《有机化学》课程教学资源（参考书籍）Schaum's Outline of Organic Chemistry 4th Ed 2009, 1806 fully solved problems

SCHAUM'S outlines Problem Organic Solved Chemistry Fourth Edition 1,806 fully solved problems Concise explanations of al course concepts Information on organic compounds, molecular structure, chemical reactivity and organic reactions, stereochemistry, and spectroscopy and structure USE WITH THESE COURSES Organic Chemistry. Physical Chemistry. Analytical Chemistry Biochemistry. Chemistry Education.Anatomy/Physiology Molecular Biology Herbert Meislich, Ph.D. . Howard Nechamkin, Ph.D. Jacob Sharefkin, Ph.D. . George Hademenos, Ph.D

Preface Eac Schauudertaken toveaclar vi of fityer Organic Chemisyhothe detailed solution of illustrative problems.Such problems make up over8%of the book.the remainder being a concise presentation of the material.Our goal is for students to learn by thinking and solving problems rather han by merely being to rt of a standard text a The second edition has been reorganized by combining chapters to emphasize the similarities of functional groups and reaction types as well as the differences.Thus.polynuclear hydrocarbons are combined with ben ene and aromaticity.Nuc eophilic a up-to-date by including solvent efects CMR spectroscopy.elbration of polymer chemistryand newer concepts of stereochemistry,among other material. JACOB SHAREFKIN GEORGE J.HADEMENOS i●

vii Preface The beginning student in Organic Chemistry is often overwhelmed by facts, concepts, and new language. Each year, textbooks of Organic Chemistry grow in quantity of subject matter and in level of sophistication. This Schaum’s Outline was undertaken to give a clear view of first-year Organic Chemistry through the careful detailed solution of illustrative problems. Such problems make up over 80% of the book, the remainder being a concise presentation of the material. Our goal is for students to learn by thinking and solving problems rather than by merely being told. This book can be used in support of a standard text, as a supplement to a good set of lecture notes, as a review for taking professional examinations, and as a vehicle for self-instruction. The second edition has been reorganized by combining chapters to emphasize the similarities of functional groups and reaction types as well as the differences. Thus, polynuclear hydrocarbons are combined with ben￾zene and aromaticity. Nucleophilic aromatic displacement is merged with aromatic substitution. Sulfonic acids are in the same chapter with carboxylic acids and their derivatives, and carbanion condensations are in a sepa￾rate new chapter. Sulfur compounds are discussed with their oxygen analogs. This edition has also been brought up-to-date by including solvent effects, CMR spectroscopy, an elaboration of polymer chemistry, and newer concepts of stereochemistry, among other material. HERBERT MEISLICH HOWARD NECHAMKIN JACOB SHAREFKIN GEORGE J. HADEMENOS

Contents CHAPTER 1 Structure and Properties of Organic Compounds CHAPTER2 Bonding and Molecular Structure CHAPTER3 Chemical Reactivity and Organic Reactions 31 CHAPTER 4 Alkanes 50 CHAPTER5 Stereochemistry 69 CHAPTER 6 Alkenes 87 CHAPTER 7 Alkyl Halides 118 CHAPTER8 Alkynes and Dienes 140 CHAPTER 9 Cyclic Hydrocarbons 162 CHAPTER 10 Benzene and Polynuclear Aromatic Compounds 189 CHAPTER 11 Aromatic Substitution;Arenes 205 CHAPTER 12 Spectroscopy and Structure CHAPTER 13 Alcohols and Thiols 6 CHAPTER 14 Ethers,Epoxides,Glycols,and Thioethers 278 CHAPTER 15 Carbonyl Compounds:Aldehydes and Ketones CHAPTER 16 Carboxylic Acids and their Derivatives CHAPTER 17 Carbanion-Enolates and Enols 373 CHAPTER 18 Amines CHAPTER 19 Phenolic Compounds 430 CHAPTER 20 Aromatic Heterocyclic Compounds 448 Index 464 ix●

