Organic chemistry 8th edition (2017) part 1

giáo trình Organic chemistry 8th edition (2017) part 1 giáo trình Organic chemistry 8th edition (2017) part 1 giáo trình Organic chemistry 8th edition (2017) part 1 giáo trình Organic chemistry 8th edition (2017) part 1 giáo trình Organic chemistry 8th edition (2017) part 1 giáo trình Organic chemistry 8th edition (2017) part 1 giáo trình Organic chemistry 8th edition (2017) part 1 giáo trình Organic chemistry 8th edition (2017) part 1

Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 NH2 NH N Amine, primary Amine, secondary Amine, tertiary O CH CH2 (CH3CH2)3N (CH3CH2)2NH CH3CH2NH2 CH3CNH2 HC CH2 CH3CH3 CH3CH Triethylamine Diethylamine Ethylamine Ethanamide (Acetamide) Ethyne (Acetylene) Ethene (Ethylene) Ethane Ethanal (Acetaldehyde) Ethanol (Ethyl alcohol) Ethanoyl chloride (Acetyl chloride) Ethanoic anhydride (Acetic anhydride) IUPAC Name Thiol Sulfide Phenol Nitro Nitrile Ketone Haloalkane Ether Ester Epoxide O C C H O CH2 S S H O N1 O OH CH3CH2SH CH3SCH3 CH3NO2 CH3 C # N CH3CCH3 C 9C#N O CH3CH2Cl CH3OCH3 CH3COCH3 H2C O CH3SSCH3 CH3COH O Example O OH X F, Cl, Br, I X O C O C S S Disulfide O O C Carboxylic acid O Functional Group* * Where bonds to an atom are not specified, the atom is assumed to be bonded to one or more carbon or hydrogen atoms in the rest of the molecule C Amide N C C Alkyne O C C H Alkene Alkane C O O Aldehyde CH3CH2OH CH3CCl O CH3COCCH3 OH C Alcohol Cl O O O C O C O Example Acid chloride Acid anhydride O Functional Group* Some Important Organic Functional Groups Ethanethiol (Ethyl mercaptan) Dimethyl sulfide Phenol Nitromethane Ethanenitrile (Acetonitrile) Propanone (Acetone) Chloroethane (Ethyl chloride) Dimethyl ether Methyl ethanoate (Methyl acetate) Oxirane (Ethylene oxide) Dimethyl disulfide Ethanoic acid (Acetic acid) IUPAC Name Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 Periodic Table of the Elements Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 Chemical Connections and Connections to Biological Chemistry, and MCAT Practice Essays Section 1.5 Section 1.7 Section 1.9 MCAT Practice: Fullerenes Connections to Biological Chemistry: Phosphoesters MCAT Practice: VSEPR and Resonance Section 2.6 Section 2.9 MCAT Practice: Tetrodotoxin Chemical Connections: Octane Rating—What Those Numbers at the Pump Mean Section 3.8  onnections to Biological Chemistry: C Chiral Drugs MCAT Practice: Amino Acid Stereochemistry Section 4.4  Connections to Biological Chemistry: The Ionization of Functional Groups at Physiological pH Section 4.6 MCAT Practice: Acid-Base Equilibria Section 5.3 Section 5.4 Chemical Connections: The Case of Iowa and New York Strains of the European Corn Borer Connections to Biological Chemistry: The Importance of Cis Double Bonds in Fats Versus Oils Section 6.6  Connections to Biological Chemistry: Trans Fatty Acids: What They Are and How To Avoid Them Section 8.5 Section 8.7 Chemical Connections: Freons MCAT Practice: Antioxidants Section 9.9 Section 9.10 MCAT Practice: Solvents and Solvation Connections to Biological Chemistry: Mustard Gases and the Treatment of Neoplastic Diseases Section 10.2 Section 10.7 Section 10.8 Connections to Biological Chemistry: The Importance of Hydrogen Bonding in Drug-Receptor Interactions MCAT Practice: Pinacol Rearrangement Chemical Connections: Blood Alcohol Screening Connections to Biological Chemistry: The Oxidation of Alcohols by NAD+ MCAT Practice: Alcohol Oxidations Section 11.9 MCAT Practice: Benzo[a]pyrene Section 13.10 Chemical Connections:  Magnetic Resonance Imaging Section 14.3 Connections to Biological Chemistry:  Mass Spectra of Biological Macromolecules Section 15.3 MCAT Practice: Inorganic Coordination Compounds Section 16.8 MCAT Practice:  Pyridoxine (Vitamin B6), a Carrier of Amino Groups Section 16.11  Connections to Biological Chemistry:  NADH— The Biological Equivalent of a Hydride Reducing Agent Section 17.3 Chemical Connections:  From Willow Bark to Aspirin and Beyond Section 17.6 Chemical Connections:  Industrial Synthesis of Acetic Acid—Transition Metal Catalysts  Chemical Connections: Esters as Flavoring Agents Section 17.8 MCAT Practice: Permethrin and Bifenthrin Section 17.9 Connections to Biological Chemistry:  Ketone Bodies and Diabetes Mellitus Section 18.1 Chemical Connections: From Cocaine to Procaine and Beyond Chemical Connections:  From Moldy Clover to a Blood Thinner Section 18.2 Connections to Biological Chemistry:  The Unique Structure of Amide Bonds Section 18.4  Chemical Connections: Mechanistic Alternatives for Ester Hydrolysis: SN2 and SN1 Possibilities Section 18.8  MCAT Practice: b-Lactam Antibiotics Section 19.4 Section 19.9 Chemical Connections:  Drugs That Lower Plasma Levels of Cholesterol MCAT Practice:  Ibuprofen— The Evolution of an Industrial Synthesis Section 20.4 Chemical Connections:  Curry and Cancer Section 21.4  MCAT Practice:  Capsaicin, “Some Like It Hot” Section 23.4 Chemical Connections:  The Poison Dart Frogs of South America Section 23.5 MCAT Practice: The Planarity of !NH2 Groups on Heterocyclic Rings Section 25.2 Section 25.3 Section 25.4 Section 25.5 Chemical Connections:  l-Ascorbic Acid (Vitamin C) Chemical Connections:  Testing for Glucose MCAT Practice: Fucose Chemical Connections:  A, B, AB, and O Blood Group Substances Chemical Connections:  