CM Chemical Spectroscopy and Applications. Final Examination Solution Manual AY2013/2014

Size: px
Start display at page:

Download "CM Chemical Spectroscopy and Applications. Final Examination Solution Manual AY2013/2014"

Transcription

1 NANYANG TECHNOLOGICAL UNIVERSITY DIVISION OF CHEMISTRY AND BIOLOGICAL CHEMISTRY SCHOOL OF PHYSICAL & MATHEMATICAL SCIENCES CM Chemical Spectroscopy and Applications Final Examination Solution Manual AY2013/2014 Prepared by EUGENE ONG November 2017

2 Question 1 (a) The assignment of each proton is shown in Figure 1. Figure 1. Assignment of olefinic protons of methyl acrylate. Coupling constants in the olefin: Trans protons: Large. Cis protons: Medium. Geminal protons: Small. (b) The answers are summarized in Table 1. Compound 2 Three types of protons (H a, H b, H c /H c' ). H c and H c' are magnetically equivalent. H a couples with H b and H c /H c'. dt expected. Compound 3 Two types of protons (H a /H a', H b /H b' ). H a and H a' are magnetically inequivalent. H b and H b' are magnetically inequivalent. Compound 4 Four types of protons (H a, H b, H c, H d ). H a couples with H b, H c, and H d. ddd expected. Table 1. Analysis of Compounds 2 to 4. (c) Resonance structures of Compound 6 are shown in Figure 2. Figure 2. Resonance structures of Compound C NMR of asterisked carbon: Higher electron density in 6 than in 5 is expected from the resonance structure A. Thus lower chemical shift of 6 is expected. IR absorption of carbonyl group: Resonance structure C indicates that C=O bond of 6 has more single bond character than that of 5. Thus C=O stretching wavenumber will be smaller for 6. 1

3 UV absorption: NOT EXAMINABLE (d) Rule of 13: 97/13 = 7 + 6/13. Base formula = C7H13. Nitrogen rule. At least one N atom. C7H13 + N CH2 = C6H11N. Does not match with 1 H NMR integration. IR absorption at 1704 cm 1 and 13 C NMR peak at indicate the presence of C=O group(s) (acid, ester, amide). C6H11N + O CH4 = C5H7NO. Still does not match with 1 H NMR integration. Further modification of formula. C5H7NO + O CH4 = C4H3NO2. Match with 1 H NMR integration. U = 4. Cannot be a benzene derivative (only 4 C atoms) but contains multiple unsaturated bonds. Two types of carbons ( 13 C NMR). One is likely from symmetric olefin (XHC=CHX; 1 H NMR 6.73 ppm, 13 C NMR ppm), another from C=O (171.4 ppm). Because there are total 4 carbons, this should contain two equivalent C=O groups. Figure 3. Partial structures leading to Compound 7. Question 2 (a) U = 6. 1 H NMR: Tetrasubstituted benzene with three OCH3 groups. Two of the OCH3 groups should be chemically equivalent (symmetric structure expected) IR: C N bond (~2200 cm 1 ) indicated. Then candidates should be A and B. Figure 4. Partial structures leading to Compound 8 and regioisomer B. 1 H chemical shift is the final key. 1 H chemical shift of the aromatic region of 8 is ca. 6.1, and is much smaller than that of its regioisomer (6.88). Aromatic 1 H chemical shift of A should be smaller than that of B due to "triple" resonance effects of A (compared to "double" resonance effect of B). 2

4 (b) 1 H NMR: -CH2-CH2- unit (triplet-triplet in 2-4 ppm). Monosubstituted benzene (multiplets in 7-8 ppm). Carboxylic acid COOH (singlet at > 12 ppm). And one isolated proton (singlet at ~8.5 ppm). 13 C NMR: Two different C=O groups of acid, ester, or amide (~165, ~175 ppm). Four peaks in ppm are from monosubstituted benzene. Two peaks in ppm are from X-CH2-CH2- Y unit. IR: Two different C=O (~1700, ~1620 cm 1 ). The former should be acid (see 1 H NMR interpretation). This is consistent with broad COOH band (3300~2600). The latter is likely amide (because of small wavenumber). Sharp band at ~3300 indicates secondary amide (RNHC=O). In addition to the above partial structures, the information that "hydrolysis of 9 gives aromatic carboxylic acid" is the final clue to determine its structure. For example, structure like 9' can be excluded. Figure 5. Partial structures leading to Compound 9. (c) 1 H NMR: Two isolated CH3 groups (singlets in ~2-2.5 ppm), which should not be directly bonded to O. Para-disubstituted benzene with different substituents (two app. d signals in ~ ppm). One isolated proton, which may be from NH or OH (but not COOH) (singlet at ~8 ppm). 13 C NMR: Two different C=O groups of ester, acid, or amide (two peaks at ~170 ppm). But acid is impossible (see 1 H NMR and IR). Four peaks in ppm are from benzene ring. Two peaks in ppm are from two different CH3 groups. IR: Two different C=O (~1750, ~1690 cm 1 ). The former may be ester (definitely not acid because of 1 H NMR and absence of broad COO-H band). The latter is likely amide. Sharp band at ~3400 indicates secondary amide (RNHC=O). You are almost complete if you get the structure 10 or 10'. The key to distinguish these two is the C=O wavenumber of amide. 10' should have even smaller C=O wavenumber because of double resonance effects (phenyl and amino groups bonded to C=O). 3

