8 CHAPTER 1 Structure and Properties of Organic Compounds and the species is an unchanged molecule: GROUP UNSHARED sHared +1/2 FORMAL NUMBER ELECTRONS ELECTRONS CHARGE Catoms 4 0 A 0 N atoms 0 4 +1 H atoms 1 0 0 Net charge on species +1 0: (c) :0一S2+0： 可 GROUP T UNSHARED +1/2 SHARED FORMAL NUMBER ELECTRONS ELECTRONS CHARGE S atoms 6 0 4 +2 each O atom 6 6 1 -1 Net charge is+2+4(-1)= These examples reveal that formal charges appear on an atom that does not have its usual covalence and does not have more than an octet of valence electrons.Formal charges always occur in a molecule or ion that can be conceived to be formed as a result of coordinate covalent bonding Problem 1.14 Show how(a)H,NBF,and(b)CH,NH can be formed from coordinate covalent bonding.Indi cate the donor and acceptor,and show the formal charges. donor acceptor (@HN:+BF3→H,N-BF (b)CH2NH2+H+ [CH3NH3] SUPPLEMENTARY PROBLEMS Problem 1.15 Why are the compounds of carbon covalent rather than ionic? With four valence electrons,it would take too much energy for C to give up or accept four electrons.There fore,carbon shares electrons and forms covalent bonds. Problem 1.16 Classify the following as(i)branched chain,(ii)unbranched chain,(iii)cyclic,(iv)multiple bonded,or(v)heterocyclic CH2 CH3 CH3 CH2 CH3一C一CH H2C-CH2一CH2 HC. HC NH HC=CH CH3 HC一CH (a) (b) (c) (e (a)(iii)and (iv);(b)(i);(c)(ii);(d)(v);(e)(iv)and (ii)
and the species is an unchanged molecule: 8 CHAPTER 1 Structure and Properties of Organic Compounds These examples reveal that formal charges appear on an atom that does not have its usual covalence and does not have more than an octet of valence electrons. Formal charges always occur in a molecule or ion that can be conceived to be formed as a result of coordinate covalent bonding. Problem 1.14 Show how (a) H3 NBF3 and (b) CH3 NH 3 can be formed from coordinate covalent bonding. Indicate the donor and acceptor, and show the formal charges. SUPPLEMENTARY PROBLEMS Problem 1.15 Why are the compounds of carbon covalent rather than ionic? With four valence electrons, it would take too much energy for C to give up or accept four electrons. Therefore, carbon shares electrons and forms covalent bonds. Problem 1.16 Classify the following as (i) branched chain, (ii) unbranched chain, (iii) cyclic, (iv) multiple bonded, or (v) heterocyclic: (a) (b) (c) (a) (iii) and (iv); (b) (i); (c) (ii); (d) (v); (e) (iv) and (ii)
TABLE 1.3 Some Common Functional Groups EXAMPLE GROUTONAL SGS出 NANERAL FORMULA IUPAC NAME COMMON NAME CHAPTER 1 None C.H2m42 Alkane CHCH Ethane Ethane C=c CxHzm Alkene H.C-CHz Ethene Ethylene -C=C- CaHzn-2 Alkyne HCCH Ethyne Acetylene -C1 R-CI Chloride CHCH.CI Chloroethane Ethyl chloride -Br R-Br Bromide CHaBr Bromomethane Methyl bromide -OH R-OH Alcohol CH,CH2OH Ethanol Ethyl alcohol 0 R-O-R Ether CHCHzOCH2CHa Ethoxyethane Diethyl ether -NHz RNH2 Amine2 CH CH CH NH, 1-Aminopropane Propylamine Structure and Properties of Organic Compounds -NRiX- RN+X- CH,(CH:)N(CH3)Cl salt chloride chloride 0 Aldehyde CH:CH2CH=O Propanal Propionaldehyde H H 2-Butanone Methyl ethyl Ketone CH:CH2C=0 ketone 0 Ethanoic acid Acetic acid -C-OH R-C-OH Carboxylic acid CH-C-OH
TABLE 1.3 Some Common Functional Groups CHAPTER 1 Structure and Properties of Organic Compounds 9
TABLE 1.3 (continued) EXAMPLE 出 &证RL FORMULA IUPAC NAME COMMON NAME Ester CH:-C -0CH5 Ethyl ethanoate Ethyl acetate Amide CH-C NHz Ethanamide Acctamide R-8。 Acid chloride Ethanoyl chloride Acetyl chloride 8.8 9 Acid anhydride Ethanoic anhvdride Acetic anhydride -C-N R-C=N Nitrile CH:C-N Ethanenitrile Acetonitrile CHAPTER 1 -NO2 R-NO2 Nitro CH3-NO; Nitromethane -SH R-SH Thiol CH3-SH Methyl mercaptan -S- R-S-R CH,-S-CH3 Dimethyl thioether Dimethyl sulfide Structure -s-s R-S-S-R Disulfide CHs-S-S-CH3 Dimethyl disulfide Dimethyl disulfide R Sulfonic acid OH and Properties 03 Sulfoxide CH;-S-CH3 Dimethyl sulfoxide Dimethyl sulfoxide O6 anic 一R -CH Dimethyl sulfone Dimethyl sulfone The italicized pe A primary (1)amine:there are als ndary (2)R NH.and tertiary(3)R,N amines. Another name is propanamine
