《有机化学》课程教学资源（参考书籍）Organic Chemistry, 8th ed 2017, Paula Yurkanis Bruice, University Of California Santa Barbara.pdf_P26-P30

xxvii ENGAGING MIXED SCIENCE MAJORS IN ORGANIC CHEMISTRY Students better understand the relevance of what they’re studying by seeing the connections between the reactions of organic compounds that occur in the laboratory and those that occur in a cell. Changes throughout this edition provide students with this much-needed “bioorganic bridge,” while maintaining the rigor of the traditional organic course. For example, we teach students about halide and sulfonate leaving groups. Adding phosphate leaving groups takes little additional time, but it introduces students to valuable information, particularly if they are taking organic chemistry because of an interest in the biological sciences. Students who are studying organic chemistry learn about tautomerization and imine hydrolysis, and students studying biochemistry learn that DNA has thymine bases in place of the uracil bases in RNA. But how many of these students are ever told that the reason for the difference in the bases in DNA and RNA is tautomerization and imine hydrolysis? 26.10 Why DNA Contains Thymine Instead of Uracil 19 PROBLEM 12 ♦ A change in which base of a codon is least likely to damage a protein? PROBLEM 13 ♦ Write the sequences of bases in the sense strand of DNA that resulted in the mRNA in Problem 9 . PROBLEM 14 List the possible codons on mRNA that specify each amino acid in Problem 9 and the anticodon on the tRN 26. 10 WHY DNA CONTAINS THYMINE INSTEAD OF URACIL In Section 23.7 we saw that dTMP is formed by methylating dUMP, with coenzyme N5 , N10 - methylenetetrahydrofolate supplying the methyl group. N 2 -deoxyribose-5-P HN + N5 ,N10-methylene-THF + dihydrofolate thymidylate synthase O O N HN O CH3 O Because the incorporation of the methyl group into uracil oxidizes tetrahydrofolate to dihydrofolate, dihydrofolate must be reduced back to tetrahydrofolate to prepare the coenzyme for another catalytic reaction. The reducing agent is NADPH. tetrahydrofolate NADP + + dihydrofolate NADPH + + H+ dihydrofolate reductase The NADP + formed in this reaction has to be reduced back to NADPH by NADH. Every NADH formed in a cell can result in the formation of 2.5 ATPs ( Section 24.10 ) . Therefore, reducing dihydrofolate comes at the expense of ATP. This means that the synthesis of thymine is energetically expensive, so there must be a good reason for DNA to contain thymine instead of uracil. The presence of thymine instead of uracil in DNA prevents potentially lethal mutations. Cytosine can tautomerize to form an imine ( Section 17.2 ) , which can be hydrolyzed to uracil ( Section 16.8 ) . The overall reaction is called a deamination because it removes an amino group. tautomerization imino tautomer N O NH2 N H N H N H HN + NH3 H2O O NH HN O O cytosine amino tautomer uracil deamination If a C in DNA is deaminated to a U, the U will specify incorporation of an A into the daughter strand during replication instead of the G that would have been specified by C, and all the progeny of the daughter strand would have the same mutated chromosome. Fortunately, there is an enzyme that recognizes a U in DNA as a “mistake” and replaces it with a C before an incorrect base can be inserted into the daughter strand. The enzyme cuts out the U and replaces it with a C. If Us were normally found in DNA, the enzyme would not be able to distinguish between a normal U and a U formed by deamination of a cytosine. Having Ts in place of Us in DNA allows the Us that are found in DNA to be recognized as mistakes. 26.13 Genetic Engineering 1177 26. 13 GENETIC ENGINEERING Genetic engineering (also called genetic modification) is the insertion of a segment of DNA into the DNA of a host cell so that the segment of DNA is replicated by the DNA-synthesizing machinery of the host cell. Genetic engineering has many practical applications. For example, replicating the DNA that codes for human insulin makes it possible to synthesize large amounts of the protein, eliminating the dependence on pigs for insulin and helping those who are allergic to pig insulin. Recall that pig insulin differs from human insulin by one amino acid ( Section 21.8 ) . Agriculture is benefiting from genetic engineering. Crops are now being produced with new genes that increase their resistance to drought and insects. For example, genetically engineered cotton crops are resistant to the cotton bollworm, and genetically engineered corn is resistant to the corn rootworm. Genetically modified organisms (GMOs) have been responsible for a nearly 50% reduction in the use of chemicals for agricultural purposes in the United States. Recently, corn has been genetically modified to boost ethanol production, apples have been genetically