5 Figure 6. Partial structures leading to Compound 10. (d) α-cleavage at the amide moiety of Compound 9 gives an acylium ion of m/z = 105, as shown in Figure 7. Figure 7. Mechanism for the fragmentation of Compound 9. Therefore, Compound 9 = Spectrum A α-cleavage at either the amide or ester moiety of Compound 10 gives an acylium ion of m/z = 43, as shown in Figure 8. Figure 8. Mechanism for the fragmentation of Compound 10. Therefore, Compound 10 = Spectrum C The remaining Spectrum B belongs to Compound 11. 4

6 Question 3 (a) The chemical shifts, coordination shifts and coupling constants of the given spectrum are determined, as shown below: 1. Chemical shifts: δp a (Cis-isomer) = [( )/2] = ppm δp a (Trans-isomer) = [( )/2] = ppm δp b (Trans-isomer) = [( )/2] = ppm δp b (Cis-isomer) = [( )/2] = ppm 2. Coupling constants: 2 JPP(Cis) = {[( ) + ( )]/2} x 161 MHz = Hz 2 JPP(Trans) = {[( ) + ( )]/2} x 161 MHz = Hz 3. Coordination shifts: Given: δligand 12 = ppm; δligand 13 = ppm. ΔP a (Cis-isomer) = (- 8.70) = ppm ΔP a (Trans-isomer) = (- 8.70) = ppm ΔP b (Trans-isomer) = (- 5.30) = ppm ΔP b (Cis-isomer) = (- 5.30) = ppm (b) Two isomers of Complex 11 are observed in solution and their structures are shown in Figure 9. Figure 9. Two isomers of Complex 11. (c) Assignment of signals: 1. Trans-isomer: δp a = ppm (doublet) δp b = ppm (doublet) 2. Cis-isomer: δp a = ppm (doublet) δp b = ppm (doublet) (d) There are two isomers in solution, the trans-isomer is the major isomer. From the given spectrum, the trans-isomer signals have larger peak area and intensity compared to the cisisomer signals. Integration of the peaks would give the isomeric ratio. 5

7 (e) Due to chelate effect, the bidentate Ligand 12 is not as labile as the monodentate triphenylphosphine ligand. Since Ligand 12 is not prone to dissociation, its P Pd bond is much more stable than the triphenylphosphine P Pd bond. (f) The expected 31 P { 1 H} NMR signals of Complex 11 at an operating frequency of 100 MHz is calculated below and the resulting spectrum is shown in Figure 10. We know that 2 JPP(Cis) = Hz and 2 JPP(Trans) = Hz. At an operating frequency of 100 MHz, the distance between the doublet peaks are ppm for the cis-isomer and ppm for the trans-isomer. We know that the chemical shifts do not change, so we can determine the position of the new signals based on the information obtained from (a) and the calculated peak distances. δp a (Cis-isomer) = ppm δp a (Trans-isomer) = ppm δp b (Trans-isomer) = ppm δp b (Cis-isomer) = ppm Let s start from the downfield region first: (0.2576/2) = ppm (0.2576/2) = ppm (4.2504/2) = ppm (4.2504/2) = ppm (4.2504/2) = ppm (4.2504/2) = ppm (0.2576/2) = ppm (0.2576/2) = ppm Figure P { 1 H} NMR spectrum of Complex 11 at an operating frequency of 100 MHz. xxx End of Solution Manual xxx 6

8 Supporting Information Disclaimer: The following spectra are identical to the given spectra in the final examination. The spectra are available through the Spectral Databases for Organic Compounds: Compound 7 7

9 8

10 9

11 Regioisomer of Compound 8 10

12 Compound 8 11

13 12

14 13 m/z Intensity (%)

15 Compound 9 14

16 15

17 16 m/z Intensity (%)

18 Compound 10 17

19 18

20 19 m/z Intensity (%)

Paper 12: Organic Spectroscopy

Paper 12: Organic Spectroscopy Subject hemistry Paper No and Title Module No and Title Module Tag Paper 12: Organic Spectroscopy 34: ombined problem on UV, IR, 1 H NMR, 13 NMR and Mass- Part 6 HE_P12_M34 TABLE OF ONTENTS 1. Learning

More information

Answers to Assignment #5

Answers 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 information

(b) How many hydrogen atoms are in the molecular formula of compound A? [Consider the 1 H NMR]

(b) How many hydrogen atoms are in the molecular formula of compound A? [Consider the 1 H NMR] CHEM 6371/4511 Name: The exam consists of interpretation of spectral data for compounds A-C. The analysis of each structure is worth 33.33 points. Compound A (a) How many carbon atoms are in the molecular