1 The italicized portion indicates the group. 2 A primary (1º) amine; there are also secondary (2º) R2NH, and tertiary (3º) R3N amines. 3 Another name is propanamine. TABLE 1.3 (continued) CHAPTER 1 Structure and Properties of Organic Compounds 10
CHAPTER 1 Structure and Properties of Organic Compounds 11 Problem 1.17 Refer to a periodic chart and predict the covalences of the following in their hydrogen compounds:(a)O;(b)S;(c)Cl;(d)C;(e)Si;(f)P;(g)Ge:(h)Br:(i)N;(j)Se. The number of covalent bonds typically formed by an element is 8 minus the group number.Thus:(a)2: (b)2:(c)1;(d)4:(e)4:(f3:(g)4:(h1:()3:(j)2. Problem 1.18 Which of the following are isomers of 2-hexene,CH,CH-CHCH,CH,CH? (a)CH.CH,CH-CHCH,CH, (b)CH,=CHCH,CH,CH,CH, (c)CH,CH,CH,CH=CHCH, (d) H CH3 (e) H2C CH2 HC-CH2 All but(c),which is 2-hexene itself. Problem 1.19 Find the formal charge on each element of :f: :Ar:B:E and the net charge on the species (BF,Ar). GROUP 厂#UNSHARED ATOM NUMBER ELECTRONS +1/2 SHARED FORMAL CHARGE ELECTRONS OF ATOM each F + 1 0 Ar 8 6 1 Problem 1.20 Write Lewis structures for the nine isomers having the molecular formula C,HO,in which C, H,and O have their usual covalences;name the functional group(s)present in each isomer. One cannot predict the number of isomers by mere inspection of the molecular formula.A logical method runs as follows.First write the different bonding skeletons for the multivalent atoms,in this case the three C's and the O.There are three such skeletons: ()C一C—C一O(i)C一O—C一C(ii0C一C—C 0 To attain the covalences of 4 for C and 2 for O,eight H's are needed.Since the molecular formula has only six H's,a double bond or ring must be introduced onto the skeleton.In (i),the double bond can be situated three ways,between either pair of C's or between the C and O.If the H's are then added,we get three isomers:(1), (2).and (3).In (ii),a double bond can be placed only between adjacent C's to give(4).In (iii),a double bond can be placed between a pair of C's or Cand O.giving (5)and(6).respectively. (1)H2C=CHCH2OH(2)CH3CH=CHOH (3)CHCH2CH2CH=0 (4)CHOCH=CH2 alkene alcohols (enols) an aldehyde an alkene ether (5)H2C=CHCH3 (6)CH;CCH3 :OH :0 an enol aketone
Problem 1.17 Refer to a periodic chart and predict the covalences of the following in their hydrogen compounds: (a) O; (b) S; (c) Cl; (d) C; (e) Si; (f) P; (g) Ge; (h) Br; (i) N; (j) Se. The number of covalent bonds typically formed by an element is 8 minus the group number. Thus: (a) 2; (b) 2; (c) 1; (d) 4; (e) 4; (f) 3; (g) 4; (h) 1; (i) 3; (j) 2. Problem 1.18 Which of the following are isomers of 2-hexene, CH3CHCHCH2 CH2CH3? (a) CH3CH2CHCHCH2CH3 (b) CH2 CHCH2CH2 CH2CH3 (c) CH3CH2CH2CHCHCH3 All but (c), which is 2-hexene itself. Problem 1.19 Find the formal charge on each element of CHAPTER 1 Structure and Properties of Organic Compounds 11 and the net charge on the species (BF3 Ar). Problem 1.20 Write Lewis structures for the nine isomers having the molecular formula C3H6O, in which C, H, and O have their usual covalences; name the functional group(s) present in each isomer. One cannot predict the number of isomers by mere inspection of the molecular formula. A logical method runs as follows. First write the different bonding skeletons for the multivalent atoms, in this case the three C’s and the O. There are three such skeletons: To attain the covalences of 4 for C and 2 for O, eight H’s are needed. Since the molecular formula has only six H’s, a double bond or ring must be introduced onto the skeleton. In (i), the double bond can be situated three ways, between either pair of C’s or between the C and O. If the H’s are then added, we get three isomers: (1), (2), and (3). In (ii), a double bond can be placed only between adjacent C’s to give (4). In (iii), a double bond can be placed between a pair of C’s or C and O, giving (5) and (6), respectively.
12 CHAPTER 1 Structure and Properties of Organic Compounds In addition,three ring compounds are possible: ()）HC-CHOH CH2 a cyclic alcohol heterocyclic ethers
In addition, three ring compounds are possible: 12 CHAPTER 1 Structure and Properties of Organic Compounds