Paper 12: Organic Spectroscopy

Size: px
Start display at page:

Download "Paper 12: Organic Spectroscopy"

Transcription

1 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

2 TABLE OF CONTENTS 1. Learning Outcomes 2. Introduction 3. Problems and their solutions

3 1. Learning Outcomes After studying this module, you shall be able to Solve problem related with electronic transitions Learn how to differentiate molecule on the basis of IR spectroscopy Correlate spectra with structure of compound Interpret the spectroscopic data 2. Introduction The knowledge and concepts of UV-visible, IR, 1 H NMR, 13 C NMR and Mass help us in solving problems based on the experimental data. It will help us in analysing the experimental data to elucidate the structure of any organic compound. While analysing the data the following point must be kept in mind: In UV-visible spectroscopy; the types of bonds and electrons plays important role in understanding the electronic transitions. UV-visible spectroscopy gives information regarding the presence of conjugation, carbonyl group etc. The IR values gives information regarding the functional group present in the molecule The 1 H NMR tells us the number and environment of neighbouring hydrogens present. The 13 C NMR helps in getting the information about the type of carbon atom(s) present in the molecule. Mass spectral data gives information about the total mass and fragmentation pattern of the molecule. By combining all the information one can find the structure of the molecule.

4 3. Problems and their solutions Q. 1: How is PMR used to establish on which carbon, monochlorination of methyl ethyl ether occurs? A. 1: The three possible products with their spectra are described as follows: a.) CH3-CH2-O-CH2Cl (a triplet, a more downfield quartet and a most downfield singlet); (t) (q) (s) b.) CH3-CHCl-O-CH3 (a doublet, a quartet downfield and a singlet between them); (d) (q) (s) c.) ClCH2-CH2-O-CH3 (a singlet and two triplets, the downfield order is: t1 < s < t2) (t2) (t1) (s)

5 Q. 2: Problem: Find the structure of an unknown compound of molecular formula C9H12, using its 1 H NMR spectum and its IR spectrum. 1H NMR Spectrum: IR Spectrum:

6 A. 2: Step 1: Determine the degrees of unsaturation. Just plug the formula C9H12 into the equation for degrees of unsaturation: Degrees of Unsaturation = [((# of Carbons x 2)+2) - # of Hydrogens]/2. [((9 x 2)+2) - 12]/2 = [20-12]/2 = 4. (For non hydrocarbon elements: we treat nitrogen atoms as ½ of a carbon in the above equation, halides as 1 hydrogen, and ignore oxygens.) There are 4 degrees of unsaturation. With four or more degrees of unsaturation, you can often expect to find an aromatic ring in your unknown compound. Step 2: Look at the IR to find what functional groups are in the molecule. You have a jumble of peaks between 1800 and 2000 cm-1 in the IR, indicative of an aromatic ring. You can confirm the presence of the ring by looking at the NMR, at the peaks around 7 ppm. Step 3: Find out how many hydrogens each set of peaks represents. To get the relative ratios, it's best to use a ruler. Since that's difficult on a computer screen, I can tell you that the relative integration for the peaks are 5:1:6 for the three sets of signals. Since this adds up to the number of hydrogens in the molecular formula (12), the peaks at around 7 ppm represents 5 hydrogens, the peak at 3 ppm represents 1 hydrogen, and the peak at around 1 ppm represents 6 hydrogens. Step 3: Find all of the compound fragments. To find the compound fragments, look at the NMR integration. First, you notice the benzene peaks at around 7.0 ppm, which is the piece you have already identified from the IR. It integrates for 5H, so from the integration you know that the benzene ring is substituted exactly once (6H for unsubstituted benzene - 1H), since one of the ring hydrogens is must be replaced by a substituent. Write down this fragment (if you haven't already).

