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FOURTH EDITION to my students TO THE STUDENT As you study organic chemistry, make certain you fulty understand each new fact that you encounter. While studying the material, you should be continuously asking yourself “"why?": Why does that reaction occur? Why is that product formed? Why is that compound more stable? If you truly understand each new piece of information, you will be creating a foundation upon which to lay subsequent information. A strong foundation will allow you to master a great deal of material with much less effort than you would have to put out if you were memorizing your way through the material. Often it is the new vocabulary that you encounter when you are first exposed to a discipline that can be the biggest hurdle to mastering the material. For that reason I have included a list of the important terms and their definitions at the beginning of each chapter in the Study Guide, Reading these is a good way to review some of the important aspects of the chapter. There are two kinds of problems in the textbook. The problems at the end of each section within a chapter are designed to let you see if you have understood the material presented in that section, and to reinforce the material. You should do these problems as you encounter them. The problems at the end of each chapter integrate the concepts in the chapter and sometimes include concepts that were mastered in previous chapters. Try to do as many of these as possible. The more problems you do, the more comfortable you will become with organic chemistry, and the more you will enjoy it. Organic chemists use curved arrows to show the bonds that break and the bonds that form in an organic reaction. When you start studying organic reactions in Chapter 3, take time to do the exercise on drawing curved arrows that you will find on page 101. There is also an exercise in model building (page 151) that will help you with the material in Chapter 5. Good luck in your study. If you have any comments or suggestions about how the Study Guide could be improved for those students who will follow you, 1 would be very happy to hear from you. Paula Yurkanis Bruice Department of Chemistry and Biachemistry University of California, Santa Barbara Santa Barbara, CA 93106 pybruice E chem.ucsb.edu CHAPTER 6 CHAPTER 7 CHAPTER 8 CHAPTER 9 CHAPTER 10 CHAPTER 11 CHAPTER 12 CHAPTER 13 CHAPTER 14 Reactions of Alkynes « Introduction to Multistep Synthesis Important Terms Solutions to Problems Practice Test Electron Delocalization, Resonance, and Aromaticity * More About Molccular Orbital Theory Important Terms Solutions to Problems Practice Test Reactions of Dienes * Ultraviolet/Visible Spectroscopy Important Terms Solutions to Problems Practice Test Reactions of Alkanes * Radicals Important Terms Solutions to Problems Practice Test Substitution Reactions of Alkyl Halides Important Terms Solutions to Problems Practice Test Elimination Reactions of Alkyl Halides Competition Between Substitution and Elimination Important Terms Solutions to Problems Practice Test Reactions of Alcohols, Ethers, Epoxides, and Sulfur-Containing Compounds * Organometallic Compounds Important Terms Solutions to Problems Practice Test Mass Spectrometry and Infrared Spectroscopy Important Terms Solutions to Problems Practice Test NMR Spectroscopy Important Terms Solutions to Problems Practice Test 194 196 211 213 215 234 237 240 262 265 266 278 280 282 300 302 303 330 332 334 367 369 371 392 394 397 420 CHAPTER 24 CHAPTER 25 CHAPTER 26 CHAPTER 27 CHAPTER 28 CHAPTER 29 CHAPTER 30 Answers to Practice Tests Catalysis Important Terms Solutions to Problems The Organic Mechanisms of the Coenzymes Important Terms Solutions to Problems Lipids Important Terms Solutions to Problems Nucleosides, Nucleotides, and Nucleic Ácids Important Terms Solutions to Problems Synthetic Polymers Important Terms Solutions to Problems Pericyclic Reactions Important Terms Solutions to Problems The Organic Chemistry of Drugs: Discovery and Design Important Terms Solutions to Problems 705 707 726 728 744 747 764 767 782 785 80 806 82 826 832 CHAPTER 1 Electronic Structure and Bonding * Acids and Bases Important Terms acid acid-base reaction acid dissociation constant acidity