High-Fructose Corn Syrup Section 26.2 Connections to Biological Chemistry:  FAD/ FADH2: Agents for Electron Transfer in Biological Oxidation–Reductions: Fatty Acid Oxidation Section 26.5 Chemical Connections:  Snake Venom Phospholipases Section 26.6 MCAT Practice:  Vitamin K, Blood Clotting, and Basicity Section 27.6 Chemical Connections:  Spider Silk Section 28.2 Section 28.3 Section 28.5 Chemical Connections:  The Search for Antiviral Drugs Chemical Connections:  The Fountain of Youth Chemical Connections:  DNA Fingerprinting Section 29.5 Chemical Connections:  Stitches That Dissolve Section 29.6 Chemical Connections:  Organic Polymers That Conduct Electricity MCAT Practice:  The Chemistry of Superglue Chemical Connections:  Recycling of Plastics Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 Organic Chemistry Eighth Edition William H Brown Beloit College, Emeritus Brent L Iverson University of Texas, Austin Eric V Anslyn University of Texas, Austin Christopher S Foote University of California, Los Angeles Chapter 29 was originally contributed by Bruce M Novak University of Texas at Dallas Australia ● Brazil ● Mexico ● Singapore ● United Kingdom ● United States Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 Organic Chemistry, Eighth Edition William H Brown, Brent L Iverson, Eric V Anslyn, Christopher S Foote Product Director: Dawn Giovanniello Product Manager: Courtney Heilman Content Developer: Peter McGahey Product Assistant: Anthony Bostler Marketing Manager: Ana Albinson Content Project Manager: Teresa L Trego Art Director: Sarah B Cole © 2018, 2014 Cengage Learning ALL RIGHTS RESERVED No part of this work covered by the copyright herein may be reproduced or distributed in any form or by any means, except as permitted by U.S copyright law, without the prior written permission of the copyright owner For product information and technology assistance, contact us at Cengage Learning Customer & Sales Support, 1-800-354-9706 For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions Further permissions questions can be e-mailed to permissionrequest@cengage.com Manufacturing Planner: Judy Inouye Production Service: MPS Limited Library of Congress Control Number: 2016952760 Photo Researcher: Lumina Datamatics Student Edition: Text Researcher: Lumina Datamatics ISBN: 978-1-305-58035-0 Copy Editor: MPS Limited Text Designer: Pier Design Company Cover Designer: Pier Design Company Cover Image: © Philippe Reichert/Getty Images Compositor: MPS Limited Loose-leaf Edition: ISBN: 978-1-305-86554-9 Cengage Learning 20 Channel Center Street Boston, MA 02210 USA Cengage Learning is a leading provider of customized learning solutions with employees residing in nearly 40 different countries and sales in more than 125 countries around the world. Find your local representative at www.cengage.com Cengage Learning products are represented in Canada by Nelson Education, Ltd To learn more about Cengage Learning Solutions, visit www.cengage.com Purchase any of our products at your local college store or at our preferred online store www.cengagebrain.com Printed in the United States of America Print Number: 01   Print Year: 2016 Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 Dedication This Eighth Edition is dedicated to the memory of our dear friend and colleague, Christopher Foote Chris’s insights, encouragement, and dedication to this project can never be replaced His kind and nurturing spirit lives on in all who are lucky enough to have known him Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 About the Authors William H Brown  is an Emeritus Professor of Chemistry at Beloit College, where he has twice been named Teacher of the Year His teaching responsibilities included organic chemistry, advanced organic chemistry, and special topics in phar-macology and drug synthesis He received his Ph.D from Columbia University under the direction of Gilbert Stork and did postdoctoral work at the California Institute of Technology and the University of Arizona Brent L Iverson  received his B.S from Stanford University and his Ph.D from the California Institute of Technology He is a University Distinguished Teaching Professor and the Dean of the School of Undergraduate Studies at the University of Texas at Austin as well as a respected researcher Brent’s research spans the interface of organic chemistry and molecular biology His group has developed several patented technologies, including an FDA-approved treatment for late-stage anthrax Eric V Anslyn  received his B.S from California State University, Northridge, and his Ph.D from the California Institute of Technology He is the Norman Hackerman Professor and a University Distinguished Teaching Professor at the University of Texas at Austin Eric’s research focuses on the physical and bioorganic chemistry of synthetic and natural receptors and catalysts Christopher S Foote  received his B.S from Yale University and his Ph.D from Harvard University His scholarly credits include Sloan Fellow; Guggenheim Fellow; ACS Baekland Award; ACS Cope Scholar; Southern California Section ACS Tolman Medal; President, American Society for Photobiology; and Senior Editor, Accounts of Chemical Research He was a Professor of Chemistry at UCL A Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 Contents in Brief Preface / xxiii