CHEM 31.1ex3 organic chemistry school requirementFull description
FAM Organic BandungFull description
Organic Chemistry FIFTH
EDITION
Marc Loudon Purdue University
ROBERTS A D COMPA
Y PUBLISHERS
Greenwood Village. Colorado
Contents Preface Reviewers and consultants About the Author
CHEMICAL BONDING AND CHEMICAL STRUCTURE 1.1
1.2
1
Introduction
1
A. What Is Organic Chemistry? B. Emergence of Organic Chemistry C. Why Study Organic Chemistry?
2
Classical Theories of Chemical Bonding A. Electrons in Atoms B. The Ionic Bond C. The Covalent Bond D. The Polar Covalent Bond
1.3
xxxi xxxvi xxxix
Structures of covalent Compounds A. Methods for Determining Molecular Geometry B. Prediction of Molecular Geometry
1
3 3 3 5 9
13 13 14
1.4
Resonance Structures
20
1.5
wave Nature of the Electron
22
1.6
Electronic structure of the Hydrogen Atom
23 23 25
A. Orbitals, Quantum Numbers. and Energy B. Spatial Characteristics of Orbitals C. Summary: Atomic Orbitals of Hydrogen
1.7 1.8
Electronic Structures of More complex Atoms
29
Another Look at the covalent Bond: Molecular Orbitals
32 32 36
A. Molecular Orbital Theory B. Molecular Orbital Theory and the Lewis Structure of H ~
1.9
28
Hybrid Orbitals A. Bonding in Methane B. Bonding in Ammonia
37 37 40
vii
Viii
CONTENTS
Key Ideas in Chapter 1 Additional Problems
ALKANES
42 43
46
2.1
Hydrocarbons
46
2.2
Unbranched Alkanes
48
2.3
conformations of Alkanes
50
A. Conformation of Ethane B. Conformations of Butane
50 53
Constitutional Isomers and Nomenclature
57 57
2.4
A. Isomers B. Organic omcndature C. Sub!.titutivc omenclarure of Alkane~ D. Highl) Conden:,ed Structures E. Classification of Carbon Substitution
58 59
64
66
2.5
cycloalkanes and Skeletal structures
67
2.6
Physical Properties of Alkanes
70
A. Boiling Points B. Melting Point~ C. Other Phy!>ical Properties
70 73 74
2.7
combustion
76
2.8
Occurrence and Use of Alkanes
78
2.9
Functional Groups, compound Classes, and the "R" Notation
81
A. Functional Group~ and Compound Cla\scs B. "R" Notation
81 82
Key Ideas in Chapter 2 Additional Problems
83 83
ACIDS AND BASES. THE CURVED-ARROW NOTATION 3.1
Lewis Acid-Base Association Reactions A. Elcctron-Dclicicnt Compounds B. Reactions of Electron-Deficient Compounds with Lewis Bases C. The Curved-Arrow Notation for Lewis Acid- Base Association and Dissociation Reactions
87 87 87 88
89
CONTENTS
3.2
Electron-Pair Displacement Reactions A. Donation of Electrons to Atom!> That Are ot Electron-Deficient B. The Curved-Arrow Notation for Electron-Pair Di~placement Reactions
3.3
Review of the Curved-Arrow Notation A. Usc of the Curved-Arrow Notation to Represent Reaction~ B. U:.c of the Curved-Arrow otation to Derive Resonance Structure~
3.4
Bronsted- Lowry Acids and Bases
ix
90 90 91 94 94 9-l
96
A. Definition of Bronsted Acid., and Ba!.c~ B. uclcophilc!.. Electrophiles. and Leaving Groups C. Strength.., of Bronsted Acid~ D. Strength-; of BrV)nsted Base~ E. Equilibria in Acid-Base Reactions
101 103 104
3.5
Free Energy and Chemical Equilibrium
106
3.6
Relationship of Structure to Acidity
108
A. The Element Effect B. The Charge Effect C. The Polar Effect
Key Ideas in Chapter 3 Additional Problems
INTRODUCTION TO ALKENES. STRUCTURE AND REACTIVITY 4.1
4.2
Structure and Bonding in Alkenes
96
98
lOR 110 Ill
116 117
122
A. Carbon Hyhridization in Alkenes B. The 7T (Pi) Bond C. Double-Bond Stereoisomer!.
122 123 125 128
Nomenclature of Alkenes
131
A. I UPAC Substitut ive Nomenclature B. Nomenclature of Double-Bond Stcreoisomers: The £.2 Sy~tem
131
134
4.3
Unsaturation Number
139
4.4
Physical Properties of Alkenes
140
4.5
Relative Stabilities of Alkene Isomers
141 141
A. Heats of Formation B. Relative Stabil ities of Alkene Isomer~
144
4.6
Addition Reactions of Alkenes
147
4.7
Addition of Hydrogen Halides to Alkenes
147 148 149
A. Regio:..electivity of Hydrogen llalidc Addition B. Carbocation Intermediate~ in Hydrogen Halide Addition
X
CONTENTS
C. Structure and Stability of Carbocationo; D. Carbocation Rearrangement in Hydrogen Halide Addition
4.8
Reaction Rates A. The Transition State B. The Energy Barrier C. Multistep Reactions and the Rate-Limiting Step D. Hammond·s Postulate
4.9
catalysis A. Catalytic Hydrogenation of Alkenes B. Hydration of Alkenes C. Enzyme Catalysis
Key Ideas in Chapter 4 Additional Problems
ADDITION REACTIONS OF ALKENES
151 154
157 158 160 162 164
166 168 169 172
172 174
178
5.1
An overview of Electrophilic Addition Reactions
178
5.2
Reactions of Alkenes with Halogens
181 181
A. Addition of Chlorine and Bromine B. Halohydrins
183
5.3
Writing Organic Reactions
186
5.4
Conversion of Alkenes into Alcohols
187 187
A. Oxymercuration- Reduction of Alkenes B. Hydroboration-Oxidation of Alkenes C. Compari!>On of Methods for the Synthesis of Alcohols from Alkenes
190 194
5.5
ozonolysis of Alkenes
196
5.6
Free-Radical Addition of Hydrogen Bromide to Alkenes
200 200
A. B. C. D. E.
The Peroxide Effect Free Radicals and the "Fishhook" Notation Free-Radical Chain Reactions Explanation of the Peroxide Effect Bond Dissociation Energies
201
202 207 211
5.7
Polymers: Free-Radical Polymerization of Alkenes
214
5.8
Alkenes in the Chemical Industry
216
Key Ideas in Chapters Additional Problems
219 220
CONTENTS
PRINCIPLES OF STEREOCHEMISTRY 6.1
226
A. Enantiomers and Chirality B. Asymmetric Carbon and Stereocenters C. Chirality and Symmetry
226 226 229 229
6.2
Nomenclature of Enantiomers: The R,S system
231
6.3
Physical Properties of Enantiomers: Optical Activity A. Polari;cd Light B. Optical Activity C. Optical Activities of Enantiomcrs
234 235 235 238
6.4
Racemates
239
6.5
stereochemical correlation
241
6.6
Diastereomers
242
6.7
Meso compounds
246
6.8
Enantiomeric Resolution
249
6.9
Chlral Molecules without Asymmetric Atoms
251
Conformational Stereoisomers A. Stcrcoisorners lnterconvened by Internal Rotations B. Asymmetric Nitrogen: Amine Inversion
253 253 255
Drawing Structures That Contain Three-Dimensional Information
257
The Postulation of Tetrahedral carbon
259
Key Ideas in Chapter 6 Additional Problems
263 263
6.10
6.11 6.12
Enantiomers, Chirality, and symmetry
xi
CYCLIC COMPOUNDS. STEREOCHEMISTRY OF REACTIONS
268
7.1
Relative Stabilities of the Monocyclic Alkanes
268
7.2
Conformations of cyclohexane
269 269 273 274
A. The Chair Conformation B. lntcrconversion of Chair Conformations C. Boat and Twist-Boat Conformations
A. Ci'>-Trans 1-.omeri m in Disubstituted Cyclohexanc~ B. Conformational Analysis C. U'>e of Planar Structure for Cyclic Compound'> D. Stereochemical Con ...equences of the Chair lntercomer,ion
7.5
cyclopentane, cyclobutane, and Cyclopropane A. Cyclopcntane B. Cyclobutane and Cyclopropane
7.6
Blcyclic and Polycyclic compounds A. Classification and Nomenclature B. Cis and Trans Ring Fusion C. Trans-Cycloalkenes and Bredt's Ru le D. Steroids
7.7
Relative Reactivities of Stereo isomers A. Relative Reactivities of Enantiomers B. Relative Reactivities of Diastereomers
7.8
Reactions That Form Stereoisomers A. Reactions of Achiral Compounds That Gi\'e Enantiomeric Product' B. Reaction' That Give Diastereomeric Products
7.9
Stereochemistry of Chemical Reactions A. Stereochemistry of Addition Reactions B. Stereochcmi<>tl) of Substitution Reactions C. Stereochemi'>Lry of Bromine Addition D. Stcreochcmi~try of Hydroboration- Oxidation E. Stereo<..:hemistry of Other Addition Reaction:-
INTRODUCTION TO ALKYL HALIDES, ALCOHOLS, ETHERS, THIOLS, AND SULFIDES
A . 1 omenclature of Alkyl Halides B. omenclature of Alcohols and Thiols C. omenclature of Ethers and Sulfide!>
8.2
Structures
8.3
Effect of Molecular Polarity and Hydrogen Bonding on Physical Properties Point~
290 290 292 294 296 298 298 300
301 301
30-f 305 305
306 308 3 12
313
323 324
Nomenclature
A. Boiling
288 288 289
314 316
Key Ideas in Chapter 7 Additional Problems
8.1
285
32-f 326 330
332
of Ethers and Alkyl
Halide~
333 333
CONTE NTS
B. Boiling Points of Alcohol\ C. H)drogen Bonding
8.4
Solvents in Organic Chemistry A. Cla...sification of Solvents B. Solubility
8.5
8.6
Applications of Solubility and Solvation Principles
x iii 335
336
339 339 3-W
346
A. Cell Membranes and Drug Solubility B. Cation-Binding Molecules
3.+6 351
Acidity of Alcohols and Thiols
355
A. Formation of Alkoxidcs and Mcrcaptidcs B. Polar Effects on Alcohol Acidity C. Role of the Solvent in Alcohol Acidity
356 358 358
8.7
Basicity of Alcohols and Ethers
359
8.8
Grignard and Organolithium Reagents
361
A. Formation of Grignard and Organolithium Reagl:nt-; B. Protonoly~is of Grignard and Organolithium Reagent!--
8.9
Industrial Preparation and use of Alkyl Halides, Alcohols, and Ethers A. B. C. D.
Free-Radical Halogenation of Alkane' Use<.. of Halogen-Containing Compound~ Production and Use of Alcohol!. and Ether' Safct) HaLards of Ether.
Key Ideas in Chapter 8 Additional Problems
THE CHEMISTRY OF ALKYL HALIDES 9.1
An overview of Nucleophilic substitution and IJ-Elimination Reactions A. uc leophi lic Substitution Reactions B. 13-Eiimination R eact ion~ C. Competition between uclcophilic Sub~tillltion and 13-Eiimination Reactions
361 362
364 364 365 368 371
372 373
377 377 377 378 380
9.2
Equilibria in Nucleophilic Substitution Reactions
381
9.3
Reaction Rates
382
A. Definition of Reaction Rate B. The Rate La\\ C. Rdation~hip of the Rate Con\tant to the Standard Free Encrg) of Acti\ at ion
9.4
The SN2 Reaction A. Rate Lm' and Mechani m of the S:--~2 Reaction
382 383 384
386 386
XiV
CONTENTS
B. Comparison of the Rates of S:-~2 Reactions and Bron. ted Acid- Base Reactions C. Stereochemistry of the SN2 Reaction D. Effect of Alkyl Halide Structure on the S-.:2 Reaction E. ucleophilicit} in tl1e S._2 Reaction F. Lea\ ing-Group Effects in the S._2 Reaction G. Summary of the S~2 Reaction
9.5 The E2 Reaction A. Rate Law and Mechanism of the E2 Reaction B. Why the E2 Reaction Is Concened C. Leaving-Group Effects on the E2 Reaction D. Deuterium Isotope Effects in the E2 Reaction E. Stereochemistry of the E2 Reaction F. Rcgioselectivity of the E2 Reaction G. Competition between the E2 and SN2 Reactions: A Closer Look H. Summary of the E2 Reaction
9.6
The SN1 and E1 Reactions A. Rate Law and Mechanism of SN I and E I Reactions B. Rate-Limiting and Product-Determining Step!. C. Reactivity and Product Di tributions in S~,I -El Reaction~ D. Stereochemistry of the S:-) Reaction E. Summary of the S, I and E1 Reactions
9.7 9 .8
388 388 390 392 398 399
400 400 400 402 402 404 406 407 41 I
412 412 414
416 418
420
summary of substitution and Elimination Reactions of Alkyl Halides
420
carbenes and carbenoids
424
A. a-Elimination Reactions B. The Simmons- Smith Reaction
424 426
Key Ideas in Chapter 9 Additional Problems
THE CHEMISTRY OF ALCOHOLS AND THIOLS
428 42 9
436
10.1
Dehydration of Alcohols
436
10.2
Reactions of Alcohols with Hydrogen Halides
440
10.3
Sulfonate and Inorganic Ester Derivatives of Alcohols
443 443 447 448
A. B. C. D.
10.4
Sulfonate E ter Derivatives of Alcohols Alkylating Agent~ Ester Derivati\·es of Strong Inorganic Acids Reaction!> of Alcohols with Thionyl Chloride and Pho!>phorll'. Tribromide
Conversion of Alcohols into Alkyl Halides: Summary
449
450
CONTENTS
10.5
10.6
XV
A. Half-Reactions and Oxidation Numbers B. Oxidizing and Reducing Agents
452 452 456
Oxidation of Alcohols
459
Oxidation and Reduction in Organic Chemistry
A. Oxidation ro Aldehydes and Ketones B. Oxidation to Carboxylic Acids
459 461
10.7
Biological Oxidation of Ethanol
462
10.8
Chemical and Stereochemical Group Relationships A. Chemical Equivalence and Nonequivalence B. Stereochemjstry of the Alcohol Dehydrogenase Reaction
465 465 469
Oxidation of Thiols
471
10.10
synthesis of Alcohols
474
10.11
Design of Organic synthesis
474
Key Ideas in Chapter 10 Additional Problems
476 477
10.9
THE CHEMISTRY OF ETHERS, EPOXIDES, GLYCOLS, AND SULFIDES 11.1
synthesis of Ethers and sulfides A. Williamson Ether Synthesis B. Alkoxymercurarion-Reducrion of Alkenes C. Ethers from Alcohol Dehydration and Alkene Addition
482 482 482 484 485
A. Oxidation of Alkenes with Peroxycarboxylic Acids B. Cyclization of Halohydrins
488 488 491
11.3
Cleavage of Ethers
492
11.4
Nucleophilic Substitution Reactions of Epoxides
495 495 497
11.2
synthesis of Epoxides
A. Ring-Opening Reactions under Basic Cond itions B. Ring-Opening Reactions under Acidic Conditions C. Reaction of Epoxides with Organometallic Reagents
11.5
Preparation and Oxidative Cleavage of Glycols A. Preparation of Glycols B. Oxidative Cleavage of Glycols
11.6
oxonium and Sulfonium Salts A. Reactions of Oxonium and Sulfonium Salts B. S-Adenosylmelhionine: Nature's Methylating Agent
500
503 503
506
508 508 509
XV i
CONTENTS
11.7
11.8
Intramolecular Reactions and the Proximity Effect
510
A. Neighboring-Group Participation B. The Proximity Effect and Effective Molarity C. Stereochemical Consequences of Neighboring-Group Pm1icipation
5 10 513 516
oxidation of Ethers and sulfides
11.9 The Three Fundamental operations of Organic synthesis 11.10
522
Key Ideas in Chapter 11 Additional Problems
527 528
Introduction to Spectroscopy A. Electromagnetic Radiation B. Absorption Spectroscopy
12.2
Infrared Spectroscopy A. The Infrared Spectrum B. Physical Basis of IR Spectroscopy
12.3
12.4
520
Synthesis of Enantiomerically Pure compounds: Asymmetric Epoxidation
INTRODUCTION TO SPECTROSCOPY. INFRARED SPECTROSCOPY AND MASS SPECTROMETRY 12.1
518
Infrared Absorption and Chemical Structure
536 536 5~6 5~8
540 540 542
A. Factors That Determine IR Absorption Position B. Factors That Determine IR Absorption Intensity
544 545 5-l8
Functional-Group Infrared Absorptions
552
A. IR Spectra of Alkanes B. IR Spectra of Alkyl Halides C. !R Spectra of Alkene!> D. TR Spectra of Alcohols and Ether.
552 552 553 556
12.5
Obtaining an Infrared Spectrum
557
12.6
Introduction to Mass Spectrometry
558 558
A. B. C. D. E.
Electron-Impact Mass Spectra Isotopic Peaks Fragmentation The Molecular Ion. Chemical-Ionization Mass Spectra The Mass Spectrometer
Key Ideas in Chapter 12 Additional Problems
560 563 566 569
571 571
CONTENTS
13
NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY
578
13.1
An overview of Proton NMR Spectroscopy
578
13.2
Physical Basis of NMR Spectroscopy
581
13.3
The NMR spectrum: Chemical Shift and Integral
583 5H3 585 586 589 591 593
A. Chemical Shift B. Chemical Shift Scales C. Relation,hip of Chemical Shift to Structure D. The umber of Ab,orption\ in an 1MR Spectrum E. Couming Protons with the Integral F. Using the Chemical Shift and Integral to Determine Unk nown Structure:-
13.4
The NMR spectrum: Spin- Spin Splitting A. The 11 + I Splitling Rule 8. Why Spliuing Occur\ C. Solving Un!..nown Structure~ with ; MR Spectra lnvoh ing Spliuing
complex NMR spectra
595 596 599 601
A. Multiplicative Splitting 8. Breakdown of the 11 + I Ruh;
603 603 607
13.6
use of Deuterium in Proton NMR
611
13.7
Characteristic Functional-Group NMR Absorptions A. NMR Spectra of Alkene~ B. NMR Spectra of Alkanei> and Cycloa l kane~ C. NMR Spectra of Alkyl Halide~ and Ethers D. MR Spectra of Alcohob
612 611 614 616 616
13.8
NMR spectroscopy of Dynamic systems
619
13.9
carbon NMR
622
13.10
Solving Structure Problems with Spectroscopy
629
13.11
The NMR Spectrometer
632
13.12
Other uses of NMR
634
Key Ideas in Chapter 13 Additional Problems
635 636
13.5
14
xvii
THE CHEMISTRY OF ALKYNES
644
14.1
Nomenclature of Alkynes
644
14.2
Structure and Bonding in Alkynes
646
XViii
CONTENTS
14.3
Physical Properties of Alkynes A. Boiling Points and Solubilities B . lR Spectroscopy of Alkyne C. MR Spectroscopy of Alkynes
649 649 649
650
14.4
Introduction to Addition Reactions of the Triple Bond
652
14.5
conversion of Alkynes into Aldehydes and Ketones
654 654 657
A. Hydration of Alkynes B. H ydroboration-Oxidation of Alkynes
14.6
Reduction of Alkynes A. Catalytic H ydrogenation of Alk yne:-. B. Reduction of Alk) ne!> with Sodium in Liquid Ammonia
14.7
Acidity of 1-Aikynes A. Acctylenic Anion!> B. Acetylenic Anion-. a~
ucJeophile~
659 659 660 662 662 665
14.8
organic synthesis Using Alkynes
666
14.9
Pheromones
668
Occurrence and Use of Alkynes
670
Key Ideas in Chapter 14 Additional Problems
671 671
14.10
DIENES, RESONANCE, AND AROMATICITY 15.1
Structure and Stability of Dienes A. Stability of Conjugated Dienes. Molecular Orbital., B. tructure of Conjugated Dienes C. Structure and Stability of Cumulated Dienes
15.2
Ultraviolet-Visible spectroscopy A. The UV- Vi Spectrum B. Phy!.ical Ba<;i:-. of UV- Vis Spectro<;copy C. UV- Vis Spectroscopy of Conjugated Alkenes
15.3
The Diels-Aider Reaction A. Reaction of Conjugated Dienes with A lkenes B. Effect of Diene Conformation on the D iets-Alder Reaction C. Stereochemistry of the DieIs- Alder Reaction
15.4
Addition of Hydrogen Halides to conjugated Dienes A. 1.2- and 1.4-Additions B. All) lie Carbocations. The Connection between Resonance and Stability C. K inetic and Thermodynamic Cont rol
676 677 677
680 682
684 68-l 686
687
690 690 694 696
700 700
702 705
CONTENTS
xix
15.5
Diene Polymers
708
15.6
Resonance
709
A. Drawing Re!.onance Su·uctures B. Relative lmpo11ance of Resonance Structures C. Use of Resonance Structures
15.7
Introduction to Aromatic compounds A. Benzene. a Puzzling ·'Alkene" B. Structure of Benzene C. Stability of Benzene D. Aromaticity and the HUcke! 4n 12 Rule E. Antiaromatic Compounds
Key Ideas in Chapter 15 Additional Problems
THE CHEMISTRY OF BENZENE AND ITS DERIVATIVES
710 711 714
716 717 7 18 721 721
728
730 731
740
16.1
Nomenclature of Benzene Derivatives
740
16.2
Physical Properties of Benzene Derivatives
743
16.3
Spectroscopy of Benzene Derivatives
743
A. rR Spectroscopy B. NMR Spcctro~copy C. "c MR Spectroscopy D. UV Spectroscopy
16.4
Electrophilic Aromatic Substitution Reactions of Benzene A. Halogenation of Benzene B. Electrophilic Aromatic Substitution C. Nitration of Benzene D. Sulfonation of Benzene E. Friedel-Craft~ Alkylation of Benzene F. Friedel-Crafts Acylation of Benzene
16.5
Electrophilic Aromatic Substitution Reactions of Substituted Benzenes A. Directing Effects of Subst itucnts B. Activating and Deactivating Effects of Substituents C. Use of Elcctrophilic Aromatic Substitution in OrganicS) nthcsis
743 744 748 748
750 751 753 754 755 756 759
762 762 768 772
16.6
Hydrogenation of Benzene Derivatives
776
16.7
Source and Industrial use of Aromatic Hydrocarbons
777
Key Ideas in Chapter 16 Additional Problems
780 780
XX
CONTENTS
ALLYLIC AND BENZYLIC REACTIVITY
788
17.1
Reactions Involving Allylic and Benzylic carbocations
789
17.2
Reactions Involving Allylic and Benzylic Radicals
793
17.3
Reactions Involving Allyllc and Benzylic Anions
798
A. A llylic Grignard Rl!agents B. E2 Elimination~ Involving Allylic or Ben1ylic Hydrogen'>
799 801
17.4
Allylic and Benzylic SN2 Reactions
802
17.5
Allylic and Benzylic Oxidation
803
A. Oxidation of Allylic and BcnL.ylic Alcohob B. Benzy lic Oxidalion of A lkylbenJene!.
17.6
Terpenes A. The Isoprene Rule B. Biosymhe'i" of T!!rpenes
Key Ideas in Chapter 17 Additional Problems
THE CHEMISTRY OF ARYL HALIDES, VINYLIC HALIDES, AND PHENOLS. TRANSITION-METAL CATALYSIS
803 805
807 807 810
81 3 814
822
Lack of Reactivity of Vinylic and Aryl Halides under SN2 conditions
823
18.2
Elimination Reactions of Vlnylic Halides
825
18.3
Lack of Reactivity of Vinylic and Aryl Halides under SN 1 conditions
826
Nucleophilic Aromatic substitution Reactions of Aryl Halides
828
Introduction to Transition-Metal catalyzed Reactions
831
18.1
18.4 18.5
A. Transition Metal' und Their Complexc!. B. Ox idation State C. The d" Notation D. Electron Counting: The 16- and 18-Electron Rule~ E. Fundamental Reactions of Tran'>ition-Metal Complexe-.
18.6
Examples of Transition-Metal-Catalyzed Reactions A. The Heck Reaction B. The Suzuki Coupling
832 835
lB6 836 839
845 845 848
CONTENTS
C. Alkene M etathesis D . Other Examples of Tran~ition-Meta i -Cata l yzed Reactions
18.7
Acidity of Phenols A. Re~onance and Polar Effect'> on the Acidit} of Phenol., B. Formation and Use of Phcn
XXi
l-!52 l-!57
858 R58 861
18.8
Oxidation of Phenols to Quinones
862
18.9
Electrophilic Aromatic Substitution Reactions of Phenols
867
Reactivity of the Aryl- Oxygen Bond
870 870
18.10
A . Lack of Rcacti' it} of the Ar) 1- 0\) gcn Bond in S, I and Sr-:2 Reaction'> B. Sub~tillltion at the Ar) 1-0\ygcn Bond: The Stille Reaction
18.11
Industrial Preparation and use of Phenol
874
Key Ideas in Chapter 18 Additional Problems
875 876
THE CHEMISTRY OF ALDEHYDES AND KETONES. CARBONYL-ADDITION REACTIONS 19.1
871
Nomenclature of Aldehydes and Ketones
888
A. Common omenclature B. Sub1.titutivc omenclature
890 890 892
19.2
Physical Properties of Aldehydes and Ketones
894
19.3
Spectroscopy of Aldehydes and Ketones
895
A. IR Spcctro...copy B. Proton MR Spectroscop) C. Carbon NM R Spcctro~copy D. UV Spectroscopy E. Ma~'> Spectrometry
895
897 898
899 901
19.4
Synthesis of Aldehydes and Ketones
903
19.5
Introduction to Aldehyde and Ketone Reactions
903
19.6
Basicity of Aldehydes and Ketones
904
19.7
Reversible Addition Reactions of Aldehydes and Ketones
907 907
A. Mcchanio,rm of Carbonyl-Addi tion Reacri on:-. B. Equilibria in Carbonyl-Addition ReactiOn\ C. Rate\ of Carbonyl-Addition Reaction~
19.8
Reduction of Aldehydes and Ketones to Alcohols
910
913
914
xxii
CONTENTS
19.9
Reactions of Aldehydes and Ketones with Grignard and Related Reagents A. Preparation and Hydrolysi of Acetal'> B. Protecting Groups
921 921 925
Reactions of Aldehydes and Ketones with Amines
926
19.10 Acetals and Their Use as Protecting Groups
19.11
918
A. Reaction with Primary An1i.nes and Other MonosubstilLitcd Derivatives of Ammonia B. Reaction with Secondary Amines
926 929
19.12
Reduction of carbonyl Groups to Methylene Groups
931
19.13
The Wittig Alkene synthesis
933
19.14
oxidation of Aldehydes to carboxylic Acids
936
19.15
Manufacture and use of Aldehydes and Ketones
938
Key Ideas in Chapter 19 Additional Problems
939 940
THE CHEMISTRY OF CARBOXYLIC ACIDS
A. Common Nomenclature B. Substitutive Nomenclature
948 948 951
20.2
Structure and Physical Properties of Carboxylic Acids
953
20.3
Spectroscopy of carboxylic Acids
955 955 955
20.1
Nomenclature of carboxylic Acids
948
A. IR Spectroscopy B. NMR Spectroscopy
20.4
Acid-Base Properties of carboxylic Acids A. Acidity of Carboxylic and Sulfonic Acids B. Basicity of Carboxylic Acid
957 957 960
20.5
Fatty Acids, soaps, and Detergents
960
20.6
synthesis of carboxylic Acids
963
20.7
Introduction to carboxylic Acid Reactions
964
20.8
conversion of carboxylic Acids into Esters
965 965 968
A. Acid-Cataly1ed E!>tcrification B. Esterification by Alkylation
CONTENTS
20.9
conversion of carboxylic Acids into Acid Chlorides and Anhydrides A. Synthesis of Acid Chlorides B. Synthc. i~ of Anhydrides
XXiii
970 970 972
20.10
Reduction of carboxylic Acids to Primary Alcohols
974
20.11
Decarboxylation of carboxylic Acids
976
Key Ideas in Chapter 20 Additional Problems
978 979
THE CHEMISTRY OF CARBOXYLIC ACID DERIVATIVES 21.1
Nomenclature and Classification of carboxylic Acid Derivatives A. Ester<. and Lactones
B. Acid
Halide~
C. Anhydrides D. itriles E. Amidcs. Lactams. and lmides F. Nomenclature of Substituent Group!. G. Carbonic Acid Derivatives
986
986 986 988 988
989 989 991 991
21 .2 Structures of carboxylic Acid Derivatives
992
21 .3
994 994 99-l
Physical Properties of carboxylic Acid Derivatives A. Esters B. Anhydrides and Acid Chloride!. C. itrile!. D. Amide<.
21.4
Spectroscopy of carboxylic Acid Derivatives A. IR Spectroscopy B. NMR Spectroscopy
995 995
996 996 997
21.5
Basicity of carboxylic Acid Derivatives
1000
21 .6
Introduction to the Reactions of carboxylic Acid Derivatives
1003
Hydrolysis of carboxylic Acid Derivatives
1004
21 .7
A. B. C. D. E.
H ydroly~i!. of Esters
Hydroly1>is of Amides Hydrolysis of Nitriles Hydrolysis of Acid Chlorides and Anhydrides Mechanisms and Reactivity in Nucleophilic Acyl Substitution Reactions
1004 1008 1009 lOll I 0 II
xxiv
CONTENTS
21 .8
Reactions of carboxylic Acid Derivatives with Nucleophiles A. Reactions of Acid Ch l oride~ with uclcophi lcl> B. Reaction<. of Anhydride!' with uclcophiles C. Reaction<. of E'tcr<. with ' ucleophilc'
21.9
Reduction of carboxylic Acid Derivatives A. Reduction or E~ t c rs ro Primary Alcohols B. Reduction or Amidcs to Amine:; C. Reduction of itrilc~ to Primary Am incs D. Reduction of Acid Chlorides to Aldehyde'> E. Relati\·e Reacti\ itie!> of CarbOn) I Compound~
21 .10
Reactions of carboxylic Acid Derivatives with organometallic Reagents A. Reaction of E~tcr:-. with Grignard Reagents B. Reaction of Acid Chlorides wi th Lithium Dial kylcupratc),
1016 10 16 1019 1020
1022 1022 1023 1025 1027 1028
1029 1029 1031
21 .11
Synthesis of carboxylic Acid Derivatives
1032
21.12
use and occurrence of carboxylic Acids and Their Derivatives
1034
A. Nylon and Polyesters B. Waxes. Fat~. and Phospholipid:-.
Key Ideas in Chapter 21 Additional Problems
THE CHEMISTRY OF ENOLATE IONS, ENOLS, AND a,p-UNSATURATED CARBONYL COMPOUNDS 22.1 Acidity of Carbonyl compounds A. Formation of Enolatc Anions B. Introduction to Reactions of Enolatc Ions
1034 1036
1037 1038
1047 1048 1048 1051
22.2
Enolization of carbonyl compounds
1053
22.3
cl'"Halogenation of Carbonyl Compounds
1057
A. Acid-Catalytcd a-Halogenation B. Halogenation of A ldehydes and Ketones in Base: The Haloform Reaction C. a- Bromination of Carboxylic Acids D. Reaction" of a-Halo Carbon yl Compounds
22.4
Aldol Addition and Aldol condensation A . Ba<;e-Catal) ted Aldol Reaction<. B. Acid-Catalyted Aldol Conden~ation C. Special Type' of Aldol Reactions D. Synthesis w ith the A ldol Conden~ation
1057 1059 I 060 I 062
1063 1063 1066 1067 1070
CONTENTS
22.5
condensation Reactions Involving Ester Enolate Ions A. B. C. D.
Claisen Condensation Dieckmann Condensation Crossed Claisen Condensation Synthesis with the Claisen Condensation
XXV
1072 1072 1076 1076 1078
22.6
Biosynthesis of Fatty Acids
1081
22.7
Alkylation of Ester Enolate Ions
1084
A. Malonic Ester Synthesis B. Direct Alkyla tion of Enolate Ions Derived from Monoesters C. Acetoacetic Ester Synthesis
22.8
conjugate-Addition Reactions A . Conjugate Addition to a.f3-Unsaturated Carbonyl Compounds B. Conjugate Addition Reactions versus Carbonyl-Group Reactions C. Conjugate Addition of Enolate Ions
22.9 22.10
1092 1095 1097
1100
Reactions of a,f3·Unsaturated carbonyl compounds with organometallic Reagents
1101 1101 1102
Organic Synthesis with Conjugate-Addition Reactions
1103
Key Ideas in Chapter 22 Additional Problems
1105 1106
THE CHEMISTRY OF AMINES 23.1
1092
Reduction of a,f3·Unsaturated carbonyl Compounds
A. Addjtion of Organol ithium Reagents to the Carbonyl Group B. Conjugate Addition of Lithium Dialkylcuprate Reagents
22.11
1084 1086 1089
Nomenclature of Amines A. Common Nomenclature B. Substitutive Nomenclature
1116 1117 1117 1117
23.2
structure of Amines
1119
23.3
Physical Properties of Amines
1120
23.4
Spectroscopy of Amines
1121
A. IR Spectroscopy B. NMR Spectroscopy
23.5
1121 1121
Basicity and Acidity of Amines
1122
A. Basicity of Amines B. Substituent Effects on Amine Basicity C. Separations Using Amine Basicity D. Acidity of A mines E. Summary of Acidity and Basicity
1122 1123 1127 1128 1129
xxvi
CONTENTS
23.6
Quaternary Ammonium and Phosphonium Salts
1129
23.7
Alkylation and Acylation Reactions of Amines
1131
A. Direct Alkylation of Amines B. Reductive Amination C. Acylation of Amines
23.8
Hofmann Elimination of Quaternary Ammonium Hydroxides
1136
23 .9
Aromatic Substitution Reactions of Aniline Derivatives
1138
Diazotization; Reactions of Diazonium Ions
1139
23 .10
A. Formation and Substitution Reactions of Diazonium Salts B. Aromatic Substitution with Diazonium tons C. Reactions of Secondary and Tertiary Amines with Nitrous Acid
23.11
synthesis of Amines A. Gabriel Synthesis of Primary Amines B. Reduction of Nitro Compounds C. Amination of Aryl Halides and Aryl Triftates D. Curtius and Hofmann Rearrangements E. Synthesis of Amines: Summary
23.12
1139 l142 1144
1145 1145 1146 1147 1150 1154
use and occurrence of Amines
1155
A. Industrial Use of Amines and Ammonia B. Naturally Occun·ing Amines
1155 1155
Key Ideas in Chapter 23 Additional Problems
24
1131 1133 1135
CARBOHYDRATES
1157 1158
1166
24.1
Classification and Properties of carbohydrates
1167
24.2
Fischer Projections
1168
Structures of the Monosaccharides
1173
24.3
A. Stereochemistry and Configuration B. Cyclic Structures of the Monosaccharides
1173 1178
24.4
Mutarotation of Carbohydrates
1183
24.5
Base-catalyzed Isomerization of Aldoses and Ketoses
1186
24.6
Glycosides
1188
24.7
Ether and Ester Derivatives of carbohydrates
1191
24.8
Oxidation and Reduction Reactions of carbohydrates
1193
A. Oxidation to Aldonic Acids
1194
CONTENTS
B. Oxidation to Aldaric Acids
C. Periodate Oxidation D. Reductio n to Alditols
24.9 24.10
1198
Proof of Glucose Stereochemistry
1199 1199 1203
Disaccharides and Polysaccharides A. Disaccharides B. Polysaccharides
1205 1205 1209
Key Ideas in Chapter 24 Additional Problems
1212 1213
THE CHEMISTRY OF THE AROMATIC HETEROCYCLES 25.1
Nomenclature and Structure of the Aromatic Heterocycles A. omenclature B. Structure and Aromaticity
25.2
Basicity and Acidity of the Nitrogen Heterocycles A. Basicity of the Nitrogen Heterocycles B. Acidity of Pyrro le and Indole
25.3
The Chemistry of Furan, Pyrrole, and Thiophene A. Electrophilic Aromatic Substitution B. Addition Reactions ofFuran C. Side-Chain Reactions
25.4
The Chemistry of Pyridine A. B. C. D. E.
25.5
25.6
1195 1196 1197
Kiliani-Fischer synthesis A. Which Diastereomer? The Fischer Proof B. Which Enantiomer? The Absolute Configuration of D-( +)-Glucose
24.11
xxvii
Electrophilic Aromatic Substitution Nucleophi lic Aromatic Substitution N-Aikylpyridinium Salts and Their Reactions Side-Chain Reactions of Pyridine Derivatives Pyridinium Ions in Biology: Pyridoxal Phosphate
A. Nucleosides and Nucleotides B. The Structures of DNA and RNA C. DNA Modification and Chemical Carcinogenesis
1245 1245 1248 1253
Other Biologically Important Heterocyclic compounds
1255
Key Ideas in Chapter 25 Additional Problems
1257 1258
xxviii
CONTENTS
AMINO ACIDS, PEPTIDES, AND PROTEINS 26.1
Nomenclature of Amino Acids and Peptides
1265 1266 1266
A. Nomenclature of Amino Acids B. Nomenclature of Peptides
1267
26.2
Stereochemistry of the a-Amino Acids
1270
26.3
Acid-Base Properties of Amino Acids and Peptides
1271
A. Zwitterionic Structures of Amino Acids and Peptides B. lsoelectric Points of Amino Acids and Pcptides C. Separations of Amino Acids and Peptides Using Acid- Base Properties
26.4
Synthesis and Enantiomeric Resolution of a-Amino Acids A. Alkylation of Ammonia B. Alky lation of Aminomalonate Derivatives C. Strecker Synthesis D. Enantiomeric Resolution of a-Amino Acids
1271 1273 1277
1279 1279 1279 J 280
128 1
26.5
Acylation and Esterification Reactions of Amino Acids
1282
26.6
Solid-Phase Peptide synthesis
1283
26.7
Hydrolysis of Peptides A . Complete Hydrolysis and Amino Acid Analysis B. Enzyme-Catalyzed Peptide Hydrolysis
1292 1292 1295
Primary Structure of Peptides and Proteins
1296
26.8
A. B. C. D.
26.9
Peptide Sequencing by Mass Spectrometry Peptide Sequencing by the Edman Degradation Protein Sequencing Posllranslational Modification of Proteins
1299
1303 1305 1305
Higher-Order Structures of Proteins
1308 1308
A. Secondary Structure B. Tertiary and Quaternary Structure
26.10
1310
Enzymes: Biological catalysts
1315 1315
A. The Catalytic Action of Enz.ymes B. Enzymes as Drug Targets: Enzyme Inhibition
1318
Key Ideas in Chapter 26 Additional Problems
1323 1324
PERICYCLIC REACTIONS 27.1
Molecular Orbitals of conjugated A. Molecular Orbitals of Conjugated A lkenes
1333 ~Electron
systems
1336 1336
CONTENTS
27.2
xxix
B. Molecular Orbitals of Conjugated Ions and Radicals
1340
C. Excited States
1}43
Electrocyclic Reactions
1343
A. Ground-State (Thermal) Electrocyclic Reactions B. Excited-State (Photochemical) Electrocyclic Reactions C. Selection Rules and Microscopic Reversibility
1343
1.146 1347
27.3
cycloaddition Reactions
1349
27.4
Thermal Sigmatropic Reactions
1353
A. Classification and Stereochemistry B. Thermal [3.31 Sigmarropic Reactions C. Summary: Selection Rules for Thermal Sigmalropic Reactions
13S 3
1360 1362
27.5
Fluxional Molecules
1364
27.6
Biological Pericyclic Reactions: The Formation of Vitamin D
1365
Key Ideas in Chapter 27 Additional Problems
1367 1368
A-1
APPENDICES Appendix 1.
substitutive Nomenclature of organic compounds
A -1
Appendix 11.
Infrared Absorptions of organic compounds
A-2
Proton NMR Chemical Shifts in Organic compounds
A -5
A. Protons within Functional Groups B. Protons Adjacent to Functional Groups
A-S A-S
13
A -7
Appendix 111.
APPENDIX IV.
C NMR Chemical Shifts in Organic compounds
A-7 A-7
A. Chemical Sh ifts of Carbons wi thin Functional Groups B. Chemical Sh i fts of Carbons Adjacent to Functional Groups
APPENDIX v. summary of synthetic Methods A. Synthesis of Alkanes and Aromatic Hydrocarbons B. Synthesis of Alkenes C. Synthesis of Alkynes D. Synthesis of Alky l. Aryl. and Vinylic Halides E. Synthesil> of Grignard Reagents and Related Organometallic F. Synthesis of Alcohols and Phenols G. Synthesis of Glycols H. Synthesis of Ethers, Acetals. and Sul!1des I. Synthesis of Epoxides J. Synthesis of Disulfides
A -8
Compound~
A-8 A-9 A-9 A-9 A-9 A- I 0 A- 10 A-1 0 A- I 0 A- 10
XXX
CONTENTS
K. Synthesis of Aldehydes L. Synthesis of Ketones M.Synthesis of Sulfoxides and Sulfones N. Synthesis of Carboxylic and Sulfonic Acids 0. Synthesis of Esters P. Synthesis of Anhydrides Q. Synthesis of Acid Chlorides R. Synthesis of Amides S. Synthesis of Nitri tes T. Synthesis of Amines U. Synthesis of Nitro Compounds
Typical Acidities and Basicities of Organic Functional Groups
A-14
A. Acidities of Groups That Ionize to Give Anionic Conjugate Bases B. Basicities of Groups That Protonate to Give Cationic Conjugate Acids
Credits Index
A-14 A-15
C-1 1-1
Preface A PREVIEW OF THE FIFTH EDITION I believe, and research in chemical education shows. that students who make the effort to learn but :-till have trouble in organic c:hcmi:-.try are in many ca:,cs trying to memorize their way through the subject. One of the keys to students' success. then. is to provide them with help in relating one part of the subject to the next- to help them see how various reactions that seem very di fferent are tied together by certain fundamentals. An ovcrarching goal of my text is to help ~tudent~ achieve a relational understanding of organic chemisll:\'. Here are ome of the ways that I have tried to help students meet this goal.
use of an Acid-Base Framework Is a Key to Understanding Mechanisms Although I have organized Orgm1ic ChemistrY 5th Edition by functional group, I have used mechanistic reasoning to help students understand the ''why"' of reactions. Mechanisms alone, however, do not provide the relational understanding that students need. Left to their own devices, many students view mechanism. as something else to mcmori1.e, and they are baffled by the ..curved-arrow.. notation. I believe passionately that an understanding of acid-base chemi try is the key that can unlock the door to a mechanistic under tanding of much organic chemi~tr). I n Or~anic Chemistry 5th Edition. I use both Lewis acid!> and bases and Bronsted acid.., and base-. a:-. the foundations for mechani!.tic reasoning. Although students have memoriLed the appropriate definitions in general chemi~try. few have developed real insight about the implications of these concepts for a broader range of chemistry. I have dedicated Chapter 3 to these fundamental acid- base concepts. The terms ..nucleophile." "electrophile.'' and "leaving group" then spring easi ly from Lewis and Br~msted acid-base concepts. and the curvedarrow notation makes sense. I have provided a substantial number of dri ll problems to test how well students have mastered these principles. I have reinforced these ideas repeatedly with each new reaction type. Free-radical reaction). arc al. o covered. but nm until the electron -pair concept are fully established.
