www.tuhoc.edu.vn bão hòa H2N-R-COOH + NaOH H2N-R-COONa + H2O Amino axit H2N-R-COOH axit H2N-R-COOR' nhóm este H2N-R-COOR' + NaOH H2N-R-COONa + R'OH RCOONH3R' + NaOH RCOONa + RNH2 + H2O cacboxylat RCOONH3R' k k 3) VD1: 4H9O2 X + NaOH Y + HCl (dư) Y + CH4O Z + NaCl A) H2NCH2CH2COOCH3 CH3CH(NH3Cl)COOH B) CH3CH(NH2)COOCH3 CH3CH(NH3Cl)COOH C) CH3CH(NH2)COOCH3 CH3CH(NH2)COOH D) H2NCH2COOC2H5 ClH3NCH2COOH CH4O CH3OH X + NaOH CH3OH (ancol) H2N–R–COOCH3, Y H2N–R–COONa Z ClH3N–R–COOH 3H9O2 A) B) C) D) htttp://tuhoc.edu.vn/blog www.tuhoc.edu.vn E2 3H7O2N Khi 3H6O2Na E2 cho 1, E2 2H4O2 1, E2 5H11O2 2H4O2NNa TOPPER Chú ý 3H7O2 2 X2 Khi cho X2 2H8O3N2 CH3CH(NH2)COOH HNO2 X H2SO 170 180 oC Y Br Z(C3H4O2Br2) htttp://tuhoc.edu.vn/blog www.tuhoc.edu.vn VD2: C2H7NO2 C2H7NO2 có k = 3COONH4 CH3COONH4 + NaOH HCOONH3CH3 + NaOH 3, dZ CH3COONa + H2O + NH3 HCOONa + H2O + NH2CH3 CH3NH2 13, 75 H2 nz = x + y = 0,2 mol MZ x y MZ = 27,5 CH3NH2 (Z = 31) | 31 27,5 | | 17 27,5 | nCH3COONa nNH3 nHCOONa nCH3NH2 0, 05 (mol) 0,15 (mol) m 14,3 (gam) 4H9NO2 B) 9,4 C) 8,2 Cho 10,3 D) 9,6 htttp://tuhoc.edu.vn/blog x = 0,05; y = 0,15 A) 10,8 27,5 NH3 (Z = 17) , HCOONH3CH3 A) 3,56 NH3 B) 5,34 C) 2,67 D) 4,45 www.tuhoc.edu.vn VD3: xHyOzNt Trong phân z)NH3 A) CH3COONH4 C) H2NCH2COOCH3 X + HCl R(Oz)NH3Cl B) H2NCH2COOH D) CH3CH(NH2)COOH MX = 14 : mO 75 42, 67 100 18, 67 100 75 32 xHyO2N MX = 12x + y + 32 + 14 = 75 12x + y = 29 x y B A) C2H5O2 C) C3H7O2 A R(Oz)NH3 2SO4 A) CH2=CHCOONH4 C) H2N(CH2)3COOH B) C3H7O2 D) C4H9O2 xHyOzNt B) H2NCH2COOH D) CH3CH(NH2)COOH htttp://tuhoc.edu.vn/blog www.tuhoc.edu.vn VD4: 3H9O2 X B) CH3COONH3CH3 D) HCOONH2(CH3)2 A) HCOONH3CH2CH3 C) CH3CH2COONH4 3H9O2N MX = 91 MRCOOR' = MR + 44 + MR' = 91 nX = 1, 82 91 MR + MR' = 47 0, 02 (mol) RCOONa + R'OH RCOONa = nRCOOR' = 0,02 mol MRC OONa 1, 64 0, 02 82 MR 67 82 MR 15 MRCOONa 3) Thay vào MR' = 32 –NH3CH3 B A) CH2=CHCOONH4 C) H2NCH2COO-CH3 B) H2NCOO-CH2CH3 D) H2NC2H4COOH C 3H7O2N A) HCOOH3NCH=CH2 C) CH2=CHCOONH4 B) H2NCH2CH2COOH D) H2NCH2COOCH3 htttp://tuhoc.edu.vn/blog www.tuhoc.edu.vn VD5: gam CO2; 4,5 gam H2O 1,12 lít khí N2 A) H2NCH2COOH C) CH3CH(NH2)COOH nCO B) CH3COONH4 D) H2NCH2COOCH3 0, (mol); nH O 0, 25 (mol); nN 0, 05 (mol) nC 0, (mol) mC 2, (mol) nH 0,5 (mol) mH 0,5 (mol) nN 0,1 (mol) mN 1, (mol) NC : NH : N N : N O mO 3, (gam) 0, : 0,5 : 0,1 : 0, nO 0, (mol) 2:5:1:2 tØ lÖ tèi gi¶n 2H5NO2 2H5NO2)n nHCl nX 2H5NO2 A D 2, nCO nH O H2O N2 A) B) C) D) E 8,8 gam CO2, 6,3 gam H2O 1,12 lít N2 A) C2H5NO2 B) C2H7NO2 C) C2H7NO2 D) C3H9NO2 htttp://tuhoc.edu.vn/blog www.tuhoc.edu.vn F A) C2H5O2N C) C3H9O2N B) C3H7O2N D) C4H9O2N G A) H2NCH2COOH H2N(CH2)2COOH B) H2N(CH2)3COOH H2N(CH2)2COOH C) H2NCH2COOH H2N(CH2)3COOH D) H2N(CH2)5COOH H2N(CH2)6COOH H CO2, 0,56 lít khí N2 2N-CH2-COONa A) H2N–CH2–COO–C3H7 C) H2N–CH2–CH2–COOH B) H2N–CH2–COO–CH3 D) H2N–CH2– 2H5 I O : mN 2, A) 13 gam C) 15 gam H2O N2 B) 20 gam D) 10 gam htttp://tuhoc.edu.vn/blog www.tuhoc.edu.vn Câu Câu A D B C C D D B E B B F A C G A B H B I A C E1 = CH3CH(NH2)COOH, E2 = H2NCH2COOCH3 CH3CH(NH2)COOH + NaOH CH3CH(NH2)COONa + H2O H2NCH2COONa + CH3OH H2NCH2COOCH3 + NaOH X = H2NCH2COOCH(CH3)2 H2NCH2COOCH(CH3)2 + NaOH (CH3)2CHOH + CuO t o H2NCH2COONa + (CH3)2CHOH (CH3)2CO + Cu + H2O X = CH2=CHCOONH4 CH2=CHCOONH4 + H2 CH3CH2COONH4 + HCl CH3CH2COOH + NH3 Ni to CH3CH2COONH4 CH3CH2COOH + NH4Cl CH3CH2COONH4 X = CH3CH2NH3NO3, Y = CH3CH2NH2 CH3CH2NH3NO3 + NaOH CH3CH2NH2 + NaNO3 + H2O X = CH3CH(OH)COOH, Y = CH2=CHCOOH, Z = CH2BrCHBrCOOH htttp://tuhoc.edu.vn/blog