Summary

This document contains problems related to organic chemistry, specifically structure and bonding topics, such as isomers (cis-trans and constitutional isomers). The problems include determining the relationship between different structural formulas.

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84    CHAPTER 1 Structure and Bonding PROBLEM 1-25 Give the relationship between the following pairs of structures. The possible relationships are:...

84    CHAPTER 1 Structure and Bonding PROBLEM 1-25 Give the relationship between the following pairs of structures. The possible relationships are: same compound constitutional isomers (structural isomers) cis-trans isomers not isomers (different molecular formula) (a) CH3CH2CHCH2CH3 and CH3CH2CHCH2CH2CH3 CH2CH3 CH3 (b) Br H Br Br C C and C C H Br H H (c) Br H Br H C C and C C H Br Br H (d) Br H H Br C C and C C H Br Br H (e) H Cl Cl Cl H C C H and H C C H Cl H H H (f) H CH3 CH3 H H C C H and H C C CH3 CH3 H H H (g) CH3 ¬ CH2 ¬ CH2 ¬ CH3 and CH3 ¬ CH “ CH ¬ CH3 (h) CH2 “ CH ¬ CH2CH2CH3 and CH3 ¬ CH “ CH ¬ CH2CH3 (i) CH2 “ CHCH2CH2CH3 and CH3CH2CH2CH “ CH2 (j) CH 3 CH 3 (k) CH 3 CH 3 and and CH 3 CH 3 Essential Terms Each chapter ends with a glossary that summarizes the most important new terms in the chapter. These glossaries are more than just a dictionary to look up unfamiliar terms as you encounter them (the index serves that purpose). The glossary is one of the tools for reviewing the chapter. You can read carefully through the glossary to see if you understand and remember all the terms and associated chemistry mentioned there. Anything that seems unfamiliar should be reviewed by turning to the page number given in the glossary listing. cis-trans isomers (geometric isomers) Stereoisomers that differ in their cis-trans arrangement on a double bond or on a ring. The cis isomer has similar groups on the same side, and the trans isomer has similar groups on opposite sides. (p. 82) constitutional isomers (structural isomers) Isomers whose atoms are connected differently; they differ in their ­bonding sequence. (p. 81) M01_WADE4255_10_GE_C01.indd 84 12/07/22 4:57 PM Essential Terms    85 covalent bonding Bonding that occurs by the sharing of electrons in the region between two nuclei. (p. 43) single bond: A covalent bond that involves the sharing of one pair of electrons. (p. 45) double bond: A covalent bond that involves the sharing of two pairs of electrons. (p. 45) triple bond: A covalent bond that involves the sharing of three pairs of electrons. (p. 45) degenerate orbitals Orbitals with identical energies. (p. 40) delocalized charge A charge that is spread out over two or more atoms. We usually draw resonance forms to show how the charge can appear on each of the atoms sharing the charge. (p. 50) dipole moment (μ) A measure of the polarity of a bond (or a molecule), proportional to the product of the charge separation times the bond length. (p. 46) double bond A bond containing four electrons between two nuclei. One pair of electrons forms a sigma bond, and the other pair forms a pi bond. (p. 68) electron density The relative probability of finding an electron in a certain region of space. (p. 39) electronegativity A measure of an element’s ability to attract electrons. Elements with higher electronegativities attract electrons more strongly. (p. 46) electrostatic potential A computer-calculated molecular representation that uses colors to show the charge distribu- map (EPM) tion in a molecule. In most cases, the EPM uses red to show electron-rich regions (most negative electrostatic potential) and blue or purple to show electron-poor regions (most positive electrostatic potential). The intermediate colors orange, yellow, and green show regions with intermediate electrostatic potentials. (p. 46) empirical formula The ratios of atoms in a compound. (p. 62) See also molecular formula. formal charges A method for keeping track of charges, showing what charge would be on an atom in a par- ticular Lewis structure. (p. 47) geometric isomers (IUPAC term: cis-trans isomers) Stereoisomers that differ in their cis-trans arrangement on a double bond or on a ring. (p. 82) Hund’s rule When there are two or more unfilled orbitals of the same