Đáp Án HSGQG hoá học 2019

19a.pdf 19b.pdf K Môn: 13/01/2019 hai thi c Câu I I.1 a b c lk ,n plk plk lk n=n -n ( lk nlk plk plk ( anion O2 d 2 O2 trình I.2 c phân li O3 a + O ·mol 1) OO2 b trang 1/12 I.1 2,0 a (0,5 Ghi ms = electron 0,25 b (0,5 - (0,25 , * HOMO c (0,5 - O O O O - ng (0,25 ( s)2( s * O ) ( z)2( x)2( y)2( x * )( * * y ) ( z ) O (0,25 Ghi *s n = * z z n = * x x n = * y y n = x y s electron x electron (0,25 d (0,5 - ( s)2( s * ) ( z)2( x)2( y)2( x*)1( y*)1 ( s)2( s*)2( z)2( x)2( y)2( x * )( O2 - ( s) ( 2O s * 2 2 ) ( z) ( x) ( y) ( x * y ) (0,25 ) ( y*)2( * * z ) (0,25 trang 2/12 I.2 1,0 a (0,5 - O3 E hc O2 N A 6, 626.10 34.3.108.6, 022.10 23 330.10 362, 74 (kJ·mol-1) (0,25 (0,25 b (0,5 Câu II II.1 m1 m2 m2 %m m2 100% ), m1 %m CaC2O4 H2 100 87,7 o (1) CaC2O4 H2O t R1 + K1 to (2) R1 R2 + K2 to (3) R2 R3 + K3 R K 1:1:1 68,5 38,4 a t oC R1, R2, R3, K1, K2 K3 200 510 780 b II.2 2+ A) A A o B B A 200 oC? a b II.1 a (2,25 - o %m C: m R1 100% m CaC2O4 H2O M R1 128 M R1 M CaC2O4 H 2O 100% M R1 100% 87, 7% (128 18) M K1 18 Vì v y: K1 H2O r n R1 CaC2O4 (0,75 m) trang 3/12 - T i nhi g n 510oC: mR %m mCaC2O4 H 2O M R 100, 01 K2 - T i nhi MR 100 MCaC2O H 2O 100 MR 100 68,5% 146 M K 28 R2 CaCO3 (0,75 o g n 780 C: mR %m mCaC2O H 2O MR3 56, 06 MR3 100% MCaC 2O H 2O M K3 K3 CO2 100% MR3 100% 38, 4% 146 44 (0,75 R3 CaO b (0,25 o H 298 II.2 (0,5 a (0,2 A câu II.1: 200oC 2O4.nH2O hoàn toàn (0,25 b (0,2 - A CaC2O4 nH2O; B CaC2O4 - ng ph 3: (0,25 Câu III (3,0 2, o III.1 o 800 2CO(k) + O2(k) III.2 III.3 III.4 2(k) ·h·kg 1) Gd2O3 (*) W = J·s -(CH2)4- NH2-(CH2)4-NH2 a b ho III.1 (0,5 2=CH-CN -(CH2)4-CN -(CH2)4-CN Go = nF Eo Eo = Go/nF = 380.103/4.96485 (J/C) = 0,9846 (V) trang 4/12 : III.2 (0,5 Catot: O2 + 4e Anot: CO + O2 III.3 (1,0 2 (0,25 (0,25 + 2e 2CO2 - q = 96485 (C) = 192970 (C) = 192970 (A s) mol CO: P = 0,9846.192970 (V A s) = 190000 W s = 52,78 W h ,0 kg CO: (52,78 1000)/28 = 1884,92 (W h kg 1) (G (0,5 ) III.4 (1,0 a -(CH2)4-CN: + CH2=CH-CN + 2H + 2e -(CH2)4-CN b -(CH2)41 mol NC-(CH2)41.M CO mCO = 0, 26 M C6 H8 N2 (0,5 100% (0,5 Câu IV (3 6H5-CH3 hay Ph-CH3 IV.1 ·L [Cl2] (mol·L ) 0,0775 0,0565 Hãy xác 0,0410 10 0,0265 ·L 1·phút 14 0,0175 1 ·L IV.2 mol·L , t ·L 1) ·L 1·phút 1) 0,050 0,00144 Cl2 2Cl 0,075 0,00265 0,100 0,00408 IV.3 IV.4 Ph-CH3 + Cl Ph-CH2 + Cl2 k2 Ph-CH2 + HCl Ph-CH2Cl + Cl -CH3 k3 tìm Ph-CH3 IV.1 (1,0 