Baixe Carey - Organic Chemistry - sgchapt01 e outras Manuais, Projetos, Pesquisas em PDF para Química, somente na Docsity! G CHAPTER 1 CHEMICAL BONDING SOLUTIONS TO TEXT PROBLEMS 1.1 The element carbon has atomic number 6, and so it has a total of six electrons. Two of these elec- trons are in the 1s level. The four electrons in the 2s and 2p levels (the valence shell) are the valence electrons. Carbon has four valence electrons. 1.2 Electron configurations of elements are derived by applying the following principles: (a) The number of electrons in a neutral atom is equal to its atomic number Z. (b) The maximum number of electrons in any orbital is 2. (c) | Electrons are added to orbitals in order of increasing energy, filling the 1s orbital before any electrons occupy the 2s level. The 2s orbital is filled before any of the 2p orbitals, and the 3s orbital is filled before any of the 3p orbitals. (d) All the 2p orbitals (2p,, 2p,, 2p.) are of equal energy, and each is singly occupied before any is doubly occupied. The same holds for the 3p orbitals. With this as background, the electron configuration of the third-row elements is derived as follows [2pº = 2p /2p,2p?]: Na (Z=11) 1522522pS3s! Mg(Z = 12) 1522522pS3s? AL(Z=13) 152522p93523p Si (Z= 14) 15252p9353p '3p, P(Z=15) 18252p%3823p 3p /3p! S (Z=16) 15252p353p23p 3p.! CHZ=17 0 18252p33p3p Bp.! Ar(Z=18) 0 15252p353p23p 3p? Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website 2 CHEMICAL BONDING 1.3 The electron configurations of the designated ions are: Number of Electrons Electron Configuration Ton Z in Ion oflon (b) Het 2 1 Is! (co) Ho 1 2 Is? (d) OT 8 9 1522822p22p 2p.! (e) FT 9 10 1522522pº (1) Cat 20 18 1522522p935º3pº Those with a noble gas configuration are H”, F”, and Ca”. 1.4 A positively charged ion is formed when an electron is removed from a neutral atom. The equation representing the ionization of carbon and the electron configurations of the neutral atom and the ion is: c —— Ct +e 192522p 2p,! 19º252p! A negatively charged carbon is formed when an electron is added to a carbon atom. The addi- tional electron enters the 2p. orbital. c + — [om 18º292p, 2p, 152252p/p 2p.! Neither C* nor C” has a noble gas electron configuration. 1.5 Hydrogen has one valence electron, and fluorine has seven. The covalent bond in hydrogen fluoride arises by sharing the single electron of hydrogen with the unpaired electron of fluorine. Combine H- and to give the Lewis structure for hydrogen fluoride H' 1.6 We are told that C,H, has a carbon-carbon bond. to write the Thus, we combinetwo -Ç- andsix H- Lewis structure of ethane There are a total of 14 valence electrons distributed as shown. Each carbon is surrounded by eight electrons. 1.7 (b) Eachcarbon contributes four valence electrons, and each fluorine contributes seven. Thus, C,F, has 36 valence electrons. The octet rule is satisfied for carbon only if the two carbons are at- tached by a double bond and there are two fluorines on each carbon. The pattern of connections shown (below left) accounts for 12 electrons. The remaining 24 electrons are divided equally (six each) among the four fluorines. The complete Lewis structure is shown at right below. (c) | Since the problem states that the atoms in C;H;N are connected in the order CCCN and all hy- drogens are bonded to carbon, the order of attachments can only be as shown (below left) so as to have four bonds to each carbon. Three carbons contribute 12 valence electrons, three hy- drogens contribute 3, and nitrogen contributes 5, for a total of 20 valence electrons. The nine Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website CHEMICAL BONDING 5 (d) This compound has the same molecular formula as the compound in part (c), but a different structure. It, too, has 26 valence electrons, and again only chlorine has unshared pairs. H H H—C—C— El: H Cl (e) The constitution of CH;NHCH,CH, is shown (below left). There are 26 valence electrons, and 24 of them are accounted for by the covalent bonds in the structural formula. The remaining two electrons complete the octet of nitrogen as an unshared pair (below right). Lodi Lodi E RN E E H H H H H H (f) Oxygen has two unshared pairs in (CH;)jCHCH=0. 