energy (degenerate orbitals), the lowest-energy configuration places the electrons in different orbitals (with parallel spins) rather than pairing them in the same orbital. (p. 42) hybrid atomic orbital A directional orbital formed from a combination of s and p orbitals on the same atom. (pp. 65–68) sp hybrid orbitals give two orbitals with a bond angle of 180° (linear geometry). sp2 hybrid orbitals give three orbitals with bond angles of 120° (trigonal geometry). sp3 hybrid orbitals give four orbitals with bond angles of 109.5° (tetrahedral geometry). ionic bonding Bonding that occurs by the attraction of oppositely charged ions. Ionic bonding usually results in the formation of a large, three-dimensional crystal lattice. (p. 43) isomers Different compounds with the same molecular formula. (p. 81) Constitutional isomers (structural isomers) are connected differently; they differ in their bonding sequence. Cis-trans isomers (geometric isomers) are stereoisomers that differ in their cis-trans arrangement on a double bond or on a ring. Stereoisomers differ only in how their atoms are oriented in space. Stereochemistry is the study of the structure and chemistry of stereoisomers. isotopes Atoms with the same number of protons but different numbers of neutrons; atoms of the same element but with different atomic masses. (p. 39) LCAO (linear combination of atomic orbitals) Wave functions can add to each other to produce the wave functions of new orbitals. The number of new orbitals generated equals the original number of orbitals. (p. 65) Lewis structure A structural formula that shows all valence electrons, with the bonds symbolized by dashes ( ¬ ) or by pairs of dots, and nonbonding electrons symbolized by dots. (p. 44) line–angle formula (skeletal structure, stick figure) A shorthand structural formula with bonds represented by lines. Carbon atoms are implied wherever two lines meet or a line begins or bends. Atoms other than C and H are drawn in, but hydrogen atoms are not shown unless they are on an atom that is drawn. Each carbon atom is assumed to have enough hydrogens to give it four bonds. (p. 60) M01_WADE4255_10_GE_C01.indd 85 12/07/22 4:57 PM 86    CHAPTER 1 Structure and Bonding H O H H OH C H C C H H C C C H H H H cyclohex-2-en-1-ol Lewis structure of cyclohex-2-en-1-ol equivalent line–angle formula lone pair A pair of nonbonding electrons. (p. 44) molecular formula The number of atoms of each element in one molecule of a compound. The empirical formula simply gives the ratios of atoms of the different elements. For example, the molecular formula of glucose is C6H12O6. Its empirical formula is CH2O. Neither the molecular formula nor the empirical formula gives structural information. (p. 62) molecular orbital (MO) An orbital formed by the overlap of atomic orbitals on different atoms. MOs can be either bond- ing or antibonding, but only the bonding MOs are filled in most stable molecules. (pp. 65–68) A bonding molecular orbital places a large amount of electron density in the bonding region ­between the nuclei. The energy of an electron in a bonding MO is lower than it is in an atomic orbital. An antibonding molecular orbital places most of the electron density outside the bonding region. The energy of an electron in an antibonding MO is higher than it is in an atomic orbital. node In an orbital, a region of space with zero electron density. (p. 40) nodal plane In an orbital, a flat (planar) region of space with zero electron density. (p. 40) nonbonding electrons Valence electrons that are not used for bonding. A pair of nonbonding electrons is often called a lone pair. (p. 44) octet rule Atoms generally form bonding arrangements that give them filled shells of electrons (noble-gas con- figurations). For the second-row elements, this configuration has eight valence electrons. (p. 43) orbital An allowed energy state for an electron bound to a nucleus; the probability function that defines the distribution of electron density in space. The Pauli exclusion principle states that up to two electrons can occupy each orbital if their spins are paired. (p. 39) organic chemistry New definition: The chemistry of carbon compounds. Old definition: The study of compounds derived from living