modified to prevent them from turning brown when they are cut, and soybeans have been genetically modified to prevent trans fats from being formed when soybean oil is hydrogenated ( Section 5.9 ) . Resisting Herbicides Glyphosate, the active ingredient in a well-known herbicide called Roundup, kills weeds by inhibiting an enzyme that plants need to synthesize phenylalanine and tryptophan, amino acids they require for growth. Corn and cotton have been genetically engineered to tolerate the herbicide. Then, when fields are sprayed with glyphosate, the weeds are killed but not the crops. These crops have been given a gene that produces an enzyme that uses acetyl-CoA to acetylate glyphosate in a nucleophilic acyl substitution reaction ( Section 15.11 ) . Unlike glyphosphate, N -acetylglyphosphate does not inhibit the enzyme that synthesizes phenylalanine and tryptophan. glyphosate an herbicide O O O− C P enzyme NH − O O− N-acetylglyphosate harmless to plants acetyl-CoA O O O C O− C P N − O O− + O C CH3 + CoASH CH3 SCoA corn genetically engineered to resist the herbicide glyphosate by acetylating it Using Genetic Engineering to Treat the Ebola Virus Plants have long been a source of drugs—morphine, ephedrine, and codeine are just a few examples ( Section 10.9 ) . Now scientists are attempting to obtain drugs from plants by biopharming. Biopharming uses genetic engineering techniques to produce drugs in crops such as corn, rice, tomatoes, and tobacco. To date, the only biopharmed drug approved by the Food and Drug Administration (FDA) is one that is manufactured in carrots and used to treat Gaucher’s disease. An experimental drug that was used to treat a handful of patients with Ebola, the virus that was spreading throughout West Africa, was obtained from genetically engineered tobacco plants. The tobacco plants were infected with three genetically engineered plant viruses that are harmless to humans and animals but have structures similar to that of the Ebola virus. As a result of being infected, the plants produced antibodies to the viruses. The antibodies were isolated from the plants, purified, and then used to treat the patients with Ebola. The experimental drug had been tested in 18 monkeys who had been exposed to a lethal dose of Ebola. All 18 monkeys survived, whereas the three monkeys in the control group died. Typically, drugs go through rigorous testing on healthy humans prior to being administered to infected patients (see page 290) . In the Ebola case, the FDA made an exception because it feared that the drug might be these patients’ only hope. Five of the seven people given the drug survived. Currently, it takes about 50 kilograms of tobacco leaves and about 4 to 6 months to produce enough drug to treat one patient. tobacco plants More Applications Than Any Other Organic Text NEW! and Updated Application boxes connect the discussion to medical, environmental, biological, pharmaceutical, nutritional, chemical, industrial, historical, and general applications and allow students to relate the material to real life and to potential future careers. 392 CHAPTER 9 Substitution and Elimination Reactions of Alkyl Halides This chapter focuses on the substitution and elimination reactions of alkyl halides—compounds in which the leaving group is a halide ion ( F- , Cl- , Br- , or I- ). alkyl halides alkyl fluoride R F alkyl chloride R Cl alkyl bromide R Br alkyl iodide R I Alkyl halides are a good family of compounds with which to start the study of substitution and elimination reactions because they have relatively good leaving groups; that is, the halide ions are easily displaced. After learning about the reactions of alkyl halides, you will be prepared to move on to Chapter 10 , which discusses the substitution and elimination reactions of compounds with poorer leaving groups (those that are more difficult to displace) as well as a few with better leaving groups. Substitution and elimination reactions are important in organic chemistry because they make it possible to convert readily available alkyl halides into a wide variety of other compounds. These reactions are also important in the cells of plants and animals. We will see, however, that because cells exist in predominantly aqueous environments and alkyl halides are insoluble in water, biological systems use compounds in which the group that is replaced is more polar than a halogen and, therefore, more water soluble (Section 10.12). The Birth of the Environmental Movement Alkyl halides have been used as insecticides since 1939, when it was discovered that DDT (first synthesized in 1874) has a high toxicity to insects and a relatively low toxicity to mammals. DDT was used widely in World War II to control typhus and malaria in both the military and civilian populations. It saved millions of