antibonding molecular orbital atomic number atomic orbital atomic weight aufbau principle base basieity bond dissociation energy bonding molecular orbital bond lengih Bronsted acid Brynsted base buffer solution carbanion carbocation condensed structure conjugate acid conjugate base a substance that donates a proton. a reaction in which an acid donates a proton to à base. a measure of the degree to which an acid dissociates. a measure of how easily a compound gives up a proton. a molecular orbital that results when two atomic orbitals with opposite signs interact. Electrons in an antibonding orbital decrease bond strength. tells how many protons (or electrons) the neutral atom has. an orbital associated with an atom. the average mass of the atoms in the naturally occurring element. states that an electron will always go into the available orbital with the lowest energy. a substance that accepts a proton. describes the tendency of a compound to share its electrons with a proton. the amount of energy required to break a bond homolytically (each of the atomns retains one of the bonding electrons) or the amount of energy released when a bond is formed. a molecular orbital that results when two atomic orbitals with the same sign interact. Electrons in a bonding orbital increase bond strength. the intenuclear distance between two atoms at minimum energy (maximum stability). a substance that donates a proton. a substance that accepts a proton. solution of a weak acid and íts conjugate base. a compound containing a negatively charged carbon. a compound containing a positively charged carbon. a structure that does not show some (or all) of the covalent bonds. the compound formed when a base accepts a proton. the compound formed when an acid loses a proton. hydrogen ion (proton) inductive electron withdrawal ionic bond ionic compound ionization energy isotopes Kekulé structure Lewis acid Leyris base Lewis structure lone-pair electrons mass number molecular orbital molecular orbital (MO) theory node nonbonding electrons nonpolar covalent bond octet rule orbital orbital hybridization organic compound Pauli exclusion principle pH pi (x) bond Chapter 1 3 a positively charged hydrogen. the pull of electrons through sigma bonds by an atom or a group of atoms. a bond formed through the attraction of two ions of opposite charges. a compound composed of a positive ion and negative ion. the energy required to remove an electron from an atom. atoms wilh the same number of protons but a different number of neutrons. a model that represents the bonds between atoms as lines. a substance that accepts an electron pair. a substance that donates an electron pair. a model that represents the bonds between atoms as lines or dots and the nonbonding electrons as dots. valençe electrons not used in bonding. the number of protons plus the number of neutrons in an atom. an orbital associated with a molecule. describes a model in which the electrons occupy orbitais as they do in atorms but the orbitals extend over the entire molecule. a region within an orbital where there is zero probability of finding an electron. valence electrons not used in bonding. a bond formed between two atoms that share the bonding electrons equalty. states that an atom will give up, accept, or share clectrons in order to achieve a filled shell. Because a filled second shell contains eight electrons, this is known as the octet rule. the volume of space around the nucleus where an electron is most likely to be found. mixing of atomic orbitals. a compound that contains carbon. states that no more than two electrons can occupy an orbital and that the two electrons must have opposite spin. the pH scale is used to describe the acidity of a solution (pH = log[H*]). a bond formed as a result of side-to-side overlap of p orbitals. 