Fall 2016
http://ww2.chemistry.gatech.edu/~collard/SpecID/fall2016.html
https://t-square.gatech.edu/portal (for grades)
CHEM 6371 – Spectroscopic Identification of Organic Compounds
CHEM 4341 – Applied Spectroscopy
office
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22-Aug
24-Aug
26-Aug
29-Aug
31-Aug
2-Sep
5-Sep
7-Sep
9-Sep
12-Sep
14-Sep
16-Sep
19-Sep
21-Sep
23-Sep
26-Sep
28-Sep
30-Sep
3-Oct
5-Oct
7-Oct
10-Oct
12-Oct
14-Oct
17-Oct
19-Oct
21-Oct
24-Oct
26-Oct
28-Oct
31-Oct
2-Nov
4-Nov
7-Nov
9-Nov
11-Nov
14-Nov
16-Nov
18-Nov
21-Nov
23-Nov
25 Nov
28-Nov
30-Nov
Intro-1
Intro-2
Intro-3
Intro-4
Intro-5 HW 1
Intro-6
LABOR DAY
Intro-7
Intro-8 HW 2
EXAM1
IR-1
Infrared
IR-2
Spectroscopy
Pavia Chap 2
IR-3
IR-4
IR-5 HW 3
EXAM 2
Mass SpectrometryMS-1
Pavia Chap 8
MS-2
MS-3
MS-4 HW 4
MS-5
DC
DC
DC
DC
DC
DC
MS-6 HW 5
MS-7
EXAM 3
1
H and 13C NMR NMR-1
NMR-2
Spectroscopy
Pavia Chap 4,5,6 NMR-3
NMR-4
NMR-5
NMR-6 HW 6
NMR-7
NMR-8
NMR-9
NMR-10 HW 7
EXAM 4
Adv-1
2D NMR
Pavia Chap 10
Adv-2
Adv-3
Adv-4 HW 8
DC
DC
Introduction
Your intro organic
textbook and
Pavia Chap 1,3
DC
DC
F-DC
F-DC
1
F-DC
H NMR and 13C NMR
Problems
Reading (4e)
1:1-5+ intro text
1:6;8.7+ intro text
2:intro -7+ intro text
3:1-9+ intro text
3:10-17+ intro text
3:19;4:14.2A+ intro text
Q3.1-12
Fundamental principles, using basic information to solve structures
JS
JS
JS
JS
JS
F- JS
F- JS
CH, CC bonds
OH, NH bonds
CO bonds
C=O and CN bonds, reporting IR data
Problems
2:8-11
2:12;14F,15
2:13
2:14,16-21
Q2.1-11
IR, plus use of basic principles of IR and NMR to solve structures
DC
DC
DC
DC
DC
F-DC
F-DC
MID-SEMESTER BREAK
F 2-Dec
M 5-Dec More problems
W 14-Dec (8:00-10:50 a.m.)
Topic
Introduction
Elemental analysis, sites of unsaturation
Mass spectrometry
Infrared spectroscopy
1
H NMR
1
H NMR
F-DC
MS techniques
Fragmentation, Hydrocarbons
Alcohols, ethers and amines: α-cleavage
C=O compounds
Halo compounds, reporting MS data
8:intro-6
8:8A-C,I-M
8:8N-O,T
8:8P-U
8:8V
Problems
Problems
Q8.1-15
MS, plus use of basic principles of IR and NMR to solve structures
LG
DC
DC
LG
LG
LG
LG
LG
LG
LG
F-DC
F-DC
Tu-LG
NMR theory
1
H Chemical Shift
1
H Chemical Shift
1
H Multiplicity
1
H Multiplicity, reporting NMR data
13
C Chemical shift, shift calculations
Decoupling, NOE, etc
Edited spectra, APT, DEPT
Problems
Problems
6:1-5
6:6-10
5:1-6
5:7-11
4:1-4,11-16
4:5-9; 6.11
4:10; 10.4-5
9:intro-21
Q9.22-43
NMR, and use of MS and IR to solve structures
LG
LG
LG
LG
Tu-LG
LG
LG
Tu-LG
Tu-LG
2D NMR Theory
COSY
NOESY
HETCOR, HSQC
10:1,6
10:7
10.10
10.8-9
HMBC
Problems
Q10.1-11
THANKSGIVING
BREAK
Adv-5
Adv-6 HW 9
EXAM 5
Problems
FINAL
Basic and Adv. NMR, MS and IR to solve
structures
LG
Tu-LG
Problems
Comprehensive
INSTRUCTORS
David M. Collard
[email protected]
Leslie Gelbaum
[email protected]
Jessie Sandridge
[email protected]
Office hours (only for times when teaching during the semester – see schedule; or by appointment)
Friday noon-1 p.m.
Tuesday 1:00 - 2:00 p.m.
Friday noon-1p.m.
MoS&E 2100J
MS&E 113
MoSE 1226
LECTURES
Come prepared to ask and answer questions!
MWF, 11:05-11:55
College of Computing (CoC) 17
WORK PROBLEMS
Work as many problems as possible, from the notes, from the book, from other sources.
REQUIRED TEXTBOOK
Introduction to Spectroscopy, 4e, Donald L. Pavia, Gary M. Lampman, George S. Kriz, and James A. Vyvyan;
Brooks Cole; ISBN-10 / ASIN: 0495114782; ISBN-13 / EAN: 9780495114789.
Note: Section numbers and problems in older and international editions vary from those in the 4th edition (U.S.)
GRADES
Graded Assignments
Exam 1
Exam 2
Exam 3
Exam 4
Exam 5
Homework
Final
Topic
Fundamental principles and using basic information to solve structures
IR, plus use basic principles of MS and NMR to solve structures
MS, plus use of basic principles of IR and NMR to solve structures
NMR, plus use of MS and IR to solve structures
Basic and Advanced NMR techn., plus use of MS and IR to solve structures
Nine HW assignments (score normalized to 100 points)
Wed Dec 14 at 8:00 am-10:50 am: Comprehensive
100 points†
100 points†
100 points†
100 points†
100 points†
100 points
200 points
The lowest score of the five mid-term exams (†) will be dropped. If you miss an exam that score (0) will be
dropped. The course grade will be determined based on your score out of 700 points.
Typical Grade Cut-offs
A: 85%+
B: 70-84.99
C: 60-69.99
D: 50-59.99
RETURNED WORK AND REGRADES
All graded assignments will be returned as soon as possible, usually within a week. If you want any work
regraded you must make a written request and return the assignment within one week. Work will not be regraded
after this deadline.
LECTURE ATTENDANCE
It is strongly recommended that you attend all lectures.
MATERIAL COVERED, KEEPING UP, WORKING PROBLEMS, STUDENT RESPONSIBILITIES
You are responsible for all material presented in lectures and in assigned readings. You are also responsible for
announcements made in class or by email. You must check your gatech.edu email account on a regular basis.
Note: there are potential problems associated with automatic forwarding of messages from your gatech mail to
other email addresses; check your gatech account even if you have it set up to forward email elsewhere.
By the end of each section you should have completed all reading associated with that section, and worked all of