1.13 (b) This compound has a four-carbon chain to which are appended two other carbons. Ge H is equivalent to CH—C—C—CH, Vich may be | written as cH, (CH;),CHCH(CH;), (c) | The carbon skeleton is the same as that of the compound in part (b), but one of the terminal carbons bears an OH group in place of one of its hydrogens. t HO. HO-—C—H . CH,0H is equivalent to ' which may be o ca CH, rewritten as CH,CHCH(CH;), | CH, (d) The compound is a six-membered ring that bears a —C(CH,), substituent. Lim H—C—CT is equivalent HZ NA which may be is equivalent to A KR Hs rewritten as C(CH;); H — C—cH C—Cc 3 o Ni H 4 HCH, 1.14 The problem specifies that nitrogen and both oxygens of carbamic acid are bonded to carbon and one of the carbon-oxygen bonds is a double bond. Since a neutral carbon is associated with four Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website 6 CHEMICAL BONDING bonds, a neutral nitrogen three (plus one unshared electron pair), and a neutral oxygen two (plus two unshared electron pairs), this gives the Lewis structure shown. Carbamic acid 1.15 (b) There are three constitutional isomers of C;H,O: a CcH,CH;CH,OH | CH;CHCH, — CH,CH,ÓCH, (c) Four isomers of C,H,,O have —OH groups: To CH,CH,CH,CH,0H CHÇHCHCH; CH,CHCH,ÕH — CH,COH :0H cH, cH, Three isomers have C—O—C units: CH,;ÓCH;CH,CH, — CH;CH,ÓCH,CH, — CH,ÓCHCH, cH, 1.16 (b) Move electrons from the negatively charged oxygen, as shown by the curved arrows. Equivalent to original structure The resonance interaction shown for bicarbonate ion is more important than an alternative one involving delocalization of lone-pair electrons in the OH group. SK Or / So Not equivalent to original structure; not as stable because of charge separation —C + O—H (c) All three oxy gens are equivalent in carbonate ion. Either negatively charged oxygen can serve as the donor atom. Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website CHEMICAL BONDING 7 (d) Resonance in borate ion is exactly analogous to that in carbonate. and 1.17 | There are four B—H bonds in BH, . The four electron pairs surround boron in a tetrahedral orien- tation. The H—B —H angles are 109.5º. 118 (b) (c) (d) 119 (b) (c) (d) Back| Forward Main Menul Nitrogen in ammonium ion is surrounded by 8 electrons in four covalent bonds. These four bonds are directed toward the corners of a tetrahedron. H Haé NH H Each HNH angle is 109.5º. Double bonds are treated as a single unit when deducing the shape of a molecule using the VSEPR model. Thus azide ion is linear. e tos CN=N=Nº The NNN angle is 180º. Since the double bond in carbonate ion is treated as if it were a single unit, the three sets of electrons are arranged in a trigonal planar arrangement around carbon. 30: CN The OCO angle is 120º. Water is a bent molecule, and so the individual O —H bond dipole moments do not cancel. Water has a dipole moment. DO HO OH Individual OH bond moments in water , Bo H H Direction of net dipole moment Methane, CH,, is perfectly tetrahedral, and so the individual (small) C—H bond dipole moments cancel. Methane has no dipole moment. Methyl chloride has a dipole moment. H “sc n7 + Direction of molecular dipole moment Directions of bond dipole moments in CH,CI TOC| Study Guide TOC] Student OLC| | MHHE Website 10 Back| (cd) 1.26 (a) (b) (c) (d) CHEMICAL BONDING elements, and thus have a formal charge of —1 in the tetrahedral anions BF, and AIH,” respectively. H Ss . 