1) A compound gives a mass spectrum with peaks at m/z = 77 (40%), 112 (100%), 114 (33%), and essentially no other peaks. Identify the compound.

Size: px
Start display at page:

Download "1) A compound gives a mass spectrum with peaks at m/z = 77 (40%), 112 (100%), 114 (33%), and essentially no other peaks. Identify the compound."

Transcription

1 1) A compound gives a mass spectrum with peaks at m/z = 77 (40%), 112 (100%), 114 (33%), and essentially no other peaks. Identify the compound. First, your molecular ion peak is 112 and you have a M+2 peak at 114. Therefore, you have a halogen. Now, your molecular ion peak and M+2 peak are in a 3 to 1 ratio. This means chlorine. So, =77 # C s 77/12=6 carbons so C 6 H 5 Cl. DOUS (2(6)+2-5-1)/2=4 Cl

2 2) While organizing the undergraduate stockroom, a new chemistry professor found a half-gallon jug containing a cloudy liquid (bp C), marked only "STUDENT PREP". She ran a quick mass spectrum, which is shown below. As soon as she saw the spectrum (without even checking the actual mass numbers), she said, "I know what it is." What compound is the "student prep"?

3 So molecular ion peak at 136 and M+2 peak at 138, so halogen present. They are in a 1:1 ratio so Br. So =57/12 = 4x12= =9, C 4 H 9 Br (2(4)+2-9-1)/2=0 Br The peaks at 107 (C 2 H 5 ) and 93(C 3 H 7 ) tell us it is a linear chain instead of a branched one.

4 3) A laboratory student added 1-bromobutane to a flask containing dry ether and Mg turnings. An exothermic reaction resulted, and the ether boiled vigorously for several minutes. Then she added acetone to the reaction mixture, and the ether boiled even more vigorously. She added dilute acid to the mixture, and separated the layers. She evaporated the ether layer, and distilled a liquid that boiled at 143 C. GC MS analysis of the distillate showed one major product with a few minor impurities. The mass spectrum of the major product is shown below. Show the structure of this major product.

5 BrMg/etherMgBrOO-H+OH The molecular ion peak should be at 116, but the loss of a carbon from the quatenary C forms a stable carbocation at 101.

6 1) One of the following compounds is responsible for the IR spectrum shown. Draw the structure of the responsible compound. 1-butene, 1-butanol, 4-hydroxy-1-butene, methyl propyl ether, butanoic acid.

7 First thing to notice is the presence of an alcohol at cm -1, which narrows our choices down to 1-butanol, 4-hydroxy-1-butene and butanoic acid.. OHO OH OH Only 1-butanol works, because there are no peaks corresponding to C=C and C=O.

8 2) One of the following compounds is responsible for the IR spectrum shown. Draw the structure of the responsible compound. phenylacetone, benzoic acid, acetophenone, benzyl alcohol, benzaldehyde

9 First you can eliminate benzoic acid and benzyl alcohol because there is no OH peak. Second you can eliminate benzaldehyde because there is no peak at 2740 cm -1 (aldehyde peak). That leaves phenylacetone and acetophenone. OO Acetophenone is the answer because the carbonyl peak is at 1700 cm -1 and a simple ketone like that on phenylacetone would absorb at a higher energy.

10 3) One of the following compounds is responsible for the IR spectrum shown. Draw the structure of the responsible compound. 2-ethynylcyclohexanone, 2-methyl-2-cyclohexenone, acetophenone, cyclohexylmethyl alcohol, 4-ethylcyclohexanone.

11 First, 2-ethynylcyclohexanone can be eliminated because there is no peak for a carbon carbon triple bond. Second, cyclohexylmethyl alcohol is eliminated because there is no OH peak present. Acetopheone can be eliminated next because there is no peak for aromatic C-H stretches. Also the carbonyl peak would be at a higher energy like around 1800 cm methyl-2-cyclohexenone can be eliminated because there is no O carbon carbon double bond peak. Leaving 4-ethylcyclohexanone.

12 (4) Determine the structure of the compound that gives rise to the following mass and IR spectra.

13 The molecular ion peak is at 162 and the M+2 peak is at 164, and they are in a 1:1 ratio, therefore there is a bromine atom =83 83/12=6 carbons and 6x12= 72 so 83-72=11 C 6 H 11 Br So 2(6) =2/2=1, which means 1 double bond or 1 ring. Looking at the IR, there is no C=C peak so that means a ring. Br

14 (5) Determine the structure of the compound that gives rise to the following mass and IR spectra.

15 So the molecular ion peak is 72. So 72/12= 6 carbons 2(6)+2=14/2=7 a bit high. So subtract 1 C and replace with 12 Hs. C 5 H 12. 2(5)+2-12=0 Not pentane, there is a carbonyl stretch.

16 So to add an O, subtract a methyl group C 4 H 8 O. 2(4)+2-8=2/2=1 Lastly there is a peak at 2740 which tells us that the carbonyl si due to an aldehyde. O

17 (6) Draw a structure consistent with the following data: The MS shows a molecular ion at 59 amu. The IR spectrum shows a double-humped strong absorbance at around 3300 cm 1 (the only absorbance in the functional group region) and a single absorbance at about 1385 cm 1. Odd molecular ion peak tells you there is a nitrogen =45 45/12=3 carbons 45-36=9 hydrogens C 3 H 9 N. 2(3)+2-9+1=0 The peak at 3300 cm 1 tells us that the N is part of an amine. NH2

18 (1) The following 1 H NMR spectrum is of a compound of molecular formula C 3 H 8 O. Propose a structure for this compound.

19 First you have a septet that integrates to 1 H and a doublet that integrates to 6 Hs. This is typical of an isopropyl group. Then the peak at 2.5ppm. Is a singlet and represents a H on an OH group. OH

20 2) Draw the structure of the compound with the 1H NMR and IR spectra shown and the formula C5H12O.

