Solving Spectroscopy Problems Part 1

1
Solving Spectroscopy Problems Part 1

• Lecture Supplement
• Take one handout from the stage

2
Solving Spectroscopy Problems

• How do I 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 use logic, and sometimes
  guess
• Check your work!

3
Sample Problem 1Steps 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)
Nine carbon atoms
m/z 122 (0.42)
No sulfur, chlorine, or bromine
Formula 120 - (9 x 12) 12 amu for oxygen,
nitrogen, and hydrogen Not enough amu for
nitrogen or oxygen
Formula C9H12
DBE DBE C - (H/2) (N/2) 1 9 -
(12/2) (0/2) 1 4
Four rings and/or pi bonds Possible benzene ring
4
Sample Problem 1Step 3 Functional Groups from
IR
Zone 1 (3700-3200 cm-1)
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
5
Sample Problem 1Step 3 Functional Groups from
IR
Zone 2 (3200-2700 cm-1)
Aryl/vinyl sp2 C-H Alkyl sp3 C-H Aldehyde
C-H Carboxylic acid O-H
Present - peaks gt 3000 cm-1 Present - peaks lt
3000 cm-1 Absent - no peak 2700 cm-1 no CO in
zone 4 Absent - not broad enough no CO in zone 4
6
Sample Problem 1Step 3 Functional Groups from
IR
Zone 3 (2300-2000 cm-1)
Absent - no peak Absent - no peak no nitrogen in
formula
Alkyne C?C Nitrile C?N
7
Sample Problem 1Step 3 Functional Groups from
IR
Zone 4 (1850-1650 cm-1)
CO
Absent - no strong peak no oxygen in formula
8
Sample Problem 1Step 3 Functional Groups from
IR
Zone 5 (1680-1450 cm-1)
Benzene ring CC Alkene CC
Present - peaks at 1610 cm-1 and 1500
cm-1 Absent - not enough DBE for alkene plus
benzene
9
Sample Problem 1Step 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
10
Sample Problem 1Step 4 C-H Skeleton from 1H-NMR
Step 4b Divide hydrogens according to integrals
Shift Splitting
Integral of H Implications
5 H 2 H 2 H 3 H
7.40-7.02 ppm multiplet 5 2.57 ppm
triplet 2 1.64 ppm sextet 2 0.94 ppm
triplet 3
Totals 12
? 12 H
11
Sample Problem 1Step 4 C-H Skeleton from 1H-NMR
Step 4c Combine splitting with number of
hydrogens to get implications
Shift Splitting
Integral of H Implications
5 H 2 H 2 H 3 H
C6H5 (benzene ring)
7.40-7.02 ppm multiplet 5 2.57 ppm
triplet 2 1.64 ppm sextet 2 0.94 ppm
triplet 3
CH2 in CH2CH2 CH2 in CHCH2CH
2 x CH in CHCH2 2 x CH in CHCHCH
CH2 or 2 x CH
two neighbors
CH2 in CH2CH5
2 x CH in CH3CHCH2 2 x CH in (CH)2CHCH3 2 x CH in
(CH2)2CHCH
CH2 or 2 x CH
CH2 in CH3CH2CH2
five neighbors
CH3 in CH3CH2
3 x CH in CHCH2 3 x CH in CHCHCH
two neighbors
CH3 or 3 x CH
Totals 12
? 12 H
12
Sample Problem 1Step 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
5 H 2 H 2 H 3 H
C6H5 (benzene ring)
7.40-7.02 ppm multiplet 5 2.57 ppm
triplet 2 1.64 ppm sextet 2 0.94 ppm
triplet 3
CH2 in CH2CH2 CH2 in CHCH2CH
2 x CH in CHCH2 2 x CH in CHCHCH
CH2 or 2 x CH
two neighbors
CH2 in CH2CH5 CH2 in CH3CH2CH2
2 x CH in CH3CHCH2 2 x CH in (CH)2CHCH3 2 x CH in
(CH2)2CHCH
CH2 or 2 x CH
five neighbors
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
13
Sample Problem 1Step 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
14
Sample Problem 1Step 6 Assembly
Now to put it all together... Pieces
C6H5 CH2 of CH2CH2 CH2 of CH2CH2CH3 CH3 of CH3CH2
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

15
Sample Problem 2Steps 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 two possible formulas
• C5H10O3
• C5H2N4

DBE 1 One ring or one pi bond
Rejected More than 2 signals in NMR
No oxygen for CO (IR zone 4)
16
Sample Problem 2Step 3 Functional Groups from
IR
Zone 1 (3700-3200 cm-1)
Alcohol O-H Amide/amine N-H ?C-H
Absent - no strong peak Absent - no peak no
nitrogen in formula Absent - no peak no C?C peak
in zone 3
17
Sample Problem 2Step 3 Functional Groups from
IR
Zone 2 (3200-2700 cm-1)
Aryl/vinyl sp2 C-H Alkyl sp3 C-H Aldehyde
C-H Carboxylic acid O-H
Absent - no peaks gt 3000 cm-1 Present - peaks lt
3000 cm-1 Absent - no peak 2700 cm-1 Absent -
not broad enough
18
Sample Problem 2Step 3 Functional Groups from
IR
Zone 3 (2300-2000 cm-1)
Absent - no peak not enough DBE Absent - no
peak no nitrogen in formula
Alkyne C?C Nitrile C?N
19
Sample Problem 2Step 3 Functional Groups from
IR
Zone 4 (1850-1650 cm-1)
CO
Present
1741 cm-1 ester (1750-1735 cm-1)
ketone (1750-1705 cm-1)
aldehyde (1740-1720 cm-1)
20
Sample Problem 2Step 3 Functional Groups from
IR
Zone 5 (1680-1450 cm-1)
Benzene ring CC Alkene CC
Absent - no peak not enough DBE Absent - no
peak not enough DBE for CO plus alkene
21
Solving Spectroscopy Problems Part 2

