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1 th 37 ICh Theoretical Examination : - ( ): ( ) - : G D L U C K 1
2 2
3 1 amu = kg N = mol -1 k = J K -1 e = C F = C mol -1 R = J K -1 mol -1 = L atm K -1 mol -1 m e = kg m n = kg m p = kg h = J s c = m s -1 (298 K ) E = E ( / n) log K k = Ae -Ea/RT ln P = - vap / RT + B λ = h / mv PV = nrt G = TS E = hv G = G + RT ln Q G = - nfe U = q + w w = - P V = πr 2 h = 4/3 πr 3 = 4 πr 2 1 = m 1 W = 1 J s -1 1 J = 1 kg m 2 s -2 1 cal = J 1 Pa = 1 kg m -1 s -2 = 1 N m -2 1 bar = 10 5 Pa 1 atm = Pa = 760 mmg (torr) 1 = kj mol -1 = Pa RT at K = kj mol -1 π =
4 Total Scores: 38 points N,N- (DMF) C N C 3 C 3 C C 3 N C N,N- ( A) N- (C 3 CNC 3, B) (C 3 C 2 CN 2, C) > > ( A, B, C ) 1-2 C= C 1715 cm -1 (a) 1660 cm -1 (b) 1660 cm -1 (c) 1740 cm -1 (d) 1740 cm -1 4
5 1-3 ( 2 N-C 2 -C) α 3 2 Gly-Gly-Gly 1-4 α L- D- L- D- 3 2 N Glycine (Gly) 2 N C 3 L-Alanine (L-Ala) 2 N 3 C D-Alanine (D-Ala) (Gly) L- (L-Ala) D- (D-Ala) D 4- ( E) 4- ( F) > > ( D, E, F ) UV-vis 5 (p ) 5
6 G + 2 Phenol concentrated 2 S o C, 5 h Phenolphthalein + Phenol Phenolphthalein concentrated 2 S Na 1-8 G 2 G (a) (b) (c) (d) (e) 6
7 Total Scores: 48 points L- (L-ribose I) C 2 Me C + A sealed tube C 2 Me s 4 B C 2 Me C 2 Me Me 2 C(Me) 2 +, C 3 CC 3 C C 2 Me C 2 Me pig liver esterase D (minor) C 2 Me C 2 + E (major) C 2 C 2 Me 3 Me Me 2 C C 2 Me Me 2 C C 2 Me C MCPBA 1 4 (1) Me/ (2) LiAl 4, then 2 F G I (L-ribose) sealed tube = pig liver esterase = (minor) = (major) = 2-1 A C A 2-2 A C True T False F 7
8 (a) A B s 4 (b) B C (c) B C (d) Me 2 C(Me) 2 C C D E 20 E [ ] D = 37.1 o 20 o [ ] D = 49.0 E 2-3 D E D/E 2-4 F - MCPBA G True T False F (a) F (b) MCPBA (c) R/S C-1 C NMR 1 NMR (CDCl3 ) δ 1.24 (s, 3), 1.40 (s, 3), 3.24 (m, 1 ), 3.35 (s, 3), 3.58 (m, 2), 4.33 (m, 1), 4.50 (d, J = 6 z, 1), 4.74 (d, J = 6 z, 1), 4.89 (s, 1). s: d: m: I C-1 C-4 R S C-1: ; C-2: ; C-3: ; C-4:. 8
9 2-7 I L- P, Q, R, S, T, U P R T C Q S U C glycosidic bond glycosidic bond = 2-8 J D- 5 pentasaccharide J derived from D-glucose D- J 9
10 3: Total Scores: 36 points Points A B Na +, K +, Cs + N N C 2 C 2 A B Binding constant (log 10 K) Metal ion Radius (pm) Compound A Compound B Na K Cs Binding constant Metal ion Radius (pm) Compound 325 nm E 10
11 3-1 C D (1) NaB 4 (2) PCl 5 /benzene C (1) pyridine S 2 Cl (2) t-bu - K + (C 2 ) 2 N(C 2 ) 2, D K 2 C 3 toluene E N C 2 benzene: pyridine: toluene: F G E, F, G 2 C N 2 C N F G 3-2 Cl (a) (b) E F (c) G (d) 3-3 E, F, G (10-5 M) (a) E (b) F (c) G 11
12 3-4 F (a) (b) (c) (d) - I I hν hν oxidation trans-stilbene heat I Phenanthrene trans-stilbene - heat oxidation Phenanthrene I cis( ) trans( ) J J K C 3 NC 1 CN CN CN hν heat C J 8 7 K heat 12