More information

Paper 12: Organic Spectroscopy

Paper 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 information

Structure solving based on IR, UV-Vis, MS, 1 H and 13 C NMR spectroscopic data. Problem solving session

Structure solving based on IR, UV-Vis, MS, 1 H and 13 C NMR spectroscopic data. Problem solving session Structure solving based on IR, UV-Vis, MS, 1 H and 13 C NMR spectroscopic data Problem solving session S. SANKARARAMAN DEPARTMENT OF CHEMISTRY INDIAN INSTITUTE OF TECHNOLOGY MADRAS CHENNAI 600036 sanka@iitm.ac.in

More information

CH 3. mirror plane. CH c d

CH 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 information

Chem 213 Final 2012 Detailed Solution Key for Structures A H

Chem 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 information

Answers to Problem Set #2

Answers to Problem Set #2 hem 242 Spring 2008 Answers to Problem Set #2 1. For this question we have been given the molecular formula, 3 5 l. Looking at the IR, the strong signal at 1720 cm 1 tells us that we have a carbonyl (we

More information

EXPT. 9 DETERMINATION OF THE STRUCTURE OF AN ORGANIC COMPOUND USING UV, IR, NMR AND MASS SPECTRA

EXPT. 9 DETERMINATION OF THE STRUCTURE OF AN ORGANIC COMPOUND USING UV, IR, NMR AND MASS SPECTRA EXPT. 9 DETERMINATION OF THE STRUCTURE OF AN ORGANIC COMPOUND USING UV, IR, NMR AND MASS SPECTRA Structure 9.1 Introduction Objectives 9.2 Principle 9.3 Requirements 9.4 Strategy for the Structure Elucidation

More information

STRUCTURE ELUCIDATION BY INTEGRATED SPECTROSCOPIC METHODS

STRUCTURE ELUCIDATION BY INTEGRATED SPECTROSCOPIC METHODS Miscellaneous Methods UNIT 14 STRUCTURE ELUCIDATION BY INTEGRATED SPECTROSCOPIC METHODS Structure 14.1 Introduction Objectives 14.2 Molecular Formula and Index of Hydrogen Deficiency 14.3 Structural Information

More information

ORGANIC SPECTROSCOPY NOTES

ORGANIC SPECTROSCOPY NOTES - 1 - ORGANIC SPECTROSCOPY NOTES Basics of Spectroscopy UV/vis, IR and NMR are all types of Absorption Spectroscopy, where EM radiation corresponding to exactly the energy of specific excitations in molecules

More information

The Final Learning Experience

The 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 information

Organic Chemistry 321 Workshop: Spectroscopy NMR-IR Problem Set

Organic Chemistry 321 Workshop: Spectroscopy NMR-IR Problem Set Organic Chemistry 321 Workshop: Spectroscopy NMR-IR Problem Set 1. Draw an NMR spectrum for each of the following compounds. Indicate each peak by a single vertical line (for example, a quartet would be

More information

IR, MS, UV, NMR SPECTROSCOPY

IR, 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 information

Lecture 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 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 information

CHEM Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W

CHEM 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 information

1. Predict the structure of the molecules given by the following spectral data: a Mass spectrum:m + = 116

1. 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 information

(Refer Slide Time: 0:37)

(Refer Slide Time: 0:37) Principles and Applications of NMR spectroscopy Professor Hanudatta S. Atreya NMR Research Centre Indian Institute of Science Bangalore Module 3 Lecture No 14 We will start today with spectral analysis.

More information

The resonance frequency of the H b protons is dependent upon the orientation of the H a protons with respect to the external magnetic field:

The resonance frequency of the H b protons is dependent upon the orientation of the H a protons with respect to the external magnetic field: Spin-Spin Splitting in Alkanes The signal arising from a proton or set of protons is split into (N+1) lines by the presence of N adjacent nuclei Example 1: Bromoethane The resonance frequency of the H

More information

ORGANIC - CLUTCH CH ANALYTICAL TECHNIQUES: IR, NMR, MASS SPECT

ORGANIC - 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 information

ORGANIC - CLUTCH CH ANALYTICAL TECHNIQUES: IR, NMR, MASS SPECT

ORGANIC - 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 information

Organic Chemistry II (CHE ) Examination I February 11, Name (Print legibly): Key. Student ID#:

Organic Chemistry II (CHE ) Examination I February 11, Name (Print legibly): Key. Student ID#: rganic hemistry II (HE 232-001) Examination I February 11, 2009 Name (Print legibly): Key (last) (first) Student ID#: PLEASE observe the following: You are allowed to have scratch paper (provided by me),

More information

Proton NMR. Four Questions

Proton NMR. Four Questions Proton NMR Four Questions How many signals? Equivalence Where on spectrum? Chemical Shift How big? Integration Shape? Splitting (coupling) 1 Proton NMR Shifts Basic Correlation Chart How many 1 H signals?