21 2(5)+2-12=0 so no double bonds or rings. Also there is no OH or C=O peaks in the IR, so it has to be an ether. Looking downfield you have a triplet and a singlet. For there to be a singlet there must be only a methyl on one side of the ether. Thus giving us the following structure. O Looking at this structure, it explains the presence of the pentet and sextet for the middle two CH 2 s. And finally the triplet upfield is for the terminal methyl group.

22 3) Draw the structure of the compound with the 1H NMR and IR spectra shown and the formula C6H12O2.

23 First 2(6)+2-12=2/2=1 And based on the carbonyl peak in the IR we know this is our degree of unsaturation. Also we know that there must also be an ether since there is no OH peak in the IR. And based on the proton NMR we have two types of protons. One has to be connected to the ether. And for the rest to be all the same, there must be an isobutyl group. OO

24 (4) Determine the structure of the following compound based on its mass, IR, and 1 H NMR spectra.

25 114/12=9 carbons =6 hydrogens C9H6 2(9)+2-6= 7 degrees of unsaturation Based on the IR we know there is a carbonyl So CH4 add O C8H2O 2(8)+2-2=16/2=8 Lets take off a C and add 12 Hs C7H14O 2(7)+2-14=2/2=1 Which accounts for the C=O

26 And now look at the proton NMR and there are 3 types of protons, this indicates a symmetrical ketone. There should be two triplets and one sextet. O

27 (5) Either 2-butanone, 2-methyl-2-nitropropane, 3-pentanone, 1- nitropropane, nitroethane, or 2-bromopropane is responsible for the 1 H NMR spectrum shown. Draw the structure of the responsible compound.

28 There are two signals in the spectra so we can eliminate 2- butanone, 2-methyl-2-nitropropane and 1-nitropropane because they have either more or less than 2 types of protons. Next you can eliminate 3-pentanone because it would have a triplet and a quartet which is not seen in the spectra. Also nitroethane can be eliminated because it would have a doublet and a quartet. While the answer is 2-bromopropane which has a doublet and a heptet. Br

29 6) Either 2-butanone, 2-methyl-2-nitropropane, 3-pentanone, 1- nitropropane, nitroethane, or 2-bromopropane is responsible for the 1 H NMR spectrum shown. Draw the structure of the responsible compound.

30 Because there are 3 signals it is either 2-butanone or 1- nitropropane. 1-nitropropane should have a doublet, a quartet and a triplet. While our answer, 2-butanone should have a singlet, a quartet and a triplet.. O

31 7) The molecular formula of a compound is C 6 H 12 O. Determine the structure of the compound based on its molecular formula and its 13 C NMR spectrum.

32 First 2(6)+2-12=2/2=1 so either a ring or double bond. No peak shows up in the double bond region, C=C or C=O. So that leaves a ring. Four peaks and with this structure we have 4 different types of carbon. OH

33 8) Identify the compound with molecular formula C 3 H 5 Cl 3 that gives the following 13 C NMR spectrum. (The resonance at 0 ppm is due to the TMS standard, not the unknown.)

34 First, 2(3)+2-5-3=0 so no double bonds or rings. Secondly there are 3 peaks, so 3 different kinds fo carbon. So that leaves two choices that are correct. ClClClClClCl

For example: (Example is from page 50 of the Thinkbook)

For example: (Example is from page 50 of the Thinkbook) SOLVING COMBINED SPECTROSCOPY PROBLEMS: Lecture Supplement: page 50-53 in Thinkbook CFQ s and PP s: page 216 241 in Thinkbook Introduction: The structure of an unknown molecule can be determined using

More information

How to Quickly Solve Spectrometry Problems

How to Quickly Solve Spectrometry Problems How to Quickly Solve Spectrometry Problems You should be looking for: Mass Spectrometry (MS) Chemical Formula DBE Infrared Spectroscopy (IR) Important Functional Groups o Alcohol O-H o Carboxylic Acid

More information

The Four Questions to Ask While Interpreting Spectra. 1. How many different environments are there?

The Four Questions to Ask While Interpreting Spectra. 1. How many different environments are there? 1 H NMR Spectroscopy (#1c) The technique of 1 H NMR spectroscopy is central to organic chemistry and other fields involving analysis of organic chemicals, such as forensics and environmental science. It

More information

HOMEWORK PROBLEMS: IR SPECTROSCOPY AND 13C NMR. The peak at 1720 indicates a C=O bond (carbonyl). One possibility is acetone:

HOMEWORK PROBLEMS: IR SPECTROSCOPY AND 13C NMR. The peak at 1720 indicates a C=O bond (carbonyl). One possibility is acetone: HMEWRK PRBLEMS: IR SPECTRSCPY AND 13C NMR 1. You find a bottle on the shelf only labeled C 3 H 6. You take an IR spectrum of the compound and find major peaks at 2950, 1720, and 1400 cm -1. Draw a molecule

More information

13C NMR Spectroscopy

13C NMR Spectroscopy 13 C NMR Spectroscopy Introduction Nuclear magnetic resonance spectroscopy (NMR) is the most powerful tool available for structural determination. A nucleus with an odd number of protons, an odd number

More information

Suggested solutions for Chapter 3

Suggested solutions for Chapter 3 s for Chapter PRBLEM Assuming that the molecular ion is the base peak (00% abundance) what peaks would appear in the mass spectrum of each of these molecules: (a) C5Br (b) C60 (c) C64Br In cases (a) and

More information

Proton Nuclear Magnetic Resonance Spectroscopy

Proton Nuclear Magnetic Resonance Spectroscopy Proton Nuclear Magnetic Resonance Spectroscopy Introduction: The NMR Spectrum serves as a great resource in determining the structure of an organic compound by revealing the hydrogen and carbon skeleton.