Tiered Topic Development Provides Reinforcement of Important Ideas I have introduced complex subjects in "tiers:· This means that students will see many concepts introduced initially in a fairl y simply way. then reviewed with another layer of complexity added, and reviewed again at a greater level of sophistication. Acid- base chemistry. discussed above. is an example of tiered development. After the initial chapter on acid- base chemistry and the curved-arrow notation. the e concepts are revisited in detail as they are used in the early examples of reaction<, and mechanisms. and again with the introduction of each new reaction type.
xxxi
xxxi i
PREFACE
The pre<,entation of stereochemi two chapter!> later. Cyclic compounds and the stercochem i~try of reaction'> follow !-.Ub~equent l y. Then the ideas of group equivalence and noncquivalcncc are introduced even later. both in the context of en;yme catalysis and NMR spectro:-.copy. The approach to organic synthesis i~ yet another example. I stan with 'imple reactions and then show <;ludcnt-. how to think about them in rever~e. Then. later. I introduce the idea of multi<;tep -.ynthe"i" using relatively !.implc two- and three-<>tep sequence'-.. Later '-.till. we have another di<,cu~-.ion in which stereochemi-.tr) come" into play. Even later. the u\e of protecting group' ic; introduced. This ··tiered pre-.cntation·· of ke) topic<, require-, some repetition. Although the repetition of key point'> might be considered inertlcient. I belien: that it ;, crucial to the learning proces~. When a topic i:-. con:-.idered after its first introduction. I have provided dewiled cro~!>-rcfcrenc ing to the original material. Student~ arc never caM adrift with terminology that has not been completely delined and reinforced.
Everyday Analogies Help Students to construct Their own Knowledge I belie\e in the cm1.Hrttcril'ist theof) of learning. \\ hich holds that \llldenl\ con-,truct learning in their own mind'> b) relating each new idea to something they alread) know. Thi' i!-. wh) the relational approach to learning organic chcmi\tr) i.., so imponant. For the -;amc rea<,on. l have provided common analogie:-. from everyda) experience for many of the di~cu\\ion-; of chemical principle~ so that students can relate a new idea to somethi ng they already know. One of many exa mples can be found i n the sidebar on p. 164.
Biological Examples Motivate students Interested In the Allied Health sciences ~any organic chemiMry clas~es are populated largely by premedical -.wdcnl\. prepharmac) c;tudemc;, and other \tudents interested in the life o;ciences. Biological example'> help to motivate these Mudent<,. I have provided a number of example from modern biochemi!-.lf) and medicine throughout the book. Amino acid!. and proteins have a dedicated chapter that has been completely rewritten in light of modem developments. Carbohydrate:-. abo have a dedicated chapter that has been moved so that it now follows carbonyl and amine chemistry. I have integrat ed many other biological example!> into discu!.sions of the rel evant chem istry. The ultimate goal of these examples is to reinforce the chemistry being di~cus~ed with material that students !thould find particularly relevant. Among these arc discussion!. of cell membranes. bioorganic <,tereochemi!.try. pheromone<,. imaging agents. nucleic acid-.. coen;yme mechanism'>. and many. many more. One of man) -.uch di-.cussions. for example. i" found in the sidebar on pp. 396-398 and the accompanying illu'>tnllion on p. 399.
students in an Introductory course Should see Examples of contemporary Organic Chemistry The ·'canon" of undergraduate organic chemi!'try necessaril y contain~ many classical reactions. but this text introduces some very modern chemistry as well. For example. the 4th edition introduced a \ection on transition-metal catalysis, a field that has literally exploded in the last few years. Thi<; section carefully explain'> the con,entions u-;ed in the field for electron counting and calculating oxidation state . Thi-; edition build-. on that introduction. which pre,iously included
PREFACE
xxxiii
the Heck and Stille reactions. by adding sections on the Suzuki coupling. alkene metathesis, and Buchwald-Hartwig amination. Asymmetric epoxidation is also introduced. and a modern approach to understanding the rate accelerations observed for many intramolecu lar reactions has been developed. There is a :-;omewhat higher level of molecu lar orbital theory than in previous edit ions. Accompanied by detai led explanations and illustrations. MO theory is related ro practical considerations. such as the meaning of resonance structures, the basis of aromaticity, and the understanding of reaction stereochemistry.
Solving Problems Is an Essential component of the Learning Process We all know that solving problems is a key to learning organic chemistry. I have provided 1672 problems. many or them mul tipart, ranging from dril l problems to problems that wi ll challenge the most astute , tudents. Many are based direct ly on material in the literature. The 840 problems within the body of the text are typically drill problems that test whether student. understand the current material. The 832 problems at the end of the chapters cover material from the entire chapter and. in many cases. integrate material from earlier chapters. Additionally. I have interspersed 123 Study Problems throughout the text. Each of these problems has a worked-out solution that carefully shows students the logic involved in the problem-solving process. Many students rely too heavily on the Solutions Manual. To help avoid this problem. r have reintroduced in this edi tion a "pai red problems"' approach. This means that the solution to about 60C/'I' of the problems are provided in the Srudy Guide and Solurions Manual that accompanies the text (see below). but the solutions to the rest of the problems are not provided to the students. Instructors will receive complete solutions to these "unsolved'" problems in PDF format. An instructor who would like to make these solutions available to his or her swdents can simply post them on a cou rse Web site or hand them out. r have experimented with different ways to use these additional solutions. For example, I have prov ided the relevant solutions to students just prior to major exams: this strategy encourages them to try these problems without immediate help from the olutions manual. but also allows them to check their answers later. A lternatively. an instructor could these additional problems as the basis for exam questions. Many of the '"unsolved"' problems are adjacent to at least one ..sol ved'" problem of the same type.
A Full-Color Presentation Improves Pedagogy With this edi tion. this textbook make~ its tirst appearance in full color. As I have redesigned the art program to make use of color. I have kept a few things uppermost in my mind: color should be u~ed solel y for functional. pedagogical purposes: and both gratuitous i llustrations and excessive color should be avoided. (1 believe students have enough to won·y about rather than trying to figure out what's important in a textbook.) The use of color. the presentation of the art. and the text design itself flow from the ideas in The Psychology (4 Even·dar ]flings. a book by Don Norman. The core idea is that these elements in the text shou ld provide subliminal cues to students that facilitate the learning process.
supplements Provide Additional Help for Both Students and Instructors I. The Study Guide and Solurions Manual presents chapter summaries, glossaries of terms. reaction summaries. elu tions to selected problems. Study Guide Links. and Further
xxxiv
PREFACE
Explorations. The Study Guide Links, wh ich are called out with mm·gin icons in the text. are additional discussions of certain topics with which, in my experience, many students require additional assistance. Example. are ·'How to Study Organ ic Reactions.'' and ''Solving Structure Problems... The Further Explorations. also called out with margin icons in the text. are short discussions that move beyond the text material. An example is "Fourier-Transform NMR.'' 2. Molecular models (Model 1013) from Maruzen International are available as a bundle with the textbook. For moi·e information and pricing. please contact Ben Roberts: bwr@ roberts-pub! isher.s.com. 3. We will provide. to adopting instructors. a PDF image of the problem solutions that are not provided in the Study Guide and Solutions Manual (see the discussion of "paired problems'' above), a well as full-color images of all figures in conventional fo rmats so that they can be used in classroom presentations. Instructors at adopting institutions who want this material should request it from the publisher at [email protected]. 4. As with the fourth edition, we will maintain. on the World-Wide Web. an up-to-date list of errata in PDF format for both the text and the Study Guide and Solutions Manual supplement. These lists of errors will be generally available to insu·uctors and students alike.
A BOOK WITH A SCHOLARLY HISTORY The first edition of Organic Chemis1ry. published in 1984. required 7~ years of development, because each topic was researched back to the original or review literature. Subsequent editions. including this one. have conrinued this scholarly development process. Almost every reaction example is taken from the literature. Each ed ition has benefited from a thorough peer review.
ACKNOWLEDGMENTS 1 am indebted to my dean. Craig Svensson. and my department head. Rick Borch. for provid-
ing a sabbatical leave in 2007- 2008. as wel l as a climate in which this edition could be completed successfully. The electronic resources of the Purdue Library have streamlined the research process for this text in a way that was unimaginable 25 years ago. and I would like to thank Emily Mobley, Purdue's Dean of Libraries Emerita, for bringing the electronic library to fruition . and Jim Mullins. the present dean, for fostering its continued improvement. Thanks to my faculty and staff col leagues at Purdue and beyond-John Grutzner. Don Bergstrom. Mark Green. Chris Rochet. Ross Weatherman. Phil Fuchs. Mark Lipton. Ei-ichi Negishi. Markus Lill, David Nichols. Mark Cushman. Arun Ghosh. John Bartmess (University of Tennessee), Bob Hammer (formerly of Louisiana State University). Karl Wood. David Allen. and Susan Holladay- for advice, assistance, and suggestions. Special thanks go to two very special teaching assistants, Lisa Bonner and Ani mesh Aditya. for their hard work, their advice. and their effective and inspiring teaching. The reviewers named in the list that follows this preface provided invaluable assistance in polishing this text. I am panicularly indebted to Prof. David Hansen of Amherst College, Prof. Paul Rablen of Swarthmore College. and Prof. Carolyn Anderson of Calvin College. who provided invaluable suggestions through all or most of the project. Prof. AJ1amindra Jain of Harvard University. a dedicated teacher and a delightful collaborator. made some very valuable suggestions in the early going. and I was very much looking forward to working with him as my coauthor on the Study Guide and Solutions
PREFACE
XXXV
Manual ~uppl ement. Sadly. Ahamindra passed away in 2008. and i), sorely missed. I would al !>o like tO thank the many student-. from all over the coun try who made !>uggcstions, offered comment!>. and reported errors. I welcome co tTespondence wi th the ~tudents using this edition. I can be reached by e-mai l at marc. loudon. I @purdue.edu. M y relationship with the professionals at Side By Side Swdios in San Francisco, Mark Ong and Susan Riley, has been particularly gratifying. ot on ly has their composition work been superb. but also their advice has been invaluable. Side By Side gets tivc l.tars! Working with Ben Robert), at Roberts and Company Publi-,her<: ha!> aho been a delight. and I hope our association continues far into the future. I very much appreciate the hard work. advice. and attention to detail of the copy editor. John Murdtck. and the proofreader. Gunder Hcfla. I would particularly like to thank those acknowledged eparately in the Credits !>ection for their w illingness to allow me to reproduce their materials. I could not have completed thi ~ project wi thout the love and support of Judy and my family, for which I am gratefu l beyond words. M y wish is that the students who usc this text will see the amazing diversity and beauty of science through their study of organic chemistry. and that they will benefit from using this book as much as I have enjoyed writing it. October 2008 West Lafayette. Indiana
M arc L oudon
1 Chemical Bonding and
Chemical Structure INTRODUCTION A. What Is Organic Chemistry? Organ ic chemistry is the branch of :-.cienc.:e that deals general ly w ith compounds of carbon. Yet the name mgc111ic seem:-. to imply a connection with living thing:-. Let\ explore this connection. for the emergence of organic chemistry as a :-.eiencc i~ linked to the early evolution of the life ..,<.:iences.
B. Emergence of Organic Chemistry A-, early a~ the sixteenth century, ~cholars seem to have had some realii' by Ji.i n!. Jacob Berzel ius (I 779 1848). Somehow. the fact that these chemi<.:al substances were organic in nature was thought to put them beyond the '>cope of the experimentalist. The logic of the time seems to have been that life i~ not understandable: organic compound:. !->pring from life: therefore. organic compound., are not undeNandable.
1
2
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
The barrier between organic (living) and inorganic (nonliving) chemi~.tr) began to crumble in 1828 becau!>e of a serendipitous (accidental) discovery by Friedrich Wohler ( 1800-1882). a German analyst originally trained in medicine. When Wohler heated ammonium cyanate, an inorganic compound, he isolated urea, a known urinary excretion product of mammals. a mmonium cyanate (CH 4 N 4 0)
an inorganic compound
heat
urea ( CH~ N 4 0 )
(1.1)
an organic compound
Wohler rccognited that he had symhesized this biological material ..,\ ithoutthe u~e of kidneys. nor an animal. be it man or dog... ot long thereafter fo llowed the synthesi<> of acetic acid b) Hermann Kolbe in 1845 and the preparation of acetylene and methane by Marcell in Berthelot in the period 1856-1863. Although "vi talism" was not so much a widely accepted formal theory as an intuitive idea that something might be special and beyond human grasp about the chemistry of living things. Wohler did not identify his urea synthesis with the demise of the vitalistic idea: rather. his work signaled the start of a period in which the synthesis of !'o-called organic compounds was no longer regarded as ~omcthing outside the province of laboratory investigation. Orgunic chemists now investigate not only molecules of biological importance. but also intriguing molecule of bizarre structure and purely theoretical intcrc-.t. Thus. organic chemistry deals with compounds of carbon regardless of their origin. Wohler o,eems to have anticipated these development when he \\rote to his mentor B erzeliu~...Organic chemistry appears to be like a primeval tropical forest. full of the most remarkable thing<.:·
c . Why study organic Chemistry? The study of organic chemistry is imponant for several reasons. First, the fi eld has an independent vitality as a branch of science. Organic chemistry is characterized by continuing development of new knowledge, a fact evidenced by the large number of journals devoted exclusively or in large pan to the subject. Second. organic chemistry lies at the heart of a l>Ub!>talllial fraction of the modem chemical industry and therefore contributes to the economic. of many nations. Third, many student~ who take organic cherni~try nowadays are planning caree r~ in the biological sciences or in allied health disciplines. !>uch as medici11e or pharmacy. Organic chemistry is immensely important as a foundation to the~e fields. and its importance is sure to increa e. One need only open modern textbooks or journals of biochemistry or biology to appreciate the sophisticated organic chemistry that is central to these ureas. Finally. even for those who do not plan a career in any of the sciences, a study or organic chemistry is important. We live in a technological age that is made possible in large part by applications of organic chemistry to industries as diverse as plastics. textiles. communications, transportation. food. and clothing. In addition. problem!. of pollution and depletion of resources are all around u~. If organic chemistry has played a part in creating these problem . it will surely have a role in their solutions. A~ a science. organic chemistry lie~ at the interface of the physical and biological sciences. Re earch in organic chemistry is a mixture of sophisticated logic and empirical observation. At its best. it takes on artistic dimensiom•. You can use the study of organic chemistry to develop and apply basic skills in problem solvi ng and. at tJ1e same time. to learn a subject of immense practical value. Thus, to develop as a chemist. to remain in the mainstream of a health profession, or to be a well-informed citizen in a technological age, you wil l find value in the study of organic chemistry. In this text we have several objectives. We' ll present the ··nuts and bolt!>..- lhc nomenclature, classi lication. structure, and prope11ie~ of organic compounds. We' II also cover the principal reaction and the syntheses of organic molecules. But, more than this, we'll develop underlying principles that allow us to understand, and ~ometimes to predict, reactions rather than
1.2 CLASSICAL THEORIES OF CHEMICAL BONDING
3
).imply mernori;ing them. We'll con).ider . orne of the organic chemi~try that is industrially importanl. Finally. we'll examine ~orne of the beautiful applications of organic chemi try in biology, such as how nature does organic chemistry and how the biological world has inspired a great deal of the research in organic chemistry.
-; 1.2
CLASSICAL THEORIES OF CHEMICAL BONDING To under..,tand organic chemistry. it is necessary to have orne understanding of the ch emi cal bond- the force!. thar hold atom!> together within molecules. FiN. we'll review some of the older. or ..classical:· ideas of chemical bonding-ideas that, despite their age. remain u eful today. Then. in the last part ofthi!-. chapter. we'll consider more modem ways of describing the chemical bond.
A. Electrons in Atoms Chemi.1try happens because of the he!ral'ior ofelecmms in atoms and molecules. The basis of this behavior is the arrangement of electrons within atom . an arrangement suggested by the periodic table. Consequently. let'-; lir"t review the organization of the periodic table (see page facing in'>idc back cover). Tht· '>haded ~lernc:nh art' M greak't importam:c tn organic chemt:-try; knowing their atomic number" and rclati1e position will be valuable later on. For the moment. however. consider the foliO\\ ing details of the periodic table because they were important in the development of the concept!> of bonding. A neutral atom of each element contnins a number of both protons and electrons equal 10 its atomic number. T he periodic aspect or the table- its organization into groups of elements with simi lar chemical propert ies- led to the idea that elect rons reside in layers, or shells, about the nucleus. The outenno~t -;hell 111' c:lec trons in an atom i~ ~ailed its valente shell 1 'and the electrons in this shell are callcp valencl' ell'ctrons. The nu111ber of l 'cl encc: electron.\ .f(Jr any nemral atom in (,111 A Kroup n( the perio(/ic' t(tble (except he I ium) equal.\ its group number. Thu~. lithium. sodium. and pota~\ium (Group I A) have one valence electron, whereas carbon (Group -lA) has four. the halogenc; (Group 7A ) have seven. and the noble gases (except helium) have cigh1. Helium has two 'valence electrons. Walter Kosscl ( 1888-1956) noted in 1916 that ~hen atom~ fl>rm ion' they (end to g1lin or lose valence clectrom. :0 a~ lo have the '>amc numb...:r of electrons as the noble gas of closes atomil.! number. Thus, potassium, with one valence electron (aJlcl 19 total electrons), tends to lose an electron to become K+. the potas~ium ion. which has the same number of electrons ( 18) as the nearest noble ga~ (argon). Ch lorine. w ith seven valence electrons (and 17 total electrons) t...:nds to accept an electron to become the 18-electron chloride ion, Cl-. which also has the same number of electron as argon. Because the noble gases have an octet of electrons (that i .... eight electrons) in their \alence ~hells. the tcndenc) of atom' ro gain or los-e-valence electron'> to form ions wi(h the noble-tws conligurarion has been called the octet rule.
B. The Ionic Bond A chem ical compound in which the cQmponent atoms exist as ions i~ called an ionic corn]JOund. Potassium chlori de. K CI. is a common ionic compound. The electronic configurations of the potassium and chloride ions obey the octet rule. The structure of crystalline KCI is shown in Fig. 1.1 on p. 4. In the KCI structure. which is typical of many ionic compound .... each po~itive ion is surrounded by negative ions. and each ncgatil·c ion i <:urrounded by po~iti1e ion~. The crystal structure is stabilitcd by an interaction between ions of opposite charge. Such a -.wbillltng interacttqn between opposite charges is
4
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
~hloridl'
ion (( I )
each K' surrounded by Cl
Figure 1. 1 Crystal structure of KCI. The potassium and chlorine are present in this substance as K• and Cl- ions, respectively. The ionic bond between the potassium ions and chloride ions is an electrostatic attraction. Each positive ion is surrounded by negative ions, and each negative ion is surrounded by positive ions. Thus, the attraction between ions in the ionic bond is the same in all directions.
• ~allcJ an elcctrolitatic a ttraction. An electro~trtlit· auraction that holds ion!. together. a' )n cry<,talline KCI. i~ ~ailed an iol)ic bond. Thu~. the Cl) '>tal ..,tructurc of KCI i~ maimained by ionic bond\ between potassium iono; and chloride ion'>. The ionic bond i'> the '>ame in all direction<,; that i'>. a po~itive ion has the <;ame attraction for each of it'> neighboring negative ions. and a negative ion ha'> the same attraction for each of it\ neighboring po'>itive ion<;. When an ionic compound such a!. KCI di<.<.ohe-. in water. it di,..,ociate~ into free ions (each <,urrounded b) water). (We ·11 consider this proce..,, further in Sec. SA.) Each potassium ion mmc~ around in solution more or le~~ indcpcndcmly of each chloride ion. The conduction of electricity by KCI solutions show that the ion<, arc prc.,cnt. Thu~. the ionic bond is broken when KCI dissolve~ in water. To :-.u11Hnaritc. (he ionic bond
''•
...
•
I . "'an dcctro:.tatk attracti
•
1. 1 How many valence electrons are found in each of the following ~pecies? Ia INa (b) Ca (c) 0 2 - (d) Br+
..
1.2 When two different species have the !>arne number of electron~. they are said to be isoelecrmnic. ame the species that satisfies each of the following criteria: Ia I the '>ingly charged negative ion isoelectronic with neon (b) the singly charged positive ion isoelectronic with neon lc I the dipositive ion isoelectronic "ith argon (d) the neon ~pecies that is isoelectronic with neutral fluorine
,....* The 'olution' to problems or problem parts labeled "ith boldface blue found in the Swdr Guide and Solwions Manual :-.upplcmcnt.
number~ or ieuer<; can be I
1.2 CLASSICAL THEORIES OF CHEMICAL BONDING
5
c . The covalent Bond M any compounds contain bond<; that are very different from the ionic bond in KCI. either th c~c compound'> nor their <;olution<> conduct electricity. Thi<; observation indicates that the!>e compound'> arc not ionic. How are the bonding force<., that hold atom<> together in '>uch compound~ differe111 from those in KCI'? In 1916. G. . Lewi~ (1875-19~6). an American phy~i cal chemist. propo. ed an electronic model for bonding in nonionic compounds. According to this modeL the chemical hond in a non ionic compound i~ a CO\ alent hond. \\ hich con<-ist' qr an dectron pair that is \'hmnl bet~ ccn bOJ)(kd atom~. Let':- c>.amine .,ome of the 1dea~ a.,so~ ciated \\ ith the covalent bond. One of the simplest examples of a covalent bond is the bond between the two hydrogen atom. in the hydrogen molecu le.
Lewis Structures
The symbob '·:"and .._ .. are both used to denote an electron pair. II ,·Jwred r:lecrron pair is til<' r:ssence qf the Covalen/ bond. Molecu lar 'tructurc' that usc this notation for the electronpair bond arc called Lewis structures. In the hydrogen molecule. an electron-pair bond holds the two hydrogen atoms together. Conceptually. the bond can be envisioned to come from the pairing or the valence electrons of two hydrogen atom'>:
H·
+ H· __... 11:11
Both electrons in the covalent bond arc <>ltared equally between the hydrogen atoms. Even though electron<, are mutual!) repulsive. bonding occurs because the electron of each hydrogen atom is attracted to both hydrogen nuclei (proton'>)
·C· +
4 H·
__...
II H :C: H H
or
I I
11 - C- 11
or
( 1.3)
II
In the previous examples. all valence electron~ of' the bonded atom~ are shared. In some cova lent compounds. such as water ( H 20). however. some valence electrons remain unshared. In the wa ter molecule. oxygen ha!-. si x valence e lectron~. Two o f these combine w ith hydrogens to make two 0-H covalent bonds: four of the oxygen valence electrons arc left over. These arc represented in the L ewis <.;tructure of water a<. electron pairs on the oxygen. In general. unshared valence electrons in Lewi-, '>lructures are depicted a'> paired dot<, and rcfen·ed to as un-
sharcd pairs.
--
unl>h.trcd p.tm
...
....,.
I I -0 -
H
Although we often write water as H - 0- H. or even H ,O. it 1 a good hahit to indicate all un<;hared pair-, with paired dots until ~(HI remember in.,tincll\'eh that they are there. The foregoing examples illustrate an important point: 7he l l /111 o(a/1 s!wrefl (llld tm~/l(lted I'll fence
electrons around each atom ill nuon·
~table c·m·a/1'111 <'tllllfHUmdv
is ei~/11 (fii'O for the
6
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
jn·dm~en
a rom 1. Thi<. is !he octet rule for covalent bonding. and it will pro1•e to be extremely imporram for understanding chemical reacti1•ity. It is reminiscent of the oclet rule for ion for-
mation (Sec. 1.2A), exceplthat in ionic compounds, valence electron~ belong completely to a particular ion. In covalent compounds, shared electrons are counted twice, once for each of the sharing partner atoms. Notice how the covalent compounds we've just considered follow the octet rule. ln the structure of methane (Eq. J.3). four . harcd pairs surround the carbon atom- that is, eight shared electron!., an octet. Each hydrogen share. two electrons, the "octet rule" number for hydrogen. Similarly. the oxygen of the wa1er molecule has four shared electron~ and two unshared pair\ for a Iota I of eight. and again !he hydrogens have two 'hared electrons. Two atom., in covalent compounds may be connected by more !han one covalent bond. The following compounds are common examples:
H
H
\ I c::c I \ H J-1
H
H
H
H
\ I C=C I \
or
ethylene
H
H
\ .. I c::o ..
\ .. C=O I ..
or
H
H formaldehyde
H-C:: C-1 1
or acetylene
Ethylene and formaldeh) de each comain a double bond-a bond consis1ing of two electron pam•. Acetylene contains a t ripl e bond- a bond invoh ing three electron pair'. Covalenl bonds are especially important in organic chemistry because all orga11ic t/l_olecul('\ collltlfll c ol'a/ent bonds.
Formal Charge The Lewis structures considered in the previous discussion are those of neutral molecules. However, many fami li ar ionic species, such a~ ISO~f-. INH4 )+, and IBF~ 1-. also contain covalent bonds. Consider the tctrafluoroborate anion, which contains covalent B-F bond~:
.. ·i=· I .. ].. I ..F:
:F - B-
[
:F:
tetrafluoroborate ion
•
-
Because the ion bears a negative charge, one or more of the a10ms w ith in the ion must be charged-but which one(s)? The rigorou~ answer is that the charge I!> shared by all of the atom!>. However, chemists have adopted a useful and important procedure for electronic bookkeeping that assigns a charge to specific atoms. The charge on each a1om thus assigned is called its formal charge. The sum of th~.: formal charge on the individual atom!> must equal tho.: lotal ..:hargc on the lon.
1.2 CLASSICAL THEORIES OF CHEM ICAL BONDING
7
Computation of formal charge on an atom invol ves dividing the total number of val ence electrons between the atom and i ts bonding partners. Each atom receives all of its unshared electrons and ha((of its bonding electrons. To ;\<:sign a formal charge,to an at()m. then. use the following pr~ edure:
I. Write ~!"own the gnmp nwnher of the atom froru i t;, col umnJJead ing ln the periodic table. This is equal to the number of valence electrons i n the newral atom. 2. Determine the l'ttlt'nce eleCtron counr for the atom by adoing the number of unshared valenc~ clec~rOIJS qn the aiOJll to the numbu or C()\ ~'\ ent bond, to the atom. Counting the covalent bonds in effect adds half the bond ing electrons- one electron for each bond . 3. Subtract the valence electron coun t from the group number. The result is the formal charge. This procedure is iUustrated i n Study Problem 1.1 . ..
r , wh ich has the
Assig11 a formal charge to each of the atoms in the tetratluoroborate ion. LBF4 structure shown above.
Solution Let's first apply the procedure outlined above to fluorine: Group lltllllber offluorine: 7 Valence-electron count: 7 (Unshared pairs contribute 6 electrons; the covalent bond contributes l electron.) Formal charge on fluorine: Group number - Valence-electron count = 7 - 7 = 0
4-r
Because all fluorine atoms in LBF are equivalent. they al l must_have the same fonnal chargezero. It follows that the boron must bear the formal negative charge. Let's compute it to be sure. Group number of boron: 3 Valence-electron coum: 4 (Four covalent bonds contribute I electron each.) Formal charge on boron: Group number - Valence-electron count = 3 - 4 = - I
STUDY GUIDE LIN K 1.1•
Formal Charge
Because the forma l charge of boron is - L. the structu re of [BF4 ]- is written with the minus charge assigned to boron:
:p: .. I .. :F-B=-F: .. I .. :f :
When indicating charge on a compound. we can show the formal charges on each atom, or we can show the formal charge on the ion as a whole. bw we should not show both
:p: .. I .. :r -s=-r: .. I .. :f :
formal charge
*
.. . ]- [* . .1.[ T .T : F-B-F :
:F- -F:
: .. f:
: .. p:
overall charge
don' t show both
..
I ..
.. I ..
Study Guide Li nks are short discuss ions in Lhe Study Guide and Solutions Manual supplement that provide extra hints or shortcuts that can help you master the material more casily.
8
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STR UCTURE
-'Rules for Writing Lewis Structures
The previous two sec ti ons can be summariLed in
the following rule for writing Lewis structures. '\
._
-
I. Hyur()g0n can. h:\re nQmore than two electron:-. 2. Th ~urn of all 9onding electrons and unsharetl pairs for afoms in the second period oC
·'·'
the periodk tablc~the row beginning \\lith llthiun'Ji:...ls ne\er grearer th an eight (octet rule). lhe!-.e atoms may. ho\\ ever. have fewer than eight elet:tron!'-. In 'om CH~es. atoms below the second period of the periodic.. table may have more than
elg!1t cle<.:trons. However. rule 2 slwuld also be C()llowed ft>r Lhese cases until exceptions arc di:; ·ussed later in the te)
Draw a Lewis strucLUre for the covalent compound methanol, Cl-l40. Assume that the octet ru le is obeyed. and that none of the atoms have fom1a l charges.
Solution For carbon to be both neutral and consistent with the octet rule, it must have four covalent bonds:
I
-c1 There is also only one way each for oxygen and hydrogen to have a formal charge of zero and simultaneously not violate ll1e octet rule:
-0-
H-
If we connect the carbon and the oxygen. and fill in the remai ning bonds with hydrogens. we obtain a,structure that meets all the criteria in the problem: H
I .. I ..
H-C-0-H H correct structure of methanol
1.3 Draw a Lewis structure for each of the following species. Show all unshared pairs and the formal charges, if any. Assume that bonding follows the octet rule in all cases. HCCI3 (b) NH3 (c) [NH~J + (d) LH,OJ+ ammonia ammonium ion
(a)
1.4 Write two reasonable structures corresponding to the formula C2H60. Assume that all bonding adheres to the octet rule, and that no atom bear a formal charge.
1.2 CLASSICAL THEORIES OF CHEMICAL BONDING
9
I.S Draw a Lewis structure for acetonitrile. c;H3N. assumi ng that aU bonding obeys the octet rule. and that no atom bears a formal charge. Acetonitrile contai ns a carbon- nitrogen triple bond.
J .6 Compute the formal charges on each awm of the following structures. In each case. what is the charge on the en tire structure? :Q: {II) H H (b)
I I
I I
.. I ..
H-B-N-1-1
1-1
..
.H
I
..
:Q :
D. The Polar covalent Bond
... •
:O-P-0-H
I
In many covalent bonds the electrons arc not ~hareJ equa) ly between two bonded atoms. Consider. for example. the covalent compou nd hydrogen chlori de. HCI. (Although HCI dissolves in water to form H 30+ and Cl- ions, in the gaseou~ state pure HCI is a covalent compound.) The electrons in the H -Cl covalent bond arc une l'('nly distributed between the two atoms: they are polarized. or "pulled." toward the chlorine and away from the hydrogen . .A L'i9.nd in wh ich electron~ are sh:.tred une,tenly is called ' indicmed b-'' its elect.ronegati vit '· The electronegat ivities of a few elements that are important in organ ic chemis try are shown in Table I. I. Notice the trends in this table. El ~c tronegativity increa<;es to the top and tQ the right of the table. T he more an atom atlracts electrons. the more electronegative it is. r luorine i,s the most electronegative elemcnL. E lcctroncgati vi ty decreases to the bottom and to the left of the periodic table. The les. an at m :lttract ~ electrons. the more electropositive i t iS. Of the common stable elements. ce ium i::; the most · lcctropo. i rive.
1(;1:1!111
Average Pauling Electronegativities of some Main-Group Elements H
2.20
t
Li 0.98
Be
B
c
1.57
2.04
2.55
N 3.04
Na 0.93
Mg 1.31
AI 1.61
Si 1.90
2.19
K 0.82
Ca 1.00
Rb 0.82 Cs 0.79
'II
p
\
·.
3.44
F 3.98
5 2.58
3.16
0
Cl
Se
Br
2.55
2.96 I
2.66
10
CHAPTER 1 • CHEMICAL BONDING AND CHEM ICAL STRUCTURE
If two bonded atoms have equal electro negativities, then the bonding electrons are shared equa lly. But if two bonded atoms have considerably different e lectronegativities. then the electrons are unequally shared , and the bond is po lar. (We mighuhink of a polar covalent bond as a covalent bond that i trying Lo become ionicl) Thus, c7palarl1o11Ii is a bond between atom.
wirh sixnificanll\lllifferent elecrronegativiries. Sometimes we indicate the polarity o f a bond in the follow ing way: 0+
6-
H -CI In this notation. the delta {8) is read as ·'partially'' or " o mewhat.'' so that the hydrogen atom of HCI is '·partia lly positive," and the chlorine atom is ''partially negati ve.'' Another more graphical way that we' ll use to show polarities is the e/ecrrosraric porenrial map. n electrostatic potentjal map (EP.M) of a molecule starts with a map of the total e lectron density. This is a picture of the spatial disu·ibution of the e lecu·ons in the molecule that comes from molecular orbi tal theory, which we'll learn about in Sec. 1.8. Think o f this as a picture of "where the e lectrons are:· Th t;;PM is a map of total electron density that has been color-c(~<:kd for regions of"local positive and negative ch~rge. Ar~as of greater negative charge arc colored red. and areas of greater positive- harge-ar co1bred blue..~eas of neutraJity are
•
~."olored
!.!n:en.
" •
'
'
'
.. .
-~~
lo~al po~itiw (harg<'
-
)o(a]
negative ..:!urge
-
This is called a "pote nlial map" because it represents the interaction of a test positive charge with the molecule at various points in the molecule. When the test positi ve charge encounters negative charge in the molecule. an allractive potential energy occurs: this is color-coded red. When the test positive charge encounters positive c harge, a repulsive potential energy results. and this is color-coded blue. The EPM o f H--cl shows the red region over the C l and the blue region over the H, as we expect trom the greater electro negativity o f C l versus H. In contras t. the EPM of dihydrogen shows the same color on both hyd rogens because the two atoms share the e lectrons equally. T he green color indicates that neither hydrogen atom bears a net charge. electron~
pulkd toward., U
\
dihydrogen (H- H)
hydrogen chloride (H- CI)
Furtller EXploration 1.1•
Dipole Moments
How can bond polarity be measured experimentally? he une.,ven electron distribtHion in a tom pounu containing covalent 9onds is measured by a quiml)ty called tqe d i)?oJe moment, which is abbreviated wi th thc..,Greek'letter p.. (mu). The dipole moment is commonl y g iven in deri ved unit called debyes. abbreviated D. and named for the physical chemist Peter Debye (1884- 1966), w ho won the 1936 Nobe l Prize in Chemistry. For example, the HC I mo lecule has a dipole moment of 1.08 D, whereas dihyd rogen (H2 ), which has a uniform electron distribution. has a di pole moment of zero.
*
Further Explorations are brief sections in the Swdy Guide and Solwions Manual supplement that cover the subject in greater depth.
1.2 CLASSICAL THEORIES OF CHEMICAL BONDING
11
fllhe dip6 lon1Qrncpt is de li ned by the f<,> llowing eqwttiqn: f..t
=--qr
( 1.4)
In this equation, ~ is the magnitu(,le of the separated charge1-a nd r is a vector fro m the site of the positive charge to the &ite oflhe negative charge. For a simple molecule like HCl, the magnitude of the vector r is merely fhe length of the HCI bond, and it is oriented from the H (the positive end of the dipole) to the Cl (the negalive end). The d ipole moment is a vector quantity, and /.t and r have the same direction- from the positive to the negative end of the dipole. As a result. the dipole moment vector for the HCI molecule is 01iented along the H-Cl bond from the H to the Cl: ..,.. .----dipole moment vector for HCI
-+--l,..• '
H - Cl
Notice that the magnitude of the dipole moment is affected not only by the amount of charge that is separated (q) but also by how far the charges are separated (r ). Consequent ly. a molecule in which a relatively small amount of charge is separated by a large distance can have a dipole moment as great as one in which a large amount of charge is separated by a small distance. Mb lecl.tles Chat have p rmaoenl Clipole moments arc called pQiar moJecules. HCI is a polar molecule, whereas H 2 is a nonpolar molecule. Some molecules contain several polar bonds. Each olar bond ha. associated with it a dipole l'no mcnl C\)lltribution. called a bond dipole. T he nb dipole moment 6f :.uch a polar molecul · is the vector ::.um or its bond dipoles. (Because HCl has only one bond. its dipole moment is equal to the H -CI bond djpole.) Dipole moments of typical polar organic molecules are in the l- 3 D range. The vectorial aspect of bond dipoles can be illustraled in a relatively simple way with ihe carbon diox ide molecu le, C0 2:
C-0 bond dipoles /
~ C=O
0=
<-----:_ I
• EPM of carbon dioxide
Because the C0 2 molecu le is linea1; the C-0 bond dipoles are oriented in opposite directions. Because they have eq ~1 a l magnitude~. the:n "xqcth· cance.f. (Two vectors of equal magnitude orie nt u in opposite directions a l way~ ~:ance l.l C911Sequent l y. C02 is a nonpolar molecule. even though it has polar bonds. In conn·ast. if a molecule contains-Se\1cral bonti dipoles that do not cancel, the v
1\
I.S~~~ D Y ~>4.s ~II H
, , ,, resulta11t dipole momelll
~v
ofHzO
vector addition diagram EPM of water
12
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
Polarity i:. an important concept because the polarity of a molec:ulc.can signi ficantly inlluen(\; , (~chem ical and physical properties. For example. a molecule\ polarity may give some indication of how it reaCis chemically. Returning to the HCI molecule. we know that HCJ in water dissociates to its ions in a manner suggested by its bond polarity.
t
H20
+
8<
I>-
H-CI
( 1.5)
We ' II find many similar examples in organic chem istry in which bond polarity provides a due to chemical reactivity. Bond polarity is also useful because it gives us some insight that we can apply to the concept of formal charge. It ':-. lmponunt to keep ii1 mind that formal char~is only a bQQkkce p~g deYi<.:c For keeping track of ch<\rge. In some cases. formal charge corresponds to the actual charge. For example, the actual negative charge on the hydroxide ion. - oH. is on the oxygen. because oxygen is much more electronegat ive than hydrogen. In this case. the locations of the formal charge and actual charge arc the same. But in other cases the rorma l charge does not correspond to the actual charge. F<1r ex~rnplc. in !he - B F4 ani6n, lfuorine i s lllllcli more electronegative th ~111 boron (Table 1.1 ). So. most of the charge should actually be situated on tht<' lluorine~ In fact. the ac!Ha! charges on the atoms of the retrafluoroborate ion are in accord with this intu ition:
F-o.:;:;
' -o
''F- -
i~f
os:;
l -11.55 actual charge on the tctralluoroboratc anion
...
Because the Lewis structure doesn 't provide a simple way of showing thi:-, distribution. we assign the charge to the boron by the formal-charge rules. An analogy m ight help. Let's say a big corporation arbitrarily chalks up all of its receipts to its sales department. This is a bookkeeping device. Everyone in the company knows that the receipts are in real ity due to a companywide effort. As long as no one forget~> the reality. the adm inistrative convenience of showing receipts in one place makes keeping track of the money a l ittle simpler. Thus. showing formal charge on single atoms makes our handling of Lewis structures much simpler. but applying wha t we kll(iv. ahout bond pbltU:ities helps U.\ to see where rhe Ch<:l rge really resides.
IQ,f,J:IiJcl~j
•••• • .....,.-
_ 1 7 Analyze the polari ty of each bond in the fo llowing organic compou nd. Which bond. other than the C-C bond, is the least polar one in the molecule? Which carbon has the most partial positive character?
H 0 I
II
H-C-C-CI
I
Cl 1.8 F'or which of the followi ng ions does the formal charge give a fai rly accurate picture of where the charge really is? Explain in each case. + (a) NH4
+
(b) H 30 :
(c)
-
=tfH2
+
(d) CHJ
1.3 STRU CTURES OF COVALE NT COMPOUNDS
13
1.9 Draw an appropriate bond dipole for the carbon- magnesium bond of dimethyl magnesium. Expl ain your reasoning.
dimethylmagnesium
1.3
STRUCTURES OF COVALENT COMPOUNDS We know the structure of a molecule containi ng covalent bonds when we know its ato111ic connectil'ity and its 111olemlar geometry. Atomic connectivity i~ the specil1cation of how atoms in a molecule are connected. For example. we specify the atomic connecti vity within the water molecule when we say that two hydrogens are bonded to an oxygen. Molecular geometry is the spec ification of how far apart the atoms are and how they are situated in space. Chem ists learned about atomic connectiv ity before they learn ed about molecul ar geometry. The concepl or covalent compounds as three-d i mensional objects emerged i n the latter part of the nineteenth century on the basi s o f indirect chemical and physica l evidence. Until the earl y part of the twentieth century. however. no one k new w hether these concepts had <111Y physical real ity. because scienti sts had no techniques for viewing m o l ecu le~ at the atomic level. B y the second decade o f the twentieth century. inves tigators could ask two que. tions: ( I ) D o organic molecules have , pecilic geometr ies and. if so. what are they? (2) H ow can molecul ar geometry be predic ted?