7 fragment worth C6H5 Now take the molecular formula and subtract out the aromatic piece you just figured out to see what you have left. The benzene substituent is a C6H5 fragment, so if you subtract that from C9H12 you get C3H7 (note that the benzene ring contains all 4 degrees of unsaturation: one for the ring and three for the three double bonds. This means you can have no more rings and no more double or triple bonds). C3H7 is what you have left. Now take a look at the other integrations. You have a set of peaks that integrates for 1H, so write that fragment down: The bonds coming off marked as lines are simply bonds not attached to hydrogens. The other peak is a doublet that integrates for 6H. Trouble you say? A carbon can't have six hydrogens coming off of it? This is really no problem. It just means that there are two symmetric 3- hydrogen groups (methyls) that are chemically equivalent in the NMR, so draw those., symmetric That's all of the fragments of the molecule. You can check to make sure all atoms are accounted for by subtracting out all of the fragments from the molecular formula. C9H12 - (Benzene fragment C6H5) - (CH fragment) - (2 CH3 fragments) = C0H0. Good, all atoms are accounted for. Step 5: Piecing it all together.

8 At this point there is only one way that the pieces can go together, and that is as shown. The 4 fragments come together to make isopropyl benzene: the four fragments come together to make Step 6: Checking the answer. You know this structure is correct because there was only one way to put the pieces together, but let's just check the peak splitting to make sure. For the tertiary carbon with one hydrogen, you would expect a septet (multiplet) that integrates for 1H. For the CH3s you would expect a doublet that integrates for 6. For the aromatic ring, you would expect several peaks at around 7 ppm that integrates for 5H. Q. 3: A compound (molecular formula: C4H10O) gave PMR spectrum consisting of two group of lines (multiplets) with relative intensities in the ratio of 3:2. The PMR spectrum of another compound having the same formula exhibited two lines with a relative area of 9:1. Identify these substances. A. 3: Diethyl ether and t-butylether Q. 4: The PMR spectra of dimethyl formamide shows two signals at 2.84 and 3.0 δ for the methyl protons at room temperature but a single sharp line appears at higher temperatures (165 C). A. 4: At the room temperature, the lone pair of electrons on the nitrogen atom increases sufficiently the double bond character of the (C-N) to restrict rotation at room temperature and the spectrometer senses two different methyl groups (cis and trans) w.r.t lone hydrogen.

9 At elevated temperatures, however, the rotation around the C-N bond is so rapid that eacg=h methyl group experiences the same time averaged environment. Q. 5: Identify the unknown compound using the following spectral information.

10 A. A. 5: Mass Spectrum Analysis: The integral molecular weight of the unknown is 88. Because this is an even number, the compound has either no nitrogen atoms or an even number of N atoms. The highest m/z peaks do not show evidence of the presence of atoms such as Cl or Br (which would have two peaks in 75:25 and 50:50 ratios, respectively). Elemental Analysis: None was given