lives, but no one realized at that time that, because it is a very stable compound, it is resistant to biodegradation. In addition, DDT and DDE, a compound formed as a result of elimination of HCl from DDT, are not water soluble. Therefore, they accumulate in the fatty tissues of birds and fish and can be passed up the food chain. Most older adults have a low concentration of DDT or DDE in their bodies. In 1962, Rachel Carson, a marine biologist, published Silent Spring, where she pointed out the environmental impacts of the widespread use of DDT. The book was widely read, so it brought the problem of environmental pollution to the attention of the general public for the first time. Consequently, its publication was an important event in the birth of the environmental movement. Because of the concern it raised, DDT was banned in the United States in 1972. In 2004, the Stockholm Convention banned the worldwide use of DDT except for the control of malaria in countries where the disease is a major health problem. In Section 12.12 , we will look at the environmental effects caused by synthetic alkyl halides known as chlorofluorohydrocarbons (CFCs). PROBLEM 1 PROBLEM 2 Methoxychlor is an insecticide that was intended to take DDT’s place because it is not as soluble in fatty tissues and is more readily biodegradable. It, too, can accumulate in the environment, however, so its use was also banned—in 2002 in the European Union and in 2003 in the United States. Why is methoxychlor less soluble in fatty tissues than DDT? O O methoxychlor Cl Cl Cl Cl Cl DDT

GUIDED APPROACH TO PROBLEM SOLVING Essential Skill Tutorials These tutorials guide students through some of the topics in organic chemistry that they typically find to be the most challenging. They provide concise explanations, related problem-solving opportunities, and answers for self-check. The print tutorials are paired with MasteringChemistry online tutorials. These are additional problem sets that can be assigned as homework or as test preparation. xxviii Preface 19.8 Organizing What We Know about the Reactions of Organic Compounds 19. 8 ORGANIZING WHAT WE KNOW ABOUT THE REACTIONS OF ORGANIC COMPOUNDS N Z benzene pyridine pyrrole, furan, thiophene Group IV Z = N, O, or S H Halo-substituted benzenes and halo-substituted pyridines are electrophiles. They undergo nucleophilic aromatic substitution reactions. These are nucleophiles. They undergo electrophilic aromatic substitution reactions. These are electrophiles. They undergo nucleophilic acyl substitution reactions, nucleophilic addition reactions, or nucleophilic addition–elimination reactions. Removal of a hydrogen from an A-carbon forms a nucleophile that can react with electrophiles. O C R Z O R Z C Group II Group III Z = C or H Z = an atom more electronegative than C These are electrophiles. They undergo nucleophilic substitution and/or elimination reactions. alkyl halide alcohol ether epoxide Group II R X R OH R OR R R CHCH R CHCH RCHCH R R CC These are nucleophiles. They undergo electrophilic addition reactions. alkene alkyne diene O R R X = F, Cl, Br, I sulfonate ester sulfonium salt quaternary ammonium hydroxide R O S O O R R SR R R R HO− NR R + + Group I We saw that t he families of organic compounds can be put in one of four groups, and that all the families in a group react in similar ways. Now that we have finished studying the families in Group IV, l et’s review how these compounds react. All the compounds in Group IV are aromatic. To preserve the aromaticity of the rings, these Porphyrin, Bilirubin, and Jaundice The average human body breaks down about 6 g of hemoglobin each day. The protein portion (globin) and the iron are reutilized, but the porphyrin ring is cleaved between the A and B rings to form a linear tetrapyrrole called biliverdin (a green compound). Then the bridge between the C and D ring is reduced, forming bilirubin (a red-orange compound). You can witness heme degradation by observing the changing colors of a bruise. Enzymes in the large intestine reduce bilirubin to urobilinogen (a colorless compound). Some urobilinogen is transported to the kidney, where it is oxidized to urobilin (a yellow compound). This is the compound that gives urine its characteristic color. If more bilirubin is formed than can be metabolized and excreted by the liver, it accumulates in the blood. When its concentration there reaches a certain level, it diffuses into the tissues, giving them a yellow appearance. This condition is known as jaundice. 