ix Contents CHAPTER 1 Structure and Properties of Organic Compounds 1 CHAPTER 2 Bonding and Molecular Structure 13 CHAPTER 3 Chemical Reactivity and Organic Reactions 31 CHAPTER 4 Alkanes 50 CHAPTER 5 Stereochemistry 69 CHAPTER 6 Alkenes 87 CHAPTER 7 Alkyl Halides 118 CHAPTER 8 Alkynes and Dienes 140 CHAPTER 9 Cyclic Hydrocarbons 162 CHAPTER 10 Benzene and Polynuclear Aromatic Compounds 189 CHAPTER 11 Aromatic Substitution;Arenes 205 CHAPTER 12 Spectroscopy and Structure 230 CHAPTER 13 Alcohols and Thiols 256 CHAPTER 14 Ethers, Epoxides, Glycols, and Thioethers 278 CHAPTER 15 Carbonyl Compounds:Aldehydes and Ketones 302 CHAPTER 16 Carboxylic Acids and their Derivatives 331 CHAPTER 17 Carbanion-Enolates and Enols 373 CHAPTER 18 Amines 400 CHAPTER 19 Phenolic Compounds 430 CHAPTER 20 Aromatic Heterocyclic Compounds 448 Index 464

CHAPTER 1 Structure and Properties of Organic Compounds 1.1 Carbon Compounds Organic chemistry is the study of carbon (C)compounds,all of which have covalent bonds.Carbon atoms can bond to each other t T网e3光里 with double bonds and triple bonds are shown in Fig.1.1(e).Cyclic compounds having at least one atom in the ring other than C(a heteroatom)are called heterocyclics.Fig.1.1(f).The heteroatoms are usually oxygen (O),nitrogen(N).or sulfur(S). Problem 11 Why are there so many compounds that contain carbon? Bonds between C's are covalent and strong,so that C's can form long chains and rings,both of which may ssoCSsesnmeRmntaikthAnkraioncRe Problem1.2 Compare and contrast the properties of ionic and covalent compounds. Ionic compounds are generally inorganic,have high melting and boiling points due to the strong electro Covalent compounds,on the other hand,are commonly organic:have relatively low melting and boiling points because of weak intermolecular forces;are soluble in organic solvents and insoluble in water:burn read ily and are thus susceptible to oxidation because they are less stable to heat,usually decomposing at tempera complex, ding higher

CHAPTER 1 1 Structure and Properties of Organic Compounds 1.1 Carbon Compounds Organic chemistry is the study of carbon (C) compounds, all of which have covalent bonds. Carbon atoms can bond to each other to form open-chain compounds, Fig. 1.1(a), or cyclic (ring) compounds, Fig. 1.1(c). Both types can also have branches of C atoms, Fig. 1.1(b) and (d). Saturated compounds have C atoms bonded to each other by single bonds, C—C; unsaturated compounds have C’s joined by multiple bonds. Examples with double bonds and triple bonds are shown in Fig. 1.1(e). Cyclic compounds having at least one atom in the ring other than C (a heteroatom) are called heterocyclics, Fig. 1.1(f). The heteroatoms are usually oxygen (O), nitrogen (N), or sulfur (S). Problem 1.1 Why are there so many compounds that contain carbon? Bonds between C’s are covalent and strong, so that C’s can form long chains and rings, both of which may have branches. C’s can bond to almost every element in the periodic table. Also, the number of isomers increases as the organic molecules become more complex. Problem 1.2 Compare and contrast the properties of ionic and covalent compounds. Ionic compounds are generally inorganic, have high melting and boiling points due to the strong electro￾static forces attracting the oppositely charged ions, are soluble in water and insoluble in organic solvents, are hard to burn, and involve reactions that are rapid and simple. Also, bonds between like elements are rare, with isomerism being unusual. Covalent compounds, on the other hand, are commonly organic; have relatively low melting and boiling points because of weak intermolecular forces; are soluble in organic solvents and insoluble in water; burn read￾ily and are thus susceptible to oxidation because they are less stable to heat, usually decomposing at tempera￾tures above 700ºC; and involve reactions that are slow and complex, often needing higher temperatures and/ or catalysts, yielding mixtures of products. Also, bonds between carbon atoms are typical, with isomerism being common