trang 5/12 v = k [Ph-CH3]a[xúc tác]b[Cl2]c -CH3 c 2] -CH3]a[xúc tác]b [Cl ] k , t ln o [Cl ]t [Cl2], mol·L (0,25 : 10 14 0,0775 0,0565 0,0410 0,0265 0,0175 0,1053 0,1061 0,1073 0,1063 k' k , (phút-1) 0,1063 -1 mol·L 1·phút 2] = 0,05 mol/L; (0,5 ) 2] = 0,1063.0,05 = 0,005315 (0,25 IV.2 (0,5 v = k[Ph-CH3]a[Cl2][xúc tác]b, mà k[Ph-CH3]a[Cl2 [Cl2 v1 v2 v = const[xúc tác]b 0, 00144 0, 00265 ( 0, 05 b ) 0, 075 ( [xúc tác]1 b ) [xúc tác]2 (0,25 b 1,5 (0,25 IV.3 (0,5 3, FeCl3 (0,25 [AlCl4]- (0,25 4] - + + - [AlCl4] trang 6/12 IV.4 (1,0 d[Cl ] 2k1[Cl2 ] k 1[Cl ]2 k [Ph CH ][Cl ] k 3[Ph CH ][Cl ] (I) dt d[Ph CH ] k [Ph CH ][Cl ] k 3[Ph CH ][Cl ] (II) dt Thay (II) [Cl ]2 [Cl ] (0,5 (I) : k1[Cl ] k1 k1[Ph CH ] k1 Ph-CH3 k1[Ph CH ] (III) k1 - 2): d[Ph CH ] dt (0,25 d[Ph CH ] dt k [Ph CH ][Cl ] k k [Ph CH ] [Ph CH ] k1 d [Ph-CH ] [Ph-CH ]3/2 k2 k1 dt k1 kdt k k2 k1 k1 k2 k1 k1 [Ph CH ] (0,25 Câu V V.1 18 electron (n-1)d10ns2np6 a X X Y khí Z Xác Y Z b 2(CO)8 ZC = 6; ZFe = 26; ZCo = 27 V.2 a M b M M ·cm 3) M 1,40 Å N có N N M ·cm M N trang 7/12 V.1 (2,0 a (1,25 3d4s n [Fe(CO)n] 8+ (0,5 [Fe(CO)5] cacbonyl X không (0,25 Y [Fe(CO)4(I)2] Y (0,25 (0,25 Z CO [Fe(CO)5] + I2 [Fe(CO)4(I)2] + CO) b (0,75 - Trong [Co2(CO)8], m -Co Co có , 2(CO)8] (0,25 2(CO)8]: - 2(CO)8 -cho* cho (0,25 - 2(CO)8]: (0,25 sp3 trang 8/12 sau V.2 (2,0 a (1,0 1 ion 1 12 ion là: c - D 4.M M a N A M là: 4.72 (4,30.10 ) 6, 022.1023 2+ - R Fe2 b (1,0 FeO (0,5 6, 015(g c m ) (0,25 : a R Fe2 Theo ra, N M, a R O2 M R O2 4,30 1, 40 0, 75 Å (0,25 N N N FexO N là: mN = D.V = 5,70 (4,30.10-8)3 = 4,532.10-22 gam - 4.16 23, 45% 4,532.10 22.6, 022.10 23 %Fe 56.x 76,55% x 0,9327 %O 16 23, 45% N Fe0,9327O (0,25 %O - II -N (0,25 III Fea Fe b O a + b = 0,9327 2a + 3b = a có: II a = 0,7981; b = 0,1346 III N Fe0,7981 Fe0,1346 O C (0,25 ) N M M N: N M, N + H2O + 3+ + 3H2O FeO + 2H+ 2O3 + 6H + 3+ 2+ + Fe + 4H2O 3O4 + 8H + 2+ 2+ + H2O trang 9/12 H B H NH2 H O + O O N H R N C OH O N H R C O H B R CH N O N H O O O H decarboxyl H NH2 O R + O R H N C H O N H H2O H N H R N C H OH N + CO2 H b Et Ce(IV) Ce(III) Et H2O Et HO Et H OH2 Ce(IV) Ce(III) O Ce(III) Et Ce(IV) O OH Et HO Et Et H OH2 trang 7/19 O O Et H2O Et H H HO Ce(IV) O Ce(III) H O Et [1,2] HO H Et OH -H OH OH Et Et Ce(IV) OH O Et Et H H O Ce(III) O H O O Ce(IV) Ce(III) OH O Et O Et H O trang 8/19 2/8 c Et O Et N Et N N Et O N to [4+2] CO2Me O CO2Me CO2Me CO2Me O H O Me Et Et N N COOMe CO2Me Me MeO O H OH O Me O H m d CHCl2 H O O O N Cl2CH C N Na H O CHCl2 O N O O O H O H O CHCl NH H O H+/H2O OH H O H O CHCl H2O CHCl2 O N BF3 N BF3 OH OH CHCl2 O N BF3 Et2O OH OH -H NH ROH H O CHCl OH CHCl2 HN OH O II.3 Me Me Me Me N O 9a Me Me Me N Me O 9b Me N H 9c Me N H Me Me 9d 9a, 9b, 9c, 9d, 10 hay 11 Me N H 10 N CO2Et Me COOEt 11 trang 9/19 thích Me NH2OH O Me Me OH N Me CH3C CH N O H2O Me Me Me HO Me Me + N H2O, OH Me Me N O C6H11NO 9a O Me 9b Me Me Me O to O Me O NH2 O Me N H 9c Me N H 9c Me N H 9d Me Me BrCH2CO2Et Me CO2Et Me N MgCl N H 10 BrCH2COOEt KHCO3 N H - H2O Me Me MeMgCl Me - H2O O Me NH O Me Me NH2 O Me Me Me Me Me NH to HN Me O Me Me N Me HN N Me Me Me Me Me N K Me Me Me N COOEt 9d 11 Câu III III.1 Nhi a ropon có O -benzotropon (12) hoá 12 b 13 metyl hoá MeI, 15 16 ( 13 14 Sau 13, 14, 15 16 G 15 16 14 CN HO , 17 enzym GSK phosphatase T Quá 14 III.2 a 12 BnO MeO Eugenol ,h O MeO 17 17 b 21 1) TBS-Cl, imidazol Eugenol 1) H2O, H 18 2) NaOH, to 2) PCC 18 III.3 Các 22 NH2 xylen, to quy - 22 PPA, to 21 2) H3O C16H15NO2 theo SOCl2, EtOH 22 1) Bu4NF 20 21 24 anhydrid sucinic NH2 PhNHNH2 19 K2CO3, DMF 23 C12H14N2O2 22 o-phenylendiamin 24 ClCH2CO2Et C16H20N2O4 24 23 o-phenylendiamin, anhydrid sucinic o HCl trang 10/19 lí X khan 24 a b Hãy c = 22 23, K2CO3 khan ClCH2CO2Et dung môi DMF 24: 22, 23 24 X 2CO3 III.1 a - khan 24 12: NH3+ NH2 1) HNO3, H2SO4 NHAc 1) H2SO4, 160 oC 2) H2/Pd-C 2) Ac2O NH3 NO2 CN NH2 1) Fe/HCl CHO CN 1) NaNO2, HCl 2) HO-/H2O - 2) H3O+, to HO3S NH2 NO2 1) HNO3, AcOH CHO 1)DIBAL 2) H3O+ 2) CuCN -oxopentanedioat: O Br HOC2H4OH Br - TsOH, 2% Br O O O 1) KCN Br + 2) H3O , t HO2C o O SOCl2 CO2H EtO2C EtOH CO2Et -benzotropon (12): CHO O CHO EtO2C CO2Et CO2Et NaOH O O to CO2Et 12 lí 12 m nhân - b Me OMe 13 Me 14 15 16 13, 15 hay 16 Cation 14 + = 10 e theo Huckel h nhóm MeO- 14 hoá 13 qua 14, trang 11/19 14 14 14 Me MeMgI OMe + [MeMg(I)OMe] 13 III.2 a 14 (10e Me 15 16 17 O 1) BnBr, K2CO3 BnO 2) KOH, to HO BnO Cl NaCN BnO CN DMF 3) O3 MeO H 1) NaBH4 2) SOCl2 MeO MeO MeO 4) Me2S CO2Et EtONa O CO2Et CO2Et CN CO2Et BnO O MeO NaBH4 HC CH MeOH CN MeO MeO t 1) H2SO4, H2O, >140 oC H3C CH2OH BrH2C CH2Br 2) Br2 1) HBr BnO CN 17 H3C CHO NaH BnO NaCl, H2O DMSO, to BuOK, to HC CH SOCl2, Pyr CO2H 2) Mg, CO2 C, to CO2Et EtOH CH2OH 1) CO, HCl CH2Br HBr 