Spectroscopy by Pavia.pdf

Pavia Lampman...

Pavia Lampman Gain an understanding of Kriz Pavia | Lampman | Kriz | Vyvyan Vyvyan t h e l a t e s t a d va n c e s i n s p e c t r o s c o p y w i t h t h e t e x t t h a t ’s s e t t h e u n r i va l e d Spec t ros co p y standard for more than 30 years. Introduction to Pavia/Lampman/Kriz/Vyvyan’s Introduction to Spectroscopy, 4e, is a comprehensive resource that provides an unmatched, sys- tematic introduction to spectra and basic theoretical concepts in spectroscopic methods that creates a practical learning re- source, whether you’re an introductory student or someone who needs a reliable reference text on spectroscopy. This well-rounded introduction features updated spectra, a modernized presentation of one-dimensional Nuclear Magnetic Resonance (NMR) spectroscopy, the introduction of biological molecules in mass spectrometry, and inclusion of modern tech- niques alongside DEPT, COSY, and HECTOR. Count on this book’s exceptional presentation to provide the comprehensive cover- age needed to truly understand today’s spectroscopic techniques. Visit us on the Web! Fourth a c a d e m i c. c e n g a g e. c o m /c h e m i s t r y Edition Introduction to For your course and learning solutions, visit academic.cengage.com Spectroscopy Four th Edition Purchase any of our products at your local college store or at our preferred online store www.ichapters.com 9780495114789_cvr_se.indd 1 40 AM 14782_FM_i-xvi pp3.qxd 2/7/08 9:11 AM Page i F O U R T H E D I T I O N INTRODUCTION TO SPECTROSCOPY Donald L. Pavia Gary M. Lampman George S. Kriz James R. Vyvyan Department of Chemistry Western Washington University Bellingham, Washington Australia Brazil Japan Korea Mexico Singapore Spain United Kingdom United States 14782_FM_i-xvi pp3.qxd 2/7/08 9:11 AM Page ii 14782_FM_i-xvi pp3.qxd 2/7/08 9:11 AM Page iii TO ALL OF OUR “O-SPEC” STUDENTS 14782_FM_i-xvi pp3.qxd 2/7/08 9:11 AM Page iv Introduction to Spectroscopy, © 2009, 2001 Brooks/Cole, Cengage Learning Fourth Edition ALL RIGHTS RESERVED. No part of this work covered by the copyright Donald L. Pavia, Gary M. Lampman, herein may be reproduced, transmitted, stored, or used in any form George S. Kriz, and James R. Vyvyan or by any means graphic, electronic, or mechanical, including but not Acquisitions Editor: Lisa Lockwood limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval Development Editor: Brandi Kirksey systems, except as permitted under Section 107 or 108 of the 1976 Editorial Assistant: Elizabeth Woods United States Copyright Act, without the prior written permission of Technology Project Manager: Lisa Weber the publisher. Marketing Manager: Amee Mosley For product information and technology assistance, contact us at Marketing Assistant: Elizabeth Wong Cengage Learning Customer & Sales Support, 1-800-354-9706 Marketing Communications Manager: Talia Wise For permission to use material from this text or product, submit all requests online at cengage.com/permissions Project Manager, Editorial Production: Further permissions questions can be e-mailed to Michelle Cole [email protected] Creative Director: Rob Hugel Art Director: John Walker Library of Congress Control Number: 2007943966 Print Buyer: Paula Vang ISBN-13: 978-0-495-11478-9 Permissions Editor: Bob Kauser ISBN-10: 0-495-11478-2 Production Service: PrePress PMG Photo Researcher: Susan Lawson Brooks/Cole Copy Editor: Kathleen Brown 10 Davis Drive Cover Designer: Dare Porter Belmont, CA 94002-3098 Cover Image: Eddie Gerald/Alamy USA Compositor: PrePress PMG Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan. Locate your local office at international.cengage.com/region Cengage Learning products are represented in Canada by Nelson Education, Ltd. For your course and learning solutions, visit academic.cengage.com Purchase any of our products at your local college store or at our preferred online store www.ichapters.com Printed in the United States of America 1 2 3 4 5 6 7 12 11 10 09 08 14782_FM_i-xvi pp3.qxd 2/7/08 9:11 AM Page v PREFACE T his is the fourth edition of a textbook in spectroscopy intended for students of organic chemistry. Our textbook can serve as a supplement for the typical organic chemistry lecture textbook, and it can also be used as a “stand-alone” textbook for an advanced undergraduate course in spectroscopic methods of structure determination or for a first-year graduate course in spectroscopy. This book is also a useful tool for students engaged in research. Our aim is not only to teach students to interpret spectra, but also to present basic theoretical concepts. As with the previ- ous editions, we have tried to focus on the important aspects of each spectroscopic technique with- out dwelling excessively on theory or complex mathematical analyses. This book is a continuing evolution of materials that we use in our own courses, both as a supple- ment to our organic chemistry lecture course series and also as the principal textbook in our upper division and graduate courses in spectroscopic methods and advanced NMR techniques. Explana- tions and examples that we have found to be effective in our courses have been incorporated into this edition. This fourth edition of Introduction to Spectroscopy contains some important changes. The discussion of coupling constant analysis in Chapter 5 has been significantly expanded. Long-range couplings are covered in more detail, and multiple strategies for measuring coupling constants are presented. Most notably, the systematic analysis of line spacings allows students (with a little practice) to extract all of the coupling constants from even the most challenging of first-order multiplets. Chapter 5 also includes an expanded treatment of group equivalence and diastereotopic systems. Discussion of solvent effects in NMR spectroscopy is discussed more explicitly in Chapter 6, and the authors thank one of our graduate students, Ms. Natalia DeKalb, for acquiring the data in Figures 6.19 and 6.20. A new section on determining the relative and absolute stereochemical con- figuration with NMR has also been added to this chapter. The mass spectrometry section (Chapter 8) has been completely revised and expanded in this edition, starting with more detailed discussion of a mass spectrometer’s components. All of the common ionization methods are covered, including chemical ionization (CI), fast-atom bombard- ment (FAB), matrix-assisted laser desorption ionization (MALDI), and electrospray techniques. Different types of mass analyzers are described as well. Fragmentation in mass spectrometry is dis- cussed in greater detail, and several additional fragmentation mechanisms for common functional groups are illustrated. Numerous new mass spectra examples are also included. Problems have been added to each of the chapters. We have included some more solved prob- lems, so that students can develop skill in solving spectroscopy problems. v 14782_FM_i-xvi pp3.qxd 2/7/08 9:11 AM Page vi vi Preface The authors are very grateful to Mr. Charles Wandler, without whose expert help this project could not have been accomplished. We also acknowledge numerous contributions made by our stu- dents who use the textbook and who provide