the end-of-chapter problems and any additional problems which have been distributed. These questions should
form the basis for discussion with your peers, and serve as a guide for the types of questions to appear on
examinations (some of these questions might even appear on the exams!) Do not submit answers to these
problems, they will not be graded.
EXAMS: SCHEDULE, MAKE-UPS AND DROPS
You must take the exam at the assigned time. All exams are closed to textbooks, class-notes and electronic
devices (unless otherwise stated prior to the exam). Tables of NMR, IR, MS data, along with a periodic table, will
be provided.
The only valid reasons for missing an exam are illness and official GA Tech business. Make-up exams can only
be given if advance notification is given or upon presentation of a doctor’s note. All make-up exams must be
administered before the exams are returned to the class (typically before the next class). Exams not made-up by
this time for any reason will receive a score of zero and will be the drop grade for the class (i.e., it will be the
lowest score).
WORK PROBLEMS
Work as many problems as possible, from the notes, from the book, from other sources.
WORKING IN GROUPS
Most learning takes place outside of the classroom. Although lectures should provide a framework for learning
and put things in perspective, working through the textbook and solving the problems is when you will come to
terms with the material. We encourage you to work together on these reading and problem assignments. For most
students, it is actually unwise to try to work alone. Although you might study in groups, remember that you are
ultimately responsible for your learning. Everybody can benefit from team work. If you are struggling with the
material you stand to learn a lot; if you are a “spectroscopy wizard” you also stand to learn from the challenge of
presenting your understanding to others - you will learn through teaching.
Office hours are available for individual instruction. No new information will be introduced during office hours.
Come prepared to ask and answer problems.
COMPETITION AND GRADING
Formal education often puts students in competition with each other for good grades. We do not believe that
competition for grades, and the exclusion of everything else, is the most effective way to foster student
development. Although grades will be assigned based on a numerical score, which judges attainment on
exams/homework, we hope that the course is structured such that if you show a desire to learn, put the effort in,
and have the intellectual ability, you can get the grade you want.
CANCELLATION OF CLASSES
If class is cancelled by Georgia Institute of Technology owing to campus closing, the entire schedule for the
course will be delayed by one lecture. This will move all exams and the homework due dates back by one lecture.
TIME COMMITMENTS
We all have extensive demands on our time. For each hour of lecture you should aim to put in at least another two
hours of your own time. You will need to spend more time preparing for exams. Some students will require more,
some less.
WORK PROBLEMS
Work as many problems as possible, from the notes, from the book, from other sources.
SOME STUDY TIPS
Work Problems. Understand and Rationalize. Read the text, prepare your own summaries. Study in groups.
Keep up to date! Ask Questions!! Work more problems.
STUDENT CLASS ACCOMMODATIONS
Students with disabilities who require reasonable accommodations to fully participate in course activities or meet
course requirements are encouraged to register with ODS-Office of Disability Services at (404)894-2564 or www.
http://disabilityservices.gatech.edu/
Contact the instructors within the first two weeks if you expect to take exams with ODS. Please send reminders
one week before each exam.
GEORGIA TECH ACADEMIC HONOR CODE
Please visit www.honor.gatech.edu
For Graded Homework Assignments: You may work with others in developing approaches to solve problems,
but submitted work must be in your own handwriting.
For Tests: Cheating from another person's exam and use of unauthorized materials are direct violations of the GT
Academic Honor Code, and will be dealt with accordingly.
For any questions involving these policies, please discuss them with the instructors or consult
www.honor.gatech.edu.
LASTLY
Work as many problems as possible, from the notes, from the book, from other sources; work hard; enjoy.
Introduction to Spectroscopy, 4e
Donald L. Pavia, Gary M. Lampman, George S. Kriz,
and James A. Vyvyan
Brooks Cole
ISBN-10 / ASIN: 0495114782
ISBN-13 / EAN: 9780495114789
The expensive 5th edition
COURSE OUTLINE
17 Aug - 04 Sept Introduction
Your undergraduate textbook and P(4,5) Chap 1,3
04 Sept
EXAM 1 - Fundamental principles and using basic information to solve
structures
9 - 21 Sept
21 Sept
Infrared Spectroscopy
P(4,5) Chap 2
EXAM 2 - IR, plus use of basic principles of IR and NMR to solve
structures
23 Sept -07 Oct
07 - Oct
Mass Spectrometry
P(4) Chap 8; P(5) Chap 3,4
EXAM 3 – MS, plus basic principles of IR and NMR to solve structures
14 Oct - 06 Nov
06 Nov
Fundamentals of 1H and 13C NMR Spectrometry P(4) Chap 3-6;P(5) 5-8
EXAM 4 NMR, plus use of MS and IR to solve structures
09 Nov - 23 Nov
23 Nov
2D, Advanced and multinuclear NMR
P(4) Chap 10; P(5) Chap 9
EXAM 5 - Basic and Advanced NMR techn., plus use of MS and IR to
solve structures