4H Nat FRA Lit H—AIZ N, N, F H Sodium tetrafluoroborate Lithium aluminum hydride Both of the tetrahedral anions have 32 valence electrons. Sulfur contributes 6 valence elec- trons and phosphorus 5 to the anions. Each oxygen contributes 6 electrons. The double negative charge in sulfate contributes 2 more, and the triple negative charge in phosphate contributes 3 more. Potassium sulfate Sodium phosphate The formal charge on each oxygen in both ionsis —1. The formal charge on sulfur in sulfate is +2; the charge on phosphorus is +1. The net charge of sulfate ion is —2; the net charge of phosphate ion is —3. Each hydrogen has a formal charge of O, as is always the case when hydrogen is covalently bonded to one substituent. Oxygen has an electron count of 5. H—O—H | Electron count of oxygen = 2 +46) = 5 H Unshared A Ne Covalently pair bonded electrons A neutral oxygen atom has 6 valence electrons; therefore, oxygen in this species has a formal charge of +1. The species as a whole has a unit positive charge. Itis the hydronium ion, H,O*. The electron count of carbon is 5; there are 2 electrons in an unshared pair, and 3 electrons are counted as carbon's share of the three covalent bonds to hydrogen. 4 Two electrons “owned” by carbon. [= one of the electrons in each C—H bond “belongs to carbon. An electron count of 5 is one more than that for a neutral carbon atom. The formal charge on carbon is —1, as is the net charge on this species. This species has 1 less electron than that of part (b). None of the atoms bears a formal charge. The species is neutral. sG—H Electron count of carbon = 1 + (6) = 4 H Unshared 4 Ne Electrons shared electron in covalent bonds The formal charge of carbon in this species is +1. Its only electrons are those in its three covalent bonds to hydrogen, and so its electron count is 3. This corresponds to 1 less electron than in a neutral carbon atom, giving it a unit positive charge. Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website Back| CHEMICAL BONDING un (e) In this species the electron count of carbon is 4, or, exactly as in part (c), that of a neutral carbon atom. Its formal charge is 0, and the species is neutral. Two unshared electrons contribute 2 4 to the electron countof carbon. E NA Half of the 4 electrons in the two covalent bonds contribute 2 to the electron count of carbon. 1.27 Oxygen is surounded by a complete octet of electrons in each structure but has a different “electron count” in each one because the proportion of shared to unshared pairs is different. (a) CH, (b) CH;ÓCH, (c) CHÉiCH, CH, Electron count Electron count Electron count =6+)=7T =4+H4)=6; formal charge = —1 formal charge = 0 1.28 (a) Each carbon has 4 valence electrons, each hydrogen 1, and chlorine has 7. Hydrogen and chlo- rine each can form only one bond, and so the only stable structure must have a carbon-carbon bond. Of the 20 valence electrons, 14 are present in the seven covalent bonds and 6 reside in the three unshared electron pairs of chlorine. Ut or a H H (b) Asin part (a) the single chlorine as well as all of the hydrogens must be connected to carbon. There are 18 valence electrons in C,H;CI, and the framework of five single bonds accounts for only 10 electrons. Six of the remaining 8 are used to complete the octet of chlorine as three unshared pairs, and the last 2 are used to form a carbon-carbon double bond. H NR or E H :ÇE (c) | All of the atoms except carbon (H, Br, CI, and F) are monovalent; therefore, they can only be bonded to carbon. The problem states that all three fluorines are bonded to the same carbon, and so one of the carbons is present as a CF, group. The other carbon must be present as a CHBrCI group. Comect these groups together to give the structure of halothane. H | — C—Cl (Unshared electron pairs omitted for clarity) H or F— T—0O— r (d) | Asin part (c) all of the atoms except carbon are monovalent. Since each carbon bears one chlorine, two CICF, groups must be bonded together. Li or E A (Unshared electron pairs omitted for clarity) F F Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website 12 CHEMICAL BONDING 1.29 Place hydrogens on the given atoms so that