organisms and their natural products. (p. 37) pi bond (π bond) A bond formed by sideways overlap of two p orbitals. A pi bond has its electron density in two lobes, one above and one below the line joining the nuclei. (p. 68) H H half of π bond 2 sp sp2 H H H H C sp 2 sp2 C C C C C H H H H sp2 sp2 h alf d H H of π b o n σ bond framework π bond ethylene (viewed from above the plane) (viewed from alongside the plane) polar covalent bond A covalent bond in which electrons are shared unequally. A bond with equal sharing of electrons is called a nonpolar covalent bond. (p. 46) resonance hybrid A molecule or ion for which two or more valid Lewis structures can be drawn, differing only in the placement of the valence electrons. These Lewis structures are called resonance forms or resonance structures. Individual resonance forms do not exist, but we can estimate their rela- tive energies. The more important (lower-energy) structures are called major contributors, and the less important (higher-energy) structures are called minor contributors. When a charge is spread over two or more atoms by resonance, it is said to be delocalized and the molecule is said to be resonance stabilized. (pp. 50–51) sigma bond (s bond) A bond with most of its electron density centered along the line joining the nuclei; a cylindrically symmetrical bond. Single bonds are normally sigma bonds. (p. 65) M01_WADE4255_10_GE_C01.indd 86 12/07/22 4:57 PM Essential Problem-Solving Skills in Chapter 1    87 stereochemistry The study of the structure and chemistry of stereoisomers. (p. 82) stereoisomers Isomers that differ only in how their atoms are oriented in space. (p. 82) structural formulas A complete structural formula (such as a Lewis structure) shows all the atoms and bonds in the molecule. A condensed structural formula shows each central atom along with the atoms bonded to it. A line–angle formula (sometimes called a skeletal structure or stick figure) assumes that there is a carbon atom wherever two lines meet or a line begins or ends. See Section 1-10 for examples. (pp. 58–61) structural isomers (IUPAC term: constitutional isomers) Isomers whose atoms are connected differently; they differ in their bonding sequence. (p. 81) triple bond A bond containing six electrons between two nuclei. One pair of electrons forms a sigma bond, and the other two pairs form two pi bonds at right angles to each other. (p. 77) valence The number of bonds an atom usually forms. (p. 45) valence electrons Those electrons that are in the outermost shell. (p. 42) vitalism The belief that syntheses of organic compounds require the presence of a “vital force.” (p. 37) VSEPR theory (valence-shell electron-pair repulsion theory) Bonds and lone pairs around a central atom tend to be separated by the largest possible angles: about 180° for two, 120° for three, and 109.5° for four. (p. 69) wave function (c) The mathematical description of an orbital. The square of the wave function (c2) is proportional to the electron density. (p. 64) Essential Problem-Solving Skills in Chapter 1 Each skill is followed by problem numbers exemplifying that particular skill. 1 Write the electronic configurations for the elements hydrogen through neon. Explain how electronic configurations determine the electronegativities and bonding properties of these elements, and how the third row elements (e.g., Si, P, and S) differ from them. Problems 1-26, 27, 28, and 30 2 Draw all possible structures corresponding to a given molecular formula. Problems 1-34, 35, and 36 3 Draw and interpret Lewis, condensed, and line–angle structural formulas. Calculate Problems 1-30, 31, 32, 33, 34, 37, formal charges. 38, 41, 42, and 45 4 Predict patterns of covalent and ionic bonding involving C, H, O, N, and the ­halogens. Identify resonance-stabilized structures and compare the relative ­importance of their Problems 1-26, 29, 36, 40, 41, 42, resonance forms. 