Acknowledgements / xxix Covalent Bonding and Shapes of Molecules  /  Things You Should Know I: General Conclusions from Quantum Mechanics / 69 Alkanes and Cycloalkanes  /  72 Stereoisomerism and Chirality  /  127 Acids and Bases  /  170 Alkenes: Bonding, Nomenclature, and Properties  /  206 Things You Should Know II: Nucleophiles and Electrophiles  /  228 Things You Should Know III: Reaction Mechanisms  /  232 Reactions of Alkenes  /  240 Alkynes / 297 Haloalkanes, Halogenation, and Radical Reactions  /  330 Things You Should Know IV: Common Mistakes in Arrow Pushing / 369 Nucleophilic Substitution and b-Elimination / 374 10 Alcohols / 437 11 Ethers, Epoxides, and Sulfides  /  492 12 Infrared Spectroscopy / 535 13 Nuclear Magnetic Resonance Spectroscopy  /  558 14 Mass Spectrometry / 606 15 An Introduction to Organometallic Compounds  /  630 16 Aldehydes and Ketones  /  652 v Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 9.5  b-Elimination   401 In the second and third illustrations, there are two nonequivalent b-carbons, each bearing a hydrogen; therefore, two alkenes are possible In each case, the major product of these and most other b-elimination reactions is the more substituted (and therefore the more stable) alkene (Section 6.6B) Formation of the more substituted alkene in an elimination is common, but it is not always the outcome When the more substituted alkene is the dominant product, the reaction is said to follow Zaitsev’s rule or to undergo Zaitsev elimination Zaitsev’s rule A rule stating that the major product of a b-elimination reaction is the most stable alkene; that is, it is the alkene with the greatest number of substituents on the carboncarbon double bond Example 9.6  b-Elimination Products Predict the b-elimination product(s) formed when each bromoalkane is treated with sodium ethoxide in ethanol If two or more products might be formed, predict which is the major product (b) (a) Solution (a) There are two nonequivalent b-carbons in this bromoalkane, and two alkenes are possible 2-Methyl-2-butene, the more substituted alkene, is the major product b b 1 2-Methyl-2-butene 3-Methyl-1-butene (b) There is only one b-carbon in this bromoalkane, and only one alkene is possible b 3-Methyl-1-butene Problem 9.6 Predict the b-elimination product(s) formed when each chloroalkane is treated with sodium ethoxide in ethanol If two or more products might be formed, predict which is the major product (a) (b) (c) Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 402  Chapter 9: Nucleophilic Substitution and b-Elimination 9.6  Mechanisms of b-Elimination There are two limiting mechanisms for b-eliminations A fundamental difference ­between them is the timing of the bond-breaking and bond-forming steps Recall that we made the same statement about the two limiting mechanisms for nucleophilic substitution reactions (Section 9.3) A.  E1 Mechanism E1 reaction A unimolecular b-elimination reaction At one extreme, breaking of the C!Lv bond to give a carbocation is complete before any reaction occurs with the base to lose a hydrogen and form the carbon-carbon double bond This mechanism is designated an E1 reaction, where E stands for Elimination and stands for unimolecular One species (in this case, the haloalkane) is involved in the rate-determining step The mechanism of an E1 reaction is illustrated here by the reaction of 2-bromo-2-methylpropane to form 2-methylpropene Mechanism 9.4 E1 Reaction of 2-Bromo-2-methylpropane Step 1:  Break a bond to give stable molecules or ions.  Rate-determining ionization of the C—Br bond gives a carbocation intermediate 9 9 1 Step 2:  Take a proton away.  Proton transfer from the carbocation intermediate to solvent (in this case, methanol) gives the alkene 1 In an E1 mechanism, one transition state exists for the formation of the carbocation in Step and a second exists for the loss of a hydrogen in Step (Figure 9.6) Formation of the carbocation intermediate in Step crosses the higher energy barrier and is the rate-determining step This reaction competes with SN1 substitution E1 and SN1 almost always occur together B.  E2 Mechanism E2 reaction A bimolecular b-elimination reaction At the other extreme of elimination mechanisms is a concerted process In an E2 ­reaction (here illustrated by the reaction of 2-bromobutane with sodium ethoxide) proton transfer to the base, formation of the carbon-carbon double bond, and ejection of the bromide ion occur simultaneously; all bond-breaking and bond-forming steps are concerted Because the base removes a b-hydrogen at the same time that the C!Br bond is broken to form a halide ion, the transition state has considerable double-bond character (Figure 9.7) Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 9.6  Mechanisms of b-Elimination   403 H C C Lv B H ␦1 H C H H C H C ␦1 Figure 9.6  An energy diagram for an E1 reaction showing two transition states and one carbocation intermediate H C2 DH H C C Lv C B H C C H C C C Figure 9.7  An energy diagram for an E2 reaction There is considerable double - bond character in the transition state Lv Lv C C Mechanism 9.5 E2 Reaction of 2-Bromobutane Take a proton away and simultaneously break a bond to give stable molecules or ions.  