Transmittance (%) 4000 3000 2000 1500 1000 Stretching frequency (cm-1) Zone 5 (1680-1450 cm-1) C5H10O3 DBE = 1 Benzene ring C=C: Alkene C=C: Absent - no peak; not enough DBE Absent - no peak; not enough DBE for C=O plus alkene

Sample Problem #2 Step 4: C-H Skeleton from 1H-NMR 1H-NMR: 4.22 ppm (triplet; integral = 1.0), 3.59 ppm (triplet; integral = 1.0), 3.39 ppm (singlet; integral = 1.5), 2.09 ppm (singlet; integral = 1.5) Step 4a: Copy NMR data to table Shift Splitting Integral # of H Implications 4.22 ppm triplet 1.0 3.59 ppm triplet 1.0 3.39 ppm singlet 1.5 2.09 ppm singlet 1.5

Sample Problem #2 Step 4: C-H Skeleton from 1H-NMR Step 4b: Divide hydrogens according to integrals Shift Splitting Integral # of H Implications 4.22 ppm triplet 1.0 3.59 ppm triplet 1.0 3.39 ppm singlet 1.5 2.09 ppm singlet 1.5 2 H 3 H Totals 5  10 H

Sample Problem #2 Step 4: C-H Skeleton from 1H-NMR Step 4c: Combine splitting with number of hydrogens to get implications Shift Splitting Integral # of H Implications 2 H 3 H CH2 in CH2CH2 CH2 in CHCH2CH 2 x CH in CHCH2 2 x CH in CHCHCH 4.22 ppm triplet 1.0 3.59 ppm triplet 1.0 3.39 ppm singlet 1.5 2.09 ppm singlet 1.5 two neighbors CH2 or 2 x CH CH2 in CH2CH2 CH2 in CHCH2CH 2 x CH in CHCH2 2 x CH in CHCHCH two neighbors CH2 or 2 x CH CH3 or 3 x CH no neighbors CH3 or 3 x CH CH3 or 3 x CH no neighbors CH3 or 3 x CH Totals 5  10 H

Sample Problem #2 Step 4: C-H Skeleton from 1H-NMR Step 4d: Select most likely implications, then total the atoms Most likely implication = least number of atoms Shift Splitting Integral # of H Implications 2 H 3 H CH2 in CH2CH2 CH2 in CHCH2CH 2 x CH in CHCH2 2 x CH in CHCHCH 4.22 ppm triplet 1.0 3.59 ppm triplet 1.0 3.39 ppm singlet 1.5 2.09 ppm singlet 1.5 two neighbors CH2 or 2 x CH CH2 in CH2CH2 CH2 in CHCH2CH 2 x CH in CHCH2 2 x CH in CHCHCH two neighbors CH2 or 2 x CH CH3 or 3 x CH no neighbors CH3 or 3 x CH CH3 or 3 x CH no neighbors CH3 or 3 x CH Totals 5  10 H CH2 + CH2 + CH3 + CH3 = C4H10

Sample Problem #2 Step 5: Checks Step 5: Check to see that all atoms and DBE are used Atom Check Formula - atoms used = atoms left over C5H10O3 - C4H10 (from 1H-NMR) - C=O (ester or ketone from IR) = 2 O DBE Check DBE from formula - DBE used = DBE left over 1 - 1 (C=O) = all DBE used

Sample Problem #2 Step 6: Assembly Pieces CH2 in CH2CH2 CH3 C=O (ester or ketone) O CH2CH2 CH3C=O CH3O Assembly Coupling suggests 2 x CH2 join to form CH2CH2 CH3 (singlet in NMR) cannot be in CH3CH2CH2 Leaves CH3O and CH3C=O Typically 3.8 ppm  Typically 2.0-2.3 ppm  3.39 ppm observed shift 2.09 ppm observed shift

Sample Problem #2 Step 6: Assembly Pieces CH2CH2 CH3O CH3C=O (ester or ketone) O Three ways to assemble these pieces: X Does not use all pieces Observed CH2 chemical shifts = 4.22 and 3.59 ppm Typical OCH2 chemical shift = 3.6 - 4.6 ppm Typical O=CCH2 chemical shift = 2.2 - 3.0 ppm More consistent with middle structure Step 7: Check your work Left as a student exercise

Sample Problem #3 Steps 1 and 2: Formula and DBE Mass spectrum m/z = 87 (M; 100%) m/z = 88 (4.90%) m/z = 89 (0.22%) Molecular mass (lowest mass isotopes) = 87 Odd number of nitrogen atoms 4.9% / 1.1% = 4.5 Four or five carbon atoms No sulfur, chlorine, or bromine Formula Usual procedure gives two formula candidates: C4H9NO DBE = 1 One ring or one pi bond C5H13N Rejected: Does not fit 1H-NMR integration (4:4:1 Integral sum = 9)

Sample Problem #3 Step 3: Functional Groups from IR 100 Transmittance (%) 4000 3000 2000 1500 1000 Stretching frequency (cm-1) Zone 1 (3700-3200 cm-1) C4H9NO DBE = 1 Alcohol O-H: Amide/amine N-H: Terminal alkyne C-H: Present? - weaker than usual Present? Absent - not enough DBE; no CC in zone 3