carbon has four bonds, nitrogen three, and oxygen two. Place unshared electron pairs on nitrogen and oxygen so that nitrogen has an electron count of 5 and oxygen has an electron count of 6. These electron counts satisfy the octet rule when nitrogen has three bonds and oxygen two. 130 (a) (b) (c) (d) (e) 00) (8) dy) ( t (a) (e) HoO-=N=H H H b) H-C=N-0—H (d) Oca H HH Species A, B, and C have the same molecular formula, the same atomic positions, and the same number of electrons. They differ only in the arrangement of their electrons. They are therefore resonance forms of a single compound. + H Net N=N: =" H H Na wW A B c Structure A has a formal charge of —1 on carbon. Structure C has a formal charge of +1 on carbon. Structures A and B have formal charges of +1 on the internal nitrogen. Structures B and C have a formal charge of —1 on the terminal nitrogen. All resonance forms of a particular species must have the same net charge. In this case, the net charge on A, B, and C is 0. Both A and B have the same number of covalent bonds, but the negative charge is on a more electronegative atom in B (nitrogen) than it is in A (carbon). Structure B is more stable. Structure B is more stable than structure C. Structure B has one more covalent bond, all of its atoms have octets of electrons, and it has a lesser degree of charge separation than C. The carbon in structure C does not have an octet of electrons. The CNN unit is linear in A and B, but bent in C according to VSEPR. This is an example of how VSEPR can fail when comparing resonance structures. 1.31 The structures given and their calculated formal charges are: (a) (b) (c) (d) (e) 00) (8) (h) ( Structure D contains a positively charged carbon. Structures A and B contain a positively charged nitrogen. None of the structures contain a positively charged oxygen. Structure A contains a negatively charged carbon. None of the structures contain a negatively charged nitrogen. Structures B and D contain a negatively charged oxygen. All the structures are electrically neutral. Structure B is the most stable. All the atoms except hydrogen have octets of electrons, and the negative charge resides on the most electronegative element (oxygen). Structure C is the least stable. Nitrogen has five bonds (10 electrons), which violates the octet rule. Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website Back| Forward CHEMICAL BONDING 15 Filling in the appropriate hydrogens gives the correct structures: CH.CH,CH,;CH, and — CH;CHCH, cH, Continue with the remaining parts of the problem using the general approach outlined for part (a). (b) CH ii CH,CH;CH,CH,CH, — CH,CHCH,CH, CH;—C—CH, cH, cH, 3 () CGHCL CH,CHCL, and — CICH,CH,CI (d) CHBr cH CH,CH,CH,CH.Br CHGHCH CH; CH,CHCH;Br | CH—C—Br Br cH, cH, (e) CHN Hs CHCHCHNH, — CHCHNHCH, CH—N CHAÇHNH, cH, cu, Note that when the three carbons and the nitrogen are arranged in a ring, the molecular formula based on such a structure is C;H,N, not C;H,N as required. H,C—CH, H,C—NH (not an isomer) 1.37 (a) All three carbons must be bonded together, and each one has four bonds; therefore, the mo- lecular formula C,H, uniquely corresponds to: Li A (CH.CH,CH,) H HH (b) With two fewer hydrogen atoms than the preceding compound, either C,;H, must contain a carbon-carbon double bond or its carbons must be arranged in a ring: thus the following structures are constitutional isomers: H,C=CHCH, and HC—CH, SZ CH, MainMenu) TOC] StudyGuideTOC| Student OLC| | MHHE Website 16 CHEMICAL BONDING (c) | The molecular formula C,H, is satisfied by the structures H,C=C=CH, HC=CCH, HC=CH - - NZ cH, 1.38 (a) The only atomic arrangements of C,H,O that contain only single bonds must have a ring as part of their structure. H,C—CHOH | H;C—CHCH, — HC—CH, SZ S4 [| CH, o O—CH, (b) Structures corresponding to C;H,O are possible in noncyclic compounds if they contain a carbon-carbon or carbon-oxy gen double bond. i í CH,CH,CH