Bond breaking and bond forming are concerted; that is, they occur simultaneously This mechanism is designated E2, where E stands for Elimination and stands for bimolecular; both the haloalkane and the base are involved in the transition state for the rate-determining step Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 404  Chapter 9: Nucleophilic Substitution and b-Elimination Although in principle any base can be made to induce an E2 reaction under a­ ppropriate experimental conditions, chemists commonly employ particularly strong bases such as hydroxide, alkoxides, and amide anions (NR22) These bases have conjugate acids with pKa’s above 11 When we use other bases whose conjugate acid pKa’s are near or below 11 (e.g., carboxylates, thiolates, and cyanide), the intention is to effect a substitution reaction via using these reactants as nucleophiles Therefore, one simplifying aspect of the competition between substitution and elimination is to consider an E2 pathway only when hydroxide, alkoxides, acetylides, and amide anions are used 9.7 Experimental Evidence for E1 and E2 Mechanisms As we examine some of the experimental evidence on which these two contrasting mechanisms are based, we consider the following questions: What are the kinetics of base-promoted b-eliminations? Where two or more alkenes are possible, what factors determine the ratio of the possible products? What is the stereoselectivity? A. Kinetics E1 Reactions The rate-determining step in an E1 reaction is ionization of the leaving group (often a halide, X) to form a carbocation Because this step involves only the haloalkane, the reaction is said to be unimolecular and follows first-order kinetics Rate d fRX g kfRX g dt Recall that the first step in an SN1 reaction is also formation of a carbocation Thus, for both SN1 and E1 reactions, formation of the carbocation is the first step and the rate-determining step E2 Reactions Only one step occurs in an E2 mechanism, and the transition state is bimolecular The reaction is second order: first order in haloalkane and first order in base Rate d fRXg kfRXgfBaseg dt B. Regioselectivity E1 Reactions The major product in E1 reactions is the more stable alkene; [i.e., the alkene with the more highly substituted carbon-carbon double bond (Zaitsev’s rule)] After the carbocation is formed in the rate-determining step of an E1 reaction, it may lose a hydrogen to complete b-elimination or it may rearrange to a more stable carbocation and then lose a hydrogen Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 9.7  Experimental Evidence for E1 and E2 Mechanisms   405 E2 Reactions For E2 reactions that use strong bases and in which the leaving group is a halide ion, the major regioisomeric product is also that formed following Zaitsev’s rule, unless special steric relations apply (Section 9.7C) Double-bond character is so highly d ­ eveloped in the transition state that the relative stability of possible alkenes commonly determines which regioisomer is the major product Thus, the transition state of lowest energy is commonly that leading to the most highly substituted alkene For similar reasons, trans double bonds predominate over cis double bonds in the products when either is possible Note that E2 elimination at a 2° carbon predominates over S N2 reaction with the strongly basic alkoxide ions 1 2-Bromohexane 2-Hexene cis 1-Hexene trans With larger, sterically hindered bases such as tert-butoxide, however, where isomeric alkenes are possible, the major product is often the less substituted alkene because reaction occurs primarily at the most accessible H atom Sterically hindered bases such as tert-butoxide are also noteworthy because the steric hindrance prevents them from reacting as nucleophiles, even with primary haloalkanes C. Stereoselectivity The stereochemistry of E2 reactions is controlled by a conformational effect The ­lowest-energy transition state of an E2 reaction is commonly the one in which the !Lv and !H are oriented anti and coplanar (at a dihedral angle of 180°) to each other The reason for this preferred geometry is that it allows for proper ­orbital ­overlap between the base, the proton being removed, and the departing leaving group Remembering the anti and coplanar geometry requirement is important b ­ ecause it allows prediction of alkene stereochemistry in E2 reactions, namely whether E or Z products are produced This is shown more clearly in a Newman projection with a bromide as the leaving group As with the required backside attack associated with an SN2 reaction, there is an orbital-based reason for the anti and coplanar arrangement of the —H and !Lv involved in an E2 reaction The following diagram shows a filled C!H Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 406  Chapter 9: Nucleophilic Substitution and b-Elimination s bonding molecular orbital aligned with the empty C!Lv s antibonding molecular orbital s s As the strong base removes the proton, we consider the two electrons in the C!H orbital filling the antibonding C!Lv orbital and thereby breaking the C!Lv bond An anti and coplanar arrangement of the C!H and C!Lv leads to proper phasing in the resulting p bond When the H and Lv are aligned as shown, a collision of the base along the C!H bond leads to the lowest-energy E2 pathway For example, treatment of 1,2-dibromo-1,2-diphenylethane with sodium methoxide in methanol gives 1-bromo-1,2-diphenylethylene