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9 Solving Spectroscopy Problems pt1

Organic Chemistry: Structure and Function

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1 Solving Spectroscopy Problems Part 1 Lecture Supplement: Take one handout from the stage
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2 Solving Spectroscopy Problems w do I deduce structure from spectra? ogical, orderly procedure ompare information between spectra when necessary onservative analysis: do not discard possibilities until 100% sure ocedure MS gives formula ormula gives DBE se formula plus DBE to guide IR analysis of functional groups MR gives skeleton ssemble the pieces: use logic, and sometimes guess heck your work!
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3 Sample Problem #1 Steps 1 and 2: Formula and DBE Mass spectrum m/z = 120 (M; 100%) Molecular mass (lowest mass isotopes) = 120 Even number of nitrogen atoms 9.8% / 1.1% = 8.9 m/z = 121 (9.8%) m/z = 122 (0.42%) Formula 120 - (9 x 12) = 12 amu for oxygen, nitrogen, and hydrogen Not enough amu for nitrogen or oxygen DBE DBE = C - (H/2) + (N/2) + 1 = 9 - (12/2) + (0/2) + 1 = 4 Formula = C 9 H 12 Four rings and/or pi bonds Possible benzene ring Nine carbon atoms No sulfur, chlorine, or bromine
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4 Sample Problem #1 Step 3: Functional Groups from IR Alcohol O-H: Amide/amine N-H: C-H: Absent - no peak; no oxygen in formula Absent - no peak; no nitrogen in formula Absent - no peak; no C C peak in zone 3 Zone 1 (3700-3200 cm -1 )
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5 Sample Problem #1 Step 3: Functional Groups from IR Aryl/vinyl sp 2 C-H: Alkyl sp 3 C-H: Aldehyde C-H: Carboxylic acid O-H: Zone 2 (3200-2700 cm -1 ) 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
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6 Sample Problem #1 Step 3: Functional Groups from IR Alkyne C C: Nitrile C N: Zone 3 (2300-2000 cm -1 ) Absent - no peak Absent - no peak; no nitrogen in formula
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7 Sample Problem #1 Step 3: Functional Groups from IR C=O: Zone 4 (1850-1650 cm -1 ) Absent - no strong peak; no oxygen in formula
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8 Sample Problem #1 Step 3: Functional Groups from IR Benzene ring C=C: Alkene C=C: Zone 5 (1680-1450 cm -1 ) Present - peaks at ~1610 cm -1 and ~1500 cm -1 Absent - not enough DBE for alkene plus benzene
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9 Sample Problem #1 Step 4: C-H Skeleton from 1 H-NMR hift Splitting Integral # of H Implications Step 4a: Copy NMR data to table 40-7.02 ppm multiplet 5 57 ppm triplet 2 64 ppm sextet 2 94 ppm triplet 3 1 H-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).
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10 Sample Problem #1 Step 4: C-H Skeleton from 1 H-NMR hift Splitting Integral # of H Implications tep 4b: Divide hydrogens according to integrals 0-7.02 ppm multiplet 5 7 ppm triplet 2 4 ppm sextet 2 triplet 3 Totals 12 12 H 5 H 2 H 2 H 3 H
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11 Sample Problem #1 Step 4: C-H Skeleton from 1 H-NMR hift Splitting Integral # of H Implications p 4c: Combine splitting with number of hydrogens to get implications 0-7.02 ppm multiplet 5 7 ppm triplet 2 4 ppm sextet 2 triplet 3 Totals 12 12 H 5 H 2 H 2 H 3 H C 6 H 5 (benzene ring) CH 2 in CH 2 CH 2 CH 2 in CHCH 2 CH 2 x CH in CH CH 2 2 x CH in CHCH CH two neighbors CH 2 or 2 x CH five neighbors CH 2 or 2 x CH two neighbors CH 3 or 3 x CH CH 2 in CH 2 CH 5 2 x CH in CH 3 CH CH 2 2 x CH in (CH) 2 CH CH 3 2 x CH in (CH 2 ) 2 CH CH CH 3 in CH 3 CH 2 3 x CH in CH CH 2 3 x CH in CHCH CH CH 2 in CH 3 CH 2 CH 2
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12 Sample Problem #1 Step 4: C-H Skeleton from 1
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