13 3-7 (a) J (b) K 3-8 K CF 3 C 2 K (a) C-2 (b) C-3 (c) C-4 (d) C-5 13
14 4: Total Score: 42 points 4A-1 4A-2 4A-3 4A-4 4A-5 4A-6 4B-1 4B-2 4B-3 4B-4 4B A Chiufen Chiufen KCN (CN - ) Au(CN) Au(s) + 8 CN (aq) + 2 (g) (l) 4 Au(CN) 2 (aq) + 4 (aq) 4A-1 Au(CN) 2 4A-2 20g KCN 3:1( ) Au(s) + N 3 (aq) + Cl (aq) AuCl 4 (aq) + N 2 (g) 4A-3 4A-4 4A-3 14
15 noble AuCl 4 Au 3+ (aq) + 3 e Au(s) E ο = V AuCl 4 (aq) + 3 e Au(s) + 4 Cl (aq) E ο = V 4A-5 25 AuCl 4 3+ K = [AuCl 4 ] / [Au ] [Cl ] 4 4A-6 Cl Cl (a) Cl (b) Cl (c) Cl (d) Cl B (AuNP) Brust-Schiffrin AuNP 1.5 nm 5.2 nm AuCl 4 -n- NaB 4 AuNP 24 AuNP 4B-1 (a) (b) 15
16 4B-2 -n- AuCl - 4 -n- (a) (b) (c) 4B-3 NaB 4 (a) (b) (c) (d) 4B-4 3 nm Au (Au 1.44 ) (a) 10 2 (b) 10 3 (c) 10 4 (d) B-5 Au (a) 20-30% (b) 40-50% (c) 60-70% (d) 80-90% 16
17 5: Total Score: 21 points (a) N 2 (b) N 3 (c) 3 (d) S S,S-, 2 SC(N 2 ) 2, N S C N 5-3 S,S- 5-4 Valence Shell Electron Pair Repulsion a (a) (b) (c) T 17
18 5-4b (a) (b) (c) T 5-4c (a) (b) (c) T S,S- S C N N 65 o N, S, C S SC(N 2 )
19 6: C 2 Total Scores: 40 points C 3, C3, and p C 3, C3, and = [C 3 ] + 2[C3 ] + [ ] [ ] p (298K) C 2(g) C 2(aq) K C2 = 3.44x10-2 C 2(aq) C 3 K 2 C 3 = 2.00x C 3 C 3 + K a1 = 2.23x C 3 C3 + K a2 = 4.69x10-11 CaC 3(s) Ca C 3 K sp = 4.50x K w = 1.00x10-14 : 6-1 ( ) C 2 [ + ] = M - 2- [ 2 C 3 ] : [C 3 ] : [C3 ] (a) : 1.00 : (b). (a) (b). 19
20 6-2 C 2 ( 1.01 x 10 5 Pa, 298K C % ( )) C 2 (aq) (mol/l) ( = 1.01 x 10 5 Pa) C 2 (aq) =1.11x10 5 M - 2- C 2 S = [C 2(aq) ] + [ 2 C 3 ] + [C 3 ] + [C3 ] 298K 1.01 x 10 5 Pa C C 2 (mol/l) x10-3 mol/l Na C K, 1.01 x 10 5 Pa CaC 3 CaC 3(s) + C 2 (aq) + 2 Ca C K eq = 5.00x CaC 3 C 2 (mg/l) Ca 2+ Ca 2+ (aq) = 40.1 mg/l 6-7 (mol/l) 20
21 6-8 CaC 3 C 2 Ca mg/l Ca 2+ C 2 (Pa) 21
22 ( 3 ) k k -1 (1) k (2) 1, -1, , 2,
23 7-3 d[]/dt = 0 d[ dt 3 ] = k 1 2k k 1 [ 2 2 [ 3 ] + k 2 ] 2 [ 3 ] ( ) CCl 2 F 2 (Freon-12) CCl 2 F 2 Cl hν CCl 2 F 2 CF 2 Cl + Cl (3) 7-4 Cl (g) + 3(g) Cl (g) + 2(g) (4) 7-23
36 th IChO : - 3 ( ) , G O O D L U C K final 1
36 th ICh - - 5 - - : - 3 ( ) - 169 - -, - - - - - - - G D L U C K final 1 1 1.01 2 e 4.00 3 Li 6.94 4 Be 9.01 5 B 10.81 6 C 12.01 7 N 14.01 8 16.00 9 F 19.00 10 Ne 20.18 11 Na 22.99 12 Mg 24.31 Periodic
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Microsoft Word - 11問題表紙（選択）.docx
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現代物理化学 2-1(9)16.ppt
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001_046_物理化学_解答_責_2刷_Z06.indd
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現代物理化学 1-1(4)16.ppt
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