The meso isomer of 1,2-dibromo-1,2-diphenylethane gives (E)-1-bromo-1,2-diphenylethylene, whereas the racemic mixture of 1,2-dibromo-1,2-diphenylethane gives (Z)-1-bromo-1,2-­ diphenylethylene We use the anti coplanar requirement of the transition state to ­account for the stereospecificity of these E2 b-eliminations 1 meso-1,2-Dibromo1,2-diphenylethane racemic 1,2 Dibromo1,2-diphenylethane 1 (E )-1-Bromo1,2-diphenylethylene 1 1 (Z )-1-Bromo1,2-diphenylethylene Because it is preferred for an E2 reaction that !H and !Lv be anti and coplanar, it is important to identify the reactive conformation of a haloalkane starting material Following is a stereorepresentation of the meso isomer of 1,2-dibromo-1,2-­ diphenylethane, drawn to show the plane of symmetry Mechanism 9.6 E2 Reaction of meso-1,2-Dibromo-1,2-diphenylethane Take a proton away and simultaneously break a bond to give stable molecules or ions.  Clockwise rotation of the left carbon by 60° brings !H and !Br into the required anti and coplanar relationship Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 9.7  Experimental Evidence for E1 and E2 Mechanisms   407 (E )-1-Bromo-1,2diphenylethylene E2 reaction on this conformation gives only the (E)-alkene, as shown below in a Newman projection The !H and !Br are anti and coplanar All bond-breaking and bond-forming steps are concerted (E )-1-Bromo-1,2diphenylethylene E2 reaction of either enantiomer of the racemic mixture of 1,2-dibromo-1,2diphenylethane gives only the (Z)-alkene as predicted by analysis of the proper anti and coplanar conformations Mechanism 9.7 E2 Reaction of the Enantiomers of 1,2-Dibromo-1,2-diphenylethane Take a proton away and simultaneously break a bond to give stable molecules or ions.  The !H and !Br are anti and coplanar All bond-breaking and bond-forming steps are concerted The (1R,2R ) Enantiomer The (1S,2S ) Enantiomer (Z )-1-Bromo-1,2diphenylethylene (Z )-1-Bromo-1,2diphenylethylene Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 408  Chapter 9: Nucleophilic Substitution and b-Elimination The required anti and coplanar transition state geometry can also be used to predict the regiochemistry of E2 elimination in halocyclohexanes such as chlorocyclohexanes In these molecules, anti and coplanar correspond to trans and diaxial Consider the E2 reaction of the cis isomer of 1-chloro-2-isopropylcyclohexane The major product is 1-isopropylcyclohexene, the more substituted cycloalkene 1 cis-1-Chloro-2-isopropylcyclohexane 1-Isopropylcyclohexene (R)-3-Isopropylcyclohexene Mechanism 9.8 E2 Reaction of cis-1-Chloro-2-isopropylcyclohexane Take a proton away and simultaneously break a bond to give stable ­molecules or ions.  In the more stable chair conformation of the cis isomer, the considerably larger isopropyl group is equatorial and the smaller chlorine is axial In this chair conformation, !H on carbon and !Cl on carbon are anti and ­coplanar Concerted E2 elimination gives 1-isopropylcyclohexene, a trisubstituted alkene, as the major product Note that !H on carbon and !Cl are also anti and ­coplanar Dehydrohalogenation of this combination of !H and !Cl gives 3-isopropylcyclohexene, a disubstituted (and therefore less stable) alkene The formation of the 1-isomer as the major product is in agreement with Zaitsev’s rule However, Zaitsev’s rule can be counteracted by the anti coplanar arrangement of !H and !Lv (Example 9.7) 1 1-Isopropylcyclohexene The factors favoring E1 or E2 elimination are summarized in Table 9.10 Table 9.10  Summary of E1 Versus E2 Reactions for Haloalkanes Alkyl Halide E1 E2 Primary RCH2X E1 not observed Primary carbocations are so unstable that they are never observed in solution E2 is favored if elimination is observed Usually requires sterically hindered strong base Secondary R2CHX Main reaction with weak bases such as H2O, ROH Main reaction with strong bases such as OH2 and OR2 Tertiary R3CX Main reaction with weak bases such as H2O, ROH Main reaction with strong bases such as OH2 and OR2 Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 9.7  Experimental Evidence for E1 and E2 Mechanisms   409 Example 9.7  Anti and Coplanar Arrangements in E2 Reactions From trans-1-chloro-2-isopropylcyclohexane, only 3-isopropylcyclohexene, the less substituted alkene, is formed Using conformational analysis, explain why this product is observed Also, will the E2 reaction with trans-1-chloro2-isopropylcyclohexane or cis-1-chloro-2-isopropylcyclohexane occur faster under the same basic conditions? trans -1-Chloro-2 isopropylcyclohexane (R)-3-Isopropylcyclohexene Solution Step 1:  In the more stable chair conformation of the trans isomer, both isopropyl and chlorine are equatorial In this conformation, the hydrogen atom on carbon is cis to the chlorine atom One of the hydrogen atoms on carbon is trans to !Cl, but it is not anti and coplanar Therefore, the reaction is not favored from this conformation In the alternative, less stable chair conformation of the trans isomer, both isopropyl and chlorine are axial In this conformation, the axial hydrogen on carbon is anti and coplanar to chlorine and E2 b-elimination can occur to give 3-isopropylcyclohexene Thus, even though the diaxial conformation is less stable, the reaction goes through this conformation because it is the only one with an anti-coplanar arrangement