More information

Solving Spectroscopy Problems

Solving Spectroscopy Problems Solving Spectroscopy Problems The following is a detailed summary on how to solve spectroscopy problems, key terms are highlighted in bold and the definitions are from the illustrated glossary on Dr. Hardinger

More information

Organic Chemistry Tenth Edition

Organic Chemistry Tenth Edition Organic Chemistry Tenth Edition T. W. Graham Solomons Craig B. Fryhle Welcome to CHM 22 Organic Chemisty II Chapters 2 (IR), 9, 3-20. Chapter 2 and Chapter 9 Spectroscopy (interaction of molecule with

More information

Proton Nuclear Magnetic Resonance ( 1 H-NMR) Spectroscopy

Proton Nuclear Magnetic Resonance ( 1 H-NMR) Spectroscopy Proton Nuclear Magnetic Resonance ( 1 H-NMR) Spectroscopy Theory behind NMR: In the late 1940 s, physical chemists originally developed NMR spectroscopy to study different properties of atomic nuclei,

More information

Chapter 13 Spectroscopy NMR, IR, MS, UV-Vis

Chapter 13 Spectroscopy NMR, IR, MS, UV-Vis Chapter 13 Spectroscopy NMR, IR, MS, UV-Vis Main points of the chapter 1. Hydrogen Nuclear Magnetic Resonance a. Splitting or coupling (what s next to what) b. Chemical shifts (what type is it) c. Integration

More information

PROTON NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY (H-NMR)

PROTON NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY (H-NMR) PROTON NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY (H-NMR) WHAT IS H-NMR SPECTROSCOPY? References: Bruice 14.1, 14.2 Introduction NMR or nuclear magnetic resonance spectroscopy is a technique used to determine

More information

Molecular Formula Determination

Molecular Formula Determination Molecular Formula Determination Classical Approach Qualitative elemental analysis Quantitative elemental analysis Determination of empirical formula Molecular weight determination Molecular formula determination

More information

electron does not become part of the compound; one electron goes in but two electrons come out.

electron does not become part of the compound; one electron goes in but two electrons come out. Characterization Techniques for Organic Compounds. When we run a reaction in the laboratory or when we isolate a compound from nature, one of our first tasks is to identify the compound that we have obtained.

More information

NMR SPECTROSCOPY A N I N T R O D U C T I O N T O... Self-study booklet NUCLEAR MAGNETIC RESONANCE. 4 3 2 1 0 δ PUBLISHING

NMR SPECTROSCOPY A N I N T R O D U C T I O N T O... Self-study booklet NUCLEAR MAGNETIC RESONANCE. 4 3 2 1 0 δ PUBLISHING A N I N T R O D U T I O N T O... NMR SPETROSOPY NULEAR MAGNETI RESONANE 4 3 1 0 δ Self-study booklet PUBLISING NMR Spectroscopy NULEAR MAGNETI RESONANE SPETROSOPY Origin of Spectra Theory All nuclei possess

More information

CHE334 Identification of an Unknown Compound By NMR/IR/MS

CHE334 Identification of an Unknown Compound By NMR/IR/MS CHE334 Identification of an Unknown Compound By NMR/IR/MS Purpose The object of this experiment is to determine the structure of an unknown compound using IR, 1 H-NMR, 13 C-NMR and Mass spectroscopy. Infrared

More information

Used to determine relative location of atoms within a molecule Most helpful spectroscopic technique in organic chemistry Related to MRI in medicine

Used to determine relative location of atoms within a molecule Most helpful spectroscopic technique in organic chemistry Related to MRI in medicine Structure Determination: Nuclear Magnetic Resonance CHEM 241 UNIT 5C 1 The Use of NMR Spectroscopy Used to determine relative location of atoms within a molecule Most helpful spectroscopic technique in

More information

Determining the Structure of an Organic Compound

Determining the Structure of an Organic Compound Determining the Structure of an Organic Compound The analysis of the outcome of a reaction requires that we know the full structure of the products as well as the reactants In the 19 th and early 20 th

More information

Identification of Unknown Organic Compounds

Identification of Unknown Organic Compounds Identification of Unknown Organic Compounds Introduction The identification and characterization of the structures of unknown substances are an important part of organic chemistry. Although it is often

More information

E35 SPECTROSCOPIC TECHNIQUES IN ORGANIC CHEMISTRY

E35 SPECTROSCOPIC TECHNIQUES IN ORGANIC CHEMISTRY E35 SPECTRSCPIC TECNIQUES IN RGANIC CEMISTRY TE TASK To use mass spectrometry and IR, UV/vis and NMR spectroscopy to identify organic compounds. TE SKILLS By the end of the experiment you should be able

More information

IR Applied to Isomer Analysis

IR Applied to Isomer Analysis DiscovIR-LC TM Application Note 025 April 2008 Deposition and Detection System IR Applied to Isomer Analysis Infrared spectra provide valuable information about local configurations of atoms in molecules.