43, 44, 45, and 46 Problems 1-26, 41, 42, 43, 44, 45, Problem-Solving Strategy: Drawing and Evaluating Resonance Forms 46, 53, and 54 5 Calculate empirical and molecular formulas from elemental composition. Problems 1-47 and 48 6 Draw the structure of a single bond, a double bond, and a triple bond. Problems 1-52, 55, 57, and 60 7 Predict the hybridization and geometry of the atoms in a molecule, and draw a Problems 1-49, 51, 52, 53, 56, 57, three-dimensional representation of the molecule. and 60 8 Draw the resonance forms of a resonance hybrid, and predict its hybridization and geometry. Predict which resonance form is the major contributor. Problems 1-43, 44, 52, 53, and 54 9 Identify constitutional isomers and stereoisomers, and predict which compounds can exist as constitutional isomers and as cis-trans (geometric) isomers. Problems 1-49, 56, 58, and 59 M01_WADE4255_10_GE_C01.indd 87 12/07/22 4:57 PM 88    CHAPTER 1 Structure and Bonding Study Problems It’s easy to fool yourself into thinking you understand organic chemistry when you actually may not. As you read through this book, all the facts and ideas may make sense, yet you have not learned to combine and use those facts and ideas. An examination is a painful time to learn that you do not really understand the material. The best way to learn organic chemistry is to use it. You will certainly need to read and reread all the material in the chapter, but this level of understanding is just the beginning. Problems are provided so that you can work with the ideas, applying them to new compounds and new reactions that you have never seen before. By working problems, you force yourself to use the material and fill in the gaps in your understanding. You also increase your level of self-confidence and your ability to do well on exams. Several kinds of problems are included in each chapter. There are problems within the chapters, providing examples and drill for the material as it is covered. Work these problems as you read through the chapter to ensure your understanding as you go along. Answers to many of these in-chapter problems are found at the back of this book. Study Problems at the end of each chapter give you additional experience using the material, and they force you to think in depth about the ideas. Problems with red stars (*) are more difficult problems that require extra thought and perhaps some extension of the material presented in the chapter. Taking organic chemistry without working the problems is like skydiving without a parachute. Initially there is a breezy sense of freedom and daring. But then, there is the inevitable jolt that comes at the end for those who went unprepared. 1-26 (a)   Draw the resonance forms for CO2 (bonded O ¬ C ¬ O). (b)   Draw the resonance forms for ozone (bonded O ¬ O ¬ O). (c)   Carbon dioxide has one more resonance form than ozone. Explain why this structure is not possible for ozone. 1-27 Name the element that corresponds to each electronic configuration. (a)   1s22s22p63s1 (b)   1s22s22p6 (c)   1s22s22p1 (d)   1s22s22p63s2 1-28 There is a small portion of the periodic table that you must know to do organic chemistry. Construct this part from ­memory, using the following steps. (a)   From memory, make a list of the elements in the first two rows of the periodic table, together with their numbers of valence electrons. (b)   Use this list to construct the first two rows of the periodic table. (c)   Organic compounds often contain sulfur, phosphorus, chlorine, bromine, and iodine. Add these elements to your ­periodic table. 1-29 For each compound, state whether its bonding is covalent, ionic, or a mixture of covalent and ionic. (a)   KCl (b)   KOH (c)    CH3CH2Li (d)   CH3Cl     (e)    KOCH3 (f)    CH3CO2Na (g)   CCl4    1-30 (a) Both PF3 and PF5 are stable compounds. Draw Lewis structures for these two compounds. (b) NF3 is a known compound, but all attempts to synthesize NF5 have failed. Draw Lewis structures for NF3 and a ­hypothetical NF5, and explain why NF5 is an unlikely structure. 1-31 Draw a Lewis structure for each species. (a) N2H4 (b) N2H2 (c) (CH3)2NH2Cl (d) CH3CN (e) CH3CHO (f)   CH3S(O)CH3 (g) H2SO4 (h) CH3NCO (i)   CH3OSO2OCH3 (j)   CH3C(NH)CH3 (k)  (CH3)3CNO 1-32 Draw a Lewis structure for each compound. Include all nonbonding pairs of electrons. (a)   CH3COCH2CHCHCOOH (b)  NCCH2COCH2CHO (c)   CH2CHCH(OH)CH2CO2H (d)  CH2CHC(CH3)CHCOOCH3 1-33 Draw a line–angle formula for each compound in Problem 1-32. 