2) NaBH4, MeOH 17 lí ) b Me O TBSO Me TBSO Me 18 19 Me OH OMe OMe N H MeO MeO OTBS N H 20 21 trang 12/19 18, 19, 20 hay 21 trang 13/19 III.3 a N N N N N H N H CO2H 22 CO2Et CO2Et CO2Et 23 24 22, 23 hay 24 b ( 22 hoá phenylendiamin), proton hoá benzo H N N HCl N H CO2H N H CO2H mà không v c K2CO3 + K2CO3 anion amid NHO - sinh anion HO gây - -be N N K2CO3, DMF N H H-OH N CO2Et CO2Et N CO2- N H HO ClCH2CO2Et H2O HOCH2CO2 2CO3 òng benzoimidazol - 1/8 2CO3 2CO3 Câu IV IV.1 25 30 trang 14/19 BnO OBn BnO t PCC OH 26 Ph3P CHCO2 Bu n NIS 27 O 28 Bu3SnH MeOH 29 AIBN MeO 26 31 1) 32, BF3 Et2O, < 25 oC O Me H2, (Ph3P)3RhCl 33 Ph3PMeBr, THF 34 t benzen, -25 oC 2) KOH 15%, MeOH 1) SeO2,tBuOOH 36 1) Ac2O, Et3N H2O2, NaOH 37 2) PDC, tBuOOH 2) NaBH4, CeCl3 NaBH4 O 38 38' MeOH, 25 oC p-TsCl, Et3N 39 O OH OsO4, NMO 40 MeOH, 25 oC t 35 BuOK, oC Me Me 38' OMe 30 IV.2 Peribysin (31) 38 O 30 BnO H 25 35 O MeOH Peribysin (31) BuOH C15H24O4 32 (E)-2-metylbut-2-enal a 32 b IV.3 31 33 40 41 to 42 43 to 44 h 45 25 oC C12H10Br2O2 1) KOH, H2O 46 Ag2CO3, Br2 n 47 Bu3SnH AIBN 2) H3O 41 C10H12 41 1: 4; 41 43 (C6H2Br2O2 47 Câu IV sp2 IV.1 OBn OBn OBn BnO CHO BnO CO2 t-Bu 26 BnO O O t-Bu H O 28 O BnO MeO 29 OMe 30 26, 27, 28, 29 30 IV.2 a O H OMe O CO2 t-Bu OMe 27 OBn I O O BnO BnO CuCl, NH4Cl H2, Pd-CaCO3 HCl Pb(OAc)2, quinoline trang 15/19 O 1) MeMgBr 2) HCOCl O HO CHO CHO axit Lewis 2) HOC2H4OH, p-TsOH (xt) HO 32 O 1) Ag2CO3 O HO2C HO2C 2) AgNO3, NH4OH, 3) NH4Cl, H2O O HO O 1) HIO4 O HO 1) OsO4 2) Br2, CCl4 lí Br Br Me 1) LiAlH4 O Me 2) H3O+ CHO 32 lí ; (2) ; (3) - nên 32 - b Me Me O Me Me O Me Me Me Me Me Me Me Me OAc OH Me O 33 Me Me 34 Me Me O 35 Me Me O OH 36 Me Me O 37 HO O OH OH O O OH 38 39 O 40 31 31, 33 40 IV.3 41 , có CTPT C10H12, suy 41 43 (C6H2Br2O2 sp2 suy 43 O Br Br O Br O O Br 42 44 O O Br Br CO2H 45 Br 46 41 Câu V (3,0 V.1 CO2H Br 47 47 -B (48 48 trang 16/19 1) (CF3CO)2O, Et3N OH H2 49 2) 2-metyl-5-nitrophenol; CuCl2 H2O, Et3N Me Me H2 52 C12H15NO4 140 C 1) RCO3H 52 2) KOH Pd(OAc)2, 130 oC 54 xt [Pd], Cs2CO3 xt PtO2 51 o xt Lindlar m-Br-C6H4-OSiiPr3 53 xylen 50 55 Cu(OAc)2, AcOH C27H41NO3Si CHO DDQ 55 Bu4NF 56 O C27H39NO3Si N H HO 49 57 (C17H18O9) 58 (C11H6O3) D-glucose V.2 59 1) BnBr, K2CO3 2) Br2, xt 2) O Me Me 56 -glycosidase 57 58 1) Mg/Et2O 60 Furoclausine-B (48) OH 61 1) H2, Pd/C 1) CO, HCl 2) SOCl2, Et3N ; 3) H3O Pd/C 62 63 