of the Cl and a b-H; consequently, the non-Zaitsev product is formed H Step 2: E2 reaction can take place now that an !H and a !Cl are anti and coplanar This reaction doesn’t follow the Zaitsev rule because the mechanism of the reaction requires the anti arrangement 2 (R )-3-Isopropylcyclohexene The rate at which the cis isomer undergoes E2 reaction is considerably greater than the rate for the trans isomer We can account for this observation in the (Continued) Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 410  Chapter 9: Nucleophilic Substitution and b-Elimination following manner The more stable chair conformation of the cis isomer has !H and !Cl anti and coplanar, and the activation energy for the reaction is that required to reach the E2 transition state The more stable chair conformation of the trans isomer, however, cannot undergo anti elimination To react, it must first be converted to the less stable chair, and the transition state for elimination is correspondingly higher in energy because of the axial isopropyl group Watch a video explanation Problem 9.7 1-Chloro-4-isopropylcyclohexane exists as two stereoisomers: one cis and one trans Treatment of either isomer with sodium ethoxide in ethanol gives 4-isopropylcyclohexene by an E2 reaction 1-Chloro-4isopropylcyclohexane 4-Isopropylcyclohexene The cis isomer undergoes E2 reaction several orders of magnitude faster than the trans isomer How you account for this experimental observation? Although much rarer than an anti and coplanar arrangement of the C—H and C—Lv bonds in an E2 reaction, a syn and coplanar arrangement of these bonds can also lead to E2 Such an arrangement means that the C—H and C—Lv bonds are eclipsed; therefore, only certain constrained ring systems have this geometry As an example, the following reaction occurs via elimination of Ha rather than Hb because the C—Ha bond is aligned with the C—Cl bond while the C—Hb bond is gauche to the C—Cl bond, that is, the latter two bonds lie at a dihedral angle of 60° 1 9.8  Substitution Versus Elimination Nucleophilic substitution and b-elimination often compete with each other, and the ratio of products formed by these reactions depends on the relative rates of the two reactions In this section, we consider factors that influence this competition 1 b 1 A. SN1 Versus E1 Reactions Reactions of secondary and tertiary haloalkanes in polar protic solvents give mixtures of substitution and elimination products In both reactions, Step is the formation Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 9.8  Substitution Versus Elimination   411 of a carbocation intermediate This step is then followed by one or more characteristic carbocation reactions: (1) loss of a hydrogen (E1) to give an alkene, (2) reaction with solvent (SN1) to give a substitution product, or (3) rearrangement followed by reaction (1) or (2) In polar protic solvents, the products formed depend only on the structure of the particular carbocation For example, tert-butyl chloride and tert-butyl iodide in 80% aqueous ethanol both react with solvent, giving the same mixture of substitution and elimination products Because iodide ion is a better leaving group than chloride ion, tert-butyl iodide reacts over 100 times faster than tert-butyl chloride Yet, the ratio of products is the same because the intermediate tert-butyl cation is the same 9 1 9 1 9 " 2 9 It is difficult to predict the ratio of substitution to elimination products for first-order reactions of haloalkanes For the majority of cases, however, SN1 predominates over E1 when weak bases are used B. SN2 Versus E2 Reactions It is considerably easier to predict the ratio of substitution to elimination products for second-order reactions of haloalkanes with reagents that act both as nucleophiles and bases The guiding principles are: Branching at the a-carbon or b-carbon(s) increases steric hindrance about the a-carbon and significantly retards SN2 reactions Conversely, branching at the a-carbon or b-carbon(s) increases the rate of E2 reactions because of the ­increased stability of the alkene product The greater the nucleophilicity of the attacking reagent, the greater the SN2to-E2 ratio Conversely, the greater the basicity of the attacking reagent, the greater the E2-to-SN2 ratio b a b a a b A second point involves a relative comparison of nucleophilicity to basicity It is often difficult to definitively predict in advance whether nucleophilicity will outcompete basicity, thereby favoring or not favoring SN2 versus E2 This competition is particularly important with secondary haloalkanes (see below) However, a general guideline is reasonably predictive for secondary haloalkanes If a nucleophile/base has a conjugate acid with a pKa below 11 and is a good nucleophile, then an SN2 reaction will dominate If the pKa of the conjugate acid of the nucleophile/base is above 11, the Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 412  Chapter 9: Nucleophilic Substitution and b-Elimination basic character will usually outcompete the nucleophilic character and an E2 reaction will dominate A particularly good example of this phenomenon is the comparison between a thiolate anion (RS2) and an alkoxide anion (RO2) Thiolates are excellent nucleophiles in polar protic media, and the pKa's of their conjugate acids (thiols, RSH) are in the range of 10 to 12 However, alkoxides are also excellent nucleophiles, but the pKa's of their conjugate acids (alcohols, ROH) are much higher—in the range of 16 to 18 Hence, for a reaction with the same haloalkane in the same polar protic solvent, the percent of SN2 will be greater for a thiolate nucleophile/base, while the percent of E2 will be greater for the reaction performed with an alkoxide nucleophile/base Temperature is another factor that influences the balance between SN2 and E2 reactions In general, higher temperatures result in increasing extents of elimination at the expense of substitution The reason derives from differences in the number of products compared to the number of reactants Elimination reactions involve the ­creation of increasing numbers of molecules because a base and an R—Lv react to give the conjugate acid of the base, an alkene, and the free leaving group In contrast, substitution reactions not change the number of molecules because a nucleophile and an R—Lv react to give the substituted product and the free leaving group The more particles formed in a reaction, the more entropically favored the reaction Hence, because elimination reactions create more particles than substitution reactions do, they are more entropically favored, which will be reflected in differences in the ­energies of the transition states for these two reactions Recall that DG DH2TDS; therefore, entropy effects become more accentuated at higher temperatures because the TDS term becomes increasingly important Thus, when two or more reactions are in competition, at higher temperatures, the reactions with the more favorable entropies will increase at the expense of those with less favorable entropies With the competition between SN2 and E2, the more favorable entropy for elimination results in an increase in elimination at higher temperatures 2 Watch a video explanation 1 9 9 1 C.  Putting It All Together In this chapter, we examined SN2, SN1, E2, and E1 mechanisms and learned how they compete with each other depending upon the alkyl group, the leaving group, the solvent, and the nucleophile We also examined solvent effects on nucleophilicity Nature does not always have clear-cut rules, but here we summarize guidelines that chemists use to predict the outcome of reactions between haloalkanes and various nucleophiles and bases Figure 9.8 shows a flowchart that allows you to predict the major product of substitution or elimination reactions Use the chart as a guide to the following discussion Alternatively, you can follow the discussion by referring to Table 9.11 We first classify the haloalkane as (a) primary (RCH2X), (b) secondary (R2CHX), or (c) tertiary (R3CX) We consider 1° carbons that are not sterically hindered in Part (a) because this covers most 1° cases If the 1° carbon is sterically hindered, such as that in the neopentyl group [R (CH3)3CCH2X, recall Section 9.3B], we treat it as if it were a secondary carbon (b) that cannot undergo an elimination reaction The flowchart does not show what happens if the alkyl group is methyl (CH3X), because the only possible outcome is an SN2 reaction irrespective of the structure of the nucleophile, the leaving group, and the solvent (Table 9.11 puts CH3X at the top) Recall that methyl cations are too unstable to form (SN1 is ruled out) and there is only one carbon, meaning that elimination to create a double bond is impossible (E1 and E2 are ruled out) Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 9.8  Substitution Versus Elimination   413 (a) (a) (a) t t t (b) (b) (b) (c) (c) (c) Figure 9.8  Flowchart for determining the experimental conditions and choice of reagents that favor SN2, SN1, E2, and E1 reactions Table 9.11  Halide Summary of Substitution Versus Elimination Reactions of Haloalkanes Reaction Comments Methyl CH3X SN2 SN1 reactions of methyl halides are never observed The methyl cation is so unstable that it is not observed in common solvents Primary RCH2X SN2 The main reaction with good nucleophiles/weak bases such as I2 and CH3COO2 E2 The main reaction with strong, bulky bases such as (CH3)3CO2 SN1/E1 Primary cations are rarely formed in solution; therefore, SN1 and E1 reactions of primary halides are unlikely SN2 The main reaction with bases/nucleophiles where pKa of the conjugate acid is 11 or less, as, for example, I2 and CH3COO2 E2 The main reaction with bases/nucleophiles where the pKa of the conjugate acid is 11 or greater, as, for example, OH2 and CH3CH2O2 SN1/E1 Common in reactions with weak nucleophiles in polar protic solvents, such as water, methanol, and ethanol E2 Main reaction with strong bases such as HO2 and RO2 SN1/E1 Main reactions with poor nucleophiles/weak bases if the solvent is polar protic SN2 SN2 reactions of tertiary halides are never observed because of the extreme crowding around the 3° carbon Secondary R2CHX Tertiary R3CX Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 414  Chapter 9: Nucleophilic Substitution and b-Elimination We next examine the structure of the nucleophile Because all nucleophiles are bases, we refer to them as Nuc/Base (a) Primary Alkyl Groups: ●● ●● ●● ●● Because primary carbocations are too unstable to form, SN1 or E1 mechanisms are not possible If the Nuc/Base is a strong base and sterically hindered, it will not