More information

How to Interpret an IR Spectrum

How to Interpret an IR Spectrum How to Interpret an IR Spectrum Don t be overwhelmed when you first view IR spectra or this document. We have simplified the interpretation by having you only focus on 4/5 regions of the spectrum. Do not

More information

Nuclear Magnetic Resonance Spectroscopy

Nuclear Magnetic Resonance Spectroscopy Nuclear Magnetic Resonance Spectroscopy Nuclear magnetic resonance spectroscopy is a powerful analytical technique used to characterize organic molecules by identifying carbonhydrogen frameworks within

More information

Proton Nuclear Magnetic Resonance Spectroscopy

Proton Nuclear Magnetic Resonance Spectroscopy CHEM 334L Organic Chemistry Laboratory Revision 2.0 Proton Nuclear Magnetic Resonance Spectroscopy In this laboratory exercise we will learn how to use the Chemistry Department's Nuclear Magnetic Resonance

More information

Nuclear Magnetic Resonance Spectroscopy

Nuclear Magnetic Resonance Spectroscopy Nuclear Magnetic Resonance Spectroscopy Introduction NMR is the most powerful tool available for organic structure determination. It is used to study a wide variety of nuclei: 1 H 13 C 15 N 19 F 31 P 2

More information

Organic Spectroscopy. UV - Ultraviolet-Visible Spectroscopy. !! 200-800 nm. Methods for structure determination of organic compounds:

Organic Spectroscopy. UV - Ultraviolet-Visible Spectroscopy. !! 200-800 nm. Methods for structure determination of organic compounds: Organic Spectroscopy Methods for structure determination of organic compounds: X-ray rystallography rystall structures Mass spectroscopy Molecular formula -----------------------------------------------------------------------------

More information

Nuclear Magnetic Resonance notes

Nuclear Magnetic Resonance notes Reminder: These notes are meant to supplement, not replace, the laboratory manual. Nuclear Magnetic Resonance notes Nuclear Magnetic Resonance (NMR) is a spectrometric technique which provides information

More information

Symmetric Stretch: allows molecule to move through space

Symmetric Stretch: allows molecule to move through space BACKGROUND INFORMATION Infrared Spectroscopy Before introducing the subject of IR spectroscopy, we must first review some aspects of the electromagnetic spectrum. The electromagnetic spectrum is composed

More information

0 10 20 30 40 50 60 70 m/z

0 10 20 30 40 50 60 70 m/z Mass spectrum for the ionization of acetone MS of Acetone + Relative Abundance CH 3 H 3 C O + M 15 (loss of methyl) + O H 3 C CH 3 43 58 0 10 20 30 40 50 60 70 m/z It is difficult to identify the ions

More information

Infrared Spectroscopy 紅 外 線 光 譜 儀

Infrared Spectroscopy 紅 外 線 光 譜 儀 Infrared Spectroscopy 紅 外 線 光 譜 儀 Introduction Spectroscopy is an analytical technique which helps determine structure. It destroys little or no sample (nondestructive method). The amount of light absorbed

More information

Chapter 11 Structure Determination: Nuclear Magnetic Resonance Spectroscopy. Nuclear Magnetic Resonance Spectroscopy. 11.1 Nuclear Magnetic Resonance

Chapter 11 Structure Determination: Nuclear Magnetic Resonance Spectroscopy. Nuclear Magnetic Resonance Spectroscopy. 11.1 Nuclear Magnetic Resonance John E. McMurry http://www.cengage.com/chemistry/mcmurry Chapter 11 Structure Determination: Nuclear Magnetic Resonance Spectroscopy 11.1 Nuclear Magnetic Resonance Spectroscopy Many atomic nuclei behave

More information

Mass Spectrometry. Overview

Mass Spectrometry. Overview Mass Spectrometry Overview Mass Spectrometry is an analytic technique that utilizes the degree of deflection of charged particles by a magnetic field to find the relative masses of molecular ions and fragments.2

More information

CHEM 51LB EXP 1 SPECTROSCOPIC METHODS: INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY

CHEM 51LB EXP 1 SPECTROSCOPIC METHODS: INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY CHEM 51LB EXP 1 SPECTRSCPIC METHDS: INFRARED AND NUCLEAR MAGNETIC RESNANCE SPECTRSCPY REACTINS: None TECHNIQUES: IR Spectroscopy, NMR Spectroscopy Infrared (IR) and nuclear magnetic resonance (NMR) spectroscopy

More information

Assessment Schedule 2013 Chemistry: Demonstrate understanding of the properties of organic compounds (91391)

Assessment Schedule 2013 Chemistry: Demonstrate understanding of the properties of organic compounds (91391) NCEA Level 3 Chemistry (91391) 2013 page 1 of 8 Assessment Schedule 2013 Chemistry: Demonstrate understanding of the properties of organic compounds (91391) Evidence Statement Q Evidence Achievement Achievement

More information

Mass Spec - Fragmentation

Mass Spec - Fragmentation Mass Spec - Fragmentation An extremely useful result of EI ionization in particular is a phenomenon known as fragmentation. The radical cation that is produced when an electron is knocked out of a neutral

More information

Unit Vocabulary: o Organic Acid o Alcohol. o Ester o Ether. o Amine o Aldehyde

Unit Vocabulary: o Organic Acid o Alcohol. o Ester o Ether. o Amine o Aldehyde Unit Vocabulary: Addition rxn Esterification Polymer Alcohol Ether Polymerization Aldehyde Fermentation Primary Alkane Functional group Saponification Alkene Halide (halocarbon) Saturated hydrocarbon Alkyne

More information

Experiment 11. Infrared Spectroscopy

Experiment 11. Infrared Spectroscopy Chem 22 Spring 2010 Experiment 11 Infrared Spectroscopy Pre-lab preparation. (1) In Ch 5 and 12 of the text you will find examples of the most common functional groups in organic molecules. In your notebook,

More information

Molecular Models Experiment #1

Molecular Models Experiment #1 Molecular Models Experiment #1 Objective: To become familiar with the 3-dimensional structure of organic molecules, especially the tetrahedral structure of alkyl carbon atoms and the planar structure of

More information

Sample exam questions for First exam CHM 2211

Sample exam questions for First exam CHM 2211 Sample exam questions for First exam CM 2211 1. The IR absorption due to the stretching of which of these carbon-hydrogen bonds occurs at the highest frequency? I II III E) V IV V 2. ow many signals would

More information

Examination of Proton NMR Spectra

Examination of Proton NMR Spectra Examination of Proton NMR Spectra What to Look For 1) Number of Signals --- indicates how many "different kinds" of protons are present. 2) Positions of the Signals --- indicates something about magnetic