11 IR Analysis: The absorption at 3000 cm 1 (and not to the left) shows only aliphatic (saturated) C H groups. The carbonyl, C=O, absorbs strongly at 1740 cm 1. The two intense peaks between cm 1 indicate the carbonyl belongs to the ester group. 13 C-NMR analysis: The NMR shows 4 different kinds of carbon atoms. The absorption at 172 ppm is typical of the carbonyl carbon. The upfield absorptions at 14 and 21 ppm are likely CH3 and/or CH2 groups (although no C H splitting patterns are given with this spectrum). The absorption at 61 ppm could be a C next to an electronegative atom. 1 H-NMR analysis: The NMR shows three different kinds of CH groups. The integration of 2:3:3 suggests a methylene ( CH2 ) and two methyls ( CH3). The CH group absorbing as a singlet cannot be next to a neighboring CH group (n+1=1 and so n=0). The upfield triplet must be next to a C bonded to 2H s (n+1=3) and the downfield quartet must be next to a C bonded to 3 H s (n+1=4). The triplet integrates for 3H s and so must be a CH3 group next to a methylene; the quartet integrates for 2 H s and so must be a CH2 next to a methyl. Thus, there is an ethyl group present, CH2CH3. Combined Analysis: The IR shows the only functional group present is an ester. There is no aromaticity or other unsaturation in the molecule (except for the C=O). This means that the absorption in the 13 C-NMR at 61 ppm cannot be due to a C=C group and the carbon causing the absorption is next to the oxygen atom of the ester. The presence of a C=O is confirmed in the 13 C-NMR. The ester group accounts for 44 amu of the 88 integral mass. The ethyl group is on one side of the ester functional group and accounts for 29 mass units. The remaining mass ( = 15 and belongs to the singlet methyl group. The methyl is on the other side of the ester functional group. Because the quartet is so far downfield in the range for methylene 1H-NMR absorption, the methylene must be bonded to the O of the ester. Unknown Identity: Thus, the unknown is identified as ethyl acetate. Q. 6: Biphenyl exhibits a very intense absorption band (Ɛmax = 19000) at 250 nm but its 2,2 derivative shows absorption almost similar to o-xylene (λmax 262, Ɛmax = 270). A. 6: Since the angle of twist in biphenyl is small, therefore conjugation between the rings is not affected. Biphenyl thus shows a very intense absorption band at 250 nm (K band). 2,2 - dimethylbiphenyl with methyl substituents in ortho positions, however, is more stable in twisted conformations as compared to biphenyl and suffers serious non-bonded compressions from the juxtaposed substituents. Thus, 2,2 - dimethylbiphenyl may be regarded as two moles

12 of o-xylene and shows absorption similar to o-xylene, due to loss of conjugation in the twisted conformation.

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

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

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

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

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

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

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

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

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

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

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

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

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

22 and Applications of 13 C NMR

22 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 information

PAPER No.12 :Organic Spectroscopy MODULE No.30: Combined problem on UV, IR, 1 H NMR, 13 C NMR and Mass - Part II

PAPER No.12 :Organic Spectroscopy MODULE No.30: Combined problem on UV, IR, 1 H NMR, 13 C NMR and Mass - Part II Subject Chemistry Paper No and Title Module No and Title Module Tag 12 : rganic Spectroscopy 30: Combined problem on UV, IR, 1 H NMR, 13 C NMR and Mass Part-II CHE_P12_M30 TABLE F CNTENTS 1. Learning utcomes

More information

Experiment 2 - NMR Spectroscopy

Experiment 2 - NMR Spectroscopy Experiment 2 - NMR Spectroscopy OBJECTIVE to understand the important role of nuclear magnetic resonance spectroscopy in the study of the structures of organic compounds to develop an understanding of

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

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

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

Chapter 20: Identification of Compounds

Chapter 20: Identification of Compounds Chemists are frequently faced with the problem of identifying unknown compounds. Environmental scientists may have to identify pollutants in soils and water, synthetic chemists may want to confirm that

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

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

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

Nuclear Magnetic Resonance Spectroscopy

Nuclear Magnetic Resonance Spectroscopy 13 Nuclear Magnetic Resonance Spectroscopy Solutions to In-Text Problems 13.1 (b) Apply Eq. 13.2b with = 360 MHz. chemical shift in Hz = δ = (4.40)(360) = 1584 Hz 13.2 (b) Follow the same procedure used

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

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

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

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

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

Spectroscopy. Empirical Formula: Chemical Formula: Index of Hydrogen Deficiency (IHD)

Spectroscopy. Empirical Formula: Chemical Formula: Index of Hydrogen Deficiency (IHD) Spectroscopy Empirical Formula: Chemical Formula: Index of Hydrogen Deficiency (IHD) A)From a structure: B)From a molecular formula, C c H h N n O o X x, Formula for saturated hydrocarbons: Subtract the