us careful and thoughtful feedback. We wish to alert persons who adopt this book that answers to all of the problems are available on line from the publisher. Authorization to gain access to the web site may be obtained through the local Cengage textbook representative. Finally, once again we must thank our wives, Neva-Jean, Marian, Carolyn, and Cathy for their support and their patience. They endure a great deal in order to support us as we write, and they deserve to be part of the celebration when the textbook is completed! Donald L. Pavia Gary M. Lampman George S. Kriz James R. Vyvyan 14782_FM_i-xvi pp3.qxd 2/7/08 9:11 AM Page vii CONTENTS CHAPTER 1 MOLECULAR FORMULAS AND WHAT CAN BE LEARNED FROM THEM 1 1.1 Elemental Analysis and Calculations 1 1.2 Determination of Molecular Mass 5 1.3 Molecular Formulas 5 1.4 Index of Hydrogen Deficiency 6 1.5 The Rule of Thirteen 9 1.6 A Quick Look Ahead to Simple Uses of Mass Spectra 12 Problems 13 References 14 CHAPTER 2 INFRARED SPECTROSCOPY 15 2.1 The Infrared Absorption Process 16 2.2 Uses of the Infrared Spectrum 17 2.3 The Modes of Stretching and Bending 18 2.4 Bond Properties and Absorption Trends 20 2.5 The Infrared Spectrometer 23 A. Dispersive Infrared Spectrometers 23 B. Fourier Transform Spectrometers 25 2.6 Preparation of Samples for Infrared Spectroscopy 26 2.7 What to Look for When Examining Infrared Spectra 26 2.8 Correlation Charts and Tables 28 2.9 How to Approach the Analysis of a Spectrum (Or What You Can Tell at a Glance) 30 vii 14782_FM_i-xvi pp3.qxd 2/7/08 9:11 AM Page viii viii Contents 2.10 Hydrocarbons: Alkanes, Alkenes, and Alkynes 31 A. Alkanes 31 B. Alkenes 33 C. Alkynes 35 2.11 Aromatic Rings 43 2.12 Alcohols and Phenols 47 2.13 Ethers 50 2.14 Carbonyl Compounds 52 A. Factors that Influence the CJO Stretching Vibration 54 B. Aldehydes 56 C. Ketones 58 D. Carboxylic Acids 62 E. Esters 64 F. Amides 70 G. Acid Chlorides 72 H. Anhydrides 73 2.15 Amines 74 2.16 Nitriles, Isocyanates, Isothiocyanates, and Imines 77 2.17 Nitro Compounds 79 2.18 Carboxylate Salts, Amine Salts, and Amino Acids 80 2.19 Sulfur Compounds 81 2.20 Phosphorus Compounds 84 2.21 Alkyl and Aryl Halides 84 2.22 The Background Spectrum 86 Problems 88 References 104 CHAPTER 3 page 122 NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY PART ONE: BASIC CONCEPTS 105 3.1 Nuclear Spin States 105 3.2 Nuclear Magnetic Moments 106 3.3 Absorption of Energy 107 3.4 The Mechanism of Absorption (Resonance) 109 3.5 Population Densities of Nuclear Spin States 111 3.6 The Chemical Shift and Shielding 112 3.7 The Nuclear Magnetic Resonance Spectrometer 114 A. The Continuous-Wave (CW) Instrument 114 B. The Pulsed Fourier Transform (FT) Instrument 116 3.8 Chemical Equivalence—A Brief Overview 120 14782_FM_i-xvi pp3.qxd 2/7/08 9:11 AM Page ix Contents ix 3.9 Integrals and Integration 121 3.10 Chemical Environment and Chemical Shift 123 3.11 Local Diamagnetic Shielding 124 A. Electronegativity Effects 124 B. Hybridization Effects 126 C. Acidic and Exchangeable Protons; Hydrogen Bonding 127 3.12 Magnetic Anisotropy 128 3.13 Spin–Spin Splitting (n + 1) Rule 131 3.14 The Origin of Spin–Spin Splitting 134 3.15 The Ethyl Group (CH3CH2I) 136 3.16 Pascal’s Triangle 137 3.17 The Coupling Constant 138 3.18 A Comparison of NMR Spectra at Low- and High-Field Strengths 141 1 3.19 Survey of Typical H NMR Absorptions by Type of Compound 142 A. Alkanes 142 B. Alkenes 144 C. Aromatic Compounds 145 D. Alkynes 146 E. Alkyl Halides 148 F. Alcohols 149 G. Ethers 151 H. Amines 152 I. Nitriles 153 J. Aldehydes 154 K. Ketones 156 L. Esters 157 M. Carboxylic Acids 158 N. Amides 159 O. Nitroalkanes 160 Problems 161 References 176 CHAPTER 4 NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY PART TWO: CARBON-13 SPECTRA, INCLUDING HETERONUCLEAR COUPLING WITH OTHER NUCLEI 177 4.1 The Carbon-13 Nucleus 177 4.2 Carbon-13 Chemical Shifts 178 A. Correlation Charts 178 B. Calculation of 13C Chemical Shifts 180 13 4.3 Proton-Coupled C Spectra—Spin–Spin Splitting of Carbon-13 Signals 181 14782_FM_i-xvi pp3.qxd 2/7/08 9:11 AM Page x x Contents 4.4 Proton-Decoupled 13C Spectra 183 4.5 Nuclear Overhauser Enhancement (NOE) 184 4.6 Cross-Polarization: Origin of the Nuclear Overhauser Effect 186 13 4.7 Problems with Integration in C Spectra 189 4.8 Molecular Relaxation Processes 190 4.9 Off-Resonance Decoupling 192 4.10 A Quick Dip into DEPT 192 4.11 Some Sample Spectra—Equivalent Carbons 195 4.12 Compounds with Aromatic Rings 197 4.13 Carbon-13 NMR Solvents—Heteronuclear Coupling of Carbon to Deuterium 199 4.14 Heteronuclear Coupling of Carbon-13 to Fluorine-19 203 4.15 Heteronuclear Coupling of Carbon-13 to Phosphorus-31 204 4.16 Carbon and Proton NMR: How to Solve a Structure Problem 206 Problems 210 References 231 CHAPTER 5 NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY PART THREE: SPIN–SPIN COUPLING 233 5.1 Coupling Constants: Symbols 233 5.2 Coupling Constants: The Mechanism of Coupling 234 1 A. One-Bond Couplings ( J) 235 B. Two-Bond Couplings (2J) 236 3 C. Three-Bond Couplings ( J) 239 4 n D. Long-Range Couplings ( J– J) 244 5.3 Magnetic Equivalence 247 5.4 Spectra of Diastereotopic Systems 252 A. Diastereotopic Methyl Groups: 4-Methyl-2-pentanol 252 B. Diastereotopic Hydrogens: 4-Methyl-2-pentanol 254 5.5 Nonequivalence within a Group—The Use of Tree Diagrams when the n + 1 Rule Fails 257 5.6 Measuring Coupling Constants from First-Order Spectra 260 A. Simple Multiplets—One Value of J (One Coupling) 260 B. Is the n + 1 Rule Ever Really Obeyed? 262 C. More Complex Multiplets—More Than One Value of J 264 5.7 Second-Order Spectra—Strong Coupling 268 A. First-Order and Second-Order Spectra 268 B. Spin System Notation 269 C. The A2, AB, and AX Spin Systems 270 D. The AB2... AX2 and A2B2... A2X2 Spin Systems 270 14782_FM_i-xvi pp3.qxd 2/7/08 9:11 AM Page xi Contents xi E. Simulation of Spectra 272 F. The Absence of Second-Order Effects at Higher Field 272 G. Deceptively Simple Spectra 273 5.8 Alkenes 277 5.9 Measuring Coupling Constants—Analysis of an Allylic System 281 5.10 Aromatic Compounds—Substituted Benzene Rings 285 A. Monosubstituted Rings 286 B. para-Disubstituted Rings 288 C. Other Substitution 291 5.11 Coupling in Heteroaromatic Systems 293 Problems 296 References 328 CHAPTER 6 NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY PART FOUR: OTHER TOPICS IN ONE-DIMENSIONAL NMR 329 6.1 Protons on Oxygen: Alcohols 329 6.2 Exchange in Water and D2O 332 A. Acid/Water and Alcohol/Water Mixtures 332 B. Deuterium Exchange 333 C. Peak Broadening Due to Exchange 337 6.3 Other Types of Exchange: Tautomerism 338 6.4 Protons on Nitrogen: Amines 340 6.5 Protons on Nitrogen: Quadrupole Broadening and Decoupling 342 6.6 Amides 345 6.7 The Effect of Solvent on Chemical Shift 347 6.8 Chemical Shift Reagents 351 6.9 Chiral Resolving Agents 354 6.10 Determining Absolute and Relative Configuration via NMR 356 A. Determining Absolute Configuration 356 B. Determining Relative Configuration 358 6.11 Nuclear Overhauser Effect Difference Spectra 359 Problems 362 References 380 CHAPTER 7 ULTRAVIOLET SPECTROSCOPY 381 7.1 The Nature of Electronic Excitations 381 7.2 The Origin of UV Band Structure 383 7.3 Principles of Absorption Spectroscopy 383 14782_FM_i-xvi pp3.qxd 2/7/08 9:11 AM Page xii xii Contents 7.4 Instrumentation 384 7.5 Presentation of Spectra 385 7.6 Solvents 386 7.7 What Is a Chromophore? 