More information

j. SO 3, SO 2, NaCl, Na 2 O (1 mark each) Total 10 a) 525 kj mol -1 per mole of Mg (2 marks) (-1 for incorrect sign or unit)

j. SO 3, SO 2, NaCl, Na 2 O (1 mark each) Total 10 a) 525 kj mol -1 per mole of Mg (2 marks) (-1 for incorrect sign or unit) ANSWERS RUND 1 1. This is a question about trends in chemistry a. Na, S 8, Al, Si b. 2, 4, N 3, Li c. Mg, Na, a, K d. Mg 2+, Na +, F -, l - e. K, Na, l, F f. Si, S, P, l g. Br 2, l 2, 2, N 2 h. XeF 4,

More information

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Exam Name SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 1) Calculate the magnetic field that corresponds to the proton resonance frequency of 300.00

More information

NMR and other Instrumental Techniques in Chemistry and the proposed National Curriculum.

NMR and other Instrumental Techniques in Chemistry and the proposed National Curriculum. NMR and other Instrumental Techniques in Chemistry and the proposed National Curriculum. Dr. John Jackowski Chair of Science, Head of Chemistry Scotch College Melbourne john.jackowski@scotch.vic.edu.au

More information

INFRARED SPECTROSCOPY (IR)

INFRARED SPECTROSCOPY (IR) INFRARED SPECTROSCOPY (IR) Theory and Interpretation of IR spectra ASSIGNED READINGS Introduction to technique 25 (p. 833-834 in lab textbook) Uses of the Infrared Spectrum (p. 847-853) Look over pages

More information

DETERMINACIÓN DE ESTRUCTURAS ORGÁNICAS (ORGANIC SPECTROSCOPY) IR SPECTROSCOPY

DETERMINACIÓN DE ESTRUCTURAS ORGÁNICAS (ORGANIC SPECTROSCOPY) IR SPECTROSCOPY DETERMINACIÓN DE ESTRUCTURAS ORGÁNICAS (ORGANIC SPECTROSCOPY) IR SPECTROSCOPY Hermenegildo García Gómez Departamento de Química Instituto de Tecnología Química Universidad Politécnica de Valencia 46022

More information

Austin Peay State University Department of Chemistry CHEM 1021 TESTING FOR ORGANIC FUNCTIONAL GROUPS

Austin Peay State University Department of Chemistry CHEM 1021 TESTING FOR ORGANIC FUNCTIONAL GROUPS TESTING FOR ORGANIC FUNCTIONAL GROUPS Caution: Chromic acid is hazardous as are many of the organic substances in today s experiment. Treat all unknowns with extreme care. Many organic substances are flammable.

More information

Unit 2 Review: Answers: Review for Organic Chemistry Unit Test

Unit 2 Review: Answers: Review for Organic Chemistry Unit Test Unit 2 Review: Answers: Review for Organic Chemistry Unit Test 2. Write the IUPAC names for the following organic molecules: a) acetone: propanone d) acetylene: ethyne b) acetic acid: ethanoic acid e)

More information

NOMENCLATURE OF ORGANIC COMPOUNDS 2010, 2003, 1980, by David A. Katz. All rights reserved.

NOMENCLATURE OF ORGANIC COMPOUNDS 2010, 2003, 1980, by David A. Katz. All rights reserved. NMENCLATURE F RGANIC CMPUNDS 2010, 2003, 1980, by David A. Katz. All rights reserved. rganic chemistry is the chemistry of carbon compounds. Carbon has the ability to bond with itself to form long chains

More information

MOLECULAR REPRESENTATIONS AND INFRARED SPECTROSCOPY

MOLECULAR REPRESENTATIONS AND INFRARED SPECTROSCOPY MLEULAR REPRESENTATINS AND INFRARED SPETRSPY A STUDENT SULD BE ABLE T: 1. Given a Lewis (dash or dot), condensed, bond-line, or wedge formula of a compound draw the other representations. 2. Give examples

More information

Chapter 5 Classification of Organic Compounds by Solubility

Chapter 5 Classification of Organic Compounds by Solubility Chapter 5 Classification of Organic Compounds by Solubility Deductions based upon interpretation of simple solubility tests can be extremely useful in organic structure determination. Both solubility and

More information

Survival Organic Chemistry Part I: Molecular Models

Survival Organic Chemistry Part I: Molecular Models Survival Organic Chemistry Part I: Molecular Models The goal in this laboratory experience is to get you so you can easily and quickly move between empirical formulas, molecular formulas, condensed formulas,

More information

Question Bank Organic Chemistry-I

Question Bank Organic Chemistry-I Question Bank Organic Chemistry-I 1. (a) What do you understand by the following terms : (i) Organic chemistry (ii) Organic compounds (iii) Catenation? [3] (b) Why are there very large number of organic

More information

EXPERIMENT 1: Survival Organic Chemistry: Molecular Models

EXPERIMENT 1: Survival Organic Chemistry: Molecular Models EXPERIMENT 1: Survival Organic Chemistry: Molecular Models Introduction: The goal in this laboratory experience is for you to easily and quickly move between empirical formulas, molecular formulas, condensed

More information

Chapter 12 Organic Compounds with Oxygen and Sulfur

Chapter 12 Organic Compounds with Oxygen and Sulfur Chapter 12 Organic Compounds with Oxygen and Sulfur 1 Alcohols An alcohol contains a hydroxyl group ( OH) that replaces a hydrogen atom in a hydrocarbon. A phenol contains a hydroxyl group ( OH) attached

More information

Ultraviolet Spectroscopy

Ultraviolet Spectroscopy Ultraviolet Spectroscopy The wavelength of UV and visible light are substantially shorter than the wavelength of infrared radiation. The UV spectrum ranges from 100 to 400 nm. A UV-Vis spectrophotometer

More information

Nuclear Magnetic Resonance

Nuclear Magnetic Resonance Nuclear Magnetic Resonance NMR is probably the most useful and powerful technique for identifying and characterizing organic compounds. Felix Bloch and Edward Mills Purcell were awarded the 1952 Nobel

More information

Laboratory 22: Properties of Alcohols

Laboratory 22: Properties of Alcohols Introduction Alcohols represent and important class of organic molecules. In this experiment you will study the physical and chemical properties of alcohols. Solubility in water, and organic solvents,

More information

Chemical Calculations: The Mole Concept and Chemical Formulas. AW Atomic weight (mass of the atom of an element) was determined by relative weights.