• Lecture Supplement
• Take one handout from the stage

22
Sample Problem 2Summary
Information so far...

• Formula C5H10O3
• DBE 1 One ring or one pi bond
• Present from IR

Alkyl sp3 C-H CO ester or ketone
23
Sample Problem 2Step 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
24
Sample Problem 2Step 4 C-H Skeleton from 1H-NMR
Step 4b Divide hydrogens according to integrals
Shift Splitting
Integral of H Implications
2 H 2 H 3 H 3 H
4.22 ppm triplet 1.0 3.59 ppm
triplet 1.0 3.39 ppm singlet
1.5 2.09 ppm singlet 1.5
Totals 5
? 10 H
25
Sample Problem 2Step 4 C-H Skeleton from 1H-NMR
Step 4c Combine splitting with number of
hydrogens to get implications
Shift Splitting
Integral of H Implications
CH2 in CH2CH2 CH2 in CHCH2CH
2 x CH in CHCH2 2 x CH in CHCHCH
2 H 2 H 3 H 3 H
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
26
Sample Problem 2Step 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
CH2 in CH2CH2 CH2 in CHCH2CH
2 x CH in CHCH2 2 x CH in CHCHCH
2 H 2 H 3 H 3 H
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
27
Sample Problem 2Step 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) - CO
(ester or ketone from IR) 2 O DBE Check DBE
from formula - DBE used DBE left over 1 - 1
(CO) all DBE used
28
Sample Problem 2Step 6 Assembly
Pieces CH2 in CH2CH2 CH2 in CH2CH2 CH3 CH3 CO
(ester or ketone) O O

• Assembly Logic and clues combined with a few
  guesses
• Coupling suggests 2 x CH2 join to form CH2CH2

• CH3 (singlet in NMR) cannot be in CH3CH2CH2
• Leaves CH3O and
  CH3CO

2.09 ppm observed shift
3.39 ppm observed shift
29
Sample Problem 2Step 6 Assembly
Pieces CH2CH2 CH3O CH3CO (ester or
ketone) O Three ways to assemble these pieces
Step 7 Check your work Left as a student
exercise
X
Does not use all pieces

• Observed CH2 chemical shifts 3.59 and 3.39 ppm
• Typical OCH2 chemical shift 3.3-4.1 ppm
• Typical OCCH2 chemical shift 2.2-3.0 ppm
• More consistent with middle structure

30
Sample Problem 3Steps 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 only one acceptable
  formula C4H9NO
• DBE 4 - (9/2) (1/2) 1
• 1
• One ring or one pi bond

31
Sample Problem 3Step 3 Functional Groups from
IR
Zone 1 (3700-3200 cm-1) Alcohol O-H Amide/amine
N-H ?C-H
Present? - weaker than usual Present? Absent -
not enough DBE no C?C in zone 3
32
Sample Problem 3Step 3 Functional Groups from
IR
Zone 2 (3200-2700 cm-1)
Aryl/vinyl sp2 C-H Alkyl sp3 C-H Aldehyde
C-H Carboxylic acid O-H
Absent - no peaks gt 3000 cm-1 Present - peaks lt
3000 cm-1 Absent - 2700 cm-1 present but no CO
in zone 4 Absent - no CO in zone 4 not broad
enough
33
Sample Problem 3Step 3 Functional Groups from
IR
Zone 3 (2300-2000 cm-1)
Absent - no peaks not enough DBE
C?C and C?N
34
Sample Problem 3Step 3 Functional Groups from
IR
Zone 4 (1850-1650 cm-1)
CO
Absent - no peaks
35
Sample Problem 3Step 3 Functional Groups from
IR
Zone 5 (1680-1450 cm-1)
Benzene ring Alkene CC
Absent - no peak 1600 cm-1 not enough
DBE Absent - no peak 1600 cm-1
36
Sample Problem 3Step 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
37
Sample Problem 3Step 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
4 H 4 H 1 H
3.67 ppm triplet 4.0 2.86
ppm triplet 4.0 2.59 ppm
singlet 1.0
Total 9.0
? 9 H
38
Sample Problem 3Step 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
4 H 4 H 1 H
3.67 ppm triplet 4.0 2.86
ppm triplet 4.0 2.59 ppm
singlet 1.0
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
39
Sample Problem 3Step 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
4 H 4 H 1 H
3.67 ppm triplet 4.0 2.86
ppm triplet 4.0 2.59 ppm
singlet 1.0
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
IR more consistent with NH than OH
no neighbors
Total 9.0
? 9 H
(2 x CH2) (2 x CH2) NH C4H9N
40
Sample Problem 3Step 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
CO, CC, CN absent in IR Therefore DBE ring
41
Sample Problem 3Step 6 Assembly
Pieces 2 x CH2 in CH2CH2 2 x CH2 in CH2CH2 NH O
(ether) one ring

• Assembly
• Splitting pattern requires two sets of 2 x CH2 to
  become two equivalent CH2CH2

• Remaining pieces can only be assembled in one way