30 Nov - 04 Dec
More problems
W 09 Dec (8:00 – 10:50)
FINAL - Comprehensive
1
GRADING
Exam 1
Exam 2
Exam 3
Exam 4
Exam 5
Homework
100
100
100
100
100
100
Final
200
drop lowest score from E1-5
Course grade out of 700 points
AN APPROACH TO DETERMINING
ORGANIC STRUCTURES
Combustion analysis
Mass spectrum
MS isotope pattern
MS exact mass
Empirical formula
Molecular weight
Molecular formula
Sites of unsaturation
(pi bonds, rings)
Infrared spectrum
1H NMR
13C NMR
Functional groups
Structural features
fragments, connectivity
2
REQUIREMENTS FOR PROOF OF STRUCTURE
AND PURITY IN YOUR RESEARCH
New Compounds
IR: assign important peaks
1H NMR: assign all peaks
13C NMR: assign all peaks, account
for all carbon atoms
Advanced NMR techniques as
necessary
m.p. (range) or b.p. (range?)
Elemental analysis
(combustion analysis  0.4%)
or
High resolution mass spectrum and
chromatographic proof of purity
Previously Reported
Compounds
At a minimum:
1H NMR
m.p range
J. Org. Chem. guidelines
http://pubs.acs.org/paragonplus/submission/j
oceah/joceah_authguide.pdf
EMPIRICAL FORMULAS, MOLECULAR
FORMULAS, SITES OF UNSATURATION &
COMBUSTION ANALYSIS
3
MOLECULAR FORMULAS
Simply considering common valancies (C 4; H and Hal 1, O and S 2, N 3),
acylic alkanes (linear or branched) have the formula CnH2n+2
Alkenes and cycloalkanes have the formula CnH2n
For each ring or pi bond in a molecule there are two fewer hydrogen atoms than
expected for a non-cyclic alkane, so:
Number of pi bonds or rings = (2#C + 2 – #H) / 2
“sites of unsaturation”, “index of hydrogen deficiency” (IHD) or “sum of double bonds and
rings” (SODAR)
The same equation is true in the presence of oxygen or sulfur atoms:
For C,H,O,S:
SODAR = (2#C + 2 – #H) / 2
Each halogen atom present replaces a hydrogen atom:
SODAR = (2#C + 2 – (#H+#Hal) / 2
4
Each nitrogen atom requires addition of one hydrogen
For C,H,O,N,S, Hal:
Number of pi bonds or rings = (2#C + 2– #H – #Hal + #N) / 2
Alternative formula for SODAR= C-(#H+#Hal)/2+#N/2+1
Problems - How many rings or pi bonds are there in each of the following
compounds?
C 6H 6
C6H7BrO
C10H22N2
C10H13NO2
C21H35ClS2
5
Some consequences of valency
1. A hydrocarbon, or CHO-containing molecule, will have an even number of
hydrogen atoms. With n carbon atoms there cannot be more than 2n+2
hydrogen atoms.
2. A molecular formula cannot have an odd number of hydrogen atoms unless
there is an odd total number of halogen and nitrogen atoms [“the nitrogen
rule”].
Differentiating pi bonds and rings
C=C and C≡C bonds hydrogenated to C–C
C=N and N=O also subject to hydrogenation
Benzene rings and C=O bonds do not undergo hydrogenation under these
conditions.
Compound A, C7H14N2O3, undergoes hydrogenation to give a product with the
formula C7H18N2O3. How many rings are there in compound A? How many 
bonds?
6
COMBUSTION ANALYSIS
xCO2 + (y/2)H2O
CxHyOz + O2
Measure mass of CO2 and H2O formed from a known mass of
compound; data cited as mass% of each element present.
Generally do not measure mass %O.
e.g.,
C
H
Mass%
54.51
9.09
Mole Ratio
=
=
Empirical formula
O
Molecular Formula
could be…
7
Elemental analyses provided ±0.4 mass%. A combustion analysis (±0.4
mass%) might represent more than one possible empirical formula. Be
careful when rounding.
e.g., C, 91.51% H, 8.44%
 C:H ratio 1.000: 1.111
If assume emprical formula
then mass % would be
C1H1  C, 92.26 H, 7.74
C10H11  C, 91.55 H, 8.45
C9H10  C, 91.47 H, 8.53
This becomes a significant problem for larger molecules
e.g, C, 79.21 H, 10.65 could be
C30H49NO2
C, 79.07 H, 10.84 (N, 3.07; m = 455.7)
C30H47NO2
C, 79.42 H, 10.44 (N, 3.09 ; m = 453.7)
C29H47NO2
C, 78.86 H, 10.73 (N, 3.17 ; m = 441.7)
C30H47N3
C, 79.58 H, 10.82 (N, 9.60 ; m = 449.7)
C30H48N2O
C, 79.59 H, 10.69 (N, 6.19 ; m = 452.7)
For an accurate determination of empirical
formula from combustion analysis
- 2-5 mg for C,H,N
- Sample must be pure, dry
- Sample must fully combust
8
Problems - With the aid of a calculator:
(i) Calculate the elemental composition of:
(a) C21H25NO5
(b) C17H20BrNO4
Use atomic masses to 3 decimal places (C, 12.011, ….)
(ii) Calculate the empirical formula of a compounds that give the following
analyses:
(a)
(b) C 68.09%
TB-III-32A
1966-06
B. Moore
T. Brooking
X
C
71.23%
H
9.35%
9-10-09
CHEM, Georgia Tech
901 State St.
Atlanta, GA 30306-0400
9-8-09
H
8.64%
N
6.59%
Cl
16.57%
C,H
X
C,H
9-11-09
Formula  %mass calculators
fluorine.ch.man.ac.uk/research/analyse.php
www.calctool.org/CALC/chem/molecular/elemental
%mass  formula calculators
http://www.chemicalaid.com/tools/empirical.php
[NOTE: be careful with syntax; only gives one possible formula!]
9
MASS SPECTROMETRY
INSTRUMENTATION
M + e– (70 eV)
M +• + 2e–
Ease of removal of an electron:
n (non-bonding, i.e., an electron in a lone pair) > -bonding > -bonding
10
m/z = H2er2/2V
m is the mass
V is the accelerating voltage
r is the radius
H is the magnetic field strength
z is the number of charges
e is the charge on the electron
M+• and empirical formula (from combustion analysis)
 molecular formula
From combustion analysis, empirical formula C2H4O
From mass spectrometry: Molecular ion: M+•, m/z =
Molecular formula is
There are a bunch of
possible structures
consistent with this
data – draw them….
11
…here!
PREVIEW: Molecular ion, fragmentation and base peak
M+• and empirical formula (from combustion analysis)  molecular formula