Chemical Calculations: The Mole Concept and Chemical Formulas. AW Atomic weight (mass of the atom of an element) was determined by relative weights. 1 Introduction to Chemistry Atomic Weights (Definitions) Chemical Calculations: The Mole Concept and Chemical Formulas AW Atomic weight (mass of the atom of an element) was determined by relative weights.

More information

Chapter 22 Carbonyl Alpha-Substitution Reactions

Chapter 22 Carbonyl Alpha-Substitution Reactions John E. McMurry www.cengage.com/chemistry/mcmurry Chapter 22 Carbonyl Alpha-Substitution Reactions The α Position The carbon next to the carbonyl group is designated as being in the α position Electrophilic

More information

Name Lab #3: Solubility of Organic Compounds Objectives: Introduction: soluble insoluble partially soluble miscible immiscible

Name  Lab #3: Solubility of Organic Compounds Objectives: Introduction: soluble insoluble partially soluble miscible immiscible Lab #3: Solubility of rganic Compounds bjectives: - Understanding the relative solubility of organic compounds in various solvents. - Exploration of the effect of polar groups on a nonpolar hydrocarbon

More information

Using Nuclear Magnetic Resonance Spectroscopy to Identify an Unknown Compound prepared by Joseph W. LeFevre, SUNY Oswego

Using Nuclear Magnetic Resonance Spectroscopy to Identify an Unknown Compound prepared by Joseph W. LeFevre, SUNY Oswego m o d u l a r l a b o r a t o r y p r o g r a m i n c h e m i s t r y publisher:. A. Neidig organic editor: Joe Jeffers TE 711 Using Nuclear Magnetic Resonance Spectroscopy to Identify an Unknown ompound

More information

Reactions of Aldehydes and Ketones

Reactions of Aldehydes and Ketones Reactions of Aldehydes and Ketones Structure Deduction using lassification Tests 1 Determination of Structure Determining the structure of an unknown organic compound is an exercise in deductive reasoning.

More information

for excitation to occur, there must be an exact match between the frequency of the applied radiation and the frequency of the vibration

for excitation to occur, there must be an exact match between the frequency of the applied radiation and the frequency of the vibration ! = 1 2"c k (m + M) m M wavenumbers! =!/c = 1/" wavelength frequency! units: cm 1 for excitation to occur, there must be an exact match between the frequency of the applied radiation and the frequency

More information

IDENTIFICATION OF ALCOHOLS

IDENTIFICATION OF ALCOHOLS IDENTIFICATION OF ALCOHOLS Alcohols are organic compounds that which considered as derivatives of water. One of the hydrogen atoms of water molecule (H-O-H) has been replaced by an alkyl or substituted

More information

Experiment 6 Qualitative Tests for Alcohols, Alcohol Unknown, IR of Unknown

Experiment 6 Qualitative Tests for Alcohols, Alcohol Unknown, IR of Unknown Experiment 6 Qualitative Tests for Alcohols, Alcohol Unknown, I of Unknown In this experiment you are going to do a series of tests in order to determine whether or not an alcohol is a primary (1 ), secondary

More information

Nuclear Shielding and 1. H Chemical Shifts. 1 H NMR Spectroscopy Nuclear Magnetic Resonance

Nuclear Shielding and 1. H Chemical Shifts. 1 H NMR Spectroscopy Nuclear Magnetic Resonance NMR Spectroscopy Nuclear Magnetic Resonance Nuclear Shielding and hemical Shifts What do we mean by "shielding?" What do we mean by "chemical shift?" The electrons surrounding a nucleus affect the effective

More information

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Ch14_PT MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Compounds with the -OH group attached to a saturated alkane-like carbon are known as A)

More information

IUPAC System of Nomenclature

IUPAC System of Nomenclature IUPAC System of Nomenclature The IUPAC (International Union of Pure and Applied Chemistry) is composed of chemists representing the national chemical societies of several countries. ne committee of the

More information

Nuclear Structure. particle relative charge relative mass proton +1 1 atomic mass unit neutron 0 1 atomic mass unit electron -1 negligible mass

Nuclear Structure. particle relative charge relative mass proton +1 1 atomic mass unit neutron 0 1 atomic mass unit electron -1 negligible mass Protons, neutrons and electrons Nuclear Structure particle relative charge relative mass proton 1 1 atomic mass unit neutron 0 1 atomic mass unit electron -1 negligible mass Protons and neutrons make up

More information

Background A nucleus with an odd atomic number or an odd mass number has a nuclear spin that can be observed by NMR spectrometers.

Background A nucleus with an odd atomic number or an odd mass number has a nuclear spin that can be observed by NMR spectrometers. NMR Spectroscopy I Reading: Wade chapter, sections -- -7 Study Problems: -, -7 Key oncepts and Skills: Given an structure, determine which protons are equivalent and which are nonequivalent, predict the

More information

NMR Spectroscopy of Aromatic Compounds (#1e)

NMR Spectroscopy of Aromatic Compounds (#1e) NMR Spectroscopy of Aromatic Compounds (#1e) 1 H NMR Spectroscopy of Aromatic Compounds Erich Hückel s study of aromaticity in the 1930s produced a set of rules for determining whether a compound is aromatic.

