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Baøi 5

Baøi 5. röôïu Phaûn öùng taùch nöôùc . GV. NGUYEÃN TAÁN TRUNG (Trung Taâm Luyeän Thi Chaát Löôïng Cao VÓNH VIEÃN). Caùc phaûn öùng taùch nöôùc. Coù 3 loaïi sau:. Taïo Anken (olefin). Taïo eâte. Taïo saûn phaåm ñaëc bieät. Hoaëc Al 2 O 3 , t o 400 o C. hôi röôïu.

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Baøi 5

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  1. Baøi 5 röôïu Phaûn öùng taùch nöôùc GV. NGUYEÃN TAÁN TRUNG (Trung Taâm Luyeän Thi Chaát Löôïng Cao VÓNH VIEÃN)

  2. Caùc phaûn öùng taùch nöôùc Coù 3 loaïi sau: • Taïo Anken (olefin) • Taïo eâte • Taïo saûn phaåm ñaëc bieät

  3. Hoaëc Al2O3, to 400oC hôi röôïu H2SO4ñ 170OC Ví duï: C2H5OH + H2O C2H4 CAÙC PHAÛN ÖÙNG TAÙCH H2O Taùch H2O taïo OLEÂFIN (ANKEN): • Ñieàu kieän röôïu: Röôïu ñôn chöùc, no, Soá C  2 • Ñieàu kieän phaûn öùng: • H2SO4ñ , 170OC • Phaûn öùng: ÑKpöù CnH2n + CnH2n+1OH H2O ( n2)

  4. X Cl2,to Br2 NaOH -H2O  B A -D G (4) (1) (2) (3) Y Bieát: -X,Y: hôïp chaát ñôn chöùc, CH2-CH-CH2 - A laø chaát khí duy nhaát Cl Br Br - G: Glyxerin D: • Aùp duïng 1:( Trích ÑHDL NNTH - 2000) Hoaøn thaønh caùc phaûn öùng theo sô ñoà sau: Glyxerin

  5. Aùp duïng 1:( Trích ÑHDL NNTH - 2000) Hoaøn thaønh caùc phaûn öùng theo sô ñoà sau: X Cl2,to Br2 NaOH -H2O  B A -D G (4) (1) (2) (3) Y Glyxerin Bieát: - X,Y: hôïp chaát ñôn chöùc CH2-CH-CH2 - A laø chaát khí duy nhaát Cl Br Br D: CH2-CH = CH2 D: A: CH3-CH = CH2 Cl

  6. Aùp duïng 1:( Trích ÑHDL NNTH - 2000) Hoaøn thaønh caùc phaûn öùng theo sô ñoà sau: X Cl2,to Br2 NaOH -H2O  B A -D G (4) (1) (2) (3) Y Glyxerin Bieát: - X,Y: hôïp chaát ñôn chöùc CH2-CH-CH2 - A: CH3-CH = CH2 Cl Br Br D: X: CH2-CH-CH3 OH CH2-CH = CH2 D: Y: CH3-CH2-CH2-OH Cl

  7. Ñun röôïu A ñôn chöùc , no vôùi H2SO4ñaëc ; thu ñöôïc chaát höõu cô B, vôùi dB/A=1,7 Tìm CTPT-CTCT cuûa A; B. • Aùp duïng 2: A ñôn chöùc , no Sai Thí Sinh: B laø olefin

  8. H2SO4ñ X Olefin to X: Röôïu ñôn, no  Olefin H2SO4 ñ Röôïu ñôn, no to EÂte • Caàn nhôù: ( Do moïi röôïu taùch nöôùc ñeàu coù theå taïo eâte)

  9. Hoaëc Al2O3, to 200oC hôi röôïu CAÙC PHAÛN ÖÙNG TAÙCH H2O Taùch H2O taïo EÂTE: • Ñieàu kieän röôïu: Moïi Röôïu • Ñieàu kieän phaûn öùng: • H2SO4ñ , 140OC • Phaûn öùng: Phuï thuoäc chöùc röôïu !

  10. CAÙC PHAÛN ÖÙNG TAÙCH H2O ÑKPÖÙ ÑKPÖÙ + R-O-R’ Taùch H2O taïo EÂTE: • Röôïu ñôn H2O + HO-R’ R-OH • Röôïu ña m R(OH)n + n R’(OH)m Rm-(O)n.m-R’n + n.m H2O

  11. CnH2n H2SO4ñ CnH2n+1OH to (CnH2n+1)2O => • Toùm laïi caàn nhô:ù ( Röôïu ñôn chöùc, no) (*) (*) => Molefin<Mröôïu<Meâte dolefin/röôïu<1 deâte/röôïu >1