Next…
Size of M+1 (M+2), …peak  molecular formula (w/o combustion analysis)
Exact mass  molecular formula (w/o combustion analysis)
Later…
Analysis of fragmentation  structural information
12
DETERMINATION OF THE MOLECULAR
FORMULA FROM THE MOLECULAR ION
Pavia 4e 8.6
Two approaches
Consider the size of the “M+1” and “M+2” peaks arising from presence of
13C, 15N, 17O, 34S, etc.
“Exact” mass determination of the molecular ion containing the most
abundant isotopes.
Typically ± 5 ppm
e.g. 100 ± 0.0005 (i.e., between 99.9995 and 100.0005)
200 ± 0.001 (199.999 to 200.001)
Weighted average of
99.985% 1H
0.015% 2H
Weighted average of
98.9% 12C
1.1% 13C
32,33,34,36
…
35,37
79,81
Pavia 4e Tables of Isotope Distributions 8.4, 8.5
13
ISOTOPE DISTRIBUTIONS
Pavia Tables 4e 8.4, 8.5
1.008
1H
C 12.011
12C
N 14.007
14N
99.63%
15N
O 15.999
16O
18O
99.76%
0.20%
17O
0.038%
F 18.998
19F
100.00%
Si 28.086
28Si
30Si
92.23%
3.10%
29Si
4.67%
P 30.974
31P
100.00%
S 32.066
32S
34S
95.02%
4.21%
33S
36S
0.75%
0.020%
Cl 35.453
35Cl
75.77%
37Cl
24.23%
Br 79.904
79Br
50.69%
81Br
49.31%
I 126.905
127I
100.00%
H
99.99%
2H
0.015 %
98.90%
13C
1.10%
0.37%
DETERMINATION OF MOLECULAR FORMULA:
ISOTOPE PATTERNS OF MOLECULAR ION
Carbon: 98.9% 12C, 1.1% 13C
C10H10 (M = 130)
1.08% of all carbon atoms are 13C.
100
10
1H
10
(100%)
14
12
relative abundance
For a compound of n carbon atoms,
1.1n % of the molecules will contain
a single 13C atom and have a molecular
weight of M+1.
12C
12C 13C 1H
9
1 10
12C 1H 2H
10 9 1
10
(11.3%)
8
6
12C 13C 1H
8
2 10
12C 13C 1H 2H
9
1 9 1
12C 1H 2H
10 8 2
4
2
0
129 130 131 132 133
For small C,H,O-containing
molecules, # carbon atoms
predicted by:
100
I(M+1)
I(M)
1.11
m/z
14
C5H6O4
C10H10 (M = 130)
100
12C
10
1H
10
14
100
100
14
12
relative abundance
??(M = 130)
12C 13C 1H
9
1 10
10
12
10
8
8
6
6
4
4
2
2
0
0
129 130 131 132 133
m/z
5.67
0.94
129 130 131 132 133
m/z
e.g.
58.6%
5.4%
Molecular formula  M+ isotope pattern
http://www.colby.edu/chemistry/NMR/IsoClus.html
15
Pavia Table 8.8
Bromine: ca. 1:1 79Br/81Br
C3H7Br
100
90
80
70
60
50
40
30
20
10
0
C3H779Br C3H781Br
C3H6Br2
100
90
80
70
60
50
40
30
20
10
0
119 120 121 122 123 124 125 126
C3H679Br81Br
C3H679Br2
C3H681Br2
199 200 201 202 203 204 205 206
m/z
m/z
1 Br, 2 peaks – 1:1
2 Br, 3 peaks – 1:2:1
3 Br, 4 peaks – 1:3:3:1
4 Br, 5 peaks – 1:4:6:4:1
Chlorine: ca. 3:1 35Cl/37Cl
C3H7Cl
100
90
80
70
60
50
40
30
20
10
0
C3H735Cl
C3H737Cl
75 76
77 79
78 80
80 82
84
76
77 78
81 83
82 83
m/z
C3H6Cl2
100
90
80
70
60
50
40
30
20
10
0
C3H635Cl2
C3H635Cl37Cl
C3H637Cl2
111 112 113 114 115 116 117 118
m/z
Nitrogen, Oxygen and Sulfur
If an odd number of N atoms are present M+• will be odd,
and for each N, M+1 will be +0.37% of M+•
(15N)
For each O present, M+2 will be +0.2% of M+• (18O)
For each S present, M+2 will be +4% of M+•
(34S)
16
MCLAFFERTY’S MAGICAL TABLE:
THE WORLD’S MOST USEFUL INSIDE FRONT COVER
C15H20O3 (% of M+• peak)
M+1: ~16.5+(30.04) = 16.6%
M+2: ~1.27+(30.2) = ~1.87%
C15H20N2O (% of M+• peak)
M+1: ~16.5+(20.37)+(10.04)=
~17.8%
M+2: ~1.27+(10.2) = ~1.47%
C8H12OS (% of M+• peak)
M+1: ~8.8+(10.04)+(10.8) =
~9.6%
M+2: ~0.34+(10.2)+(14.4)=
~4.9%
17
Compound X revisited:
Size of M+1, M+2, …peak  clues about molecular formula
(do not need combustion analysis)
h(M) = 30%; h(M+1) = 1.3%
Molecular formula:
Exercises
(i) Using McLafferty’s table, determine the mass and approximate relative
abundance of the M+•, M+1 and M+2 peaks for following molecules
(a) C14H18N2O5
(b) C7H5NO3S
(ii) Determine the most likely molecular formulas of the molecules that give the
following molecular ion isotope clusters (relative abundance in parentheses)
(a) 162 (100.00); 163 (11.08); 164 (0.97)
(b) 99 (100.000); 100 (5.59); 101 (4.64)
Pavia: chapter 8, questions 3, 4, 5,
18
Mendelev’s Magic Table
Weighted average of
99.985% 1H (M=1.007825)
0.015% 2H (M=2.014102)
Weighted average of
98.9% 12C (M=12.000000)
1.1% 13C (M=13.003355)
…
ISOTOPE DISTRIBUTIONS
Pavia Tables 8.4, 8.5
Isotopic Abundances and Exact Masses
H
C
N
O
1.008
12.011
14.007
15.999
1H
12C
14N
16O
18O
F 18.998
Si 28.086
19F
28Si
30Si
P 30.974
S 32.066
31P
32S
34S
Cl 35.453
Br 79.904
I 126.905
35Cl
79Br
127I
1.007825
12.000000
14.003074
15.994915
17.999159
18.998403
27.976928
29.973772
30.973763
31.972072
33.967868
34.968853
78.918336
126.904477
99.99% 2H
98.90% 13C
99.63% 15N
99.76% 17O
0.20%
100.00%
92.23% 29Si
3.10%
100.00%
95.02% 33S
4.21% 36S
75.77% 37Cl
50.69% 81Br
100.00%
2.014102
13.003355
15.000109
16.999131
0.015 %
1.10%
0.37%
0.038%
28.976496
4.67%
32.971459
35.967079
36.965903
80.916290
0.75%
0.02%
24.23%
49.31%
19
DETERMINATION OF MOLECULAR FORMULA
FROM ACCURATE MASS
nominal
mass
C2H6N2S
C2H6N2O2
C3H6O3
C3H10N2O
C4H10O2
C4H14N2
C4H10S
CH6N4O
C7H6
90
90
90
90
90
90
90
90
90
most common
isotopes
accurate
mass
Mexp = 90.045722
12C
90.025170
90.042928
90.031695
90.079313
90.068080
90.115698
90.050322
90.054161
90.046950
228
31
155
373
248
777
51
93
13
2
1H
12C
2
1H
32S
2
14N 16O
6
2
2
1H
16O
6
3
1H
14N 16O
10
2
1H
16O
10
2
1H
14N
14
2
1H
32S
10
1H
14N 16O
6
4
1H
6
12C
3
12C
3
12C
4
12C
4
12C
4
12C
1
12C
7
6
14N
 (ppm)
ACS: “Found values should be close enough to the Calculated values … to
exclude alternative plausible formulas.”  (ppm) = 106  (Mexp – Mtheor)/ Mtheor
Exact mass  possible molecular formula
http://www-jmg.ch.cam.ac.uk/tools/magnus/EadFormW.html
Pavia Appendix 11
20
Compound X, yet again
Spectral Database for Organic Compounds (SDBS)
Accurate mass  molecular formula
(do not need combustion analysis)
exact
mass
Experimental accurate mass: M+•, m/z = 88.0515
Molecular formula is:
Exercises
(i) Determine the most likely molecular formulas corresponding to the following
experimental exact masses, and determine  (ppm) for each.
(a) 60.0568
(b) 59.0595
(ii) Calculate the exact masses of the following.
(a) C16H13ClN2O (valium)
(b) C14H19NO2 (ritalin)
Pavia: chapter 8: questions 1, 2
(and calculate )
21
INFRARED SPECTROSCOPY
SPECTROSCOPY: INTERACTION OF
LIGHT AND MATTER
electronic
transitions
ionization
Interaction with nuclear
magnetic moment in a
magnetic field
molecules
rotate
bonds
vibrate
c = 
c: speed of light (m/s)
(lamda): wavelength (m)
 (nu): frequency (Hz)
WREK
FM91.1
energy per photon
E = h
h is Planck’s constant
22
SPECTROSCOPY: INTERACTION OF
LIGHT AND MATTER
Stretching Vibrations