More information

Infrared Spectroscopy

Infrared 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 information

1.1 Is the following molecule aromatic or not aromatic? Give reasons for your answer.

1.1 Is the following molecule aromatic or not aromatic? Give reasons for your answer. Page 1 QUESTION ONE 1.1 Is the following molecule aromatic or not aromatic? Give reasons for your answer. 1.2 List four criteria which compounds must meet in order to be considered aromatic. Page 2 QUESTION

More information

Module 20: Applications of PMR in Structural Elucidation of Simple and Complex Compounds and 2-D NMR spectroscopy

Module 20: Applications of PMR in Structural Elucidation of Simple and Complex Compounds and 2-D NMR spectroscopy Subject Chemistry Paper No and Title Module No and Title Module Tag Paper 12: Organic Spectroscopy Module 20: Applications of PMR in Structural Elucidation of Simple and Complex Compounds and 2-D NMR spectroscopy

More information

Using NMR and IR Spectroscopy to Determine Structures Dr. Carl Hoeger, UCSD

Using NMR and IR Spectroscopy to Determine Structures Dr. Carl Hoeger, UCSD Using NMR and IR Spectroscopy to Determine Structures Dr. Carl Hoeger, UCSD The following guidelines should be helpful in assigning a structure from NMR (both PMR and CMR) and IR data. At the end of this

More information

OAT Organic Chemistry - Problem Drill 19: NMR Spectroscopy and Mass Spectrometry

OAT 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 information

ORGANIC - EGE 5E CH UV AND INFRARED MASS SPECTROMETRY

ORGANIC - EGE 5E CH UV AND INFRARED MASS SPECTROMETRY !! www.clutchprep.com CONCEPT: IR SPECTROSCOPY- FREQUENCIES There are specific absorption frequencies in the functional group region that we should be familiar with EXAMPLE: What are the major IR absorptions

More information

Chapter 15 Lecture Outline

Chapter 15 Lecture Outline Organic Chemistry, First Edition Janice Gorzynski Smith University of Hawaii Chapter 5 Lecture Outline Introduction to NMR Two common types of NMR spectroscopy are used to characterize organic structure:

More information

Organic Chemistry II (CHE ) Examination I March 1, Name (Print legibly): _KEY _. Student ID#: _

Organic Chemistry II (CHE ) Examination I March 1, Name (Print legibly): _KEY _. Student ID#: _ Organic Chemistry II (CHE 232-002) Examination I March 1, 2007 Name (Print legibly): _KEY _ (last) (first) Student ID#: _ PLEASE observe the following: You are allowed to have scratch paper (provided by

More information

Organic Chemistry II* (CHE ) Examination I March 1, Name (Print legibly): KEY. Student ID#:

Organic Chemistry II* (CHE ) Examination I March 1, Name (Print legibly): KEY. Student ID#: Organic Chemistry II* (CHE 232-002) Examination I March 1, 2007 Name (Print legibly): KEY (last) (first) Student ID#: PLEASE observe the following: You are allowed to have scratch paper (provided by me),

More information

Basic Concepts of NMR: Identification of the Isomers of C 4 O 2. by 1 H NMR Spectroscopy

Basic Concepts of NMR: Identification of the Isomers of C 4 O 2. by 1 H NMR Spectroscopy Basic Concepts of NM: Identification of the Isomers of C H 8 O by H NM Spectroscopy Objectives NM spectroscopy is a powerful tool in determining the structure of compounds. Not only is it able to give

More information

Objective 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. 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 information

Table 8.2 Detailed Table of Characteristic Infrared Absorption Frequencies

Table 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 information

PAPER No. 12: ORGANIC SPECTROSCOPY. Module 19: NMR Spectroscopy of N, P and F-atoms

PAPER No. 12: ORGANIC SPECTROSCOPY. Module 19: NMR Spectroscopy of N, P and F-atoms Subject Chemistry Paper No and Title Module No and Title Module Tag Paper 12: Organic Spectroscopy CHE_P12_M19_e-Text TABLE OF CONTENTS 1. Learning Outcomes 2. 15 N NMR spectroscopy 3. 19 F NMR spectroscopy

More information

CHAPTER 3 SOLUTIONS TO EXERCISES

CHAPTER 3 SOLUTIONS TO EXERCISES CHAPTER 3 SOLUTIONS TO EXERCISES Solution 3.1. 600.13 MHz 1 H spectrum (CDCl3): 6.921 (tt, J = 7.3, 1.1 Hz, 1H). The chemical shift is taken from the central line: 4153.66 Hz / 600.13 MHz = 6.9213 ppm.