More information

Infrared Spectroscopy

Infrared Spectroscopy Infrared Spectroscopy 1 Chap 12 Reactions will often give a mixture of products: OH H 2 SO 4 + Major Minor How would the chemist determine which product was formed? Both are cyclopentenes; they are isomers.

More information

1. The functional group present in carboxylic acids is called a A) carbonyl group. B) carboxyl group. C) carboxylate group. D) carbohydroxyl group.

1. The functional group present in carboxylic acids is called a A) carbonyl group. B) carboxyl group. C) carboxylate group. D) carbohydroxyl group. Name: Date: 1. The functional group present in carboxylic acids is called a A) carbonyl group. B) carboxyl group. C) carboxylate group. D) carbohydroxyl group. 2. Which of the following statements concerning

More information

Alcohols An alcohol contains a hydroxyl group ( OH) attached to a carbon chain. A phenol contains a hydroxyl group ( OH) attached to a benzene ring.

Alcohols An alcohol contains a hydroxyl group ( OH) attached to a carbon chain. A phenol contains a hydroxyl group ( OH) attached to a benzene ring. Chapter : rganic Compounds with xygen Alcohols, Ethers Alcohols An alcohol contains a hydroxyl group ( H) attached to a carbon chain. A phenol contains a hydroxyl group ( H) attached to a benzene ring.

More information

Chemistry 307 Chapter 10 Nuclear Magnetic Resonance

Chemistry 307 Chapter 10 Nuclear Magnetic Resonance Chemistry 307 Chapter 10 Nuclear Magnetic Resonance Nuclear magnetic resonance (NMR) spectroscopy is one of three spectroscopic techniques that are useful tools for determining the structures of organic

More information

CHEM 51LB: EXPERIMENT 5 SPECTROSCOPIC METHODS: INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY

CHEM 51LB: EXPERIMENT 5 SPECTROSCOPIC METHODS: INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY CHEM 51LB: EXPERIMENT 5 SPECTROSCOPIC METHODS: INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY REACTIONS: None TECHNIQUES: IR, NMR Infrared (IR) and nuclear magnetic resonance (NMR) spectroscopy are

More information

Benzene and Aromatic Compounds

Benzene and Aromatic Compounds Benzene and Aromatic Compounds Benzene (C 6 H 6 ) is the simplest aromatic hydrocarbon (or arene). Benzene has four degrees of unsaturation, making it a highly unsaturated hydrocarbon. Whereas unsaturated

More information

F321 THE STRUCTURE OF ATOMS. ATOMS Atoms consist of a number of fundamental particles, the most important are... in the nucleus of an atom

F321 THE STRUCTURE OF ATOMS. ATOMS Atoms consist of a number of fundamental particles, the most important are... in the nucleus of an atom Atomic Structure F32 TE STRUCTURE OF ATOMS ATOMS Atoms consist of a number of fundamental particles, the most important are... Mass / kg Charge / C Relative mass Relative Charge PROTON NEUTRON ELECTRON

More information

EXPERIMENT 9 Dot Structures and Geometries of Molecules

EXPERIMENT 9 Dot Structures and Geometries of Molecules EXPERIMENT 9 Dot Structures and Geometries of Molecules INTRODUCTION Lewis dot structures are our first tier in drawing molecules and representing bonds between the atoms. The method was first published

More information

Determination of Molecular Structure by MOLECULAR SPECTROSCOPY

Determination of Molecular Structure by MOLECULAR SPECTROSCOPY Determination of Molecular Structure by MOLEULAR SPETROSOPY hemistry 3 B.Z. Shakhashiri Fall 29 Much of what we know about molecular structure has been learned by observing and analyzing how electromagnetic

More information

Organic Functional Groups Chapter 7. Alcohols, Ethers and More

Organic Functional Groups Chapter 7. Alcohols, Ethers and More Organic Functional Groups Chapter 7 Alcohols, Ethers and More 1 What do you do when you are in Pain? What do you do when you are in a lot of pain? 2 Functional Groups A functional group is an atom, groups

More information

4. It is possible to excite, or flip the nuclear magnetic vector from the α-state to the β-state by bridging the energy gap between the two. This is a

4. It is possible to excite, or flip the nuclear magnetic vector from the α-state to the β-state by bridging the energy gap between the two. This is a BASIC PRINCIPLES INTRODUCTION TO NUCLEAR MAGNETIC RESONANCE (NMR) 1. The nuclei of certain atoms with odd atomic number, and/or odd mass behave as spinning charges. The nucleus is the center of positive

More information

Organic Spectroscopy: a Primer

Organic Spectroscopy: a Primer EM 03 rganic Spectroscopy: a Primer INDEX A. Introduction B. Infrared (IR) Spectroscopy 3. Proton Nuclear Magnetic Resonance ( NMR) Spectroscopy A. Introduction The problem of determining the structure

More information

4/18/2011. 9.8 Substituent Effects in Electrophilic Substitutions. Substituent Effects in Electrophilic Substitutions

4/18/2011. 9.8 Substituent Effects in Electrophilic Substitutions. Substituent Effects in Electrophilic Substitutions 9.8 Substituent effects in the electrophilic substitution of an aromatic ring Substituents affect the reactivity of the aromatic ring Some substituents activate the ring, making it more reactive than benzene

More information

Element of same atomic number, but different atomic mass o Example: Hydrogen

Element of same atomic number, but different atomic mass o Example: Hydrogen Atomic mass: p + = protons; e - = electrons; n 0 = neutrons p + + n 0 = atomic mass o For carbon-12, 6p + + 6n 0 = atomic mass of 12.0 o For chlorine-35, 17p + + 18n 0 = atomic mass of 35.0 atomic mass