1-34 Draw Lewis structures for (a)  two compounds of formula C4H10 (b)  two compounds of formula C2H6O (c)  two compounds of formula C2H7N (d)  three compounds of formula C2H7NO (e)  three compounds of formula C3H8O2 (f)   three compounds of formula C2H4O 1-35 Draw a complete structural formula and a condensed structural formula for (a)  three compounds of formula C3H8O (b)  five compounds of formula C3H6O 1-36 Some of the following molecular formulas correspond to stable compounds. When possible, draw a stable structure for each formula. CH2 CH3 CH4 CH5 C2H2 C2H3 C2H4 C2H5 C2H6 C2H7 C3H3 C3H4 C3H5 C3H6 C3H7 C3H8 C3H9 Propose a general rule for the numbers of hydrogen atoms in stable hydrocarbons. M01_WADE4255_10_GE_C01.indd 88 12/07/22 4:57 PM Study Problems    89 1-37 Draw complete Lewis structures, including lone pairs, for the following compounds. OH H O OH (a) (b) N (c) O (d) O phenol pyrrole dioxane O O (e) (f) OHC (g) (h) H2N C O CH3 H 1-38 Give the molecular formula of each compound shown in Problem 1-37. 1-39 1. From what you remember of electronegativities, show the direction of the dipole moments of the following bonds. 2. In each case, predict whether the dipole moment is relatively large (electronegativity difference >0.5) or small. (a)  C ¬ Cl (b)  C ¬ H (c)  C ¬ Li (d)  C ¬ N (e)  C ¬ O (f)  C ¬ B (g)  C ¬ Mg (h)  N ¬ H (i)  O ¬ H (j)  C ¬ Br 1-40 For each of the following structures, 1. Draw a Lewis structure; fill in any nonbonding electrons. 2. Calculate the formal charge on each atom other than hydrogen. (a)        CH3NO (b)    (CH3)3NO (c)    [N3]- (nitrosomethane) (trimethylamine oxide) (azide ion) (d)  [(CH3)3O]+ (e)  CH3NC (f)  (CH3)4NBr 1-41 Determine whether the following pairs of structures are actually different compounds or simply resonance forms of the same compounds. + O O O O (a) and     (b)   and + H – – + O + O O OH (c)   and       (d)   and _ _ + O O H3CH2C NH2 H3CH2C NH2 (e)   and          (f)   C and C NH2 NH2 + + CH3 CH (g) and (h) CH3CH2CH    CHCH3 and (CH3)2C CHCH3 C + C H3CH2C CH H3C CH2CH3 _ O H OH O O _ (i) CI C C H and CI C C H    (j) H C C H and H C C H H H H H H − − O O O + O O OH + (k) H2N C O and H2N C O     (l) and H H M01_WADE4255_10_GE_C01.indd 89 12/07/22 4:57 PM 90    CHAPTER 1 Structure and Bonding 1-42Draw the important resonance forms to show the delocalization of charges in the following ions. In each case, indicate the major resonance form(s). O O (c) + − − CH2 (a) CH3 C CH2 (b) H C CH CH CH2      (d) (e) (f) O− + NH + (i) + (g) + (h) CH2 CH CH CH CH CH3 − O O + + (j) CH3 CH CH CH CH CH2 CH2 (k) CH2 CH CH O CH CH2 1-43 In the following sets of resonance forms, label the major and minor contributors and state which structures would be of equal energy. Add any missing important resonance forms. (a) − − CH3 CH C N CH3 CH C N (b) O− O− + CH3 C CH CH CH3 CH3 C CH CH CH3 + (c) O O O− O − CH3 C CH C CH3 CH3 C CH C CH3 − − (d) CH3 CH CH CH NO2 CH3 CH CH CH NO2 (e) NH2 NH2 + CH3 CH2 C NH2 CH3 CH2 C NH2 + 1-44 For each of these ions, draw the important resonance forms and predict which resonance form is likely to be the major contributor. (a) + (b) O (c) + O CH2 CH2 HN − CH2 + (e) (f) (d) – N – O 1-45 For each pair of ions, determine which ion is more stable. Use resonance forms to explain your answers. + + (a) CH3 CH CH3 or CH3 CH OCH3 (b) CH3 N CH3 CH3 CH CH3 or CH3 C CH3 CH3 C CH3           + + + + − − (c) CH2 CH3 or CH2           (d) CH CH CH CH2 CH2 CH2 CH3 or CH2 C N (e) + + (f) – O O CH2 CH2 or or –                M01_WADE4255_10_GE_C01.indd 90 12/07/22 4:57 PM Study Problems    91 1-46 Use resonance structures to identify the areas of high and low electron density in the following compounds: O O NH O (a) CH3 C (b) CH3 H    (b) C (c) CH3 NH2    (c) C H    (d) CH3 (d) C OCH3 (e) O (f) NH (g) CN (e) CH3O C NH2 O NH2 (h) (h) O (i) H O (j) O N O C H CH3O 1-47 Compound X, isolated from lanolin (sheep’s wool fat), has the pungent aroma of dirty sweatsocks. A careful analysis showed that compound X contains 62.0% carbon and 10.4% hydrogen. No nitrogen or halogen was found. (a) Compute an empirical formula for compound X. (b) A molecular weight determination showed that compound X has a molecular weight of approximately 117. Find the molecular formula of compound X. (c) Many possible structures have this molecular formula. Draw complete structural formulas for four of them. *1-48 In 1934, Edward A. Doisy of Washington University extracted 3000 lb of hog ovaries to isolate a few milligrams of pure estradiol, a potent female hormone. Doisy burned 5.00 mg of this precious sample in oxygen and found that 14.54 mg of CO2 and 3.97 mg of H2O were generated. (a) Determine the empirical formula of estradiol. (b) The molecular weight of estradiol was later determined to be 272. Determine the molecular formula of estradiol. 