More information

4. NMR spectra. Interpreting NMR spectra. Low-resolution NMR spectra. There are two kinds: Low-resolution NMR spectra. High-resolution NMR spectra

4. NMR spectra. Interpreting NMR spectra. Low-resolution NMR spectra. There are two kinds: Low-resolution NMR spectra. High-resolution NMR spectra 1 Interpreting NMR spectra There are two kinds: Low-resolution NMR spectra High-resolution NMR spectra In both cases the horizontal scale is labelled in terms of chemical shift, δ, and increases from right

More information

Experiment 11: NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY

Experiment 11: NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY Experiment 11: NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY Purpose: This is an exercise to introduce the use of nuclear magnetic resonance spectroscopy, in conjunction with infrared spectroscopy, to determine

More information

SPECTROSCOPY MEASURES THE INTERACTION BETWEEN LIGHT AND MATTER

SPECTROSCOPY 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

Look for absorption bands in decreasing order of importance:

Look for absorption bands in decreasing order of importance: 1. Match the following to their IR spectra (30 points) Look for absorption bands in decreasing order of importance: a e a 2941 1716 d f b 3333 c b 1466 1.the - absorption(s) between 3100 and 2850 cm-1.

More information

ORGANIC - BROWN 8E CH NUCLEAR MAGNETIC RESONANCE.

ORGANIC - BROWN 8E CH NUCLEAR MAGNETIC RESONANCE. !! www.clutchprep.com CONCEPT: 1 H NUCLEAR MAGNETIC RESONANCE- GENERAL FEATURES 1 H (Proton) NMR is a powerful instrumental method that identifies protons in slightly different electronic environments

More information

Technical Note. Introduction

Technical Note. Introduction Technical Note Characterization of Eleven 2,5-Dimethoxy-N-(2-methoxybenzyl)- phenethylamine (NBOMe) Derivatives and Differentiation from their 3- and 4- Methoxybenzyl Analogues - Part II Patrick A. Hays*,

More information

12. Structure Determination: Mass Spectrometry and Infrared Spectroscopy

12. 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 information

CHEMISTRY 213 XMAS 04 1 ANSWER ON THE GREEN COMPUTER ANSWER SHEET PROVIDED USING A PENCIL CHOICES MAY BE USED MORE THAN ONCE

CHEMISTRY 213 XMAS 04 1 ANSWER ON THE GREEN COMPUTER ANSWER SHEET PROVIDED USING A PENCIL CHOICES MAY BE USED MORE THAN ONCE CHEMISTRY 213 XMAS 04 1 ANSWER ON THE GREEN COMPUTER ANSWER SHEET PROVIDED USING A PENCIL CHOICES MAY BE USED MORE THAN ONCE Using the molecules: A: CH 3 CH CHCO 2 CH 3 B: CH 3 CO 2 CH CHCH 3 C: CH 3 CH

More information

CHEM 213 FALL 2018 MIDTERM EXAM 2 - VERSION A

CHEM 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 information

AQA A2 CHEMISTRY TOPIC 4.10 ORGANIC SYNTHESIS AND ANALYSIS TOPIC 4.11 STRUCTURE DETERMINATION BOOKLET OF PAST EXAMINATION QUESTIONS

AQA A2 CHEMISTRY TOPIC 4.10 ORGANIC SYNTHESIS AND ANALYSIS TOPIC 4.11 STRUCTURE DETERMINATION BOOKLET OF PAST EXAMINATION QUESTIONS AQA A2 CHEMISTRY TOPIC 4.10 ORGANIC SYNTHESIS AND ANALYSIS TOPIC 4.11 STRUCTURE DETERMINATION BOOKLET OF PAST EXAMINATION QUESTIONS 1 1. Consider the following reaction sequence. CH 3 CH 3 CH 3 Step 1

More information

Exemplar for Internal Achievement Standard. Chemistry Level 3

Exemplar for Internal Achievement Standard. Chemistry Level 3 Exemplar for Internal Achievement Standard Chemistry Level 3 This exemplar supports assessment against: Achievement Standard 91388 Demonstrate understanding of spectroscopic data in chemistry. An annotated

More information

Chapter 14. Nuclear Magnetic Resonance Spectroscopy

Chapter 14. Nuclear Magnetic Resonance Spectroscopy Organic Chemistry, Second Edition Janice Gorzynski Smith University of Hawai i Chapter 14 Nuclear Magnetic Resonance Spectroscopy Prepared by Rabi Ann Musah State University of New York at Albany Copyright

More information

Structure Determination. How to determine what compound that you have? One way to determine compound is to get an elemental analysis

Structure Determination. How to determine what compound that you have? One way to determine compound is to get an elemental analysis Structure Determination How to determine what compound that you have? ne way to determine compound is to get an elemental analysis -basically burn the compound to determine %C, %H, %, etc. from these percentages

More information

ORGANIC - BROWN 8E CH INFRARED SPECTROSCOPY.