387 7.8 The Effect of Conjugation 390 7.9 The Effect of Conjugation on Alkenes 391 7.10 The Woodward–Fieser Rules for Dienes 394 7.11 Carbonyl Compounds; Enones 397 7.12 Woodward’s Rules for Enones 400 7.13 a,b-Unsaturated Aldehydes, Acids, and Esters 402 7.14 Aromatic Compounds 402 A. Substituents with Unshared Electrons 404 B. Substituents Capable of p-Conjugation 406 C. Electron-Releasing and Electron-Withdrawing Effects 406 D. Disubstituted Benzene Derivatives 406 E. Polynuclear Aromatic Hydrocarbons and Heterocyclic Compounds 409 7.15 Model Compound Studies 411 7.16 Visible Spectra: Color in Compounds 412 7.17 What to Look for in an Ultraviolet Spectrum: A Practical Guide 413 Problems 415 References 417 CHAPTER 8 MASS SPECTROMETRY 418 8.1 The Mass Spectrometer: Overview 418 8.2 Sample Introduction 419 8.3 Ionization Methods 420 A. Electron Ionization (EI) 420 B. Chemical Ionization (CI) 421 C. Desorption Ionization Techniques (SIMS, FAB, and MALDI) 425 D. Electrospray Ionization (ESI) 426 8.4 Mass Analysis 429 A. The Magnetic Sector Mass Analyzer 429 B. Double-Focusing Mass Analyzers 430 C. Quadrupole Mass Analyzers 430 D. Time-of-Flight Mass Analyzers 432 8.5 Detection and Quantitation: The Mass Spectrum 435 8.6 Determination of Molecular Weight 438 8.7 Determination of Molecular Formulas 441 14782_FM_i-xvi pp3.qxd 2/7/08 9:11 AM Page xiii Contents xiii A. Precise Mass Determination 441 B. Isotope Ratio Data 441 8.8 Structural Analysis and Fragmentation Patterns 445 A. Stevenson’s Rule 446 B. The Initial Ionization Event 447 C. Radical-site Initiated Cleavage: a-Cleavage 448 D. Charge-site Initiated Cleavage: Inductive Cleavage 448 E. Two-Bond Cleavage 449 F. Retro Diels-Adler Cleavage 450 G. McLafferty Rearrangements 450 H. Other Cleavage Types 451 I. Alkanes 451 J. Cycloalkanes 454 K. Alkenes 455 L. Alkynes 459 M. Aromatic Hydrocarbons 459 N. Alcohols and Phenols 464 O. Ethers 470 P. Aldehydes 472 Q. Ketones 473 R. Esters 477 S. Carboxylic Acids 482 T. Amines 484 U. Selected Nitrogen and Sulfur Compounds 488 V. Alkyl Chlorides and Alkyl Bromides 492 8.9 Strategic Approach to Analyzing Mass Spectra and Solving Problems 496 8.10 Computerized Matching of Spectra with Spectral Libraries 497 Problems 498 References 519 CHAPTER 9 COMBINED STRUCTURE PROBLEMS 520 Example 1 522 Example 2 524 Example 3 526 Example 4 529 Problems 531 Sources of Additional Problems 586 14782_FM_i-xvi pp3.qxd 2/7/08 9:11 AM Page xiv xiv Contents CHAPTER 10 NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY PART FIVE: ADVANCED NMR TECHNIQUES 587 10.1 Pulse Sequences 587 10.2 Pulse Widths, Spins, and Magnetization Vectors 589 10.3 Pulsed Field Gradients 593 10.4 The DEPT Experiment 595 10.5 Determining the Number of Attached Hydrogens 598 A. Methine Carbons (CH) 598 B. Methylene Carbons (CH2) 599 C. Methyl Carbons (CH3) 601 D. Quaternary Carbons (C) 601 E. The Final Result 602 10.6 Introduction to Two-Dimensional Spectroscopic Methods 602 10.7 The COSY Technique 602 A. An Overview of the COSY Experiment 603 B. How to Read COSY Spectra 604 10.8 The HETCOR Technique 608 A. An Overview of the HETCOR Experiment 608 B. How to Read HETCOR Spectra 609 10.9 Inverse Detection Methods 612 10.10 The NOESY Experiment 613 10.11 Magnetic Resonance Imaging 614 10.12 Solving a Structural Problem Using Combined 1-D and 2-D Techniques 616 A. Index of Hydrogen Deficiency and Infrared Spectrum 616 B. Carbon-13 NMR Spectrum 617 C. DEPT Spectrum 617 D. Proton NMR Spectrum 619 E. COSY NMR Spectrum 621 F. HETCOR (HSQC) NMR Spectrum 622 Problems 623 References 657 ANSWERS TO SELECTED PROBLEMS ANS-1 APPENDICES Appendix 1 Infrared Absorption Frequencies of Functional Groups A-1 1 Appendix 2 Approximate H Chemical Shift Ranges (ppm) for Selected Types of Protons A-8 Appendix 3 Some Representative 1H Chemical Shift Values for Various Types of Protons A-9 14782_FM_i-xvi pp3.qxd 2/7/08 9:11 AM Page xv Contents xv 1 Appendix 4 H Chemical Shifts of Selected Heterocyclic and Polycyclic Aromatic Compounds A-12 Appendix 5 Typical Proton Coupling Constants A-13 1 Appendix 6 Calculation of Proton ( H) Chemical Shifts A-17 13 Appendix 7 Approximate C Chemical-Shift Values (ppm) for Selected Types of Carbon A-21 13 Appendix 8 Calculation of C Chemical Shifts A-22 13 Appendix 9 C Coupling Constants A-32 1 Appendix 10 H and 13C Chemical Shifts for Common NMR Solvents A-33 Appendix 11 Tables of Precise Masses and Isotopic Abundance Ratios for Molecular Ions under Mass 100 Containing Carbon, Hydrogen, Nitrogen, and Oxygen A-34 Appendix 12 Common Fragment Ions under Mass 105 A-40 Appendix 13 A Handy-Dandy Guide to Mass Spectral Fragmentation Patterns A-43 Appendix 14 Index of Spectra A-46 INDEX I-1 14782_FM_i-xvi pp3.qxd 2/7/08 9:11 AM Page xvi 14782_01_Ch1_p001-014.pp3.qxd 1/25/08 10:11 AM Page 1 C H A P T E R 1 MOLECULAR FORMULAS AND WHAT CAN BE LEARNED FROM THEM B efore attempting to deduce the structure of an unknown organic compound from an exami- nation of its spectra, we can simplify the problem somewhat by examining the molecular formula of the substance. The purpose of this chapter is to describe how the molecular for- mula of a compound is determined and how structural information may be obtained from that for- mula. The chapter reviews both the modern and classical quantitative methods of determining the molecular formula. While use of the mass spectrometer (Section 1.6 and Chapter 8) can supplant many of these quantitative analytical methods, they are still in use. Many journals still require that a satisfactory quantitative elemental analysis (Section 1.1) be obtained prior to the publication of research results. 1.1 ELEMENTAL ANALYSIS AND CALCULATIONS The classical procedure for determining the molecular formula of a substance involves three steps: 1. A qualitative elemental analysis to find out what types of atoms are present... C, H, N, O, S, Cl, and so on. 2. A quantitative elemental analysis (or microanalysis) to find out the relative numbers (per- centages) of each distinct type of atom in the molecule. 3. A molecular mass (or molecular weight) determination. The first two steps establish an empirical formula for the compound. When the results of the third procedure are known, a molecular formula is found. Virtually all organic compounds contain carbon and hydrogen. In most cases, it is not neces- sary to determine whether these elements are present in a sample: their presence is assumed. However, if it should be necessary to demonstrate that either carbon or hydrogen is present in a compound, that substance may be burned in the presence of excess oxygen. If the combustion produces carbon dioxide, carbon must be present; if combustion produces water, hydrogen atoms must be present. Today, the carbon dioxide and water can be detected by gas chromatographic methods. Sulfur atoms are converted to sulfur dioxide; nitrogen atoms are often chemically re- duced to nitrogen gas following their combustion to nitrogen oxides. Oxygen can be detected by the ignition of the compound in an atmosphere of hydrogen gas; the product is water. Currently, all such analyses are performed by gas chromatography, a method that can also determine the rel- ative amounts of each of these gases. If the amount of the original sample is known, it can be entered, and the computer can calculate the percentage composition of the sample. Unless you work in a large company or in one of the larger universities, it is quite rare to find a research laboratory in which elemental analyses are performed on site. It requires too much time to set up the apparatus and keep it operating within the limits of suitable accuracy and precision. Usually, samples are sent to a commercial microanalytical laboratory that is prepared to do this work routinely and to vouch for the accuracy of the results. 