HW_1 solutions

HW_1 solutions - CHAPTER 1 1.1 (See Figure 1.1 for...

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Unformatted text preview: CHAPTER 1 1.1 (See Figure 1.1 for equations) th 1.20.10—2 E=‘= =120kJ/1nolIR 2. 10‘4 ( ) I -2 = 91L = 1.2-101m /mol(radio) 104 _ l c 3- 101° 30 z = _. = = cm microwave V 109 ' —2 E = z 4 -10'41d /mol(mjcrowave) 30-10 1.2 a) same properties 7 (b) same properties (c) same properties ((1) same properties (e) mirror image properties, therefore distinguishable (f) mirror image properties, therefore distinguishable ‘ Order shown below: 962$ 21 A B C ‘3c 1 signal 2 signals 1 signal UV yes no no 1.4 88.2% c gig—2'2: 7.35 =>1 E.F. = C5Hg MW of 68 1.7% H Eli—7: 11.7 => 1.59 (CH1_59)n x 5 EW = 68 Hence W: C10H16 MW =136 1.5 60.0 _ C E —5 10 different 8 0 carbons H ‘— 8 E-F-=C5H802 E W2C10H1604 ll Oi H 1 2 162 12 0.35 6.67 _ éfl - H T *667 0.35 1 16.82_ #5 z O 16 ‘105:> 0.35 3 4.91 _ £5 :1 ~——1 —0.35 035 MS says MW = 285 So C17H19NO3 1.7 (a) (b) secondary alcohol d. (d) (e) 3°=C1, C2 2°=C3 1°=C10 (f) R*/S* (relative stereochemistry) 1.8 Five isomers are: Br 3 \ Z >——Br A W131- Br/\/ r/fi 1.82 1.8b 1.86 1 8d 1.86 (a) UN = 1 (b) 1.83 (c) 1.8d & 1.8e are geometric isomers (d) 1.8b (e) 1.83 vs 1.8b etc. C15H1302B1‘Cl (a) UN based on “H20” = 6 (b) Mult bonds = 2 + 2 = 4 so rings = 2. Note the revised structure Shown below in the literature has only one ring but three double bonds (one is telrasubstituted). Note the different conclusion in the cited literature. 1.10 a) UN = 1 o o 0 Hut 0 r (b) /U\OCH3 \AOH ’ OH H, G a a '0 ,etC OH (0) Look at Aldrich and Merck Index C :> ———=2.41 34:1 12 2.41 6 49 6.49 H:>~=6.49, :>——=269 1 2.41 O :> 100 ~(29.0 + 6.49) = 64.51: = 4.03, :> = 1.67 16 2.41 (3) n=3 (C1H2.6901.67)n 3 QHsos (blcsHssos, C3H8SZO (C) C3H8320 ((1) SH OH I.sz 1.1 1 1.12 (e) Compound 1.11 is not symmetrical while 1.12 has an internal 1 plane. 1.12 C4H6O UN=2 (@ o /—-OH 0_ —_ __ , .— 4—— I O OH / E \ B (b) l *0 0 etc 3H5NO, UN = 2 /OH O NH O N H H /U\/ y 0 9 NH 3 7 HO Mao/N er) 0 \/\\\N , ’ H y /\ N/O N (B \c W /\o_CEN /\O~NE \\0 Z 1 /NH2 HN (only compound that fits formula from Merck Index). k HZN NH (8) C:>l322594:>fl=4.99 12 1.19 H:>—9'—5—9-=9.59:>&~52=8.06 l 1.19 O _> 100 -(7l.39 +9.59) 3 19.02 _ 1.18 _1 16 *1.19 ‘ EF=MF=C5H30 UN=2 0CD (b) Only three compounds in Merck Index. 1.1521 1.15b 1.150 only 1.153 fits data. I =C4H4N202 UN = W = 4 (a) do not forget malic hydrazide NH 1 NH O/ko 1.16b NH 2 //', o 0 Z NH N 0 Z Z N 0 Eli \ H Z=N202 \ 1.16c 1.162: O t O N/km H 1.16d (b) Its URACIL! 1.16:: above. Note: 0 NH does not I I fit functional NH group list 0 mp>300o 1.16f NH ' A 0 NH malic hydrazide mp 260 1.16e OH OH OH O OH OH original structure contains conjugated ester 1.18 0 C1 0 0e :_—_‘ ® Me AN/ AN/ eq Me's I 1 all Me‘s different all Me's different 1.19 /(a) CQHISNO If 0 _ . UN=9x2+2 ISM-1:11 2 2 unpaired e‘ is present in this nitroxide. (b) H Fe H H C5H5 H H __ UN: 5x2+2 5:31 2 2 1 ring Zdb le‘ is aromatic (67re—, is a 4m: + 2) (c) (CH3)3OBF4 note B behaves as N soUN — [((3 x 2) +2) _ 9+1" 4] _ ‘4 _ -2 2 2 calculation does not work because this compound is a salt. (CH3)3O+ BF4' (d) (CH30)5P CH5H1 5051’ note P behaves as N UN: (5 x2)+2—15+1:_1 2 calculation does not work because P is pentacoordinate (e) (CH3)4NCI = C4H12NC1 (4 x 2) + 2 ~ 1 9 UN = “a = ~— = 4 5 2 2 so must be a salt Me4N+ Cl’ 1.20 UN R norbornane C7H12 5 W 2 W 7x2+2-12 = 2 R: 1+ 1/2 [(5)Q+2x3)]-7= 2 2 prismane C6H6 6 7772:“ 6x2+2—6 2 4 R=1+ 1/; (626)—6 =4 2 “MEN = 3 R=1 + 1/2 (4)3}4 = 3 we 6 QR“ Sflgz-S = 5 R=1 + 1/2 (8x3)—8 = 5 C60 UN: 60 x22+ 2 - 0 = 61 6O sp2 carbons = 60 R=1+ 1/2(6Ox3) =90-60+ 1 =31 Finally, double bonds =UN-R=61-31 =30 1.22 4:1 ratio 0 O O mai<KMe>2 W (325% M km 0 OHi 0 min KMe KMC Egg TEE” M enol 1.23 ...
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This note was uploaded on 08/06/2008 for the course CHEM 124 taught by Professor Amagata during the Winter '08 term at UCSB.

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HW_1 solutions - CHAPTER 1 1.1 (See Figure 1.1 for...

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