C8H8O2 58 to 2) Ac2O, AcONa, to 59 (C6H6O2 57 (C11H8O4 a b V.3 64 -glycosidase 64 65 58 58 63 57 66) 57 64 65 58 66 - 66 20,42% C, 2,57 65 aldopyranose (67 68 68 -NCS) 69 2:1:2 equatorial 69 68 thành 69 C = 12,01; H = 1,01; N = 14,01; O =16,00; S = 32,06 67 a b 66 V.1 Me O Me Me O Me Me Me Me OH OH Me Me O Me Me NO2 NO2 49 NO2 50 Me NO2 51 52 Me O Me O OH Me Me TIPSO OH Me Me NH NH2 53 54 (TIPS: iPr3Si-) trang 17/19 Me CHO O O OH N H TIPSO OH Me Me N H TIPSO 55 Me Me 56 49 c 56 V.2 a Do 59 (C6H6O2 59 là: HO OH O OH OH Br BnO OBn BnO 60 OBn 61 O 62 O O O O 63 O 58 57 là: 58 O HO CH2 H O HO HO O HOOC H H OH 57 63 57 b 57 64 65 58: COOH H2O, to O 57 COOH O Glucose O 64 COOH enzym O O Glucose OH 65 H3O+, to O O O -H2O O COOH OH 58 Thí sin 57 64 65 58 V.3 a 67 equatorial 67 là: trang 18/19 OH O HO HO OH OH 66, 68 69 là: 66: C10H16KNO9S2 68: C4H6KNO4S2 69: C4H5NS Isothiocyanat 69 69 là: N C S 69 tách KHSO4, suy 68 68 là: HS N O SO3K 66 là: OH O HO HO OH S N O SO3K 66, 68 hay 69 trúc 66 69 68 thành 69: b H S KHSO4 S N H O SO3K N C S N O SO3K 69 68 Câu VI VI.1 a 3, , b So sánh c pha khí d PCl5 ion [PCl4]+ [PCl6] PCl3, NCl3 3, PBr5 PCl5 + 4] [PCl6] V PCl3 NCl3 trang 19/19 , VI.2 2+ (1) Hb + O2 (2) Hb + CO a G b HbO2 HbCO : K1 = 60 K2 = 1,5.104 B = 5, ZN = 7, ZP = 15, ZCl = 17, ZFe = 26, ZBr = 35; atm VI.1 (20/8 a Chóp tam giác b So = 120o P-Cl, NCl3 AX3E1 Chóp tam giác (6/8 3, PCl3, NCl3 ) -Cl (4/8 rP = 1,07 Å; rCl = 1,02 Å, rBr = 1,20 Å) (4/8 (Bá 4] ) chóp tam giác c d (6/8 AX3E0 3 + [PCl6] Cl Cl ) Cl Cl P - Cl Cl P Cl Cl Cl Cl [PCl4] [PCl6] VI.2 (12/8 a (4/8 +P +P + b (8/8 (1) Hb + O2 (2) Hb + CO nên HbO2 HbCO K1 = 60 K2 = 1,5.104 trang 20/19 2] = b CO: o [HbCO] = 0,25a; [Hb] = 0,75a-b K1 K2 b (0, 75a - b)PO2 0, 25a (0, 75a - b)PCO %CO 0, 030% b 0, 69485a pCO 3, 022.10 (atm) (4/8 ) (2/8 ) (2/8 ) 0,5 C * trang 21/19 ... Kw h (*) h h=5,4875.10-9 = 10-8,26M hay pH = 8,26 S2019 = 5,14.10-4 (mol·L 1) S1950 = 4,80.10-4 (mol·L 1) S1900 = 4,71.10-4 (mol·L 1) - (0,25 -2019: S1= 3,4.10-5 tan thêm: n = S.V = 3,4.10-5 5.1010...K Môn: 13/01 /2019 hai thi c Câu I I.1 a b c lk ,n plk plk lk n=n -n ( lk nlk plk plk ( anion O2 d 2 O2 trình... 5.1010 = 1,70.106 (mol) 6 (0,25 tan thêm: m = 1,7.10 100 (g) = 170.10 (g) = ii) Giai -1 ( km-3) - -2019 -1950 tan trung bình là: 3,4.10-5.100.1012-6/69 = 49,3 (0,25 tan trung bình: 9.10-6.100.1012-6/50