cH.CCH, — CH.CH=CHOH CH,OCH=CH, cHç =cH, H,C=CHCH,0H OH 1.39 The direction of a bond dipole is governed by the electronegativity of the atoms it connects. In each of the parts to this problem, the more electronegative atom is partially negative and the less elec- tronegative atom is partially positive. Electronegativities of the elements are given in Table 1.2 of the text. (a) -Chlorine is more electronegative (d) | Oxygen is more electronegative than than hydrogen. hydrogen. +— H—cl Sa H H (b) -Chlorine is more electronegative (e) | Oxygen is more electronegative than than iodine. either hydrogen or chlorine. +— o 11 Sã H cal (c) Todine is more electronegative than hydrogen. +—— HI 1.40 The direction of a bond dipole is governed by the electronegativity of the atoms involved. Among the halogens the order of electronegativity is F > CI > Br > I Fluorine therefore attracts electrons away from chlorine in FCI, and chlorine attracts electrons away from iodine in ICI. <—+ F—CI I—CI u=09D — u=07D Chlorine is the positive end of the dipole in FCI and the negative end in ICI. 1.41 (a) Sodium chloride is ionic; it has a unit positive charge and a unit negative charge separated from each other. Hydrogen chloride has a polarized bond but is a covalent compound. Sodium chloride has a larger dipole moment. The measured values are as shown. Na* CI ismorepolarthan H—CI u94D uLID Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website CHEMICAL BONDING 7 (b) | Fluorine is more electronegative than chlorine, and so its bond to hydrogen is more polar, as the measured dipole moments indicate. +— +— H—F is more polar than H—CI n1ID nlID (c) Boron trifluoride is planar. Its individual B—F bond dipoles cancel. It has no dipole moment. +— it HP lar th b — Is more ar than P PR F NF E 1ID “OD (d) | A carbon-chlorine bond is strongly polar; carbon-hydrogen and carbon-carbon bonds are only weakly polar. u21D nO1D (e) A carbon-fluorine bond in CCIF opposes the polarizing effect of the chlorines. The carbon-hydrogen bond in CHCI, reinforces it. CHCI, therefore has a larger dipole moment. is more polar than Ç Pa n1OD n05D (f) Oxygen is more electronegative than nitrogen; its bonds to carbon and hydrogen are more polar than the corresponding bonds formed by nitrogen. 270% H,C H is more polar than n1ID (g) The Lewis structure for CH;NO, has a formal charge of +1 on nitrogen, making it more electron-attracting than the uncharged nitrogen of CH;NH,. ++ 42 Ds A H;C>N, is more polar than H;C>N, No N, o H u31D u13D 1.42 (a) There are four electron pairs around carbon in :CH,; they are arranged in a tetrahedral fashion. The atoms of this species are in a trigonal pyramidal arrangement. Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website 20 CHEMICAL BONDING 1.45 (a) Carbonis sp”-hybridized when it is directly bonded to four other atoms. Compounds (a) and (d) in Problem 1.43 are the only ones in which all of the carbons are sp”-hybridized. SA LA (a) (dy (b) Carbonis sp?-hybridized when it is directly bonded to three other atoms. Compounds (7), (g), and (j) in Problem 1.43 have only sp-hybridized carbons. H HH Ho 4 Q H H H H : . H Br N : H H HH HW o N Br H H H H o H O (9) D None of the compounds in Problem 1.43 contain an sp-hybridized carbon. 1.46 The problem specifies that the second-row element is sp*-hybridized in each of the compounds. Any unshared electron pairs therefore occupy sp*-hybridized oribitals, and bonded pairs are located in o orbitals. (a) | Ammonia (e) Borohydride anion H Hai aa Four o bonds formed V s, sp Hybrid by sp-s overlap H orbital Three o bonds formed by spi-s overlap (b) Water (1) -Amide anion 5 ; a “uso. mL Two sp” Ed SN | hybrid De orbitals Two sp? hybrid orbitals Two o bonds formed by spi-soverlap Two o bonds formed by sp*s overlap (c) | Hydrogen fluoride (g) Methyl anion VTN Three H HE spº hybrid ERÊ De orbitals ca es CD sp: Hybrid orbital One o bond formed by H sp-s