ORGANIC - BROWN 8E CH 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 information

Unit 2 Organic Chemistry. 2.3 Structural Analysis Part 2:

Unit 2 Organic Chemistry. 2.3 Structural Analysis Part 2: CFE ADVANCED HIGHER Unit 2 Organic Chemistry 2.3 Structural Analysis Part 2: Mass Spectroscopy Infra-red Spectroscopy NMR Proton Spectroscopy Answers to Questions in Notes Learning Outcomes Exam Questions

More information

Spectroscopy in Organic Chemistry. Types of Spectroscopy in Organic

Spectroscopy in Organic Chemistry. Types of Spectroscopy in Organic Spectroscopy in Organic Chemistry Spectroscopy Spectrum dealing with light, or more specifically, radiation Scope to see Organic Spectroscopy therefore deals with examining how organic molecules interact

More information

CHEM 242 NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY CHAP 14B ASSIGN

CHEM 242 NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY CHAP 14B ASSIGN CHEM 242 NUCLEAR MAGNETIC RESNANCE SPECTRSCPY CHAP 14B ASSIGN 1. A proton NMR spectrum is observed to contain following the pattern below; what do you conclude? A. This must be a quartet that is part of

More information

ORGANIC - BRUICE 8E CH MASS SPECT AND INFRARED SPECTROSCOPY

ORGANIC - 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 information

PAPER No.12 :Organic Spectroscopy MODULE No.29: Combined problem on UV, IR, 1 H NMR, 13 C NMR and Mass - Part I

PAPER No.12 :Organic Spectroscopy MODULE No.29: Combined problem on UV, IR, 1 H NMR, 13 C NMR and Mass - Part I Subject Chemistry Paper No and Title Module No and Title Module Tag 12: rganic Spectroscopy 29: Combined problem on UV, IR, 1 H NMR, 13 C NMR and Mass - Part I CHE_P12_M29 TABLE F CNTENTS 1. Learning utcomes

More information

Structure Determination

Structure Determination There are more than 5 million organic compounds, the great majority of which are colourless liquids or white solids. Identifying or at least characterising determining some of its properties and features

More information

Unit 3 Organic Chemistry. 3.3 Structural Analysis Part 2:

Unit 3 Organic Chemistry. 3.3 Structural Analysis Part 2: Unit 3 Organic Chemistry 3.3 Structural Analysis Part 2: Mass Spectroscopy Infra-red Spectroscopy NMR Proton Spectroscopy Answers to Questions in Notes Learning Outcomes Exam Questions & Answers MODIFIED

More information

Chem 2320 Exam 1. January 30, (Please print)

Chem 2320 Exam 1. January 30, (Please print) Chem 2320 Exam 1 January 30, 2006 Name: (first) (last) (Please print) Last 4 digits of I.D. I. Multiple Choice ( /20) Score /60 II /15 III /25 Total score /100 I. Multiple choice questions. (3 points each).

More information

Chapter 24. Amines. Based on McMurry s Organic Chemistry, 7 th edition

Chapter 24. Amines. Based on McMurry s Organic Chemistry, 7 th edition Chapter 24. Amines Based on McMurry s Organic Chemistry, 7 th edition Amines Organic Nitrogen Compounds Organic derivatives of ammonia, NH 3, Nitrogen atom with a lone pair of electrons, making amines

More information

CHEM 3.2 (AS91388) 3 credits. Demonstrate understanding of spectroscopic data in chemistry

CHEM 3.2 (AS91388) 3 credits. Demonstrate understanding of spectroscopic data in chemistry CHEM 3.2 (AS91388) 3 credits Demonstrate understanding of spectroscopic data in chemistry Spectroscopic data is limited to mass, infrared (IR) and 13 C nuclear magnetic resonance (NMR) spectroscopy. Organic

More information

General Infrared Absorption Ranges of Various Functional Groups

General 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 information

NUCLEAR MAGNETIC RESONANCE AND INTRODUCTION TO MASS SPECTROMETRY

NUCLEAR MAGNETIC RESONANCE AND INTRODUCTION TO MASS SPECTROMETRY NUCLEAR MAGNETIC RESONANCE AND INTRODUCTION TO MASS SPECTROMETRY A STUDENT SHOULD BE ABLE TO: 1. Identify and explain the processes involved in proton ( 1 H) and carbon-13 ( 13 C) nuclear magnetic resonance

More information

Introduction to Organic Spectroscopy

Introduction to Organic Spectroscopy Introduction to rganic Spectroscopy Chem 8361/4361: Interpretation of rganic Spectra 2009 2013 Andrew Harned & Regents of the University of Minnesota What is spectroscopy?? From Wikipedia Spectroscopy:

More information

11. Proton NMR (text , 12.11, 12.12)

11. Proton NMR (text , 12.11, 12.12) 2009, Department of Chemistry, The University of Western Ontario 11.1 11. Proton NMR (text 12.6 12.9, 12.11, 12.12) A. Proton Signals Like 13 C, 1 H atoms have spins of ±½, and when they are placed in

More information

Closed book exam, no books, notebooks, notes, etc. allowed. However, calculators, rulers, and molecular model sets are permitted.