A. Methods for Determining Molecular Geometry A mong the greatest developments of chemical ph y~ ics i n the earl y twentieth century were the discoveri es of ways to deduce the stru cwres o f molecule . Such techniques include various types of spectroscopy and mass spectrometry. which we'll consider in Chapters 12- 15. As i mport ant as these techniques are. they are used primaril y to provide i nfor mation about atomic connecti vi ty. Other physical methods. however. permit the determination of molecular structures that arc complete in every detai l. M ost complete structures today come from three sources: X-ray crystallography. electron dif fraction. and microwave spectroscopy. The arrangement of atoms in the crystalline sol id stare can be determined by X-ray crystallography. This technique. inven ted in 1915 an,d subsequent ly revolutionized by the avai labi l ity of high-speed computers, uses the fact that X-rays are di ffracted from the atoms of a crystal in precise pallern s that can be mathematically deciphered Lo give a molecular structure. I n 1930. electroll dijji·actio11 wa!> developed. With this technique. the diffraction of electron!> by molecules of gaseous substances can be interpreted in term s of the arrangements of atoms in molecules. Following the development or rad ar in Worl d War II came microwm•e spec1roscopy. in which the absorp tion of microwave rad iation by mol ecules in the gas phase provides detailed structural i nformation. M ost of the spati al details of molecul ar structure i n this book are deri ved from gas-phase methods: electron di ffraction and microwave spectroscopy. For molecules that are not readily sllldied i n the gas phase. X-ray crys tallography is the most i mportant source of structural information. No methods of comparable prec ision exi st for molecules in solution. a fac t that i~ unfortunate because most chemical reactions take place in sol ution. The consistency of gasphase and crys tal structures su gge~ts . however, that molecular structures in sol ution probably differ li n le from those of molecules in the sol id or gaseous state.
14
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
B. Prediction of Molecular Geometry The way a molecule reacts is determined by the characteristics of its chemical bonds. The characteristics of the chemical bonds, in turn, are closely connected to molecular geometry. Molecular geomeuy is important. then. because it is a starting point for understanding chemical reactivity. Given the connectivity of a covalent molecule, what else do we need to describe its geometry? Let's start with a simple diatomic molecule, such as HCI. The structure of such a molecule is completely defined by the bond length, the distance between the centers of the bonded nuclei. Bond length is usual ly given in angstroms; I A = I 10 m = 10-8 em = I 00 pm (picometers). Thus, the structure of HCJ i completely specified by the H - CI bond length.
o-
1.274 A.
When a molecule ha more than two aroms. understand ing its structure requires knowledge of not only each bond length, but also each bond angle, the angle between each pair of bonds to the same atom. The structure of water (H 20) is completely determined, for example. when we know the 0 - H bond lengths and the H-0- H bond angle.
,
O/ """-bond length
/"~)
,
H '-...__.../ H bond angle We can general ize much of the information that has been gathered about molecular structure into a few principles that allow us to analyze trends in bond length and to predict approximate bond angles.
Bond Length The following three generalizations can be made about bond length. in decreasing order of importance.
I. Bond I~Jlgths i'flcrease sign(focamly toward /1igher p~riods {rows) of the periodic tablrf: This trend is illustrated in Fig. 1.2. For example. the H-S bond in hydrogen sulfide is longer than the other bonds to hydrogen in Fig. 1.2: sulfur is in the third period of the periodic table. whereas carbon. nitrogen. and oxygen are in the second per iod. Similarly. a C-H bond is sho11er than a C - F bond, which is shorter than a C-Cl bond. These effects all reflect atomic size. Because bond length is the distance between the centers of bonded atoms, larger atoms form longer bond~.
H
H""......... c ~
I
H
H
..
H 11""""/~rH H
.. /0~ ..f ~6'
H
H
/"' s
H methan e
am monia
water
H
hydrogen sulfide
Figure 1.2 Effect of atomic size on bond length. (Within each structure. all bonds to hydrogen are equivalent. Dashed bonds are behind the p lane of page, and wedged bonds are in front.) Compare the bond lengths in hy· drogen sulfide with those of the other molecules to see that bond lengths increase toward higher periods of the period ic table. Compare the bond lengths in methane, ammonia, and water to see that bond lengths decrease toward higher atomic number within a period (row) of the periodic table.
1.3 STRUCTURES OF COVALENT COMPOUNDS
H
\.c
'j
H
,H 1.536 A
H',.......
...········•
c,___,.-
H
\
\
I
1-1 U30A
ethane
I
C=C
H
H
H
15
\
acetylene
H
ethylene
Figure 1.3 Effect of bond order on bond length. As the carbon- carbon bond o rder increases, the bond length decreases.
2. Bond lengths decrease with increasing bond mder. Bond order describes the number of covalent bonds shm·cd by two atoms. ~or example. a C-C bond has a bond order of I. a C=C bond ha. a bond order of 2, and a C:=C bond has a bond order of 3. The decrease of bond length with increasing bond order is illustrated in Fig. 1.3. Notice that the bond lengths for carbon-carbon bonds are in the order C-C > C=C > c~c.
3.
Bond~
of a oivcm order dr;rrease. in lenglh ICJ.!mrd big/1q a/UnJic number (lhm is, to the Compare, for example, the H-C, H- N. and H- 0 bond lengths in Fig. 1.2. Likewise, the C-F bond in H 1C-F. at 1.39 A, is shorter than the bond in H3C-CH3. at 1.54 A. Because atoms.on the right of the periodic table in a given row are smaller, this trend. like that in item I. also results from differences in atomic size. However. this effect is much less significant than the differences in bond length observed when atoms of different periods ru·e compared. ri~f11) along a gil·en row (period) oL !he periodic !able.
c-c
Bond Angle The bond angles within a molecule determine its shape. For example. in the case of a triatomic molecule such as H 20 o r BeH 2, the bond angles determine w hether it is bent or linear. To predict approxi mate bond angles. we rely on valence-shell electron-pa ir rep ulsion theory , or VSEPR theory, which you may have encou ntered in general chemistry. According 10 VSEPR theory. both the bondi ng electron pairs and the unshared valence electron pairs have a spatial requirement. The fundamenral idea of VSEPR theory is that bonds and eleC!ron crre arranged abou/ a central a/om so !hC/1 the bonds are as far apart as possible. This arrangement minimizes repulsions between electrons in the bonds. Let's first apply YSEPR theory to three situations involving bonding electrons: a central atom bound to four, three, and two groups. respectively. When four groups arc bondt!d to a central atom. the bonds are larthest apart when the central atom has tetrahedral geometry. This means that the four bound groups lie at the vertices of a tetra hedron . A tetrahedron is a three-dimensional object with fou r triangular faces (Fig. 1.4a, p. 16). Methane, C H~ . has tetrahedral geometry. The central atom is the carbon and the four groups are the hydrogens. The C- H bonds of methane are as far apart as possible when the hydrogens lie at the vertices of a tetrahedron. Because the fou r·C-H bonds of methane are identical. the hydrogens lie at the vertices of a regular lelrahedron, a tetrahedron in which all edges are equal (Fig. 1.4b). The tetrahedral shape of methane requires a bond angle of I09.5° (Fig. 1.4c). fn applying VSEPR theory fo r the purpose of predicting bond angles. we regard all groups as identical , For example, the groups that surround carbon in C H3C I (methyl chloride) are treated as if they were identical, even though in reality the C-CI bond is considerably longer
16
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
H
H
~
H
H
H
H H
H
a regular tetrahedron (a)
(b )
methane
(c)
Figure 1.4 Tetrahedral geometry of methane.(a) A regular tetrahedron. (b) In methane, the carbon is in the middle of a tetrahedron and the four hydrogens lie at the vertices. (c) A ball-and-stick model of methane. The tetrahedral geometry requires a bond angle of 109.5°.
than the C-1-1 bonds. Although the bond angle show minor deviations from the exact tetrahedral bond angle of I 09.5°. methy l chloride in fact has the general tetrahedral shape. Because you' l l see tetrahedral geometry repea tedl y, it is worth the effort to become familiar with it. Tetrahedral carbons are often represented by line-and- wedge structures, as illustrated by the following structure of methylene chloride, CH 2C1 2•
11' \ in the plane of the page
behind the page
H· ....
/----..c~
H STUDY GUIDE LINK 1.2
Structure-Drawing conventions
t
in front of the page The carbon, the two chlorines. and the C- CI bonds are in the plane of the page. The C-CI bonds are represented by lines. One of the hydrogens is behind the page. The bond to th is hydrogen recedes beh ind the page from the carbon and is represemed by a dashed wedge. The remaining hydrogen is in fron t of the page. The bond to this hydrogen emerges from the page and is represented by a solid wedge. Several possible line-and-wedge structures are possible for any given molecule. For example. we could have drawn the hydrogens in the plane of the page and the chlorines in the out-of-plane positions. or we could have drawn one hydrogen and one chlorine in the plane and the other hydrogen and chlorine out-of-plane. A good way to become fam il iar with the tetrahedral shape (or any other aspect of molecular geometry) is to use m ol ecular m odel s, which are commerciall y available scale models from which you can cons truct simple organic molecules. Perhaps your instructor has required that you purchase a set of models or can recommend a set to you. A/mosr all beginning sw-
derlls require models. ar /easr inirially. 10 visualize rhe rhree-dimensional aspecrs of organic chemisny. Some of the types of models available are shown in Fig. 1.5. I n this text, we use ball-and-stick models (Fig. 1.5a) to visualize the directionality of chemical bonds. and we use space-fill ing model (Fig. 1.5c) to see the consequences of atomic and molecular volumes. You should obtain an inexpensive set of molecular models and use them frequently. Begin
usinf? them by building a model o.frhe merhy!ene chloride molecule discussed above and re!aring irro rhe line-and-wedge srructure.
1
{ a)
3 STRUCTURES OF COVALENT COMPOUNDS
(b )
17
( c)
Figure 1 5 Molecular models of methane. (a) Ball·and·stick models show the atoms as balls and the bonds as connecting sticks. Most inexpensive sets of student models are of this type. (b) A wire-frame model shows a nu· deus (in this case, carbon) and it s attached bonds. (c) Space-filling models depict atoms as spheres with radii pro· portional to their covalent or atomic radii. Space-filling models are particularly effective for showing the volume occupied by atoms or molecules.
Molecular Modeling by computer In recent years. scientists have used computers to depict molecular models. Computerized molecular modeling is particularly useful for very large molecules because building real molecular models in these cases can be prohibitively expensive in both time and money. The decreasing cost of computing power has made computerized molecular modeling increasingly more practical. Most of the models shown in this text were drawn to scale from the output of a molecular-modeling program on a desktop computer.
When t/IJ'(>p groups surround an atom. the bonds are as far apart as possible when all bonds 120°. Thi~ is. for example. the geometry of boron trilie in the same plane with bond anglt:~ fluoride:
or
F
F
boron trifluoride
In such a situation the surrounded atom (in this case boron) is said 10 have trigonal planar geometry. When an atom is surrounded by Mo group<;, maximum separation of the bond-; demands a bond angle of 180°. This i the situation\\ ith each carbon in acetylene. 1-l - C::::;::C-H. Each carbon b -.urrounded by two group'>: a hydrogen and another carbon. otice that the triple bond (a'> well as a double bond in other compounds) i. considered as one bond for purposes of YSEPR tht:ory. because all rhree bontl~ connt:ct the same two atoms. Atom~ with 180° bond angles arc ~aid to have linear geometry. Thus. acetylene is a fi11ear molct:ulc.
H
H
acetylene
18
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
ow let's consider how unshared valence electron pairs are treated by VSEPR theory. An
unshared l'alence electron pair is treared as if it were a bond \t'itlwut a nucleu~ at one end. For example. in VSEPR theory. the nitrogen in ammonia. :N H 3. is surroundeo by four "bonds": three N- 1-1 bonds and the unshared valence electron pair. These ''bonds" are directed to the vertices of a tetrahedron so that the hydrogens occupy three of the four tetrah edral vertices. This geometry is called trigonal pyramidal becau!-e the three N-H bond~ al~o lie along the edges of a pyramid.
VSEPR theory also po tulates that unshared l'alence electron pairs occupy more space than an ordinary bond. It' as if the electron pair ''spreads out'' because it il>n 't con trained by a second nucleus. As a resuh. the bond angle between the un~harcd pair and the other bonds are somewhat larger than tetrahedral. and the r - H bond angles are corrc~pondi ngly smaller. In fact. the H- N - H bond angle in ammonia is 107.3° . an unshared electron pair '~-==----l occupies more space than
a bonding electron pair
H
11
ammonia
Estimate each bond angle in the following molecule. and order the bonds according to length. beginning with 1he shortest. 5
0 !
I
J
II
6
H-C===C-C-CI ,,
~
d
Solution Because carbon-2 is bound to two groups (H and C). its geometry is linear. Similarly. carbon-3 a lso has linear geometry. The remaining carbon (carbon-4) is bound to three groups (C, 0, and Cl); therefore. it has approximately trigonal planar geometry. To arrange the bonds ih order of length. recall the order of importance of tl1e bond-length rules. The major influence on length i!> the row in the periodic table from which the bonded atoms arc taken. Hence, the H-C bond is shorter than all carbon-<:arbon or carbon-oxygen bonds, which are shorter than the C-CI bond. The next major effect is the bond order. Hence, the C=C bond is shorter than the C=O bond. which is shorter than the C-C bond. Putting these conclusion~ together. therequired order or bond lengths is (a)< (b) <(e) <(c) <(d)
14ii.1:10Wtl •••• • •••• ••-
J.lo
Predict the approximate geometry in each of the following molecules. ldl HJC-C=N : (a) water lbl [BF,J- (c) H2C=O formaldehyde
acetonitrile
1.3 STRUCTURES OF COVALENT COMPOUNDS
1.11
19
Es1ima1e each of the bond angles and order !he bond lcng1hs (smalle'l fir.a) for each of the following molecules. For molecule (b). slalc any poim~ of ambigui1y and explain. I h) (a) :Q: H 1-1
I ~
H
C-C-H
\ C=C • /t
;, H
/H
<\
""' H
\/H s· ..
\ /H
H-e
\
;•-cl= C=C
'/
I
H
=~.r:
\
:CI:
Dihedral Angles To completely describe the shapes of' molecules that are more complex than the ones we've just discussed. we need to specify not only the bond lengths and bond angles. but also the spatial relationship of the bonds 011 adjac<~111 atoms. To illustrate this problem. con~ider the molecule hydrogen peroxide. H-g~·g-H. Both 0 - 0-H bond angles are 96.5°. However. knowledge of these bond angles is not sufficient to describe completely the shape or the hydrogen peroxide molecule. To understand why. imagine two planes, each contain ing one of the oxygen~ and its two bonds (Fig. 1.6). To completely describe the structure or hydrogen peroxide. we need to know the angle between these two planes. This angle is called the dihedral angl e or tor sion angl e. Three possibilities for the dihedral angle are shown in Fig. 1.6. You can also' i\ualit.e these dihedral angle u ing a model of hydrogen peroxide by holding one 0 - H bond fixed and rotating the remaining oxygen and its bonded hydrogen about the 0 -0 bond. (The actual dihedral angle in hydrogen peroxide is addressed in Problem I .43. p. 44.) Molecules containing many bonds typically contain many dihedral angles to be ~pecified. We'll begin to learn !,Ome of the principles that allo'" U'- to predict dihedral angles in Chapter 2. Let's summarit.e: The geometry of a molecule i-, completely determined by three elements: it~ bond lengths. its bond angles. and its dihedral angles. The geometries of diatomic molecules arc completely determined by their bond length~. The geometries of molecules in which a central atom is surrounded by two or more other atom!. arc dctem1ined by both bond lengths and bond angle!.. Bond lengths. bond angles. and dihedral angles arc required to specify the geometry of more complex molecules.
r
r•IIJI<' unl' pl.mc
~llolh
dihedral angle
=0°
dihedral angle = 90°
dihedral angle = 180°
Figure 1.6 The concept of dihedral angle illustrated for 1he hydrogen peroxide molecule, H-0-0-H. Knowledge of !he bond angles does not define the dihedral angle. Three possibililies for the dihedral angle (o•. 90•, and 180°) are shown.
20
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
14;i.J:IO~~~~
•••• • ••••••-
. 1 12 The dihedral angle~ in ethane. H,C-CH,. relate the plane-. conHtining the H-C-C bonds centered on the two carbons. Prepare models of ethane in which thi'> dihedral angle is (a) 0°: (b) 60°: (c) 180°. Two of thei>e models are identical; explain.
1.13 (a) Give the H-C=O bond angle in methyl fom1ate. :Q:
II
..
H-C-
(b) One dihedral angle in methyl fonnate relates the plane containing the O=C-0 bonds to the plane containing the C-0-C bonds. Sketch two structures of methyl fonnate: one in which this dihedral angle is 0° ~mel the other in which it is 180°.
RESONANCE STRUCTURES Some compounds are not accurately described by a <;inglc Lcwi~ :-.tructure. Consider. for example. the structure of nitromethane. H 1C- 0~.
nitromethane Thb Lewi\ ... tructure ~hows an l -0 single bond and an = 0 double bond. From the preceding <.,ection. we e.xpect double bondc, to be '>honer than -,ingle bond'>. Howe\ cr. it is found experimentally that the two nitrogen-oxygen bond-. of nitromcthanc have the :-.ame length. and thi:-. length i~ intermediate between the lengths of <>inglc and double nitrogen- oxygen bonds found in other molecules. We can convey this idea by writing the \lructure of nitromethane a!> follows:
( 1.6)
The double-headed arTow (~)means that nitromcthane is a single compound that is the ··average'' of both structures: nitromethane is said to be a r esonance hybr·id of these two -;tructure~. ote carefully that the double-headed arrow ill different from the arrows u\ed in chemical equilibria. . The two ~o,tructurcl> for nitrornethane are nor rapidl y interconverting and they are nor in equilibrium. Rather. they arc alternative representations of one molecule. In thi!l text. resonance structures \\ill be enclo~o,cd in bracl--ets to emphasize this point. Resonance !>tructures are necessary because of the inadcqUent nitromethane accurately. The two re!>onance Mructures in Eq. 1.6 arc jic1irious. but nitromethane is a real molecule. Becau~e we have no way to describe nitromethane accurate I) \\ ith a '>ingle Lewis Mrucrure. we rnu-.t de!'>cribe it as the hybrid of t\\O fictitiou<., '-tructurc\. An analogy to thi~ situation is a de'>cription of Fred Flarfoot. a real detective. Lacking word-. to de.,cribe Fred. we picture him as a re~onance hybrid of two jic1ional characters: Fred Flatfoot
= [Sherl ock Holme., -
Jame!> Bond I
1 .4 RESONANCE STRUCTURES
21
This sugge~ts that Fred is a da~hing, violin-playing. pipe-smoking. highly intelligem British agent with an assistant named Watson. and that Fred likes hi ~ martinis shaken. not stirred. When two re-.onance structure" arc identical. a-. thcy arc for nitrornethane. they are equall) important in dc-,cribing the molecule. We can think nitromcthane a-. a I: I average of the '-lructures in Eq. 1.6. For example. each oxygen bear-, half a negative charge. and each nitrogen-oxygen bond b neither a single bond nor a double bond. but a bond halfway in between. If two rei>onancc ~ tructures arc not identical. then the molecule they represent is a ll'eigllled a1•erage of' the two. That is. one the structures is more important than the other in describing the molecule. Su<.:h is the <.:asc. for example, with the mcthoxymcthy l <.:ation:
or
or
[H/:-Q-C11 3
•
•
H~C={j-CH 3 ]
( 1.7)
methoxymethyl cation It tllrrh out that the :,tructure on the ri ght is a better dc~c ripti on of this cation because al l atom:, have complete octets. Hence. the C - 0 bond has significant double-bond character. and most of the formal po-.itivc charge reside-. on the oxygen. Formal charge has the same limitation' as a bool..keeping de\ ice in resonance 'tructure\ that it t.lo.:' in other \tmcture~. Although 1110\l of thc.fomw/ charge 111 the mcthoxymeth) I cation reside-, on 0"\.ygen. the ('II , ~arbon bear~ more of I he acTual po'-oili\C charge. hl:c;m~c oxygen i' more clcclroncgative than carhon. The hybrid character of some molecules is sometimes conveyed in a ~ing l e structure in which dw;hed lines are used to rcprc~cnt partial bond~. For example. nitromethane can be reprc~cnted in thi-. notation in either of the following way-.:
In the hybrid '> tructure on the left. the locations of the ~hared negative charge arc shown e'<.plicitly with partial charge!'.. ln the h) brid structure on the right. the location<> of the shared negati\e charge arc not ~hown. Although the use of h) brid \tructure!> is '>ometimes convenient. it is difficult to apply electron-counting rules and the curved-arrow notation w them. To avoid confusion, wc' lluse conventional resonance s tructure~ in these s ituati on~. Ano ther very important aspect resonance structures j.., that they have implications for the stability of the molecule they reprc'>cnt. A molecule represented by resonance structures is more swhle tlwn it.\ fictional re.Hmann• contrihwor.\ . For example. the actual molecule nitromethanc is more stable than either one of the lictional molecules de~cribed by the contributing re~ouancc '>tntctures in Eq. 1.6. itromethane i' tllll'> '>aid to be a rewmance-stabili:ed molecule. as i!- the methoxymethyl cation. How do we know when to u~c rc~onance structures, how to draw them. or how to assess their relati ve importance? In Chapter 3, we'll leam a technique for deri ving resonance structures. and in C hapter 15. we'll return to a more detailed study of the other a..,pects of resonance. I n the meantime. we'll dr:l\~ rc'>onance structure:, for you and tell you when they're important. Ju~t tr) to remember the following point\:
or
1. Re~onancc '>tructures are u~cd for compounds that arc not adequate!y described by a ~in glc Lcwi~ structure. 2. Resonance structure:- arc not in equ ilibrium : that i ~. the compound they describe is not one rc..,onancc ~truc ture part o f the time and the other resonance structure part of the time. but rather a single '>tructure.
22
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
3. The structure of a molecule is the weighted average of its resonance structures. When resonance structures are identical. they arc equally important descriptions of the molecule. 4. Re onance hybrid are more <.table than any of the fictional structures uc;ed to describe them. Molecules described by re~onance stmciUre-. are c;aid to be resonance-stabili::.ed.
1. 14 (a) Draw a resonance structure for the allyl anion that shows, along with the following srrucLUre, that the two CH2 carbons are equivalent and indistinguishable. [ H 2C=CH-CH 2 -
]
allyl anion
lb) According to the resonance !.tructures. how much negative charge is on each of the CH2 carbons? (c) Draw a single hybrid structure for the allyl anion that shows hared bonds as dashed lines and charges as partial charges. 1.15 The compound ben::.ene has only one type of carbon-carbon bond, and this bond has a length intermediate between that of a single bond and a double bond. Draw a resonam.-e structure of benzene that, taken with the following structure. account:. for the carbon-carbon bond length.
benzene
WAVE NATURE OF THE ELECTRON You've learned that the covaJent chemical bond can be viewed as the sharing of one or more electron pair between two atom~. Although this implc model of the chemical bond is very useful, in some l.ituations it i~ inadequate. A deeper in),ight into the nature of the chemical bond can be obtained from an area of l.Cience called qualllwn mechanics. Quantum mechanics deals in detail with. among other things. the behavior of electrons in at01m and molecules. Although the theory involves some l.Ophisticated mathematic!.. we need not explore the mathematical detail to appreciate some general conclusions of the theory. The starting point for quantum mechanic!. i'> the idea that small particles such a .1 electrons also hare the character ofwm•es. How did thil. idea evolve? As the twentieth century opened. it became clear that certain things about the behavior of electrons could not be explained by conventional theori es. There seemed to be no doubt that the electron was a particle; after all. both its charge and ma~~ had been measured. However. electrons could also be diffracted like light. and diffraction phenomena were a~~ociated with waves. not panicles. The traditional views of the physical world treated particle<> and wave a~ unrelated phenomena. In the mid-1920 . thi!'. mode of thin'-ing was changed by the advent of quantum mechanic'>. This theory holds that. in the submicroscopic world of the electron and other small particles, there is no real distinction between particles and waves. The behavior of small panicles !.Ltch as the electron can be described by the physics of wave. . In other words, maller can be regarded as a wave-particle duality.
1 .6 ELECTRONIC STRUCTURE OF THE HYDROGEN ATOM
23
How doe~ this wave-particle duality require us to aher our thinking about the electron? In our everyday lives. we're accu~tomcd to a deterministic world. That i~. the po!>ition of any familiar object can be measured preci'icly. and its velocity can be determined. for all practical purposes. to any desired degree of accuracy. For example. we can point to a baseball resting on a table and o,tatc \\ ith confidence. ·That ball is at rest (it!. velocity is zero). and it i~ located exactly I foot from the edge of the table:· othing in our experience indicates that we couldn't make similar mca~urement'> for an electron. The problem is that humans. chemi. try books. and baseballs arc of a certain scale. Electrons and other tiny objects are of a much smaller scale. A ccnu·al principle of quantum mechanics. the Heisenber g uncertainty principl e, tells us that
the accuracy ll'ith ll'hich 11·e can determine the position and l'elocity of a panicle is inherently limited. For "large·· object~. !>uch a-. basketballs. organic chemistry textbooks. and even molecules. the uncertainty in po. ition i'> 'mall rclati\'e to the o,i7e of the object and is inconsequential. But for very small objects such as electron!>. the uncertainty is ignificant. As a result. the position of an electron becomes '·futt.y.'· According to the Heisenberg unccr1aint y principle, we are limited to stating the probability tha t an electron is occupying a certain region of space. ln summary: I. ElectrOn'> have wavelike properties. 2. The exact po~ition of an electron cannot be '>pecified: only the probability that it occupies a certain region of '>pace can be pecified.
ELECTRONIC STRUCTURE OF THE HYDROGEN ATOM To understand the implication~ of quanlllm theory for covalent bonding. we must first understand what the theory ays about the electronic structure of atoms. Thb -,cction pre ents the applications of quantum theory to the simplest atom. hydrogen. We deal with the hydrogen atom because a very detailed description of its electronic structure has been developed. and because this dc~cripti on has direct applicability to more complex atoms.
A. Orbitals, Quantum Numbers, and Energy In an earlier model of the hydrogen atom. the electron wa~ thought to circle the nucleus in a well-defined orbit. much as the earth circles the sun. Quantum theory replaced the orbit with the orbiwl. which. despite the similar name. is something quite different. An atomic orbital is a description of the wave properties of an electron in an atom. We can think of an atom ic orbital of hydrogen as an al!oiVed slate- that is. an allowed wave motion-of an electron in the hydrogen atom. An atomic orbital in physics i described by a mathematical function called a wavefunction. As an analogy. you might de-,cribe a ine wave by the function 1/J = sin x. This is a simple wave function that covers one spatial dimension. Wavefunctions for an electron in an atom arc conceptually sim ilar. except that the wavefunctions cover three spatial dimensions. and the mathematical functions are different. Many possible orbital s. or state!.. are available to the electron in the hydrogen atom. This means that the wave propertie
24
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
number-. have mathematical significance in the wave equation<; of the electron. for U\ they serve a!> labeh. or designators. for the variou~ orbi tals. or wave motion-;. an\ilable to the electron. The~c quantum numbers can have only certain values. and the value-. of -.ome quantum numbers depend on the values of other<.. The principal quantum number. abbreviated 11. can have any integral value greater than zero-that is. 11 = I. 2. 3.... The angular momentum quantum number. abbreviated /.depends on the value of 11. The I quantum number can have any integral value from zero through 11 I. that i'>. I = 0. I. I. So that the) are not confu-.ed \\ ith the principal quantum number. the values of 2..... 11 I are encoded a'> letter!>. To I = 0 is tt!>'>igned the letrer s: to I = I. the letter p: to I = 2. the letter d: and to I = 3.the letter/ The 'alue., of I are ummarized in Table 1.2. It follow~ that there can be only one orbital. or wavefunction. with 11 = I: thi~ i~ the orbital with I = 0-a Is orbital. llowever. two values of I. that i~. 0 and I. are allowed for 11 = 2. Con~equcntl y. an electron in the hydrogen atom can exist in either n 2,1 or a 2p orbital. The magnetic quantum number, abbreviated m1• is the third orbital quantum number. Its values depend on the value of l. The m1 quantum number can be 1cro as well a~ both positive and nega~ive integer., up to ±!-that is. 0. ± I. ± 2..... ± 1. Thu-.. for I = 0 (an s orbital). 1111 can on I) he 0. For I = I (ap orbital). m1 can ha'e the \alue~ -I. 0. and + I. In other word!>, there is one ~orbi tal "ith a giYen principal quantum number. but (for 11 > I} there are three p orbitab with a given principal quanwm number. one corre~ponding to each ,·alue of 1111• Becau~c of the multiple po-;sibilities for I and 1111• the numher of orbitals become'> incrl.!a.,ingl) large :b 11 increa<;es. Thb poilll b illu~trated in Table 1.2 up to 11 = 3. Ju!>t :.t\ an l.!lcctron in the hydrogen atom can exist only in certain ~tate~ . or orbitals. it can also ha'e only certain allowed energic\. Each orbital is associated with a characterist ic electron energy. The energy of WI electron in o hydrogen atom is determined l1y 1he principol quantum llltlllher 11 t?f i1s orbiwl. This i~ one of the central ideas of quantum theory. The energy of the electron i~ ~aid to be quanri-;.ed. or limited to certain values. This feature of the atomic electron is a direct consequence of ih wave propenie!.. An ch.:ctron in the hydrogen atom re.,ide~ in an orbital "ith n = I (a I.\ orbital) and remain., in that '>tate unle..,, the atom i!. ~ubjected to the exact amount of encrg) {sa). from light) required to incrca-,e the energy of the electron to a '>late with a higher 11. sa) 11 = 2:
radi.Hion wi1h cncrg) t.F
II =
I
II ::::
I
Tf that h appen~. the electron absorbs energy and instantaneously assume., the new. more energetic. wave motion characteristic of the orbital with 11 = 2. (Such energy-absorption experiments gave the fir~t clue~ to the quanti led nature of the atom.) An analogy to thi), may be familiar. If you have eYer blown across the opening of a ~oda-pop boule (or a Outc. which is a more -;ophisticatcd example of the ame thing). you know that only a certain pitch can be produced
1.6 ELECTRONIC STRUCTURE OF THE HYDROGEN ATOM
lt!j:j!Jfl n 0 (ls)
25
Relationship Among the Three Orbital Quantum Numbers
m,
n
0
2
m,
n
0 (2s)
0
3
1 (2p)
- 1
m, 0 {3sl
0
1 (3p)
- 1
0
0
+1
+1 2(3d)
- 2 - 1 0 +1 +2
by a bottle of a given ~ i Lc. If you blow harder. the pitch doc~ not ri se. but on ly becomes louder. However. i~ you blow hard enough. the ~ound ~uddenly jumps to a note of higher pitch. The pitch i ~ quallliLcd: only cenain sound frequencie!-> (pi lche~) an.: allowed. Such phenomena are ob~crved becau!-e ~ound i a wm·e motion of the air in the boule, and only certain pile he can cxi'-t in a cavi ty of given dimension~ wi thout canceling thcrmclve.., out. The progressively higher pitches you hear as you blow harder (c.alled m·erw,u•s of the lowest pitch) are analogous to the progressive!) higher energ) state~ (orbital-.) of the electron in the atom. Just a each overtone in the boule i~ described b) a wa,efunction "ith higher "quantum number:· each orbital of higher energy i~ de. cribed by a wavefunction of higher principal quantum number 11.
B. Spatial Characteristics of Orbitals One of the mo~t imponant aspects of atomic ~tructurc for organic chemi~try is that each or-
hiwl is characteri~ed by a three-dimemional region t!f" \f){ICl' in which the electron is most likely fO exist. That is. orbitals have spatial charactcrisric<>. The .1i::e of an orbital is governed mainl y by it~ principal quanlum number n: the larger 11 i~. the greater !he regio n or space oc-
Further ExplOration 1.2
Electron Density Distribution In Orbitals
cupied by the corresponding orbital. The shape of an orbiral i~ governed by its angular momenlum quanltlm number f. The directionality of an orbital i ~ governed by il~ magnelic quantum number 1111• These points are best illustrated by example. When an electron occupies a Is orbital , it is most likely to be found in a sphere surrounding the atomic nucleus (Fig. 1.7. p.26) . We cannot say exactly ll"here in thm stJhere the elec1ron is by the uncertainty principle: localing the electron i~ a ma11er of probabil ity. The mathematic!' of quamum theory indicates that the probabi lily is about 90(/~' that an electron in a Is orbilal wil l be found w ithin a sphere of' radiu-> 1.4 A about the nucleu-.. This "90'n probability level·· is taken a'5 1he approx imate size of an orbital. Thu!-. we can depic1 an electron in a Is orbital as a smear of electron demi(r. most of which i!> within I A A of the nucleu .... Becau\e orbital' are aciUall~ mathematical function' of three 'Patial dimen.,ion~. it would take a founh dimension to plot the value of the orbital (or the electron probabtlit} J at each point in space. hO\\ each orbital a'> a geometric figure that enclo-.e' 'ome fntction (in our ca\c. 90C'() of the electron probabilit). The detailed quantitative distribution of electron probabilit) \\ithin each figure i!> not show·n. When an electron occupies a 2s orbital. il al ~o lie-. in a -.phere. but the sphere i. considerably larger- about three times the radiu-. of the 1.\ orbital (Fig. 1.8. p. 16). A 3s orbital is even
26
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
z
- - - - - - - .\
Figure 1.7 A 1s orbital. Most (90%) of the electron density lies in a sphere within 1.4 A of the nucleus.
1.1 out~r ball
A-
z
of elntnm d~n'''' ,,,1\l
)'
trough .........
/
,sphcncal nod~
/
annab.tll
of cl" tron
nc,Jk
Figure 1.8 A 2s orbital in a cutaway view, showing the positive (peak-containing) region of the electron wave in blue and the negative (trough-containing) region in green. Th is orbital can be described as two concentric spheres of electron density. A 2s orbital is considerably larger than a 1s orbital; most (90%) of the electron density of a 2s orbital lies within 3.9 A of the nucleus.
larger !>till. The iLe of the orbital reflect!> the fact that the electron has greater energy; a more energetic electron can escape the auraction of the po-.itive nudeu-; to a greater extent. The 1s orbital al o illustrates a node. another \ery •mportam ~patial aspect of orbital . You may be familiar with a simple wave motion. -;uch a'> the wa,·e in a vibrating ~tring. or wave!> in a pool of water. lf so. you know that waves have peaks and trough~. regions where the waves arc at their maximum and minimum heights. respective!). As you knov. from trigonometry. a ~imple ~inc wave 1/1 = sin x has a positive ~ign at it'> peal-.. and a negative ~ign at it trough (Fig. 1.9). Becau~e the wa'e is continuous. it has to have a ;ero \aluc somewhere in berween the peal-.. and the trough. A node is a point or. in a three-dimensional ,-.ave. a swface. at which the wave is zero.
1.6 ELECTRONIC STRUCTURE OF THE HYDROGEN ATOM
'< ~
·;;;; II
+
~
27
peak
~
.2u
0
t::
..2 t.J >
;e "'
Figure 1 .9 An ordinary sine wave (a plot of ~{1 = sin x) showing peaks, troughs, and nodes. A peak occurs in the region in which is positive, and a trough occurs in a region in which is negative. The nodes are points at wh ich
w
w
1/1= 0.
As you can see from Fig. 1.8 and the subsequent figures, the peaks and troughs are colorcoded: peaks are blue and troughs are green. When the nodal properties of orbitals aren't important for the discussion, the orbita ls will be colored grey. Pay very careful attention to one point of potential confusion. The sign of the wavefunction for an electron is not the same as the charge on the electron. Electrons always bear a negative charge. The sign of the wavefunction refers to the sign of the mathematical expression that describes the wave. By convention. a wave peak (color-coded blue) has a positive (+)sign and a wave trough (color-coded green) has a negative (-) sign.
As shown in Fig. 1.8. the 2s orbital has one node. This node separates a wave peak near the nucleus from a wave trough further out. Because the 2s orbital is a three-dimensional wave. its node is a swface. The nodal surface in the 2s orbital is an infinitely thin sphere. Thus. the 2s orbital has the characteristics of two concentric balls of eleftron density. In the 2s orbital, the wave peak corresponds to a positive value in the 2s wavefunction, and the wave trough conesponcls to a negative value. The node- the spherical shell of zero e lectron density- lies between the peak and the trough. Some students ask. "'If the electron cannot exist at the node. how does it cross the node?" The answer is that the electron is a wave, and the node is pan of its wave motion. just as the node is part of the wave in a vibrati ng string. The electron is not analogous to the string; it is analogous to the wave in the string. We turn next to the 2p o rbitals (Fig. 1. 10. p. 28), w hich are especially important in organic chemistry. The 2p orbital illustrates how the I quantum number governs the shape of an orbital. All s orbitals are spheres. In contra<;l. a ll p orbitals have dumbbell shapes and are directed in space (that is, they lie along a particular axis). One lobe of the 2p orbital corresponds to a wave peak, and the other to a wave trough: the electron density. or probability of fi ndi ng the electron. is identical in con·esponding parts of each lobe. Note that the two lobes are parts of the same orbital. The node in the 2p orbital, which passes through the nucleus and separates the two lobes. is a plane. T he s ize of the 2p orbita l. like that of other orbitals. is governed by its principal quan tum number: it extends abou t the same distance from the nucleus as a 2s orbitaJ. Fig. 1. JOb illustrates a drawing convention for 2p orbitals. Quite often the lobes of these orbi tals are drawn in a less rounded, "teardrop" shape. (This shape is derived from the square of the wavefunction. which is proportional to the actual electron density.) This shape is usefu l becau e it emphasizes the directionality of the 2p orbital. This convention is so commonly adopted that we ·11 often use it in this text.
28
CHAPTER 1 • CHEMICAL BONDING A ND CHEMICAL STRUCTURE
z
_1__________ _ 4.2A
_j_______~ ---~
(a }
(b)
(c)
Figure 1.10 (a} A 2p orbital. Notice the planar node that separates the orbital into two lobes. Most (90%} of the electron density lies within 4.2 A of the nucleus. (b) A widely used drawing style for the representation of 2p orbitals. (c) The three 2p orbitals shown together. Each orbital has a different value of the quantum number m1•
Recall (Table 1.2) that there are three 2p orbitals, one for each allowed val ue of the quanlllm number m1• The three 2p orbitals i llustrate how the 1111 quantum number govern~ the direclionality of orbitals. The axes along which each of the 2p orbitals ··points·· are mutuall y perpendicular. For this reason. the three 2p orbitals are sometimes di fferentialed with the labels 2Pr· 2p,. and 2P:· The three 2p orbitals are shown superimposed in Fig. l.lOc. Let's examine one more atomic orbital, the 3p orbital (Fig. 1.1 I ). First. notice the greater si ze of this orbital, which is a consequence of its greater principal quantum number. The 90o/r probability level for this orbital is al most I 0 A from the nucleus. Next. notice the shape of the 3p orbital. ll is generally lobe-shaped. and it consists of four regions ~eparated by node:.. The two inner regions resemble the lobes of a 2p orbital. The outer regions. however. are large and diffuse. and resemble mushroom cap. . Finally. notice the number and the character of the nodes. A 3p orbital contains two nodes. One node is a plane through the nucleus. much like the node of a 2p orbital. The other is a spherical node, shown in Fig. 1.1 1b, that separates the inner part of each lobe from the larger outer part. A11 orbitalll'ith principal quantum number n has n - I nodes. Because the 3p orbital has n = 3, it has (3 - I ) = 2 nodes. The greater number of node!. in orbitals with higher n i s a reflection of their higher energies. Again, the analogy to sound waves is striking: overtones of higher pitch have larger number~ of nodes.
c.
summary: Atomic Orbitals of Hydrogen Here are the important poims about orbitals in the hydrogen atom: I. An orbital is an allowed state for the electron. It is a desciiption of the wave motion of the electron. The mathematical description of an orbital is called a wcll·e_fcmction. 2. Electron densi ty within an orbital is a matter of probabil ity. by the Heisenberg uncertainty principle. We can think of an orbi tal as a "smear.. of electron density. 3. Orbitals are described by three quantum numbers: a. The principal quantum number n governs the energy of an orbital: orbi tals of higher n have higher energy. b. The angular momentum quamum number I governs the shape of an orbital. Orbitals with I = 0 (s orbitaJs) are spheres: orbitals wi th I = I (p orbitals) have lobes oriented along an axis. c. The magnetic quantum number m1 governs the orientation of an orbi tal.
29
1. 7 ELECTRONIC STRUCTURES OF M ORE COMPLEX ATO MS
z
/
z
plan.tr nnt!c
X
X
'Phai(al node
(a )
(b)
Figure 1.11 (a) Perspec tive representation of a 3p o rbi tal; o nly the p lanar nod e is shown. There are three such o rb ital s, and they are mut ually perpendicular. Notice t hat a 3p orbital is mu ch larger tha n a 2p orb ital in Fig 1.10. Most (90%) of t he electron density lies w ithin 9.75 A of the nucleus; about 60% o f t he electron density lies in the large outer lo bes. (b) Schematic p lanar representation of a 3p orbital showing both the planar and the spherical nodes.
4. Orbitals with 11 > I contai n nodes. which arc surfaces of zero electron density. The nodes separate peaks o f electron density from troughs. or. equivalentl y. region in which the wavefunction describing an orbi tal has opposite sign. Orbi tals with principal quantum number 11 have 11 - I node!.. 5. Orbital size increases with increasing 11.
t/1 = sin 1u for the domain 0 :::; x:::; 7T for n = 1, 2. and 3. W hat is the relationship between Lhe " quantum number" n and Lbe number of nodes in the wavefunction?