2● CHAPTER 1 Structure and Properties of Organic Compounds HHHH HH H-C-C-C-C-H H- HH H HHHH H H H H n8c (c) (d) HH Hc-H H H-C=C-H H H Ethene(Ethylene) Cyclopentene have double bonds (e) Figure 1.1 1.2 Lewis Structural Formulas Molecular formulas merely include the kinds of atom and the each in )St e number of electrons are included.the latter are called Lewis (electron-dot)structures see Fig.Covalences of the common elements-the numbers of covalent bonds they usually form-are given in Table 1.1:these help us to wis structures.Multicovalent clements such a C.O.and N may have multiple bonds,as shown in om to dens they are atta CH.CH(CH . Problem 1.3 (a)Are the covalences and group numbers(numbers of valence electrons)of the elements in 1.1 relate ()Why aren Grup lements incluhrone d in (a)Yes.For the elements in Groups 4 through 7,Covalence=8-(Group number). (b)No.The elements in Groups 4 through 7 do attain the octet,but the elements in Groups 2 and 3 have less octet of ence clectrons.d andr enods.suandmay acheverea (c)They fom dson c rather than covalent bond Most c th by the C atom.The spatial relationship is indicated as in Fig.1.2(a)(Newman projection)or in Fig.1.2(b) xcept for ethene,which is planar,and ethyne,which is linear,the structures in Fig.1.1 are ulas ur s Thi of isomerism is exemplified by isobutane and n-butane Fig1.1(a)and (The number of isomers increases as the number of atoms in the organic molecule increases

1.2 Lewis Structural Formulas Molecular formulas merely include the kinds of atoms and the number of each in a molecule (as C4H10 for butane). Structural formulas show the arrangement of atoms in a molecule (see Fig. 1.1). When unshared electrons are included, the latter are called Lewis (electron-dot) structures [see Fig. 1-1(f)]. Covalences of the common elements—the numbers of covalent bonds they usually form—are given in Table 1.1; these help us to write Lewis structures. Multicovalent elements such as C, O, and N may have multiple bonds, as shown in Table 1.2. In condensed structural formulas, all H’s and branched groups are written immediately after the C atom to which they are attached. Thus, the condensed formula for isobutane [Fig. 1-1(b)] is CH3CH(CH3)2. Problem 1.3 (a) Are the covalences and group numbers (numbers of valence electrons) of the elements in Table 1.1 related? (b) Do all the elements in Table 1.1 attain an octet of valence electrons in their bonded states? (c) Why aren’t Group 1 elements included in Table 1.1? (a) Yes. For the elements in Groups 4 through 7, Covalence 8 (Group number). (b) No. The elements in Groups 4 through 7 do attain the octet, but the elements in Groups 2 and 3 have less than an octet. (The elements in the third and higher periods, such as Si, S, and P, may achieve more than an octet of valence electrons.) (c) They form ionic rather than covalent bonds. (The heavier elements in Groups 2 and 3 also form mainly ionic bonds. In general, as one proceeds down a group in the periodic table, ionic bonding is preferred.) Most carbon-containing molecules are three-dimensional. In methane, the bonds of C make equal angles of 109.5º with each other, and each of the four H’s is at a vertex of a regular tetrahedron whose center is occupied by the C atom. The spatial relationship is indicated as in Fig. 1.2(a) (Newman projection) or in Fig. 1.2(b) (“wedge” projection). Except for ethene, which is planar, and ethyne, which is linear, the structures in Fig. 1.1 are all three-dimensional. Organic compounds show a widespread occurrence of isomers, which are compounds having the same molec￾ular formula but different structural formulas, and therefore possessing different properties. This phenomenon of isomerism is exemplified by isobutane and n-butane [Fig. 1.1(a) and (b)]. The number of isomers increases as the number of atoms in the organic molecule increases. 2 CHAPTER 1 Structure and Properties of Organic Compounds H C C H H CC C C H H HH O H H H H n-Butane unbranched, open-chain (a) Isobutane branched, open-chain (b) Cyclopropane unbranched, cyclic (c) Methylcyclopropane branched, cyclic (d) Ethene (Ethylene) have double bonds Cyclopentene (e) Ethyne (Acetylene) has a triple bond Ethylene oxide heterocyclic (f ) H H H C H H C H H C H H C H H H C C H C C C H H H H H H H H C H H H H H C H H H H H HH H C C H H H H H H H C C C C C C H Figure 1.1