Closed book exam, no books, notebooks, notes, etc. allowed. However, calculators, rulers, and molecular model sets are permitted. Massachusetts Institute of Technology Organic Chemistry 5.13 Friday, September 26, 2003 Prof. Timothy F. Jamison Hour Exam #1 Name (please both print and sign your name) Official Recitation Instructor

More information

CHEMISTRY 341. Final Exam Tuesday, December 16, Problem 1 15 pts Problem 9 8 pts. Problem 2 5 pts Problem pts

CHEMISTRY 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 information

Terms used in UV / Visible Spectroscopy

Terms used in UV / Visible Spectroscopy Terms used in UV / Visible Spectroscopy Chromophore The part of a molecule responsible for imparting color, are called as chromospheres. OR The functional groups containing multiple bonds capable of absorbing

More information

Name: Student Number: University of Manitoba - Department of Chemistry CHEM Introductory Organic Chemistry II - Term Test 2

Name: Student Number: University of Manitoba - Department of Chemistry CHEM Introductory Organic Chemistry II - Term Test 2 Name: Student Number: University of Manitoba - Department of Chemistry CHEM 2220 - Introductory Organic Chemistry II - Term Test 2 Thursday, March 12, 2015; 7-9 PM This is a 2-hour test, marked out of

More information

EXPT. 7 CHARACTERISATION OF FUNCTIONAL GROUPS USING IR SPECTROSCOPY

EXPT. 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 information

NMR Spectroscopy. Chapter 19

NMR Spectroscopy. Chapter 19 NMR Spectroscopy Chapter 19 Nuclear Magnetic Resonance spectroscopy is a powerful analytical technique used to characterize organic molecules by identifying carbon-hydrogen frameworks within molecules.

More information

Can First Year Organic Students Solve Complicated Organic Structures?

Can First Year Organic Students Solve Complicated Organic Structures? an First Year rganic Students Solve omplicated rganic Structures? Tools of the trade (for problems in this discussion): 1. Mass Spec information limited data (M+, M+1, M+2, M+3, etc.) 2. I information

More information

Page 2. Q1.Consider the five cyclic compounds, A, B, C, D and E. The infrared spectra of compounds A, B, C and D are shown below.

Page 2. Q1.Consider the five cyclic compounds, A, B, C, D and E. The infrared spectra of compounds A, B, C and D are shown below. Q1.Consider the five cyclic compounds, A, B, C, D and E. (a) The infrared spectra of compounds A, B, C and D are shown below. Write the correct letter, A, B, C or D, in the box next to each spectrum. You

More information

More information can be found in Chapter 12 in your textbook for CHEM 3750/ 3770 and on pages in your laboratory manual.

More 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 information

Interpretation of Organic Spectra. Chem 4361/8361

Interpretation 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 information

Chapter 1 Introduction, Formulas and Spectroscopy Overview

Chapter 1 Introduction, Formulas and Spectroscopy Overview hapter 1 Introduction, Formulas and Spectroscopy verview Problem 1 (p 1) - elpful equations: c = ( )( ) and = (1 / ) so c = ( ) / ( ) c =. x 1 8 m/sec =. x 1 1 cm/sec a. Which photon of electromagnetic

More information

Other problems to work: 3-Chloropentane (diastereotopic H s), 1- chloropentane.

Other problems to work: 3-Chloropentane (diastereotopic H s), 1- chloropentane. Let s look at some specific examples. Dichloroacetaldehyde, l 2 HHO, has two inequivalent toms, H1 and H2. We expect to see two resonances, one at around δ 10.5 ppm and one around δ 5.5 ppm. (The H2 resonance

More information

Chemistry 14C Winter 2017 Exam 2 Solutions Page 1

Chemistry 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 information

Group 2 compounds: C, H hydrocarbons = alkanyl (=R), alkenyl, alkynyl and aromatic. Answer similarly.

Group 2 compounds: C, H hydrocarbons = alkanyl (=R), alkenyl, alkynyl and aromatic. Answer similarly. Simulated I spectra Group 1 compounds: Functional group patterns are listed below. Match each structure with one of the 4 simulated I spectra that follow. The wave numbers listed in each spectrum are intended

More information

4) protons experience a net magnetic field strength that is smaller than the applied magnetic field.

4) protons experience a net magnetic field strength that is smaller than the applied magnetic field. 1) Which of the following CANNOT be probed by an spectrometer? See sect 15.1 Chapter 15: 1 A) nucleus with odd number of protons & odd number of neutrons B) nucleus with odd number of protons &even number

More information

NMR = Nuclear Magnetic Resonance

NMR = Nuclear Magnetic Resonance NMR = Nuclear Magnetic Resonance NMR spectroscopy is the most powerful technique available to organic chemists for determining molecular structures. Looks at nuclei with odd mass numbers or odd number

More information

HWeb27 ( ; )

HWeb27 ( ; ) HWeb27 (9.1-9.2; 9.12-9.18) 28.1. Which of the following cannot be determined about a compound by mass spectrometry? [a]. boiling point [b]. molecular formula [c]. presence of heavy isotopes (e.g., 2 H,

