anic chemistry.we have to determine whether two structures are the same or In order for two structures to be the sa the groups mus amc sideite thwith you compound.) (a)different compounds;H and CH,on one carbon of the double bond,and CH3 and CH2CH,on the other carbo the second structure,so they are DIFFERENT compounds HC.' CH2CH HC these are DIFFERENT H .CH .CH2CH; compar on thehtheh cron abeenoed ut heon (c)different compounds:H and Br on one carbon.F and Cl on the other carbon in both structures:H and (d)same compound:in the structure on the right,the right carbon has been rotated 120 2.9 CH3~N: a) H H -C=N. CH,一CH CONVERTIBLE H sp wo CHa's o aer mae side (c)the CH. the n is on the sa :N-CH drawn-only one possible structure 2.10 H F (a) and c=c (0 H H C=C cis trans H (b)no cis-trans isomerism] H )cm and c=C CH e c=C =C xample H H and H CH3 cis trans 29
2-8 Very commonly in organic chemistry, we have to determine whether two structures are the same or different, and if they are different, what structural features are different. In order for two structures to be the same, al l bonding connections have to be identical , and in the case of double bonds, the groups must be on the same side of the double bond in both structures. (A good exercise to do with your study group is to draw two structures and ask if they are the same; or draw one structure and ask how to draw a different compound.) (a) different compounds; H and CH3 on one carbon of the double bond, and CH3 and CH2CH3 on the other carbon-same in both structures; drawing a plane through the p orbitals shows the H and CH3 are on the same side of the double bond in the first structure, and the H and the CH2CH3 are on the same side in the second structure, so they are DIFFERENT compounds compare H e3 ' " CH \ / 3 - - - - - - - e ::E-C- - - - - - - - - - I \ H" " CH2CH3 .. ... .. ... .. compare these are DIFFERENT (b) same compound; in the structure on the right, the right carbon has been rotated, but the bonding is identical between the two structures (c) different compounds; H and Br on one carbon, F and Ci on the other carbon in both structures; H and CI on the same side of the plane through the C=C in the first structure, and H and F on the same side of the plane through the C=C in the second structure, so they are DIFFERENT compounds (d) same compound: in the structure on the right, the right carbon has been rotated 1 200 2-9 (a) H H H " , H-C-C=N- C-H (b) ".. CH3 :N " CH3 ",.. C ....... H NOT INTERCONVERTIBLE CH3 - N : " CH3 ",.. C ....... H /, �I , '-. 3 H \ H sp sp 2 two CH3's on opposite sides of the C=N two CH3's on the same side of the C=N (c) the CH3 on the N is on the same side as another CH3 no matter how it is drawn-only one possible structure 2-10 (a) and ".. CH3 :N " ",.. C" CH3 CH3 H F " / C=C / "- F H trans . , . . two IdentIcal groups on one (b) no cis-trans iSOmeriSm} . . (c) no cis-tram Isomensm . . . car on b f 0 th e ou e on d bl b d (d) no cis-trans Isomensm (e) Q ... CH3 C=C ... , H H CIS and QC � c, H ... "- H CH3 trans 29 "cis " and "trans " not defined for this example
rans, ond is cis )onomh the renhe cons somer the seco rent same carbon in the second structure same compoun -just flipped over same compound- just rotated (g)not isomer different molecular formulas constitutional somer -the double bond has changed position the CH. groups are in different relative positions (k)constitutional isomers-the double bond is in a different position relative to the CH3 2.12 (a)2.4D= 0positive chargs on carbon and ive chrge on oe 6 0: R R R一NR A B were the 2.13 In Nont ohrn te rtion of away from the nitrogen;the three bond polarities cancel the lone pair polarity,so the net result is a very small molecular dipole moment. H parentheses. Eachom has thee monbonded clctron pairs.not shon below)The Cs usually consic (b) H H十 H large dipole (1.54) large dipole (1.81) net net 30