CHAPTER 1 Structure and Properties of Organic Compounds 3 TABLE 1.1 Covalences of H and Second-Period Elements in Groups 2 through 7 Group 1 2 4 5 6 1 Lewis Symbol H .Be B F Covalence 1 2 3 4 3 2 1 H-B-H H H-N-H HH HReH H H-F. Hydrogen Beryllium Boron H HH Hydrogen hydride hydride* Methane Ammonia Water fluoride Exists as B.He TABLE 1.2 Normal Covalent Bonding BONDING FOR C BONDING FOR N BONDING FOR O =C= -C≡ -N- -N= -0- 0= as in as in as in as in as H -H :0=C=0:H-C=C-H H-N-H H-0-N=0::N=C-H H-0-H 0= H H Methane Ethene Ammonia Nitrous Water Formaldehyde (Ethylene) dioxide (Acetylene) acid cyanide Problem 1.4 Write str ural and condensed formulas for (a)three isomers with molecular formula CHand (b)two isomers with molecular formulaC.H (a)Carbon forms four covalent bonds:hydrogen forms one.The carbons can bond to each other in a chain: -H (structural formula) CH:(CH2)CHs (condensed formula) n-Pentane or there can be"branches"(shown circled in Fig.1.3)on the linear backbone (shown in a rectangle) (b)We can have a double bond or a ring H H CHC-CH2 H Propene(Propylene) Cyclopropane

Problem 1.4 Write structural and condensed formulas for (a) three isomers with molecular formula C5 H12 and (b) two isomers with molecular formula C3H6. (a) Carbon forms four covalent bonds; hydrogen forms one. The carbons can bond to each other in a chain: CHAPTER 1 Structure and Properties of Organic Compounds 3 Group Lewis Symbol Covalence Compounds with H 1 H. 1 2 Be.. 2 3 B. . . 3 * Exists as B2 H6 . 45 67 C. . . . N. . . . . O. . . . . . F: . . . . . 43 21 H—H Hydrogen H—Be—H Beryllium hydride H—B—H | Boron hydride* H | H—C—H | H Methane . . H—N—H | H Ammonia . . H—O—H . . Water . . H—F: . . Hydrogen fluoride H BONDING FOR C as in as in as in as in as in as in as in as in as in BONDING FOR N BONDING FOR O C C C C H H C H H . . . . C C O O C H H H H C H H Methane Ethene (Ethylene) Carbon dioxide Ethyne (Acetylene) Ammonia Nitrous acid Hydrogen cyanide Water Formaldehyde . . . . . . H . . O O: N . . HOH . . . . :N C H . . O . . H N H H H H C C . . N . . . . O . . . . O . . N . . N : : or there can be “branches” (shown circled in Fig. 1.3) on the linear backbone (shown in a rectangle). (b) We can have a double bond or a ring. TABLE 1.1 Covalences of H and Second-Period Elements in Groups 2 through 7 TABLE 1.2 Normal Covalent Bonding

4 CHAPTER 1 Structure and Properties of Organic Compounds H/'s project toward viewer Hs project away Irom view H H Figure 1.3 Problem1.5 Write Lewis structures for(a)hydrazine,N,H(b)phosgene.COCl:and(c)nitrous acid,HNO In general,first bond the multicovalent atoms to each other and then,to achieve their normal covalences bond them to the univalent atoms (H,Cl,Br.Iand F).If the number of univalent atoms is insufficient for this structure sh (a)N needs three covalent bonds,and H needs one.Each N is bonded to the other N and to two H's. (b)Cis bonded to Oand to each Cl.To satisfy the tetravalence of C and the divalence of a double bond is placed between C and O. the N. and the other O.(Convince yourself that bonding the Hto the N would not lead structure.) H-i-i-6:

Problem 1.5 Write Lewis structures for (a) hydrazine, N2 H4 ; (b) phosgene, COCl2 ; and (c) nitrous acid, HNO2 . In general, first bond the multicovalent atoms to each other and then, to achieve their normal covalences, bond them to the univalent atoms (H, Cl, Br, I, and F). If the number of univalent atoms is insufficient for this purpose, use multiple bonds or form rings. In their bonded state, the second-period elements (C, N, O, and F) should have eight (an octet) electrons but not more. Furthermore, the number of electrons shown in the Lewis structure should equal the sum of all the valence electrons of the individual atoms in the molecule. Each bond represents a shared pair of electrons. (a) N needs three covalent bonds, and H needs one. Each N is bonded to the other N and to two H’s. 4 CHAPTER 1 Structure and Properties of Organic Compounds Figure 1.2 Figure 1.3 (b) C is bonded to O and to each Cl. To satisfy the tetravalence of C and the divalence of O, a double bond is placed between C and O. (c) The atom with the higher covalence, in this case the N, is usually the more central atom. Therefore, each O is bonded to the N. The H is bonded to one of the O atoms, and a double bond is placed between the N and the other O. (Convince yourself that bonding the H to the N would not lead to a viable structure.)

CHAPTER 1 Structure and Properties of Organic Compounds Problem 16 Why are none of the following Lewis structures for COClcorrect? @:g-c-0-0:(b):g-C=0-a:(c:C=C=6-年(d):C=C=0-a: The total number of valence electrons that must ar in the Lewis structure is 24.from 12 x 71(2CI's)+ 4(C)+6().Structures()and (c)can be rejected because they each show only 2 electrons.Furthermore in(b).O has 4 rather than 2 bonds,and in (c),one Cl has 2 bonds.In (a).C and O do not have their normal covalences.In (d),O has 10 electrons,though it cannot have more than an octet. Problem 17 Use the Lewis-Langmuir octet rule to write Lewis electron-dot structures for:(a)HCN,(b)CO (e)CCl.and (d)C.H,O. (a)Attach the H to the C.bec than N.The normal covalences of Nand Care me with a triple is the correct Lew s structure (b)The Cis bonded to eachby double bonds to achieve the normal covalences 0=C=Q: (c)Each of the four Cl'sis singly bonded to the tetravalent Cto give: (d) of 4 and 2.respectively. e get two c Ethanol Dimethyl ether Problem 1.8 Determine the positive or negative charge,if any,on: (a)H-C-: (d)H-N--H (e) -: The charge on a species is numerically equal to the total number of valence electrons of the unbonded atoms minus the total number of electrons shown (as bonds or dots)in the Lewis structure. (a)The sum of the valence electrons(6 for O.4 for C.and 3 for three H's)is 13.The electron-dot formula shows 14 electrons.The net charge is 13-14=-1,and the species is the methoxide anion,CH3O:

Problem 1.6 Why are none of the following Lewis structures for COCl2 correct? CHAPTER 1 Structure and Properties of Organic Compounds 5 (c) Each of the four Cl’s is singly bonded to the tetravalent C to give: (d) The three multicovalent atoms can be bonded as C—C—O or as C—O—C. If the six H’s are placed so that C and O acquire their usual covalences of 4 and 2, respectively, we get two correct Lewis structures (isomers): The total number of valence electrons that must appear in the Lewis structure is 24, from [2  7](2Cl’s)  4(C)  6(O). Structures (b) and (c) can be rejected because they each show only 22 electrons. Furthermore, in (b), O has 4 rather than 2 bonds, and in (c), one Cl has 2 bonds. In (a), C and O do not have their normal covalences. In (d), O has 10 electrons, though it cannot have more than an octet. Problem 1.7 Use the Lewis-Langmuir octet rule to write Lewis electron-dot structures for: (a) HCN, (b) CO2 , (c) CCl4 , and (d) C2 H6 O. (a) Attach the H to the C, because C has a higher covalence than N. The normal covalences of N and C are met with a triple bond. Thus, H—C ⎯N: is the correct Lewis structure. (b) The C is bonded to each O by double bonds to achieve the normal covalences. Problem 1.8 Determine the positive or negative charge, if any, on: The charge on a species is numerically equal to the total number of valence electrons of the unbonded atoms minus the total number of electrons shown (as bonds or dots) in the Lewis structure. (a) The sum of the valence electrons (6 for O, 4 for C, and 3 for three H’s) is 13. The electron-dot formula shows 14 electrons. The net charge is 13 14 1, and the species is the methoxide anion, .