1.1 6 Sketch a plot of tbe wavefunction
1. 17 Use the trend s in orbi ~al shapes you've just learned to descri be the general features of (a) a 3s orbit al (b) a 4s orb ital
1 .7
ELECTRONIC STRUCTURES OF MORE COMPLEX ATOMS The orbitals avai lable to electrons in atoms wi th atomic number greater than I arc. to a useful approx i mation. essentially like those of the hydrogen atom. T his similarit y includes the shapes and nodal properties of the orbitals. There is. however, one importan t di f ference: In atoms other than hydrogen, electrons w ith the same principal quantum number n but w ith different values or I have di fferent energies. For example. carbon and oxygen, like hydrogen, have 2s and 2p orbitals, but. un like hydrogen. electrons in Lhese orbitals di ffer in energy. T he ordering of energy level s for atoms with more than one electron is illustrated schematically in Fig. 1.1 2. p. 30. As this fi gure shows. the gaps between energy levels become progressively
30
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
3d 3p --s----3s
2p
2s
ls -------------- --- ---~I I
'
Figure 1.12 The relative energies of different orbitals illustrated for the first three principal quantum numbers of the carbon atom. The ls orbital energy on this scale is almost five page lengths below the 2s energy1 The energy separations differ for different atoms. The 4s orbital, not shown, has about the same energy as the 3d orbital, and the 4s energy drops below the 3d energy for atoms of higher atomic number.
smaller as the principal quantum number increases. Furrhermore, the energy gap between orbitals that differ in principal quanrum number is greater than the gap between two orbi tals within the same principal quantum level. Thus. the difference in energy between 2s and 3s orbitals is greater than the difference in energy between 3s and 3p orbitals. Atoms beyond hydrogen have more than one electron. Let's now consider the el ectronic configurations of these atoms-that is. the way their electrons are distributed among their atomic orbitals. To describe electronic configurations we need to introduce the concept of el ectron spin, which is a magnetic property of the elecLron. An electron can have only two values of spin. sometimes described as "up" and ''down." Spin is characterized by a fourth quantum number ms. w hich. in quantum theory. can have the values ("up'') and ("down''). Four quantum numbers. then, are associated with any electron in an atom: the three and ll1e spin quantum number ms. orbital quantum numbers n, I. and The aufbau principle (German. meaning ''bui ldup principle'') tells us how to determine electJonic configurat ions. According to this principle. electrons are placed one by one into orbitals of the lowest possible energy in a manner consistent with the Pauli exclusion principle and Hund 's rules. The Pauli exclusion principle states that no two electrons may have all four quantum numbers the same. As a consequence of this principle, a maximum of two electrons may be placed in any one orbital, and these electrons must have different spins. To illustrate. consider the electronic configuration of the helium aLom. which contains two electrons. Both electrons can be placed into the Is orbital as long as they have differing spin. Consequently, we can write the electronic configuration of helium as follows:
+1
-3
'"f.
helium, He: ( l s) 2 This notation means that helium has two electrons of differing spin in a Is orbital. To illustrate Hund·s rules, consider the electron ic configuration of carbon, obviously a very important element in organ ic chemistry. A carbon atom has six electrons. The first two electrons (with opposite spins) go into the Is orbital; the next two (also with opposite spins) go into
1.7 ELECTRONIC STRUCTURES OF MORE COMPLEX ATOMS
31
the 2s orbital. Hund's rules tell us how to distribute the remaining two electrons among the three equivalent 2r orbitals. Hund 's rules state, first. that to distribute electrons among identical orbitals of equal energy. single electrons are placed into separate orbitals before the orbitals are filled: and second. that the spins of these unpaired electrons are the same. Representing electrons as ruTows. and letting their relative directions correspond to their relative spins. we can show the electronic configuration of carbon as follows:
2p
1
-r-
- I,-
-)
,,1kn<.:..:
2s
ls
,, I •
.:b. tro n~
*
In accordance with Hund's rules, the electrons in the carbon 2p orbitals are unpaired wi th identical spin. Placing two electrons in different 2p orbitals ensures that repulsions between electrons ru·c minimized, because electrons in different 2p orbitals occupy different regions of space. (Recall from Fig. l .l Oc that the three 2p orbitals ru·e murually perpendicu lar.) A s shown above. we can aJ o write the electronic configuration of carbon more concisely as ( I s) 2(2s) 2(2p,) 1(2p,}' . which shows the two 2p electrons in different orbitals. (The choice of x and y as subscripts is arbitrary; 2p, and 2pz, or other combinations. are equally valid: the important point about this notation is that the two half-popu lated 2p orbitals are different. ) Let's now re-define the term 1•alence electrons. first defined in Sec. 1.2A, in light of what we've learned about quantum theory. The valence electrons of an atom are the electrons that occupy the orbitals with the highest principal quantum number. (Note carefully that this definition applies only to elements in the "A .. groups- that is. the nontransition groups--of the periodic table.) For example, the 2s and 2p electrons of carbon are its valence electrons. A neutral carbon atom, therefore, has four valence electrons. The valence orbitals of an atom ru·e the orbitals that contain the valence electrons. Thus, the 2s and 2p orbitals are the valence orbitals of cru·bon . It is important to be able to identify the valence electrons of common atoms because chemical intemctions be/1Veen atoms invo!l•e their valence electrons and valence orbitals.
study Problem 1.
II
Describe the electronic configuration of the sulfur atom. Identify the valence electrons and valence orbitals.
Solution Because sulfur has an atomic number of 16. a neutral sulfur atom has 16 electrons. Following the autbau principle, the first two electrons occupy the Is orbital with opposite spins. The next two, again with opposite spins, occupy the 2s orbital. The next six occupy the three 2p orbitals, with each 2p orbital containing two electrons of opposite spin. The next two electrons go into the 3s orbital with paired spin . The remaining four electrons are distributed among the three 3p orbitals. Taking Hund 's rules into account. the first three of these electrons are placed. unpaired and with identical spin, into the three equivalent 3p orbitals: 3p,, 3p>'. and 3p~. The oneremaining electron is then placed, with opposite spin. into the 3px orbital. To sunm1arize:
32
CHA PTER 1 • CHEMICAL BONDIN G AN D CHEM ICAL STR UCTURE
3p 3s
2p
-!4-- -t-#-
+
} va lence electrons
-it-* -it-
2s
*
Is
*
As shown in the diagram. the 3s and 3p electrons are the va lence electrons of sul fur; the 3s and 3p orbitals are the va lence orbitals.
1.1 8 Give the e lectronic configurations of each of the following aLOms and ions. Identify the vale nce electrons and valence orbi tals in each. (a) ox ygen mom (b) chloride ion, Cl- (c) potassium ion, K + (d) sod iu m atom
1.8
ANOTHER LOOK AT THE COVALENT BOND : MOLECULAR ORBITALS A. Molecular Orbital Theory One way to think about chemical bonding is to assume that a bond consists of two electrons localized between two specific atoms. This is the simplest view of a Lewis electron-pair bond. As useful as U1is picture is. it is sometimes too restrictive. When atoms combine into a molecule, the electrons contributed to the chemical bonds by each atom are 1,10 longer localized on individual atoms but ''belong'' to the entire molecule. Consequently. atom ic orbitals are no longer appropriate description for the state of electrons in molecules. Instead. molecular orbitals, nicknamed MOs.whk h are orbitals for the entire molecule. are used. Determi ning the electronic configuration of a molecule is a lot like determining the electronic configuration of an atom, except that molecular orbitals are used instead of atomic orbitaJs. The following four steps summarize conceptually how we start with two isolated hydrogen atoms and end up with the electronic configuration of the dihydrogen molecule, H2.
Step 1.
Start with the isolated atoms of the molecule and bring them together to the positions that they have in the molecule. Their valence atomic orbitals will overlap. For H2 , this means bringing the two hydrogen atoms together until the nuclei are separated by the length of the H- H bond (Fig. l. l3a). At this distance. U1e Is orbitals of the atoms overlap.
Step 2.
Allow the overlapping valence atomic orbitals to interact to form molecular orbitals (MOs). This step implies that MOs of H2 are derived by combining the Is atomic orbitals of the two hydrogen atoms in a certain way. Conceptually, this is reasonable: molecules result from a combination of atom!.. so molecular orbitals result from a combination of atomic orbitals. We'll shortly learn the process for combining atomic orbitals to form molecular orbitals.
1.8 ANOTHER LOOK AT THE COVALENT BOND: MOLECULAR ORBITALS
I
I orbtt,tls
\
overlapping orbit als in teract
hrint: ~tunh to~<·th~r to th" bonding dist.m«
33
molecular orbitals are fo rmed
nuclei (a )
I' orbll.tl'
I ' bonding molecular orbital fo r H2
(b)
J'lolk .:h,Hl!,!CJ ro trou~h
I, nr btt,tl'
,'
/planar nndc
I
change one peak 10 a trough
add
antibonding mo lecular orbital for H 1
(c)
Figure 1.13
Formation of H2 molecular orbitals. (a) Two hydrogen atoms are brought to the H- H bonding distance so that the ls orbitals overlap. Interaction of these atomic orbitals forms the molecular orbitals. (b) To form the bond ing MO of H 2, add the1s orbital wavefunctions of the interacting hydrogen atoms. (c) To form the anti bonding MO of H2,subtract thels orbital wavefunctions by changing one peak to a trough and then add ing. This process results in a node in the antibonding MO. r
Step 3 .
Arrange the MOs in order of increasing en ergy. Steps I and 2 will yield two MOs for 1-1 2 that differ in energy. We'll also learn how to determine relative energies of these MOs.
Step 4 .
Determine the electronic configuration of the molecule by redistributing the electrons from the constituent atoms into the MOs in order of increasing MO energy; the Pauli principle and Hund's rules are used. We redistribute the two electrons (one from each starting hydrogen atom) into the MOs of 1-12 to g ive the electronic configurution of the molecule. How to carry out steps 2 and 3 is the key to understanding the formation of molecu lar orbitals. Quantum theory gives us a few simple rules that allow us to derive the essemial fea tures of molecular orbi tals without any calcu lations. We' ll state these ru les as they apply to 1-12 and other cases involving the overlap of two atomic orbitals. (The~e rules will require only s light modification for more complex cases.)
34
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
Rules.for.forming molecular orbircil.~: I. The combination of IWO atomic orbitals gives two molecular orbitals. For H2 • this rule means that the overlap of two ls orbitals from the constituent hydrogen atoms gives two molecular orbitals. Later. we'll have situations in which we combine more than two atomic orbitals. When we combine} atomic orbitals. we always obtain} molecular orbitals.
2. One molecular orbital is derived by the addilion of the llVo atomic orbilals in the region of overlap. To apply this to 1-:12 , remember that the Is orbital is a wave peak. When we add two wave peaks. they reinforce. When we add two Is orbitals in the overlap region. they reinforce to form a continuous orbital that includes the region between the two nuclei (Fig. 1.13b). This molecular orbital is called a bonding molecular orbital, or bonding MO. The reason for the name is that, when electrons occupy this MO, they are attracted to both nuclei simultaneously. In other words, the electrons occupy not only the region around the nuclei but also /he region between the nuclei, thus providing "electron cement" that holds the nuclei together, just as mo11ar between two bricks holds the bricks together. 3. The other molecular orbital is derived by subtraCiion of the two atomic orbilals in !he region of overlap. To subtract the two Is orbitals, we change one of the Is orbitals from a peak to a trough. (This is equivalent to chang ing the mathematical sign of the ls wavefunction.) Then we add the two resulting orbitals. This process i illustrated in Fig. l.l3c. Add ing a wave peak to a wave trough resu lts in cancellation of the two waves in the region of overlap and formation of a node-a region in which the wave is zero. In this case. the node is a plane. The resulting orbital is called an anti bonding molecular orbital or a n ti bonding MO. Electrons that occupy this MO decrease bonding because the region between the nuclei contains no electron density. 4. The two molecular orbitals have different enerf?ies. Orbital energy increases IVith the
number of nodes. The bonding MO has a lower energy than 1he isolaled Is orbitals and 1he aruibonding MO has a higher energy 1han I he isolated Is orbitals. The orbi tal energ ies are summari zed in an orbital interaction diagram, shown in Fig. 1.14. This diagram is a plot of orbital energy versus the position of the two interacting nuclei. The isolated atomic orbitals and their energies are shown on lhe left and right sides of the diagram, and the molecular orbitals and their energies are shown in the center, where the separation of the atoms corresponds to the bond length. The number of nodes tells us the relmive energies oflhe MOs: the more nodes an MO has, the higher is its energy. The bonding MO bas no nodes and therefore has the lower energy. The antibonding MO has one node and has the higher energy. Notice that the energies of the two MOs "spread'' about the energy of the isolated Is orbitals- the energy of the bond ing MO is lowered by a certain amount and the energy of the antibonding MO is raised by the same amount. Now that we've described how to form the MOs and rank their energies, we're ready to populate these MOs with electrons. We apply the autbau principle. We have two electronsone from each hydrogen atom-to redistribute. Both can be placed in the bonding MO with opposite spins. Electron occupancy of the bonding MO is also shown in Fig. 1.14. When we talk about the energy of an orbital, what we are really talking about is the energy of an electron that occupies the orbital. It follows. then, that the electrons in the bonding MO have lower energy than two electrons in their parent ls orbitals. In other words. chemical bonding is an energelically favorable process. Each electron in the bonding MO of H2 contributes about half to the stability of the H-H bond. It takes about 435 kJ ( I04 kcal) to dissociate a mole of H 2 into hydrogen atoms, or about 2 18 kJ (52 kcal) per bonding electron. This a lot of
1.8 ANOTHER LOOK AT THE COVALENT BONO: MOLECULAR ORBITALS
35
ANTI BO Dl G MOLECULAR ORBITAL
I
I
I
,---- ----, \
\
\
I
\
I I
/
I
\
ls orbital
\\ \
---~\+
energy of i'olatcd lsorbitah
\
\
'
\
+ (__ I
\
'
\ \
\
I
/
\,_ _* _ _ electron occupancy
/ /'
/
Is orbital
1/
BONDrNG MOLECULAR ORBITAL
0 nuclear position
+oo
Ftgure 1.14 An orbital interaction diagram for the formation of the H2 molecular orbitals from interacting 1s orbitals of two hydrogen atoms. The dashed lines show schematically how the two 1s orbitals interact as the internuclear distance changes from very large ( +co) to the H- H bond length. The bonding MO has lower energy than the 1s orbitals and the anti bonding MO has higher energy. Both electrons occupy the bonding MO.
energy on a chem ical scale- more than enough to raise the temperature of a ki logram of water from freezing to boil ing. According to the picture j ust developed. the chemical bond in a hydrogen molecule results from the occupancy of a bonding molecular orbital by Jwo electrons. You may wonder why we concern ourselves with the antibonding molecular orbital if it is not occupied. The reason is that it can be occupied! I f a third electron were introduced into the hydrogen molecule. then the antibonding molecular orbital would be occupied. The resulting three-electron species is the hydrogen molecule anion. H~ (see Prob. 1.19b. p. 36). H ~ exists because each electron in the bonding molecular orbital of the hydrogen molecule comribute<; equally to the stability of the molecule. The third electron in H 2. the one in the antibonding molecular orbital. has a high energy that offsets the stabilization afforded b) one of the bonding electrons. However. the 'tabilization due to the econd bonding electron remain'>. Thu .... H ~ i'> a c;Lable species. but only about half a~ Mable~ the h)drogen molecule. In term., of our brick-and-mortar analogy. if electron~ in a bonding MO bind the two nuclei together a., mortar bind!. two bricks. then electrons in an antibonding MO act as "anti-mortar": not only do they 1101 bind the two nuclei together. but they oppose the binding effect of the bonding electrons. The importance of the antibonding MO is particularly evident when we attempt to con~truct diatomic helium. He2, as ~hown in Study Problem 1.5.
36
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
u~e
molecular orbital them; to explain why Hez doc~ not exi't. The molecular orbital'> of He2 are formed in the same way as those of H2•
Solution The orbital interaction diagram for the MO~ of lie_ i-. conceptual!) the ~a me as for H2 CFig. 1.14). However. He1 contains four eleCLron~-two from each lie atom. According to the aufbau principle. two electrons are placed into the bonding MO. but the other two must occupy the antibonding MO. Any stability contributed by the bonding electron\ i offset by the in~tability contributed by the antibonding electron\. Hence. formation of Hc 2 ha\ no energetic advantage. As a result. He is monatomic.
Molecular orbital!>, like atomic orbitab. have shape~ that cotTe,pond to regions of significant electron density. Consider the shape of the bonding molecular orbital of H2• shown i-n bmh Figs 1. 13b and 1.14. In this molecular orbital the electrons occupy an e ll ipsoidal region or ~pace. No matter how we turn the hydroge n mokculc about a line joining the two nuclei. its electron density looks the same. This i~ another way or saying that the bond in the hydrogen molecule has cylindrica l symmetr y. Other cylindrically ~y mmt.: trical objects arc shown in Fig. 1. 15 Bond-.. in which the electron den~ity is cylindrically\) mmetrical allout the internuL'Icar axi-. arc called sigma bonds [abbreviated crbon(h). The bond in the hydrogen molecule ts thu~" 1r bond. The lower-case Greek. lener 'igma wa' cho'>Cll to de~cribe the bonding molecular orbital of hydrogen becau!.e it is the Greek lcuer equivalent ol s. the letter used to de~c ribc the atomic orbital of lowe~t e nergy.
1.19 Ora\\ an orbital interaction diagram corre~ponding to Fig. 1.1-t for each of the following species. Indicate which are likely to exist a' diatomic ..,pec te~ . and which would dissociate into monatomic fragment~. Explain. (a} the Hej ion (b) the Hi ion (c} Li 2 (d) the II~- ion (cl the Hj ion 1.211 The bond di sociation energy of H1 i' 4351J mol- 1 (104 kcal mol - 1) : that is, it takes this amount of energy to dissociate H~ into its atoms. Estimate the bond di~soc iation energy of H; and explain your answer.
B. Molecular Orbital Theory and the Lewis structure of H2 Let\ now relate the quantum-mechanical description of' H ~ to the concept of the Lewis electron-pair bond . In the Lewis strucwre of H2 • the bond is represe nted by an electron pair shared between the two nuclei. In the quantum-mechanical de ...cription. the bond i ~ the result of the presence of two electrons in a bonding molecular o rbital and the resulting electron density between the two nuclei. Both electrons are auracted to each nuc leus. and these e lectrons thu · serve a~ the "cement" that holds the nuc le i together. Thu'>. for H1• 1he Lewis elecm m -pair lxmd i.\ equil'alelll 10 !he quanltmt-meclwnical idea of a bollllin,t: molecular orbital occupied by a pair of elecmms. The Lewis picture place\ the electron\ ...quare ly between the nuclei. QuantumtheOr) '> pace. Molecular orbital theory show<;. ho\\ e\ cr. that a chemical bond need not be an electron pair. For C\ample. Hj (the hydrogen molecule cation. "hich '' c might repre~e nt in the Lewis -.cn~e as HtH ) io, a ~table species in the ga.., pha<,e (',ee Pro b. 1. 19e). It i-, not ' o stable as the hydrogen molecule itself because the ion has only one electron in the bonding molecular orbital. rather than the two found in a neutral hydrogen molecuh.:. The hydrogen molecule anion.
1.9 HYBRID ORBITALS
37
bonding molecular orbital of the h)rdrogen molecule
a barrel
a top
a goblet
Figure 1.15 Some cylindrically symmetrica l objects. Objects are cylindrically symmetrical when they appear the same no matter how they are rotated about their cylindrical axis (black line).
1-12. discussed in the previous section. might be considered to have a three-electron bond consisti ng of two bonding electrons and one anti bonding electron. The electron in the antibonding orbital is also shared by the two nuclei. but shared in a way that red uce~ the energetic advantage or bonding. (1-12 is not so stable as 1-12 : Sec. 1.8A.) This example demonstrates that the sharing of electron s between nuclei in some cases does not contribute to bonding. Nevertheless. the most stable an·angement of electrons in the clihydrogen molecular orbitals occurs when the bonding MO contains two electrons and the antibonding MO is empty-in other ' words. when there is an electron-pair bond.
HYBRID ORBITALS A. Bonding in Methane We ultimately want to describe the chemical bonding in organic compounds. t~nd our first step in this direction is to understand the bonding in methane. CH 4. Before quantum theory was applied to the bonding problem, it was known experimentally that the hydrogens in methane. and thus the bonds to these hydrogens, were oriented tetrahedrally about the cen tral carbon. The valence orbiuLis in a carbon atom. however. are nor directed tetrahedrally. The 2s orbital. as you've learned. is spherically symmetrical (see Fig. 1.8), and the 2p orbirals are perpendicular (see Fig. 1.1 0). If the valence orbitals of carbon aren't directed tetrahedrally. why is methane a tetrahedral molecule? The modern solution to this problem is to apply molecul ar orbital theory. You can't do this with just the simple rules that we applied to H 2 • but it can be done. The result is that the combination of one carbon 2s and three carbon 2p orbitals with four tetrahedrally placed hydrogen Is orbitals gives four bonding MOs and four antibonding MOs. (The combination of eight atomic orbitals give eight molecular orbitals: rul e I, p. 34, with j = 8.) Eight electrons (four from carbon and one from each of the four hydrogens) are just sufficient to fill the four bonding MOs with electron pairs. This molecular orbital description of methane accounts accurately for its electronic properties. The conceptual difficulty with the molecular orbital description of methane is that we can't associate a given pair of electrons in the molecule wi th any one bond. Instead, the electrons from all of the atoms arc redistributed throughou t the entire molecule. We can' t even tell where atoms begin or end! If we add up all of the contributions of the electrons in the four bonding MO. of methane. we obtain a picture of the total electr on density- that is, the prob-
38
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
ability of finding electrons in the methane molecule. (The electrostatic potential maps. or EPMs. introduced in Sec. 1.20, are superimposed on such pictures of total electron density.)
methane total electron density
methane total electron density with imbedded model
d.ihydrogen total electron density with imbedded model
In this picture of total electron density, methane looks like an "electron pudding·· containing the nuclei. Although this electron density has a tetrahedral shape, there are no discrete C-H bonds. In conu·ast. because H2 has only one bond, we can associate the total electron density in H2 with the H-H bond. as shown in the previous section. Historically, chemists have liked to think that molecules are made up of atoms connected by individual bonds. We like to build models, hold them in our hands. and manipulate chemical bonds by plucking off certain atoms and replacing them by others. A~ though the molecular orbital description of methane certainly describes bonding, it suggests that the discrete chemical bond between individual atoms is something rationally conceivable but not rigorously definable (except perhaps for simple molecules like H2 ). Nevertheless. the concept of the chemical bond is so useful in organic chemistry that we can't ignore it! The problem, then. is this: Is there an electronic theory of bonding in methane that allows us to retain the notion of discrete C-H bonds? To use a bonding theory that we know isn't quite right may seem inappropriate, but science works this way. A theory is a framework for unifying a body of knowledge in such a way that we can u e it to make useful predictions. An example you're probably familiar with is the ideal gas Jaw. PV = nRT. Most real gases don't follow this Jaw exactly. but it can be used to make some useful predictions. For example. if you're wondering what will happen to the pre~sure in your automobile tires when the temperature drops in winter. this Jaw gives a perfectly useful answer: the pressure drops. If you're interested in calculating ex
1.9 HYBRID ORBITALS
39
"s-p-cubed"). The six carbon electrons in this orbital picture are distributed between one ls orbital and four equivalent sp 3 hybrid orbitals in quantum level 2. This mental transformation can be summarized as follows:
n_l_,~.:urhitab . ?. :. ~.:.:.l .:.:1l_l2_J'-
- - ) __
2( ~P')
_L,_
--f-
, Jl hybrid +- -nrhitab
2~
un.1! 1nl«l
h
In hnd11. 11un ,._ .:..:.,;.;..;.;..::.:.:...:.;.:...;..;.;.. /
I·
unhybridized carbon
hybridized carbon ( as in methane)
This orbital mix ing can be done mathematically, and it yields the perspective drawing of an sp 3 hybrid orbital shown in Fig. 1. 16a on p. 40. A simpler representation used in most texts is shown in Fig. I. 16b. As you can see from these pictures. an sp3 orbital consists oft wo lobes separated by a node. much like a 2p orbital. However. one of the lobes is very small. and the other i s very large. In other words. the electron densi1y in w1 sp3 hybrid orbital is highly directed in space. This directional character is ideal for bond formation along the axis of the large lobe. The number of hybrid orbitals (four in this case) is the same as the number of orbitals that are mixed to obtain them. (One s orbital + three p orbitals= four sp 3 orbitals.) It turns out that the large lobes of the four carbon sp 3 orbitals are directed to the corners of regular tetrahedron. as shown in Figure 1.1 6c. In hybridization theory. each of the four electron-pair bonds in methane results from the overlap of a hydrogen Is orbital containing one electron wi th a carbon sp3 orbital, also containing a single electron. The resulting bond is a cr bond.
hydrogen nucleus
carbon nucleus
.
nvcrh•r
carbon sp3 hybrid orbital con taining one electron
hydrogen Is orbital containing one electron
overlapping
a C-H
sp3 and Is orbitals
electron-pair bond (a crbond )
This overlap looks a lot like the overlap of two atomic orbitals that we canied out in constructing the molecular orbiwls of H2 . However. the hybrid orbital treatment is not a molecular orbi tal treatment because it deals with each bond in isolation. The hybrid orbital bonding picture for methane is shown in Fig. 1.1 6d. The hybridization of carbon itsel f actually costs energy. (If this weren't so. carbon atoms would ex ist in a hybridized configuration.) Remember. though. that this is a model for carbon in methane. H ybridization allows carbon to form four bonds to hydrogen that are much stronger than the bonds that would be formed without hybJidjzation, and the strength of these bonds more than offsets the energy required for hybridization. Why does hybridization make these bonds stronger? First. the bonds are as far apart as possible. and repul sion between electron pairs in the bonds is therefore minimized. The pure sand p orbitals available on nonhybridized carbon. in contrast, are not directed tetrahedrally. Second, in each hybridized orbital.
40
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
C tT
H )l)
nucleu\
m>d.tl ,Uif.ld.' (b )
Ca l
(c)
(d )
Figure 1 16 (a} Perspective representation of a carbon sp1 hybrid orbital.(b) A more common representation of an sp 3 orbital used in drawings.(c)The four sp 3 orbitals of carbon shown together. (d) An orbital p icture oftetrahedral methane showing the four equivalent u bonds formed from the overlap of carbon sp 3 and hydrogen 1s or· bitals. The rear lobes of the orbita ls shown in (c) are omitted in (d) for clarity.
1he bulk of the electron density is directed toward the bound hydrogen. This directional character provide'> more electron ··cement" between the carbon and hydrogen nuclei. and thi re'-Uit\ in stronger (that is. more stable) bond'-.
B. Bonding in Ammonia The hybrid orbital picture is readily extended to compound'> containing un-.hared electron pair\. '>Uch as ammonia. : 'H 3. The valence orbitaJ... of nitrogen in ammonia are. like the carbon in methane. hybridized to yield four .\p' h) brid orbital-.: howe,er. unlike the corresponding carbon orbital<;, one of these hybrid orbital<. i' fully occupied" ith a pair of electrons.
-+- -+- 1 --"-llx;;.;..l~'d.; ;Is.nd~2-f'_.,
2p
01h11
-
t
lll'hrid mhital'
-H-
unh r bridized nitrogen 1
hybridi'lcd nitrogen (a'> in .1mmonia)
Each of the sp- orbitals on nitrogen containing one electron can overlap with the Is orbital of a hydrogen atom. also containing one electron. to give one of the three - H (1 bonds of ammonia. The electron' in the filled ~P 1 orbital on nitrogen become the um.hared electron pair in ammonia. The unshared pair and the three 1 - H bond'-. becau-.c they are made up of \fl1 hybrid orbital\. are directed to the corners of a regular tetrahedron (Fig. 1.17). The advantage of orbital hybridi7ation in ammonia i'> the ..ame a-; in carbon: hybridi1ation accommodate" the maximum '>eparation of the un hared pair and the three h) drogens and. at the same time. pro' ide~ 'trong. directed 1 -H bonds. You may recall from Sec. I .3B that the H-N - H bond angle in ammonia is I 07.3v. a little "nailer than tetrahedral ( I 09.5° ). Our hybrid-orbital pit:turc can accommodate this struc-
1. 9 HYBRID ORBITALS
41
..
,,... N..........._
J-Ill'"' /
. . . . . . . I-I
1-I Lcwi5 structure
orbital picw re
Figure 1.17 The hyb rid o rb ital description o f ammo ni a, :NH 3• As in Fig.1.16, the sm all rear lo bes of the hybrid o rbitals are omitted for clarity.
tural refinement as well. Unshared electron pairs prefer s orbitab. because s orbit al s have lower energy than p orbital s. Or. to look at it another way. there's no energeti c advantage to putting an unshared pair in a spatiall y directed orbital if it's not going to be invol ved in a chemical bond. But if the unshared pair w ere left in an un hy bridized 2s orbital. each bond to hydrogen would have to be deri ved from a pure nitrogen 2p orbital. I n such a bond. half o f the electron density ("'e lectron cement..) would be <.lirected away f rom the hydrogen, and the bond would be weak. I n such a case. the H-N- H bond ang le woul<.l be 90° , the same as the angle between the 2p orbitals used to form the bonds. The actual geometry of ammonia is a compromise between the preference of unshared pairs for orbitals of high s character and the preference of bonds for hy brid character. The orbital containing the unshared pair has a little mores character than the bonding orbit als. Because s o rbital cover an entire sphere (see Fig . 1.8. p. 26 ). orbital s w ith mores character occupy more space. Hence. unshared pairs have a greater spatial requirement than bonds. Hence, the ang le between the unshared pair and each o r the N - 1-1 bonds i ~ somewhat greater than tetrahedral , and the bond ang les between the N- H bond~. as a consequence. are ~o mew hm /ess than tetrahedral. This is the same concl usion we obtained fro m the applicat ion of V SEPR theory to ammonia (p. 18).
A connection exists between the hyhridization t~{an atom and the a rrangemelll in space of the bonds around that a/Om. Atom s surrounded by four groups ( incl uding unshared pairs) in a tetrahedral arrangement are .sp·l.hybrid ized . The converse is al1'o true: sp 3-hybridized atoms always have tetrahedral bonding geometry. A trigonal p lanar bonding arrangement is associ ated w ith a di ffe rent hybridi zation. and a linear bonding arrangement w ith yet a third ty pe of hy bridi zation. (These types o f hy brid izat ion are d i scu!>~ed in Chapters 4 and 14.) In other word s,
hybridization and molecular geometry are closely correlated. The hy bridization picture of covalent bond ing also dr ives home o ne of the most important differences between the ionic and covalent bond: the covalent bond has a definite direction in space. w hereas the io nic bond is the same in all directions. The directionality of' covalent bonding is responsible fo r molecular shape; and. as we shall see. mo lecular shape has some very important chemical consequences.
1.21 (a) Construct a hybrid orbital 'picture for the water molecule using oxygen sp 3 hybrid orbitals. (b) Predjct any departures from tetrahe(jral geometry that you might expect from the presence of two unshared elecu·on pairs. Explain your answer. 1.22 (a) Construct a hybrid orbital picture for the hydronium ion (Hp +) using oxygen sp 3 hybrid orbital s. (b) How would you expect the H- 0 - H bond angles in hydronium ion to compare with those in water (larger or smaller)? Explain.
42
CHAPTER 1 • CHEMICAL BONDING AND CHEMICAL STRUCTURE
KEY IDEAS IN CHAPTER 1 •
Chemical compounds can contain two types of bonds: ionic and covalent. In ionic compounds, ions are held together by electrostatic attraction (the attraction of opposite charges). In covalent compounds, atoms are held together by the sharing of electrons.
•
Both the formation of ions and bonding in covalent compounds tend to follow the octet rule: Each atom is surrounded by eight valence electrons (two electrons for hydrogen).
•
The formal-charge convention assigns charges within a given species to its constituent atoms. The calculation of formal charge is given in Study Problem 1.1. Formal charge is a bookkeeping device. In some cases the actual charge on an atom and the formal charge do not correspond.
•
•
•
In covalent compounds, electrons are shared unequally between bonded atoms with different electronegativities. This unequal sharing results in a bond dipole moment. The dipole moment of a molecule is the vector sum of its individual bond dipole moments. The local charge distribution in a molecule can be described graphically with an electrostatic potential map (EPM). The structure of a molecule is determined by its connectivity and its geometry. The molecular geometry of a molecule is determined by its bond lengths, bond angles, and dihedral angles. Bond lengths are governed, in decreasing order of importance, by the period of the periodic table from which the bonded atoms are derived; by the bond order (whether the bond is single, double, or triple); and by the column (group) of the periodic table from which the atoms in the bond are derived. Approximate bond angles can be predicted by assuming that the groups surrounding a given atom are as far apart as possible. A complete description of geometry for complex molecules requires a knowledge of dihedral angles, which are the angles between bonds on adjacent atoms when the bonds are viewed in a planar projection. Molecules that are not adequately described by a single Lewis structure are represented as resonance
hybrids, which are weighted averages of two or more fictitious Lewis structures. Resonance hybrids are more stable than any of their contributing resonance structures. •
As a consequence of their wave properties, electrons in atoms and molecules can exist only in certain allowed energy states, called orbitals. Orbitals are descriptions of the wave properties of electrons in atoms and molecules, including their spatial distribution. Orbitals are described mathematically by wavefunctions.
•
Electrons in orbitals are characterized by quantum numbers, which, for atoms, are designated n,/, and m1• Electron spin is described by a fourth quantum number m, .The higher the principal quantum number n of an electron, the higher its energy. In atoms other than hydrogen, the energy is also a function of the I quantum number.
•
Some orbitals contain nodes, which separate the wave peaks of the orbitals from the wave troughs. An atomic orbital of quantum number n has n - 1 nodes.
•
The distribution of electron density in a given type of orbital has a characteristic arrangement in space governed by the I quantum number: all s orbitals are spheres, all p orbitals contain two equal-sized lobes, and so on.The orientation of an orbital is governed by its m1 quantum number.
•
Atomic orbitals and molecular orbitals are both populated with electrons according to the aufbau principle.
•
Covalent bonds are formed when the orbitals of different atoms overlap. In molecular orbital theory, covalent bonding arises from the filling of bonding molecular orbitals by electron pairs.
•
The directional properties of bonds can be understood by the use of hybrid orbitals. The hybridization of an atom and the geometry of the atoms attached to it are closely related. All sp3-hybridized atoms have tetrahedral geometry.
ADDITIONAL PROBLEMS
43
ADDITIONAL PROBLEMS 1.23 In each of the fo llowing set~. specify the one compound that is likely 10 have completely ionic bond!> in its sol id state. HCI 1 aAt 1:11 CCI~ HF BF3 (bl cs ~ CsF 1.:!4 Whtch of the atom!> in each of the following ~pecies ha'> a complete octet? What i'> the formal charge on each? A~~ume all unshared valence electron' arc hown.
tal CH, (d) BH,
UO Which of the following orbitals is (arc) not permitted by the quan tum theory of the hydrogen awm? Explain.
2s 6.1· 5d
2d 3p
1.31 Predict the approximate bond angle\ in each of the fol· lowing molecules. rat :CH ~ {b) BeH ~
Cdl :~I~Si
~~
H'l
+ell \
Q=g-g:
(b) :NH 3 It' I :C H, tel :')': (f) B H~
o:wnc
1.2S Drnw one Lewis structure for cat.:h of the fo llowing compounds: show all unshared electron pairs. None of the atoms in the compound'> bear:- a forma l charge. and :111 atom~ ha\'e octet~ (hydrogen\ ha\'c duets).
{C ive 11 - C -Cand C-C-C angle~.) a Ilene
C 2H tCI {b) ketene. C1 H1 0. which ha' a carbon-carbon double bond
( a)
1.26 Draw two Lewis structures for a compound with the fonnula C4 H 10. No atom bear!. a charge. and all carbons have complete octets.
1.27 Give the forma l charge on each mom and the net charge on each ~pecie~ in the following ~trucwre~. All un~hared valence electron\ are ..hown.
:Q:
tal
the percem .1· character of ( I ) the orbital containing a Jon..: pair and (2) the orbital used to form the bond to hydrogen in each of the following compounds. State any a~~umptions. (Him: What would be the angle between bonds containing Q<;r 1 character'!) a1, H ,, H - H bond angle 107.3 (b) H10. H-0-H bond angle 105.5°
.. I .. :O-CI-0: .. I .. :o:
perchlorate
1.32 The percents character describes the hybridization of an orbital. For example. an ~p' orbital has 25% s character. Given the bond angles in each case. calculate
trimethylamine
1..'.\ The allyl mtion can be repre,cntcd by the following re~onance structure,.
oxid e
It')
q.. -:P
0 , .. q:
(t•J
allyl catio n meth ylene
o7.one
I
II,C-C·
\
(a) What is the bond order of each carbon-carbon bond in the allyl cation·~ tb) How much positive charge resides on each carbon of
H hypoch lorite
H
eth yl radical 1.2~
G ive the electronic configuration of (a) the chlorine atom: (b) the ch loride ion: (c) the argon atom; (d) the magnesium atom.
1.29 Gi\e the electronic configuratton of \ilicon (Si). Indicate the valence electron'>.
the allyl cation? (c) Although the preceding ~tructures are reasonable de~cription~ of the allyl cation. the fo llowing cation 1'0111101 be described by analogous resonance structures. Explain why the structu re on the right is not a reasonable resonance Mructurc.
l
+ I H H1C-C=NH 3
44
CHAPTER 1 • CHEMICAL BONDING A ND CHEMICAL STRUCTUR E
1.39 Account for the fact that H,C- CI (dipole moment 1.94 Dl and li ,C F (dipole moment 1.82 0) have ahno\1 identical dipole moment-.. e1 en though ftuorine i-. con,iderabl> more clcctronegatiw than chlorine. angle~ do not permit a dbtinction bet\\ecn the folio\\ ing l\\O conceivable form\ of ethylene.
1.40 1l1c principle' for predicting h
Hm' much negmiH! charge i~ on each ox)gen of the carbonate ion'? Cb) What i~ the bond order of each carbon-o"
1.3fl The \hapc of one of the five energetically equi1 alent 3d orbital\ follov' <;, From your an-.wer to Problem 1.35. \ketch the node' of thi\ 3d orbital. and a~o;ociate a "ave p.;:ak or a wave trough with each lobe of the orbital. {Him : It doe-.n't matter" here you put your first peak: you 'hould be concerned only \\ith the relative position~ of peak\ and trough\.) )'
1.37 Orbital\ "ith I - 3 are called f orbitab. Cal Hov' man) energetically equi,alcntforbitab are thcre·l Cbl ln "hat pnncipal quantum level do f orbital\ fir:,t appear'! (c l HOI\ man) node-. doe' a ~f orbital have'! I.J~
Sketch a 4p orbital. Sho" the node, and the region~ of wave peak-. and "a\e trough\. CH i11t: u~e Fig. I. II and the de,cnption:. of node~ in Problem 1.35a.)
II
II
II
\ I c- c I \
1.35 (a) Two type' of node' occur in atomic orbitals: spheri-
cal ~urface~ and plane!.. Examine the nodes in 2.1. '2p. and 3p orbita b. and ~ h ow that they ag ree with the following ~ t at cmcnts: I. An orbital of prim:ipal quantum number n has 11 I node,. 2. The 1aluc of I give' the nu mber of planar nodes. {b) How many \phcrical node~ doe~ a S.1 orbital ha,•e'l A 3d orbital'? I low many node~ of all type~ does a 3d orbiwl ha,·e'?
II
plunar
I ·c = c
H
H.... II /
\ 1-1 s taggered
The dipole moment nf ethyle ne i~ t ero. Doc!' thi ~ experimental fac t provide a ciL1c to the preferred dihedral an· gles in ethylene? Why or why nnt? 1.41 A wcll-l.rlO\\ n ehcmi,t. llavno St.enl\. ha' heard you appl) the ru le~ for predicting molecular geometry to water: you ha"c propo\ed (Problem I. lOa. p. 18) a bent geometl') for thi' compound. Dr. Stem~ i' unconvinced by your argument' and conunue-. to propo-.e that water is a linear molecule. He demand\ that you debme the i~sue with him before a di'iingui,hed acadl!my. You muM therefore come up \\ llh 1'.\fll'rimenwl data that \\ill prm e to an objectile txxl~ ol -.cienti\b that water indeed ha!> a bent geometl'). E\plain why the dipole moment of water. 1.84 D. could be U\ed to ~uppon ) our ca!>e. 1.42
U~C }OUr know!l.:dge of \CCIOr~ to el\plain why. e1·en though the C- CI bond dipole h large. the dipole moment of c;~rbon tetrachloride. CCI 1• b tero. (H int : Take the rcsultam ol' any two C-CI bond dipoles: then take the re~u h a n t of the other two. Now add the two re~ ultants to get the dirolcmoment of the molecule. Use models!)
J..B Three possihlc di hedral angb fur li P~ (0°. 90 . and 180 ) an.: \huwn in Fig. 1.6 un p. IY. (a) A~\ume that the li p~ molecule ex ists predominant!) in one of the'c :~rrangcmenl\. Which of the dihedral angle' can he ruled out hy the fact that 11p 1 ha" a large dipole moment <2. 13 0)7 Explain. (b) The bond dipole moment of the 0 - H bond h tabulated"' 1.52 D. L,-.c thi-. fact and the merall dipole moment ol H,0 _in pan Cal to decide on the preferred dihedral angle-. in 11~0~. Take the H- 0 - 0 bond angle to bc the known 'aluc <96.5 l. ( H im: Appl) the lav\ of co,ine-..J
ADDITIONAL PROBLEMS
1.44 Gh en the dipole mon1t:nt of ''mer 1 1.8-1 D>and the 11-0-H bond angle ( 10-1.-15 ). jtNify the stmement in Problem 1.43(b) that the bond dipole moment of the 0 -H bond is 1.52 D. 1.45
Con~ i der two 2p orbitab. one on each of two atom~. oriemed head-to-head a~ in rigun: p 1.-15. Imagine bringing the nuclei clo-..cr together until the t\\O wa\e peak\ !the blue lobe') of the orbitalqu\t m erl;tp. a~ ~ho'' n in the figure . A new ') -..tem of molecular orbitab is fonned h) thi~ merlap. (a) Sl-etch the ~hapc of the re-..ulting bonding and antibonding molecular orbital\. (b) Identify the node\ in each molecular orbital. (C) Construct an orbital interaction uiagram for mo lecular orbita l formation. (d) If two e lectron' occupy the bonding molecular orbital. i., the re~ulting bond a IT bond'! Explain.
1.-16 Con,ider two 2p orbital\. one on each of two different c~o:up) the bond111g molecular orbital. i' the re,ulting bond a IT bond? Explain.
Figure P1 .45
Figure P1 .46
1.~7
45
When a h)drogen molecule ah\orb\ light. an electron jump' from the bonding molecular mbital to the antibonding molecular orbital. Explain why thi't light ab~orption can lead to the di,~ociation of the hydrogen molec ule into two hydrogen atom~. (This process. called photodissociation. can sometime\ be used to initiate chemical reaction\.)