1 14782_01_Ch1_p001-014.pp3.qxd 1/25/08 10:11 AM Page 2 2 Molecular Formulas and What Can Be Learned from Them Before the advent of modern instrumentation, the combustion of the precisely weighed sample was carried out in a cylindrical glass tube, contained within a furnace. A stream of oxygen was passed through the heated tube on its way to two other sequential, unheated tubes that contained chemical substances that would absorb first the water (MgClO4) and then the carbon dioxide (NaOH/silica). These preweighed absorption tubes were detachable and were removed and reweighed to determine the amounts of water and carbon dioxide formed. The percentages of carbon and hydrogen in the original sample were calculated by simple stoichiometry. Table 1.1 shows a sample calculation. Notice in this calculation that the amount of oxygen was determined by difference, a common practice. In a sample containing only C, H, and O, one needs to determine the percentages of only C and H; oxygen is assumed to be the unaccounted-for portion. You may also apply this practice in sit- uations involving elements other than oxygen; if all but one of the elements is determined, the last one can be determined by difference. Today, most calculations are carried out automatically by the computerized instrumentation. Nevertheless, it is often useful for a chemist to understand the fun- damental principles of the calculations. Table 1.2 shows how to determine the empirical formula of a compound from the percentage compositions determined in an analysis. Remember that the empirical formula expresses the simplest whole-number ratios of the elements and may need to be multiplied by an integer to obtain the true molecular formula. To determine the value of the multiplier, a molecular mass is required. Determination of the molecular mass is discussed in the next section. For a totally unknown compound (unknown chemical source or history) you will have to use this type of calculation to obtain the suspected empirical formula. However, if you have prepared the compound from a known precursor by a well-known reaction, you will have an idea of the structure of the compound. In this case, you will have calculated the expected percentage composition of your TA B L E 1. 1 CALCULATION OF PERCENTAGE COMPOSITION FROM COMBUSTION DATA CxHyOz + excess O2 ⎯→ x CO2 + y/2 H2O 9.83 mg 23.26 mg 9.52 mg 23.26 mg CO2 millimoles CO2 = ᎏᎏ = 0.5285 mmoles CO2 44.01 mg/mmole mmoles CO2 = mmoles C in original sample (0.5285 mmoles C)(12.01 mg/mmole C) = 6.35 mg C in original sample 9.52 mg H2O millimoles H2O = ᎏᎏ = 0.528 mmoles H2O 18.02 mg/mmole (2 mmoles H ) (0.528 mmoles H2O) ᎏᎏ = 1.056 mmoles H in original sample 1 mmole H2O (1.056 mmoles H)(1.008 mg/mmole H) = 1.06 mg H in original sample 6.35 mg C % C = ᎏᎏ × 100 = 64.6% 9.83 mg sample 1.06 mg H % H = ᎏᎏ × 100 = 10.8% 9.83 mg sample % O = 100 − (64.6 + 10.8) = 24.6% 14782_01_Ch1_p001-014.pp3.qxd 1/25/08 10:11 AM Page 3 1.1 Elemental Analysis and Calculations 3 TA B L E 1. 2 CALCULATION OF EMPIRICAL FORMULA Using a 100-g sample: 64.6% of C = 64.6 g 10.8% of H = 10.8 g 24.6 g 24.6% of O = ᎏᎏ 100.0 g 64.6 g moles C = ᎏᎏ = 5.38 moles C 12.01 g/mole 10.8 g moles H = ᎏᎏ = 10.7 moles H 1.008 g/mole 24.6 g moles O = ᎏᎏ = 1.54 moles O 16.0 g/mole giving the result C5.38H10.7O1.54 Converting to the simplest ratio: C⎯ ⎯ H 10.7 5.38 ⎯— = C3.49H6.95O1.00 ⎯— O1.54 1.54 1.54 1.54 which approximates C3.50H7.00O1.00 or C7H14O2 sample in advance (from its postulated structure) and will use the analysis to verify your hypothesis. When you perform these calculations, be sure to use the full molecular weights as given in the peri- odic chart and do not round off until you have completed the calculation. The final result should be good to two decimal places: four significant figures if the percentage is between 10 and 100; three figures if it is between 0 and 10. If the analytical results do not agree with the calculation, the sam- ple may be impure, or you may have to calculate a new empirical formula to discover the identity of the unexpected structure. To be accepted for publication, most journals require the percentages found to be less than 0.4% off from the calculated value. Most microanalytical laboratories can eas- ily obtain accuracy well below this limit provided the sample is pure. In Figure 1.1, a typical situation for the use of an analysis in research is shown. Professor Amyl Carbon, or one of his students, prepared a compound believed to be the epoxynitrile with the struc- ture shown at the bottom of the first form. A sample of this liquid compound (25 μ L) was placed in a small vial correctly labeled with the name of the submitter and an identifying code (usually one that corresponds to an entry in the research notebook). Only a small amount of the sample is re- quired, usually a few milligrams of a solid or a few microliters of a liquid. A Request for Analysis form must be filled out and submitted along with the sample. The sample form on the left side of the figure shows the type of information that must be submitted. In this case, the professor calcu- lated the expected results for C, H, and N and the expected formula and molecular weight. Note that the compound also contains oxygen, but that there was no request for an oxygen analysis. Two other samples were also submitted at the same time. After a short time, typically within a week, the 14782_01_Ch1_p001-014.pp3.qxd 1/25/08 10:11 AM Page 4 4 Molecular Formulas and What Can Be Learned from Them n a l y t i c al Microa ny, Inc. i c a l Compa a l y t Microanny, Inc. REQUEST FOR ANALYSIS FORM Date: October 30, 2006 Compa Report To: Professor Amyl Carbon Department of Chemistry November 25, 2006 Western Washington University Professor Amyl Carbon Bellingham, WA 98225 Department of Chemistry Sample No: PAC599A P.O. No : PO 2349 Western Washington University Report By: AirMail Phone Email (circle one) [email protected] Bellingham, WA RESULTS OF ANALYSIS Elements to Analyze: C, H, N Other Elements Present : O Sample ID Carbon (%) Hydrogen (%) Nitrogen (%) X Single Analysis Duplicate Analysis Duplicate only if results are not in range PAC599A 67.39 9.22 11.25 PAC589B 64.98 9.86 8.03 M.P. B.P. 69 ˚C @ 2.3 mmHg PAC603 73.77 8.20 Sensitive to : Weigh under N? Y N Dry the Sample? Y N Details: Dr. B. Grant Poohbah, Hygroscopic Volatile Explosive Ph.D. THEORY OR RANGE Director of Analytical Services %C 67.17 Amount Provided L Microanalytical Company, Inc PAC599A PAC603 %H 8.86 Stucture: O PAC589B %N 11.19 CN %O %Other Comments: C7H11NO Mol. Wt. 125.17 F I G U R E 1. 