be a good nucleophile and E2 is the major pathway A common example is the use of tert-butoxide ion as the Nuc/Base Amide anions are exceptions Although they are not hindered, they are so basic that E2 dominates If the Nuc/Base is a strong base and not sterically hindered, we next consider w ­ hether it is a good nucleophile Examples of strong bases that are also good nucleophiles are hydroxide, acetylide, and methoxide Weak bases are defined as bases that have conjugate acids with pKa’s below 11 Examples of weak bases that are good nucleophiles are thiolate (RS2), cyanide (NC2), iodide (I2), and azide (N32) anions Even moderate nucleophiles that are weak bases, such as unhindered amines (NR3) and phosphines (PR3), participate in efficient SN2 reactions Hence, with any of the basic or weakly basic good-to-moderate nucleophiles, we find products predominantly from SN2 pathways However, some accompanying E2 mechanism is likely with strong bases Finally, if the Nuc/Base is neither a good nor moderate nucleophile, we are likely to get SN2 and E2 in a ratio that is difficult to predict or to get no reaction Examples are water, alcohols, and carboxylic acids (b) Secondary Alkyl Groups: ●● ●● ●● ●● If the Nuc/Base is a strong base, whether or not it is hindered, E2 will dominate Strong bases are defined as bases that have conjugate acids with pKa’s above 11, such as hydroxide, alkoxides, acetylides, and H2N2 With a weak base that is a good to moderate nucleophile, SN2 will dominate Examples are those nucleophiles that have conjugate acids with pKa’s below 11 However, because the alkyl group is secondary, the SN2 reaction may be sluggish, and SN1 and E2/E1 elimination pathways may compete to a small extent When the nucleophile is not good, we need to examine the solvent In a polar protic solvent, often with gentle warming, we can induce SN1 and E1 pathways The extent that substitution or elimination occurs is hard to predict Finally, if the solvent is neither polar nor protic and the Nuc/Base is neither a strong base nor a good nucleophile, all four reaction pathways are possible, and it is difficult to predict which will dominate or whether a reaction will occur at all (c) Tertiary Alkyl Groups: ●● ●● ●● ●● ●● We start by noting that an SN2 mechanism cannot occur on a tertiary alkyl group If the Nuc/Base is a strong base, E2 will dominate Because SN2 is not possible, we not have to consider whether the Nuc/Base is a strong or weak nucleophile Hence, this question is not relevant to predicting the dominant reaction pathway Therefore, instead we check the solvent If the solvent is polar and protic, we can induce SN1 and E1 pathways, often by applying heat Whether substitution or elimination dominates is hard to predict Finally, if the Nuc/Base is not a strong base and the solvent is not polar and protic, SN1, E2, and E1 reaction pathways are possible, and it is difficult to predict which will dominate or whether a reaction will occur at all Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 9.9  Analysis of Several Competitions Between Substitutions and Eliminations   415 9.9 Analysis of Several Competitions Between Substitutions and Eliminations Now that we have outlined a step-by-step process with which to make predictions about relative extents of SN2, SN1, E2, and E1 mechanisms, we’ll examine specific examples as we did in Section 9.4 Following are five examples of reactions between a haloalkane and a Nuc/Base in specific solvents, along with an analysis of the factors that favor the various mechanistic pathways We start with examples that have clear-cut predictions and move to those that are more challenging In some cases, we also cover predictions for the reactions under slightly different experimental conditions Competition The haloalkane is primary; therefore, S N1 and E1 cannot occur The Nuc/ Base is a strong base because the pKa of its conjugate acid is several units above 11 (pKa HOEt 15.9) But due to its basicity and because it is not sterically hindered, ethoxide is also a good nucleophile Hence, SN2 will dominate over E2 Competition D The haloalkane is tertiary; therefore, SN2 cannot occur The Nuc/Base is a weak base The solvent is polar protic; therefore, SN1 and E1 mechanisms will occur D If this same reaction were performed with NaOEt in the ethanol, E2 would have been the dominant pathway because ethoxide is a strong base Competition The haloalkane is secondary Because the Nuc/Base has a pKa of its conjugate acid far below 11 (pKa of HOAc 4.7), it is a weak base; hence, there will be little to no E2 However, acetate is a moderate nucleophile Hence, SN2 is the best prediction Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 ... 10 98 24.4 Catalytic Allylic Alkylation  11 04 24.5 Palladium-Catalyzed Cross-Coupling Reactions  11 07 24.6 Alkene Metathesis  11 12 Study Guide  11 15       Problems  11 19 25 Carbohydrates 11 34... Connections  High-Fructose Corn Syrup 11 55 25.6 Glucosaminoglycans 11 56 Study Guide  11 58       Problems  11 63 26 Lipids 11 70 26 .1 Triglycerides 11 71 26.2 Soaps and Detergents  11 74 Copyright 2 018 Cengage... Oxidation (Section 16 .10 A) 16 .13 Sodium Borohydride Reduction of an Aldehyde or a Ketone (Section 16 .11 A) 16 .14  Wolff-Kishner Reduction (Section 16 .11 E) 16 .15  Acid-Catalyzed a-Halogenation of