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.
|
|
- Adele Logan
- 7 years ago
- Views:
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)
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 informationHow 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 informationThe 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 informationHOMEWORK 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 information13C 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 informationSuggested 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 informationProton 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 informationSolving 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 informationOrganic 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 informationProton 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 informationChapter 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 informationPROTON 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 informationMolecular Formula Determination
Molecular Formula Determination Classical Approach Qualitative elemental analysis Quantitative elemental analysis Determination of empirical formula Molecular weight determination Molecular formula determination
More informationelectron 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 informationNMR 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 informationCHE334 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 informationUsed 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 informationDetermining 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 informationIdentification 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 informationE35 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 informationIR 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 informationHow 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 informationNuclear 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 informationProton 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 informationNuclear 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 informationOrganic 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 informationNuclear 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 informationSymmetric 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 information0 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 informationInfrared 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 informationChapter 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 informationMass 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 informationCHEM 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 informationAssessment 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 informationMass 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 informationUnit 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 informationExperiment 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 informationMolecular 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 informationSample 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 informationExamination 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 informationj. 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 informationSHORT 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 informationNMR 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 informationINFRARED 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 informationDETERMINACIÓN 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 informationAustin 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 informationUnit 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 informationNOMENCLATURE 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 informationMOLECULAR 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 informationChapter 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 informationSurvival 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 informationQuestion 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 informationEXPERIMENT 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 informationChapter 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 informationUltraviolet 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 informationNuclear 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 informationLaboratory 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 informationChemical 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 informationChapter 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 informationName 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 informationUsing 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 informationReactions 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 informationfor 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 informationIDENTIFICATION 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 informationExperiment 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 informationNuclear 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 informationMULTIPLE 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 informationIUPAC 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 informationNuclear 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 informationBackground 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 informationNMR 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 informationInfrared 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 information1. 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 informationAlcohols 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 informationChemistry 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 informationCHEM 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 informationBenzene 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 informationF321 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 informationEXPERIMENT 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 informationDetermination 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 informationOrganic 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 information4. 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 informationOrganic 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 information4/18/2011. 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 informationElement 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 informationNuclear 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 informationDetermination 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 informationORGANIC 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 informationAlcohols. 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 informationChemistry 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 informationOrganic 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 informationAROMATIC 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 informationChapter 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 informationCh17_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 informationCHM220 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 informationOrganic 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 informationPage 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 informationProperties 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 informationCalculating 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 informationIn 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