6 CHAPTER 1 Structure and Properties of Organic Compounds (b)There is no charge on the formaldehyde molecule,because the 12 electrons in the structure equals the number of valence electrons-that is,6 for O.4 for C,and 2 for two H's. (c)This species is neutral,because there are 13 electrons shown in the formula and 13 valence electrons: 8 from two C's and 5 from five H's. (d)There are 15 valence electrons:6 fromO.5 from N.and 4 from four H's.The Lewis dot structure shows 14 electrons.It has a charge of 15-14=+1 and is the hydroxylammonium cation.[H,NOH]. (e)There are 25 valence electrons,21 from three Cl's and 4 from C.The Lewis dot formula shows 26electrons. It has a charge of 25 -26 =-I and is the trichloromethide anion.:CCl. 1.3 Types of Bonds Covalent bonds,the mainstays of organic compounds,are formed by the sharing of pairs of electrons.Sharing can occur in two ways: 1)A,+,B4A·B In method (1),each atom brings an electron for the sharing.In method (2),the donor atom (B:)brings both electrons to the"marriage"with the acceptor atom(A):in this case,the covalent bond is termed a coordinate covalent bond e me mar he新 n he th ise by alant hond (a)NHj:(b)BF(c)(CH),OMgCl:and (d)Fe(CO)s acceptor (a)H+ +:NH3 NH+(All N-H bonds are alike.) (b)F3B +:月： BF (All B-F bonds are alike.) (o):g-Mg-:+:0-CH, →(CH326-MgC12 CH: (d)Fe +5:C=0: →Fe(C=O)5 Notice that in each of the products there is at least one element that does not have its usual covalence-this is typical of coordinate covalent bonding. Rcnm心o@二ooe5边o总t品h器co bonds,it some Problem 1.10 Show how the ionic compound Li'F-forms from atoms of Li and F. Li.+.F:Lit:F:(or simply LiF) 1.4 Functional Groups Hydrocarbons contain only Cand hydrogen(H).H's in hydrocarbons can be replaced by other atoms or groups ctional group nmolecules.The al groups ne common onal groups are given n Table 13

(b) There is no charge on the formaldehyde molecule, because the 12 electrons in the structure equals the number of valence electrons—that is, 6 for O, 4 for C, and 2 for two H’s. (c) This species is neutral, because there are 13 electrons shown in the formula and 13 valence electrons: 8 from two C’s and 5 from five H’s. (d) There are 15 valence electrons: 6 from O, 5 from N, and 4 from four H’s. The Lewis dot structure shows 14 electrons. It has a charge of 15 14 1 and is the hydroxylammonium cation, [H3NOH]. (e) There are 25 valence electrons, 21 from three Cl’s and 4 from C. The Lewis dot formula shows 26 electrons. It has a charge of 25 26 1 and is the trichloromethide anion, :CCl– 3 . 1.3 Types of Bonds Covalent bonds, the mainstays of organic compounds, are formed by the sharing of pairs of electrons. Sharing can occur in two ways: 6 CHAPTER 1 Structure and Properties of Organic Compounds (1) A ⋅+⋅B → A:B (2) A + :B → A : B coordinate covalent acceptor donor In method (1), each atom brings an electron for the sharing. In method (2), the donor atom (B:) brings both electrons to the “marriage” with the acceptor atom (A); in this case, the covalent bond is termed a coordinate covalent bond. Problem 1.9 Each of the following molecules and ions can be thought to arise by coordinate covalent bond￾ing. Write an equation for the formation of each one and indicate the donor and acceptor molecule or ion: (a) NH 4 ; (b) BF 4 ; (c) (CH3 )2 OMgCl2 ; and (d) Fe(CO)5 . Notice that in each of the products there is at least one element that does not have its usual covalence—this is typical of coordinate covalent bonding. Recall that an ionic bond results from a transfer of electrons (M·  A· M  :A). Although C usually forms covalent bonds, it sometimes forms an ionic bond (see Section 3.2). Other organic ions, such as CH3 COO (acetate ion), have charges on heteroatoms. Problem 1.10 Show how the ionic compound LiF forms from atoms of Li and F. These elements react to achieve a stable noble-gas electron configuration (NGEC). Li(3) has one electron more than He and loses it. F(9) has one electron less than Ne and therefore accepts the electron from Li. (a) (b) (c) (d) 1.4 Functional Groups Hydrocarbons contain only C and hydrogen (H). H’s in hydrocarbons can be replaced by other atoms or groups of atoms. These replacements, called functional groups, are the reactive sites in molecules. The C-to-C double and triple bonds are considered to be functional groups. Some common functional groups are given in Table 1.3.