overlap Three o bonds formed by sp'-s overlap (d) | Ammonium ion H N— H Four o bonds formed bysp3-s overlap Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website CHEMICAL BONDING 21 147 (a) The electron configuration of N is 15/252p,'2p,'2p.!. If the half-filled 2p,, 2p,, and 2p, orbitals are involved in bonding to H, then the unshared pair would correspond to the two electrons in the 2s orbital. (b) The three p orbitals 2p,, 2p,, and 2p, have their axes at right angles to one another. The H—N—H angles would therefore be 90º. 1.48 A bonding interaction exists when two orbitals overlap “in phase” with each other, that is, when the algebraic signs of their wave functions are the same in the region of overlap. The following orbital is a bonding orbital. It involves overlap of an s orbital with the lobe of a p orbital of the same sign. Eis) (0) (onding) On the other hand, the overlap of an s orbital with the lobe of a p orbital of opposite sign is antibonding. ef) (b) (antibonding) Overlap in the manner shown next is nonbonding. Both the positive lobe and the negative lobe of the p orbital overlap with the spherically symmetrical s orbital. The bonding overlap between the s orbital and one lobe of the p orbital is exactly canceled by an antibonding interaction between the s orbital and the lobe of opposite sign. (a) (nonbonding) 1.49-1.55 Solutions to molecular modeling exercises are not provided in this Study Guide and Solutions Man- ual. You should use Learning By Modeling for these exercises. SELF-TEST PART A A-1. Write the electronic configuration for each of the following: (a) - Phosphorus (b) -Sulfide ion in Na,S A-2. Determine the formal charge of each atom and the net charge for eachof the following species: :6: l HC=NH, A-3. Write a second Lewis structure that satisfies the octet rule for each of the species in Prob- lem A-2, and determine the formal charge of each atom. Which of the Lewis structures for each species in this and Problem A-2 is more stable? Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website 22 AS, A-6. A-7. CHEMICAL BONDING Write a correct Lewis structure for each of the following. Be sure to show explicitly any unshared pairs of electrons. (a) -Methylamine, CH;NH, (b) -Acetaldehyde, C,H,O (the atomic order is CCO; all the hydrogens are connected to carbon.) What is the molecular formula of each of the structures shown? Clearly draw any unshared electron pairs that are present. o (a) CA. (9) SO oH (b) (a) 2 N >Br Which compound in Problem A-5 has (a) Only sp*-hybridized carbons (b) Only sp>-hybridized carbons (c) - Asingle sp-hybridized carbon atom Account for the fact that all three sulfur-oxy gen bonds in SO, are the same by drawing the appropriate Lewis structure(s). The cyanate ion contains 16 valence electrons, and its three atoms are arranged in the order OCN. Write the most stable Lewis structure for this species, and assign a formal charge to each atom. What is the net charge of the ion? Using the VSEPR method, (a) -Describe the geometry at each carbon atom and the oxygen atom in the following molecule: CH,OCH=CHCH,. (b) Deduce the shape of NCl,, and draw a three-dimensional representation of the molecule. Is NCI, polar? . Assign the shape of each of the following as either linear or bent. (a) CO, (b) NO (0) NOW « Consider structures A, B, C, and D: (a) Which structure (or structures) contains a positively charged carbon? (b) Which structure (or structures) contains a positively charged nitrogen? (ce) Which structure (or structures) contains a positively charged oxygen? (d) Which structure (or structures) contains a negatively charged carbon? (e) Which structure (or structures) contains a negatively charged nitrogen? (/). Which structure (or structures) contains a negatively charged oxygen? (g) Which structure is the most stable? (h) Which structure is the least stable? Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website