225 ESSENTIAL SKILL TUTORIAL DRAWING CURVED ARROWS This is an extension of what you learned about drawing curved arrows on pp. 199 – 201 . Working through these problems will take only a little of your time. It will be time well spent, however, because curved arrows are used throughout the book and it is important that you are comfortable with them. (You will not encounter some of the reaction steps shown in this exercise for weeks or even months, so don’t worry about why the chemical changes take place.) Chemists use curved arrows to show how electrons move as covalent bonds break and/or new covalent bonds form. ■ Each arrow represents the simultaneous movement of two electrons (an electron pair) from a nucleophile (at the tail of the arrow) toward an electrophile (at the point of the arrow). ■ The tail of the arrow is positioned where the electrons are in the reactant; the tail always starts at a lone pair or at a bond. ■ The head of the arrow points to where these same electrons end up in the product; the arrow always points at an atom or at a bond. In the following reaction step, the bond between bromine and a carbon of the cyclohexane ring breaks and both electrons in the bond end up with bromine. Thus, the arrow starts at the electrons that carbon and bromine share in the reactant, and the head of the arrow points at bromine because this is where the two electrons end up in the product. Br + Br− + Notice that the carbon of the cyclohexane ring is positively charged in the product. This is because it has lost the two electrons it was sharing with bromine. The bromine is negatively charged in the product because it has gained the electrons that it shared with carbon in the reactant. The fact that two electrons move in this example is indicated by the two barbs on the arrowhead. Notice that the arrow always starts at a bond or at a lone pair. It does not start at a negative charge. CH3CHCH3 + Cl CH3CHCH3 − Cl + In the following reaction step, a bond is being formed between the oxygen of water and a carbon of the other reactant. The arrow starts at one of the lone pairs of the oxygen and points at the atom (the carbon) that will share the electrons in the product. The oxygen in the product is positively charged, because the electrons that oxygen had to itself in the reactant are now being shared with carbon. The carbon that was positively charged in the reactant is not charged in the product, because it has gained a share in a pair of electrons. CH3CHCH3 + CH3CHCH3 + + H2O OH H PROBLEM 1 Draw curved arrows to show the movement of the electrons in the following reaction steps. (The answers to all problems appear immediately after Problem 10 .) Cl Br Br− + a. CH3CH2C + CH3 CH3 CH3CH2C+ CH3 CH3 Cl− b. + Enhanced by TUTORIAL ESSENTIAL SKILL Organizing What We Know About the Reactivity of Organic Compounds This organization emphasizes the unifying principles of reactivity and offers an economy of presentation while discouraging memorization. Students learn that • organic compounds can be classified into families and that all members of a family react in the same way. • the families can be put into one of four groups and that all the family members in a group react in similar ways. The Organizing What We Know table builds as students work sequentially through the four groups. Group I: electrophilic addition reactions Group II: nucleophilic substitution reactions and elimination reactions Group III: nucleophilic acyl substitution reactions, nucleophilic addition reactions, and nucleophilic addition–elimination reactions Group IV: electrophilic (and nucleophilic) aromatic substitution reactions

Preface xxix Emphasis on the Strategies Needed to Solve Problems and Master Content Passages explaining important problem-solving strategies are clearly labeled with a LEARN THE STRATEGY label. Follow-up problems that require students to apply the just-learned strategy are labeled with a USE THE STRATEGY label. These labels, which are implemented throughout the text, allow students to easily find important content and practice its use. Designing a Synthesis This recurring feature helps students learn to design multi-step syntheses and facilitates the development of complex problem-solving skills. Many problems include the synthesis of well-known compounds such as Novocain®, Valium®, and Ketoprofen®. 836 CHAPTER 17 Reactions at the a-Carbon PROBLEM 42 SOLVED Starting with methyl propanoate, how could you prepare 4-methyl-3-heptanone? methyl propanoate 4-methyl-3-heptanone ? CH3 O CH3CH2 OCH3 C O CH3CH2 CHCH2CH2CH3 C Because the starting material is an ester and the target molecule has more carbons than the starting material, a Claisen condensation appears to be a good way to start this synthesis. The Claisen condensation forms a b -keto ester that can be easily alkylated at the desired carbon because it is flanked by two carbonyl groups. Acid-catalyzed hydrolysis forms a 3-oxocarboxylic acid that decarboxylates when heated. DESIGNING A SYNTHESIS V 17. 