n=4.12

where m=m1m2/m1+m2
Analogy to masses and springs:
Spring constant (strength) , Frequency, 
Masses  , 
Only bonds that undergo a change dipole moment upon vibration
will show a IR peak (m = q.d)
23
Electromagnetic Radiation
For molecular vibrations,
Frequency, n = 6 x 1012 to 1.25 x 1014 Hz
Define wavenumber,
a measure of frequency,
Wavelength,  = 50 to 2.5 mm
n 
1
 (in cm)
= 200 to 4000 cm-1
Instrumentation
Spectrum
sample
IR source
reference
detector
4000
wavenumber / cm-1
600
Selected Infrared Absorptions
Functional Group
sp3 C─H
sp2 C─H
sp C─H
C═C
C≡C
N─H
O─H (H-bonded)
O─H (carboxylic acid)
C─O
C=O
C≡N
Range, cm-1
2850-2960
3010-3190
about 3300
1620-1660
2100-2260
3300-3500
3200-3550
2500-3000
1050-1150
1630-1780
2200-2260
Intensity and shape
medium to strong; sharp
medium to strong; sharp
medium to strong; sharp
weak to medium; sharp
weak to medium; sharp
medium; broad
strong; broad
medium; very broad
medium to strong; sharp
medium to strong; sharp
medium; sharp
24
Infrared spectroscopy indicates the presence of particular bonds in a sample. The
combination of bonds indicates which functional groups are present.
You do not need to memorize the data in Table 2.7
O-H
C=O C-O
(or the previous slide) for this class. However, you


Alcohol
must develop experience at interpreting IR spectra

Ether
to be able to determine the presence of particular

Aldehyde or ketone
functional groups.