More information

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

CM Chemical Spectroscopy and Applications. Final Examination Solution Manual AY2013/2014 NANYANG TECHNOLOGICAL UNIVERSITY DIVISION OF CHEMISTRY AND BIOLOGICAL CHEMISTRY SCHOOL OF PHYSICAL & MATHEMATICAL SCIENCES CM 3011 - Chemical Spectroscopy and Applications Final Examination Solution Manual

More information

Chapter 9. Nuclear Magnetic Resonance. Ch. 9-1

Chapter 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 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

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

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

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

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

Module 13: Chemical Shift and Its Measurement

Module 13: Chemical Shift and Its Measurement Subject Chemistry Paper No and Title Module No and Title Module Tag Paper 12: Organic Spectroscopy CHE_P12_M13_e-Text TABLE OF CONTENTS 1. Learning Outcomes 2. Introduction 3. Shielding and deshielding

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

CHEMISTRY Organic Chemistry Laboratory II Spring 2019 Lab #5: NMR Spectroscopy

CHEMISTRY Organic Chemistry Laboratory II Spring 2019 Lab #5: NMR Spectroscopy Team Members: Unknown # CHEMISTRY 244 - Organic Chemistry Laboratory II Spring 2019 Lab #5: NMR Spectroscopy Purpose: You will learn how to predict the NMR data for organic molecules, organize this data

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

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

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

CHEM311 FALL 2005 Practice Exam #3

CHEM311 FALL 2005 Practice Exam #3 EM311 FALL 2005 Practice Exam #3 Instructions: This is a multiple choice / short answer practice exam. For the multiple-choice questions, there may be more than one correct answer. If so, then circle as

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

CHEM1102 Worksheet 4 Answers to Critical Thinking Questions Model 1: Infrared (IR) Spectroscopy

CHEM1102 Worksheet 4 Answers to Critical Thinking Questions Model 1: Infrared (IR) Spectroscopy CEM1102 Worksheet 4 Answers to Critical Thinking Questions Model 1: Infrared (IR) Spectroscopy 1. See below. Model 2: UV-Visible Spectroscopy 1. See below. 2. All of the above. 3. Restricted to the identification

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

Nuclear spin and the splitting of energy levels in a magnetic field

Nuclear 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 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 16.1 Chapter 16: 1 A) nucleus with odd number of protons & odd number of neutrons B) nucleus with odd number of protons &even number

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

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

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

16.1 Introduction to NMR. Spectroscopy

16.1 Introduction to NMR. Spectroscopy 16.1 Introduction to NMR What is spectroscopy? Spectroscopy NUCLEAR MAGNETIC RESNANCE (NMR) spectroscopy may be the most powerful method of gaining structural information about organic compounds. NMR involves

More information

Chapter 14 Spectroscopy

Chapter 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 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 16.1 Chapter 16: 1 A) nucleus with odd number of protons & odd number of neutrons B) nucleus with odd number of protons &even 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

16.1 Introduction to NMR Spectroscopy. Spectroscopy. Spectroscopy. Spectroscopy. Spectroscopy. Spectroscopy 4/11/2013

16.1 Introduction to NMR Spectroscopy. Spectroscopy. Spectroscopy. Spectroscopy. Spectroscopy. Spectroscopy 4/11/2013 What is spectroscopy? NUCLEAR MAGNETIC RESONANCE (NMR) spectroscopy may be the most powerful method of gaining structural information about organic compounds. NMR involves an interaction between electromagnetic

More information

10-1 You might start this exercise by drawing all of the isomers of C7H16 of which there are nine:

10-1 You might start this exercise by drawing all of the isomers of C7H16 of which there are nine: Copyright 2010 James K Whitesell 10-1 You might start this exercise by drawing all of the isomers of C7H16 of which there are nine: Pick one with both secondary and tertiary carbon atoms and simply add

More information

CHEM311 FALL 2005 Practice Exam #3

CHEM311 FALL 2005 Practice Exam #3 CHEM311 FALL 2005 Practice Exam #3 Instructions: This is a multiple choice / short answer practice exam. For the multiple-choice questions, there may be more than one correct answer. If so, then circle