I AS Supp<>'>e you take a trip to a di,tant univer\C and lind
that the periodic table there i' deri,cd from an arrangement of quantum numbcf' different from the one on earth. The rule, in that uni' cr,e arc I. 2.... (as on I. principal quantum number 11 canhl: 2. angul:u· momentum quan tum number I = 0. I. 2. . ... n - I (as on earth): 3. magnetic quantum number 1111 - 0. I. 2..... I (that i~. only pmitil't' integer\ up to and including I are allowed 1: and 4. \pin quantum number m - - I. 0. + I (that i!.. three allowed \'alue' ol 'pin). (a ) "'\uming that the Pauli C\clu,ion principle remain' Yalid. what i~ the ma>.imum number of electron~ that can populate a given orbital'! (b) Write the electronic ..:on nguration of the c lement with atomic number 8 in the periodic wble. (l') What i-. the atomic number of the \econd noble ga~'! (d) What rule replace ~ the octet rule?
2 Alkanes Organic compounds all comain carbon. but the) can al<.o contain a wide variety of other element\. Before we can appreciate such chemical di\'er'>ity. however. we have to begin at the beginni ng. with the ~im plest organic compound~. the hydmcarhom. H ydrocarbons are compound'> that contain only the elements carbon and hydrogen.
HYDROCARBONS Methane. CH~. is the simplest hydrocarbon. As you learned in Sec. 1.38. all of the hydrogen atoms of methane are equivalent. occupying the corner~ o f a regular tetrahedron. Imagine now. that in~tead of being bound only to hydrogens. a carbon atom could be bound to a second carbon with enough hydrogens to fulfill the octet rul e. The resulting compound is erhane.
HH H : ~:~: H
Lewis structures of ethane:
HH
H
H
I
II
I
I-1 - C- C- H
I
H
I
H
H H
H
H
H
ball-and-stick model of ethane
46
space-filling model of ethane
2 1 HYDROCARBONS
47
Ftgure 2 1 Hybrid-orbital description of the bonds in ethane. (The small rear lobes of the carbon sp3 orbitals are omitted for clarity.) The component sp 3 and ls orbitals are shown with dashed lines for the two labeled bonds.
In ethane. the bond between the two carbon atom is longer than a C- H bond. but. like the C - H bonds. it is a covalent bond in the Lewis sense. In term!. of hybrid orbitals. the carbon-carbon bond in ethane con'>i'>t!. of two electron in a bond formed by the overlap of 1 t'' o sp h) brid orbitals. one from each carbon. Thus. the carbon-carbon bond in ethane is an 1 3 sp -.\p (T bond (Fig. 2.1 ). The C- H bonds in ethane are like those of methane. They con ist 1 of covalo..:nt bond~. each of which i~ formed by the overlap of a carbon sp orbi tal with a hy1 drogen b orbital: that is. they are sp - l s (Tbonds. Both the H -C-C and H -C-H bond angles in ethane are approximately tetrahedral because each carbon bears four groups. We can go on to envision other hydrocarbons in which any number o f carbons are bonded in this way to form chains or carbon~ bearing their associated hydrogen atoms. Indeed, the abi Iity of a carbon to form stable bond~ to other carbons is what give:. rise to the tremendou~ number of known organic compound-.. The idea of carbon chains. a revolutionary one in the early days of chemistry. was developed independently by the German chemist August Kekule (1829- 1896) and the Scotsman Archibald Scott Couper ( 1831 - 1892) in about 1858. Kekule's account of hi!. inspiration for thi'> idea il> amu<;ing. During my stay in London 1 resided for a con~iderable time in Clapham Road in the neighborhood of Clapham Common .... One tine ~ummer evening I was returning by the last bus. "outside'' as usual. through the deserted streets of the city that are at other times so full of life. I fell into a reverie. and lo. the atOms were gamboling before my eyes. Whenever. hitherto. these diminutive beings had appeared to me they had alway:. been in motion. Now. however. J :.aw how. frequently. two smaller atom~ united to form a pair.... I 'aw how the larger ones formed a chain. dragging the smaller one~ after them but onl) at the end\ of the chain .... The cry of the conductor, "Clapham Road:· awakened me from Ill) dreaming. but I ~pent a part of the night putting on paper m least sketche~ of these dream form\. Thi s was the origin of the "Structure Theory...
Hydrocarbons are divided into two broad cla!.ses: aliphatic hydrocarbons and aromatic hydrocarbons. (The term aliphatic come. from the Greek aleiplwtos. which means ''fat." Fat~ contain long carbon chains that. as you will learn. are aliphatic group~.) The aliphatic hydr ocarbons consist of three hydrocarbon families: alkanes. alkene.1·. and afkynes. We 'II begin our study of aliphatic hydrocarbons with the al kanes, which are sometimes known as par affins. Alkanes are hydrocarbons that contain only single bonds. Methane and ethane are the simplest alkane. . Later we'll consider the al kcnes, or oletins, hydrocarbons that contain carbon-carbon double bonds: and the alkynes, or acetyl enes. hydrocarbons that contain carbon-carbon triple
48
CHAPTER 2 • ALKANES
bonds. The la'>t hydrocarbons we'll ~IUd) are the aromatic hydrocarbons. which include benLene and itc, ~ubMituted derivatives.
II
I
11 -........ _.......c~ _.......H
II
I
II
H
I
\
H
I
C=C I \
11-C-C- H I I
H
II
an a lkane
H- C==C-H
c c II I c c 1-r·,... -........c ~ -.. . . . H
an alkyne
II
H H
I
an alkene
benzene till
aromatic l1ydrocarbon
aliphatic l1ydrocarbons
UNBRANCHED ALKANES Carbon chaine, take many forms in the alkane.,: they rna) be branched or unbranched. and they can even cxi'>t ac., rings (cyclic alkanes). All-anec., v.ith unbranched carbon chain., arc ~ometimes called normal alkanes, or n-aJkanes. A fc\\ of the unbranched alkanes are !>hO\\ n in Table 2.1. along with some of their physical propenie'>. You ~hould learn the name" of the fir~ttwelve unbranched alki.lnes because they are the ba-.is for naming many other organic compounds. The names methane. ethane. propane. and btllalle have their origins in the early hi!'ltory of organic chem istry, but the names of the higher alkanes arc derived from the corresponding Greek numerical names: pentane (pe11t
lf;i:l!fJI
= five): hexane (ltex = six ): and so on.
The unbranched Alkanes
Compound name
Molecular formula
Condensed structural formula
Melting point ("C)
methane
CH4
CH 4
- 182.5
161.7
ethane
ClH6
CH 3CH 1
183.3
88.6
propane
ClHs
CH 3CH1CH3
- 187.7
42.1
butane
C, H10
CH 3 (CH1)2CH 3
- 138.3
0.5
pentane
CsH12
CH 3(CH 2hCH,
- 129.8
36.1
0.6262
hexane
c~.H,,
CH 3 (CH 2) 4CH 3
- 95.3
68.7
0.6603
heptane
C,H,6
CH 3 (CH 2lsCH 3
- 90.6
98.4
0.6837
octane
c, H,,
CH 3 (CH 2)6CH 3
- 56.8
125.7
0.7026
nonane
C9H2o
CH 3 (CH 2hCH 3
- 53.5
150.8
0.7177
decane
C,oH22
CH 3(CH 2)8CH 3
29.7
174.0
0.7299
undecane
c ,,H2•
CH 3(CH1) 9CH 1
- 25.6
195.8
0.7402
dodecane
c ,2H26
CH 3 (CH 2) , 0 CH 3
- 9.6
216.3
0.7487
eicosane
C20H42
CH 3(CH 2)18CH 3
+ 36.8
343.0
0.7886
'The densities tabulated in this text are of the liquids at 20 C unless otherwise noted.
Boiling point (°C)
Density* (gml- 1 )
2.2 UNBRANCHED ALKANES
49
Organic molecu les are represented in different way~. which we·IJ illustrate using the alkane hexane. The m ol ecular formula of a compound (for example. C6H 14 for hexane) gives its atomic compo!.ition. All noncyclic alkanes (alkanes without rings) have the general formula C,. H ~,+ 1 . in which 11 is the number of carbon atom!.. The structuraJ formul a of a molecule j, it~ Lewi'> '>tructurc. which show.; the connectivity of it., atoms-that is. the order in which it~ atoms are connected. For example. a structural formula for hexane is the following:
H
J-1
J-1
ll
II
H
H
H
H
H
I I I I I I H-C -C-C-C-C -C-H I I I I I I H
H
hexane
otice that this type of formula doc!. 1101 portray the molecular geometry.) Writing each hydrogen atom in this way is very time-consuming, and a simpler representation of this molecule. ca lled a condensed structural formula, conveys the same information.
hexane
In such a ~tructure. the hydrogen atom<; are understood to be connected to carbon atoms with single bonds. and the bonds shown explicitly are ho11d.1 henree11 carho11 moms. Sometimes even these bonds are omitted, so that hexane can also be written CH 1CH 2CH 2CH 2CH 2CHy The structural formula may be further abbreviated as shown in the third column of Table 2.1. Ln thb type of formula. for example. <<; H 1 >~ mean<; -CH 1CH 2CH 2CH 2- . and he.xane can thuo; be written CH l(Cl l 2 ) 1CH 3.
two other representations of hexa ne
The family of unbranched alkanes forms a series in which successive members differ from one another by one -CH! - group (m ethylene group ) in the carbon chain. A series of compound~ that differ by the addition of methylene group!> ill called a hom ologous series. Thu~. the unbranched alkane), constitute one homologous ~cries. Generally. phy),ical properties within a homologou~ series vary in a regular way. An examination of Table 2.1. for example. reveals that the boi ling points and densitie:-. of the unbranched alkanes vary regularly wi th increasing number of carbon atoms. This variation can be usefu l for quickly estimating the properties of a member of the :-.erie!> whose properties are not known. The French chcmi.t Charle~ Gerhardt ( 1816-1856) made an important chemical observation in 1845 about members of homologous series. Hi' ob~ervarion till ha~ !.ignificant implications for learning organic cherni!.try. He wrote. "These (related) !>UbstanCCll undergo reactions according to the same equation~. and it is only necessary to know the reactions of one in order to predict the reactions o f the others.'· What Gerhardt was saying. for example, is that we can study the chemical reactions of propane with the confidence that ethane, butane. or dodecane will undergo analogou:, reactions.
14;1.1:14441
•••• • , ....,..
2.1 (al How many hydrogen atoms arc in the w1branched alkane witl1 18 carbon atoms?
(b) Is there an unbranched alkane containing 23 hydrogen atoms'? lf so, give its structural formu la; if not, explain why not. 2.2 Give the ~tructural formula and c~timate the boiling point of tridecane. CpH 2g.
50
CHAPTER 2 • ALKANES
2.3
CONFORMATIONS OF ALKANES In Section 1.38, we learned that understanding the struc tu re~ of many molecules requires that we specify not only their bond lengths and bond angles but abo their dihedral angle'>. In this section, we' ll use the simple alkanes ethane and butane to de,elop some widely applicable simple principles that will allow us to predict the dihedral angles in more complex molecu le~.
A. conformation of Ethane To specify the dihedral angles in ethane. we must define the relation hip between the C-H bonds on one carbon and those on the other. A convenient way to do this is to view the molecule in a Newman projection. A Newman projection is a I) pe of planar projection along one bond. which we"ll call the projected bond. For ex ample, suppose we wish to view the ethane molecule in a Newman projection along the carbon-carbon bond, as shown in Fig. 2.2. In thi s projection, the carbon-carbon bond is the projected bond. To draw a Newman projection. start with a circle. The remaining bonds to the nearer atom in the projected bond arc drawn to the center of the circle. The remaining bonds to the farther atom in the projected bond are drawn to the periphery of the circle. In the ewman projection of ethane (Fig. 2.2c), the three C- H
ball-and-stick models:
~
---- --
!HUJI.'CtcJ bond
\
li11e-and-wedge formulas:
H \ ~
H
ll
\ ' _______ H•""jc-c H
II
(a) viewing a model of ethane from one end
(b) end-on view
(c) Newman projection (II ~.hhniral anglo:!
Figure 2.2 How to derive a Newman projection for ethane using ball-and-stick models (top) and line-and· wedge formulas (bottom). (The hydrogens and C- H bonds farthest from the observer are shown in blue.) First view the ethane molecule from the end of the bond you wish to project. as in part (a). The resulting end-on view is shown in part (b). This is represented as a Newman projection (c) in the plane of the page.ln the Newman projection, the bonds drawn to the center of the circle are attached to the carbon closer to the observer; the bonds drawn to the periphery of the circle (blue) are attached to the carbon farther from the observer. The projected bond (the carbon- carbon bond) is hidden.
2.3 CONFORMATIONS OF ALKANES
51
bonds drawn tO the center of the circle are bonds to the front carbon. The C- H bond to the periphery of the circle are the bonds to the rear carbon. The projected bond itself, ll'hich is the
fourth bond to each carbon. is hidden.
e
The Newman projection of ethane make1> it very easy to sec the dihedral Gll!!,les between i ts C- H bonds. When we have specified all of the dihedral angles in a molecule. we have specified its cm({Ormarion. Thu ~. the conform ation of a molecule is the spatial arrangement of its atoms '"hen all of it dihedral angle are ..,pecified. We canal o refer to confom1ation of parts of molecules. for example. conformations about individual bonds. Two limiting possibilities for the conformation of ethane can be seen from its Newman projections; these arc termed the SWKgered COI(/(mnation and the eclipsed conformation. H=60
II
/
·~
H
H
staggered conformation of ethane
eclipsed conformation of etha ne
In the staggered conformation, a C- H bond of one carbon bisects the angle between two C- H bond<. of the other. The -.mullest dihedral angle in the \taggered conformation is () = 60 . (The other dihedral angle~ are 8 = 180 and () = 300°.) In the eclip ed conformation, the C-H bond~ on the respective carbons arc superimpo1>ed in the ewman projection. The
e
·'
smallest dihedral angle is 8 = 0°. (The other dihedral angles are = 120° and 0 = 240°.) Of course. conformations intermediate between the staggered and eclipsed con formations are po~sible. but these two conformations will prove to be of central importance. Which i~ the preferred conformation of ethane'! The encrgie~ of the ethane conformation!> can be described by a plot of relative energy ver~m. dihedral angle. which is shown in Fig. 2.3 on p. 52. ln this figure. the dihedral angle is the angle between the bonds to the colored hydrogens on the different carbons. To liCe the relationships in Fig. 2.3. build a model of ethane. hold either carbon fixed. and turn the mher carbon about the C-C bond. As the angle of rotation changes. the model passe alternately through three identical Maggered and three identical eclipsed conformations. A~ sho"'n by Fig. 2.3. identical conformations have identical energies. The graph also shows that the eclipsed conformation is characterized by an energy maximum. and the staggered conformation is characteri zed by an energy minimum. The staggered conformation is therefore the more swbfe conformation of ethane. The graph hows that the '>taggered conformation is more o;table than the eclip ed conformation by about 12 kJ mol- 1 (about 2.9 kcal mol 1). This means that it would take about 12 kJ of energy to convert one mole of staggered ethane into one mole of ecl ipsed ethane. The reason for the relative stability of the !>taggered conformation has been debated for years. One theory is that the staggered form is more stable because there is a favorable interaction between the bonding and antibonding molecular orbitals associated with the C-H bonds on the two carbons that lie at a dihedral angle of 180°. Thi~ dihedral angle, and hence the stabilizing interaction. is on ly possible in the staggered form. A second theory holds that there is repulsion between the electrons in the C- H bonds on the two carbons. Because the bond~ are closer when they have a dihedral angle of this repubion is greater in the eclipsed form. Thb repulsion is termed torsional strain. otice that the repulsion i!> not between the hydrogens but between the electrons i n the bonds themselve)>. An assessment in 2007 of the two effects indicated that tOrsional strain accounts for about 75% of the energy difference
oo.
52
CHAPTER 2 • ALKANES
0
120
60
i~
L~ll H
..__/II
180 dihedn1 l Jngle, degree~
H~U H
II
If
II
H
H
II
Hxp: H H H
II
II
II II
II~H H~H
H
I
HA
360 (=0 )
11
II
300
240
II
H
H
H~H II H
Figure 2.3 Variat1on of energy with dihedral angle about the carbon-carbon bond of ethane. In this diagram. the rear carbon is held fixed and the front carbon is rotated, as shown by the green arrows. The dihedral angle plotted is the angle between the bonds to the red and blue hydrogens. Note that the staggered conformations are at the energy minima. and the eclipsed conformations are at the energy maxima.
between the eclipsed and staggered forms. and lhat the favorable orbital interaction account'> for about 25%. One staggered conformation of e1hane can convert into another by rotation of either carbon relative to the other about the carbon-carbon bond. Such a rotation about a bond is called an internal r otation (to differentiate it from a rotation of the entire molecule). When an internal rotation occur<.,, an ethane molecule must bri efly pa'>'> through the e<.:lip ed conformation. To do '>0. it must acquire the additional energy of the eclip~ed conformation and then lo~e it again. What i ~ the source o f this energy? At t e mperature~ above absol ute 7ero. molecules are in constant motion and therefore have kinetic energy. Heat i'> a manife<.tation of lhis kinetic energy. In a sample of ethane. the molecules move about in a random manner. much a' people might mill about in a large crowd. The~c moving molecules frequently collide. and molecules can gain or lose energ) in such collisions. (A n analogy is the collision of a bat wi th a ball: some o f the kinetic energy of the bat i:, lost to the ball.) When an ethane molecule gai ns sufficient energy from a coll ision. it can undergo imernal rotation. passing through lhe more energetic eclipsed conformation into another '>taggered conformation. Whether a given ethane molecule acquires ufficient energ) to undergo an internal rotation is ~trictly a maHer or probability (random chance). llowever. an internal rotati on i!-> more probable at higher temperature becau~e warmer molecu les have greater kinetic energy. The probabilit) that elhane undergoes internal rotation i~ reflected as its rate of rotation: hO\\ man) times per <,econd the molecule convert., from one '> taggcred conformation into another. Thi<; rate is determined by hov, much energy mu\t be acquired for the rotation to occur: the more energy requ ired. the ~mai ler the rate. In the case or ethane, 12 kJ mol - 1 (2.9 kcal energy is small enough that the internal rotation erhane mol - 1) is required. This amount
or
or
2.3 CONFORMATIONS OF ALKANES
53
i.., very rapid even at very low tempera tures. At 25 °C. a typical ethane molecule undergoes a rotation from one '>taggered conformation to another at a rate of about I 0 11 time~ per second! This means that the interconver.,ion between ~taggercd conformations lake'> place about once every 10- 11 second. Despi te this '>hort lifetime for any one ..,taggered con format ion, an ethane molecule spends most of i t~ ti me i n its staggered conformations. passing only transiently through its eclip'>cd conformation'>. Thus. an internal rotation i-; best characteri.ced not a~ a continuous -..pinning but a a con\tant succe'>'>ion of jump'> from one -..taggered conformation to another.
B. conformations of Butane Butane contain'> t\\O distingui'>hable type'> of carbon-carbon bonds: the two terminal C -C (blue), and the central bond (red}.
c-c
bond~
H-'C- CH 2- CI1 2- CH.1 butane
two type~ ofC-C bonds We'll consider internal rotation about the central C-C bond. Thi'> rotation i~ a bit more complex than the ethane case. but examinat ion of thi'> rotation lead~ to important new i n~ ights about molecul ar conformation. As w ith ethane. we use Newman projection'>. as shown in Fig. 2.4. Remember again that the projected bond-the central C- C bond in this case-is hidden in the ewman projection.
bnll-a111i stick models:
ii11c-aml wedge formulas:
1
CH3 1 \ l
lH
II ...-II
'
~ -----H""'' jc-c., II
t
H
~
(a) viewing a model of butane from one end
\_
I
4
(b) end-on view
(c) Nc\\'man projection
of the .:cntral c.:arbon-<:arbon bond Figure 2.4 How to derive the Newman projection of the central carbon- carbon bond in butane using ball-andstick models (top) and line-and-wedge formulas (bottom). The bonds and groups on the rear carbon of the projected bond are shown in blue. (Only one of the butane conformations is shown.)
54
CHAPTER 2 • ALKANES
1
kl mor __________ t _ 2.!1 ~j!!£~~ElJ _________ _ 0
I I
(
60
~ f
~J
(!II ~HH ._../
120
II
H:q:x. (II, H
H H
180 dihedral angle, degrees
lll
gauche
300
360 (=0)
H
H
H~H H
240
H
11(H~11II
H
H:e::H lll anti
l'
ll,l ¥
H H
H H
c
II
JiH~HH
gauche
Figure 2 5 Variation of energy with dihedral angle about the central carbon-carbon bond of butane. In this di· agram, the rear carbon is held fixed and the front carbon is rotated, as shown by the green arrows. The dihedral angle plotted is the one between the bonds to the two CH 3 groups.
The graph of encrg) as a function of dihedral angle in butane is gi,en in Fig. 2.5. ote once again that the variou!. rotational po sibilitie!. are generated with a model by holding either carbon fixed (the carbon away from the observer in Fig. 2.5) and rotating the other one. Figure 2.5 shows that the staggered conformations of butane. like those of ethane, are at energy minima and are thus the stable conformations of butane. However. not all of the <;taggered conformations (nor all of the eclipsed conformations) of butane arc alike. The different staggered conformations have been given special names. The conformations with a dihedral angle of 60° and 300° in Fig. 2.5 (or ±60°) between the two C- CH3 bonds are called gauche conformations (from the French gauchiJ: meaning ··to turn aside..): the form in which the dihedral angle i!> 180° is called the anti conformation. ()
''
CI
Ill
~"
H M CI-1
II~H H gauche conformation 0=60
:=e=SJ H () u
·-
anti conformation II JHO'
II l
((>(H
II ~ H H gauche conform atio n (I .~00' hO )
The relationship between bond'> also can be described with the terms gauche and anti. Two bonds thai have a dihedral relationship of ±60° are said to be gauche bonds. Two bonds that
2.3 CONFORMATIONS OF ALKANES
Further EXploration 2.1
Atomic Radii and van der waals Repulsion
55
have a dihedral relationship of 180° are said to be an ti bonds. mice that the~e terms refer to bonds on adjacent carbon . Figure 2.5 l.hOW'i that the gauche and anti conformation!. of butane have different energies. The anti conformation is the more stable of the two by 2.80 1-.J mol- ' (0.67 kcal mol 1) . The gauche conform at ion is the less stable of the two because the Cll 1 groups are very close together-so close that the hydrogen ~ on the two groups occupy each other"s space. You can sec this with the aid of the pace-filling model in Fig. 2.6a. Thi problem can be discu~!.ed more preci!-ely in terms of atomic size. One measure of an atom's effective !--ite is its van dcr Waals radius. Energy is required to force two 110nbonded atoms together more closely than the sum of their van der Waals radii . Because the van der Waals radiu!> of a hydrogen atom i~ about 1.2 A. forcing the center<; of two nonbonded hydrogem. to be closer than twice thi\ distance require.., energy. Furthamore. the more the two hydrogens are pu~hed together. the more energy i~ required. The extra energy required to force two nonbonded atoms wi thin the !>Um of their van der Waals radii is called a van der Waal s repul si on. Thus. to attain the gauche confor mation. butane must acquire more energy. In other words. gauclte-blllane is destabili:ed by \'Onder Waals repulsions betH'een nonbonded hydrogens on the two CH1 groups. Such van der WaaJ<. repulsions are ab~ent in ami-butane ( ee Fig. 2.6h). Thu . ami-butane is more '\table than !(lllll'he-bULane. As with ethane. the ecli pscd conformations of butane arc deswbilized by torsional strain. Bu t. i n the conformation in which the two C-CH, bonds arc eclipsed. the major source of in'>tabi li ty is van der Waal repulsions between the methyl hydrogens (Fig. 2.6c). which are forced to be even closer than the) are in the gauche conformation. otice that thi<; is the mo~t un<,table of the eclipsed conformations (8 = in Fig. 2.5). It is important to understand the relative energies of the butane conformution~ becau!.e. when different !--table conform ations are in equi librium. the 11/os/ stable coJ((ormation-1/ie conformation oflou·est energy- is present in f(reatest amou/11. Thus. the anti conformation of butane is the predominant conformation of butane. At room temperature. there arc about twice as many molecule of butane in the anti conformation as there are in the gauche confonnation.
oo
11- H Jistann· j, lc" th.tn the 'um ol \an Jcr \\Jab r.1dii I
I
1-
1 I
no van der Waals repulsions
II-II di, tancc' .ln.' bs than thl· sum ohan dl·r \\'a.tb radii
''I
(a) gauche- butane
(b)
a111i-but.1ne
(c) butane with
C-CH3
bond~ cclip:.ed
Figure 2.6 Space-filling models of di fferent butane conformations with the methyl hydrogens shown in color. (a) Gauche-butane. A hydrogen atom from one CH 3 group is so close to a hydrogen at om of the other CH3 group that these hydrogens, shown in pink, violate each other's van der Waals radii. The resulting van der Waals repulsions cause gauche-butane to have a higher energy than anti-butane, in which this interaction IS absent. (b) Antibutane. This conformation is most stable because it contains no van der Waals repulsions. (c) Butane with the C- CH 3 bonds eclipsed. In this conformation, van der Waals repulsions between the hydrogens of the two CH1 groups (pink) are even greater than they are in gauche-butane.
56
CHAPTER 2 • ALKANES
The gauche and anti conformation~ of butane interconvert rapidly at room tcmperaturca!lno-.t a rapid!) tl'- the ...raggered form'> of ethane. Becau~c the cdip.,ed conformation<; lie at energy maxima and arc unstable. they do not exi'>t to an) mca!-.urablc extent. The investigation of molecular conformation!- und their relative energie~ is called <·onformntionaJ analysis. In this section, we· ve le<~rncd ~ome import ant pri nciplc!-. of conformational analy<,is that we'll
be
able to apply to more complex molecule!.. Here is a summary of the~e
principle:-,:
I. Staggered conformation!. about -.inglc bond' are favored. atom~) occur when ::~tom:-. are "squee;,ed" closer together than the sum of their van der WaaJ~; rad ii.
2. Van dcr Waals repulsion!-- (repulsions between nonbond..:d
3. Conformation!-. containing van der Waal'> repulsion. are lc's 'table than conformation<. in which \uch repul ion~ are ab'>ent.
4. Rotation about
c-c
single bonds i!> so rapid that it i!> hard 10 imagine <,eparating conformations except at very low temperature.
Ora\\ a Newman projection for the anli conformation about the C3-C4 bond of 2-methylhexane, viewing the bond so that C3 is nearest the ob~erver.
H 3C- CI I-CH 2- CI 12-CH 2-CI I, I i • CHJ I I
c
c..·
2-methylhexane
Solution
Fir!>t draw a "blank" Newman projection to represent the projected bond. Remember that the projected bond itself (the C3-C4 bond) is invi<.ible in the projection. Either template below can be U\ed.
We arbitrarily pick the template on the left. In the view dictated by the problem. the front carbon is C3. Identify the three groups mwchcd to C3 with bonds other than the projected bond. These groups are H, H. and the H 3C- y H - group. Put thc~e on the front carbon of the Newman Cll _, projection. It doesn'tmaller 11hich bonds go to 1rhich groups as long a.\ all groups are on the frrmt carbon. It'' important to undcf\land that, bccau-.e we are not e"amining the bonds within the large group. we can condense this group to (CH 3) 2CH- or even C3H 7- .
We then identify the groups attached to the back carbon (C4) by bonds other than the projected bond. These groups are H. H. and -CH 2CH 3. ow it does matter'' here we put the~e group • becau'>e we are a~ked for the ami conformation. The -CH~CH 3 group must be placed anti to the (CH1 ) 2CH- group. Remember that "anti" means a dihedral angle of 180°.
2 4 CONSTITUTIONAL ISOMERS AND NOMENCLATURE
57
2-meth ylhexane anti conformation about C3-C4 bond
Remember that Newman projection~ are u\ed to examine confonnations about a particular bo11d. If we want to examine the confonnation!. about <,everal different bond-.. we must draw a different ~et of cwman projections for each bond.
IQ;ie):JI¥tJ
•••• • •••m• •
2.3 (a) Draw a Newman projection for each conformation about the C2-C3 bond of isopentane, a compoun d containing a branched carbon chain.
isopentanc
Show both staggered and
ec l ip~ed
conformations.
(b) Sketch a curve of potentia l energy versus dihedral ang le for isopcntane. similar to that of but ane in Fig. 2.5. Label each energy maximum and m inimum with one of the conformations you drew in part (a). (c) Which conformatioru. are likely to be present in greatest amount in a sample of isopentane? Explain.
2.4 Repeat the analysis in Problem 2.3 for either one of the tenni nal bond<, of butane.
2 .4
CONSTITUTIONAL ISOMERS AND NOMENCLATURE A. Isomers When a carbon atom in an alkane is bound to more than two other carbon atom!>, a branch in the carbon chain occurs at that position. The '>mallest branched alkane ha-. four carbon atom. A.., a re\ult. there arc two four-carbon alkane-.: one i!-. bwnne. and the other i-. iwlmume.
b ut ane
bp - 0.5° isobutane 11.7<-
bp
These arc di!Terent compound<; with different properties. For example. the boiling point of buYet both ha' e the -.a me molecular fortane i-. 0.5 C. \\'hereas that of isobutane i-. I I. 7
oc.
58
CHAPTER 2 • ALKANES
mula. C4 H 10. Different compound ~ that have the same molecular formula arc said to be isomers or isomeric com pounds. There are different types of i somer~. bomer. that differ in the connectivity of their atoms. '>uch as butane and il>obutane. arc called constitutional isomers or st ructura l isomers. Recall (Sec. 1.3) that connectil·ity i. the order in which the atom<; of the molecule are bonded.The atomic connectivities of butane and isobutanc differ becauo;e in isobutane a carbon is auached to three other carbons. whereas in butane no carbon is attached to more than two other carbons.
Which of the following four structure\ repre~ent constjtutional i~omers. which repre~ent the same molecule. and which one is neither i'>omeric nor identical to the others? Explain your an wers. CH3
I
CH3CHTHCH2CII, CH, A
B
D
Solution Compound~ must have the same molecular formula to be either identical or i'>omeric. Structure A ha~ a different molecular formula (C6 H14) from the other st ructure~ (C 7 H1h). and hence structure A represents a molecule that ib neither identical nor isomeric to the others. To ~ol ve the rest of the problem. we must understand that Le~t·is stru£·111re.1 show COII/Iectivi(l' only. They do not represent the actual shape of molecule\ unless we stan adding ~patial elements such a~ wedges and dashed wedge'>. This means tJ1at we can draw a gi1'l'll .urucwre many differentll·a,rl. Have you ever heard the old spiritual. "The foot-bone ·s connected to the ankle-bone ... '"? That's a song about connectivity of the typical human body. If the de cription fi ts you. its validity doesn·r change whether you are siuing down. ~tanding up. standing on your head. or doing yoga. Sintilarly, the connectivity of a molecule doesn't change whether it is drawn forwards, backwards, or upside-down. With that in mind, let's trace the connectivity of each structure above. Consider structures B and C. Each ha rwo CH, groups connected to a CH. and thatCH is connected to another CH, which in tum is connected to both a CH3 group and a CH!CH, group. In B. this connectivity panem ~tam on the left: in C. it \tans on the bouom. But II', the same in both. Because both -;tructures have identical connect ivitie~. they represent the same molecule. Structures D and 8 (or D and Cl have the same molecular formula C7H16• but. as you should verify. their connectivities are different. so they are constitutional isomers.
Butane and i'obutane are the on I) constitutional isomer'>\\ ith the formula CJH 111• However. more constitutional isomers arc possible for alkanes with more carbon atorm. There arc nine constitutional isomers of the heptane~ (C 7 H 111). 75 constitut ional isomers of the decanes (C 10 H12 ) and 366.319 constitutional isomers of the cicosane~ (C 20 Hd 1 From these few examples. it is easy to understand that mi llions of organic compounds are known and mi llions more are conceivable. It follows that organiting the body of chemical knowledge requires a system of nomenclature that can provide an unambiguous name for each compound.
B. Organic Nomenclature An organized effort to standardit.c organic nomenc lature dates from proposals made m Geneva in 1892. From those proposals the Internationa l Union of Pure and Applied Chemistry
2 4 CONSTITUTIONAL ISOMERS AND NOMENCLATURE
59
(IUPAC). a profe!>sional association of chemists. developed and ~anctioned several accepted systems of nomenclature. The most widely applied system in use toda) is called substitutive nomenclature. The IUPAC rules for the nomenclature of alkanes form the basis for the substitutive nomenclatu re of most other compound classes. Hence. it is important to learn these rules and be able to apply them.
c. substitutive Nomenclature of Alkanes Alkane'> are named by applying the following I 0 rules in order. Thi'> mean<., that if one rule doc\n 't unambiguously determine the name of a compound of interc'>t. we proceed down the list ill order until we find a rule that doc<.,.
I. The 1111hranched alkanes are 11amed according to !he 1111111ber of carhons. as shown in Tahle 2. 1.
2. For alkanes con1aini11g branched cor/)()// chainJ. de!ermi11e tlte principal elwin. The principal chain is the longest con tinuouc; carbon chain in the molecule. To illustrate:
When identifying the principal chain. take into account that the condensed ~lmcture of a gil·en molecule may he drawn in se1·eral d(f!erent ways (Study Problem 2.2). Thus, the following structures represent the same molecule. with the principal chain shown in red:
(Be '>ure to \'erify that these structure., have identical conncctivitics and thus represent the same molecule. )
3. ({ fii'O or more chains ll'ithin a \tntCillre hm·e the same length. choose as the principal elwin tlu• one ll'ith the greater 11111nher f~{ branches. The following strucwre is an example
or ~ uch a ~ituation:
Cfl
C I I2 -CH3
( II
I
Cl 13
''" c.1rl>o • 'l,un
0 1
I
II C · CH-Cil-< 11 1 -<'112-CH ,
- - 1
CH_1
MX
c.ulwn chain
"11h on• ' and1 ( thi~ •~
the pmpcr choice for principal chain)
The correct choice of principal chain i~ the one on the rig ht, because il has two branches: the choice on the left has on ly one. (It makes no difference that the branch on the lert is larger or that it ha~ addit ional branching w ithin itself.)
4. Numlu!r the carbons o.fthe principal elwin consecwil•elyjlvm one end to the other in the direction/hat gil·es the l01rer number to !he first branching poilll.
60
CHAPTER 2 • ALKANES
or
In the following ~ tructure. the carbon~ the principa l chain are numbered con!>ccuti vely from one end to give the lower number to the carbon at the - CH , branch. t
1'"'1"' numbering
- -
I hC-CH, -CH,-CH - CH, -CH ,
.
-
- I
-
CHI
5. Name each branch and identifv the carbon nttm!Jer occurs.
(~/'the
principal chain at "·hich it
In the pre\'iou.., example. the branching group is a -CII , group. This group, called a methyl group, is located at carbon-3 of the principal chain. Branching group~ are in general called s ubstitu ent~. and 'ub<;tituent:- derived from alkane\ are called alkyl gr oups. An alkyl gmup may contain an) number of carbon~. The name of an unbranched alkyl group is derived from the name of the unbranched all-.ane with the same number of carbons by dropping the final ane and adding y/. methyl ( = meth#¢
-CH , -C H ~C H 3
or
ethyl (
-CHlCH1CH,
STUDY GUIDE LINK 2 .1
Nomenclature of Simple Branched Compounds
+ yl)
cth#s( + )ll
prop) I
Alkyl substitucnts themselves muy be branched. The most common brunched alkyl group~ have special name~. given i n Table 2.2. These should be learned becau~e they will be encountered frequentl y. Notice that the "i<;o" prefix is used for 'lllNituents containing two methyl groups at the end of a carbon chain. Al"o notice carefully the difference bcl\~ccn an isobutyl group and a ~ec-but}l group: the~c t\\0 group'> are frequent!) confused.
lt.):j!fJJ
Nomenclature of some Short Branched-Chain Alkyl Groups
Group structure
Condensed structure
Written name
Pronounced name
CH-
(CH 1) 1CH
isopropyl
isopropyl
CHCH1 -
(CH 3)~CHCH~-
isobutyl
isobutyl
sec-butyl
secondary butyl or ·sec-butyl•
tert·butyl (or t·butyl)
tertiary butyl or "tert· butyl"
neopentyl
neopentyl
HsC
\
H3C H3C
\
I
H3C
CH 3CH~CH-
CH 3 CH 3
I I
H3C-C-
(CH1) 3C-
CHJ CH 3 Hl C- C- CHI CH 3
(CH 3) 1CCH 1
-
2.4 CONSTITUTIONAL ISOMERS A ND NOMENCLATURE
61
6. Comtmct the name by writin~ the carbon number of the principal elwin at 11·hich rile .\uf>.\fifUe/11 occurs. a hyphen. the nf/11/e
1111/IIC:
3-mcthylhcxanc
'
, pJI chain
number and name of alkyl >ubstituenl Notice that the name o f the branch and tht! name of the principal c..: hain arc wriuen together as one word. Notice also that the name it ~e lf has no relationship to the name o f the isomeri c unbranched alkane: that is. the preceding compound is a constitwional isom er ofheprane because it h a~ ~even carbon atoms. but it is named as a deril'(ltil•e of hexane, because its principal chain contains six carbon atom-..
arne the following compound. and giYe the name of the unbranched alkane of which it is a constitutional isomer.
Solution Because the principal chain ha~ \even carbons. the compound io; named a<. a ~ub~tituted
heptane. The branch i~ at carbon-4. and the substituent group at thi' branch is
Table 2.2 shows that this group is an isopropyl g roup. Thus. the name of the compound is 4-isopropylhcptane:
.1lkrl
f Irom Tahlt> 2.2
~mup ll.llllC
Because this compound has the molecular formula C 10H22 , it is a constitutional isomer of the unbranched alkane decane.
62
CHAPTER 2 • ALKANES
7. If the principal chain conwins multiple substitue/11 groups, each su/)l{ituent receil·es it.f own number: The prefixes di. tri. tetra. and so 011. are used to indicate the m mtber of ide111ical suhstituem s.
Which two of the following structures represelll the same compound? Name the compound. H3C- CH,
I I H,C-CH , -CH . - I
CH - CH ,
.
CH 3
c
B
A
Solution The connectivities of both A and C are the same: [CH 3, CH 2, (CII connected to CH3) . (CH connected to CH ~). CH 2, CH 3] . The compound represented by these wucturcs ha<, six carbon~ in it!> principal chain and is therefore named a~ a hexane. There are methyl branche., at carbons 3 and ..t. Hence the name is 3.4-dimethylhcxane. (You should name compound B after you study the next rule.)
8. If substituent groups occur at more than one carhon o,{ the principal elwin. altem atil'e llt11111Jeri11g schemes are compared TILimher hy nuntber. and the one is chosen that gil•es the smaller number at the first point o f difference. This is one o f the trickiest nomenclature rule~. but i t is easy to handle if we are systematic. To apply th is rul e. wri te the two possible numberin g schemes deri ved by numbering from either end of the chain. In the following example. the two scheme' arc 3.3.5- and 3.5.5-.
Cl h
I .
I
H 3C- CH , - C-C H,-CH-CH, - CH 1
-
-
C ll ~
I
-
.
Cl-13
possible n ~ me;: 3,3.5 trimcthylhcptan e (correct) t•irncthdhl·ptanc 'i'1l •r•l· 1'
·'· <;, ~
A ded!>ion between the two numbering <.chcme!-> i;, made by a pairwise comparison of the numher !.Ct'> (3.3.5) and (3.5.5 ).
2.4 CONSTITUTIONAL ISOMERS AND NOMENCLATURE
63
!-loll' 10 do a painrise comparison:
""''"~ '"'''~~~ l'
wu1pare tl1rse fir;t
r
"
(3,3,5)
(3,5,5) rompt~re
these /t1st
Because the firH poim of difference in thc;.c '-Ct;. occurs at the second pair- 3 versus 5-the dcci'>ion i<; made at thi. point. and the fir'>t scheme is chosen. becau!>e 3 is smaller than 5. If there are difference!> in the remaining numbers. they are ignored. The c;um of the numbers is also irrelevant. Finally, it makes no difference whether the names of the J>ubstituents are the same or different; only the numerical locations arc used. The next rule deals with the order in which subsrituents m·e listed. or "cited." in the name. Don't confuse the citation order of' a substituent with its numerical prefix; they aren 't necessarily the same.
9. Substituent groups are cited in alphabetical order in the IUIIIIe reJ
H 3 C-CH,- CH - C II , -CH,-CH - CH ~
- I
•
- I
Cll,
.
CH 3
I -
Cll 1 5-eth yl-2-mcthylheptane (ethyl b cited hcforc methyl even though it h,t\ a higher number)
Cll
I I
J
CH 3CH,CJI 2- C - CI [, -CH - CH 2CH 1
.
CH 3
- I
.
CH 1CH 3
3-eth yl-5,5-dimcthyloctane (note th.u dimethyl beg.ins with the letter 111 for of citation)
purpo~cs
I 0. If the 1111111bering of different group.1 is not resoll·ed by the other rules. the first-cited group receil'es the loll'eslnwnbeJ: In the l'ollowing compound, ru le!> l - 9 do not dictate a choice between the names 3-ethyl5-methylheptanc and 5-ethyl-3-methy lhepwne. Because the ethy l group is ci ted first in the name (ru le 9). it receives the lower number. by rule 10.
64
CHAPTER 2 • ALKANES
3-cth yl-5-methylheptane Some situa tions of greater complexity arc not covered by these I 0 rule~: however, the~e will suffice for most cases.
rule~
+ij;J.l=iiM'
2.5
ame the following compound~. (al
Cll 1
I .
(b) compound 8 in Study Problem 2.-t on p. 62
CH 3
I
CII 3CIICHCH,CHCH 3
I
-
CH 3 (t')
CH2CH2CH3
I
CII 3 CH 2 CH 2 ~HCHCH 1 CH 2 Ci l1
(d)
CH 3
l
Cll 1
l .
CH 3CH 2~CH2 CH 2 Cl ICI 13
CH1CH3
Cl-13
o. Highly condensed structures When ~pace i-. at a premium, parcnthcsc\ arc ...ometimes used to form highly tures that can be written on one line. a~ in the fo llowing example. Cll
I
or
means
conden~ed
struc-
.I
H 1C-C-CH 1
.
