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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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slide1

Baøi 5

röôïu

Phaûn öùng taùch nöôùc

GV. NGUYEÃN TAÁN TRUNG

(Trung Taâm Luyeän Thi Chaát Löôïng Cao VÓNH VIEÃN)

slide2

Caùc phaûn öùng taùch nöôùc

Coù 3 loaïi sau:

  • Taïo Anken (olefin)
  • Taïo eâte
  • Taïo saûn phaåm ñaëc bieät
slide3

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)

slide4

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

slide5

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

slide6

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

slide7

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

slide8

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)

slide9

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 !

slide10

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

slide11

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

slide12

Vôùi ñôn chöùc, no

H2SO4ñ

Röôïu X

Saûn phaåm Y

  • Neáu dY/X <1  Y: olefin
  • Neáu dY/X >1 Y: ete
slide13

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

slide14

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

slide15

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

slide16

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

slide17

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

slide18

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

slide19

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

slide20

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

slide21

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.
slide22

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

slide23

GK: C2H5OH

C2H5ONa

C2H5Cl

CH3CHO

C2H4

C2H5OH

CH3-COOC2H5

C2H3COOH

CH2=CH-CH=CH2

Glucozô

slide24

Y

X

Z

D

C2H6O

I

A

B

E

G

slide25

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

slide26

=> 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)