4000


Ester


Carboxylic acid
600
Wavenumber / cm-1
Hexane
Transmittance (%)
100
50
0
4000
3000
Wavenumber (cm-1)
2000
1500
1000
500
25
1-Hexene
CH2
C
H
Transmittance (%)
100
50
0
4000
3000
Wavenumber (cm-1)
2000
1500
1000
500
1000
500
1-Hexyne
C
CH
Transmittance (%)
100
50
0
4000
3000
Wavenumber (cm-1)
2000
1500
26
Toluene
CH3
Transmittance (%)
100
50
0
4000
3000
Wavenumber (cm-1)
2000
1500
1000
500
2000
1500
1000
500
Butyl methyl ether
O
Transmittance (%)
100
50
0
4000
3000
Wavenumber (cm-1)
27
1-Hexanol
OH
Transmittance (%)
100
50
0
4000
3000
Wavenumber (cm-1)
2-Hexanone
2000
1500
1000
500
2000
1500
1000
500
O
Transmittance (%)
100
50
0
4000
3000
Wavenumber (cm-1)
28
Hexanoic Acid
O
OH
Transmittance (%)
100
50
0
4000
3000
Wavenumber (cm-1)
2000
1500
1000
500
2000
1500
1000
500
Methyl Propionate
Transmittance (%)
100
50
0
4000
3000
Wavenumber (cm-1)
29
C4H8O2
Transmittance (%)
100
50
0
4000
3000
Wavenumber (cm-1)
2000
1500
1000
500
Structure contains:
There are only nine possible structures
consistent with this data – draw them…
…here!
30
Important infrared adsorptions
COOH
3100
sp2C–H
1600 C=C
3500, O-H
1000-1200 C–O
1700
2900 sp3C–H
4000
600
Wavenumber / cm-1
Problem
Which of the following
structures is consistent
with the data provided?
Transmittance (%)
100
Molecular formula: C8H8O
50
0
4000
3000
2000
1500
Wavenumber (cm-1)
1000
500
O
O
O
HO
1
2
3
4
31
Problem
Which of the following
structures is consistent
with the data provided?
Transmittance (%)
100
Molecular formula: C4H8O2
50
0
4000
3000
Wavenumber (cm-1)
O
O
2000
1500
500
O
OH
1
1000
OH
2
O
3
4
Problem
C8H8O2. What functional group(s) is(are) present? What other structural features?
Transmittance (%)
100
50
0
4000
3000
2000
1500
1000
500
Wavenumber (cm-1)
32
Problem
Determine the structure of the compound with the following empirical
formula and IR spectrum.
Empirical formula: C4H8O
IR; peak at 1720 cm-1; no broad peak at 3300, no strong peaks at 1000
to 1200 cm-1, no strong peaks at 1600-1650 cm-1
Problems
33
34
NUCLEAR MAGNETIC RESONANCE
SPECTROSCOPY
35
NUCLEAR MAGNETIC RESONANCE
Some atomic nuclei possess angular momentum also referred to
as spin. The spin quantum number I can have integer or half
integer values.
Atomic
Mass
Atomic
Number
I
odd
odd
even
even
odd
even
odd
even
1/2, 3/2, ...
1/2, 3/2, ...
1, 2, 3,
0
Examples
1H, 19F, 31P
13C, 17O, 29Si
2H, 14N,10B
12C, 16O
When a nucleus is placed in a magnetic field the energy
splits based on the magnetic quantum # m where m goes
from I to –I in steps of 1 so that there are 2I +1 levels.
+1/2
I = 1/2
E
Bo=0
Bo
-1/2
+1
E
0
Bo = 0
Bo
I=1
-1
36
The energy difference between the levels is:
E = g(h/2)B0 = hn
where,
B0 is the magnetic field
g is the magnetogyric constant
- constant for a type of nucleus,
1H, 42.576 MHzT −1
13C, 10.705 MHzT −1
So:
n= (g/2)B0
- the Larmor relationship
The first 1H NMR spectrum
...not much chemical information
Later:
Why are there 3 peaks?
How can we use this information to provide informatin about structure?
37
Today….
… lots of chemical information
Not all 1H (or 13C, ….) in a molecule are equal (equivalent). The
inequivalent protons Ha and Hb absorb irradiation with different
frequencies as a given applied field.
E
Ha
Hb
The spectrum
Hb
Ha
B
E=hν
38
Preview: Types of Information available from a 1H NMR
spectrum
A 1H nuclear magnetic resonance spectrum contains information about the:
(a) number of different types of proton
(b) relative number of each type of proton
Cl
(c) proximity to functional groups
H H
C C H
H H
(d) the number of adjacent protons
(a) The number of peaks indicates the number of types of
hydrogen
1H
NMR spectrum of methane
CH4
one peak  one type of proton
E=hν
1H
NMR spectrum of ethane
CH3–CH3
one peak  one type of proton
E=hν
39
1H
nuclear magnetic resonance spectrum of methanol…
CH3–OH
E=hν
two peaks  two types of proton
Protons that are related to each other by a rotation or plane of symmetry are
identical: chemical shift equivalent
40
C 4 H 8 O2
SODAR = 1
IR: 1720 cm-1 » C=O
no peak at 3300 cm-1 » no O-H
Which of the following 9 structures can you now
exclude based on the number of signals?
1H
O
O
NMR
O
O
H
O
H
O
O
O
H
H
O
O
O
O
H
R/S O
O
O
O
Problem
How many signals are there in the 1H NMR spectrum of
1,4-dichlorobutane?
41
Problem
Which of the following gives 3 peaks in the 1H NMR spectrum
and does NOT have a peak at 1700-1750 cm-1 in the IR
spectrum?
O
H
O
O
O
Br
1
2
3
4
(b) The relative area of each peak (the integration)
corresponds to the relative number of each type of proton
1H
NMR spectrum of methanol
H3C OH
integral ratio
:
 ratio of types of proton
E=hν
H
1H
NMR spectrum of p-xylene
H
H3C
CH3
H
integral ratio
:
H
 ratio of types of proton
E=hν
42
C 4 H 8 O2
SODAR = 1
IR: 1720 cm-1 » C=O
no peak at 3300 cm-1 » no O-H
Which of the remaining structures can you now
exclude based on the integrals of signals?
1H