I
Cll 1 When \UCh -.tructure~ are comple\. it i'> sometim e~ not immediate!) ob\ iou\. particular!) to the beginner. which atom inside the parenthe~es b connected to the atom ouhide the parentheses. hut a l ittle analysis will generall) ~ohe the problem. Usually the 'tructure b drawn so that one of the parentheses i ntervene~ between the atoms that are connected (except for attached hydrogen~). However. if in doubt, look for the atom within the parenthc~cs tha t is m issing i t~ u~ua l number of bonds. When the group inside the parenthe~cs is C ll ~. as in the previous example, the carbon has only three bond~ (to the 1-h). Hence. it must be bound to the atom out<>ide the parentheses. Consider as another example the CHpl I groups in the following structure.
means
Because the oxygen is bound to a carbon and ton hydrogen, it has i t!'> full complement of two bonds (the two unshared pairs are under~tood). The carbon of each CH 20H group, however. is bound to only three groups (two 1-1'., and the 0): hence, it is the atom that i-. connected to the carbon out!>ide the parentheses.
2 .11 CONSTITUTIONAL ISOMERS AND NOMENCLATURE
65
If the mean ing of a conden~ed ~trucwre is not immediate ly dear. uTile it out in less condellsed.fonn. I r you wi ll take the time to do this in a few cases. it should not be long before the interpretation of condensed Re~earch
~tructurc~
become more routine.
in 'ludcnt learning 'tratcgic' ha, ~hO\\ n that '>tudcnt
'ucce~., in organic chcmi'tl') b highly intermediate -.tep~ in a problem. Such step~ in man) ca"c' imohc '' riting 'tructure' and partial ;,tructures. Student\ ma) he tempted to 'kip such '>tep' hccau'>c they 'ee their profc\\Or). working thing'> out quickly in their hca(J... and perltap;, feel that they arc expected to do the ~ame. Profes.,ors can c.lu this hecause they have year.. of experience. Mo;,t of them probably gained their expertise th rough 'h.:p-by-step problem ~olvi ng. In ~orn e ca~es. the temptation tn skip ... rep' may be a con;,equencc of time pressure. If you arc tempted in thi;, direction. remember that a 'tep-by . ,rep approach applied to rdati\ dy fe,, problem' i., a bc11er e:-.pcnditure of time than ru\hing through man:r problem\. Stud) Problem 2.5 illu~tratc\ a ,tep-by-step approach to a nomenclature problem.
correlmed \\ ith '' hether a ~tudent take' the time to
uTile 0111
Study Problem 2.5 Write the Lewis structure of 4-sec-butyl-5-ethyl-3-methyloctane. Then write the structure in a conden\ed form.
Solution To thi~ point. we've been giving name!> to ~tructurc\. This problem now requires that we work ..in rcvcN.e.. and con~truct a '>tructure from a name. Don't try to write out the ~tructure immediately: ruther, take a l>ystem:uic. stepwise approach involving intermediate struclllre~. First. write the principal chain. Because the name ends in ocume. the pr~n c i pa l chain contains eight carbons. Draw the principal chain without its hydrogen atom~:
c-c-c-c-c-c- c- c ext. number the chain from either end and attach the branche~ indicated in the name at the appropriate po~itions: a sec-butyl group at carbon4. an ethyl group at carbon-5. and a methyl group at carbon-3. (U~e Table 2.2 to learn or relearn the structure of a sec-buryl group. if necessary.) H,C
(II,
( 11
CH,-
I
,.., burrl group
c-c-c-c-c- c-c-c I I I CH ,
CH 2-CH 3
Finally. fill in the proper number of hydrogens at each carbon of the principal chain so that each carbon ha~ a total of four bonds:
4-sec-butyl-5-ethyl-3-methyloctane To write the structu re in condensed form, put like groups atlac hed to the same carbon within parentheses. Nmicc that the structure contains within it two sec-butyl groups (red in the following structure). even though onJy one i' mentioned in the name: the other consi~t'> of a methyl branch and part of the principal chain.
66
CHAPTER 2 • ALKANES
Nomenclature and Chemical Indexing The world's chemical knowledge is housed in the chemica l literature, which is the collection of books,journals,patents, technical reports. and reviews that constitute the published record of chemical research. To find out what, if anything, is known about an organic compound of interest, we have to search the entire chemical literature. To carry out such a search, organic chemists rely on two major indexes. One is Chemical Abstracts, published by the Chemical Abstracts Service of the American Chemical Society, which has been the major index of the entire chem ical literature since 1907. The second index is Beilstein's Handbook of Organic Chemistry, known to all chemists simply as Beil-
stein, which has published detailed information on organic compounds since 188l.lnitially, a search of these indexes was a laborious manual process that could require hours or days in the library. Today, however, both Chemical Abstracts and Beilstein have efficient search engines that enable chemists to search for chemical information from a personal computer. Nomenclature plays a key role in locating chemica l compounds, particularly in Chemical Abstracts, but it is also possible to search for a compound of interest by submitting its structure. A search of Chemical Abstracts yields a short summary, called an abstract, of every article that references the compound of interest, along with a detailed reference to each article. Beilstein yields not only the appropriate references but also detailed summaries of compound properties.
2.6 Draw structures for all isomers of (Il l heptane and (b) hexane. Give their systematic names. 2.7 Name the following compound.\ . 13c sure to designate the principal chain properly before constructing the name. Cal CH,CH 2-CH--CH -CH2CH 2CH3
.
I
I
CH2
CH 3
I
CH2-CH~-C1 ! 3
(b)
CH3
G ll5
I I2 . CII 3CII 2CI12CH-C- - CH -CH 2CH,CH 3 I I CH 3 CH 3
2.8 Draw a structure for (CH3CH 2CH 2h CHCH(CH 2CH 3h in which all carbon-carbon bonds are shown explicitly: then name the compound. 2.9 Draw the structure of 4-isopropyl-2.4,5-trimethylheptanc.
E. Classification of Carbon Substitution When we begin our swdy of chemical reactions, it will be important to recognite different types of carbon substitution in branched compounds. A carbon is aid to be primary, secondary, tertiary. or quaternary when it is bonded to one, two, three. or four other carbons, respectively.
2.5 CYCLOALKANES AND SKELETAL STRUCTURES
Likewise. the hydrogens bonded tiary hydrogens. respective ly.
10
67
each type of carbon are ca lled primary, secondary, or ter-
· CII
I
Cl l 1
I
li ,C-CJ I - CII. - CH - C -Ci l, 1
\ v 'nond.tr\'
h)drn~cn'
1 ~I 1 ~prim.Jn hvdro!:cm
tertiary hydrogen
14;1.1:1041 •••• • •••• ••-
2. I 0 ln the stmcture of 4-isopropyl-2,4,5-trimethylheptane (Problem 2.9) (a) ldentify the primary, secondary, tertiary, and quaternary carbons. (b) Identify the primary, secondary. and tertiary hydrogens. (c) Circle one example of each of the following groups: a methyl group; an ethyl group; an isopropyl group; a sec-butyl group; an isobutyl group. 2 .11 Identify the ethyl groups and the methyl groups in tl1e stmcture of 4-sec-butyl-5-ethyl-3-
methyloclane, the compound discussed in Study Problem 2.5. Note that these groups are not necessari ly confined to those specifically mentioned in the name.
2.5
CYCLOALKANES AND SKELETAL STRUCTURES Some alkane conta in carbon chains in closed loops, or rings: these are called cycloalka nes. Cycloalkanes are named by adding the prefi x cycfo to the name of the alkane. Thus, the sixmem bered cycloalkane is called cvclohexane.
cydohexane The names and some physical properties of the simple cycloalkanes are given in Table 2.3 . The general formula for an alkane containing a single ring has two fewer hydrogens than that of the open-chai n alkane w ith the same num ber of carbon ato ms. For example, cyclohexane has the fo rmula C 6 H 12 , whereas hexane has the formula C 6 H 14 • The general formula for the cycloalka nes wi th one ring is C,H2,. Because o f the tetrahedra l configuration of carbon in the cycloalkanes. the carbon skeletons of the cycloalkanes (except fo r cyclopropane) are not planar. We 'II s tudy the conformatio ns of cycloalkanes in C ha pter 7. For now. remember only that planar condensed structures for the cycloalkanes convey no information about thei r conformations.
Skeletal Structures An important structure-drawing convention is the use of s keletal structures, w hich are structures that show only the carbon-carbon bo nds. In th is notation, a cyc loalkane is draw n as a c losed geometric fi gure. In a s ke letal structure, it is unders tood that
68
CHAPTER 2 • ALKANES
lij1:1!fJI
Physical Properties of some Cycloalkanes Boiling point (0 ( )
Compound
Melting point (0 (
)
Density (g ml- 1)
- 32.7
127.6
cyclobutane
12.5
- 50.0
cyclopentane
49.3
- 93.9
0.7457
cyclohexane
80.7
6.6
0.7786
cycloheptane
118.5
- 12.0
0.8098
cyclooctane
150.0
14.3
0.8340
cyclopropane
a carbon is located at each vertex of the figure. and that enough hydrogens are present on each carbon to fulfill its tetravalence. Thus, the skeletal str ucture of cyclohexane is drawn as follows:
0
----a carbon and two hydrogens are at each vertex
Skeletal structures may also be drawn for open-chain alkanes. For example, hexane can be
indicated this way: for When drawing a skeletal structure for an open-chain compound, don "t fo rget that carbons are not onl y at each vertex. but also at the mds of the stmcture. Thu~>. the six carbons of hexane in the preceding structure are indicated by the four vertices and two ends o f the skeletal structure. Here are two other examples of skeletal structures:
6 5
8 7
o-<
10 9
2,6-dirncthyldccane
isopropylcyclopcntane
3,3,4-triethylhexane
Nomenclature of Cycloalkanes
The nomenclature of cycloalkanes follows essentially the same rules used for open-chain alkanes.
methyk yclobutane
1,3-dimethylcydobutane
1-ethyl-2-methylcyclohcxanc
(Note alphabetical citation, rule 9.)
2 .5 CYCLOALKANES AND SKE LETAL STRUCTURES
69
The numerical prelix 1- i ~ not necessary for monosubstilllted cycloalkanes. Thus. the tlrst compound is methy lcyclobutane. not 1- methylcycloburane. Two o r mo re substitucnts. however. must be number ed LO indicate th eir re lative positions. The lowest number is assigned in accordance wi th the usual rul es. Most of the cyclic compounds in this text, like those in the preceding examples, involve rings with sm all a lk y l b ranches. In such cases, the ring is treated as the principal c hain. H ow-
a no ncyclic carbo n c hain contains more carbons than an at.tached ring. the ring is treated as the substituent.
ever, w hen
1-cyclopropylpentane (not pentylcyclopropane)
II
Study Problem 2.6
Name the foll owing compound.
Solution
This problem. in addition
lO
illustrating the nomenclature of cyclic alkanes. is a good
illustration of' rule 8 for nomenclature, the ·'first point of difference" rule (p. 62). The compound is
a cyclopentane with two methyl substituents and one ethyl substituent. If we number the ring carbons consecutively. the following numbering schemes (and corresponding names) are possible, depending on which carbon is designated as carbon- I :
I ,2.4-
4-ethyl- 1.2-dimethylcyclopcn tane
I ,3.4-
1-ethyl-3.4-dimethylcyclopentane
1.3.5-
3-ethyl-1 ,5-dimethylcyclopentane
•The correct name is decided by nomenclature rule 8 usi ng the numbering schemes (nor the names themselves). Because all numbering schemes begin w ith I. the second number must be used to decide on the correct numbering. The scheme 1.2,4- has the lowest number at this point. Consequently, the correct name is .;l-ethyl-1.2-dimethylcyclopentane.
Draw a skeletal structure of 1ert-butylcyclohexanc.
Solution
The real questi on in this problem is how to represent a rerr-bury l group with a skeletal
structure. The branched carbon in this group has four other bonds. three of which go to CH 1 groups. Hence:
CJ-h
Q-?-cl, CH1
skeletal structure of tert-butylcydohexane
70
CHAPTER 2 • A LKA NES
IQ;{e]:J04J
•••• • ••• .,,.- 2.12 Represent each of the following compounds with a skeletal stnrcture. CH~
(a )
I
CH3CH2CH 2CH - CH - C(CH3h
I
CH 3 (b) ethylcyclopentane
2. 13 Name the following compounds. (a )
(b)
2. 14 How many hydrogens are in an alkane of n carbons containing (a ) two rings? (b) three rings? (c) m rings? 2. 15 How many rings does an alkane have if its formula is (a) C8H 10? (b) C 7H 12? Explain bow you know.
2.6
PHYSICAL PROPERTIES OF ALKANES Each time we come to a new fami ly of organic compounds. we' ll consider the trends in thei r melting poims. boiling points. densities. and solubilities. collectively referred to as thei r phvsical properties. The physical properti es of an organic compound are important because they determine the condjtions under which the compound is handled and used. For example, the form in which a drug is manufactured and dispensed is af fec ted by its physi ca l properties. In commerci al agriculture, ammonia (a gas at ordinary temperatures) and urea (a crystall ine solid) are both very important sources o f nitrogen. but their phy ical properties dictate that they are handled and dispensed i n very different ways. Your goal should not be Lo memori ze physical properties of individual compounds. but rather to Jeam to predict trends in how physical properties vary with structure.
A. Boiling Points The boiling point is the temperature at which the vapor pressure of a substance equal s atmospheric pressure (which is typically 760 mm Hg). Table 2. 1 shows that there is a regular change in the boiling points of the unbranched alkanes w ith increasing number o f carbons. T his trend of boiling point w ithin the series of unbranched alkanes is particularl y appar ent in a plot of boiling point against carbon number (Fig. 2.7). The regular increase in boiling point of
20- 30 °C per carbon atom within a series is a general trend observed .fur IIICIII." types of organic compounds . What is the reason for this i ncrease'? The key point for understanding this trend is that boiling points are a crude measure o f the attracti ve forces among molecules-intermolecula r at tractions- in the liquid state. The greater ar e these intermolecu Jar attractions. the more energy (heat. higher temperature) it takes to overcome them bO that the molecules escape into the gas phase. in which such attractions do not exist. The greater are the intermolecular a1tractio11s within a liquid, the greater is the hailing point. Now, it is impon ant to understand that there are no covalent bonds between molecules. and furthermore, that i ntermolecular attractions
2.6 PHYSICAL PROPERTIES OF ALKANES
71
250 200 ISO-
too r
u 0
50
c:
·o
•
c:
·o ..0
• •
Ot-
•
- so r - 100
r
- ISO
r
- 200
0
•
•
•
I-
Q.
efJ
•
•
•
•
• I
I
2
8 number of carbon atoms 4
I
6
J
I
I
10
12
Figure 2.7 Boiling points of some unbranched alkanes plotted against number of carbon atoms. Notice the steady increase with the size of the alkane, which is in the range of 20- 30 •c per carbon atom.
have nothing to do with the strengths or the covalen t bonds within the molecules themselves. What, then. is the origin of these intermolecular a!lractions? In Chapter I. we learned that electrons in bonds are not confined between the nuclei but rather reside in bonding molecular orbitals that surround the nuclei. We can think of the total electron distribution a). an ..electron cloud.'' Electron clouds are rather •·squishy" and can undergo distortions. Such distortions occur rapidly and at random. and when they occur. they result in the temporary for mation of regions of local positive and negative charge; that is, these distortions cause a temporary djpole moment within the molecu le (Fig. 2.8, p. 72). When a second molecule is located nearby. its electron cloud distorts to form a complementary di pole, called an in duced dipol e. The posi tive charge in one molecule is attracted to the negative charge in the other. The attraction between temporary dipoles. called a van der Waals attract ion or a di spersi on i nteraction. is the cohesive interaction that must be overcome to vaporize a liquid. Alkanes do not have sign i ficant permanent dipole moments. The dipoles discussed here are temporary. and the presence of a temporary dipole in one molecu le induces a temporary dipole in another. We might say. ·'Nearness makes the molecules grow fonder." Now we are ready to understand why larger molecules have higher boil ing points. Van der Waals attractions increase with the sUJface areas of the interacting elecu·on cloud. . T hat is, the larger the interacting surfaces, the greater the magnitude of the induced dipoles. A larger molecule has a greater surface area of electron clouds and therefore greater van der Waals interactions with other molecules. ll follows, then, that large molecules have higher boiJjng points. The shape of a molecule is also important in determining its boi ling poi nt. For example. a comparison of the boiling point of the highly branched alkane neopentane (9.4 °C) and its unbranched isomer pentane (36.1 °C) is particularly striking. Neopentane has four methyl groups di sposed in a tetrahedraJ arrangement about a central carbon. As the followi ng spacefilling models show, the molecu le almost resemble~ a compact ball. and could fit readi ly
72
CHAPTER 2 • ALKANES
t~: 11:
molecules moving at ra ndom approach each other
the electron cloud of one molecule distorts randomly to form a temporary dipole
l 14:
temporary dipoles dissipate
13:
the dipole in one molecule induces a complementary dipole in the other
weak attractions develop between opposite charges (van der Waals attractions)
Figure 2.8 A stop-frame cartoon showing the origin of van derWaals attraction. The frames are labeled c,, t 2, and so on, for successive points in time. The t ime scale is about 10-•o s. The colors represent electrostatic potential maps (EPMs). The green color of the isolated molecules (I 1 and t4 ) shows the absence of a permanent dipole moment. As the molecules approach (t 1), the electron cloud of one molecule undergoes a random distortion (t2) that produces a temporary dipole, indicated by the red and blue colors. This dipole induces a complementary charge separation (induced dipole) in the second molecule (t3 ), and attractions between the two dipoles result. Through random fluctuations of the electron clouds (t4 ), the temporary dipoles vanish. Averaged over time, this phenomenon results in a small net attraction. This is the van der Waals attraction.
within a sphere. On the other hand. pentane is rather extended. is e llipsoidal in shape. and would not fit within the same sphere.
neopenlane: compact, nearly spherical
pentane: extended, ellipsoidal
The more a molecule approaches spherical proportions. the less surface a rea it presents to other molecules. because a sphere is the three-dimensional object with the minim um surfaceto-volume ratio. Because neopentane has less surface area at which van der Waals interactions with other neopentane molecu les can occur. it has fewer cohesive interactions than pentane. and thus. a lower boil ing point.
2.6 PHYSICAL PROPERTIES OF ALKANES
73
In summar'). by analy'iis of the boiling point'> of alkane!'.. wt: have learned two general trend<; in the ,·ariation or boiling point with structure:
I. Boiling point~ increa!>e "ith incrca..,ing molecular weight '' ithin a homologou~ !>Cries!) picall) 20-30 < C per carbon atom. Thi~ incrca~c i!-. due to the greater \'an dcr Waab attraction<; between larger molecule .... " Boiling point<.,tend to be lower for highly branched molecules that approach <;pherical propor1ion., becau!'.e they ha'c le~s molecular ..,urfacc a\'ailable for \an der Waal<> anractions.
B. Melting Points The melting point of a substance is the tempera ture above wh ich it is transformed spontaneou~ l y and completely from the sol id to the l iquid ~o,wte. The melt ing poi nt is an especially importanl physical property in organic chcmi~try hccau'c i1 i.., u~ed b01h to identify organic compound<; and to assess Lheir purity. M elting po i n l ~ arc u~u•dly dcprc~sed. or lowered. by impurities. Moreover. the melting range (the range or tcmpcraw re over which a substance melts). usually quite narrow for a pure substance, i!-. substantially broadened by impurities. The melting point largely reflects the stabi lit.ing in termolecular interaction~ between molecules in the crystal as well as the molecular symmetry. which determine' the number of indistinguishable way~ in which the molecule fit~ into the crystal. The higher the melting point. the more stable i~o, the crystal ~tructure relative to the liquid )>tate. Although rno~t all-.anes are liquid~ or gases at room temperature and have relatively low melting point\. their melting points nevertheless illu'>trate trends that are ob!-.ened in the melting point!-. of other types of organic compounds. One such trend is that melting point~ tend to increase\\ ith the number of carbon\ (Fig. 2.9). Another trend i~ that the melting point'> of unbranched all-.anes with an e\en number of carbon atom'> lie on a separate. higher cune from tho!-.e of the alkane~ with an odd number of carbon'>. Thi-. rencct~ the more effectiYe packing of the eYcn-carbon all-.anc!. in the cr) 'talline <;olid ~tate. I n other \\ ord!>. the odd-carbon alkane molecule-. do not "tit together"
dd carbons
10 8 6 number of carbons 4
12
Frgure 2 9 A plot of melting points of the unbranched alkanes against number of carbon atoms. Notice the general~ncrease of melting point with molecular size. Also notice that the alkanes with an even number of carbons (red} lie on a different curve from the alkanes with an odd number of carbons (blue). This trend is observed in a
number of different types of organic compounds.
74
CHAPTER 2 • ALKANES
Branched-chain hydrocarbons tend to have IO\\Cr melting points than linear ones because the branching imerferes with regular packing in the cryMal. When a branched molecule ha~ a . ubstantial symmetry. however. its melting point is typically relatively high because of rhe ea~c wi th which symmerrical molecule~ fit 10gether within the cry,tal. For example. the melting point of the very S) mmetrical molecule neopentane. - 16.8 °C. is con~iderabl) higher than that of the less ~ymmetrical pentane. -129.8 C. (Sec model!, on p. 72.) Compare also the melting poinl of the compact and ~ymmetrical molecule cyclohexane. 6.6 oe, and the extended and less symmetrical hexane. - 95.3 oe, In summary. melting points show the following general trend~: I. M elting points tend to increa~e with increasing molecular mass wi thin a series. 2. Many highly symmetrical molecules have unusually high melting poims. 3. A sawtooth pattern of melling point behavior (sec Fig. 2.9) i~ observed within ma11y homologous series.
iij;\.J:i!i4i
2. 16 Match each of the following isomer& with the correct boiling points and melting points. Explain your choices. Compounds: 2,2.3,3-tetramethylbutane and octane Boiling points: !06.5 °C, 125.7 Melting points: - 56.8 °C.
oc
+ 100.7 oc
2.17 Which compound has Ia) the greater boiling point? (b) the greater melting point? Explain. (Hint: What is the geometry of benLene?)
H
H
H
H
*H benune
c.
toluene
Other Physical Properties Among the other significant physical propertie!l of organk compound~ arc dipole momenrs. sol ubilities, and densities. A molecu le's dipole moment (Sec. 1.20) determines i ts polarity. which, in turn, affects i ts physical propertie~. Because carbon and hydrogen differ little in their , clcctroncgativitics, alkanes have negligible dipole moment~ and arc therefore nonpolar molecttles. We can see this graphically by comparing the EPM ~ of ethane, with a dipole moment of 1.ero. and lluoromethane, a polar molecule with a dipole moment of 1.82 D.
EPM of ethane
EPJ\1 oftluoromrthanc (H 1C-F)
Solubilities are imponant in determining which ~olvcnt), can be used to form :-olutions: mo~t reaction. are carried out in solution. Water '>Oiubility is panicularl) important for several
2.6 PHYSICAL PROPERTIES OF ALKANES
75
Ftgure 2 10 The lower density of hydrocarbons and their insolubility in water allows an oil spill in flood waters to be contained by plastic tubes at a Texas refinery in the aftermath of Hurricane Rita in 2005.
rea'>on\. For one thing. water is the ~olvent in biological ') -.tcm\. For thil> rea!>on, water solubility il. a crucial factor in the activity of drug'> and other biologically important compounds. There hal. abo been an increa ing ill!erest in the u-.c of water a\ a '>OI\'ent for large-scale chemical procel>'e' as part of an effort to control em iron mental pollution by organic solvent!>. The water '>Oiubility of the compounds to be u~ed in a \\Uter-ba-.ed chemical process is crucial. (We'll deal in greater depth with the important que:-tion of <,olubility and l>olvents in Chapter 8.) The alkane~ are. for all practical purpo~e-.. in\oluble in water-thll'., the !>aying. "Oil and water don't mix ... (Alkanes are a major con~tituent of crude oil.) The density of a compound is another property. like boiling point or melting point. that determines how the compound is handled. For example. whether a water-insoluble compound is more or less dense than water determines whether it w ill appear a~ a lower or upper layer when mixed with water. Alkanes have considerably lower den~ i tie!l than water. For this reason. a mixture of an alkane and water will separate into two distinct layers w ith the less dense alkane layer on top. An oil slick is an example of this behavior (Fig. 2. 10).
14if,]:JI4Wtl •••• • •••• ••-
2.18 Gasoline consists mostly of alkanes. Explain why water is not usually very effective in extinguishing a gasoline fire. 2. J9 Ia 1 Into a separatory funnel is poured 50 mL of CH ,C H~Br (bromoethane), a water-insoluble compound wilh a density of 1.460 g/mL, and 50 mL of water. The funnel is stoppered and lhe mixture is shaken vigorously. After standing, two layer.. separate. Which substance is in which layer'! Explain. (b) Into the same funnel is poured carefully 50 mL of he.\ane (den~ity = 0.660 g/mL) so that the olher two layers are not diMurbcd. TI1c hexane form~ a third layer. The funnel i stoppered and lhe mixture is shaken vigorou~ly. After \landing. two layers separate. Which compound(s) are in which layer? Explain.
76
CHAPTER 2 • ALKANES
COMBUSTION Alkane~ are among the least reactive type!-> of organic compound~. They do not react with common acid~ or ba~cs, nor do they react with common oxidi;:ing or reducing agents. A lkane). do. however, share one type of reactivity with many other t ype~ of organic compound-;: they arc flammable. This mean~ that they react rapidly with oxygen to give carbon dioxide and water. provided that the reaction i'> initiated by a suitable heat '>Ourcc. !>uch as a flame or the '>park from a spark plug. Thi'> reaction i~ called comhwHion. For example. the combu-.tion of methane. the major alkane in natural ga<;. i<; "'ritten a.;, follm\ '·
(2.1) This reaction i~ an example of complete combustion: combu!>tion in which carbon dioxide and water are the only products. Under condition'> of oxygen deficiency. incomplete combustion may also occur with the formation of ~uch byproducts as carbon monoxide. CO. Carbon monoxide is a dead ly poison because it bonds to. and displaces oxygen from. hemoglobin. the protein in red blood cells that transports oxygen to tissues. It is also colorless and odorless. and is therefore difllcult to detect without special equipment. The fact that we can carry a container of gimple mixing of aII-ane-. and OX) gen doe~ not initiate combu..,tion. Howe,er. once a <,mall amount of heat is applied (in the form of a flame or a ~park from a ....park plug). the combu.,tion reaction proceeds vigorou~l) \\ ith the liberation of large amounh of energy. Combu..,tion i-, of tremendou!. commercial importance becau ...e it liberate~ energy that can be u~ed to keep U!. warm. generate electricity, or move motor vehicle<,. I lowe\ Cr. combustion alo;;o liberates large amounts of carbon dioxide and water.
octane or its isomers
carbon dioxide
(2.2)
A;, Eq. 2.2 illu!>trate-.. every carbon atom of a h)drocarbon combine..,\\ ith two atOm!. of oxygen to generate a molar equivalent of carbon dioxide. and c\er) pair of hydrogens combine" with one o\ygen atom to generate a molar equivalent of water. The atmo'>pherc can hold a relative!) small amount of water. and when that i<. exceeded. water return-. to the Earth a. rain or snow. However. natural processes of removing carbon dioxide from the atmo'>phere are limited. After eons in which the C01 content of the atmosphere remained relatively constant at about 290 parts per million (ppm), the amount of C0 1 in the Eanh 's atmosphere began to ri se dramatically with the advent of the industrial age. The C0 2 level nO\\ approaches -WO ppm. an increase of more than one-third (Fig. 2.11 ). Mo~t of this increase ha'> taken place in the last30 )Car~. Bccau~c <.O much of it is produced. carbon dioxide i<. the mo..,t -;ignificant of several compound.., kno\\ n to be greenhou:-,e ga\e\, atmo'>pheric compound<, that act a<, a heat-reflective blanket over the Earth. Mo\t scienti"t" arc nO\\ com inced that the temperature of the Earth i., being increased by the effect of grecnhou<.e ga'>C!->: thi-. phenomenon i., known as global warming. The<,e scientists beliC\C that global warming i!> beginning to ha' c ~ignificant advcr\C em ironmcntal consequence~. such a-. an increa~e in the intcn~ity of hurricane~. the rapid receding of glaciers. and the lo'>~ of an imal and plant specie~ at an incrca~cd rate. Global warming predict-. that ultimately the ocean lcvcb wil l rise as polar icc melt:-, and coa~ta l noocling that wi II displace hunLireds of m iI I ion:- of people wi II likely occur. A., a result of these concern'>. along with concerns about the political in:-tability of the oil-producing regions of the world. the tlc\elopmcnt of alternative fuel'> ha'> become increao,ingl) urgent. Ideally. the goal is tO produce cheap and abundant fuel<, that \\-ill not. on COillbU<;tion. incrCa\e the net CO~ content of the atmo.,pherc. (This i5.sue is con'>idcrcd further in Sec. 8.9C.l
Ftgure 2.11 Atmospheric C02 levels for the past 1000 years. The data prior to 1958 were obtained from air bubbles trapped in dated ice core samples. More recent data were obtained from air sampling towers on Mauna Loa, Hawaii, by the Scripps Institute of Oceanography ( 1958- 1974) and the National Oceanic and Atmospheric Administration (NOAA. 1974-present).The i nset shows data obta ined since 1975 in more detail. These data show the seasonal fluctuations normally observed in C02 levels. Notice the continuous rise in C02 levels since the nineteenth centu ry.
Combu!>1ion find~ a minor bu1 importanl u!>e H). an analytical tool for the determination of molecular formulas. ln this type of analysi!-, I he ma~\ co~ produced in the combustion of a known mass of an organic compound i~ used to calculate the amount of carbon in the sample. Similarly. the mass of H 20 produced is used 10 ca lcula te the amount of hydrogen in the samother elements.) Comple. (Procedures have been developed for the comhu~tion analy~is bu'ltion analysis is illustrated in Problems 2.43 and 2.44 on p. 86.
or
or
2.20 Give a general balanced reaction for (al the complete combustion of an alkane (formula c . l12nd· (b) the complete combustion of a cycloalkanc containing one ring (formula C.,H2.). 2.21 Calculate the number of pounds of C02 relea\ed into the atmosphere when a I5-gallon tank of gasoline in burned in an aULomobile engine. Assume complete combustion. Also assume that gasoline i a mixture of octane i ome~ and that the density of gasoline is 0.692 g mL-•. (This assumption ignores about 10 volume percent of oxygenated additives.) Useful conver~ion factors: I gallon= 3.785 L: I kg = 2.204 lb. 2.22 Carv and Oi Oxhide drive their famil) car about 12.000 mile per year. Their car gets about 25 mile. per gallon of gru.oline. What il. the ··carbon footprint"' (pounds of C02 released imo the atmosphere) of the Oxhide family car O\'er one year? (U~c the assumption~ and conversion factors in Problem 2.21.)
78
CHAPTER 2 • ALKANES
OCCURRENCE AND USE OF ALKANES Mo!>t alkane!> come from petroleum, or crude oil. (The word petroleum comes from the Latin words for ··rock.. and ··air·: thus. ··oil from rock._:·) Petroleum i'> a dark. viscous mixture composed mo'>tl} of all-.ane and aromatic h) drocarbonc., (bentenc and it<., deri,:lli\eS) that are -;eparated b) a technique called fractional dist illation. In fractional distillation. a mixture of compound!> is slowly boiled: the vapor is then collected, cooled. and recondensed to a liquid. Because the compounds with the lowest boiling point' vaporitc mo'>t readily, the condensate from a fractional distillation i enriched in the more 'olatile components of the mixture. A~ distillation continues. components of progressively higher boiling point appear in the conden'>atC. A <;tudcnt who takes an organic chemistry laboratory cour<,c will almost certainly become acquainted with this technique on a laboratory scale. Industrial fractiona l distillations arc carried out on a large scale in fractionating towers that arc several stories ta ll (Fig. 2.1 2). The typical fractions obtained from distillation of petroleum arc shown in Fig. 2.13. Another important alkane source is natural gas. which ill mostly methane. Natural gas comes from gas wells of various types. Sign ificunt biological ~ources of methane aJso exist thttt cou ld someday be exploited commercially. For example. methane i); produced by the action of certain anaerobic bacteria (bacteria that function without oxygen) on decaying organic matter (Fig. 2.1-l, p. 80). This type of process. for example. produces ..marsh gas:· as methane wa~ known before it wa<; characterized by organic chemi~h. This ~arne biological process can be used for the production of methane from animal and human wa'>te. Methane produced lhi way is becoming practical a!> a local ~ource of power (Fig. 2.14 ). The methane is burned to produce heat that i!. convened into electricity. Although thi~ proccs~ generate carbon dioxide. the source of the carbon i the food eaten b) the human~ and animal-.. and the carbon in that food comes from atmospheric carbon dioxide by photOS) nthesi-.. In other words. the co! produced in thi<, proceso; is ..recycled col.. and doe~ not contribute tO a net increase in atmospheric co1.
Figure 2 12 Fractionating towers such as these are used in the chemical industry to separate mixtures of compounds on the basis of their boiling points.
2 .8 OCCURRENCE AND USE OF ALKANES
I raction n.tmc
Fract inn temperature
\umber of .:arbon'
79
l}pkal li\C
bottled gns
petroleum gas
gasoline
20-70 oc
C;-CIO
automobile fuel
naphtha
70- 120 oc
Cs-Ct1
chemical feedstock
kerosene
120-240 oc
Cto-Cto
jet fuel, paraffin
light fuel oil
240-320 °C
Cts-Clo
diesel fuel
heavy oil
320-500 °C
C! I-Cls
lubricants. heating oil
asphalt, tar
rond swfacing
rrn heat
Figure 2.13 Schematic view of an industrial fractionating column. The temperature of the column decreases from bottom to top. Crude o il is introduced at the bottom and heated. As the vapors rise, they cool and condense to liquids. Fractions of progressively lower boiling points are collected from bottom to top of the column. The figure shows typical fractions, their boi ling points, the number of carbons in the compounds collected, and the typical uses of each fraction after further processing.
A l kanes of low molecular mass are in great demand for a variety of purposes-especially as motor fuels-and alkanes available directly ti·om wells do not satisfy the demand. The petroleum industry has developed methods (called caralytic cracking) for converting alkanes of high molecular mass into alkanes and alkenes of lower molecular mass (Sec. 5.8). The petroleum industry has also developed processes (called reforming) for converting unbranched alkanes into branched-chain ones. which have superior ignition properties as motor fuels. Typically. motor fuels. fuel oils. and aviation fuel account for most of the world's hydrocarbon consumption. AJl AJ·abian oil minister once remarked. " Oil is too precious to burn." He was undoubtedly referring to the important use!' for petroleum other than as fuels. Petroleum wil l remain for the foreseeable future the principal source of carbon. from which organi<.; starting materials are made for such diverse produ<.;ts as plastics and pharmaceuticals. Petroleum is thus the basis for organic chemical feedstocks-the basic organic compounds from which more complex chem ical substances are fabricated.
80
CHAPTER 2 • ALKANES
Figure 2.14 Manure fermenters at Fair Oaks Farms, a large commercial dairy farm in northwestern Indiana, are used to produce methane from cow manure. Electricity produced from burning the methane is fed into the local power grid. The power produced is frequently sufficient to support a large fraction of the farm's power requ irements. Such fermentations are carried out by methanogens (methane-producing bacteria). The inset shows Mechonococcus volcae, one of many known methanogen strains. This bacterium was named for Alessandro Volta, who, in 1776, collected "combustible air" (methane) that was liberated while he was exploring a marshy section of Lake Maggiore in northern Italy.
Alkanes as Motor Fuels Alkanes vary significantly in their quality as motor fuels. Branched-chain alkanes are better motor fuels than unbranched ones. The quality of a motor fuel relates to its rate of ignition in an internal combustion engine. Premature ignition results in "engine knock, • a condition that indicates poor eng ine performance. Severe engine knock can result in significant engine damage. The octane number is a measure of the quality of a motor fuel:the higher the octane number, the better the fuel. The octane number is the number you see associated with each grade of gasoline on the gasoline pump. Octane numbers of 100 and 0 are assigned to 2,2,4-trimethylpentane and heptane, respectively. Mixtures of the two compounds are used to define octane numbers between 0 and 100. For example, a fuel that performs as well as a 1:1 mixture of 2,2,4-trimethylpentane and heptane has an octane number of SO. The motor fuels used in modern automobiles have octane numbers in the 87- 95 range. Various additives can be used to improve the octane number of motor fuels. In the past, tetraethyllead, (CH 3CH 2) 4Pb, was used extensively for this purpose, but concerns over atmospheric lead pollution and the use of catalytic converters (which are adversely affected by lead) resu lted in a phase-out of tetraethyllead over the period 1976-1986 in the United States and in the European Union by 2000. This was followed by the use of methyl tert-butyl ether (MTBE, (CH 3) 3C-0-CH 3] as the major gasoline additive. After a meteoric rise in MTBE production, this compound became an object of environmental concern when its leakage from storage vessels into groundwater was dis· covered in several communities in the mid-1990s. Because MTBE has shown some carcinogenic (cancer-causing) activity in laboratory animals, many cities and states have enacted a phase-out of MTBE usage as a gasoline additive. Ethanol (ethyl alcohol) can be used as a substitute for MTBE, and ethanol is produced by the fermentation of sugars in corn. Farming interests in the United States have advocated the use of ethanol for fuel, and it appears that the MTBE in gasoline w ill soon be replaced by ethanol in the United States. The demand for ethanol is so great that the price of corn has escalated sharply. This increase, in turn, has had a noticeable impact on the price of foods that depend on corn as an animal food source (for example, milk, chicken, and beef). MTBE, ethanol, and other oxygen-containing additives are collectively referred to as oxygenates within the fuel industry.
81
2. 9 FUNCTIONAL GROUPS, COMPOUND CLASSES, AND THE "R" NOTATION
or
In the 1970'>. the world experienced a period in \\ hich a relati\C ~carcit) petroleum products wa.., cau~ed largely by political force!>. Tile resulting dramatic effec t ~ on energy price~ and the co n~cquc nce~ in all sector:" of the economy afforded a tiny forcta~tc of a chaotic '"'orld in which energy i~ in truly short !-.upply. In 2005. oil prices and natural gas price'> (t he Iauer a<; aresu lt of I lurTicanc Katrina) again reached record highs. There is no doubt that eventually the world will exhaust its natural petroleum reserve!>. It i ... thu~ imponant that scienti:-.1\ develop new ~ource!> of energy. which may include new way., of producing petroleum from renewable sources.
FUNCTIONAL GROUPS, COMPOUND CLASSES, AND THE " R" NOTATION A. Functional Groups and compound Classes A lkancs arc the conceptual "'rootstock" o f organ it: chemistry. Rcplat:ing C- H bond~ of alkanes give~ the many functional gro up~ of organic t:hcmistry. A fu nctiona l g roup is a charactcri!-tit:ally bonded group of atom~ that ha' about the ~a me chemical reactivity whenever it occur~ in a varie ty of compounds. Compound'> that contain the <,ame functional group comprise a compound class. Con~ider the foliO\\ ing examples:
11 _,c {
H 3C
(
I
\
H
II
I <.. I
011
IhC
Oil
H
H
isobutylene
fundtorul gwup:
t)
ethyl alco hol
I
1
II c \
acetic acid
I
tunt lltm.ll g•oup: - ( - OH
II ·'" l
\ p
!
compo und cl.1~~: c;uboxylic acid compound cia)~: alkene
compound ci.Js):
For example. the functional group that i-, characteri'>tic of the aiJ...ene compound class is the carbon-carbon double bond. Most alkene' undergo the same I) pe~ of reaction),, and these rem:tion., occur at or near the double hond. Similarly. all compound'> in the alcohol compound cia!>~ contain an -OH group bound to the carbon atom of an alkyl group. The characteristic reactions of alcohols occur at lhe - 011 group or the directly allached carbon, and this func tional group undergoes the same general chcmic.:altransforrnation' regnrdlc!-.~ of the structure of the remainder or the molecule. Needle~~ to <>ay. :-.ome compound~ can contain more than one functional group. Such compounds belong to more than one compound cia'>s.
0
II II,C=CH-C \
OH
acrylic acid contain~ both C =C .md C0 211 a nd i) thu~ bo th an alke ne and
functional groups a carboxylic acid
The organitation of this text i!-. centered for the most part on the common functional groups and corre~ponding compound cla<,<,C'>. Although you will study in detail each major functional group in -,ub-.equent chapter<;. ) ou '>hould learn to recogni7e the common functional group!-and compound da<,<;e), now. The'>e arc <,hown on the inside front co,er.
82
CHAPTER 2 • ALKANES
B. "R" Notation Sometimes we'll want to use a general -.tructure to represent an entire class of compounds. In such a case. we can use the R notation, in which an R is used to repre. ent ull alkyl groups (Sec. 2.4C). For example, R- CI can be used to represent an al kyl chlorid e.
R-CI
II <.
could represent
H,C \
CH
Cl
R- = C II
CH
Cl
II ( li ( -
R
I
Cl
_) I{
l\dohcx'l-
Ju!lt a!. alkyl groups such as methyl. ethyl. and isopropyl are ~ub!ltilllent groups derived from alkanes. aryl groups are substituent groups derived from ber11enc and its derivatives. The simplest aryl group is the phenyl group, abbreviated Ph- , which is derived from the hydrocarbon benzene. Notice that each ring carbon of an aryl group not joined to another group bears a hydrogen atom that is not shown. (This is the usual convention for !-.keletaJ structures: see Sec. 1.5.)
H *l H
H
II
H
benzene
0 skclct
_......-- PI
can be written n lCJl'
1r" ..,
Other aryl groupi> are de ignated by Ar-. Thu~. Ar-OH could refer to any one of the following compounds. or to many others.
CJ Ar
0 II
cou ld represent
or
0
OH
Cl where \ r - =
II< - { ) -
Ar
0-
Although you will not study benzene and itil derivatives until Chapter I 5, before then you will see many example!. in wh ich phenyl and nry l groups are used as substituent groups.
2.23 Draw a wuctural formula for each of the following compound~. (Several po~~iblc in each case.) Ia) a carbo,.ylic acid with the molecular formula C2H~02 (b) an alcohol with the molecular fonnula C 5 H 100
formula~
may be
ADDITIONAL PROBLEMS
83
2.24 A cenain compound was found to have the molecular formula CsH110 1. To which of the following compound classes could the compound belong'? Give one example for each positive answer. and explain any negative rcspon cs. an amide
an ether
a carbox) lie acid
a phenol
an alcohol
an eMer
KEY IDEAS IN CHAPTER 2 •
Alkanes are hydrocarbons that contain only carbon-carbon single bonds; alkanes may contain branched chains, unbranched chains, or rings.