2-11 Models will be helpful here. (a) cis-trans isomers-the first is trans, the second is cis (b) constitutional isomers-the carbon skeleton is different (c) constitutional isomers-the bromines are on different carbons in the first structure, on the same carbon in the second structure (d) same compound-just flipped over (e) same compound-just rotated (f) same compound-just rotated (g) not isomers-different molecular formulas (h) constitutional isomers-the double bond has changed position (i) same compound-just reversed (j) constitutional isomers-the CH3 groups are in different relative positions (k) constitutional isomers-the double bond is in a different position relative to the CH3 2- 12 (a) 2.4 D = 4.8 x 8 x 1 .21 A 8 = 0.41, or 41 % of a positive charge on carbon and 41 % of a negative charge on oxygen (b) :0: I .. .. ..... C ....... R ..... + R B Resonance form A must be the major contributor. If B were the major contributor, the value of the charge separation would be between 0.5 and 1.0. Even though B is "minor" , it is quite significant, explaining in part the high polarity of the C=O. 2- 13 Both NH3 and NF3 have a pair of nonbonding electrons on the nitrogen. In NH3, the direction of polarization of the N-H bonds is toward the nitrogen; thus, all three bond polarities and the lone pair polarity reinforce each other. In NF3, on the other hand, the direction of polarization of the N-F bonds is away from the nitrogen; the three bond polarities cancel the lone pair polarity, so the net result is a very small molecular dipole moment. polarities reinforce; large dipole moment 1 O· � N� H .... \\ ' H H polarities oppose; small dipole moment 2- 14 Some magnitudes of dipole moments are difficult to predict; however, the direction of the dipole should be straightforward, in most cases. Actual values of molecular dipole moments are given in parentheses. (Each halogen atom has three non bonded electron pairs, not shown below.) The C-H is usual ly considered non-polar. (a) LCi H it" / H � C� large dipole (1 .54) XX: CI I .. net 30 (b) H H�, C +--F / H I .. net large dipole (1.81)
2-14 continued net dipole=0 large dipole(1.70) net 0+→ composite of two H一C三N? small dipole(0.52) large dipole (2.95) large dipole arge dipole (2.72) H H mall dipole (0.67) net large dipole (1.45) (0)++ (m) H Cl-Be-Cl net dipole =0 net dipole =0 net dipole =0 In(k)through(m),the symmetry of the molecule allows the individual bond dipoles to cancel. large net dipole net dipole=0
2- 14 continued (c) �/ LF �� "c �F net dipole = 0 (e) 1 0�, o 'x� .�0 T o" o� (d) each end oxygen has one-half negative charge as it is the composite of two resonance forms; see ro or net solution to 1-38(b) smal l dipole (0.52) (g) 0;13 (h) �1 1 large dipole (2.72) C H/ 'CH3 (i) 1 \) (k) /N,� H3C ,/ \," 'CH 3 CH3 F \\� Il-F F net dipole = 0 net 1 net small dipole (0.67) (I) ----+ � CI - Be - CI net dipole = 0 1 large dipole (1 .70) net (f) �� H-C NCD I .. net large dipole (2.95) \ large dipole net / net large dipole (l .45) (m) � N tt� " " H"'" \\ " H H net dipole = 0 In (k) through (m), the symmetry of the molecule allows the individual bond dipoles to cancel. 2- 15 With chlorines on the same side of the double bond, the bond dipole moments reinforce each other, resulting in a large net dipole. With chlorines on opposite sides of the double bond, the bond dipole moments exactly cancel each other, resulting in a zero net dipole. large net dipole C� �CI C==C / '" H H 1 31 net dipole = 0
2-16 (a) 8+ 6 CH:CH2-O Hw>Q 8 cHcH,N CH2CH3 H8+ (hydrogen bonds shown as wavy bond) H+ 2-17 CHCHCH),hcn nd oherme an 岛，8 mo om yro bo点a n bonds and will boil at a much higher temperature than bonds at both ends and has no branching:it will boil at a bonds than A.a secondary amine with only one N-H bond. H A 〈N-H 218 &ClS86t.SncmboeRnsanansmoesohcten cH.COmameanpge0C8a2eemetuaa CH3CH2CH2CH2OH,however.the hydrogen bonding from only one OH group cannot camry a four-carbon chain into the water;this substance is only slightly soluble in water. (d)Both co nds form hy n but only the smalle