NMR
O
O
O
O
H
O
O
O
H
O


H
O

O
O
H
H
O
R/S O
O
O
O
(c) The chemical shift indicates the environment
of the proton
1H
NMR spectrum of methanol
chemical shift,  / ppm
43
Shielding and deshielding
hν
E
Bo-Bind
Ha
Bo
shielded
B
Electrons “shield” the nucleus from Bo:
The field at the nucleus is
Resonance is achieved at
hν
+1/2
TMS
E
Bo=0
-1/2
7.0
4.7
3.3`
Bo/T
Field Strength inst / Hz
B0 / T
2.33
100 x 106
4.66
200 x 106
7.00
300 x 106
CH4 – instr. / Hz
CH4 / ppm
23
46
69
- The resonance frequency (in Hz) depends on magnet strength.
- The chemical shift () is independent of magnet strength.
44
The -scale
The frequency at which a proton resonates is measured relative to the
frequency for the resonance of protons of tetramethylsilane, TMS (which
resonates at a relatively low frequency), and cited on the  (delta) scale in parts
per million of the frequency at which TMS resonates.
nproton - nTMS

x 106 [ppm]
nTMS
e.g., CH3CH3
125 Hz

x 106 [ppm]
6
100 x 10 Hz
 = 1.25
Effect of Structure on Chemical Shift ( scale, ppm)
CH3-CH3
1.2
CH3- N(CH3)2
2.2
CH3-OCH3
3.2
CH3-F
4.3
CH3-Cl
3.1
hν
E
CH3-Br
2.7
Ha
Bo-Bind
shielded
Bo
Bo-Bind
deshielded
CH3-I
2.2
B
CHCl3
7.3
CH2Cl2
5.3
CH3Cl
3.1
45
Typical 1H NMR chemical shifts
-
Type of proton
(CH3 )4 Si
3
CH3 -C-R (sp 3)
-CH2 -C-R (sp3 )
-CH-C-R (sp )
H-C-N
H-C-O
H-C-Cl
H-C-Br
H-C-C=O
H-C-C=C
H-C-Ar
2
H-C=O (sp2 )
H-C=C (sp
)
2
H-Ar (sp )
H-C C (sp)
H-N (amine)
H-OR (alcohol)
H-OAr (phenol)
H-O2 CR (acid)
chemical shift ()
0.00
0.9 - 1.8
1.1 - 2.0
1.3 - 2.1
2.2 - 2.9
3.3 - 3.7
3.1 - 4.1
2.7 - 4.1
2.1 - 2.5
1.6 - 2.6
2.3 - 2.8
9 - 10
4.5 - 6.5
6.5 - 8.5
2.5
1-3
0.5 - 5
6-8
10 - 13
You will be provided with a copy of Table 9.1 on exams. This provides approximate ranges for values of
chemical shifts for particular types of protons. Remember that protons adjacent to two (or more) electron
withdrawing groups will appear further downfield than a proton adjacent to only one.
Other peaks in spectrum:
CHCl3
10
 / ppm
H2O
5
TMS
0
46
Problem
Which of the following hydrogen atoms gives the peak furthest
downfield in the 1H NMR spectrum?
O
H
H
O
O
Br
H
O
H
1
2
3
4
Problem
Which of the following compounds (C4H7ClO, with a peak at ca. 1700 cm-1 in
the IR spectrum corresponding to a C=O) gives the following set of peaks in
the 1H NMR spectrum?
1H
peak at 4.3 ppm; 3H peak at 2.3 ppm; 3H peak at 1.6 ppm
O
O
Cl
Cl
1
2
O
O
Cl
3
Cl
4
H
47
(d) The multiplicity of a signal indicates the
number of adjacent protons
1H
NMR spectrum of ethanol
Spin-spin coupling
The magnetic field experienced by a
proton is effected by the magnetic
field generated by each adjacent
proton.
E
Bo+
Bo
↓
Bo+
Ha
B
If field from Hb spin , need to apply
larger E (=hn) to achieve resonance of
Ha
If field from Hb spin , need to apply
smaller E (=hn) to achieve resonance
of Ha
48
Ha
Hb
Bo
Hb
C
C
To achieve resonance of Ha
If Hb spins  + , need to apply higher n
If Hb spins  + , or,  +  signal appears at a
If Hb spins  + , need to apply lower n
The extent of interaction between protons, the “coupling constant” (J), is
measured in Hz, and is independent of the field (i.e., the instrument).
Coupling is only observed between non-equivalent protons.
The signal for a proton coupling to a set of N protons will be split into a multiplet
consisting of N+1 lines. The relative area of each peak within a multiplet can
be determined from Pascal’s triangle.
This is often referred to as the “N+1 rule” – but this only works for nuclei with
nuclear spin quantum numbers, I = + ½ and –½ (e.g., 1H)
This is a special case of a more general “2NI+1 rule” (for all values of I).
49
Problem
Which combination of peaks appear in the 1H NMR spectrum of
1,2-dichloroethane?
1 a singlet
2 a triplet
3 two singlets
4 two triplets
Problem
Which combination of peaks appear in the 1H NMR spectrum of
1,4-dichlorobutane?
1 two singlets
2 two triplets
3 a triplet and a pentet
4 two pentets
50
Some Common Sets of Multiplets
Et–X
i-Pr–X
X
X
a b
CH2 CH3
a
b
CH3 a
C H b
CH3 a
a
b
CH3 a
t-Bu–X
X
C CH3 a
a
CH3 a
H3C
CHCl2
51
How can you get a pentet?
Summary: Types of information available from 1H NMR
spectrum
Number of signals
Number of types of proton
Integral of signals
Relative number of each type of proton
Chemical Shift
Electronic environment
Multiplicity
Number of adjacent protons
52
Problem
Predict the appearance (chemical shift, integral and multiplicity) of the 1H
NMR signals of the following compounds.
(a) 2-butanone, CH3COCH2CH3
(b) methyl 3-chloropropanoate, ClCH2CH2CO2CH3
Problem
Provide structures consistent with the following data.
(a) Compound A: C2H4Cl2, one singlet in 1H NMR spectrum
(b) Compound B: C5H12O2, two singlets in 1H NMR spectrum
53
Problem
Which isomer of C5H10O gives the following 1H NMR
spectrum?
O
O
1
2
O
O
H
3
H
4
C 4 H 8 O2
SODAR = 1
IR: 1720 cm-1 » C=O
no peak at 3300 cm-1 » no O-H
Now, exclude further structures based on
multiplicities and chemical shifts
1H
 