4 Chapter 1 PKa polar covalent bond potential map describes the tendency of a compound to lose a proton (pKa = log Ka, where Ka is the acid dissociation constant). a bond formed between two atoms that do not share the bonding elecirons equally. a map that allows you to sec how electrons are distributed in a molecule. (electrostatic potential map) proton proton transfer reaction quantum mechanies radical (free radical) resonance resonance contributors resonance hybrid sigma (0) bond single bond tetrahedral bond angle tetrahedral carbon trigonal planar carbon triple bond valence electrons valence shell electron pair repulsion (VSEPR) model wave equation wave functions a positivcly charged hydrogen; a positively charged atamic particle. a reaction in which a proton is transferred from an acid to a base. the use of mathematical equations to describe the behavior of electrons in atoms or molecules. a species with an unpaired electron. having delocalized electrons. structures with localized electrons that approximate the true structure of a compound with delocalized electrons. the actual structure of a compound with delocalized electrons. a bond with a symmoetrical distribution of electrons about the internuclear axis. a single pair of electrons shared between two atoms. the bond angle (109.5º) formed by an sp3 hybridized central atom. an sp3 hybridized carbon; a carbon that forms covalent bonds using four sp3 hybrid orbitals. an sp? hybridized carbon. composed of a sigma bond and two pi bonds. an electron in an outermost shell. a model that combines the concepts of atomic orbitals and shared electron pairs with minimization of electron repulsions. an equation that describes the behavior of each electron in an atom or a molecule. a series of solutions of a wave equation. Chapter 1 d. magnesium (atomic number = 12): 152 252 2p9 352 calcium (atomic number = 20): 152 252 2p$ 352 3p6 452 The polarity of a bond can be determined by the difference in the electronegativities (given in Table 1.3) of the atoms sharing the bonding electrons a, KCl has the most polar bond (3.0 - 0.8 = 2.2), whereas it is 1.8 for LiBr, 1.6 for Nal, and O for Cb. b. Cl has the least polar bond because the two chlorine atoms share the bonding electrons equally. a. LiH and BF are polar (they have a red end and a blue end). b. A potential map marks the edges of the molecule*s electron cloud. The electron cloud is largest around the H in LiH because that H has more electrons around it than the Hºs in the olher molecules. c. Because the hydrogen of HF is blue, we know that that compound has the most positively charged hydrogen. Solved in the text. To answer this question, compare the clectronegativities of the two atoms sharing the bonding electrons using Table 1.3 on page 10 of the text. (Note that if the atoms being compared are in the same row of the periodic chart, the atom on the right is the more electronegative; if the atoms being compared are in the same column, the onc closer to the top of the column is the more clectronegative.) ô- 5 8 8. 8. 5+ . O & a HO—H e H;C—NH, e HO-Br g I-cI 5. 5+ 5+ 8. 3. d+ d+ 6. b. F—Br d. HC—cl f. HC-MpBr h. H;N—0OH By answering this question you will sec that a formal charge is a bookkeeping device. It does not necessarily tell you which atom has the greatest electron density or is the most electron deficient. oxygen oxygen (it is the more red) oxygen hydrogen (it is the most blue) encp Notice that in hydroxide ion, the atom with the formal negative charge is the atom with the greater electron density. In the hydronium ion, however, the atom with the formal positive charge is not the most electron deficient atom. 7 Chapter | 10. HE, g H:CiN:H je Na” EO H õ :õ HH or or or or *CI: H H, :N H + :CiCêH k H H or or :IN=N: N-O: :0= 1. or 8 12. 13. 14, 15. E E Nu H-C-N-C-C-H H-G— EG HG——C-H HHHH H H-C-H H H-Ç-H H H a Tito to! H—C o C-C—C C C-H à uca d dada É H He, has three electrons. Two are in a bonding orbital and one is in an antibonding orbital. Because there are more electrons in the bonding orbital than in the antibonding orbital, He," exists. a. n* | side-to-side overlap forms a 1 bond; the overlap of opposite phase (colors) orbitals forms an antibonding (*) orbital. b. o* | end-on overlap forms a o bond; the overlap of opposite phase (colors) orbitals forms an antibonding (*) orbital. e o* | end-on overlap forms a 6 bond; the overlap of opposite phase (colors) orbitals forms an antibonding (*) orbital. d. o | endon overlap forms a & bond; the overlap of same phase (colors) orbitals forms an antibonding (*) orbital. The carbon-carbon bonds form as a result of sp3-—sp3 overlap. The carbon-hydrogen bonds form as a result of sp'—s overlap. 16. x, 18. 19. 