1 Sample microanalysis forms. Shown on the left is a typical submission form that is sent with the samples. (The three shown here in labeled vials were all sent at the same time.) Each sample needs its own form. In the background on the right is the formal letter that reported the results. Were the results obtained for sample PAC599A satisfactory? results were reported to Professor Carbon as an email (see the request on the form). At a later date, a formal letter (shown in the background on the right-hand side) is sent to verify and authenticate the results. Compare the values in the report to those calculated by Professor Carbon. Are they within the accepted range? If not, the analysis will have to be repeated with a freshly purified sam- ple, or a new possible structure will have to be considered. Keep in mind that in an actual laboratory situation, when you are trying to determine the molec- ular formula of a totally new or previously unknown compound, you will have to allow for some variance in the quantitative elemental analysis. Other data can help you in this situation since in- frared (Chapter Two) and nuclear magnetic resonance (NMR) (Chapter Three) data will also sug- gest a possible structure or at least some of its prominent features. Many times, these other data will be less sensitive to small amounts of impurities than the microanalysis. 14782_01_Ch1_p001-014.pp3.qxd 1/25/08 10:11 AM Page 5 1.3 Molecular Formulas 5 1.2 DETERMINATION OF MOLECULAR MASS The next step in determining the molecular formula of a substance is to determine the weight of one mole of that substance. This may be accomplished in a variety of ways. Without knowledge of the molecular mass of the unknown, there is no way of determining whether the empirical formula, which is determined directly from elemental analysis, is the true formula of the sub- stance or whether the empirical formula must be multiplied by some integral factor to obtain the molecular formula. In the example cited in Section 1.1, without knowledge of the molecular mass of the unknown, it is impossible to tell whether the molecular formula is C7H14O2 or C14H28O4. In a modern laboratory, the molecular mass is determined using mass spectrometry. The details of this method and the means of determining molecular mass can be found in Section 1.6 and Chapter 8, Section 8.6. This section reviews some classical methods of obtaining the same information. An old method that is used occasionally is the vapor density method. In this method, a known volume of gas is weighed at a known temperature. After converting the volume of the gas to standard temperature and pressure, we can determine what fraction of a mole that volume represents. From that fraction, we can easily calculate the molecular mass of the substance. Another method of determining the molecular mass of a substance is to measure the freezing-point depression of a solvent that is brought about when a known quantity of test substance is added. This is known as a cryoscopic method. Another method, which is used occasionally, is vapor pressure osmometry, in which the molecular weight of a substance is determined through an examination of the change in vapor pressure of a solvent when a test substance is dissolved in it. If the unknown substance is a carboxylic acid, it may be titrated with a standardized solution of sodium hydroxide. By use of this procedure, a neutralization equivalent can be determined. The neutralization equivalent is identical to the equivalent weight of the acid. If the acid has only one carboxyl group, the neutralization equivalent and the molecular mass are identical. If the acid has more than one carboxyl group, the neutralization equivalent is equal to the molecular mass of the acid divided by the number of carboxyl groups. Many phenols, especially those substituted by electron-withdrawing groups, are sufficiently acidic to be titrated by this same method, as are sulfonic acids. 1.3 MOLECULAR FORMULAS Once the molecular mass and the empirical formula are known, we may proceed directly to the molecular formula. Often, the empirical formula weight and the molecular mass are the same. In such cases, the empirical formula is also the molecular formula. However, in many cases, the empir- ical formula weight is less than the molecular mass, and it is necessary to determine how many times the empirical formula weight can be divided into the molecular mass. The factor determined in this manner is the one by which the empirical formula must be multiplied to obtain the molecular formula. Ethane provides a simple example. After quantitative element analysis, the empirical formula for ethane is found to be CH3. A molecular mass of 30 is determined. The empirical formula weight of ethane, 15, is half of the molecular mass, 30. Therefore, the molecular formula of ethane must be 2(CH3) or C2H6. For the sample unknown introduced earlier in this chapter, the empirical formula was found to be C7H14O2. The formula weight is 130. If we assume that the molecular mass of this substance was determined to be 130, we may conclude that the empirical formula and the molecular formula are identical, and that the molecular formula must be C7H14O2. 14782_01_Ch1_p001-014.pp3.qxd 1/25/08 10:11 AM Page 6 6 Molecular Formulas and What Can Be Learned from Them 1.4 INDEX OF HYDROGEN DEFICIENCY Frequently, a great deal can be learned about an unknown substance simply from knowledge of its molecular formula. This information is based on the following general molecular formulas: alkane CnH2n+2 ⎫ ⎬ Difference of 2 hydrogens cycloalkane or alkene CnH2n ⎭ ⎫ Difference of 2 hydrogens alkyne CnH2n−2 ⎬⎭ Notice that each time a ring or p bond is introduced into a molecule, the number of hydrogens in the molecular formula is reduced by two. For every triple bond (two p bonds), the molecular for- mula is reduced by four. This is illustrated in Figure 1.2. When the molecular formula for a compound contains noncarbon or nonhydrogen elements, the ratio of carbon to hydrogen may change. Following are three simple rules that may be used to predict how this ratio will change: 1. To convert the formula of an open-chain, saturated hydrocarbon to a formula containing Group V elements (N, P, As, Sb, Bi), one additional hydrogen atom must be added to the molecular formula for each such Group V element present. In the following examples, each formula is correct for a two-carbon acyclic, saturated compound: C2H6, C2H7N, C2H8N2, C2H9N3 2. To convert the formula of an open-chain, saturated hydrocarbon to a formula containing Group VI elements (O, S, Se, Te), no change in the number of hydrogens is required. In the following examples, each formula is correct for a two-carbon, acyclic, saturated compound: C2H6, C2H6O, C2H6O2, C2H6O3 –2H C C C C H H (also compare CHOH to C O) H H –4H C C C C H H CH2 CH2 H 2C CH2 H H2C CH2 –2H H2C CH2 H H2C CH2 CH2 CH2 F I G U R E 1. 