More information

E35 SPECTROSCOPIC TECHNIQUES IN ORGANIC CHEMISTRY

E35 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 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

CHEM 241 UNIT 5: PART A DETERMINATION OF ORGANIC STRUCTURES BY SPECTROSCOPIC METHODS [MASS SPECTROMETRY]

CHEM 241 UNIT 5: PART A DETERMINATION OF ORGANIC STRUCTURES BY SPECTROSCOPIC METHODS [MASS SPECTROMETRY] CHEM 241 UNIT 5: PART A DETERMINATION OF ORGANIC STRUCTURES BY SPECTROSCOPIC METHODS [MASS SPECTROMETRY] 1 Introduction Outline Mass spectrometry (MS) 2 INTRODUCTION The analysis of the outcome of a reaction

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

Nuclear Magnetic Resonance Spectroscopy (NMR)

Nuclear Magnetic Resonance Spectroscopy (NMR) OCR Chemistry A 432 Spectroscopy (NMR) What is it? An instrumental method that gives very detailed structural information about molecules. It can tell us - how many of certain types of atom a molecule

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

CHEM Review for test 1 (ch12-15). F17. Stafford

CHEM Review for test 1 (ch12-15). F17. Stafford CHEM Multiple Choice Identify the choice that best completes the statement or answers the question. 1. What range of wavelengths corresponds to the infrared region of the electromagnetic spectrum? a. 200-400

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

Name: Student Number: University of Manitoba - Department of Chemistry CHEM Introductory Organic Chemistry II - Term Test 1

Name: Student Number: University of Manitoba - Department of Chemistry CHEM Introductory Organic Chemistry II - Term Test 1 Name: Student Number: University of Manitoba - Department of Chemistry CHEM 2220 - Introductory Organic Chemistry II - Term Test 1 Thursday, February 12, 2015; 7-9 PM This is a 2-hour test, marked out

More information

In a solution, there are thousands of atoms generating magnetic fields, all in random directions.

In a solution, there are thousands of atoms generating magnetic fields, all in random directions. Nuclear Magnetic Resonance Spectroscopy: Purpose: onnectivity, Map of - framework Process: In nuclear magnetic resonance spectroscopy, we are studying nuclei. onsider this circle to represent a nucleus

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

Department of Chemistry SUNY/Oneonta. Chem Organic Chemistry I. Examination #4 - December 7, 1998*

Department of Chemistry SUNY/Oneonta. Chem Organic Chemistry I. Examination #4 - December 7, 1998* Department of hemistry SUNY/neonta hem 221 - rganic hemistry I Examination #4 - December 7, 1998* INSTRUTINS --- This examination is in multiple choice format; the questions are in this Exam Booklet and

More information

Nuclear Magnetic Resonance Spectroscopy: Purpose: Connectivity, Map of C-H framework

Nuclear Magnetic Resonance Spectroscopy: Purpose: Connectivity, Map of C-H framework Nuclear Magnetic Resonance Spectroscopy: Purpose: Connectivity, Map of C- framework Four Factors of Proton NMR (PMR OR NMR):. Symmetry: Number of chemically different protons (symmetry) as shown by number

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

In a solution, there are thousands of atoms generating magnetic fields, all in random directions.

In a solution, there are thousands of atoms generating magnetic fields, all in random directions. Nuclear Magnetic Resonance Spectroscopy: Purpose: onnectivity, Map of - framework Process: In nuclear magnetic resonance spectroscopy, we are studying nuclei. onsider this circle to represent a nucleus

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

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

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

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

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

Structural Determination Of Compounds

Structural Determination Of Compounds EXPERIMENT 10 Mass Spectroscopy Structural Determination Of Compounds. Introduction - In mass spectrometry, a substance is bombarded with an electron beam having sufficient energy to fragment the molecule.