More information

1.1. IR is part of electromagnetic spectrum between visible and microwave

1.1. 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 information

Hour Examination # 4

Hour Examination # 4 CHEM 346 Organic Chemistry I Fall 2014 Exam # 4 Solutions Key Page 1 of 12 CHEM 346 Organic Chemistry I Fall 2014 Instructor: Paul Bracher Hour Examination # 4 Wednesday, December 3 rd, 2014 6:00 8:00

More information

Nuclear Magnetic Resonance Spectroscopy: Tools for Structure Determination

Nuclear Magnetic Resonance Spectroscopy: Tools for Structure Determination Nuclear Magnetic Resonance Spectroscopy: Tools for Structure Determination Chung-Ming Sun Department of Applied Chemistry National Chiao Tung University Hualien 300, Taiwan Introduction NMR (Nuclear Magnetic

More information

Your Name: Question 1. 2D-NMR: C 6 H 10 O 2. (20 points)

Your Name: Question 1. 2D-NMR: C 6 H 10 O 2. (20 points) Question 1. 2D-NMR: C 6 H 10 O 2. (20 points) Integrations show signals 3H 1 & 5, 2H for signal 4, and 1H each for signals 2 and 3. - Draw the structure. - Assign the hydrogens to signals 1 5 (that is,

More information

NMR Spectroscopy: Determination of Molecular Structures

NMR Spectroscopy: Determination of Molecular Structures Experiment 2 NMR Spectroscopy: Determination of Molecular Structures Reading: Handbook for Organic Chemistry Lab, chapters on NMR Spectroscopy (Chapter 18) and Identification of Compounds (Chapter 20).

More information

Infra-red Spectroscopy

Infra-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 information

Organic Chemistry II KEY March 25, a) I only b) II only c) II & III d) III & IV e) I, II, III & IV

Organic Chemistry II KEY March 25, a) I only b) II only c) II & III d) III & IV e) I, II, III & IV rganic Chemistry II KEY March 25, 2015 Exam 2: VERSIN A 1. Which of the following compounds will give rise to an aromatic conjugate base? E a) I only b) II only c) II & III d) III & IV e) I, II, III &

More information

Chapter 12 Mass Spectrometry and Infrared Spectroscopy

Chapter 12 Mass Spectrometry and Infrared Spectroscopy Organic Chemistry, 6 th Edition L. G. Wade, Jr. Chapter 12 Mass Spectrometry and Infrared Spectroscopy Jo Blackburn Richland College, Dallas, TX Dallas County Community College District 2006, Prentice

More information

Calculate a rate given a species concentration change.

Calculate 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 information

CH 320/328 N Summer II 2018

CH 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 information

(a) Name the alcohol and catalyst which would be used to make X. (2)

(a) Name the alcohol and catalyst which would be used to make X. (2) 1 The chemical X is an ester with formula CH 3 COOC(CH 3 ) 3 which occurs in raspberries and pears. It can be prepared in the laboratory by refluxing ethanoic acid with an alcohol in the presence of a

More information

CHEM 213 FALL 2016 MIDTERM EXAM 2 - VERSION A

CHEM 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

UCF - ORGANIC CHEMISTRY 2 - PROF. GERASIMOVA UCF PROF. GERASIMOVA EXAM REVIEW 1.

UCF - ORGANIC CHEMISTRY 2 - PROF. GERASIMOVA UCF PROF. GERASIMOVA EXAM REVIEW 1. UCF PROF. GERASIMOVA EXAM REVIEW 1 www.clutchprep.com 1 PRACTICE: Determine the index of hydrogen deficiency (degrees of unsaturation) for the following molecule. Antipsychotic - Haloperidol = C 21 H 23

More information

CHM 233 : Fall 2018 Quiz #9 - Answer Key

CHM 233 : Fall 2018 Quiz #9 - Answer Key HM 233 : Fall 2018 Quiz #9 - Answer Key Question 1 M25c How many different signals would you expect to see in a proton-decoupled carbon nmr spectrum of the following compound? A 3 B 4 6 D 8 3 1 2 carbons

More information

Name: 1. Ignoring C-H absorptions, what characteristic IR absorption(s) would be expected for the functional group shown below?

Name: 1. Ignoring C-H absorptions, what characteristic IR absorption(s) would be expected for the functional group shown below? Chemistry 262 Winter 2018 Exam 3 Practice The following practice contains 20 questions. Thursday s 90 exam will also contain 20 similar questions, valued at 4 points/question. There will also be 2 unknown

More information

Vibrations. Matti Hotokka

Vibrations. 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 information

Welcome to Organic Chemistry II

Welcome to Organic Chemistry II Welcome to Organic Chemistry II Erika Bryant, Ph.D. erika.bryant@hccs.edu Class Syllabus 3 CHAPTER 12: STRUCTURE DETERMINATION 4 What is this solution Soda Tea Coffee??? 5 What is this solution Soda Tea

More information

Chemistry 3720 Old Exams. Practice Exams & Keys

Chemistry 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 information