2 Formation of rings and double bonds. Formation of each ring or double bond causes the loss of 2H. 14782_01_Ch1_p001-014.pp3.qxd 1/25/08 10:11 AM Page 7 1.4 Index of Hydrogen Deficiency 7 3. To convert the formula of an open-chain, saturated hydrocarbon to a formula containing Group VII elements (F, Cl, Br, I), one hydrogen must be subtracted from the molecular formula for each such Group VII element present. In the following examples, each formula is correct for a two-carbon, acyclic, saturated compound: C2H6, C2H5F, C2H4F2, C2H3F3 Table 1.3 presents some examples that should demonstrate how these correction numbers were de- termined for each of the heteroatom groups. The index of hydrogen deficiency (sometimes called the unsaturation index) is the number of p bonds and/or rings a molecule contains. It is determined from an examination of the molecu- lar formula of an unknown substance and from a comparison of that formula with a formula for a corresponding acyclic, saturated compound. The difference in the number of hydrogens between these formulas, when divided by 2, gives the index of hydrogen deficiency. The index of hydrogen deficiency can be very useful in structure determination problems. A great deal of information can be obtained about a molecule before a single spectrum is examined. For example, a compound with an index of one must have one double bond or one ring, but it can- not have both structural features. A quick examination of the infrared spectrum could confirm the presence of a double bond. If there were no double bond, the substance would have to be cyclic and saturated. A compound with an index of two could have a triple bond, or it could have two double bonds, two rings, or one of each. Knowing the index of hydrogen deficiency of a substance, the chemist can proceed directly to the appropriate regions of the spectra to confirm the presence or absence of p bonds or rings. Benzene contains one ring and three “double bonds” and thus has an index of hydrogen deficiency of four. Any substance with an index of four or more may contain a benzenoid ring; a substance with an index less than four cannot contain such a ring. To determine the index of hydrogen deficiency for a compound, apply the following steps: 1. Determine the formula for the saturated, acyclic hydrocarbon containing the same number of carbon atoms as the unknown substance. 2. Correct this formula for the nonhydrocarbon elements present in the unknown. Add one hydrogen atom for each Group V element present and subtract one hydrogen atom for each Group VII element present. 3. Compare this formula with the molecular formula of the unknown. Determine the number of hydrogens by which the two formulas differ. 4. Divide the difference in the number of hydrogens by two to obtain the index of hydrogen deficiency. This equals the number of p bonds and/or rings in the structural formula of the unknown substance. TA B L E 1. 3 CORRECTIONS TO THE NUMBER OF HYDROGEN ATOMS WHEN GROUP V AND VII HETEROATOMS ARE INTRODUCED (GROUP VI HETEROATOMS DO NOT REQUIRE A CORRECTION) Group Example Correction Net Change V C —H S C —NH2 +1 Add nitrogen, add 1 hydrogen VI C —H S C — OH 0 Add oxygen (no hydrogen) VII C —H S C — CI –1 Add chlorine, lose 1 hydrogen 14782_01_Ch1_p001-014.pp3.qxd 1/25/08 10:11 AM Page 8 8 Molecular Formulas and What Can Be Learned from Them The following examples illustrate how the index of hydrogen deficiency is determined and how that information can be applied to the determination of a structure for an unknown substance. EXAMPLE 1 The unknown substance introduced at the beginning of this chapter has the molecular formula C7H14O2. 1. Using the general formula for a saturated, acyclic hydrocarbon (CnH2n+2, where n = 7), calculate the formula C7H16. 2. Correction for oxygens (no change in the number of hydrogens) gives the formula C7H16O2. 3. The latter formula differs from that of the unknown by two hydrogens. 4. The index of hydrogen deficiency equals one. There must be one ring or one double bond in the unknown substance. Having this information, the chemist can proceed immediately to the double-bond regions of the infrared spectrum. There, she finds evidence for a carbon–oxygen double bond (carbonyl group). At this point, the number of possible isomers that might include the unknown has been narrowed considerably. Further analysis of the spectral evidence leads to an identification of the unknown substance as isopentyl acetate. O CH3 C O CH2 CH2 CH CH3 CH3 EXAMPLE 2 Nicotine has the molecular formula C10H14N2. 1. The formula for a 10-carbon, saturated, acyclic hydrocarbon is C10H22. 2. Correction for the two nitrogens (add two hydrogens) gives the formula C10H24N2. 3. The latter formula differs from that of nicotine by 10 hydrogens. 4. The index of hydrogen deficiency equals five. There must be some combination of five p bonds and/or rings in the molecule. Since the index is greater than four, a benzenoid ring could be included in the molecule. Analysis of the spectrum quickly shows that a benzenoid ring is indeed present in nicotine. The spec- tral results indicate no other double bonds, suggesting that another ring, this one saturated, must be present in the molecule. More careful refinement of the spectral analysis leads to a structural formula for nicotine: N CH3 N 14782_01_Ch1_p001-014.pp3.qxd 1/25/08 10:11 AM Page 9 1.5 The Rule of Thirteen 9 EXAMPLE 3 Chloral hydrate (“knockout drops”) is found to have the molecular formula C2H3Cl3O2. 1. The formula for a two-carbon, saturated, acyclic hydrocarbon is C2H6. 2. Correction for oxygens (no additional hydrogens) gives the formula C2H6O2. 3. Correction for chlorines (subtract three hydrogens) gives the formula C2H3Cl3O2. 4. This formula and the formula of chloral hydrate correspond exactly. 5. The index of hydrogen deficiency equals zero. Chloral hydrate cannot contain rings or double bonds. Examination of the spectral results is limited to regions that correspond to singly bonded structural features. The correct structural formula for chloral hydrate follows. You can see that all of the bonds in the molecule are single bonds. OH Cl3C C H OH 1.5 THE RULE OF THIRTEEN High-resolution mass spectrometry provides molecular mass information from which the user can determine the exact molecular formula directly. The discussion on exact mass determination in Chapter 8 explains this process in detail. When such molar mass information is not available, how- ever, it is often useful to be able to generate all the possible molecular formulas for a given mass. By applying other types of spectroscopic information, it may then be possible to distinguish among these possible formulas. A useful method for generating possible molecular formulas for a given molecular mass is the Rule of Thirteen.1 As a first step in the Rule of Thirteen, we generate a base formula, which contains only carbon and hydrogen. The base formula is found by dividing the molecular mass M by 13 (the mass of one carbon plus one hydrogen). This calculation provides a numerator n and a remainder r. M r ᎏᎏ = n + ᎏᎏ 13 13 The base formula thus becomes CnHn+r which is a combination of carbons and hydrogens that has the desired molecular mass M. The index of hydrogen deficiency (unsaturation index) U that corresponds to the preceding for- mula is calculated easily by applying the relationship (n − r + 2) U = ᎏᎏ 2 1 Bright, J. W., and E. C. M. Chen, “Mass Spectral Interpretation Using the ‘Rule of 13,’” Journal of Chemical Education, 60 (1983): 557. 