•
Alkanes have sp 3-hybridized carbon atoms with tetrahedral geometry. They exist in various staggered conformations that rapidly interconvert at room temperature. The conformation that minimizes van der Waals repulsions has the lowest energy and is the predominant one. In butane, the major conformation is the anti conformation; the gauche conformations exist to a lesser extent.
•
Isomers are different compounds with the same molecular formula. Compounds that have the same molecular formula but differ in their atomic connectivities are called constitutional isomers.
•
Alkanes are named systematically according to the substitutive nomenclature rules of the IUPAC. The name of a compound is based on its principal chain, which, for an alkane, is the longest continuous carbon chain in the molecule.
•
The boiling point of an alkane is determined by van der Waals attractions between molecules, which in turn depend on molecular size and shape. Large mol-
ecules have relatively high boiling points; highly branched molecules have relatively low boiling points. The boiling points of compounds within a homologous series increase by 20- 30 •c per carbon atom. •
Melting points of alkanes increase with molecular mass. Highly symmetrical molecules have particularly high melting points.
•
Combustion is the most important reaction of alkanes. It find s practical application in the generation of much of the world's energy.
•
Alkanes are derived from petroleum and are used mostly as fuels; however, they are also important as raw materials for the industrial preparation of other organic compounds.
•
Organic compounds are classified by their functional groups. Different compounds containing the same functional groups undergo the same types of reactions.
•
The "R" notation is used as a general abbreviation for alkyl groups; Ph is the abbreviation for a phenyl group, and Ar is the abbreviation for an aryl (substituted phenyl) group.
ADDITIONAL PROBLEMS 2.25 Gi,·cn the boiling poim of the fil".t compound in each ~ct. c'timatc the boiling point nf the ~econd. taiCll Cll C H ,Cfl ,CH .CH ~Br (bp 155 °C) Ol ,CII:C JI ,(:II ,CII ,CH~CII ! Br
(b)
(l' l
0
II
CH .CCII,:Cll -( II
C ll ~CH,
(bp 152 •c)
0
II
CH,Cit ,CCII -Cll CH ,CH3
0
II
Cl h<
(bp 118 •q
0
II
CII ,CCI--1 ,Cf( ,C H ,CH !CH !CH,
2.26 Dnt" the 'tructure' and gi"e the name~ of all of octane \\ ith tal five carbon' (bl carbons in their principal chain\.
'i'
isomer~
84
CHAPTER 2 • ALKANES
2."!.7 Label each carbon in the following molecule~ a-.. pri-
'"' cb (b(~
the correct name for an) compound' that are not named correct! y. raJ 2-ethyi-2.4.6-Lrimeth) I heptane (b) 5-neopemyldecane (cl 1-cyclopropyl-3,4-dimcthylcyclohexane
mary, \econdar). teniary. or quaternary.
(c..l) 3-hut) 1-2.2-dimethylhexanc 2.2~
Dra\v the \tructure of an all..ane or cycloalkane that meet\ ..:aeh of the following crneria. fa I a compound that ha~ more than three carbon~ and onl) primal) hydrogen\. (b) a compound that ha~ fiyc carbon' and onl) 'ec-
ondary hydrogen~. lt'l a compound that hal> only tertiary hydrogen~. (d) a compound that has a molecu lar mttS\ o f 84.2. 2.29
2.32 Although compound~ are indexed b) their IUPAC ~ub \lituti\e names. sometime~ chemi'" giYe whim\ical names to compounds that the) di,cm·cr. A'l<;i\t these two chemists b) pro' iding \Uh,lltUII\e name'> for their C0111iXlund'>. lal Chemi~t Val Lo~ipede i\olated an alkane with the fol lov. ing skeletal structure from asphalt scrapings fo llowing a bicycle race and named it '·Tourdefran~ane ...
ame each of the followi ng compound-. u~ing IUPAC \Ub,titutiYe nomenclature.
ra1
CH,
I .
CH~ -Cl-1 -CH ,
!1 3C-C- - CH -CH -CH ,
I
CH~-CH~-CHJ
(h)
(h) Chemist Slim Pickin~ i~olutcd
a compound with the following structure from the floor of a henhouse and
dubbed it "pullane'' (p11ll11s. Latin for chick).
(l')~
(d)~
2.33 Within each ~et. which tw o \tructurc" repre,em the <,;~me compound? (a )
11 1C......._
CH3 ll y\-yCl-13
fa 1 .t-i~obutyl-2.5-dimeth) Iheptane
(bl 2.3.5-trimelhyl-+propylheptane
II~CH:CH.l Ul
Cll 3
R
A (~keletal
structure)
I t' 1 5-sec-butyl-6-rert-butyl-2.2-di methylnonane
2..\1 The fo llowing labels were found on bottles of liquid hydrocarbons in the laboratory of Dr. lmu Turkey following hi-. di:,appearance under my'>leriou' circumstances. Although each name defines a 'tructure unambiguously. \Orne are not correct IUPAC \Ub,tituti\'e name,. GiYe
/ CH .1
lly\-yl-1
II ,C~C2H; 2.30 Dr;l\'. <;tructures that corre,p
Cll
CH, C2 H5 y f - y 11
H~CI I 1 CllzCII, (
85
ADDITIONAL PROBLEMS
bond and unusual!) large C- C- C bond angles. com-
(b)
pared with the similar parameter\ for compound B ( i\obutane ).
II
H
I
.. c /
II
:\
"- c( CH d_~
(CJJ ,)!C''/
c
1.6 11 A
C(CH ~lJ
::!.34 tal Dra11 a <.keletal '>tructure ol the wmpound in part (a) of Problem 2.33 that i'> different from the other til o compound\. and name the compound. Chi Dr;l\\ a Newman projection lor the mo\t <;table conformation of the compound in part
4.. c c-c
2•.'7 The anti conformation of 1.2-llichlorocth
CH,
4. c c-c
llo0
110.1\
B
Explain wh) the indicated hond length and bond angle arc larger for compound A.
2.4(1 Which of the fo llowing compound). ~hould have the larger energy barrier to internal rotation about the indicated bond? Explain your reasoning carefully. CCHd,C-C(Cil .l 1
(\II .J ,Si - SiC CH .J,
A
I!
2.4 I From v. hat you learned in cc. 1.3B about the relative
c -c
tation abom the carbon-carbon bond of chloroethane. II 1 C- CH,-CI. The magnitude of the energy b;uTier 1 to Internal ;otation i~ 15.5 k.f mol - (3.7 kcal mol- 1). Lahel thb barrier on your diagram. 2.36 Explain how you would expect the diagram of potential cncrg} 1er'u~ dihedral anglt: .1bout the C2-C3 (centra() carb
I 1.535 A 11 c··;. . c / ' Cl-1 ,~
c- o
length'> of and bomb. predict which of the lollm\ ing compound' -.hou ld have the lnrger energy dif/en•llce between gauche and anti conformation~ about the indicated bond. Explain.
2.-'2
! <~ I
U 1,0 - CH.:CH -'
CII ,CJI ,-CH.:CH,
A
u
\\'hat 1alue i!> expected for the dipole moment of the anti conforn1ation of 1.2-dihmnmcthane. Br- CH.: -CH, -Br'! !::.., plain.
(b) The dipole moment J.L of an) compound that undergoe' internal rotation can be cxpre,>cd a_., a weightec.l average of the dipole moments or each of it'> conformation~ by the following equation: /.l = lllN l
+ l.l!N' + J.L,N ,
(a) $(..etch a graph of potential energ) \Cr~u~ dihedral angle about the carbon-carbon bond. Show the energy difference' on }OUr graph and label each minimum and ma.\imum 11 llh the appropriate conformation of 1.2-dichloroethane.
in "hich /.l; i'> the dipole moment of conformation i. and tV, i'> the mole fraction ol conlormauon i. (The mole fraction of an) conformation i j, the number ol moles of i di1 1ded b) the total mole~ of all conformations.) There arc aboutll2 mole percent of anti
Chi Which conformation of thi-. compound i~ present in greatest amount? Explain.
conformation and about 9 mole percent of each gauche conformation present at equilibrium in 1.2dihromoethane. and the oh,ervcd dipole moment f.k of 1.2-dibromoethane i' 1.0 D. U\ing the preceding equation and the an~wer to part (a 1. calculate the dipole moment of a gauche conformation of 1.2-dibromocthane.
2..\l! Predict the most stable cunl'onnation of hexane. Build a model of thi~ conformation. (/Iiiii: Consider the con lorrnauon about each carbon-carbon bond separate!).) 2.39 \\hen the '>truciUre of compound A wa\ detennined in 1972. it
\\a~
found to hale an unu...uall) long
c-c
86
CHAPTER 2 • ALKANES
2A3 Thi\ problem illustrates how combu\tion can
be used to
Lure\. Then build model<. (or dra'' wedge-dashed wedge fonnula!>) for each compound. Does this alter your :111,wer in any way'? Explain. (b) Provide the same analysis a~; 111 part (a) for the same
determine the molecular fomutla of an unknown compound. A compound X (8.00 mgJ undergoe~ combustion in :t Mrcam of oxygen to give 24.60 rng of CO~ and
11.5 I mg of l-Ip.
reaction carried out on 2.2-dimcthylbutane.
(a) Calculate the mass of carbon and hydrogen in X.
(b) How many mole~ of H arc prc,cnt in X per mole of
2.46 To which compound cia~~ doc' each of the following
C? Express this a!> a formula C 111 ,.
compound' belong?
(c) Multiply thi!- fonnula b) \UCCC"I\C integer<> until
lal CJI ,CH,
\.
the amount of H i' al\o an mtcger. Thi' i\ the molecular fomwla of X.
I
!b )~
C= :-.!
C=O
Cll~CH ~
2.44 A hydrocarbon Y is found by combu~tion analysis to
(d)
contain 87 .171# carbon and 12.83% hydrogen by mass. (a I U8e tile procedure in Problem 2.43(b) and (c) to determine the molecular formula of Y. (Him: Remem-
0
o 'o
ber that all alkanes and cycloalkanc' mu\t contain et·e11
2.47 Org:tnic compounds can contain man) different func-
numbers of hydrogen,.)
tional group\. Identify the functional group~ (aside from
!bl Draw the structure of an aiJ..ane (which ma} contain one or more rings) con~iqent with the analyl>is given 111
the alJ..anc carbons) present in acebutolol (Fig. P2.47). a drug that block\ a certain pan ol the nervous system.
part (aJ that has two tental) carbons and all other
, amc the compound cia~' belong,.
carbon' \econdary. (More than one correct answer is po,~ible.)
w '' lm:h each group
1''1 Draw the structure of an alkane (which may contain one or more rings) consi~tcnt w ith the ana l ysi ~ given
2.48 Ia) Two am ides are constituti omtl
formula C~H 9 NO. and each contains an isopropyl
in part (a) that has no primary hydrogens. no tertiary
group as part of its structure. Give structures for
carbon atoms. and one quaternary carbon atom. (More than one correct an,wer i<. pos~ible.)
these two isomeric amide<.. (b) Draw the strucUJrc nf two other amide' with the formula c.~'\0 that donm contain i'>oprop) I groups.
2.45 lmagme a reaction that can replace one h}drogen atom It)
of an aiJ..ane at random with a chlorine atom.
I I
Draw the structure of a compound X that i<> a constitutional isomer of the amide~ in pan ... (a) and (b). but
I I
i~ 1101 an amide. and contains both ttn amine and an alcohol functional group.
- c - H - - C- CI
(d) Could a compound with the formula C 4 HqNO con-
tal If pentane were subjected to such a reaction. how
Lain a nitrile functional group'! Explain.
many different compou nds with the formula C 5 H 11 C I would be obtained? Give their Lewis struc-
acebutolol
Figure P2.47
i~omer~ and have the
3 Acids and Bases. The Curved-Arrow Notation r
Thi!> chapter concentrate!> on acid-ba..,e reactions. a topic that you have studied in earlier c h e mi~try course:-.. Acid-base reactions arc worth pecial attention in an organic chemistry course, because. first. many organic reactions are themselves acid- base reactions or are close analogs of common inorganic acid- base reactions with which you arc famili ar. Thi~ means that if ynu undcr~tand the principle' hchin(l implc acid- base reaction~. you al"o understand the prinl..'lpk' behind 1he analogou' organ ll..' rl.!action'. Seumd. acid ha-..e n:al..'t ions Pr9.vid -;jmp! c C\ chapter about the cun·etf-arr01r notation. a powerful de\ ice to help you folio\\, undcrc,tand. and even predict organic reaction . Final)) . acid-ba<..c reactions provide useful example' for di,ct•~s •on ol ...umc principles of chemical e~u!libnum.
3.1
LEWIS ACID-BASE ASSOCIATION REACTIONS A. Electron-Deficient compounds In Sec. 1.2C. you learned that covalent bonding in many ca<,e-. conform~ to the octet rule. '' hich !'. that the sum of the bonding and un<>hared valence electron'> ~urrounding a given atom equals eight (two for hydrogen). The octet nile rur ~ ·duct" rule in the ca'>c nf hydrogen) hold' without exception f0r covalent I) l1ondcu atom~ fron the lir-.. t and '-l'<.:ond periods of tbe periodic table. A lthough the electronic octet can be exceeded when atom:- from period 3 and higher arc involved in covalent bond~. the rule is often obeyed for main-group elements in these period:- a!. well. The octet rule ~tipulat es the maxin111111 number of electrons. but it I" p<.l'-"ihlt for an atom to ha\'C jeu'q than an QCte\ of electron'-. 11 particular. some compound' contain atom<> that are ,·hart uj w1 octet by one or mrm· cfet 'tmn pwr.\ . Such species are called cl ectTon-deficient comp()unds. 6ne example of an clectron-ddicient compound is boron trifluoride:
87
88
CHAPTER 3 • ACIDS AND BASES. THE CURVED-ARROW NOTATION
:'~ :
.. I .. : F-B-F: ..
..
boron trifluoride Boron trilluoridc i~ electron-delicient becau~e the boron. with six electrons in it~ valence shell. i!. two electron~. or one electron pair. '>hon of an octet.
B. Reactions of Electron-Deficient Compounds with Lewis Bases Electron-deficient compounds ha1·e a tendency to undel~f?O chenucal reactiom that complete their ,·a/C'nce-.\he/1 octels. ln such reaction\. an electron-deticient compound react~ with a !.pccie-, that ha:-. one or more un~hared valence electron pair).. An example i!, the association of boron trinuoride and fluoride ion:
h··rnn " ~ k~''"''
dlll•ll'nl
:F:
or ~uc h a reaction
:F:
.. ' I .. :f.-1 -f.: + boron trifluoride
.. :f-':-
~
fluoride ion
..
.. I _ ..
: F- 1 ~ -F:
. J;- . ~.o,.Jt.ddectronpair
(3.lal
tctrafluoroborate ion
In ~uch reaction~. 11 e electron-delldcnt cpmp<>und act<. as a Le11·i1 and. A L ewis acid i<; a tl1at acccpl'- an electron pair to form a new bond in a chemical reaction . Boron trifluoride is the Lewi~ acid in Eq. 3.1 a becau~c it accept~ an electron pair from the fluoride ion to form a new B-F bond in the product, tctrulluoroborate anion. The ~pccic<. thtll donate$ ~he electron pai r t() a l.ew i~ atiil t 1 form u new bon I ~~called a Lewis base. t-=l uoride ion is the L ew1s ba~c in Eq. 3.1 a. When an electron-deficient Lewis acid and a Lcwi~ base combine to give a -,inglc product. as in this example. th~ l'caction i' called a L C\\ is acid- base association ~rcc1e-.
..' •
rca(·tiim.
:F:
.. I .. :f .. -1 - ..F: , ] ., ' WI
I\
d
(dc\lnlll .IClCJ'IOf l
:F:
.. I_ ..
+ <1 l.ewts bas.: (elect ron donor)
:F- -F: ,----,-....J L...,.--,--,
(3.1b)
boron
h.t~
a
complete octet
Otice that a' a re~ult of this association reaction. each atom in the producttctranuoroborate ion ha" a complete octet. ln fact. completion of the octet pro' ides the major drh ing force for thi~ reaction.
A pcculiant) tn the octet-counting prot·edurc i~ e' ident in Eq .... J.l a and J. I b. The fluoride ion hao, an octet. After it ~hare:. an electron pctet in the produc1 - sFJ. You might a... t.. ...Ho'' can fluorine have an m:tct both before and after it .,hare' electron.,T The answer is that we count un:.hared pairs of electron~ in the fluoride ion. but in - B1· 1• '~ e a~ ... ign to the fluorine 1hc electrons in it~ unsharcd pair' as well a ... both electrons in the newly formed chemical bond. An apt analogy to thi~ "ituation is a roor person P marrying a wealthy person W. Before the marriage. P b poor and W is wealthy: ai'Jcr the marriage. W i~ still wealthy. and P. lillilicaJion for thi1> prac1ice nt counting electron\ twice is that it prm 1de!> an extremely u~eful frame\\ orJ.. for predicting chemical reacth it). ote again that the pn>cedure U\ed in counting electron-. for the octet differ:. from the one u-.ed 111 calculating formal ch:1rge bee Sec. 1.2Cl.
89
3 .1 LEWIS ACID-BASE ASSOCIATION REACTIONS
The rcvcr.,c of a Lewb acid- base reaction can be called a Le'' is acid-base di sociation. Hence. the di~~ociation of fluoride ion from - sFJ to give BF~ and F- - that i:-. the rever~e reaction in ~ll"· 3.1 a and 3. I h-i ... an example of a Lewis acid- base tlis-;ocialion.
Which of the following compound~ can react wi th the Lewis base Cl- in a Lewis acid-base association reaction?
:CI:
H
I
H -C
.. I .. =9-AI-9:
II
I
II
aluminum chloride
methane
Solution For a compound to react as a Lewis acid in an association reaction. i t must be able to accept an electron pair from Lhe Lewil> b:1se Cl- . In ;lluminum chloride, th e aluminum is short of an octet by one pair. Hence. aluminum chloride is an electron-deficient compound and can readily accept
an electron pair from chloride ion in an as\ociation reaction. as follows:
:CI:
:CI :
.. I .. =9-AI-9:
..
:~1:-
-t-
Lewis acid (dectron-dcficicnt compound )
.. I .. :CI-At=-CI :
-
..
I
..
: ~):
Lewis base
In contrast. every atom in methane ha\ the nearest noble-gas number
or electrons (carbon has
eight, hydrogen has two). Hence, methane i~ not electron-deficient and cannot undergo a Lewis acid- base "'~ociation reaction.
c. The curved-Arrow Notation for Lewis Acid-Base Association and Dissociation Reactions Organic chemi!lts have developed <1 !'!ymholic device for keeping track of electron pairs in chemical reactions: this device i~ called the curved-a rrow notation. ;h thi<., notation ~~ applied tq the reactionSOT !ye,vi~ b desc;ribed b) a "l11111. .. ol clcl'th lll'- tim11 til<' electron donor (Lewis base) to t(u' e/e('tJ'OJI (/('U'Jnor ( Lewi aci ~ l ·'1'h1-.. "clct·u·nn llo\\ .. i-., indicaled b) a curved arrow drawn from ·the electron .Hmrce to the elec trr)ll CIJ.'Cepror. Thi~ notation i~ applied to the reaction of Eq. 3.1 a in the following way: ui."Ctron s ur
llt'" h
for'lled bond
..
:F:
:1-: "
' /
-:F :
I .. ,..., B- F: /' I .. :F :
dt:dlllll
dcsl
..
I .. I
:F - B::.... F:
(3.2 )
: F:
IJ.IIIO 1
The red curved arrow indicate!> that an un-,harcd electron pair on the fluoride ion becomes the shared electron pair in the ne\\ ly formed bond of - sF~.
90
..
CHAPTER 3 • ACIDS AND BASES. THE CURVED-ARROW NOTATION
The correct application of the curved-arrow notation also involve~ computing and properly assigning the formal charge to the products. For each reaction involving the curved-arrow notation, thl' al,r.:ehrcik S!/111 o} tlw ch111:~<' ' r nthtrrf/ortmlls must equal the al~e!Jraic .511111 ofthe cfiorge1 on tl~e 7Jroduct . In other words, /r>lal t'ltZu:~e is conserPed. Thus, in Eq. 3.2, the reactants have a net charge of - 1: henct.:, the products must have the same nt.::l charge. By calculating the formal charge on boron and fluorine. we determine that the charge must reside on boron. To illustrate the application of the cuned-arrow notation to a Lewis acid- ba!>e dissociation reaction. let\ consider the di sociation of the ion - BF4 to give BF, and F- : thi'> reaction is the rever\e of Eq. 3.2. The curved-arro\\ notation for this reaction is ac; follow!>: :f:
:f :
.. I .: :f-B-=-F: .. I ..
~
.. I .. :F-8 + : F:.. I ..
:f :
:f :
Because the B- F bond breaks in thi!> reaction. is transferred to a fluorine to give fluoride ion.
14it.l:I0$1 -••• • •••m-
3.1
(3.3)
thi~
bond is the source of the electron pair that
u~e the cur\'ed-arrow notation to derive a l>tructure for the product of each of the following Lewis acid-base association reactions: be sure to a sign formal charges. Label the Lewis acid and the Lewis base. and identify the atom that donates electrons in each case.
la J
CH 3
I
I
:F:
..
+ H20 : -
ll,c-c+
.
(b)
..
Nl-1 3
CH 3
I ..
+ B-F: I ..
-
:p:
ELECTRON-PAIR DISPLACEMENT REACTIONS ...
A. Donation of Electrons to Atoms That Are Not Electron-Deficient In some react ions. an clectrbn pair is donated to
of donated
dcctron patr
II
H- +IN
I
H
H
H ammonium ion
+
1 :-:
:OH h ydroxide ion
c
•
I I
H-N H
+
f ..
H- 0-H
(3.4)
water
a mmo nia
In this reaction, a hydrogen of the ammonium ion receives an electron pair from the hydroxide ion. As a re!.ult. this hydrogen become'> bonded to the oxygen to give water. and the electron pair in the - 1-l bond of the ammonium ion becomes the unsharcd pair in the product ammonia. If the latter electron pair had not depaned. hydrogen would have ended up\\ ith more electrons than allowed by the octet rule.
3.2 ELECTRON·PAIR DISPLACEMENT REACTIONS
\reaction -..uch
91
thi<;. in whil:h on..: ckctron p;ur I'- lh-..plaL·cd Irom an atom tin thi-. eL'. clc~.otron p;ur I rom another alom. j, called an clcrt ron-pair dis placement reacticm. In man) '>Uch reaction\. an <~ton• '' trun-.,lcrrc:tl between t\\ o other atom<;. In thi~ example. a proton b tran~fcrrcd from the nitrogen of the ammonium ion to the oxygen of the hydroxide ion. fron t1 h)drug
ll)
U!'>
b) the donation ol anothL'I
B. The curved-Arrow Notation for Electron-Pair Displacement Reactions The curved-arrow notation is particularly useful for following electron-pair displacement reactions. This usage can be illustrated with the reaction or Eq. 3.~. In this case. nm arrow,, are require~!, one for the donated electron pair and (Inc for the di,placed electron pair: don.ll.:d ~kdrnn p.•u
I
H
+I H-N I
H
II
/ 'l :-:
:OH
•
•
I
11-N
I
H
+ H - 0-H
(3.5)
II
'Vot1ce that in all u~e~ of the. cun cd-arrm' notatHlll. each cun eLl !uTu\\ ori&jnates ~lt the of e lectrons-an unsbared pair or a bond- .Jnd tenmnate-.. at the ci<'.Hincuion of the e lectron pair. oticc al<.o the cnmen·aTion of rOtalcflar~e on each \ide of the equation. as discussed in Sec. 3.1 C. The algebraic sum of the charge~ on the left .,ide i~ tero: hence. the net charge of all ~pccic:-. on the right side mu~t also be tero. The donah::d electron pair!> can nrig111atc I rom hond' "' ''ell ,1, un-.h;,ued pairs. Thi is illu!>tratcd by the reaction of - sH~ with water to gi'e dihydrogcn (H~) and hydroxide. \Otlrce
STUDY GUIDE LINK 3 .1
The Curved·ArTOW
Notation
do ),Jt.:tl< •dr<> .• 1 .111
H
!
_I /.-:-" .. 11-B - H H - OH I H
__..
I
H
+
11 -13
\
I
11- H
+-
OH
(3.6)
H
In this notation. the bond corresponding to the donated electrons is "hinged" at the tTansferred atom (the H of the B-H bond): it swing~ away from the boron and towards the atom that receive~ the electrons (the H of water). The acceptor atom can be an atom other than h)drogen. a\ "hown in Study Problem 3.2. Study Problem 3.
II
Gi"e the cun·ed-arrow notation for lhe following reaction. H
I .. .. H -C-CJ: + - :OJ-1 I .. .. H
II -
I .. I
II
..
.. + - :c..J:
H -C-OH
92
'
CHAPTER 3 • ACIDS AND BASES THE CURVED-ARROW NOTATION
•
Solution In thi~ reaction, an un~hared electron pair from the oxygen of - oil displacel. the electron p:tir from the C-CI bond onto the chlorine; the carbon atom i~ tmn~ferred from the Cl to the oxygen. Because this is an electron-pair displacement reaction. two arrows are required. Remember that a curved an-ow is drawn f'rom the wurce of an electron pair to it~ desrirwrion. The soult'l' of the donated electron pair is the - oH ion. The desrination of the donated electron pair i ~ the carbon atom. Hence. one curved arrow goes from an electron pair of the - oi l (any one of the three pair,) to the carbon atom. Becau~e carbon can have only eight electron,. it mu~tlo~e a pair of electron~ to the chloride ion. which i~ formed in the reaction. Hence. the source of thil. electron pair i~ the C- CI bond: its destination i~ the chlorine. The curved-arro" notation for this reaction is a, folio"'': H 1
H
n.
I ..
H-C-CI : -
1-l-C-OH
1\ ..
..
+ :CJ:-
1
H "-_
H
Ho :(Be
sure to read Study Guide Link 3.1 about the different ways that cul'\ed am>ws can be drawn.)
Study Problem 3.2 shows how to'' rite the curved-arrow notation for a completed reaction. Study Prohlem 3.3 !>how" how to complete a reaction for which the curved-arrow notation i!> given.
•
Given the following two reactants and the curved-arrow notation for their reaction. draw the :.tructure of the product. II
..~ /'. H,N . I C=O: ..
-
CHI
SolutiOn The bond."> or unshared electron pairs at the tails of the arrows are the ones that will not be in the ~a me place in the product. The hcnd~ or the arrows point to the places at which new bonds or unshared pairs exist in the product. Use the following steps to draw I he product.
Step 1
Redraw all atoms just as they were in the reactants: H
step 2
Put in the bonds and electron pairs that do not change: H
\ c-o: I .. Cll 3
Step 3
Draw the new bonds or electron pairs indicated by the curved-arrow notation:
3.2 ELECTRON· PAIR DISPLACEMENT REACTIONS
93
new electron pair
I
H
\ .. H/cN-C-0: I .. CH 3 new bond
step 4
STUDY GUIDE LINK 3 2
Rules for use of the Curved-Arrow
Complete the formal charges to give the product. The algebraic sum of the formal charges in the reactants and products mu t be the same-zero in this case.
lv. ay-. r(H Jcn1bcr that curved unows s_hQv. the. flow o electron pair.~. not the movement of nuclei. Beginning students sometimes forget thi point. For example, the proton transfer from HCI to 0 11 might be incorrecrly written as follows:
Notation
1:---l
•••
I . -... I
Ho: ..
..
1i~cl=
HO-H
~
+
-
..
:CI :
(3.7)
This is inl:orrcct because it s hows the movement o f the proton rather than the fl ow of electron pairs. Someone accustomed to using the notation correctl y would take this to imply the transfer of H- to OH. an impos ible reaction! The correcr use of the curved-arrow notation s hows the flow of electron pair. . as follows:
H~.F/'H4.1:
----+-
HQ -H
-!-
:~1 :
( nour:
r
(3.8)
A-. you learn to use the curved-arrow notation. you will find the additional as istance in Study Guide Links 3.1 and 3.2 to be very useful. Be s ure to read and study these carefully.
3.2 For each of the fo llowing cases, give the product(s) of the transformation indicated by the curved-arrow notation. tal IIO;Y"CH2-Lc-J:
..
I
..
CH 3
- ~ (),. (cl H 3Al-H H }C-Br: ..
(d)
HC) o·\-. 2
II
o:"'
I+ '-=g:..
H2C
3.3 Provide a curved-arrow notation for each of the following reactions in the left-to-right direction.
+
H3N-CH3 :8~:-
94
CHAPTER 3 • ACIDS AND BASES THE CURVED-ARROW NOTATION
REVIEW OF THE CURVED-ARROW NOTATION A. use of the curved-Arrow Notation to Represent Reactions In thi:, chapter, you 've learned two types of reactions that can be described by the curvedarrow notation:
I. the a!l<.ociation reactions of Lewi~
base~
with electron-deficient compound!~ (and their
re,·cr-,c di!I!IOCiation reactions): and '> electron-pair di!~placement reaction<..
£rer_1 rcauiw1 wu .\ ludy
inroh·in~
ot 111 c1 cmnhilllllion
eleclron pain can he cma(\";,ed as one ofllwse 11ro reac;iun
r~f f/1(!¥/1_
In other words. all reactions involving electron pairs can be dissected ultimately into only two fundamental processes! Because both of these fundamental processes can be described wi th curved arrows. it follows that any reoclion involving e/ec1ron pair.1cw1 he described wi1h the mrl'ed-arroll' nowrion. Becau~c the l'tl.\lllll~jontv of reaction-, in organic rhemistr~ in\'olve the mm'\.'mcm of electron pair~. it follows that these reaction., arc but variations on a theme: the are extelhiom. of the . mastery of these '>imple inorganic example., i'> a necessary first tep in understanding the reactions of organic chemim). To -,ummariLe: 'Bccau-,e of ih fun[himcntal 11np\1rtam:e _the cun ed-arrO\\ notation. if proper!) u-,cd. ~·an he a tool of great po\\cr for foll\1wing. under,tanding. '>llllplll}lllg. and e\'en pn:tlll'ting tlw rc.tl"!IOil'- of organl
'
•
B. use of the curved-Arrow Notation to Derive Resonance Structures In Sec. 1.4, you teamed that re. onance structures are used when the structure of a compound is not adequately represented by a single Lewi'i structure. Rt''''nwl!'t' 1/rucrun'\ al11 a.n differ on/ b.' rile 1/Wicnu.:m of dc.J:./J:J.Jm. In the \'a t majorit) of the re~onance .,tructures you '11 encounter. the electron~ are moYed in pair\. Because the cun·ed-aiTO\\ notation i., used to trace the flO\\ of electron pairs. it follow~ that thi.., notation can al o be u-,ed to derire resonance <>tructure..,- in other words. to show hO\\ one re..,onance su·ucturc can be obtained from another. Study Problem 3.4 illustrates this point with two resonance-stabilited molecules that were db-.cussed in Sec. 1A.
In each of the follow ing sets, show how the \Ccond fir~t b) the curYed-arrow notation.
methoxymethyl cation
nitromethane
re~onance
structure can be derived from the
3.3 REVIEW OF THE CURVED· ARROW NOTATION
95
Solution (a) In the ~tmcturc on the left. the po~itively charged carbon is electron-deficient. The structure on the right is derived by the donation of an unshared pair from the oxygen to this carbon.
This tran~formation resemble~ a Lewi\ acid-base association reaction. :md the !:ame curvedarrow notation is used: a ~ingle cuncd arrow showing the donation of the un~hared pair of electrons to the electron-deficient carbon. (b) To deri,·e the structure on the right from the one on the left, an unshared electron pair on the upper oxygen must be used to fonn a bond to the nitrogen. and a bond to the lower oxygen muM be used to form an unshared electron pair on the lower oxygen. a~ follows:
•
Two arrows are required because the formation of the new bond require' the displacement of another. Thus. we use the curved-arrow notation for electron-pair displacements. In both of the preceding examples, the curved-arrow notation i~ applied in the left-to-right direction. This notation can be applied to either ~tmcture to derive the other. Thus. for part (a) in the right-to-left direction, the curved-arrow notation is as follows:
You should draw the curved arrow for pan (b) in the right-to-left direction.
An important point is worth repeating: hen though the u:-...: of c.:uncd arrow for deriving re<,onimcc tru<.:turc' i1-. identical to tha for dl!-.c.:ribin!! a reaction. the intcrcon' ..:r,tott of reso nan ·e tructurcs is not n reactiqn. The atom-. involved In rc)onmlc<: 'trucltlrc' do IUJI movs.,. The two -. t ructurc~ are. taken ogcther. a r ·pr..:J->l'ntat ion of a. ing/e /1/ole~·ule.
(a l Using the curved-arrow notation. derive a resonance structure for the allyl cation (shown
here) which shows that each carbon-carbon bond has a bond order of 1.5 and thatlhe positive charge is shared equally by both terminal carbon atom . (A bond with a bond order of 1.5 ha~ the character of a single bond plus one-half of a double bond.) [ H 2C-CII =CH2
~!-----;;~~~
..
? ]
allyl cation
(b) Using the curved-an·ow notation, derive a resonance structure for the allyl anion (shown here) which shows that the two carbon-carbon bonds have an identical bond order of 1.5
96
CHAPTER 3 • ACIDS AND BASES. THE CURVED-ARROW NOTATION
and that the unshared electron pair (and negative charge) b minal carbons. [ H!C-CH =CII!
....l(f - - - - ,jJJ)IJ-
?
~hared
equally by the two ter-
J
allyl anion (C) U~ing
the curved-arrow notation, derive a rc~onance MruCture for benzene ( hown here) which shows that all carbon-carbon bonds are identical and have a bond order of 1.5.
[0 -·
jJJ
l
benzene
3 .4
BR0NSTED-LOWRY ACIDS AND BASES A. Definition of Brensted Acids and Bases Although les general than the Lewis concept. the Bmnsted UJII'I:r acid-hase concept pro' ide., another way of thinking about acids and bases that i~ extremely impo1tant and useful in organic chemistry. The Br~nsted-Lowry definition of acid'> and ba~e' was published in 1923, the c.,ame year that Lewi formulated hi!. ideas of acidity and ba ...icit) . A 'pecie-.. that donates a pnlton in a chcmn.:al reaction i called a Bronsted acid: a <;pecie'> that , Cl:epts a proton il a chemical reacticm.is a Bnms ted base. The reaction of ammonium ion with hydroxide ion (<..ee Eq. 3.5) is an example of a Br~m \tcd acid-base reaction.
H
II
•
+1(\
',
H- N -
•
~- .. + :QH
I
I
H-N:
I
+
'
1 - 0H
(3.9a)
II
H
ammonium ion (a Bronsted acid)
~
hydroxide ion (a Bmnsted base)
ammo nia (a Br0nMl'd base)
water (a Bronsted acid)
On the left side of this equation. the ammonium ion is acting as a Brf)nsted acid and the hydroxide ion is acting as a Br0nsted base; looking at the equation from right to left, water is acting aJ> a Br0n<,tcd acid. and ammonia as a Bn;)nsted bac.,c. The "cla!.J>ical .. definition of a Bron!>ted acid-base reaction given above focu es on the mo,ement of a proton. But in organic chemistry. we arc alwa)!> going to focu!> on the mol'eme/11 of electrons. As Eq. 3.9a illustrates. any Brrm.Ht'd acid hme reaction cafl be described wi1h 1he curred-arrow notation for electron-pair di~p/aceme/11 reactions. A Bronsted acid-hnse reaction 1 nol))ing more than an electron-pair displacement reaction on a hydrovcn. It's the action of the electrons that causes the net tranc.,fer of a proton from the Bronsted acid to the Bronsted base. In terms of electrons. a Bmnsted ba'>c. then. i-, merely a special case of a Lewi~ base: a Bron ted base i<> a Lewis ba-..c that donate~ ih electron pair to a proton. A Bronsted acid is the specie that pro' ide~ a prown to the base.
3.4 BRONSTED-LOWRY ACIDS AND BASES
97
Bronsted acid- base reaction: tm electron-pair displacemmt rl'action 011 lzydroge11 If
II II
II
:--.. :
+
I
'
II
OH
(3.9b)
Bron,ted base: a
lC\\IS ba,e
that donate'
an elrctron pa1r to a pruton
When a Bronsted acid l q:.c~ a proton. it!. conjugatt! base j, formed : \\hen'' Hronsted ga~o gain' a proton. it:-. conjugate add i~ fo rmed. When a l3ron, tccl adcllo"l''- a proton. it become<. a 6 r0pf>tcd !;lase: this acid a11d the resulling puse con ~t i t u le a conj ugal ~ acid- base J>air. In any Brvmstetl acid- base reaction there are two conjugate acid-base pair~. Hence. in Eq. 3.9. +NH 4 and NH 1 arc one conjugate ac id-ba~e pair. and H10 and - oH are the other.
• t
I,onjug:uc ac10
base
P:illl----t
'
H
H I
(3.10)
I
H
..
STUDY GUIDE LINK 3.3
Identification of Acids and Bases
Notice that the conjugate acid- ba!.c relationship is across rite equilihri11111 arrm~·s. For example. + HJ and H3 are a conjugate acid-ba.'>e pai r. but + H 4 and - o il are nor a conjugate acid- base pair. Till' idem(ficmlon oT a collliJOIIIIJI m em and or tiWil' dt:pt:lld~ 011 1!(1\1' it hehm·es ina .1pe. cific du·mical reaction. Water. for e\ample. can act as either an acid or a base. Compl)Und\ that can 'tct a" either acio. or ba~c~ an: called a mphoteric compounds. Water is the archetypal e.xumpk of an amphoteric compound. For example. in Eq. 3.1 0. water i~ the conjugate acid in the acid- base pair Hp /-OH: in the fo llowing reaction. water i-. the conjugate base in the acid-ba~c pair Hp+ / H10: H II
I +:\- H
II
+ : )
. ...
II
I
11-t\ :
H
H
II
(3.11)
I{
l~id
+§;i.l:iiiMi
+
ba'e
3.5 In the following reactions, label the conjuga1e acid-base
pair~.
Then draw the curved-arrow
notati on for these reactions in the left-to-right direction. (a)
Nil>+ -:oH ..
(b) NH 3
+
.. ..
NH3
.- _. ~
-:NH 2 + ..
- :N H2
+
HzO=
+
N H4
q:- act as conju-
3.6 Write a Brl}nsted acid-ba!>C reaction in which 11 2{:?/-:QH and CH 3QH/CH 3 gate acid-ba e pairs.
98
CHAPTER 3 • ACIDS A ND BASES. THE CURVED -ARROW NOTATION
B. Nucleophiles, Electrophiles, and Leaving Groups The Br~nstcd-Lowry acid-base concept is important for organic chemi~ try because many reaction'> are Br~>n~ted acid- base reactions, and many others have clo~c analogy to BrV)nsted acid- base reactions. Let's look
!].
HO:
~
H 3C- Br:
-~
Hq-CH_, + :Br:
(3. 12a)
:B·r~
(3.1:2bl
··
An analogou BronMed acid-base reaction i' HO:
H---4-r:
-~
'
\
~
HO-H
+
These reactions arc very similar: Both reactions are electron-pair displacements initiated by the Lew i ~ base hydroxide. In Eq . 3. 12a, the hydroxide base donates electrons to a carbon. In Eq. 3. 12b, the hydroxide base donates electrons to a hydrogen. The process in Eq. 3.12a is a well-known organic reaction: the procc~s in Eq. 3. 12b is a well-known Bn~nsted acid- base reaction. The Bron~ted acid-ba<;e reaction is much faster than the reaction at carbon. but, other than that difference. the e processes arc es<.,entially the same. We'll sec analogie~ like this all the time in organic chemistry. What make~ them particularly useful is that we know a lor about Bronstcd acid- base reaction . We can appl) what we know about thc~c reaction ro understanding the analogous organic reaction~. This analogy leads to additional widely u~ed terminology that we have to master. As we learned in the previous section, a Lcwi~ ba~u (such as hydroxide on the left of Eq. 3. 12b) that d(lnatc' an l.' lectron pair to a prqtqn IS called a Brftnsted base. 1\ Lcwi<. base (~ uc h as hydroxide on the left of Eq. 3. 12a) thnt donate~ an de~o:tron pair to an atom other than hydrogen is cal led a nucleophile (from the Greek philos. loving; nucleophile = "nucleus-loving·'). Thu!>. hydroxide is a Bronsted ba e in one reaction and a nucleophile in the other. elec11vn-pair donation to hydrogen: \
!].
HO:
H - Br :
~
HO
H
..
+
:Br~
+
:Br~
f~
(3.13a)
.tnd ,, Hnmstt·d IIIN'
electron-pair donation to an atom otha than hydrogen: HO :
!].
H ,C-Br :
-f '-._ .
..
~
H(_)_- CH 3
..
(3.13b)
I
,, l l':\\ j, b.t,<: ,md J
nudt·oplulc
The atom that receives a pair of electrons from the Lewis
ba~e
is called an electrophile
r· ·lcctro n ~loving' "). An electrophile is a type of Lewis acid. An electrophilc can be a proton or it ~:an be another atom. What makes it an elcctrophilc is the same thing that makes it a Lewis acid: it receives a pair of electrons from a Lewis base. whether that base is a nucleophile or a Bronsted base.
3.4 BRONSTED-LOWRY ACIDS AND BASES
I\
t.
fl.
II () :
·~
··
d
I ,I
I I
' fl.
HO:
H 1• -Br:
l
f,
99
1- Br:
(3.14)
·~
a Bronstcd acid The group that receives electron). from the breaking bond. in this ca~e the -Br. is called a lem·ing wrmp. A group become-. a leaving group when one of it'> atom.., accepts an electron pair from a breaking bond. One of the author'<., studems has put it le!>" formall). but perhaps more descriptively: a leal'ing group i~ a group that takes a pair of bonding electrons and runs.
HO:
;;.
ll ,C-Br:
·~
(3.1 5)
·t'
~ lt'/11'111,\C ~rtlltf'
A leaving group in one direction of a reaction becomes a nucleophile in the other direction, as the following example demonstrate<.,. d,
u.uvphut. Ul uu;
.t
{11nmrd dircCiion I
'
I ~ ~~~
ltm wg group in th~
n., eT$C dJrn.uon
H 3C
'---
Br:
I ' .':'"\
JI.O- CJ-1 3 -+
+
: Iii :
(3.16)
!