20 When planning the synthesis of a compound that requires the formation of a new carbon– carbon bond: ■ locate the new bond that needs to be made and perform a disconnection—that is, break the bond to produce two fragments. ■ determine which of the atoms that will form the new bond should be the electrophile and which should be the nucleophile. ■ choose a compound with the desired electrophilic and nucleophilic groups. Example 1 The new bond that needs to be made in the synthesis of the following b -diketone is the one that makes the second five-membered ring: O O O OH O H3C H3C H3C H3C new bond ? It is easy to choose between the two possibilities for the electrophile and nucleophile because we know that a carbonyl carbon is an electrophile. H3C H3C H3C H3C or O O + − O O + − nucleophile electrophile electrophile nucleophile Δ 1. CH3O− 2. H3O+ 1. CH3O− 2. CH3CH2CH2Br HCl, H2O Δ O CH3CH2 OCH3 C CH3 O CH3CH2 CH CH3 C O OCH3 C O CH3CH2 C CH3 CH2CH2CH3 C O OCH3 C O CH3CH2 C CH3 CH2CH2CH3 C O OH C O CH3CH2 CHCH2CH2CH3 C 696 CHAPTER 15 Reactions of Carboxylic Acids and Carboxylic Acid Derivatives acyl substitution reaction is the p bond, so this bond breaks first and the leaving group is eliminated in a subsequent step. an SN2 reaction CH3CH2 + Z− Y CH3CH2 + Y− Z PROBLEM-SOLVING STRATEGY Using Basicity to Predict the Outcome of a Nucleophilic Acyl Substitution Reaction What is the product of the reaction of acetyl chloride with CH3O-?The p Ka of HCl is –7 ; the p Ka of CH 3 OH is 15.5. To identify the product of the reaction, we need to compare the basicities of the two groups in the tetrahedral intermediate so we can determine which one will be eliminated. Because HCl is a stronger acid than CH 3 OH, Cl − is a weaker base than CH 3O− . Therefore, Cl − is eliminated from the tetrahedral intermediate and methyl acetate is the product of the reaction. + + CH3 CH3 CH3O− Cl C O CH3 OCH3 C O C Cl Cl − O− acetyl chloride OCH3 methyl acetate LEARN THE STRATEGY 15. 5 THE RELATIVE REACTIVITIES OF CARBOXYLIC ACIDS AND CARBOXYLIC ACID DERIVATIVES We just saw that there are two steps in a nucleophilic acyl substitutions reaction: formation of a tetrahedral intermediate and collapse of the tetrahedral intermediate. The weaker the base attached to the acyl group (Table 15 .1), the easier it is for both steps of the reaction to take place. Cl− < < ≈ − OR − NH2 − OH relative basicities of the leaving groups weakest base strongest base Therefore, carboxylic acid derivatives have the following relative reactivities: > > ≈ relative reactivities of carboxylic acid derivatives carboxylic acid R OH amide R NH2 acyl chloride ester most R OR′ reactive least reactive R Cl C O C O C O C O PROBLEM 7 ♦ a. What is the product of the reaction of acetyl chloride with HO - ? The p Ka of HCl is -7; the p Ka of H 2O is 15.7. b. What is the product of the reaction of acetamide with HO - ? The p Ka of NH 3 is 36; the p Ka of H 2O is 15.7. PROBLEM 8 ♦ What is the product of an acyl substitution reaction—a new carboxylic acid derivative, a mixture of two carboxylic acid derivatives, or no reaction—if the new group in the tetrahedral intermediate is the following? a. a stronger base than the substituent that is attached to the acyl group b. a weaker base than the substituent that is attached to the acyl group c. similar in basicity to the substituent that is attached to the acyl group USE THE STRATEGY 17.20 Making New Carbon–Carbon Bonds 837 If we know what the starting material is, we can use it as a clue to arrive at the desired compound. For example, an ester carbonyl group would be a good electrophile for this synthesis because it has a group that would be eliminated. Moreover, the a -hydrogens of the ketone are more acidic than the a -hydrogens of the ester, so the desired nucleophile would be easy to obtain. Thus, converting the starting material to an ester ( Section 15.18 ) , followed by an intramolecular condensation, forms the target molecule. H3C H3C 1. SOCl2 1. CH3O− 2. H3O 2. CH + 3OH O O O O O OH O OCH3 H3C H3C H3C H3C Example 2 In this example, two new carbon–carbon bonds must be formed. CH3O OCH3 O O CH3O OCH3 O O ? new bond new bond After identifying the electrophilic and nucleophilic sites, we see that two successive alkylations of CH3O OCH3 O O CH3O OCH3 1. CH3O− CH3O− 2. Br Br O O O O Br CH3O OCH3 O O CH3O OCH3 − + + − electrophile nucleophile Example 3 The diester given as the starting material suggests that a Dieckmann condensation should be used to obtain the cyclic compound: O O O CH3CH2 O O CH3OC nucleophile new bond electrophile ? + CH3O − OCH3 The Dieckmann condensation is followed by alkylation of the a -carbon (that is flanked by two carbonyl groups) of the cyclopentanone ring. Hydrolysis of the b -keto ester and decarboxylation forms the target molecule. C 1. CH3O− 2. HCl 1. C2H5O HCl, H2O − 2. CH3CH2Br CH3O O O CH3OC O O CH3CH2 CO2 CH3OC O O Δ + + CH3OH O CH3OC CH3CH2