fall 2017 chem 6371 – spectroscopic identification of...

Fall 2017 http://ww2.chemistry.gatech.edu/~collard/SpecID/fall2017.html https://t-square.gatech.edu/portal (for grades)

CHEM 6371 – Spectroscopic Identification of Organic Compounds CHEM 4341 – Applied Spectroscopy

Office Topic Reading (4e) M 21-Aug Introduction Intro-1 DC Tu-DC Introduction W 23-Aug Your intro organic Intro-2 DC Elemental analysis, sites of unsaturation 1:1-5+ intro text F 25-Aug textbook and Intro-3 DC Mass spectrometry 1:6;8.7+ intro text M 28-Aug Pavia Chap 1,3 Intro-4 DC Tu-DC Infrared spectroscopy 2:intro -7+ intro text W 30-Aug Intro-5 HW 1 DC 1H NMR 3:1-9+ intro text F 1-Sep Intro-6 DC 1H NMR 3:10-17+ intro text M 4-Sep LABOR DAY Tu-DC W 6-Sep Intro-7 DC 1H NMR and 13C NMR 3:19;4:14.2A+ intro text F 8-Sep Intro-8 HW 2 DC Problems Q3.1-12 M 11-Sep EXAM1 Tu-DC Fundamental principles, using basic information to solve structures W 13-Sep Mass Spectrometry MS-1 DC MS techniques F 15-Sep Pavia Chap 8 MS-2 DC Fragmentation, Hydrocarbons M 18-Sep MS-3 DC Tu-DC Alcohols, ethers and amines: α-cleavage W 20-Sep MS-4 HW 3 DC C=O compounds F 22-Sep MS-5 DC Halo compounds, reporting MS data M 25-Sep MS-6 HW 4 DC Problems W 27-Sep MS-7 DC Problems F 29-Sep EXAM 2 MS, plus use of basic principles of IR and NMR to solve structures M 2-Oct Infrared IR-1 LG M-LG CH, CC bonds 2:8-11 W 4-Oct Spectroscopy IR-2 LG OH, NH bonds 2:12;14F,15 F 6-Oct Pavia Chap 2 IR-3 HW 5 (MS) DC CO bonds 2:13 M 9-Oct FALL RECESS Tu-DC W 11-Oct IR-4 DC C=O and CN bonds, reporting IR data 2:14,16-21 F 13-Oct IR-5 HW 6 DC Problems Q2.1-11 M 16-Oct EXAM 3 Tu-DC IR, plus use of basic principles of IR and NMR to solve structures W 18-Oct 1H and 13C NMR NMR-1 LG NMR theory F 20-Oct Spectroscopy NMR-2 LG 1H Chemical Shift 6:1-5 M 23-Oct Pavia Chap 4,5,6 NMR-3 LG M-LG 1H Chemical Shift 6:6-10 W 25-Oct NMR-4 LG 1H Multiplicity 5:1-6 F 27-Oct NMR-5 LG 1H Multiplicity, reporting NMR data 5:7-11 M 30-Oct NMR-6 HW 7 LG M-LG 13C Chemical shift, shift calculations 4:1-4,11-16 W 1-Nov NMR-7 LG Decoupling, NOE, etc 4:5-9; 6.11 F 3-Nov NMR-8 LG Edited spectra, APT, DEPT 4:10; 10.4-5 M 6-Nov NMR-9 LG M-LG Problems 9:intro-21 W 8-Nov NMR-10 HW 8 LG Problems Q9.22-43 F 10-Nov EXAM 4 NMR, and use of MS and IR to solve structures M 13-Nov 2D NMR Adv-1 LG M-LG 2D NMR Theory 10:1,6 W 15-Nov Pavia Chap 10 Adv-2 LG COSY 10:7 F 17-Nov Adv-3 LG NOESY 10.10 M 20-Nov Adv-4 HW 9 LG M-LG HETCOR, HSQC 10.8-9 W 22-Nov THANKSGIVING F 24 Nov BREAK M 27-Nov Adv-5 LG M-LG HMBC W 29-Nov Adv-6 HW 10 LG Problems Q10.1-11 F 1-Dec EXAM 5 Basic and Adv. NMR, MS and IR to solve structures

M 4-Dec More problems Problems DC Tu-DC Problems W 8-Dec (8:00-10:50 a.m.) FINAL Comprehensive

http://ww2.chemistry.gatech.edu/%7Ecollard/SpecID/fall2017.html

https://t-square.gatech.edu/portal

INSTRUCTORS David M. Collard Leslie Gelbaum [email protected] [email protected] Tuesday 1:00-2:00 p.m. Monday 3:00 - 4:00 p.m. MoS&E 2100J MS&E 113 Office hours are only for weeks in which the particular instructor is teaching (see schedule); or by appointment Throughout the semester, Ashley Johns ([email protected]) and Bronson Cox ([email protected]) may present solutions to problems in class. They may be consulted regarding these approaches. LECTURES Come prepared to ask and answer questions! MWF, 11:15-12:05 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” This text is currently in its fifth edition. You may choose to buy a new copy or a used copy, or a used copy of the fourth edition. The syllabus (above) indicates section numbers in the 4e. 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

Topic Exam 1 Fundamental principles and using basic information to solve structures 100 points† Exam 2 IR, plus use basic principles of MS and NMR to solve structures 100 points† Exam 3 MS, plus use of basic principles of IR and NMR to solve structures 100 points† Exam 4 NMR, plus use of MS and IR to solve structures 100 points† Exam 5 Basic and Advanced NMR techn., plus use of MS and IR to solve structures 100 points† Homework Ten HW assignments (score normalized to 100 points) 100 points Final Wed Dec 14 at 8:00 am-10:50 am: Comprehensive 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. 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.

1

Introduction to Spectroscopy, 4eDonald L. Pavia, Gary M. Lampman, George S. Kriz,

and James A. VyvyanBrooks ColeISBN-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 Infrared Spectroscopy P(4,5) Chap 2

21 Sept EXAM 2 - IR, plus use of basic principles of IR and NMR to solve structures

23 Sept -07 Oct Mass Spectrometry P(4) Chap 8; P(5) Chap 3,4

07 - Oct EXAM 3 – MS, plus basic principles of IR and NMR to solve structures

14 Oct - 06 Nov Fundamentals of 1H and 13C NMR Spectrometry P(4) Chap 3-6;P(5) 5-8

06 Nov EXAM 4 NMR, plus use of MS and IR to solve structures

09 Nov - 23 Nov 2D, Advanced and multinuclear NMR P(4) Chap 10; P(5) Chap 9

23 Nov 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

2

GRADING

Exam 1 100

Exam 2 100

Exam 3 100 drop lowest score from E1-5

Exam 4 100

Exam 5 100

Homework 100

Final 200

Course grade out of 700 points

AN APPROACH TO DETERMINING

ORGANIC STRUCTURES

Combustion analysis Empirical formula

Mass spectrum Molecular weight

Molecular formula

Sites of unsaturation

Infrared spectrum Functional groups1H NMR13C NMR

MS isotope patternMS exact mass

(pi bonds, rings)

Structural featuresfragments, connectivity

3

REQUIREMENTS FOR PROOF OF STRUCTURE

AND PURITY IN YOUR RESEARCH

New Compounds

IR: assign important peaks1H NMR: assign all peaks13C 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. guidelineshttp://pubs.acs.org/paragonplus/submission/joceah/joceah_authguide.pdf

EMPIRICAL FORMULAS, MOLECULAR

FORMULAS, SITES OF UNSATURATION &

COMBUSTION ANALYSIS

4

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

5

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?

C6H6

C6H7BrO

C10H22N2

C10H13NO2

C21H35ClS2

6

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?

7

COMBUSTION ANALYSIS

CxHyOz + O2 xCO2 + (y/2)H2O

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 O

Mass% 54.51 9.09

Mole Ratio

=

=

Empirical formula Molecular Formula

could be…

8

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

9

TB-III-32ACHEM, Georgia Tech

901 State St.

Atlanta, GA 30306-0400

T. Brooking 9-8-09

X

C,H

C,H

X

B. Moore

1966-06

C 71.23%

H 9.35%

9-10-09 9-11-09

C 68.09%

H 8.64%

N 6.59%

Cl 16.57%

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)

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!]

http://fluorine.ch.man.ac.uk/research/analyse.php

http://www.calctool.org/CALC/chem/molecular/elemental

http://www.chemicalaid.com/tools/empirical.php

10

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

11

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….

12

…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

13

DETERMINATION OF THE MOLECULAR

FORMULA FROM THE MOLECULAR ION

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)

Pavia 4e 8.6

…

Weighted average of

99.985% 1H

0.015% 2H

Weighted average of

98.9% 12C

1.1% 13C

Pavia 4e Tables of Isotope Distributions 8.4, 8.5

32,33,34,36

35,37

79,81

14

ISOTOPE DISTRIBUTIONS

H 1.008 1H 99.99% 2H 0.015 %

C 12.011 12C 98.90% 13C 1.10%

N 14.007 14N 99.63% 15N 0.37%

O 15.999 16O 99.76% 17O 0.038%18O 0.20%

F 18.998 19F 100.00%

Si 28.086 28Si 92.23% 29Si 4.67% 30Si 3.10%

P 30.974 31P 100.00%

S 32.066 32S 95.02% 33S 0.75%34S 4.21% 36S 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%

Pavia Tables 4e 8.4, 8.5

DETERMINATION OF MOLECULAR FORMULA:

ISOTOPE PATTERNS OF MOLECULAR ION

Carbon: 98.9% 12C, 1.1% 13C

1.08% of all carbon atoms are 13C.

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.

For small C,H,O-containing

molecules, # carbon atoms

predicted by:

0

2

4

6

8

10

12

14

129 130 131 132 133

m/z

C10H10 (M = 130)

12C101H10 (100%)

12C913C1

1H10 (11.3%)12C10

1H92H1

12C813C2

1H1012C9

13C11H9

2H112C10

1H82H2

rela

tive

ab

un

dan

ce

100

I(M+1)

I(M)

1.11

100

15

0

2

4

6

8

10

12

14

129 130 131 132 133

m/z

100

??(M = 130)

0

2

4

6

8

10

12

14

129 130 131 132 133

m/z

C10H10 (M = 130)

12C101H10

12C913C1

1H10

rela

tive

ab

un

dan

ce

100

C5H6O4

5.67

0.94

100

e.g.

58.6%

5.4%

Molecular formula M+ isotope patternhttp://www.colby.edu/chemistry/NMR/IsoClus.html

http://www.colby.edu/chemistry/NMR/IsoClus.html

16

Bromine: ca. 1:1 79Br/81Br

0102030405060708090

100

199 200 201 202 203 204 205 206

m/z

0102030405060708090

100

119 120 121 122 123 124 125 126

m/z

C3H7Br C3H6Br2

C3H779Br C3H7

81Br C3H679Br81Br

C3H679Br2 C3H6

81Br2

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

Pavia Table 8.8

Chlorine: ca. 3:1 35Cl/37Cl

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)

C3H7Cl C3H6Cl2C3H735Cl C3H6

35Cl2

C3H635Cl37Cl

C3H737Cl

C3H637Cl2

0102030405060708090

100

111 112 113 114 115 116 117 118

m/z

0102030405060708090

100

75 76 77 78 80 82 83 84

m/z

76 77 78 79 80 81 82 83

17

MCLAFFERTY’S MAGICAL TABLE:

THE WORLD’S MOST USEFUL INSIDE FRONT COVER

C15H20O3 (% of M+• peak)M+1: ~16.5+(30.04) = 16.6%

M+2: ~1.27+(30.2) = ~1.87%

C15H20N2O (% of M+• peak)M+1: ~16.5+(20.37)+(10.04)=

~17.8%

M+2: ~1.27+(10.2) = ~1.47%

C8H12OS (% of M+• peak)M+1: ~8.8+(10.04)+(10.8) =

~9.6%

M+2: ~0.34+(10.2)+(14.4)=~4.9%

18

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,

19

…

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)

Mendelev’s Magic Table

ISOTOPE DISTRIBUTIONS

Isotopic Abundances and Exact Masses

H 1.008 1H 1.007825 99.99% 2H 2.014102 0.015 %

C 12.011 12C 12.000000 98.90% 13C 13.003355 1.10%

N 14.007 14N 14.003074 99.63% 15N 15.000109 0.37%

O 15.999 16O 15.994915 99.76% 17O 16.999131 0.038%18O 17.999159 0.20%

F 18.998 19F 18.998403 100.00%

Si 28.086 28Si 27.976928 92.23% 29Si 28.976496 4.67% 30Si 29.973772 3.10%

P 30.974 31P 30.973763 100.00%

S 32.066 32S 31.972072 95.02% 33S 32.971459 0.75% 34S 33.967868 4.21% 36S 35.967079 0.02%

Cl 35.453 35Cl 34.968853 75.77% 37Cl 36.965903 24.23%

Br 79.904 79Br 78.918336 50.69% 81Br 80.916290 49.31%

I 126.905 127I 126.904477 100.00%

Pavia Tables 8.4, 8.5

20

DETERMINATION OF MOLECULAR FORMULA

FROM ACCURATE MASS

nominal most common accurate Mexp = 90.045722

mass isotopes mass (ppm)

C2H6N2S 90 12C21H6

14N232S 90.025170 228

C2H6N2O2 90 12C21H6

14N2 16O2 90.042928 31

C3H6O3 90 12C31H6

16O3 90.031695 155

C3H10N2O 90 12C31H10

14N216O 90.079313 373

C4H10O2 90 12C41H10

16O2 90.068080 248

C4H14N2 90 12C41H14

14N2 90.115698 777

C4H10S 90 12C41H10

32S 90.050322 51

CH6N4O 90 12C11H6

14N416O 90.054161 93

C7H6 90 12C71H6 90.046950 13

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 formulahttp://www-jmg.ch.cam.ac.uk/tools/magnus/EadFormW.html

Pavia Appendix 11

21

Compound X, yet again

Accurate mass molecular formula

(do not need combustion analysis)

Experimental accurate mass: M+•, m/z = 88.0515

Molecular formula is:

Spec

tral

Dat

abas

e fo

r O

rgan

ic C

om

po

un

ds

(SD

BS)

exact mass

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 )

22

INFRARED SPECTROSCOPY

SPECTROSCOPY: INTERACTION OF

LIGHT AND MATTER

c =

c: speed of light (m/s)

(lamda): wavelength (m)

(nu): frequency (Hz)

energy per photon

E = h

h is Planck’s constant

bondsvibrate

moleculesrotate

Interaction with nuclear magnetic moment in a

magnetic field

WREKFM91.1

electronictransitions

ionization

23

Stretching Vibrations

Analogy to masses and springs:

Spring constant (strength) , Frequency,

Masses ,

n=4.12 where m=m1m2/m1+m2

SPECTROSCOPY: INTERACTION OF

LIGHT AND MATTER

Only bonds that undergo a change dipole moment upon vibration

will show a IR peak (m = q.d)

24

Electromagnetic Radiation

For molecular vibrations,

Frequency, n = 6 x 1012 to 1.25 x 1014 Hz Wavelength, = 50 to 2.5 mm

Define wavenumber,

a measure of frequency, = 200 to 4000 cm-1

Instrumentation

Spectrum

IR source

sample

referencedetector

n 1

(in cm)

wavenumber / cm-14000 600

Selected Infrared Absorptions

Functional Group Range, cm-1 Intensity and shape

sp3 C─H 2850-2960 medium to strong; sharp

sp2 C─H 3010-3190 medium to strong; sharp

sp C─H about 3300 medium to strong; sharp

C═C 1620-1660 weak to medium; sharp

C≡C 2100-2260 weak to medium; sharp

N─H 3300-3500 medium; broad

O─H (H-bonded) 3200-3550 strong; broad

O─H (carboxylic acid) 2500-3000 medium; very broad

C─O 1050-1150 medium to strong; sharp

C=O 1630-1780 medium to strong; sharp

C≡N 2200-2260 medium; sharp

25

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

(or the previous slide) for this class. However, you

must develop experience at interpreting IR spectra

to be able to determine the presence of particular

functional groups.

O-H C=O C-O

Alcohol

Ether

Aldehyde or ketone

Ester

Carboxylic acid

Wavenumber / cm-14000 600

Hexane

4000 3000 2000 1500 1000 500

Wavenumber (cm-1)

Tra

nsm

itta

nce (

%)

100

50

0

26

1-Hexene

4000 3000 2000 1500 1000 500

Wavenumber (cm-1)

Tra

nsm

itta

nce

(%

)100

50

0

CH

CH2

4000 3000 2000 1500 1000 500

Wavenumber (cm-1)

Tra

nsm

itta

nce (

%)

100

50

0

1-Hexyne

CCH

27

4000 3000 2000 1500 1000 500

Wavenumber (cm-1)

Tra

nsm

itta

nce

(%

)100

50

0

Toluene

CH3

Butyl methyl ether

4000 3000 2000 1500 1000 500

Wavenumber (cm-1)

Tra

nsm

itta

nce (

%)

100

50

0

O

28

4000 3000 2000 1500 1000 500

Wavenumber (cm-1)

Tra

nsm

itta

nce

(%

)100

50

0

1-Hexanol

OH

4000 3000 2000 1500 1000 500

Wavenumber (cm-1)

Tra

nsm

itta

nce (

%)

100

50

0

2-HexanoneO

29

4000 3000 2000 1500 1000 500

Wavenumber (cm-1)

Tra

nsm

itta

nce

(%

)100

50

0

Hexanoic Acid

OH

O

4000 3000 2000 1500 1000 500

Wavenumber (cm-1)

Tra

nsm

itta

nce (

%)

100

50

0

Methyl Propionate

30

4000 3000 2000 1500 1000 500

Wavenumber (cm-1)

Tra

nsm

itta

nce

(%

)

100

50

0

C4H8O2

Structure contains:

There are only nine possible structures

consistent with this data – draw them…

…here!

31

Important infrared adsorptions

Wavenumber / cm-14000 600

1700

1000-1200 C–O

2900 sp3C–H

3100

sp2C–H

3500, O-H

COOH

1600 C=C

Which of the following

structures is consistent

with the data provided?

Molecular formula: C8H8O

1 2 3 4

4000 3000 2000 1500 1000 500

Wavenumber (cm-1)

100

50

0

O

O

O

HO

Tra

nsm

itta

nce (

%)

Problem

32

Which of the following

structures is consistent

with the data provided?

Molecular formula: C4H8O2

1 2 3 4

4000 3000 2000 1500 1000 500

Wavenumber (cm-1)

100

50

0Tra

nsm

itta

nce (

%)

O

OH

O

O

O

OH

Problem

4000 3000 2000 1500 1000 500

Wavenumber (cm-1)

Tra

nsm

itta

nce (

%)

100

50

0

C8H8O2. What functional group(s) is(are) present? What other structural features?

Problem

33

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

Problem

Problems

35

NUCLEAR MAGNETIC RESONANCE

SPECTROSCOPY

36

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 Atomic I Examples

Mass Number

odd odd 1/2, 3/2, ... 1H, 19F, 31P

odd even 1/2, 3/2, ... 13C, 17O, 29Si

even odd 1, 2, 3, 2H, 14N,10B

even even 0 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.

Bo

EBo=0

-1/2

+1/2

-1

0

+1

I = 1/2

I = 1Bo = 0E

Bo

37

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?

38

… lots of chemical information

Today….

Ha

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.

B

E

E=hν

The spectrumHb

Hb Ha

39

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

(c) proximity to functional groups

(d) the number of adjacent protons

H

C

H

Cl C

H

H

H

(a) The number of peaks indicates the number of types of

hydrogen

1H NMR spectrum of methane CH4

one peak one type of proton

1H NMR spectrum of ethane CH3–CH3

one peak one type of proton

E=hν

E=hν

40

1H nuclear magnetic resonance spectrum of methanol…

CH3–OH

two peaks two types of proton

E=hν

Protons that are related to each other by a rotation or plane of symmetry are

identical: chemical shift equivalent

41

C4H8O2

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 NMR O H

O

O

O

O H

O

O

O

HO

O

H

O

O

O

O

HO

O

R/S

How many signals are there in the 1H NMR spectrum of 1,4-dichlorobutane?

Problem

42

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?

1 2 3 4

O

Br

O H

O

O

Problem

(b) The relative area of each peak (the integration)

corresponds to the relative number of each type of proton

1H NMR spectrum of methanol

integral ratio : ratio of types of proton

1H NMR spectrum of p-xylene

integral ratio : ratio of types of proton

CH3H3C

HH

H H

H3C OH

E=hν

E=hν

43

C4H8O2

SODAR = 1

IR: 1720 cm-1 » C=O


Which of the remaining structures can you now

exclude based on the integrals of signals?

1H NMR

R/S

O H

O

O

O

O H

O

O

O

HO

O

H

O

O

O

O

HO

O

1H NMR spectrum of methanol

chemical shift, / ppm

(c) The chemical shift indicates the environment

of the proton

44

Shielding and deshielding

Ha

B

E

Bo-BindBo

shielded

hν

hν

Electrons “shield” the nucleus from Bo:

The field at the nucleus is

Resonance is achieved at

Field Strength inst / Hz CH4 – instr. / Hz CH4

/ ppm

B0 / T

2.33 100 x 106 23

4.66 200 x 106 46

7.00 300 x 106 69

- The resonance frequency (in Hz) depends on magnet strength.

- The chemical shift () is independent of magnet strength.

TMS

Bo/T

E Bo=0

7.0

4.7

3.3

`

-1/2

+1/2

45

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.

x 106 [ppm]

e.g., CH3CH3

x 106 [ppm]

= 1.25

nproton - nTMS

nTMS

125 Hz

100 x 106 Hz

Effect of Structure on Chemical Shift ( scale, ppm)

CH3-CH3 CH3- N(CH3)2 CH3-OCH3 CH3-F

1.2 2.2 3.2 4.3

CH3-Cl

3.1

CH3-Br

2.7

CH3-I

2.2

CHCl3 CH2Cl2 CH3Cl

7.3 5.3 3.1

Ha

B

E

Bo-BindBo

shielded

hν

Bo-Binddeshielded

46

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.

-Typical 1H NMR chemical shifts

Type of proton chemical shift ()

(CH3)4Si 0.00CH3-C-R (sp

3

) 0.9 - 1.8-CH2-C-R (sp

3

) 1.1 - 2.0-CH-C-R (sp

3

) 1.3 - 2.1H-C-N 2.2 - 2.9H-C-O 3.3 - 3.7H-C-Cl 3.1 - 4.1H-C-Br 2.7 - 4.1H-C-C=O 2.1 - 2.5H-C-C=C 1.6 - 2.6H-C-Ar 2.3 - 2.8H-C=O (sp

2

) 9 - 10H-C=C (sp

2

) 4.5 - 6.5H-Ar (sp

2

) 6.5 - 8.5

H-C C (sp) 2.5

H-N (amine) 1 - 3H-OR (alcohol) 0.5 - 5H-OAr (phenol) 6 - 8H-O2CR (acid) 10 - 13

/ ppm10 5 0

TMSH2OCHCl3

Other peaks in spectrum:

47

Which of the following hydrogen atoms gives the peak furthest

downfield in the 1H NMR spectrum?

1 2 3 4

O Br O

OH

HH

O

H

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

1 2

3 4 H

O

Cl

Cl

O

O

O

Cl

Cl

Problem

48

1H NMR spectrum of ethanol

(d) The multiplicity of a signal indicates the

number of adjacent protons

Spin-spin coupling

The magnetic field experienced by a proton is effected by the magnetic field generated by each adjacent proton.

Bo

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

Ha

B

E

Bo+↓ Bo+

49

Bo

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

C C

Hb

Hb

Ha

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).

50

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

Problem

51

Some Common Sets of Multiplets

Et–X a

b

i-Pr–X a

b

t-Bu–X a

CH2 CH3

C

CH3

CH3

H

C CH3

CH3

CH3

a b

a

b

a

a

a

a

X

X

X

H3C CHCl2

52

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

53

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

54

Which isomer of C5H10O gives the following 1H NMR

spectrum?

1 2

3 4

O

O

H

O

O

H

Problem

C4H8O2

SODAR = 1

IR: 1720 cm-1 » C=O


Now, exclude further structures based on

multiplicities and chemical shifts

O H

O

O

O

O H

O

O

O

HO

O

H

O

O

O

O

HO

O

1H NMR

R/S

55

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 = hnin a 7.04 T magnet

g: 1H, 42.576 MHzT−1 n = 300 MHz13C, 10.705 MHzT−1 n = 75 MHz

- Larger range of chemical shifts (0 to >200 ppm)

56

/ ppm

200 100 50 0150

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

57

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.

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C4H5

IR 100

50

0

Wavenumber / cm-1

Mass Spec

2-lines

Elemental Analysis: C, 90.38; H, 9.47 Problem A

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C5H7

Mass Spec: M+ m/e= 134

IR 100

50

0

Wavenumber / cm-1

Problem B

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C5H5O


IR 100

50

0

Wavenumber / cm-1

H2O

Problem C

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C5H10O2


IR 100

50

0

Wavenumber / cm-1

H2O

Problem D

4000 3000 2000 1500 1000 500

13C NMR 1H NMR



IR 100

50

0

Wavenumber / cm-1

Problem E

4000 3000 2000 1500 1000 500

13C NMR 1H NMR



IR 100

50

0

Wavenumber / cm-1

Problem F

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C3H8O2


IR 100

50

0

Wavenumber / cm-1

Problem G

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C7H9N


IR 100

50

0

Wavenumber / cm-1

Problem H

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C7H9N


IR 100

50

0

Wavenumber / cm-1

Problem I

4000 3000 2000 1500 1000 500

13C NMR 1H NMR



IR 100

50

0

Wavenumber / cm-1

2-lines

Problem J

4000 3000 2000 1500 1000 500

13C NMR 1H NMR



IR 100

50

0

Wavenumber / cm-1

Problem K

fall 2017 chem 6371 – spectroscopic identification of...

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