More information

Nuclear Magnetic Resonance Spectroscopy

Nuclear Magnetic Resonance Spectroscopy Chapter 8 Nuclear Magnetic Resonance Spectroscopy http://www.yteach.co.uk/page.php/resources/view_all?id=nuclear_magnetic _resonance_nmr_spectroscopy_spin_spectrometer_spectrum_proton_t_pag e_5&from=search

More information

Determination of Equilibrium Constants using NMR Spectrscopy

Determination of Equilibrium Constants using NMR Spectrscopy CHEM 331L Physical Chemistry Laboratory Revision 1.0 Determination of Equilibrium Constants using NMR Spectrscopy In this laboratory exercise we will measure a chemical equilibrium constant using key proton

More information

ORGANIC COMPOUNDS IN THREE DIMENSIONS

ORGANIC COMPOUNDS IN THREE DIMENSIONS (adapted from Blackburn et al., Laboratory Manual to Accompany World of hemistry, 2 nd ed., (1996) Saunders ollege Publishing: Fort Worth) Purpose: To become familiar with organic molecules in three dimensions

More information

Alcohols. Copyright 2009 by Pearson Education, Inc. Copyright 2009 Pearson Education, Inc. CH 3 CH 2 CH 2 OH 1-propanol OH

Alcohols. Copyright 2009 by Pearson Education, Inc. Copyright 2009 Pearson Education, Inc. CH 3 CH 2 CH 2 OH 1-propanol OH Chapter 12 rganic Compounds with xygen and Sulfur 12.1 Alcohols, Thiols, and Ethers Alcohols An alcohol contains a hydroxyl group ( ) attached to a carbon chain. A phenol contains a hydroxyl group ( )

More information

Chemistry Diagnostic Questions

Chemistry Diagnostic Questions Chemistry Diagnostic Questions Answer these 40 multiple choice questions and then check your answers, located at the end of this document. If you correctly answered less than 25 questions, you need to

More information

Organic Spectroscopy

Organic Spectroscopy 1 Organic Spectroscopy Second Year, Michaelmas term, 8 lectures: Dr TDW Claridge & Prof BG Davis Lectures 1 4 highlight the importance of spectroscopic methods in the structural elucidation of organic

More information

AROMATIC COMPOUNDS A STUDENT SHOULD BE ABLE TO:

AROMATIC COMPOUNDS A STUDENT SHOULD BE ABLE TO: A STUDENT SHULD BE ABLE T: ARMATIC CMPUNDS 1. Name benzene derivatives given the structures, and draw the structures given the names. This includes: Monosubstituted benzenes named as derivatives of benzene:

More information

Chapter 5 Organic Spectrometry

Chapter 5 Organic Spectrometry Chapter 5 Organic Spectrometry from Organic Chemistry by Robert C. Neuman, Jr. Professor of Chemistry, emeritus University of California, Riverside orgchembyneuman@yahoo.com

More information

Ch17_PT MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Ch17_PT MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Ch17_PT MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Which molecule is a carboxylic acid? A) 1) B) C) D) E) CH3 CH2 CH2 NH2 2) Which molecule

More information

CHM220 Nucleophilic Substitution Lab. Studying S N 1 and S N 2 Reactions: Nucloephilic Substitution at Saturated Carbon*

CHM220 Nucleophilic Substitution Lab. Studying S N 1 and S N 2 Reactions: Nucloephilic Substitution at Saturated Carbon* CHM220 Nucleophilic Substitution Lab Studying S N 1 and S N 2 Reactions: Nucloephilic Substitution at Saturated Carbon* Purpose: To convert a primary alcohol to an alkyl bromide using an S N 2 reaction

More information

Organic Chemistry, 5e (Bruice) Chapter 17: Carbonyl Compounds II

Organic Chemistry, 5e (Bruice) Chapter 17: Carbonyl Compounds II Organic Chemistry, 5e (Bruice) Chapter 17: Carbonyl Compounds II 1) Which of the following compounds is an aldehyde? A) I B) II C) III D) IV E) V D Section: 17-1 2) Which of the following compounds is

More information

Page 1. 6. Which hydrocarbon is a member of the alkane series? (1) 1. Which is the structural formula of methane? (1) (2) (2) (3) (3) (4) (4)

Page 1. 6. Which hydrocarbon is a member of the alkane series? (1) 1. Which is the structural formula of methane? (1) (2) (2) (3) (3) (4) (4) 1. Which is the structural formula of methane? 6. Which hydrocarbon is a member of the alkane series? 7. How many carbon atoms are contained in an ethyl group? 1 3 2 4 2. In the alkane series, each molecule

More information

Properties of Alcohols and Phenols Experiment #3

Properties of Alcohols and Phenols Experiment #3 Properties of Alcohols and Phenols Experiment #3 Objectives: To observe the solubility of alcohols relative to their chemical structure, to perform chemical tests to distinguish primary, secondary and

More information

Calculating the Degrees of Unsaturation From a Compound s Molecular Formula

Calculating the Degrees of Unsaturation From a Compound s Molecular Formula Calculating the Degrees of Unsaturation From a Compound s Molecular Formula Alkanes have the molecular formula C n. Alkanes are saturated hydrocarbons because each member of the family has the maximum

More information

In the box below, draw the Lewis electron-dot structure for the compound formed from magnesium and oxygen. [Include any charges or partial charges.

In the box below, draw the Lewis electron-dot structure for the compound formed from magnesium and oxygen. [Include any charges or partial charges. Name: 1) Which molecule is nonpolar and has a symmetrical shape? A) NH3 B) H2O C) HCl D) CH4 7222-1 - Page 1 2) When ammonium chloride crystals are dissolved in water, the temperature of the water decreases.

More information