  12. Vôùi ñôn chöùc, no H2SO4ñ Röôïu X Saûn phaåm Y • Neáu dY/X <1  Y: olefin • Neáu dY/X >1 Y: ete

  13. Ñun röôïu A ñôn chöùc , no vôùi H2SO4ñaëc ; thu ñöôïc chaát höõu cô B, vôùi dB/A=1,7 Tìm CTPT-CTCT cuûa A; B. • Aùp duïng 2: A ñôn chöùc , no B:EÂte Thí Sinh: B laø olefin

  14. PP tìm CTPT döïa treân pöù B1.Ñaët CTTQ B2.Vieát pöù B3.Laäp pt (*) B4.Giaûi (*) • Toùm taét: H2SO4 ñ (B) Röôïu (A) (Ñôn, no) to dB/A=1,7 (A); (B) ? Vì dB/A=1,7 >1 (B):EÂte ÑaëT CTTQ (A): CnH2n+1OH

  15. H2SO4ñ to=140 MB dB/A= = 1,7 MA Phaûn öngù: CnH2n+1OH (CnH2n+1)2O + H2O (1) 2 (1) (B): (CnH2n+1)2O Theo ñeà baøi ta coù: (14n + 1).2 +16 1,7 =  14n +18  n = 3 Vaäy :(A):C3H7OH ; (B): C3H7O-C3H7

  16. Ñun röôïu A ñôn chöùc , no vôùi H2SO4ñaëc ; thu ñöôïc chaát höõu cô B, vôùi dB/A=0,7 Tìm CTPT-CTCT cuûa A; B. • Aùp duïng 3: A ñôn chöùc , no B: olefin

  17. PP tìm CTPT döïa treân pöù B1.Ñaët CTTQ B2.Vieát pöù B3.Laäp pt (*) B4.Giaûi (*) • Toùm taét: H2SO4 ñ (B) Röôïu (A) (Ñôn, no) to dB/A=0,7 (A); (B) ? Vì dB/A=0,7 < 1 (B):olefin Röôïu (A): (Ñôn, no) ÑaëT CTTQ (A): CnH2n+1OH

  18. H2SO4ñ to=170 MB dB/A= = 0,7 MA Phaûn öngù: CnH2n+1OH CnH2n + H2O (1) 2 (1) (B): CnH2n Theo ñeà baøi ta coù: 14n 0,7 =  14n +18  n = 3 Vaäy :(A):C3H7OH ; (B): CH3-CH=CH2

  19. Ñun röôïu A coù MA<120 ñvC. vôùi H2SO4ñaëc ; thu ñöôïc chaát höõu cô B, vôùi dB/A=1,419 Tìm CTPT-CTCT cuûa A. • Aùp duïng 4: B:EÂte

  20. PP tìm CTPT döïa treân pöù ÑaëT CTTQ (A): R(OH)n B1.Ñaët CTTQ B2.Vieát pöù B3.Laäp pt (*) B4.Giaûi (*) • Toùm taét: H2SO4 ñ (B) Röôïu (A) MA<120 to dB/A=1,419 (A) ? Vì dB/A=1,419 > 1 (B):eâte H2SO4ñ (1) 2R-(OH)n R-On- R nH2O (B) 2R+16n dB/A= 1,419 R+17n

  21. Ñun 132,8 g hh X:AOH;BOH;ROH vôùi H2SO4 ñ ôû 140oC ta thu ñöôïc 11,2g hh goàm 6 eâte coù soá mol baèng nhau. Maëc khaùc ñun noùng hh X vôùi H2SO4 ñ ôû 170oC thì thu ñöôïc hh Y chæ goàm coù 2 Olefin khí (ôû ñieàu kieän thöôøng). • Xaùc ñònh CTPT-CTCT cuûa caùc röôïu, (H=100%) • Tính % (theo m) cuûa hh X. • Tính %(theo m) cuûa hh Y.

  22. Ñun röôïu A vôùi H2SO4ñ; thu ñöôïc chaát höõu cô B, vôùi dB/A=0,6086 Tìm CTPT-CTCT cuûa A; B. Bieát MA 90 ñvC

  23. GK: C2H5OH C2H5ONa C2H5Cl CH3CHO C2H4 C2H5OH CH3-COOC2H5 C2H3COOH CH2=CH-CH=CH2 Glucozô

  24. Y X Z D C2H6O I A B E G

  25. H2SO4 ñ Röôïu A (B) to dB/A=1,419 MA<120 (A); (B) ? Vì dB/A=1,419>1 => B: EÂte Ñaët CTTQ (A): R-(OH)n H2SO4ñ (1) + 2R-(OH)n R-On- R nH2O (B) 2R+16n = dB/A= 1,419 R+17n

  26. => R = 14n < MA = R+17n 120 => n < 3,87 => n = 1;2;3 n=2 => A: C2H4(OH)2 GV. NGUYEÃN TAÁN TRUNG (Trung Taâm Luyeän Thi Chaát Löôïng Cao VÓNH VIEÃN)

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