20. PAR 22, Chapter 1 9 CH4 with O lone pairs has bond angles of 109.5º H5»0 with 2 lone pairs has bond angles of 104.5º The bond angle decrcases as the number of lone pairs increases because a lone pair is more diffuse than a bonding pair. , Therefore, H30* with 1 lonc pair has bond angles in between those two; its bond angles will be less than 109.5º and greater than 104.5º. The hydrogens of the ammonium ion are the bluest atoms. Therefore, they have the least electron density. In other words, they have the most positive (least negative) electrostatic potential. Water is the most polar-—it has a deep red area and the most intense blue area. Methane is the least polar—it is all the same color with no red or blue areas. Bonding electrons in shells farther from the nucleus form longer and weaker bonds due to poorer overlap of the bonding orbitals. Therefore: a, relative lengths of the bonds in the halogens are: Br > Cly. relative strengths of the bonds are: Cl > Bra. b. relative lengths: HBr > HCl > HF relative strengths: HF > HC] > HBr We know the a bond is stronger than a 7 bond because the 6 bond in cthane has a bond dissociation energy of 88 kcal/mol, whereas the bond dissociation energy of the double bond (5 + 7) in ethene is 152 kcal/mole, which is loss than twice as strong. Because the q bond is stronger, we know that it has more effective orbital-orbital overlap. The carbon-carbon sigma bond formed by spi—sp? overlap is stronger because an sp” orbital has 33.3% s character, whereas an sp" orbital has 25% s character. Because electrons in an s orbital are closer on average to the nucleus than those in à p orbital, the greater the s character in the interacting orbitals, the stronger (and shorter) bond. sp? sp2 sp q ) a. CH;CHCH=CHCH,C=CCH, CH, sp3 sp3 a oieee Chapter 1 1 s sp2 M =) sp2 p' sp? NT 5 O” SO band angles = 120º e. CCL The carbon in CCl4 is bonded to four atoms, so it uses four sp? hybrid orbitals. Each carbon-chlorine bond is formed by the overlap of an sp3 orbital of carbon with a p orbital af chiorine. Because the four sp? orbitals orient themselves to get as far away from each other as possible, the bond angles are all 109,5º. bond angles = 109.5º d. CO, : The carbon in CO is bonded to two atoms, so it uses two sp hybrid orbitals. Each carbon- oxygen bond is a double bond. One of the bonds of cach double bond is formed by the overlap of an sp orbital of carbon with an sp? orbital of oxygen. The second bond of the double bond is formed as a result of side-to-side overlap of a p orbital of carbon with a p orbital of oxygen. Because the two sp orbitals orient themselves to get as far away from each other as possible, the bond angle in CO» is 180º. 0=C=0 bond angle = 180º e. The double-bonded carbon and the double-bonded oxygen in HCOOH use sp? hybrid orbitals. The single-bonded oxygen uses sp? hybrid orbitals and each hydrogen uses an s orbital. the 6 bond is formed by sp2-sp? overlap the 7 bond is formed by p-p overlap sp2-s averlap Fi ) c Lá H fo ; spº-s overla| sp2-sp3 overlap Ro á á 12 25. 26. 27. Chapter 1 f. The triple bond consists of one o bond and two 7 bonds. Each nitrogen uses an sp hybrid orbital to form the o bond and a p orbital to form each of the two 7 bonds. N==N theo bond is formed el sp-sp overlap each 7 bond is formed by p-p overlap The electrostatic potential map of ammonia is not symmetrical in shape because it is not symmetrical in the distribution of the charge—the nitrogen end is more electron rich than the three hydrogens. The electrostatic potential map of the ammonium ion is symmetrical in shape because it is symmetrical in the distribution of the charge, Its symmetry results from the fact that the nitrogen atom forms a bond with each of the four hydrogens and the four bond angles are all the same. ae gh í CcI—Be—ct Cry ud N CI a Because Be is sp hyridized, the bond angle in BeCl, is 180º and the compound does not have a dipole mome Because C is sp3 hyridized, the bond angles in CH,Ch, are 109.5º and the compound has a dipole moment. a. (1) +NHY (2) HCI (3) HO (9 Hot b. (1) NH; (BD Br G) NO (4) HO- if the lonc pairs are not shown: + a. CH,OH asanacid CHOH + NH; ==> CHO + NH, CH,OH as à base CH,0H + HCl ==> CH;0H + Cr H b. NH, as an acid NH, + HO” ==> “NH, + HO + NH as a base NH, + HBr == NH + Br 