CHAPTER 1 Structure and Properties of Organic Compounds 7 es havine similar operties with increasing molecular weight. Problem 1.11 Methane.CH:ethane.C.H:and propane,C.Hs are the first three members of the alkane homologous series.By what structural unit does each member differ from its predecessor These members differ by aCand two H's;the unit is-CH-(a methylene group) Problem 1.12 (a)Write possible Lewis structural formulas for (1)CH,O:(2)CH,O:(3)CH,O:(4)CH N: and (5)CH,SH.(b)Indicate and name the functional group in each case. The atom with the higher valence is usually the one to which most of the other atoms are bonded H H (a)()H -0-H2② =ǒ：3)H (4)H -H (5)H-C -SH (b)(I)一9H (2) =6:3)-C6-H4)-NH2 (⑤)一3 alcohol aldchyde carboxylic acid amine thiol 1.5 Formal Charge The formal charge on a coval 1 atom equals the numk er of ence el nof the unb onded atom (h of sha us the t The sun in a molecule equals the charge on the species.In this outline.formal charges and actual ionic charges (e.g..Na) are both indicated by the signs and Problem 1 13 Determine the formal charge on each atom in the following species:(a)H NBF:(b)CH NH nd (c)SO2 H: (a) H一Nt一B一： :F: GRR-「+2a]-a :atoms The sum of all formal charges equals the charge on the species.In this case.the +1 on N and the-1 on B cancel HH (b例

Compounds with the same functional group form a homologous series having similar characteristic chemical properties and often exhibiting a regular gradation in physical properties with increasing molecular weight. Problem 1.11 Methane, CH4; ethane, C2H6; and propane, C3H8, are the first three members of the alkane homologous series. By what structural unit does each member differ from its predecessor? These members differ by a C and two H’s; the unit is —CH2 — (a methylene group). Problem 1.12 (a) Write possible Lewis structural formulas for (1) CH4O; (2) CH2O; (3) CH2O2; (4) CH5N; and (5) CH3 SH. (b) Indicate and name the functional group in each case. The atom with the higher valence is usually the one to which most of the other atoms are bonded. CHAPTER 1 Structure and Properties of Organic Compounds 7 1.5 Formal Charge The formal charge on a covalently bonded atom equals the number of valence electrons of the unbonded atom (the group number) minus the number of electrons assigned to the atom in its bonded state. The assigned number is one half the number of shared electrons plus the total number of unshared electrons. The sum of all formal charges in a molecule equals the charge on the species. In this outline, formal charges and actual ionic charges (e.g., Na) are both indicated by the signs  and . Problem 1.13 Determine the formal charge on each atom in the following species: (a) H3 NBF3 ; (b) CH3 NH 3 ; and (c) SO2 4 . (a) (b) The sum of all formal charges equals the charge on the species. In this case, the 1 on N and the 1 on B cancel (b) (a)