More information

Chapter 13 Structure t Determination: Nuclear Magnetic Resonance Spectroscopy

Chapter 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 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 NMR 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

More information

Name: Student Number:

Name: Student Number: Page 1 of 5 Name: Student Number: l University of Manitoba - Department of Chemistry CEM 2220 - Introductory rganic Chemistry II - Term Test 1 Thursday, February 14, 2008 This is a 2-hour test, marked

More information

Exam (6 pts) Show which starting materials are used to produce the following Diels-Alder products:

Exam (6 pts) Show which starting materials are used to produce the following Diels-Alder products: Exam 1 Name CHEM 212 1. (18 pts) Complete the following chemical reactions showing all major organic products; illustrate proper stereochemistry where appropriate. If no reaction occurs, indicate NR :

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

Structure Determination: Nuclear Magnetic Resonance Spectroscopy

Structure 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 information

Name: Student Number: University of Manitoba - Department of Chemistry CHEM Introductory Organic Chemistry II - Term Test 1

Name: Student Number: University of Manitoba - Department of Chemistry CHEM Introductory Organic Chemistry II - Term Test 1 Name: Student Number: University of Manitoba - Department of Chemistry CHEM 2220 - Introductory Organic Chemistry II - Term Test 1 Thursday, February 11, 2016; 7-9 PM This is a 2-hour test, marked out

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

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

Organic Chemistry 1 CHM 2210 Exam 4 (December 10, 2001)

Organic 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 information

Chem 360 Jasperse Chapter 13 Answers to in-class NMR Spectroscopy Problems

Chem 360 Jasperse Chapter 13 Answers to in-class NMR Spectroscopy Problems Chem 360 Jasperse Chapter 13 Answers to in-class NMR Spectroscopy Problems 1 1. 2. Cl integraton says CH2 beside Cl splitting says Cl-CH2 is beside another CH2 splitting says CH3 is beside a CH2. integraton

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

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

5. Carbon-13 NMR Symmetry: number of chemically different Carbons Chemical Shift: chemical environment of Carbons (e- rich or e- poor)

5. Carbon-13 NMR Symmetry: number of chemically different Carbons Chemical Shift: chemical environment of Carbons (e- rich or e- poor) Qualitative Analysis of Unknown Compounds 1. Infrared Spectroscopy Identification of functional groups in the unknown All functional groups are fair game (but no anhydride or acid halides, no alkenes or

More information

NMRis the most valuable spectroscopic technique for organic chemists because it maps the carbon-hydrogen framework of a molecule.

NMRis the most valuable spectroscopic technique for organic chemists because it maps the carbon-hydrogen framework of a molecule. Chapter 13: Nuclear magnetic resonance spectroscopy NMRis the most valuable spectroscopic technique for organic chemists because it maps the carbon-hydrogen framework of a molecule. 13.2 The nature of

More information

Principles of Molecular Spectroscopy: Electromagnetic Radiation and Molecular structure. Nuclear Magnetic Resonance (NMR)

Principles of Molecular Spectroscopy: Electromagnetic Radiation and Molecular structure. Nuclear Magnetic Resonance (NMR) Principles of Molecular Spectroscopy: Electromagnetic Radiation and Molecular structure Nuclear Magnetic Resonance (NMR) !E = h" Electromagnetic radiation is absorbed when the energy of photon corresponds

More information

13.24: Mass Spectrometry: molecular weight of the sample

13.24: Mass Spectrometry: molecular weight of the sample hapter 13: Spectroscopy Methods of structure determination Nuclear Magnetic Resonances (NMR) Spectroscopy (Sections 13.3-13.19) Infrared (IR) Spectroscopy (Sections 13.20-13.22) Ultraviolet-visible (UV-Vis)

More information

William H. Brown & Christopher S. Foote

William H. Brown & Christopher S. Foote Requests for permission to make copies of any part of the work should be mailed to:permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida 32887-6777 William H. Brown

More information