The term /eal'ing group can al-.o be u'>cd in Lewis acid-b~ e dissociation reactions. ole the absence of a nucleophile in a Lewi.., acid- base dissociation. and the ab'>ence of a leaving group in a Lcwi<, acid- base association.
:F: ·· I .. :f-: .. '---------· BI ..F:
t
.1
:J
rca(.tlon;
tlu·n.~ I ' 11n '''="\ ing groun
I_ .. I ..
I
B-F: (3.17)
:F:
:F :
nuckapJulc 111 th~
ciS.OCttiiiCin
:F: j
.1
lcm·tug grauf' 111 1he
dt('<'l tllltllll react ion; I here«""
n(l.
ll'm,hit.
The term<, you've learned here are important because the) arc used throughout the world of organic chemistry. Your instructor will u-.c them, and we'll usc them throughout this book. To summari7c: I. A Lell'is base can act a!- either a Br0nstcd base or a nucleophilc. A Brv;nsted base is a Lewi~ base that donate:-. a pair of electrons to a hydrogen and re rnovc~ the hydrogen as a proton. A nucfeop!lile i!> a Lcwi)> ba~e that donates a pair of electron.., to an atom other than hydrogen.
100
CHAPTER 3 • ACIDS AND BASES. THE CURVED-ARROW NOTATION
2. An e/ecrrophi/e is the atom that receive~ a n electron pair from a Lc'' i' ba-;e. 3. A leaving group is a group containing an ato m that accepts an electron pair from one of its bonds. which is broken as a re~uh. This sectio n is about analyzing the roles of the va rio us specie~ in reac tion\. The definition~ developed he re describe these roles. You will lind that most of the rcaclion:-. you will study can be analyzed in te rms of these roles. and hence an understanding of thi ~ ~ect i on "'il l prove to be crucial in helping you to understand and even predict reactio ns. The lir'>l <>li.!p in this under-
standing is to apply the e definitions in the analysis of reactions. SIUd) Problem 3.5 points you in this direction.
Following is a series of acid-base reaction~ that represent the individual \tcp-. in a ~nown organic transformation. the replacement of - Br by - OH at a carbon bearing three alkyl g roup~. Considering only the forward direction, classify each reaction as a Br0n~tcd acid base reaction or a Lewis acid-base association/dissociation. Classify each labeled species (or n group within each species) with one or more of the following terms: Br¢nsted ba."Je. Bronsted acid. Lewis base, Lewis acid, nucleophile. elecrrophile. ancVor leaving gtvup. CH 3 1 H 3C-C-Br:
n.
I
..
.
Cll 3 ..
I I
Hc- c•
)
Cll 3
CH3 A
c
8
CH3
H
CH 3 H
I .~ I
H C-C
)
(3. 18a)
:Br:
..
CH 3
I
I+
H C-C-0-11
:0 - H -
I
)
I
D
(3. 18b)
..
CH3 E
8
CH3 H
I I+ ...---..... :OH H C-C-0-H 3 I ·u .. 2 CH3 E
F
CH 3 1-1 •
.,
I I
H C-C3
Cf-13
I .. +
0:
+
..
.
II- OJI,
(3.18c)
II
G
Solution Classify each reaction first, and then analyze the role of each \pecic-.. Reaction 3.18a is a Lewis acid-base dissociation. ( otice that a single cuned arrO\\ de~cribe~ the dissociation.) In compound A. Br is the leaving group. Reaction 3. 18b is a Lewis acid-base as~ociation reaction. Cation 8 ('>pccifically. it\ electrondeficient carbon) is a Lewis acid and an clcctrophile. Water molecule D is a LC\\ i' ba ...c and a nucleophile. Reaction 3.18c is a Br!i'lDSted acid- base reaction. Ion £and compound G con..,tilUte a conjugate B r~nsted acid- base pair, and compound F and compound H arc a conjugate Bmnsted ba~e-acid pair. The wa1er molecule F is both a Lewis base and a Br0nsted base. The proton of E that receives an electron pair from water is a Lewis acid and is also an electrophiIc. The part of E that becomes G is a leaving group.
3.4 BRONSTED- LOWRY ACIDS AND BASES
14if,]:i04i •••• • , •• ,,._
101
3.7
Work Study Problem 3 .5 for the re1·erse of each reaction 3.18a-c.
3.8
In each of the following processe:.. complete the reaction w.ing the curved arrow given: classify the process as a Brpn ted acid-base reaction or a Lcwi' acid-ba'e a~\ociation/dissociation: and label each specie (or part of each !>pecies) with one or more of the following terms: Bronsted base. 8TJ)nsted acid, Leu·is base. Lewis acid. nucleophile. electrophile. and/or lem•ing group. In each part, once you complete the forward reaction. draw the curved arrow(!>) for the reverse reaction and do the same exercise for it a\ well.
~
lal H 1C=CH 2
()..
H-~.r:-
OH (b)
.. .....----...... 1 11,0: B-OH I
-
~
OH
3.9
For each of the following reactions, give the curved-arrow notation and write the analogous Brc.msted acid-base reaction.
3.10
Thi~
problem refers to the reactions shown in Eq~ 3.12a and 3.12b. When an equal number of of hydroxide ion. H-Br. and H,C-Br are placed in <.olution. ''hat products are formed? Explain. ( HiJu: Which reaction i<. faster'?) mole~
c.
Strengths of Bronsted Acids Another a ... pect of acid- base chemistry th at can be widcl) applied to under5tanding organic reactions i!o the strengths of Br0nsted acid:-. and ha!>e'i. The \trength of a Brc:\nsted ac id is de te rmined by how well it transfers a pro ton 10 a swndard Br0n,ted ba~e. The s tandard base traditi o nally used for comparison is water. The transfer of a proton from a general acid. HA. to water i~ indicated by the following eq uil ibrium: (3.19) The l.!quil ibrium constant for this reaction i!> g iven by
(3.20> (The quantitie!-> in brackets are molar concentratiOn!> at equilibrium.) Because v.atcr is the soi\Cnt. and its concentration remai n!> effectiYel) corNant. regardlc!->\ of the concentrations of the other '>pecic~ in the equilibrium. we multiply Eq. 3.20 through by IH ~OI and thus define ano ther con<.,tant K.,. called the dissociation cons tant:
=K
K •
<"!
[H,Q] _ IA:IJH,O+I -
IHAI
0.21 )
102
CHAPTER 3 • ACIDS AND BASES. THE CURVED-ARROW NOTATION
Each acid has it~ own unique dissociation con~tant. The larger the di~sociation con rant of an acid, the more Hp+ ions are fom1ed when the acid is di'>sol\'ed in water at a gi\'en concentration. Thul>. the 'trcn!!th ofa Bronsfed acid i.1 mell\ured by the magnitude of its dissociattOII ('(Jif ~({////.
B ecau~e the di,.,ociation con<;tant<. of different Bmn'>ted acid\ CO\'Cr a range of man) power!> or I 0. it is useful to express acid strength in a logarithmic manner. Uc.,ing p as an abbreviation for negative logarithm. we can write the following definition~;:
pK. =- log K.
(3.22a)
pH = - log 111 ,0+1
(3.22b)
Some pK0 values of several Bronsted acids are given in Table 3. 1 in order of decreasing pK•. Becau!>e stronger acids have larger K. values. it followc., from Eq. 3.22a that stronger acids /Jcll'e 1'1/loflerpK; values. Thus, HCN (pK. = 9.4) is a ~tronger acid than water (pKa = 15.7). In other words, the strengths of acids in the flr~t column of Table 3. 1 increase from the top to the bouom of the table.
STUDY GUIDE LINK 3 4
The Difference between pK0 and pH
It i\ important 10 understand the difference between pi I and pK•. The p H i~ a measure of proton concentration and is an experimentally alterable property of a o.,oltttiun. TJ1c pK. i~ a fixed property of a Bmn~tcd acid. lf this difference isn't ~ecure in your mind. be 'urc to con~uh Study Guide Link 3.4 for further help.
,\.l l What is the pK. of an acid Lhat has a dissociation con tant of {a)
10-'
(b) 5.8 X l0- 6
(c)
50
3.12 What is Lhe dissociation constant of an acid that ha' a pK. of (a) 4 (b) 7.8 (c) -2
3.13 a l Which acid is the strongest in Problem 3.11 '? (b) Which acid is the strongest in Problem 3.12?
Three points about the pK. values in Table 3.1 are worth '>pecial emphasis. The first has to do with the pK. values for very strong and very weak acid.,. The direct pK. determination of an acid in aqucou<; ~olution i:-. limited to acid, that :u·c lr~' acidic than I I 10+ and more acinic th;.tn 11 ,0. The reason is that H30+ is the strongest acid that can ex i-;t in water. If we dissolve a ~trongcr acid in water. is imn~ediatel y ionit.es to 1-I p+. Similarly. -oH is the strongest base that can ex ist in water. and stronger base~ react instant ly with water to for m -oH. However, pK3 values for very strong and very weak acids can be measured in other solvents. and through various methods these pK. values can in many cases be u~cd to estimate aqueous pK. value . This is the basis for the estimates of the aciditic~ or \trong acids such as HCI and very weak acid!> such as NH 3 in Table 3.1. These approximate pK. value~ will suffice for many of our applications. The ~econd point is that much important organi~ chemi\11') i'> carried out in nonaqueous '1\ln·nh. In nonaqueou solvent . pK, value~ typicall) dtffer '>Ub!>tantially from pK, values of tbc ..am· acid determined in water. However. in '>Ome of the.,e -;oh·ent<;. the relatin: pK. yalues are rough!} the arne as the) are in water. In other nonaqueou~ <,olvents. though. even the relative order of pK. values is different. (We'lllearn about '>oh·cnt c!Tcctc, in Chapter 8.) De'>pite these differences. aqueou pK. value<; such a<; tho. e in Table 3.1 are the most readily available and comprehensive data on which to base a discus~;ion of acidity and basicit). The last point hac; to do with the K. of water. which i'> Io· 1 ~ 7• from which we obtain pK. "i= 15.7. Don't confuse this with the ion-product C0/1.\tallf of water. which i~ defined by the
3.4 BR0NSTED-LOWRY ACIDS AND BASES
lt.j;!!¥11
103
Relative Strengths of Some Acids and Bases
Conjugate acid
Conjugate base
pK.
- 35 1
NH3 (ammonia)
- :NH 2 (a mide)
RQH(alcohol)
15-19"
Rq:- (alkoxide)
HQH (water)
15.7
Hg:- (hydroxide)
R~H (thiol)
10-12*
R~:- (thiolate)
R3NH (trialkylammonium ion)
9-11'
R3N: (trialkylamine)
NHJ (ammonium ion)
9.25
H3N: (ammonia)
HCN (hydrocyanic acid)
9.40
- :eN (cyanide)
Hz~ (hydrosulfuric acid)
7.0
H~:- (hyd rosu lfide)
+
0 II
0
..
II ..
R- C- gH (carboxylic acid)
4-5*
R-C -g:- (carboxylate)
Hf.: (hydrofluoric acid)
3.2
:f.:- (fluoride)
-o-
H3C
{p·toluene S01H sulfonic acid)
-o-
H3C
- 1
H30~ (hydronium ion)
-1.7
H2S04 (sulfuric acid)
- 3'
H~l: (hydrochloric acid)
- 6to - 7
H~[: (hydrobromic acid)
- 9.5to
HCI04 (perchloric acid)
-10(?)1
(p -toluene· sulfonate,or *tosylate")
HzQ (water) HSO; (bisulfate)
:~1:- (chloride)
1
- 8to 9.5
H.l.: (hydroiodic acid)
so;
:~(:- (bromide)
1
10
=:(:-
1
(Iodide)
CIO; (perchlorate)
*Precise value varies with the structure of R. 'Estimates; exact measurement is not possible.
expression K" =I H30+][-0Hl = I is deli ned by the expression
o-
1 •
M 2, or - log K" = l 4. The ionit.ati on constant of water
Thi!. expression has the concentration of'' ater itself in the denominator. and thus differs from the ion-product constant of water by a factor of 1/55.5. The logarithm of thi ~ factor. -1.7. account'> for the difference between pK. and pK": pK" of H 20 = - log K,. - log (1/55.5) = 15.7
D. Strengths of Bronsted Bases The stret1[Hh of a Br0nsted bnse i ~ Wrcc t l)' rel ated to the pK., of its conj ugate acid. Thus. the base strength or fluoride ion is indicated by the pKU of its conjugate acid. HF; the base SLrength of ammonia is indicated by the pK~ or its conjugate acid. the ammonium ion. + H~. That is. when we <.ay that a base is weak. we arc also saying that its conjugate acid is sLrong: or. if a
104
CHAPTER 3 • ACIDS AND BASES THE CURVED-ARROW NOTATION
ba~e
STUDY GUIDE LINK 3 .5
Basicity constants
i'> -.trong. it~ conjugate acid i' weal-.. Thu-,. it i-, ea~y to tell \\hich of two ba~e~ i" -.tronger by loo!..ing at the pK. ,·alues of their conjugate acid'>: the stronger baw lw.1 the conjugate acid ll'ith lhe.,'Rn•l/11'1' (orlesr ncgari1·e) pK, For example. -c . the conjugate ba\e of HCN. is a weaker base than -oH. the conjugate ha~\! or water. because the pK,, or HCN i... lCS\ than that or water. Thus. the \trengths o l' the ba~c\ in the third co.lumn of Table 3. I increase from the bottom to the top of the table.
E. Equilibria in Acid-Base Reactions When a Bm1Ned acid and ba~c react. we can tell immcdiatcl) ''bet her the equilibrium lie~ to the right or let't b) comparing the pK. 'alue-. of the two acid<, invol,cd. The equilthrit/111 itt 1/te 1'('/1( 11011 oJ wt a( icl and a ba.·e lthrays faror.\ the ::,ide ll' ilh the n·eakl•r add and ll'eaker bttse-. For example. in the following acid-ba~e reaction. the equilibrium lie~ well to the right. becau!->e H,O i~ the weaker acid and -c is the weaker base. (stronger (wc;lkcr base) base)
+
HCN
(3.:!3)
OH -
pK. = 9.4
pK. - 15 .7
( ~tro nga acid )
('' eakcr ,JCid )
We'll frequent)) find it u~efulto C\timate the equilibrium constant\ of acid- ba-.e reactions. The equilibrium con\tant for an acid ba-,e reaction can be calculated in a \traightforward way from the pK,, \alue~ of the two acid~ involved. To do this calculation. \Ubtract the pK. of the acid on the left "ide of the equation from the pK,, of the acid on the right and tal-..e the antilog of the rc!-.u lting number. That il>. for an ucid-ba~c reaction (3.14)
in which the pK. of AH calculated by
i~
pK \Hand the pK. of BH
i~
pKsH· the equilibrium
con~tant
can be C3.::!5a )
or (3.25b) Thi~
procedure i~ illustrated for the reuction in Eq. 3.13 in Study Problem .\.6. and is justil'ied in Problem 3.46 at Lhe end of the chapter.
Calculate the equilibrium constant for the reuction of HC with hydroxide ion (\ce Eq. 3.22).
Solution FiN idcmif) the acid~ on each ..ide of the equation. The acid on the left i!> HC • beit lose., a proton to give cyanide CC ). and the acid on the right il> HP becau<,e it loses a proton to g1ve hydroxide COH). Before doing any calculation. ask whether the equilibrium should lie to the left or right. Remember that the .\ lnmger acid and stronger base tii'C' alll'ays on one .1ide ofthe equation. and the weaker acid and weaker base are on the other side. The equi/ibriumalll'ay.l .f(t1'ors the weaker add and ll'eaker base. Thb means that the right side nf Eq. 3.23 is favored and. therefore. thatlhe equilibrium con~tant in the left-to-right direction is > I. Thil. provide<. a quick check on whether your calculation b reasonable. Next. apply Eq. 3.25a. Subtracting cau~e
3 4 BRONSTED- LOWRY ACIDS AND BASES
105
the pK. of the acid on the left of Eq. 3.23 CHCN) from the one on the right CHP>gi,·es the logarithm of the desired equilibrium con~tant K,-q· CThe rclc' ant pK. 'alue' come from Table 3.1.) log K<>l- 15.7
9.4
= 6.3
The equilibrium constant for this reaction i' the antilog of thi\ number: K<>l = 10"' - 2 X 10" Thi~
large number means thai the equilibrium of Eq. 3.23 lie~ fitr Ill the right. That is. if we dis'ohc HCN in an equimolar solution of NaOH. a reaction occur, to give a ~olution in \\hich there i~ much more -eN than either -oH or HCN. Exactly 1!011' mtwh of each specie~ is present could be determined by a detailed calculation using the equilibrium-constant expres~ion. but in a case like thi-.. 'uch a calculation i~> unnccc~sary. Th..: equilibrium con,tant i~ \O large that. even with water in large excess as the ~>olvent. the reaction lie~ far to the right. Thi\ al1>o means that if we dis~olve NaCN in water, only a minuscule amount of -eNreact~ with the Hp to give - o H and HCN .
.. 14it.l:lh$1 -••• • .....,_
3. 14 Using the pK. values in Table 3. 1, calculate the equilibrium constant for each of the following reaction!>. (a ) NH.1 acting as a base toward the acid HC (b) F- acting as a base toward the acid HC.
Sometime., \tudent\ confuse acid 'trcngth and ha'c \trength when the) encounter an ampluneric contpound (\ee p. 97 ). \:Vater pre~ent' thi' '-Orl of problem. According to the definition'> ju't de' eloped. the base \fren~t;fh ()·water l' tndit:ated b) !he pK of its conjugate acid. 11,0+. "hcr.eu'> the acid \trem?,tlt O/lriltt'r Cor th~ ba'c \trcngth oflt<; conjugate ba<; hydroxide) i., indicatell b) the pK,oi'H~O ii'CIL fhc~e two quantitie' refer to 'ery different reaction~ of 'vnllcr: ~~aer acting
os a base: (3.26a) pi\,,
\~1ter
- 1.7
acting m w1 acid: (3.26b) pK,,
l :'i.7
Write an equation for each of the following equilibria. and U\e Table 3.1 to identify the pK. ,·alue a\sociated with the acidic \pccie. in each equilibrium. (a) ammonia acting as a base toward the acid \\ater (b) ammonia acting as an acid toward the ba~e water Which of these reactions has the larger K,-q and therefore i' more imponam in an aqueous solution of ammonia?
1 06
CHAPTER 3 • ACIDS AND BASES. THE CURVED-ARROW NOTATION
FREE ENERGY AND CHEMICAL EQUILIBRIUM As you leamc<.l in the prcviou section. the equilibrium constant for a reaction tells us which :.pecies in a chemical equil ibri um are present in highest concentration~;. l n1hi~ ~ecti on . we're going to examine the connection between the equ il ibri um constant for a process and the relalil'e Slabi/ilies of the reactants and products. Let\ stan with a specific example- the di!.\Ociation equilibrium of hydrofluoric acid. a relatively weak acid: (3.27)
o-
From Table 3.1. 1he pKa of HF i:. 3.2. Hence. the dissociation constam K. of HF is I 3·1 , or 6.3 X 1 The !>mall magnitude of this equilibrium constant mean-. thai HF is dissociated to only a small extent in aqueous solution. For example, in an aqueous solution containing 0.1 M HF, a detailed calculation using the equ il ibrium-constant expression:-.how~ that on ly about 8'k of the acid i~ dissociated to fluori de ions and hyclrared protons. The dissociation constant is related to the swndard ji-ee-energy diff'erence bel ween products and reactants in the following way. If K,, is the dissociation constant as defined in Eq. 3.21. then the <;tandard free energ} of dissociation i.: defined by
o-.t.
..lG~
==
RT 'ih K
<3.28)
where In indicates natural (base-e) logarithm!.. log indicates common (base- l 0) logarithms. R is the molar gas constant (8.314 X 10-' kJ K- 1 mol- 1 or 1.987 X 10- ~ kcal K- 1 mol- 1 ). and Tis the absolute temperature in kelvins {K). Because - log K. i~ by definition the pK. {Eq. 3.22a). then Eq. 3.28 can be rewritten as ~ G~
'
2.3RT (pK,)
(3.29)
In terms of the HF ionization. the standard free energy of dis~ociation ~G~ in Eq. 3.29 is equal to the difference between the standard free energies of the ionitation products (H 30+ and F-) and the un-ionized acid {HF). The <,tandard free energy of the ~olvent {and reference base) water. because it is the same for aJI acidc;. is arbitrarily c;et to Lero (that i\. ignored). I ntroducing the pK. of HF (= 3.2) into Eq. 3.29. we find. at 25 °C (298 K). that ~ G~
= 18.2 kJ mol - 1 (4.36 kcal mol- 1 )
The meaning of this standard free-energy change is that the product~ of 1he dissociation equilibrium, H 10+ and F-. have 18.2 kJ mol- 1 (4.36 kcal mol- 1) more free energy than the undissociated ,;cid H F: that is. the. product~ arc le.\ s swhle than lh' reactants by 18.2 kJ mol - 1 (4.36 kcal mol- 1) under standard conditions. usually taken to be 1 atm pressure (for gases) or I mole per liter for liquid solutions. Phy~ically. this means that if we could \Omehow couple a free-energy .,ource. such as a battery. 10 the HF ionization reaction. this battery would have to provide 18.2 kJ (4.36 kcal) of energy to con,ert one mole per liter of HF completely into one mole per li1er of hydrated protons and one mole per liter of fluoride ion.,. Or. we can turn the idea around: if we could somehow generate a <>olution containing one mole per liter of hydrated proton., and one mole per liter of fluoride ions. this solution would release 18.2 kJ mol- 1 (4.36 kcal mol- 1) of free energy if the two reacted completely to give water and one mole per l iter of HF. Let's now generalize this result for a reaction in which the starting male rial isS and the product is P. The equilibrium constant Kc4 for the interconversion of Sand P is related to the standard free-energy difference (G~ - G~ ) between P and S as follows: ~ Go
= G~- G s = - 2.3RT log KCll
(3.30)
3.5 FREE ENERGY AND CHEMICAL EQUILIBRIUM
107
Rearranging .
•
(3.3la) or (3.3lb) 0
Notice the exponential dependence of K,~ on ~C • lh1' nk·an' th:ll '-mall ~·hange<. in ~ co rc <.ult in 1<.\rg.e <:hangcs in K~·t' Table 3.2 shows Lhi' relation-,hip numerically. This table shows that a change of 5.7 kJ mol- 1 or 1.4 kcal mol- 1 change~ the equilibrium constant K<~ by one order of magnitude. This free-energy change ha<; the ..ame effect on the product:reactant ratios at equilibrium. In making conversions between equilibrium constant and Mandard free energy, the quantity 2.3RT will appear frequently. At :25 6 C (29 Kl. 2.31l.V?T..:qual, '\.7 1 ".lrnol - 1 or U6 kcalmol- 1. These t(l;e. hand) number' 10 [Cmcmbcr.
Suppose that tJ.C 0 is negative. This means that S has a greater standard free energy than P. or the product Pis more stable than the starting materialS. When Sand P come to equilibrium. P will be present in greater amount. This foll ow~ from Eq. 3.31 b: v. hen ~Go i1- nega(ive. the c ·p 1. Suppose. on the other hand. that tJ.C 0 is positive. This mean!- that S has a smaller standard free energy than P- that i-;. the product P is less stable than the <;tarting material S. When Sand P come to equilibrium. Swill be present in greater amount. Again. this follows from Eq ..L~ Ib: "IK·n ~(; j, po,Jt i' c. the exponent is negative. and K cq ~ I. This is the situation in the H- F ionitation di'>CU,!.cd earlier. The ionization products of HF ( 1-1 1 0+ and F-) are les-.. '>table than HF. Hence. the equilibrium constant for their formation. K3 : i!- very small C10-' 1).
l@:!llfl
The Relationship between Standard Free·Energy Changes, Equilibrium constants, and Relative Equilibrium concentrations at 25 •c (298 K) .lG•
CHAPTER 3 • ACIDS AND BASES THE CURVED-ARROW NOTATION
Let"!> '>Ummari;c the important point., of thi' ...ection. I. Chcmicall'quilibrium favor!> the -,pccic' of lower '>tandard free encrg.). '1
The more t\Hl compounds differ in .,lill1dard free energ)- the grea11;r the diiTerencc in their concentration:- at equilibrium. Each 5. 7 1 I<~ rnQ] - 1 0 .)6 k.cal Ulol- 1 l increment in J.(;o atlcct' the equilibrium concentration ratio by a factor of lO.
II follow:- from thc),e two point), that if we can analyte relative ~tabilitic~ of molecule~. we can then predict equilibrium constant), by applying Eq. 3.31 b. Olice carefully the implication ofthis statement: Knoll'/edge of molecular stabilitie\ can lead to an wulersrmulin~ f~/ c·hemical phenomelw- in thi., case. chemical equilibrium. Molecular '\tabilitie., \\ill form the ba'i' for our under... tanding of other chemical propertie., U\ well- in particular. chemical n.!acti\ it). For thic, reason. we'll devote a lot of auention throughout thi., text to the rclati\C '>tabilitie' of molecule),.
3.16 tal A reaction has a standard free-energy change of - 14.6 k..J mol
1
(- 3.5 kcal mol- 1).
Calcu late the equilibrium constant for the reaction at 25 °C.
'
(b) Calculate the standard free-energy di ffcrcnce between starting materials and products for a reaction that has an equilibrium con~rant of 305.
3.17 (alA reaction A+ B C ha'> a '>lantlard free-energy change of - 2.93 kJ mol- 1 1 ( - 0.7 kcal mol ) at25 °C. What are the concentrations of A. B. and C l/1 equilibrium if. at the beginning of the reaction. their concentrations are 0.1 M. 0.2 M. and 0 M, rc,pecti\'ely?
(b) Without making a calculation. tell in a qualitati\'e sense ho\\ you would expect your an•mer for part (a) to change if the reaction has instead a standard free-energy change of + 2.93 kJ mol 1 (+0.7 kcal mol 1). 3.t8 Complete each of Lhc following statcmcntl> with a number. Assume that the temperature is 25 °C (291! K). (a) Two reactions have equilibrium cons tams that di lfer by a factor of I0. Their standard free cnergie~ differ by kJ mol 1• (hi For every I kJ mor 1 in standard free energy that two reaction'> differ. their equilibrium con'>tanl\ differ by a factor of _ __
RELATIONSHIP OF STRUCTURE TO ACIDITY The goal of thi), section is to he lp you learn to use the Slrucrures of' compounds to predict trend~ in their chemical properties. The chem ical property we are going to deal with here is Bron<,ted acidity. but what you learn can be brought to bear on other chemical properties. This section ~ill answer the following que),tion: How can we predict the relati\ c \trength!-. of Brs:}nsted acid'>\\ ithin a .,eries? Your ability to deal with que'>tions like this\\ ill require that you use all that you ha,·c learned in the pre' iou-, \ection-..
A. The Element Effect Onc pf thl' llHI'>t importanuh ings thaLdctermim:'> the acidity of a Bronsted acid i:-. rite ide111itr
ethanol ar
ethanethiol
11,.
pK. = 15.9
pK, = 10.5
3.6 RELATIONSHIP OF STRUCTURE TO ACIDITY
109
These two compounds are structurall y similar: the sole difference between them is the element (color) to which the acidic proton is attached. The elements come from the same group in the periodic table, yet the acidity of the thiol is almost a million limes that of the alcohol (which is about as acidic as wa ter). Another important example of the same trend is the rel ative acidities of the hydrogen halides. HI is the strongest of these acids: HF i s the weakest. (The relevant pK.. data are fou nd in Table 3.l. ) These dat a illustrate an important trend: Br~lnsted acidi~v increases as the atom trl "'hich the uddic hrdrogen is (tfl.ached has a greater atomic 1111111her ll'ilhin a column (group) (~/'the P<' rtOdic wbl~:. Now let\ ~ee how acidities vary across the peri odic table wi thin the same row. or period: H-CI-1.1
H-NI-1 1
H-OI-I
H-F
""55
= 35
15.7
3.2
(3.32)
(The pK" values of methane and ammonia are so high that they are not known with certainty.) These data demonstrate another important trend : Brqnstcd...I!cidi~y il/(,;rea.~·es a.\· the a191/l to ll'ltich th e a ·idic h_rdmgen is ttltached is farther to !he right ll'ilftin a row (period} ofthe peri-
odit table. T he effect of' changing tl~e direuly til/ached atom A on the acid ity of H'-A is called the clement effect oo ac idity. U!Jderstand ing the element effect starts by dividing the ionization process or a typical acid 1-1-A into three steps, as shown in Eqs. 3.33a-c. We are allowed to do this by the first law of thermodynam ics. wh ich states simply that the energy difference between two compounds doesn't depend on the pathway used to interconvert them. just as the height of a buildi ng doesn't depend on how one gets to the top to measure it. Bond breaking
Loss of an elecmm .fi·om H• Electron lran~f'er to A • Sum:
~
H· + A·
(3.33a)
11 ·
~
I-J+ + e-
(3.33b)
e- + A·
~
A:-
(3.33c)
H-A
~
H+
H+ A
+ A:-
(3.33d)
The fi rst step (Eq. 3.33a) is the breaking of the H-A bond '·in half.'' wi th one bonding electron going to each atOm. The 1.'11 rg y rcyuircd for thi.., st~p i.., C the diretl 111easure of boml s1reng1h. When we compare different bonds. the greater the bo~lcn d i ss()ciat i9n. cr\crgy. th stronger the bond. Because acid dissociation invol ves breaking the bond to hydrogen. smaller bond energies promote increased acidity. The second step of acid dissociation (Eq. 3.33b) is loss of an electron from the hydrogen atom. The energy rt:qui red for this 'll'P i\ the ionization potential of hydrogen. Because this is the same for all Bn~nsted acids. it does not enter into a comparison of different acids. The third step of ucid dissociation
1 10
CHAPTER 3 • ACIDS AND BASES. THE CURVED-ARROW NOTATION
Figure 3.1 Factors affecting Br0nsted acidity. The acidities of some Br0nsted acids H- A are organized by the position of the elements A in the periodic table. The purple arrows show the trends in acidity along rows and columns of the periodic table. The solid arrows indicate the more important factor in each trend, and the dashed arrows show the less important factor.
H- F. Even though fluorine is a much more potent "electron attractor" than iodine, the dom-
'
inant effect governing acidity is bond strength: the H-I bond is much weaker than the H - F bond. Across a row of the periodic table, bond ~trengths change relatively little (dashed arrow), hut electron aifinities clumRe sj gui/Tcanrly. The increase in acidity across a row of the periodic table, then, is mainly controlled by the electron affinity of the elements to which the acidic hydrogen is bonded. To summarize what we've learned about the element effect: 1. The m:ldi!ie:-; ofEr0nsted a'tjds H~A increase 10ward higher atomic number o f atom A within a group of the pcri()clic table. T he main source ~!}is increase is the decreasing .
or
..
B. The Charge Effect Another important influence on acidity is the effect of charge on the atom bonded to the acidic hydrogen. For example, the pK3 of Hp+ is -1.7 and the pKa or H20 is 15.7. The major factor responsible for this difference is that a positively charged oxygen attracts electrons much more than a neutral oxygen. Because the bond to the acidic hydrogen in both cases is an 0 - H bond. bond strength is not an important factor. Such charge effects are quite general.
3.6 RELATIONSHIP OF STRUCTURE TO ACIDITY
111
3.19 In each pair. which compound has the greater Br¢nsted acidity (smaller pK8 )? +
(a)
..
or NH4
:NH 3
(b) CH3?.H
+ CH 3~.H2
or
3.20 (a) Rank the foUowing four acids in order of increasing Br¢osted acidity. +
-r.
+ CH 3~H2
H2F:
CH 39H
CH30CH 3
c
H
B
A
I
D (b) Rank the following compounds in order of increasing basicity. (Him: Think about the acidity of the conjugate acids and the relationship between the acidities of acids and the basicities of their conjugate bases.)
v ''
c.
The Polar Effect The previous two sections discussed effects on acidity that are associated with changes in the atOm to which the acidic hydrogen is directly attached. ln l his section, we consider the effect on acidi ty that results from substitution at a more remote location in an acidic molecule. Our case study will involve carboxylic acids. A l though ~arboXy l i c acid!> are classified as weak acids. they are more cidic than m(lSt o ther types of organic wmpouod~ . The typical carboxylic acid in aqueous solution undergoes a small degree of ionization to give its conjugate base, a carboxy/are ion. •n. ld •.,.
hhhl>~Cil
II
~
..
'
.·o··..-
r :o·
:Q :
..
[
R- C - Q- 11 + H2Q
.
R- c..f:q:- ~ R-C=Q : II ..
general structure of a carboxylic acid
I ..
l
(3.34)
general structu re of a carboxylate ion
As shown in Eq. 3.34, carboxylate ions are resonance-stabi lized. For convenience, we'll someti mes use the following hybrid structure for car boxylate ions, which shows the sharing of double-bond character and negative charge with dashed lines and partial charges:
os//
R-C(
\, s-
hybrid structure of a carboxylate ion
o
Consider the trend in acidity indicated by the following data for acetic acid and some of its substitut d derivatives:
0
II
0
II
0
II
0
II
H:>C-C-0-H
FCH 2 -C-O-H
F2CH-C-O-H
F3C-C-O- H
acetic acid pK3 = 4.76
fluoroacetic acid pK. = 2.66
difluoroacctic acid pK, = 1.24
trifl uoroacetic acid pK,
= 0.23
(3.35)
11 2
CHAPTER 3 • ACIDS AND BASES. THE CURVED-ARROW NOTATION
Within thi s seri es, the onl y structural d ifference from compound to compound is that hydrogens have been substituted by fluorines several atoms away from the acidic hydrogen. The more fluorines there are. the stro nger the acid. A similar effect i s observed when other electronegati ve atoms or groups are substituted into a carboxylic acid molecule. The following data ill ustrate the same type of effect:
0
II
y H 2CH1CH 2 - C- O- H butanoic acid pKa = 4.82
Cl 4-chlorobutanoic acid pKa = 4.52
0
0
II
CH CHCH - C- 0 - H
31
2
II
CH CH )CH-C-0-H 3
C1
- ,
(3.36)
Cl
3-chlo robutanoic acid
2-chlorobutanoic acid pKa = 2.84
pK3 = 4.06
These data show that the closer the electro negative group is to the acidic hydrogen. the greater its effect on acidity. To understand these effects, we start with the standard free energy of the ionization process. Recall (Eq. 3.29. p. I06) that the standard free energy of ioniza tion ~c: is related to the dissociation constant K, of an acid by the equation ~G~
= 2.3RT (pK)
0 .37 )
Rearran ging this equation. we have
pK,,
~ (;~
'2.3 RT
(3.3Rl
These equations how that ~G and pK3 are directly proponional. Remember £Mat the star)dard free energy of ionizat ion fo r a carboxyl ic ac.id is e_qual to the dijference between the prod\lt:l~ of ionizati n (th conjugate base and Hp+) and the carboxylic acid itsel f'. Th i~ is shown in Fig. 3.2a. Look at thi part of the figure and think about what would happen to the pK" of a cmboxylic acid if we did something to increase the relative stabil ity (that is. lower the relative standard free energy) of its conjugate base. This is shown in Fig. 3.2b. Loll'ering rhe .-;randcml f ree energy (~f'a conjugate base de(lrease !1G and. /J..'' Eq. J.J,B. also lowers the pK,. of!lu: arid. ln other words. lmn•ring rhe stt111lfl/ld fi'ee energv oj a cor!ill$ale base makes the <'Onjugate l tciclmore addic:.. f::leclnmegal (l'e . ul1slituent g roups s..u ch as hafo.!Jens increase !he C1cidities of carb(uylic (Jcid~ by ~1abildng their conjugme-base carboxy/we. ions. This stabilization originates in the polarity of the carbon- halogen bond- that is. irs bond dipok:. To visual ize this idea. consider the electrostatic interaction (interaction between charges) or the negatively charged carboxylate oxygens in fluoroaeetic acid with the nearby carbon- halogen bond dipole:
3.6 RELATIONSHIP OF STRUCTURE TO ACIDITY
1 13
n:pubtl'e inll'ractinn
hdiH'l'll lik,• ~ehargc' d, t.lhililc' the i1>11 gr~atcr d•~t.mct•, lc.~s
F
Important
'
c-F bond dipole: - -- -- \
...
~~o
w·····r -c ~o
(3.39)
H
T his interaction is governed by the fo llowi ng equation. called the electrostatk Ia\\':
•
(3.40)
In this equation. q 1 and q,_ are charges./.. and e are COIN\tnh. and r i' the dbtancc between the charges. According to this law. l·hargc-. of opposi te :-.ign interact to give a negp tivc (~t abi l izing) contribution to the total ener~y. T his means that the interactions between the negative charges on the oxygens and the positive end or the C-F bond dipole stabilize the inn. ow. the electro tatic law abo go\'Cnl'> the inreraction<, of the carboxylate oxygcns with the negati1•e end of the
os-ll
R1C(
, H Jo+
":·O.s-
I
-lf;~;;~;;;,;;;;,-;;:;~;
1:'
R2C(
----------
M.r~ ~J
Ilk l ·•
0
II 1
\:\
'
(2)=
uG.~(2) I H'T J -·-
\
0
II
R2C-OH
R C-OH weaker add (larger pK3 )
(a)
, H3o+
os
pK ( I l
....
Os--
mr/loxylate ion
strouger acid (smaUer pK3 ) (b )
Figure 3 2 (a) The pK. of an acid is proportional to the standard free energy difference between an acid and its conjugate base. (b) Lowering the standard free energy of a conjugate base reduces the pK1 of the acid and in creases its acidity. The two un-ionized carboxylic acids have been arbitrarily placed at the same standard free en ergy to focus attention on the relative free energies of t he conjugate bases.
114
CHAPTER 3 • ACIDS AND BASES THE CURVED· ARROW NOTATION
FUrther EXplOration 3.1
Inductive Effects
' •. \
C-F bond dipole. and this interaction b destabiliLing. However. lhi-. interaction occurs across a larger distance and therefore makes a le s important contribution to the energy of the ion because a larger r in the denominator of Eq. 3.40 reduces the magnitude of the interaction. Hence. the net interaction of the carboxylate oxygen and the nearby C-F bond dipole is an attractive. stabilizing one. As you can see from Fig. 3.2b, this stabilization lowers the pKa of an acid, or strengthens the acid. Because the carboxylic acid itself is uncharged. the effect of a fluorine substituent on its stability is much less important and can be ignored. In ummary. interaction of lhe bond dipole of the C- F bond with the negative charge on lhe oxygen l.tabili1.es lhe carboxylate ion and thu. increases lhe acidity of the carboxylic acid. An effect on chemical propenle_ caused by the interactions bet\\een charges. djpoles. o(..both is ~:ailed a polar effect. {It is also known a<; an inductive effect.) Thu • in the pre~ent examples. halogens (or other electronegative substituents) have an acid-strengthening polar effect on the acidity of carboxylic acids. As the series in Eq. 3.35 shows, the more halogens there are, the greater the effect on acidity. In fact, triftuoroacctic acid borders on being a strong acid. Another way to describe the polar effect of halogens and olher electronegative groups is to say lhat lhey exert an electron -withdra\\~ng polar effect ~cause they pull electrons toward them>.ei\C'- and away from the C.. The inver. e relation hip between the interaction energy E and distance r in Eq. 3.40 means lhat the magnitude of lhe interaction between charges dimini ~hes as the distance between the interacting groups increases. Hence. polar effects should be smaller for compounds in which the two interacting group are separated by greater distances (more bonds). Indeed. within the series of Eq. 3.36, you can see that the inOucnce of a chlorine on the pK, decreases significantly as the chlorine is more remote from the carboxylate oxygen. The idea that the chemical properties of a compound follow from its structure is one of the major ideas of chemistry. In this section. you've begun ro learn how structures of molecules affect their energies, and lhus how structures affect chemical properties. We've focused on the chemical property of Br~n ted acidity. To summarize: I. The ell'lll<'nt effect is rhe effect on acidity of changing the atom to "hich the acidic hydrogen i~ hooded. The element effect ha" 11' origins in the bond energie~ to acidic h) drogen), and the electron affinities of the d0ment<.. auachcd to the acidic hydrogen . Trend~ in acidity hased on the clement effect can be predicted from the relation hip of the elcmt.d]ts in the periodic tabl . 2. The clwr~f:. e,Q'e.ct l.s the increase in a<.:tdi(y that results fi·qm increa!ling the positive charge on the atom bonded to the acidic hydrogen. The element effect and charge effect arc large effect~. 3. The polar <'{feet. which i<; typically_ <;mailer in magnitude than the element and cJlarge effcch. I'- the effect on acidit)' that polar groups in an acid hm·e through their interacti ns \\ ith the c:harge species in the ac.id- ba\e equilibrium. We'll learn about other factors that affect acidity in Chapters 14 and 18.
Rank the following compounds in order of increasing basicity. 0
0
0
II H c-c-o-
H 3C- C- NH
H3N-CH2- c- o -
acetate ion
acetamide anion
glycine (an amino .lcid)
A
B
c
3
II
II
+
3.6 RELATIONSHIP OF STRUCTURE TO ACIDITY
115
Solution Fir<>t recognize that a pt•oblem in relative basicity is equivalent to a problem in relative acidity. If you can rank the acidi tie~ of the conjugate acid~. you've solved the problem. The relevant conjugate acids are
acetic acid
acetamide
glycine conjugate acid
AH
BH
CH
Both AH and CH arc carboxylic acids; in both cases, the acidic hydrogen is bound to an oxygen. The acidic hydrogen in compound BH is bound to a nitrogen. The difference in acidities of BH and the other two compounds is therefore due primarily to the element effecr along the first row of the periodic table, and this is the mosl important effect. This effect predicts that the 0 -H group should be more acidic than a comparably substituted N- 1-l group, because oxygen is more electron-attracting than nitrogen. Thus. the acidities of both AH and CH are greater than !he acidity of BH. The difference in the acidities of AH and CHis due to the polar effect of the H1 - group in compound CHon the acidity of the nearby carboxylic acid group. The full positive charge on the nitrogen ha~ a favorable interaction with the negatively charged carboxylate oxygen. As shown in Fig. 3.2, thi~ interaction stabilizes the conjugate base and thus enhances the acidity of CH. Hence, the final order of acidity is CH > AH > BH. Because stronger acids have weaker conjugate bases, the basicity order of the conjugate bases is C
iij;\.J:Iii4i