1-49 If the carbon atom in CH2ClBr were flat, there would be two stereoisomers. The carbon atom in CH2ClBr is actually ­tetrahedral. Make a model of this compound, and determine whether there are any stereoisomers of CH2ClBr. H H H C Cl Br C Cl Br H 1-50 Cyclopropene (C3H4, a three-membered ring) is more reactive than most other cycloalkenes. (a) Draw a Lewis structure for cyclopropene. (b) Compare the bond angles of the carbon atoms in cyclopropene with those in an acyclic (noncyclic) alkene. (c) Suggest why cyclopropene is so reactive. 1-51 For each of the following compounds, 1. Give the hybridization and approximate bond angles around each atom except hydrogen. 2. Draw a three-dimensional diagram, including any lone pairs of electrons. (a) H3O+ (b) -OH (c) CH2CHCN (d) (CH3)3N (e) [CH3NH3]+ (f) CH3COOH (g) CH3CHNH (h) CH3OH (i) CH2O 1-52 For each of the following compounds and ions, 1. Draw a Lewis structure. 2. Show the kinds of orbitals that overlap to form each bond. 3. Give approximate bond angles around each atom except hydrogen. (a) [NH2]- (b) [CH2OH]+ (c) CH2 “ N ¬ CH3 (d) CH3 ¬ CH “ CH2 (e) HC ‚ C ¬ CHO (f) H2N ¬ CH2 ¬ CN O (h) O (i) (g) CH3 C OH (h) O (i) + O 1-53 In most amines, the nitrogen atom is sp3 hybridized, with a pyramidal structure and bond angles close to 109°. In urea, both nitrogen atoms are found to be planar, with bond angles close to 120°. Explain this surprising finding. (Hint: Consider resonance forms and the overlap needed in them.) O H2N C NH2 urea M01_WADE4255_10_GE_C01.indd 91 12/07/22 4:57 PM 92    CHAPTER 1 Structure and Bonding 1-54 Predict the hybridization and geometry of the carbon and nitrogen atoms in the following molecules and ions. (Hint: Resonance.) (a) O − + - (a) CH3 C CH2 (b) H2N ¬ CH “ CH ¬ CH2 (c) CH2 ¬ C ‚ N (d) (e) (e) + (f) + O N – N N H (g) O (h) O (i) O NH N N H H 1-55 Draw orbital pictures of the pi bonding in the following compounds: (a) CH3COCH3 (b) HCN (c) CH2 “ CH ¬ CHCHCN (d) CH3C ‚ CCHO (e) CH3CH “ C “ CHCH3 (f) CH3 ¬ CH “ N ¬ CH “ C “ O 1-56 (a) Draw the structure of cis- CH3 ¬ CH “ CH ¬ CH2CH3 showing the pi bond with its proper geometry. (b) Circle the six coplanar atoms in this compound. (c) Draw the trans isomer, and circle the coplanar atoms. Are there still six? (d) Circle the coplanar atoms in the following structure: CH3 1-57 In pent-2-yne (CH3CCCH2CH3), there are four atoms in a straight line. Use dashed lines and wedges to draw a three-dimensional representation of this molecule, and circle the four atoms that are in a straight line. 1-58 Which of the following compounds show cis-trans isomerism? Draw the cis and trans isomers of the ones that do. (a) CH3CH2CH “ CHCH3 (b) HC ‚ CH (c) (CH3)2C “ CHCH3 (d) (e) CH3CH2CH CCH3 (f) CH3CH2CH N cyclopropene, CH2CH3 H 1-59 Give the relationships between the following pairs of structures. The possible relationships are as follows: same ­compound, cis-trans isomers, constitutional (structural) isomers, and not isomers (different molecular formula). (a) CH3CHCHCH3 and (CH3)3CH             (b) CH2CClCH3 and CH2CHCH2Cl (c) H CH2CH3 H CH3 (d) H H H CH3 and and H3CH2C CH3 H3CH2C CH2CH3 H3CH2C CH3 H3CH2C H (e) H CH3 H3C CH3 (f) H CH2CH3 H3C CH3 and and H3CH2C CH3 H3CH2C H      H3C CH3 H3CH2C H and (g) (h) and                 O O *1-60 Dimethyl sulfoxide (DMSO) has been used as an anti-inflammatory rub for race horses. DMSO and acetone appear to have similar structures, but the C “ O carbon atom in acetone is planar, while the S “ O sulfur atom in DMSO is ­pyramidal. Draw Lewis structures for DMSO and acetone, predict the hybridizations, and explain these observations. O O CH3 S CH3 CH3 C CH3 DMSO acetone M01_WADE4255_10_GE_C01.indd 92 12/07/22 4:57 PM

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