Sample Problem #3 Step 3: Functional Groups from IR 100 Transmittance (%) 4000 3000 2000 1500 1000 Stretching frequency (cm-1) Zone 2 (3200-2700 cm-1) C4H9NO DBE = 1 Aryl/vinyl sp2 C-H: Alkyl sp3 C-H: Aldehyde C-H: Carboxylic acid O-H: Absent - no peaks > 3000 cm-1 Present - peaks < 3000 cm-1 Absent - ~2700 cm-1 present but no C=O in zone 4 Absent - no C=O in zone 4; not broad enough

Sample Problem #3 Step 3: Functional Groups from IR 100 Transmittance (%) 4000 3000 2000 1500 1000 Stretching frequency (cm-1) Zone 3 (2300-2000 cm-1) C4H9NO DBE = 1 CC: CN: Absent - no peaks; not enough DBE

Sample Problem #3 Step 3: Functional Groups from IR 100 Transmittance (%) 4000 3000 2000 1500 1000 Stretching frequency (cm-1) Zone 4 (1850-1650 cm-1) C4H9NO DBE = 1 C=O: Absent - no peaks

Sample Problem #3 Step 3: Functional Groups from IR 100 Transmittance (%) 4000 3000 2000 1500 1000 Stretching frequency (cm-1) Zone 5 (1680-1450 cm-1) C4H9NO DBE = 1 Benzene ring: Alkene C=C: Absent - no peak ~1600 cm-1; not enough DBE Absent - no peak ~1600 cm-1

Sample Problem #3 Step 4: C-H Skeleton from 1H-NMR 1H-NMR: 3.67 ppm (triplet; integral = 4.0), 2.86 ppm (triplet; integral = 4.0), and 2.59 ppm (singlet; integral = 1.0). Step 4a: Copy NMR data to table Shift Splitting Integral # of H Implications 3.67 ppm triplet 4.0 2.86 ppm triplet 4.0 2.59 ppm singlet 1.0

Sample Problem #3 Step 4: C-H Skeleton from 1H-NMR 1H-NMR: 3.67 ppm (triplet; integral = 4.0), 2.86 ppm (triplet; integral = 4.0), and 2.59 ppm (singlet; integral = 1.0). Step 4b: Divide hydrogens according to integrals Shift Splitting Integral # of H Implications 3.67 ppm triplet 4.0 2.86 ppm triplet 4.0 2.59 ppm singlet 1.0 4 H 1 H Total = 9.0  9 H

Sample Problem #3 Step 4: C-H Skeleton from 1H-NMR 1H-NMR: 3.67 ppm (triplet; integral = 4.0), 2.86 ppm (triplet; integral = 4.0), and 2.59 ppm (singlet; integral = 1.0). Step 4c: Combine splitting with number of hydrogens to get implications Shift Splitting Integral # of H Implications 2 x CH2 in CH2CH2 2 x CH2 in CHCH2CH 4 x CH in CHCH2 4 x CH in CHCHCH 3.67 ppm triplet 4.0 2.86 ppm triplet 4.0 2.59 ppm singlet 1.0 4 H 1 H two neighbors 2 x CH2 or 4 x CH 2 x CH2 in CH2CH2 2 x CH2 in CHCH2CH 4 x CH in CHCH2 4 x CH in CHCHCH two neighbors 2 x CH2 or 4 x CH CH or NH or OH no neighbors Total = 9.0  9 H

Sample Problem #3 Step 4: C-H Skeleton from 1H-NMR 1H-NMR: 3.67 ppm (triplet; integral = 4.0), 2.86 ppm (triplet; integral = 4.0), and 2.59 ppm (singlet; integral = 1.0). Step 4d: Select most likely implications, then total the atoms Most likely implication = least number of atoms Shift Splitting Integral # of H Implications 2 x CH2 in CH2CH2 2 x CH2 in CHCH2CH 4 x CH in CHCH2 4 x CH in CHCHCH 3.67 ppm triplet 4.0 2.86 ppm triplet 4.0 2.59 ppm singlet 1.0 4 H 1 H two neighbors 2 x CH2 or 4 x CH 2 x CH2 in CH2CH2 2 x CH2 in CHCH2CH 4 x CH in CHCH2 4 x CH in CHCHCH two neighbors 2 x CH2 or 4 x CH X CH or NH or OH IR more consistent with NH than OH no neighbors Total = 9.0  9 H (2 x CH2) + (2 x CH2) + NH = C4H9N

Sample Problem #3 Step 5: Checks Step 5: Check to see that all atoms and DBE are used Atom Check Formula - atoms used = atoms left over C4H9NO - C4H9N (from 1H-NMR) = one oxygen Not part of a functional group that appears in IR or 1H-NMR: An ether DBE Check DBE from formula - DBE used = DBE left over 1 - 0 = one DBE C=O, C=C, C=N absent in IR Therefore DBE = ring

Sample Problem #3 Step 6: Assembly Pieces 2 x CH2 in CH2CH2 NH O (ether) one ring 2 x CH2CH2 Assembly Splitting pattern requires two sets of 2 x CH2 to become two equivalent CH2CH2 Remaining pieces can only be assembled in one way: