Tables. For. Organic Structure Analysis
|
|
- Berniece Stokes
- 5 years ago
- Views:
Transcription
1 Tables For Organic Structure Analysis prepared by Prof. Marcel Jaspars
2 Magnetic properties of commonly studied species. Nucleus Natural abundance (%) Approximate sensitivity at constant Bo for natural abundance 1 Resonance frequency at 2.35 T (100 Mz) Relative magnetic moment ( ) Relative quadrupole moment (Q) I = 1/2 [2] * *13C x N a x *19F Si a x *31P Free electrona 65, I = 0 [0] 12C O Si Si S S 4.4 I = 1 [3] x N x I = 3/2 [4] 11B Na S x Cl x Cl x Br Br I = 5/2 [6] 17O a x I I = 3 [7] 10B x [ ] Number of energy levels. a Negative magnetic moment. * Most useful. 1 Data from Bruker Almanac adjusted for natural abundance
3 Some useful NMR solvents Solvent 1 (ppm) (mult.) 13C (ppm) (mult.) (1) 20.0 (7) Liquid Range ( C) Dielectric Constant OD (ppm) in 1 NMR Acetic Acid-d (1) 2.04 (5) Acetone-d (5) (13) (7) Acetonitrile-d (5) (1) (7) Benzene-d (1) (3) Chloroform-d 7.27 (1) 77.2 (3) Cyclohexane-d (1) 26.4 (5) D2O Dichloromethane-d (3) 54.0 (5) p-dioxane-d (m) 66.7 (5) DMF-d (1) 2.92 (5) 2.75 (5) (3) 34.9 (7) 29.8 (7) DMSO-d (5) 39.5 (7) Methanol-d (1) 49.2 (7) (5) Pyridine-d (1) 7.58 (1) 7.22 (1) (3) (3) (5) TF-d (1) 1.73 (1) Toluene-d (m) 7.00 (1) 6.98 (m) 2.09 (5) 67.6 (5) 25.4 (1) (1) (3) (3) (3) 20.4 (7) TFA-d (1) (4) (4)
4 1 NMR shifts of common impurities in various solvents ( ppm (mult)) Impurity Chloroform-d DMSO-d6 Pyridine-d5 Benzene-d6 D2O Acetic Acid 2.13 (s) 1.95 (s) 2.13 (s) 1.63 (s) 2.16 (s) Acetone 2.17 (s) 2.12 (s) 2.00 (s) 1.62 (s) 2.22 (s) Acetonitrile 1.98 (s) 2.09 (s) 1.85 (s) 0.67 (s) 2.05 (s) Benzene 7.37 (s) 7.40 (s) 7.33 (s) 7.30 (s) 7.44 (s) t-butanol 1.28 (s) 1.14 (s) 1.37 (s) 1.06 (s) 1.23 (s) Chloroform 7.27 (s) 8.35 (s) 8.41 (s) 6.41 (s) ns Cyclohexane 1.43 (s) 1.42 (s) 1.38 (s) 1.40 (s) ns Dichloromethane 5.30 (s) 5.79 (s) 5.62 (s) 4.46 (s) ns Diethyl Ether 3.48 (q) 3.42 (q) 3.38 (q) 3.27 (q) 3.56 (q) 1.20 (t) 1.13 (t) 1.12 (t) 1.10 (t) 1.17 (t) DMF 8.01 (s) 7.98 (s) 7.91 (s) 2.95 (s) 2.92 (s) 2.72 (s) 2.40 (s) 3.00 (s) 2.88 (s) 2.76 (s) 2.66 (s) 1.98 (s) 2.86 (s) DMSO 2.62 (s) 2.52 (s) 2.49 (s) 1.91 (s) 2.70 (s) p-dioxane 3.70 (s) 3.61 (s) 3.61 (s) 3.38 (s) 3.75 (s) Ethanol 3.72 (q) 3.49 (q) 3.86 (q) 3.39 (q) 3.64 (q) 1.24 (t) 1.09 (t) 1.29 (t) (t) 1.16 (t) Ethyl Acetate 4.12 (q) 4.08 (q) 4.06 (q) 3.91 (q) 4.14 (q) 2.04 (s) 2.02 (s) 1.94 (s) 1.68 (s) 2.08 (s) 1.25 (t) 1.21 (t) 1.10 (t) 0.94 (t) 1.23 (t) Methanol 3.48 (s) 3.20 (s) 3.57 (s) 3.09 (s) 3.35 (s) Petroleum Ether 1.28 (bs) 1.28 (bs) 1.20 (bs) 1.22 (s) ns 0.90 (t) 0.89 (t) 0.86 (t) 0.89 (t) i-propanol 4.03 (m) 4.16 (m) 3.76 (m) 1.22 (d) 1.06 (d) 1.29 (d) 1.01 (d) 1.18 (d) n-propanol 3.60 (t) 3.75 (t) 3.76 (t) 3.61 (t) 1.60 (m) 1.45 (m) 1.70 (m) 1.40 (m) 1.57 (m) 0.93 (t) 0.87 (t) 0.97 (t) 0.80 (t) 0.89 (t) Pyridine 8.60 (m) 8.61 (m) 8.71 (m) 8.50 (m) 8.50 (m) 7.69 (m) 7.83 (m) 7.58 (m) 7.05 (m) 7.90 (m) 7.28 (m) 7.21 (m) 7.21 (m) 6.70 (m) 7.47 (m) TF 3.74 (m) 3.63 (m) 3.67 (m) 3.01 (m) 3.75 (m) 1.85 (m) 1.78 (m) 1.64 (m) 0.87 (m) 1.88 (m) Toluene 7.19 (m) 7.22 (m) 7.22 (m) 7.10 (m) ns 2.34 (s) 2.32 (s) 2.22 (s) 2.13 (s) Triethylamine 2.56 (q) 2.47 (q) 2.43 (q) 2.40 (q) 2.59 (q) 1.03 (t) 0.99 (t) 0.96 (t) 0.95 (t) 1.02 (t) ns=not soluble From: J. Org. Chem. 1997, 62,
5 Characteristics of Important Spectrometric Methods in Common Use -1 C-13 MS IR/RAMAN UV-VIS ORD/CD X-RAY Radiation type RF RF Not relevant IR UV to visible UV to visible X-ray Spectral scale 0-15 ppm ppm amu cm nm nm Not relevant Average sample mg 5 mg 1 mg 1 mg 1 mg 1 mg Single Crystal Molecular formula Partial Partial Yes No No No Yes Functional groups Yes Yes Limited Yes Very Limited Very Limited Yes Substructures Yes Limited Yes Limited Limited No Yes Carbon connectivity Yes Yes No No No No Yes Substituent regiochemistry Substituent stereochemistry Analysis of isomer mixtures Yes Yes No Limited No No Yes Yes Yes No Limited No No Yes Yes Yes Yes (by GC/MS LC/MS) Yes (by GC/IR) No No Yes (if separate) Purity information Yes Yes Yes Yes Limited Limited Limited What is measured Peak areas Chemical shifts Coupling Relaxation Chemical shifts Coupling Relaxation Singly or multiple charged ions Vibrational transitions Electronic transitions [ ] Relative atom positions R/S absolute stereochemistry Typical units (ppm) (ppm) m/z cm -1 nm nm - 5 E A 6 Typical Representations log 4 3 K B ppm (nm) -1, nm ORTEP ppm 1
6 1 NMR Chemical Shifts in Organic Compounds ppm Phenol -O Alcohols-O Thiols-S Amines-N 2 Amides Carboxylic acids-o Aldehydes eteroaromatics O RCONR O 2 C - CO- RCON 2 N Aromatics Z Alkenes Z=O, N, S RC=CR Z 2 2 C=CR 2 -C-O- ó -C-X Alcohols, alogens Amines R-N 2 R -C-N- ó -C 2 -N -C 2 -O- Alkynes ó- C C Methyl groups (R-Me) C 2 -X C 3 -O- C 3 -N C 3 -S- C 3 -Ph C 3 -C-X C 3 -CO- C 3 -C=C- Methylenes (-C 2 -) Ph-C 2 - O -CO-C 2 - -C 2 -C=C- R-C 2 -R 2 C Cyclopropyl Metals (M-C 3 ) M-C 3 ppm Relative to TMS (0 ppm) R O R N 2 N Z Z=O, N, S S R
7 13 C NMR Chemical Shifts in Organic Compounds sp 2 sp sp 3 4 o 3 o 2 o 1 o ppm alogen (C Tertiary) 230 C=O Ketone C=O Aldehyde C=O Acid C=O Ester, Amide C=S Thioketone C=N - Azomethine C=N - eteroaromatic C=C Alkene C=C Aromatic C=C eteroaromatic C C Alkyne C N Nitrile C Quat. C-O- C-N- C-S C-X C-C C-O- C-N- C-S- C-X alogen C 2 -C (C Secondary) C 2 -O- C 2 -N- C 2 -S- C 2 -X alogen C 3 -C (C Primary) C 3 -O- C 3 -N- C 3 -S- C 3 -X alogen ppm Cyclopropanes Epoxides *Relative to TMS (0 ppm) 4 o Quaternary, 3 o Tertiary, 2 o Secondary, 1 o Primary
8 15 N NMR Chemical Shifts in Organic Compounds Aliphatic amine Amonium salt Dimethyleneamine Anilinium ion Aniline ppm Piperidine Cyclic enamine Aminophosphine Guanidine Urea, carbamate, lactame Amide ppm Primary Secondary Tertiary Primary Secondary Tertiary Thiourea Thioamide Nitramine Indole, pyrrole idrazone, triazene Imide Nitrile, isonitrile Pyrrazole Diazocompound Diazonium salt Pyridine Imine Oxime Nitrocompound Nitrosocompound Ar-NO S-NO *Nitrometane as reference (380 ppm) N-NO N=N-N Pyridine Pyrazine Imine Terminal Tertiary Internal Pyrimidine Imidazole Primary Secondary NO 2 Imine Iso N Amine Amine
9 31 P NMR Chemical Shifts in Organic Compounds ppm Z-P=P-Z Z 2 N-P=NR ArP=C=Z 2 ArP=C=Ar ArP=C=Ar 2 ArP=C=O X=P(Y)=Z Z 3 P (TMS) 3 P Cl 3 P (RO) 3 P (Ar) 3 P (Me) 3 P 3 P Z-C P TMS-C P R-C P FC P R 3 -P-R R 2 P-PR 2 Z 3 P=O R 3 P=CR 2 Z 5 P R 2 P - M + Z 6 P - R(OR) 2 P=O Cl(OR) 2 P=O (OR) 5 P R 3 P=O RO 3 P=O Cl 2 RP=O R 2 (OR) 3 P R 4 P ppm *Reference 3 PO 4 85%. Adapted from A Complete Introduction to Modern NMR Spectroscopy. R. S. Macomber. J. Wiley and Sons
10 Symmetry Effects for omotopic; Enantiotopic, and Diasterotopic Groups GROUP A/B Relationships Methylenes Example Appearance 1 RMN Achiral solvent Appearance 1 RMN Chiral solvent omotopic Enantiotopic ppm ppm Diastereotopic ppm ppm Methyls ppm ppm omotopic Enantiotopic ppm ppm Diastereotopic ppm ppm ppm ppm
11 First-order Splitting Patterns of Some Common Spin Systems. Entry Structure Notation and Pattern for a Entry Structure a multiplets J ab =J ac d dd 1 7 b b' J ab > J ab' t t 2 8 ab > 0; b = b' J ab = J ab' t dd 3 9 J ac > J ab q dt 4 10 J ab >J ac q ddd 5 11 J ab >J ab' >J ac sp dq 6 12 J ab > J ac
12 Common Coupled Spin Systems Entry No. spins Spin Type J values by inspection Examples 1 2 AX Yes 2 2 AB Yes 3 3 AMX ABX ABC Yes J AB and J (AX+BX) None 4 3 A 2 X Yes 5 3 A 2 B Yes 6 4 A 3 X A 3 B Yes Yes 7 4 AMX 2 Yes 8 4 ABX 2 J AB rare ABCX ABCD None None X Y X Y 9 4 A 2 X 2 A 2 B 2 Yes No 10 4 AA'XX' J AX when J AA' or J XX' is small No when large 11 4 AA'BB' No 12 5 A 3 X 2 A 3 B A 3 MX A 3 XY Yes Yes Yes J XY
13 Examples of Magnetic and Chemical Equivalence. Type 1 Group Symmetry Chemical Equivalence Magnetic Equivalence Spectral Type 1 Spectral Appearance C 2 Cl 2 omotopic Yes Yes A ppm C 2 F 2 omotopic Yes Yes A 2 X ppm omotopic Yes Yes A ppm omotopic Yes No AA'XX' ppm omotopic Yes Yes A ppm Two homotopic sets Yes No AA'BB' ppm omotopic vinyl s Yes Yes A 2 X ppm Two homotopic sets Yes No AA'BB' ppm
14 13 C sp 3 Chemical Shifts Increment Values (ppm) a Primary Secondary b Z Tertiary c Z ' Cyclohexyl Z Base values [Z=] CN 3 3-3±1 0± ±2 3± C C ±1-4±2-2±1 C=C ±3 4±1-3±2 0 C 2 C I ±2-2± C 3 9 8±1-1±1-1±1 5±1 9±1-4± ring COO ±1 0±2 0±2 0 14±1 2± C ±1 0 17± ±1 5± CO N ±1-1±2 22±1 7±2-4± COC ±1-3± Br ±2-2±2 28±1 11±1-3±1-1± COCl ±1-1± NR ±1-1± Cl ±1-2± NR ±1-2± O 49±1 10-4±2-1±2 42±3 10±3-4±2-1±1 38± OCOR OR ±2-1±2 48±3 6± F a Some values based on calculated s; b : use 23.2 as base value and -increment; c, : use 29.8 as base values and -increment
15 1 NMR Substituted Double Bond Substituent Increment Values (ppm) a R t =R trans ; R c =R cis ; R g =R geminal Substituent gem cis trans C Alkyl Alkyl ring b C C 2 C C=C 2 (isol) C=C 2 (conj) c C C CN CO COR (isol) COR (conj) c COO (isol) COO (conj) c COCl COOR (isol) COOR (conj) c CON OR OC 6 5, O-conj OCOR C 2 O NR N(C 6 5 ) 2, N(conj) C 2 N NO F C 2 F Cl C 2 Cl Br C 2 Br I C 2 I a Calculate shifts using (C=C) = gem. + cis + trans b Based when the double bond is part of a ring c Used when the substituent or double bond in question is further conjugated to sp 2 systems C. Pascual, J. Meier, W. Simon elv. Chim. Acta 1966, 49, 164
16 13 C NMR Substituted Double Bond Chemical Shift Increment Values (ppm) Substituent X (base value) C C C C C C=C C C CN CO COC COO COCl COOC OC OC OCOC C 2 O N(C 3 ) NO F Cl C 2 Cl Br C 2 Br I C 2 I E. Prestch, A. Fürst, W. Robien Anal. Chim. Acta 1991, 248, 415.
17 1 NMR Benzene Ring Shift Increment Values (ppm) Substituent X (base value) C C C(C 3 ) C(C 3 ) C 2 Cl CF 3 Cl C 2 O C C=C C C CN CO COC COC COC(C 3 ) COO COCl COOC COOPh CON O OC OC OCOC OCOPh OSO 2 C N N(C 3 ) N(C 3 ) NCOC NMeCOC NN NO N=NPh S SC SPh SiMe PO(OMe) F Cl Br I
18 13 C NMR Benzene Ring Shift Increment Values (ppm) X Substituent X C-1 C-2 C-3 C-4 (base value) C C 2 C C 2 C 2 C 2 C C(C 3 ) C(C 3 ) Cyclopropyl Ring C C=C C C CF CCl CN C 2 CN C 2 COO C 2 O C 2 N C 2 Cl C 2 Br C 2 SMe C 2 SOMe C 2 Br CO COC COCF COC COO COO COCl COOC CON CON(C 3 ) O ONa OC OC OC=C OCOC OCN OSiMe OPO(OPh) N N(C 3 ) N(C 3 ) NC N(C 6 5 ) NCOC NO NC NCS S SMe SCMe SPh SOC SO 2 C SO 2 Cl SO SO 2 OMe Si SiMe SnMe F Cl Br I Pretsch, E., Furst, A.; Robien, W. Anal. Chim. Acta 1991, 248,
19 Summary of 2 J Values
20 Summary of 3 J Values J sp 3 carbons Type J(z) Type J(z) Type J(z) C 2 C 3 R OR C 3 C 2 Cl C 3 C 2 Br Cl OR RO OR O N O cis 7 trans 6 cis 5 trans 3 cis 6 trans 4 cis 10 trans 5 cis 9 trans 7 ax eq eq ax ax eq O ax exo endo exo endo cis 6-9 trans 5-7 cis 2-6 trans 7-10 ax - ax =12 ax - eq =4 eq - eq =4 ax - ax =11 ax - eq =4 exo - exo =10 endo - endo =13 endo - endo =4 sp 2 carbons Type J(z) Type J(z) Type J(z) Type J(z) O c c O N b a b a J ab =1.8 J bc =3.4 J ab =2.6 J bc =3.5 c c c c N b a b a b a b a J ab =5.5 J bc = J ab =9.6 J bc =5.1 J ab =7.5 J bc =7.5 J ab =5.5 J bc =7.5 N N O N O N S N c b a J ab =6.8 J bc =8.0 R 1 R 1 X X R 2 R 2 R 1 R X=alogen 7 X=alogen
21 Summary of 3 J Values (cont.)
22 Summary of 4 J and 5 J Coupling Constants
Tables. For. Organic Structure Analysis
Tables For Organic Structure Analysis Magnetic properties of commonly studied species. Nucleus Natural abundance (%) Approximate sensitivity at constant Bo for natural abundance 1 Resonance frequency at
More informationInterpretation of Organic Spectra. Chem 4361/8361
Interpretation of Organic Spectra Chem 4361/8361 Characteristics of Common Spectrometric Methods H-1 C-13 MS IR/RAMAN UV-VIS ORD/CD X- RAY Radiation type RF RF Not relevant IR UV to visible UV to visible
More informationChapter 9. Nuclear Magnetic Resonance. Ch. 9-1
Chapter 9 Nuclear Magnetic Resonance Ch. 9-1 1. Introduction Classic methods for organic structure determination Boiling point Refractive index Solubility tests Functional group tests Derivative preparation
More informationTable 8.2 Detailed Table of Characteristic Infrared Absorption Frequencies
Table 8.2 Detailed Table of Characteristic Infrared Absorption Frequencies The hydrogen stretch region (3600 2500 cm 1 ). Absorption in this region is associated with the stretching vibration of hydrogen
More informationIntroduction to NMR spectroscopy
Introduction to NMR spectroscopy Nuclei of isotopes which possess an odd number of protons, an odd number of neutrons, or both, have a nuclear spin quantum number, I, such that, I = 1/2n, where n is an
More informationGeneral Infrared Absorption Ranges of Various Functional Groups
General Infrared Absorption Ranges of Various Functional Groups Frequency Range Bond Type of Compound cm -1 Intensity C Alkanes 2850-2970 Strong 1340-1470 Strong C Alkenes 3010-3095 Medium 675-995 Strong
More informationInfrared Spectroscopy An Instrumental Method for Detecting Functional Groups
Infrared Spectroscopy An Instrumental Method for Detecting Functional Groups 1 The Electromagnetic Spectrum Infrared Spectroscopy I. Physics Review Frequency, υ (nu), is the number of wave cycles that
More informationE35 SPECTROSCOPIC TECHNIQUES IN ORGANIC CHEMISTRY
E35 SPECTRSCPIC TECNIQUES IN RGANIC CEMISTRY Introductory Comments. These notes are designed to introduce you to the basic spectroscopic techniques which are used for the determination of the structure
More informationC h a p t e r S i x t e e n: Nuclear Magnetic Resonance Spectroscopy. An 1 H NMR FID of ethanol
0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 C h a p t e r S i x t e e n: Nuclear Magnetic Resonance Spectroscopy An 1 NMR FID of ethanol Note: Problems with italicized numbers
More informationAnswers to Assignment #5
Answers to Assignment #5 A. 9 8 l 2 5 DBE (benzene + 1 DBE) ( 9 2(9)+2-9 8+1+1 = 10 ˆ 5 DBE) nmr pattern of two doublets of equal integration at δ7.4 and 7.9 ppm means the group (the δ7.9 shift) IR band
More informationInfrared Characteristic Group Frequencies
Infrared Characteristic Group Frequencies Tables and Charts Second Edition GEORGE SOCRATES Brunei, The University of West London, Middlesex, United Kingdom JOHN WILEY & SONS Chichester New York Brisbane
More informationNuclear spin and the splitting of energy levels in a magnetic field
Nuclear spin and the splitting of energy levels in a magnetic field Top 3 list for 13 C NMR Interpretation 1. Symmetry 2. Chemical Shifts 3. Multiplicity 13 C NMR of C 3 O 1 NMR of C 3 O 13 C NMR of C
More informationTo Do s. Answer Keys are available in CHB204H
To Do s Read Chapters 2, 3 & 4. Complete the end-of-chapter problems, 2-1, 2-2, 2-3 and 2-4 Complete the end-of-chapter problems, 3-1, 3-3, 3-4, 3-6 and 3-7 Complete the end-of-chapter problems, 4-1, 4-2,
More informationSPECTROSCOPY MEASURES THE INTERACTION BETWEEN LIGHT AND MATTER
SPECTROSCOPY MEASURES THE INTERACTION BETWEEN LIGHT AND MATTER c = c: speed of light 3.00 x 10 8 m/s (lamda): wavelength (m) (nu): frequency (Hz) Increasing E (J) Increasing (Hz) E = h h - Planck s constant
More information+ + CH 11: Substitution and Elimination Substitution reactions
C 11: Substitution and Elimination Substitution reactions Things to sort out: Nucleophile Electrophile -- > substrate Leaving Group S N 2 S N 1 E 1 E 2 Analysis Scheme Kinetics Reaction profile Substrates
More informationTo Do s. Answer Keys are available in CHB204H
To Do s Read Chapters 2, 3 & 4. Complete the end-of-chapter problems, 2-1, 2-2, 2-3 and 2-4 Complete the end-of-chapter problems, 3-1, 3-3, 3-4, 3-6 and 3-7 Complete the end-of-chapter problems, 4-1, 4-2,
More informationOAT Organic Chemistry - Problem Drill 19: NMR Spectroscopy and Mass Spectrometry
OAT Organic Chemistry - Problem Drill 19: NMR Spectroscopy and Mass Spectrometry Question No. 1 of 10 Question 1. Which statement concerning NMR spectroscopy is incorrect? Question #01 (A) Only nuclei
More informationChapter 13: Nuclear Magnetic Resonance (NMR) Spectroscopy direct observation of the H s and C s of a molecules
hapter 13: Nuclear Magnetic Resonance (NMR) Spectroscopy direct observation of the s and s of a molecules Nuclei are positively charged and spin on an axis; they create a tiny magnetic field + + Not all
More informationNuclear Spin States. NMR Phenomenon. NMR Instrumentation. NMR Active Nuclei. Nuclear Magnetic Resonance
Nuclear Magnetic Resonance NMR Phenomenon µ A spinning charged particle generates a magnetic field. A nucleus with a spin angular momentum will generate a magnetic moment (!). E Nuclear Spin States aligned
More informationCHEM 203. Final Exam December 15, 2010 ANSWERS. This a closed-notes, closed-book exam. You may use your set of molecular models
CEM 203 Final Exam December 15, 2010 Your name: ANSWERS This a closed-notes, closed-book exam You may use your set of molecular models This test contains 15 pages Time: 2h 30 min 1. / 16 2. / 15 3. / 24
More informationCHEM 203. Midterm Exam 1 October 31, 2008 ANSWERS. This a closed-notes, closed-book exam. You may use your set of molecular models
CEM 203 Midterm Exam 1 ctober 31, 2008 Your name: ANSWERS This a closed-notes, closed-book exam You may use your set of molecular models This exam contains 8 pages Time: 1h 30 min 1. / 15 2. / 16 3. /
More informationCHEMISTRY 341. Final Exam Tuesday, December 16, Problem 1 15 pts Problem 9 8 pts. Problem 2 5 pts Problem pts
CEMISTRY 341 Final Exam Tuesday, December 16, 1997 Name NAID Problem 1 15 pts Problem 9 8 pts Problem 2 5 pts Problem 10 21 pts Problem 3 26 pts Problem 11 15 pts Problem 4 10 pts Problem 12 6 pts Problem
More informationLecture 11. IR Theory. Next Class: Lecture Problem 4 due Thin-Layer Chromatography
Lecture 11 IR Theory Next Class: Lecture Problem 4 due Thin-Layer Chromatography This Week In Lab: Ch 6: Procedures 2 & 3 Procedure 4 (outside of lab) Next Week in Lab: Ch 7: PreLab Due Quiz 4 Ch 5 Final
More informationPHARMACEUTICAL CHEMISTRY EXAM #1 Februrary 21, 2008
PHARMACEUTICAL CHEMISTRY EXAM #1 Februrary 21, 2008 1 Name SECTION B. Answer each question in this section by writing the letter corresponding to the best answer on the line provided (2 points each; 60
More informationChapter 13: Molecular Spectroscopy
Chapter 13: Molecular Spectroscopy Electromagnetic Radiation E = hν h = Planck s Constant (6.63 x 10-34 J. s) ν = frequency (s -1 ) c = νλ λ = wavelength (nm) Energy is proportional to frequency Spectrum
More information1. Which compound would you expect to have the lowest boiling point? A) NH 2 B) NH 2
MULTIPLE CICE QUESTINS Topic: Intermolecular forces 1. Which compound would you expect to have the lowest boiling point? A) N 2 B) N 2 C) N D) E) N Ans: : N 2 D Topic: Molecular geometry, dipole moment
More informationInfrared Spectroscopy
Infrared Spectroscopy IR Spectroscopy Used to identify organic compounds IR spectroscopy provides a 100% identification if the spectrum is matched. If not, IR at least provides information about the types
More informationExam I 19 April 2004 Name:
Chem 317 S04 Page 1 of 7 Kantorowski Exam I 19 April 2004 Name: This exam contains 6 pages of questions confirm this once you begin The last page is a spectral data table that can be removed You will have
More informationθ-1) B z dθ = 0 r 3 θ=0
207 NMR Spectroscopy and rganic Structure Determination Scott Virgil California Institute of Technology SPIN-SPIN CUPLING IN NMR SPECTRSCPY Interaction of Nuclear Spins in NMR Spectroscopy A. Dipolar Coupling
More information1. LiAlH4 :.. :.. 2. H3O +
Ch 24 Amines Description of Amines - An amine is a compound with a nitrogen atom that has single bonds to carbon and hydrogen atoms. - An uncharged nitrogen atom normally has three bonds and a lone pair.
More informationChemistry 343- Spring 2008
Chemistry 343- Spring 2008 27 Chapter 2- Representative Carbon Compounds: Functional Groups, Intermolecular Forces and IR Spectroscopy A. ydrocarbons: Compounds composed of only C and Four Basic Types:
More informationObjective 4. Determine (characterize) the structure of a compound using IR, NMR, MS.
Objective 4. Determine (characterize) the structure of a compound using IR, NMR, MS. Skills: Draw structure IR: match bond type to IR peak NMR: ID number of non-equivalent H s, relate peak splitting to
More informationChapter 9. Organic Chemistry: The Infinite Variety of Carbon Compounds. Organic Chemistry
Chapter 9 Organic Chemistry: The Infinite Variety of Carbon Compounds Organic Chemistry Organic chemistry is defined as the chemistry of carbon compounds. Of tens of millions of known chemical compounds,
More information22 and Applications of 13 C NMR
Subject Chemistry Paper No and Title Module No and Title Module Tag 12 and rganic Spectroscopy 22 and Applications of 13 C NMR CHE_P12_M22 TABLE F CNTENTS 1. Learning utcomes 2. Introduction 3. Structural
More informationChapter 16 Nuclear Magnetic Resonance Spectroscopy
hapter 16 Nuclear Magnetic Resonance Spectroscopy The Spinning Proton A spinning proton generates a magnetic field, resembling that of a small bar magnet. An odd number of protons in the nucleus creates
More informationTuesday, January 13, NMR Spectroscopy
NMR Spectroscopy NMR Phenomenon Nuclear Magnetic Resonance µ A spinning charged particle generates a magnetic field. A nucleus with a spin angular momentum will generate a magnetic moment (μ). If these
More informationUV-Vis Spectroscopy. Chem 744 Spring Gregory R. Cook, NDSU Thursday, February 14, 13
UV-Vis Spectroscopy Chem 744 Spring 2013 UV-Vis Spectroscopy Every organic molecule absorbs UV-visible light Energy of electronic transitions saturated functionality not in region that is easily accessible
More informationAll Classes of Organic Compounds
Amines All Classes of Organic Compounds ydrocarbons Functionalized ydrocarbons F,Cl,Br O,S, Alkanes Alkenes Alkynes Aromatics alides -O- -S- -- Alcohols Phenols Ethers Thiols Dissulfides Amines O C O C
More informationInfrared Spectroscopy
Infrared Spectroscopy Introduction Spectroscopy is an analytical technique which helps determine structure. It destroys little or no sample. The amount of light absorbed by the sample is measured as wavelength
More informationCH 3. mirror plane. CH c d
CAPTER 20 Practice Exercises 20.1 The index of hydrogen deficiency is two. The structural possibilities include two double bonds, a double do 20.3 (a) As this is an alkane, it contains only C and and has
More informationChemistry 14C Winter 2017 Exam 2 Solutions Page 1
Chemistry 14C Winter 2017 Exam 2 Solutions Page 1 Statistics: High score, average, and low score will be posted on the course web site after exam grading is complete. Some questions have more than one
More informationTo Do s. Read Chapter 3. Complete the end-of-chapter problems, 3-1, 3-3, 3-4, 3-6 and 3-7. Answer Keys are available in CHB204H
Read Chapter 3. To Do s Complete the end-of-chapter problems, 3-1, 3-3, 3-4, 3-6 and 3-7 Answer Keys are available in CB204 NMR Chemical Shifts Further Discussion A set of spectral data is reported when
More informationFerdowsi University of Mashhad
Spectroscopy in Inorganic Chemistry 2 Diatomic molecule C v and D h HCN H-H 3 contribution orbital electron Σ 0 σ 1 Π 1 π 1 Δ 2 δ 1 Φ 3 δ 1 Σ + Σ - 4 Linear molecule NO 2s+1 2 Π A 1 =Σ + 0 A 2 =Σ - 0 E
More informationChapter 13 Structure t Determination: Nuclear Magnetic Resonance Spectroscopy
John E. McMurry www.cengage.com/chemistry/mcmurry Chapter 13 Structure t Determination: ti Nuclear Magnetic Resonance Spectroscopy Revisions by Dr. Daniel Holmes MSU Paul D. Adams University of Arkansas
More informationCARBOXYLIC ACIDS AND THEIR DERIVATIVES. 3. Predict the relative acidity and basicity of compounds and ions. Important criteria include:
CARBXYLIC ACIDS AND TEIR DERIVATIVES A STUDENT SULD BE ABLE T: 1. Give the IUPAC name given the structure, and draw the structure given the name, of carboxylic acids and their metal salts, acyl chlorides,
More informationCHEMISTRY 263 HOME WORK
Lecture Topics: CHEMISTRY 263 HOME WORK Module7: Hydrogenation of Alkenes Hydrogenation - syn and anti- addition - hydrogenation of alkynes - synthesis of cis-alkenes -synthesis of trans-alkenes Text sections:
More informationMore information can be found in Chapter 12 in your textbook for CHEM 3750/ 3770 and on pages in your laboratory manual.
CHEM 3780 rganic Chemistry II Infrared Spectroscopy and Mass Spectrometry Review More information can be found in Chapter 12 in your textbook for CHEM 3750/ 3770 and on pages 13-28 in your laboratory manual.
More informationModule9. Nuclear Magnetic Resonance Spectroscopy Nuclear Magnetic Resonance (NMR) spectroscopy - Chemical shift - Integration of signal area
1 CHEMISTRY 263 HOME WORK Lecture Topics: Module7. Hydrogenation of Alkenes The Function of the Catalyst - Syn and anti- addition Hydrogenation of Alkynes - Syn- addition of hydrogen: Synthesis of cis-alkenes
More informationIR, MS, UV, NMR SPECTROSCOPY
CHEMISTRY 318 IR, MS, UV, NMR SPECTROSCOPY PROBLEM SET All Sections CHEMISTRY 318 IR, MS, UV, NMR SPECTROSCOPY PROBLEM SET General Instructions for the 318 Spectroscopy Problem Set Consult the Lab Manual,
More informationMassachusetts Institute of Technology Organic Chemistry Hour Exam #1. Name. Official Recitation Instructor
Massachusetts Institute of Technology rganic Chemistry. Friday, September 0, 00 Prof. Timothy F. Jamison Hour Exam # Name (please both print and sign your name) fficial Recitation Instructor Directions:
More informationEXPT. 7 CHARACTERISATION OF FUNCTIONAL GROUPS USING IR SPECTROSCOPY
EXPT. 7 CHARACTERISATION OF FUNCTIONAL GROUPS USING IR SPECTROSCOPY Structure 7.1 Introduction Objectives 7.2 Principle 7.3 Requirements 7.4 Strategy for the Interpretation of IR Spectra 7.5 Practice Problems
More informationSpectroscopy in Inorganic Chemistry. Electronic Absorption Spectroscopy
Spectroscopy in Inorganic Chemistry Diatomic molecule C v and D h NO H-H 2 contribution orbital Σ 0 σ Π 1 π Δ 2 δ Φ 3 δ 3 Linear molecule NO 2s+1 2 Π A 1 =Σ + 0 A 2 =Σ - 0 E 1 =Π 1 E 2 =Δ 2 E 3 =Φ 3 4
More informationN_HW1 N_HW1. 1. What is the purpose of the H 2 O in this sequence?
N_HW1 N_HW1 Multiple Choice Identify the choice that best completes the statement or answers the question. There is only one correct response for each question. 1. What is the purpose of the H 2 O in this
More informationORGANIC - CLUTCH CH ANALYTICAL TECHNIQUES: IR, NMR, MASS SPECT
!! www.clutchprep.com CONCEPT: PURPOSE OF ANALYTICAL TECHNIQUES Classical Methods (Wet Chemistry): Chemists needed to run dozens of chemical reactions to determine the type of molecules in a compound.
More informationUNIVERSITY OF CALGARY FACULTY OF SCIENCE MIDTERM EXAMINATION CHEMISTRY 353 READ ALL THE INSTRUCTIONS CAREFULLY
WEDNESDAY MARCH 9th, 2016 UNIVERSITY OF CALGARY FACULTY OF SCIENCE MIDTERM EXAMINATION CHEMISTRY 353 Version 1 Time: 2 Hours READ ALL THE INSTRUCTIONS CAREFULLY PLEASE WRITE YOUR NAME, STUDENT I.D. NUMBER
More informationCH 320/328 N Summer II 2018
CH 320/328 N Summer II 2018 HW 1 Multiple Choice Identify the choice that best completes the statement or answers the question. There is only one correct response for each question. (5 pts each) 1. Which
More informationORGANIC - CLUTCH CH ANALYTICAL TECHNIQUES: IR, NMR, MASS SPECT
!! www.clutchprep.com CONCEPT: PURPOSE OF ANALYTICAL TECHNIQUES Classical Methods (Wet Chemistry): Chemists needed to run dozens of chemical reactions to determine the type of molecules in a compound.
More informationChapter 13 Nuclear Magnetic Resonance Spectroscopy
William. Brown Christopher S. Foote Brent L. Iverson Eric Anslyn http://academic.cengage.com/chemistry/brown Chapter 13 Nuclear Magnetic Resonance Spectroscopy William. Brown Beloit College Two Nobel Prizes
More informationMASS SPECTROMETRY: BASIC EXPERIMENT
http://science.widener.edu/svb/massspec/ei.html relative abundance Pavia 8.1-8.5 MASS SPECTROMETRY: BASIC EXPERIMENT scienceaid.co.uk -e Molecule Molecule +. + 2e base peak [Fragments] +. fragment peaks
More informationCalculate a rate given a species concentration change.
Kinetics Define a rate for a given process. Change in concentration of a reagent with time. A rate is always positive, and is usually referred to with only magnitude (i.e. no sign) Reaction rates can be
More informationChemistry 11 Hydrocarbon Alkane Notes. In this unit, we will be primarily focusing on the chemistry of carbon compounds, also known as.
1 Chemistry 11 Hydrocarbon Alkane Notes In this unit, we will be primarily focusing on the chemistry of carbon compounds, also known as. Why is organic chemistry so important? Many of the compounds that
More informationInfrared Spectroscopy
x-rays ultraviolet (UV) visible Infrared (I) microwaves radiowaves near I middle I far I λ (cm) 8 x 10-5 2.5 x 10-4 2.5 x 10-3 2.5 x 10-2 µ 0.8 2.5 25 250 ν (cm -1 ) 13,000 4,000 400 40 ν (cm -1 1 ) =
More informationChem 213 Final 2012 Detailed Solution Key for Structures A H
Chem 213 Final 2012 Detailed Solution Key for Structures A H COMPOUND A on Exam Version A (B on Exam Version B) C 8 H 6 Cl 2 O 2 DBE = 5 (aromatic + 1) IR: 1808 cm 1 suggests an acid chloride since we
More informationMOLECULAR REPRESENTATIONS AND INFRARED SPECTROSCOPY
MOLEULAR REPRESENTATIONS AND INFRARED SPETROSOPY A STUDENT SOULD BE ABLE TO: 1. Given a Lewis (dash or dot), condensed, bond-line, or wedge formula of a compound draw the other representations. 2. Give
More informationChapter 24 From Petroleum to Pharmaceuticals
hapter 24 From Petroleum to Pharmaceuticals 24.1 Petroleum Refining and the ydrocarbons 24.2 Functional Groups and Organic Synthesis 24.3 Pesticides and Pharmaceuticals IR Tutor and Infrared Spectroscopy
More information1.1. IR is part of electromagnetic spectrum between visible and microwave
CH2SWK 44/6416 IR Spectroscopy 2013Feb5 1 1. Theory and properties 1.1. IR is part of electromagnetic spectrum between visible and microwave 1.2. 4000 to 400 cm -1 (wave numbers) most interesting to organic
More informationNuclear Magnetic Resonance Spectroscopy Chem 4010/5326: Organic Spectroscopic Analysis Andrew Harned
Nuclear Magnetic Resonance Spectroscopy Chem 4010/5326: Organic Spectroscopic Analysis 2015 Andrew Harned NMR Spectroscopy NMR Spectroscopy All nuclei have a nuclear spin quantum number (I) I = 0, 1/2,
More informationChem Final Examination August 7, 2004
Chem 281 2004-2 Final Examination August 7, 2004 Name: Student Number: Note: You are allowed to use models for this exam. Notes, textbooks and calculators are strictly prohibited. Write your final answers
More information12. Structure Determination: Mass Spectrometry and Infrared Spectroscopy
12. Structure Determination: Mass Spectrometry and Infrared Spectroscopy Determining the Structure of an Organic Compound The analysis of the outcome of a reaction requires that we know the full structure
More information2016 Pearson Education, Inc. Isolated and Conjugated Dienes
2016 Pearson Education, Inc. Isolated and Conjugated Dienes 2016 Pearson Education, Inc. Reactions of Isolated Dienes 2016 Pearson Education, Inc. The Mechanism Double Bonds can have Different Reactivities
More informationOrganic Chemistry 1 CHM 2210 Exam 4 (December 10, 2001)
Exam 4 (December 10, 2001) Name (print): Signature: Student ID Number: There are 12 multiple choice problems (4 points each) on this exam. Record the answers to the multiple choice questions on THIS PAGE.
More informationCHEM Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W Short Answer 1. For a nucleus to exhibit the nuclear magnetic resonance phenomenon, it must be magnetic. Magnetic nuclei include: a. all
More informationChemistry 3720 Old Exams. Practice Exams & Keys
Chemistry 3720 ld Exams Practice Exams & Keys 2015-17 Spring 2017 Page File 3 Spring 2017 Exam 1 10 Spring 2017 Exam 1 Key 16 Spring 2017 Exam 2 23 Spring 2017 Exam 2 Key 29 Spring 2017 Exam 3 36 Spring
More informationAlkanes 3/27/17. Hydrocarbons: Compounds made of hydrogen and carbon only. Aliphatic (means fat ) - Open chain Aromatic - ring. Alkane Alkene Alkyne
Alkanes EQ 1. How will I define Hydrocarbons? 2. Compare and contrast the 3 types of hydrocarbons (Alkanes, alkenes, alkynes). Hydrocarbons: Compounds made of hydrogen and carbon only. Aliphatic (means
More informationThe Final Learning Experience
Chemistry 416 Spectroscopy Fall Semester 1997 Dr. Rainer Glaser The Final Learning Experience Monday, December 15, 1997 3:00-5:00 pm Name: Answer Key Maximum Question 1 (Combination I) 20 Question 2 (Combination
More informationLearning Guide for Chapter 14 - Alcohols (I)
Learning Guide for Chapter 14 - Alcohols (I) I. Introduction to Alcohols and Thiols II. Acid/base Behavior of Alcohols, Phenols, and Thiols III. Nomenclature of Alcohols IV. Synthesis of Alcohols Previous
More informationCan you differentiate A from B using 1 H NMR in each pair?
Can you differentiate A from B using 1 H NMR in each pair? To be NMR active any nucleus must have a spin quantum number, different from zero (I 0) As in 1 H, the spin quantum number (I) of 13 C is 1/2
More informationQ.8. Isomers are the compounds that must have same
Choose the single correct answer for each of the following questions: (1 Mark each). Q. 1. ow many lone pairs are present in C 3 --C 3 i) 3 ii) 2 iii) 1 iv) 4 Q.2. When any hydrocarbon is added to water,
More informationKOT 222 Organic Chemistry II
KOT 222 Organic Chemistry II Course Objectives: 1) To introduce the chemistry of alcohols and ethers. 2) To study the chemistry of functional groups. 3) To learn the chemistry of aromatic compounds and
More informationChemistry 605 (Reich)
Chemistry 60 (Reich) SECD UR EXAM Sat. April 9, 0 Question/oints R-A /0 R-B / R-C /0 R-D / R-E /0 R-F /0 Total /00 Average 6 i 9 Mode Median 6 AB 7 BC Distribution from grade list (average: 6.; count:
More informationPaper 12: Organic Spectroscopy
Subject Chemistry Paper No and Title Module No and Title Module Tag Paper 12: Organic Spectroscopy 31: Combined problem on UV, IR, 1 H NMR, 13 C NMR and Mass - Part III CHE_P12_M31 TABLE OF CONTENTS 1.
More information1. What is the major organic product obtained from the following sequence of reactions?
CH320 N N_HW1 Multiple Choice Identify the choice that best completes the statement or answers the question. There is only one correct response for each question. Carefully record your answers on the Scantron
More information1. Predict the structure of the molecules given by the following spectral data: a Mass spectrum:m + = 116
Additional Problems for practice.. Predict the structure of the molecules given by the following spectral data: a Mass spectrum:m + = IR: weak absorption at 9 cm - medium absorption at cm - NMR 7 3 3 C
More informationElectrophilic Aromatic Substitution (Aromatic compounds) Ar-H = aromatic compound 1. Nitration Ar-H + HNO 3, H 2 SO 4 Ar-NO 2 + H 2 O 2.
Electrophilic Aromatic Substitution (Aromatic compounds) Ar- = aromatic compound 1. Nitration Ar- + NO 3, 2 SO 4 Ar- + 2 O 2. Sulfonation Ar- + 2 SO 4, SO 3 Ar-SO 3 + 2 O 3. alogenation Ar- + X 2, Fe Ar-X
More informationCHEM 322 Laboratory Methods in Organic Chemistry. Introduction to NMR Spectroscopy
EM 322 Laboratory Methods in Organic hemistry Introduction to NMR Spectroscopy What structural information does NMR spectroscopy provide? 1) hemical shift (δ) data reveals the molecular (functional group)
More informationCarbon Compounds. Chemical Bonding Part 2
Carbon Compounds Chemical Bonding Part 2 Introduction to Functional Groups: Alkanes! Alkanes Compounds that contain only carbons and hydrogens, with no double or triple bonds.! Alkyl Groups A part of a
More informationORGANIC - BRUICE 8E CH MASS SPECT AND INFRARED SPECTROSCOPY
!! www.clutchprep.com CONCEPT: PURPOSE OF ANALYTICAL TECHNIQUES Classical Methods (Wet Chemistry): Chemists needed to run dozens of chemical reactions to determine the type of molecules in a compound.
More informationCHEM 213 FALL 2016 MIDTERM EXAM 2 - VERSION A
CHEM 213 FALL 2016 MIDTERM EXAM 2 - VERSIN A Answer multiple choice questions on the green computer sheet provided with a PENCIL. Be sure to encode both your NAME and Registration Number (V#). You will
More information*Assignments could be reversed. *
Name Key 5 W-Exam No. Page I. (6 points) Identify the indicated pairs of hydrogens in each of the following compounds as (i) homotopic, (ii) enantiotopic, or (iii) diastereotopic s. Write the answers as
More informationInfra-red Spectroscopy
Molecular vibrations are associated with the absorption of energy (infrared activity) by the molecule as sets of atoms (molecular moieties) vibrate about the mean center of their chemical bonds. Infra-red
More informationStructure Determination: Nuclear Magnetic Resonance Spectroscopy
Structure Determination: Nuclear Magnetic Resonance Spectroscopy Why This Chapter? NMR is the most valuable spectroscopic technique used for structure determination More advanced NMR techniques are used
More informationAldehydes and Ketones Reactions. Dr. Sapna Gupta
Aldehydes and Ketones Reactions Dr. Sapna Gupta Reactions of Aldehydes and Ketones Nucleophilic Addition A strong nucleophile attacks the carbonyl carbon, forming an alkoxide ion that is then protonated.
More informationVibrations. Matti Hotokka
Vibrations Matti Hotokka Identify the stuff I ve seen this spectrum before. I know what the stuff is Identify the stuff Let s check the bands Film: Polymer Aromatic C-H Aliphatic C-H Group for monosubstituted
More informationACETONE. PRODUCT IDENTIFICATION CAS NO EINECS NO MOL WT H.S. CODE Oral rat LD50: 5800 mg/kg
ACETONE www.pawarchemicals.com PRODUCT IDENTIFICATION CAS NO 67-64-1 EINECS NO. 200-662-2 FORMULA (CH3)2C=O MOL WT. 58.08 H.S. CODE 2914.11 TOXICITY SYNONYMS Oral rat LD50: 5800 mg/kg Dimethyl ketone;
More informationCHEM 261 HOME WORK Lecture Topics: MODULE 1: The Basics: Bonding and Molecular Structure Text Sections (N0 1.9, 9-11) Homework: Chapter 1:
CHEM 261 HOME WORK Lecture Topics: MODULE 1: The Basics: Bonding and Molecular Structure Atomic Structure - Valence Electrons Chemical Bonds: The Octet Rule - Ionic bond - Covalent bond How to write Lewis
More informationBasic One- and Two-Dimensional NMR Spectroscopy
Horst Friebolin Basic One- and Two-Dimensional NMR Spectroscopy Third Revised Edition Translated by Jack K. Becconsall WILEY-VCH Weinheim New York Chichester Brisbane Singapore Toronto Contents XV 1 The
More informationChapter 14 Spectroscopy
hapter 14 Spectroscopy There are four major analytical techniques used for identifying the structure of organic molecules 1. Nuclear Magnetic Resonance or NMR is the single most important technique for
More informationCHEM 213 FALL 2018 MIDTERM EXAM 2 - VERSION A
CEM 213 FALL 2018 MIDTERM EXAM 2 - VERSIN A Answer multiple choice questions on the green computer sheet provided with a PENCIL. Be sure to encode both your NAME and Registration Number (V#). You will
More informationAromatic Hydrocarbons
Aromatic Hydrocarbons Aromatic hydrocarbons contain six-membered rings of carbon atoms with alternating single and double carbon-carbon bonds. The ring is sometimes shown with a circle in the center instead
More informationPart I. Multiple choice. (4 points each.) Choose the one best answer and mark your answer on the ScanTron sheet.
Test 3 Chemistry 2321 April 25, 2003 McMurry, Chapters 9-11 103 Total Points Name Please Print Part I. Multiple choice. (4 points each.) Choose the one best answer and mark your answer on the ScanTron
More information