14782_01_Ch1_p001-014.pp3.qxd 1/25/08 10:11 AM Page 10 10 Molecular Formulas and What Can Be Learned from Them Of course, you can also calculate the index of hydrogen deficiency using the method shown in Section 1.4. If we wish to derive a molecular formula that includes other atoms besides carbon and hydrogen, then we must subtract the mass of a combination of carbons and hydrogens that equals the masses of the other atoms being included in the formula. For example, if we wish to convert the base for- mula to a new formula containing one oxygen atom, then we subtract one carbon and four hydro- gens at the same time that we add one oxygen atom. Both changes involve a molecular mass equivalent of 16 (O = CH4 = 16). Table 1.4 includes a number of C/H equivalents for replacement of carbon and hydrogen in the base formula by the most common elements likely to occur in an organic compound.2 To comprehend how the Rule of Thirteen might be applied, consider an unknown substance with a molecular mass of 94 amu. Application of the formula provides 94 3 ᎏᎏ = 7 + ᎏᎏ 13 13 According to the formula, n = 7 and r = 3. The base formula must be C7H10 The index of hydrogen deficiency is (7 − 3 + 2) U = ᎏᎏ = 3 2 TA B L E 1. 4 CARBON/HYDROGEN EQUIVALENTS FOR SOME COMMON ELEMENTS Add Subtract Add Add Subtract Add Element Equivalent ⌬U Element Equivalent ⌬U 35 C H12 7 Cl C2H11 3 H12 C −7 79 Br C6H7 −3 79 O CH4 1 Br C5H19 4 O2 C2H8 2 F CH7 2 O3 C3H12 3 Si C2H4 1 N CH2 ⎯1⎯ P C2H7 2 2 N2 C2H4 1 I C9H19 0 S C2H8 2 I C10H7 7 2 In Table 1.4, the equivalents for chlorine and bromine are determined assuming that the isotopes are 35Cl and 79Br, respectively. Always use this assumption when applying this method. 14782_01_Ch1_p001-014.pp3.qxd 1/25/08 10:11 AM Page 11 1.5 The Rule of Thirteen 11 A substance that fits this formula must contain some combination of three rings or multiple bonds. A possible structure might be CH3 H H C7H10 H H U=3 H H H If we were interested in a substance that had the same molecular mass but that contained one oxygen atom, the molecular formula would become C6H6O. This formula is determined according to the following scheme: 1. Base formula = C7H10 U=3 2. Add: +O 3. Subtract: − CH4 4. Change the value of U: ΔU = 1 5. New formula = C6H6O 6. New index of hydrogen deficiency: U=4 A possible substance that fits these data is OH C6H6O U=4 There are additional possible molecular formulas that conform to a molecular mass of 94 amu: C5H2O2 U=5 C5H2S U=5 1 C6H8N U= 3⎯2⎯ CH3Br U=0 As the formula C6H8N shows, any formula that contains an even number of hydrogen atoms but an odd number of nitrogen atoms leads to a fractional value of U, an unlikely choice. Any compound with a value of U less than zero (i.e., negative) is an impossible combination. Such a value is often an indicator that an oxygen or nitrogen atom must be present in the molecular formula. When we calculate formulas using this method, if there are not enough hydrogens, we can subtract 1 carbon and add 12 hydrogens (and make the appropriate correction in U). This procedure works only if we obtain a positive value of U. Alternatively we can obtain another potential molecu- lar formula by adding 1 carbon and subtracting 12 hydrogens (and correcting U). 14782_01_Ch1_p001-014.pp3.qxd 1/25/08 10:11 AM Page 12 12 Molecular Formulas and What Can Be Learned from Them 1.6 A QUICK LOOK AHEAD TO SIMPLE USES OF MASS SPECTRA Chapter 8 contains a detailed discussion of the technique of mass spectrometry. See Sections 8.1–8.7 for applications of mass spectrometry to the problems of molecular formula determination. Briefly, the mass spectrometer is an instrument that subjects molecules to a high-energy beam of electrons. This beam of electrons converts molecules to positive ions by removing an electron. The stream of positively charged ions is accelerated along a curved path in a magnetic field. The radius of curvature of the path described by the ions depends on the ratio of the mass of the ion to its charge (the m/z ratio). The ions strike a detector at positions that are determined by the radius of curvature of their paths. The number of ions with a particular mass-to-charge ratio is plotted as a function of that ratio. The particle with the largest mass-to-charge ratio, assuming that the charge is 1, is the particle that represents the intact molecule with only one electron removed. This particle, called the molecular ion (see Chapter 8, Section 8.5), can be identified in the mass spectrum. From its position in the spectrum, its weight can be determined. Since the mass of the dislodged electron is so small, the mass of the molecular ion is essentially equal to the molecular mass of the original molecule. Thus, the mass spectrometer is an instrument capable of providing molecular mass information. Virtually every element exists in nature in several isotopic forms. The natural abundance of each of these isotopes is known. Besides giving the mass of the molecular ion when each atom in the molecule is the most common isotope, the mass spectrum also gives peaks that correspond to that same molecule with heavier isotopes. The ratio of the intensity of the molecular ion peak to the intensities of the peaks corresponding to the heavier isotopes is determined by the natural abundance of each isotope. Because each type of molecule has a unique combination of atoms, and because each type of atom and its isotopes exist in a unique ratio in nature, the ratio of the intensity of the molecular ion peak to the intensities of the isotopic peaks can provide information about the number of each type of atom present in the molecule. For example, the presence of bromine can be determined easily because bromine causes a pattern of molecular ion peaks and isotope peaks that is easily identified. If we identify the mass of the molecular ion peak as M and the mass of the isotope peak that is two mass units heavier than the molecular ion as M + 2, then the ratio of the intensities of the M and M + 2 peaks will be approxi- mately one to one when bromine is present (see Chapter 8, Section 8.5, for more details). When chlorine is present, the ratio of the intensities of the M and M + 2 peaks will be approximately three to one. These ratios reflect the natural abundances of the common isotopes of these elements. Thus, isotope ratio studies in mass spectrometry can be used to determine the molecular formula of a substance. Another fact that can be used in determining the molecular formula is expressed as the Nitrogen Rule. This rule states that when the number of nitrogen atoms present in the molecule is odd, the molecular mass will be an odd number; when the number of nitrogen atoms present in the molecule TA B L E 1. 