Rượu - Phản ứng tách nước

8/13/2019 Ru - Phn ng tch nc

1/26

Bai 5

GV. NGUYEN TAN TRUNG(Trung Tam Luyen Thi Chat Lng Cao VNH VIEN)


2/26

Tao Anken (olefin)Tao eteTao san pham ac biet

Co 3 loai sau:


3/26

CAC PHAN NG TACH H2O

Tach H2O tao OLEFIN (ANKEN):ieu kien ru:

Ru n chc, no, So C 2ieu kien phan ng: H2SO4, 170OC

Phan ng:

CnH2n+1OH CnH2n + H2O

Kp

Hoac Al2O3, to 400oChi ru

( n 2)V du: C2H5OH C2H4+ H2O

H2

SO4

170OC


4/26

Hoan thanh cac phan ng theo s o sau:

Biet:

-X,Y: hp chat n chc,- A la chat kh duy nhat

- G: Glyxerin

A

X

Y

-H2O

B

Cl2,to Br2 NaOH

-D G(1) (2) (3) (4)

Ap dung 1:( Trch HDL NNTH - 2000)

Glyxerin

CH2-CH-CH2

Cl Br Br

D:


5/26

- X,Y: hp chat n chc


Biet:

- A la chat kh duy nhat

A

X

Y

-H2O

B

Cl2,to Br2 NaOH

-D G(1) (2) (3) (4)


Glyxerin

CH2-CH-CH2Cl Br Br

D:

ClCH2-CH = CH2D: CH3-CH = CH2A:


6/26

- X,Y: hp chat n chc


Biet:

- A: CH3-CH = CH2

A

X

Y

-H2O

BCl2,to Br2 NaOH

-D G(1) (2) (3) (4)


Glyxerin

CH2-CH-CH2Cl Br Br

D:

ClCH2-CH = CH2D: OH

CH2-CH-CH3X:

Y: CH3-CH2-CH2-OH


7/26

un ru A n chc , novi H2SO4ac ; thu c

chat hu c B,vi dB/A=1,7Tm CTPT-CTCT cua A; B.

Ap dung 2:

A n chc , no

Th Sinh: B la olefin Sai


8/26

X H2SO4to Olefin

X: Ru n, no

Ru n, no

Olefin

EteH2SO4

to

( Do moi ru tach nc euco the tao ete)

Can nh:


9/26


Tach H2O tao ETE:ieu kien ru:

Moi Ruieu kien phan ng: H2SO4, 140OC

Phan ng:

Hoac Al2O3, to 200oChi ru

Phu thuoc chc ru !


10/26

Tach H2O tao ETE:


Ru a

R(OH)n

R(OH)m

+

Rm-(O)n.m-Rn+ H2On.m

m n

R-OH HO-R+ KP +R-O-R H2O

Ru n

KP


11/26

CnH2n+1OHCnH2n

(CnH2n+1)2O

H2SO4

to

Tom lai can nh: ( Ru n chc, no)

(*) =>Molefin


12/26

u X San pham YH2SO4 Neu dY/X 1 Y: ete

Vi n chc, no


13/26



Ap dung 2:

A n chc , no

Th Sinh: B la olefin

B:Ete


14/26

Ru (A)(n, no)

H2SO4 to (B)

(A); (B) ? dB/A=1,7V dB/A=1,7 >1

(B):EteaT CTTQ (A):CnH2n+1OH

Tom tat: PP tm CTPT da tren p

B1.at CTTQB2.Viet pB3.Lap pt (*)B4.Giai (*)


15/26

CnH2n+1OH2 (CnH2n+1 )2O + H2O (1)H2SO4to=140

(1) (B): (CnH

2n+1)2O

Theo e bai ta co:

dB/A=M

BMA = 1,7

(14n + 1).2 +16

14n +18= 1,7

n = 3

Vay :(A):C3H7OH ; (B): C3H7O-C3H7

Phan ng:


16/26



Ap dung 3:

A n chc , no

B: olefin


17/26

Ru (A)(n, no)

H2SO4 to (B)

(A); (B) ? dB/A=0,7V dB/A=0,7 < 1Ru (A): (n, no)

(B):olefin

aT CTTQ (A):

CnH2n+1OH




18/26

CnH2n+1OH2 CnH2n + H2O (1)H2SO4to=170

(1) (B): CnH2nTheo e bai ta co:dB/A=

MBMA = 0,7

14n14n +18

= 0,7

n = 3

Vay :(A):C3H7OH ; (B): CH3-CH=CH2

Phan ng:


19/26

un ru A co MA


20/26

Ru (A)MA 1(B):eteaT CTTQ (A):R(OH)n



2R-(OH)n R-On- R nH2OH2SO4 (1)

(B)

dB/A= 2R+16nR+17n1,419


21/26

un 132,8 g hh X:AOH;BOH;ROH viH2SO4 140oC ta thu c 11,2g hhgom 6 ete co so mol bang nhau. Mac

khac un nong hh X vi H2SO4 170oC th thu c hh Y ch gom co 2Olefin kh ( ieu kien thng).

a. Xac nh CTPT-CTCT cua cac ru,(H=100%)

b. Tnh % (theo m) cua hh X.

c. Tnh %(theo m) cua hh Y.


22/26

un ru A vi H2SO4; thu c

chat hu c B, vi dB/A=0,6086Tm CTPT-CTCT cua A; B. Biet

MA 90 vC

GK: C2H5OH


23/26

GK: C2H5OHC2H5ONa

C2H5OH CH3-COOC2H5C2H4C2H5Cl CH3CHO

C2H3COOHGlucozCH2=CH-CH=CH2


24/26

C2H6O I

GE

D

B A

XY

Z

H2SO4


25/26

Ru A (B)H2SO4 to

MA1

=> B: Eteat CTTQ (A): R-(OH)n

2R-(OH)n R-On- R+nH2O

H2SO4

(1)(B)

dB/A=2R+16nR+17n = 1,419

=> R = 14n


26/26

=> R = 14nMA = R+17n < 120

=> n < 3,87=> n = 1;2;3

n=2 => A: C2H4(OH)2

GV. NGUYEN TAN TRUNG(Trung Tam Luyen Thi Chat Lng Cao VNH VIEN)

Documents

Rượu - Phản ứng tách nước