O
O
NMR
O
O
H
O
O
O
H
O


H
O
O
H
O

R/S O
O
H
O
O
O
54
13C
NMR SPECTROMETRY
Introduction
13C
(not 12C) has nuclear spin
However,
13C
is only present at 1.1% abundance
- Signals are weak
- Spectra usually acquired without multiplicity information
E = g(h/2)B0 = hn
g:
in a 7.04 T magnet
42.576 MHzT−1
13C, 10.705 MHzT−1
1H,
n = 300 MHz
n = 75 MHz
- Larger range of chemical shifts (0 to >200 ppm)
55
200
150
100
50
0
 / ppm
Types of Information available from a 13C NMR spectrum
A 13C nuclear magnetic resonance spectrum contains information about the:
(a) number of different types of carbon
- Each peak corresponds to a different type of carbon
(b) type of carbons and proximity to functional groups
- Chemical shift provides information about the type
of carbon present
Unlike 1H NMR spectra, simple 13C NMR spectra do not provide information
about the:
relative number of each type of proton
or
number of adjacent protons or carbons
56
USING DATA TO DETERMINE STRUCTURE
1. Write down conclusions from each individual technique
- Use combustion analysis to determine empirical formula, and mass
spectrometry to give molecular weight (and molecular formula)
- Calculate number of rings and double bonds from molecular formula
(SODAR)
- Determine presence of functional groups present from infrared
spectroscopy.
- Use 1H and 13C NMR spectroscopy to identify other structural features.
2. Use these conclusions to determine the structure
3. Check your answer (predict the spectra; do your predicted spectra match the
data provided?)
WORK PROBLEMS!!
In the remaining lectures we will cover problem solving approaches for determining
organic structures from spectral (and other) information. Additional problem
appear in the following sources:
- Your undergraduate organic textbook
- web.centre.edu/muzyka/organic/jmol10/table/JcampUnknownsFrames.htm
- www.nd.edu/~smithgrp/structure/workbook.html
For now, just do the “easy” problems from this site. The answers to these
problems are NOT available from this site; do not bother Prof. Smith. The
answers to the easy problems, along with answers to the problems in these
notes are posted on the course web site.
- www.chem.ucla.edu/~webspectra
For now, do the “beginning” problems that do not have DEPT spectra, and
COSY spectrum.
57
Problem A
Empirical Formula:
Elemental
Analysis:CC,
H, 9.47
4H90.38;
5
IR
100
Mass Spec
50
0
4000
13C
3000
2000
1500
Wavenumber / cm-1
1000
500
1H
NMR
2-lines
NMR
Problem B
Empirical Formula: C5H7
Mass Spec: M+ m/e= 134
IR
100
50
0
4000
13C
NMR
3000
2000
1500
Wavenumber / cm-1
1000
500
1H
NMR
Problem C
Empirical Formula: C5H5O
Mass Spec: M+ m/e= 162
IR
100
H 2O
50
0
4000
13C
NMR
3000
2000
1500
Wavenumber / cm-1
1000
500
1H
NMR
Problem D
Empirical Formula: C5H10O2
Mass Spec: M+ m/e= 102
IR
100
H 2O
50
0
4000
13C
NMR
3000
2000
1500
Wavenumber / cm-1
1000
500
1H
NMR
Problem E
Empirical Formula: C4H10O
Mass Spec: M+ m/e= 74
IR
100
50
0
4000
13C
NMR
3000
2000
1500
Wavenumber / cm-1
1000
500
1H
NMR
Problem F
Empirical Formula: C6H14O
Mass Spec: M+ m/e= 102
IR
100
50
0
4000
13C
NMR
3000
2000
1500
Wavenumber / cm-1
1000
500
1H
NMR
Problem G
Empirical Formula: C3H8O2
Mass Spec: M+ m/e= 76
IR
100
50
0
4000
13C
NMR
3000
2000
1500
Wavenumber / cm-1
1000
500
1H
NMR
Problem H
Empirical Formula: C7H9N
Mass Spec: M+ m/e= 107
IR
100
50
0
4000
13C
NMR
3000
2000
1500
Wavenumber / cm-1
1000
500
1H
NMR
Problem I
Empirical Formula: C7H9N
Mass Spec: M+ m/e= 107
IR
100
50
0
4000
13C
NMR
3000
2000
1500
Wavenumber / cm-1
1000
500
1H
NMR
Problem J
Empirical Formula: C4H4O
Mass Spec: M+ m/e= 136
IR
100
50
0
4000
13C
3000
2000
1500
Wavenumber / cm-1
1000
500
1H
NMR
2-lines
NMR
Problem K
Empirical Formula: C4H5O
Mass Spec: M+ m/e= 138
IR
100
50
0
4000
13C
NMR
3000
2000
1500
Wavenumber / cm-1
1000
500
1H
NMR