5 PRECISE MASSES FOR SUBSTANCES OF MOLECULAR MASS EQUAL TO 44 amu Compound Exact Mass (amu) CO2 43.9898 N2O 44.0011 C2H4O 44.0262 C3H8 44.0626 14782_01_Ch1_p001-014.pp3.qxd 1/25/08 10:11 AM Page 13 Problems 13 is even (or zero), the molecular mass will be an even number. The Nitrogen Rule is explained further in Chapter 8, Section 8.6. Since the advent of high-resolution mass spectrometers, it is also possible to use very precise mass determinations of molecular ion peaks to determine molecular formulas. When the atomic weights of the elements are determined very precisely, it is found that they do not have exactly inte- gral values. Every isotopic mass is characterized by a small “mass defect,” which is the amount by which the mass of the isotope differs from a perfectly integral mass number. The mass defect for every isotope of every element is unique. As a result, a precise mass determination can be used to determine the molecular formula of the sample substance, since every combination of atomic weights at a given nominal mass value will be unique when mass defects are considered. For exam- ple, each of the substances shown in Table 1.5 has a nominal mass of 44 amu. As can be seen from the table, their exact masses, obtained by adding exact atomic masses, are substantially different when measured to four decimal places. PROBLEMS *1. Researchers used a combustion method to analyze a compound used as an antiknock additive in gasoline. A 9.394-mg sample of the compound yielded 31.154 mg of carbon dioxide and 7.977 mg of water in the combustion. (a) Calculate the percentage composition of the compound. (b) Determine its empirical formula. *2. The combustion of an 8.23-mg sample of unknown substance gave 9.62 mg CO2 and 3.94 mg H2O. Another sample, weighing 5.32 mg, gave 13.49 mg AgCl in a halogen analysis. Determine the percentage composition and empirical formula for this organic compound. *3. An important amino acid has the percentage composition C 32.00%, H 6.71%, and N 18.66%. Calculate the empirical formula of this substance. *4. A compound known to be a pain reliever had the empirical formula C9H8O4. When a mixture of 5.02 mg of the unknown and 50.37 mg of camphor was prepared, the melting point of a portion of this mixture was determined. The observed melting point of the mixture was 156°C. What is the molecular mass of this substance? *5. An unknown acid was titrated with 23.1 mL of 0.1 N sodium hydroxide. The weight of the acid was 120.8 mg. What is the equivalent weight of the acid? *6. Determine the index of hydrogen deficiency for each of the following compounds: (a) C8H7NO (d) C5H3ClN4 (b) C3H7NO3 (e) C21H22N2O2 (c) C4H4BrNO2 *7. A substance has the molecular formula C4H9N. Is there any likelihood that this material contains a triple bond? Explain your reasoning. *8. (a) A researcher analyzed an unknown solid, extracted from the bark of spruce trees, to determine its percentage composition. An 11.32-mg sample was burned in a combustion apparatus. The carbon dioxide (24.87 mg) and water (5.82 mg) were collected and weighed. From the results of this analysis, calculate the percentage composition of the unknown solid. (b) Determine the empirical formula of the unknown solid. 14782_01_Ch1_p001-014.pp3.qxd 1/25/08 10:11 AM Page 14 14 Molecular Formulas and What Can Be Learned from Them (c) Through mass spectrometry, the molecular mass was found to be 420 g/mole. What is the molecular formula? (d) How many aromatic rings could this compound contain? *9. Calculate the molecular formulas for possible compounds with molecular masses of 136; use the Rule of Thirteen. You may assume that the only other atoms present in each molecule are carbon and hydrogen. (a) A compound with two oxygen atoms (b) A compound with two nitrogen atoms (c) A compound with two nitrogen atoms and one oxygen atom (d) A compound with five carbon atoms and four oxygen atoms *10. An alkaloid was isolated from a common household beverage. The unknown alkaloid proved to have a molecular mass of 194. Using the Rule of Thirteen, determine a molecular formula and an index of hydrogen deficiency for the unknown. Alkaloids are naturally occurring organic substances that contain nitrogen. (Hint: There are four nitrogen atoms and two oxygen atoms in the molecular formula. The unknown is caffeine. Look up the structure of this substance in The Merck Index and confirm its molecular formula.) *11. The Drug Enforcement Agency (DEA) confiscated a hallucinogenic substance during a drug raid. When the DEA chemists subjected the unknown hallucinogen to chemical analysis, they found that the substance had a molecular mass of 314. Elemental analysis revealed the presence of carbon and hydrogen only. Using the Rule of Thirteen, determine a molecular formula and an index of hydrogen deficiency for this substance. (Hint: The molecular formula of the unknown also contains two oxygen atoms. The unknown is tetrahydrocannabinol, the active constituent of marijuana. Look up the structure of tetrahydrocannabinol in The Merck Index and confirm its molecular formula.) 12. A carbohydrate was isolated from a sample of cow’s milk. The substance was found to have a molecular mass of 342. The unknown carbohydrate can be hydrolyzed to form two isomeric compounds, each with a molecular mass of 180. Using the Rule of Thirteen, determine a molecular formula and an index of hydrogen deficiency for the unknown and for the hydrolysis products. (Hint: Begin by solving the molecular formula for the 180-amu hydrolysis products. These products have one oxygen atom for every carbon atom in the molecular formula. The unknown is lactose. Look up its structure in The Merck Index and confirm its molecular formula.) *Answers are provided in the chapter, Answers to Selected Problems REFERENCES O’Neil, M. J., et al., eds. The Merck Index, 14th ed., scale Approach, 4th ed., Belmont, CA: Brooks-Cole Whitehouse Station, NJ: Merck & Co., 2006. Thomson, 2007. Pavia, D. L., G. M. Lampman, G. S. Kriz, and R. G. Engel, Shriner, R. L., C. K. F. Hermann, T. C. Morrill, D. Y. Introduction to Organic Laboratory Techniques: A Small Curtin, and R. C. Fuson, The Systematic Identification Scale Approach, 2nd ed., Belmont, CA: Brooks-Cole of Organic Compounds, 8th ed., New York, NY: John Thomson, 2005. Wiley, 2004. Pavia, D. L., G. M. Lampman, G. S. Kriz, and R. G. Engel, Introduction to Organic Laboratory Techniques: A Micro- 14782_02_Ch2_p015-104.pp2.qxd 1/25/08 10:27 AM Page 15 C H A P T E R 2 INFRARED SPECTROSCOPY A lmost any compound having covalent bonds, whether organic or inorganic, absorbs various frequencies of electromagnetic radiation in the infrared region of the electromagnetic spectrum. This region lies at wavelengths longer than those associated with visible light, which range from approximately 400 to 800 nm (1 nm = 10−9 m), but lies at wavelengths shorter than those associated with microwaves, which are longer than 1 mm. For chemical purposes, we are interested in the vibrational portion of the infrared region. It includes radiation with wavelengths (l) between 2.5 mm and 25 mm (1mm = 10−6 m). Although the more technically correct unit for wavelength in the infrared region of the spectrum is the micrometer (mm), you will often see the micron (m) used on infrared spectra. Figure 2.1 illustrates the relationship of the infrared region to others included in the electromagnetic spectrum. Figure 2.1 shows that the wavelength l is inversely proportional to the frequency n and is governed by the relationship