(hydrogen bonds shown as wavy bond) 2- 17 (a) (CH3hCHCH2CH2CH(CH3h has less branching and boi ls at a higher temperature than (CH3hCC(CH3h . (b) CH3(CH2)sCH20H can form hydrogen bonds and will boi l at a much higher temperature than CH3(CH2)6CH3 which cannot form hydrogen bonds. (c) HOCH2(CH2)4CH20H can form hydrogen bonds at both ends and has no branching; it will boi l at a much higher temperature than (CH3hCCH(OH)CH3 . Cd) (CH3CH2CH2hNH has an N-H bond and can form hydrogen bonds; it will boil at a higher temperature than (CH3CH2hN which cannot form hydrogen bonds. (e) The second compound shown (B) has the higher boi ling point for two reasons: B has a higher molecular weight than A; and B, a primary amine with two N-H bonds, has more opportunity for forming hydrogen bonds than A, a secondary amine with only one N-H bond. 2- 18 (a) CH3CH20CH2CH3 can form hydrogen bonds with water and is more soluble than CH3CH2CH2CH2CH3 which cannot form hydrogen bonds with water. (b) CH3CH2NHCH3 is more water soluble because it can form hydrogen bonds; CH3CH2CH2CH3 cannot form hydrogen bonds. (c) CH3CH20H is more soluble in water. The polar O-H group forms hydrogen bonds with water, overcoming the resistance of the non-polar CH3CH2 group toward entering the water. In CH3CH2CH2CH20H, however, the hydrogen bonding from only one OH group cannot carry a four-carbon chain into the water; this substance is only slightly soluble in water. (d) Both compounds form hydrogen bonds with water at the double-bonded oxygen, but only the smaller molecule (CH3COCH3) dissolves. The cyclic compound has too many non-polar CH2 groups to dissolve. 32
2.19 HHHHH H HH (a)H-C-C-C-C-C-H (b)H-C-C=C-C-C-H (c)H-C-C三C-C-C-C-H HHHHH HHHHH HHH alkane alkene alkyne H H HHHH H HH (d)H-C-C=C-C-H (e)H-c-c-c-c-H (H.c-Csc-C=C-H H-O H-C-Q H H HC.c-C.H HHHH cycloalkyne cycloalkane aikbtocabon H HHH H (g) H-c-Csc-C-C-C-H se- ()H HH、HH c-C-c-C-c-H -C-H H HCC-C-H HCsc-Cc-C.H HH HH cycloalkene alkyne,alkene,cycloalkane rocarbon and cycloalkene 2-20 HH O (a)H-C-C-C-H (b)H-C-C-C-C-H (c)H-C-C-C-C-H HH HHHH H HH aldehyde alcohol ketone HHHH HH (d)H-C-C-O-C-C-H (e)H HO -c-Ccc-o-H HHHH H-c.0. (0 H -H ether H-C.c-C-H .C.C. H carboxylic acid ether 0 H H (g)H HC-0-H c:0 (h)H.H.c.C-H H-c-C ()H .H HC-C.H HH HH ketone aldehyde alcohol
2-19 H H H H H I I I I I (a) H-C-C-C-C-C-H I I I I I H H H H H alkane (Usual ly, we use the tenn "alkane" H H H I I I (b) H - C - C = C - C - C - H I I I I I H H H H H alkene only when no other groups are present.) H H I I (d) H - C - C = C - C - H (g) 2-20 I I H-C C-H '" / H C-C H H' I I 'H H H cycloalkyne H H H H H I I I I , C C-C-C-H H-C � " C � I I I I I H C H H - C .. . � C H ' I • H H C I H I H H cycloalkene H H 0 I I " (a) H-C-C-C-H (d) (g) I I H H aldehyde H H H H I I I I H-C-C- O-C-C-H I I I I H H H H ether ketone H H H H I I I I (e) H-C-C-C-C-H I I I I H-C-C H H I I ' H H H cyc10alkane H H H H "H I (h) , � C .. , H-C C-C::C-C-H I I I H-C C .. H H • H " C � ... H '" H C=C-H I I H H alkyne, alkene, cycloalkane H H H 0 H I I I I (b) H - C - C - C - C - H I I I I H H H H alcohol H H 0 H " H " (e) , � C .. , H-C C-C-O-H I I H-C C-H , .. c � . H ,. H H H carboxylic acid o H H C" -H H " (h) " C, / H ' c; �. H . C - C . H I I H H H aldehyde 33 H H H H I I I I (c) H-C-C ::C-C-C-C-H I I I I H H H H alkyne H H H I I I (f) H, � C .. � C=C-H C " C " I H , C " C � � C ' H I H aromatic hydrocarbon and alkene I;I H H (i) H C 'c ' , H ' C o:. " C � " C-H I II I H ' C "c � C " C �_ C , H I I H H aromatic hydrocarbon and cycloalkene H 0 H H I " I I (c) H-C-C -C-C-H I I I H H H ketone (f) H 0 H H -' C � " C '- H I I H-C C-H H , ... c � . ,. H H H ether alcohol