bài tập Axit-Este

7/30/2019 bi tp Axit-Este

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Li ni u

Tp Bi tp axiteste c nhm bin son nhm mc ch emn cho bn c kin thc vphng php gii cc loi bi tp trongphn ni dung axit v este, ng thi gip cc bn rn luyn k nng gii

bi tp ca mnh.

Bi tp trong ti liu ny c chia thnh 2 loi: trc nghim v tlun, ph hp vi nhu cu chung hin nay, vi h thng bi tp c phndng v hng dn gii, hy vng gip cc bn trong qu trnh luyntp.

Nhm chng ti c gng kim tra v khc phc nhng sai st,

tuy nhin, mt s li nh vn l iu kh lng trnh khi, do , hy vngnhn c s gp ca cc bn.

Nhm tc gi


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PHN I: AXIT CACBOXYLICDNG 1:HON THNH S PHN NG

V d 1:

Gii:

1) C3H8Cracking CH4 + C2H4

(X) (B) (A)

2) C2H4 + Br2 C2H4Br2

(A) (A1)

3) C2H4Br2 + 2NaOH CH2 - CH2 + 2NaBr

(A1) OH OH(A2)

4) CH2 - CH2 + 2CuO CHO + 2Cu + 2H2O

OH OH CHO

(A3)

5) CHO + 4[Ag(NH3)2]OH to COONH4 + 4Ag + 6NH3 + 2H2O

CHO COONH4

(A4)

Hoc vit:

CHO + 4Cu(OH)2 + 2NaOH to COONa + 2Cu2 + 6H2O


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1500oC

dd KMnO4

as

as

CHO COONa

6) COONH4 + 2H2SO4 COOH + 2(NH4)2SO4

COONH4 COOH

(A5)

7) 2CH4 lm lnh nhanh C2H2 + 3H2

(B) (B1)

8) C2H2 + 4[O] COOH

COOH

V d 2:

Gii:

1) C2H5COOH +Cl2 CH3 - CH - COOH + HCl

Cl (A)

2) C2H5COOH +Cl2 CH2 - CH2 - COOH + HCl

Cl (B)

3) CH3 - CH - COOH + NaOH CH3CHCOONa + H2O + NaCl

Cl (A) OH (A1)

4) 2CH3 - CH - COONa + H2SO4 2CH3 - CH - COOH + Na2SO4

OH OH (A2)


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H2SO4

H2SO4

5) CH3 - CH - COOH H2 = CH - COOH = H2O

OH (C)

6) CH2 - CH2 - COOH + 2NaOH CH2 - CH2 - COONa + NaCl + H2O

Cl (B) OH (B1)

7) 2CH2 - CH2 - COONa + H2SO4 2CH2 - CH2 - COOH + Na2SO4

OH (B1) OH (B2)

8) CH2 - CH2 - COOH CH2 = CH - COOH + H2O

OH (B2) (C)

9) CH2 = CH - COOH + ROH CH2= CH - COOR + H2O(C) (D)

10) nCH2 = CH - COOR -CH2 -CH-

COOR n

DNG 2:IU CH CC AXIT HU C

V d: a) Vit s cc phn ng iu ch CH3COOH, Cl2CHCOOH, ClCH2COOH v

BrCH2COOH i t vi, than , mui n... Sp xp cc axit trn theo trnh t tng dn tnh

axit v gii thch.

b) Ngi ta iu ch cht dit c 2,4, 5 Cl3C6H2OH2COOH (hay 2, 4, 5 T) bng cch cho

1,2, 4, 5 Cl4C6H2 vo dung dch NaOH (trong ru) ri cho thm Cl - CH2COOH. Dng cng

thc cu to vit s cc phn ng. V sao hn hp da cam (thnh phn chnh l 2, 4, 5T v

2, 4 D) m M dng trong chin tranh Vit Nam gy tc hi rt ln n con ngi?

Gii:

a) * S iu ch:

NaCl .p.n.c1

2Cl2 + Na


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* Sp xp theo trt ttnh axit tng dn:

Cl

H - CH2COOH < CH2 - COOH < CH2 - COOH < CH - COOH

Br Cl Cl

Gii thch:

- Khc nhau vm in: H< Br < Cl

- Khc nhau v s nguyn t clo.

b) * S phn ng:


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* c hi ch yu v cht dioxin

DNG 3: NHN BIT CC AXIT HU C

V d:

1. Cn dng cc phn ng ha hc g phn bit 4 cht:

CH3COOH, HCOOH, CH2=CH-COOH, CH3-CHO

2. Cn dng cc phn ng g phn bit 6 cht lng: ru propylic, ru isopropylic,

axetanehit, glixrin, ete etylic v axit axelic.Gii:

1. a) Ly mi cht mt t cho tc dng vi mui cacbonat (v d Na2CO3), cht no khng

cho kh thot ra l anehhit CH3CHO.

b) Ly mi axit mt t cho tc dng vi AgNO3 trong amonlac, ch c axit fomic cho

phn ng trng gng:

HCOOH + 2AgNO3 + 3NH3+ H2O (NH4)2CO3 + 2NH4NO3 + 2Ag

c) Cho mt t nc brom vo 2 axit cn li; axit acrylic lm mt mu nc brom:

C=2 = CH - COOH + Br2 CH2Br - CHBr - COOH

2. Trc ht dng qu tm hoc mui cacbonat hoc kim loi nhn bit axit axetic.

- Qu tm bin thnh .


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- Na2CO3 + 2CH3COOH 2CH3COONa + H2O + CO2

- 2CH3COOH + Fe Fe(CH3COO)2 + H2

- Dng Cu(OH)2 bit glixerin:

- Dng phn ng trng gng nhn bit axetanehit

CH3 - CHO + 2AgNO3 + 3NH3 + H2O CH3 - COONH4 + 2NH4NO3 + 2Ag

- Cho Na kim loi vo 3 cht cn li, ete khng tc dng vi Na nn khng c kh thot

ra:

2CH3 - CH2 - CH2OH + 2Na 2CH3 - CH2 - CH2ONa = H2

2CH3 - CH - OH + 2Na 2CH3 - CH - Ona + H2

CH3 CH3

- phn bit 2 ru, trc ht cn oxi ha chng bng CuO, sau ly sn phm thc

hin trng gng:

CH3 - CH2 - CH2OH + CuO CH3 - CH2 - CHO - Cu + H2O

CH3 - CH - OH + CuO CH3 - C = O + Cu + H2O

CH3 CH3

CH3 - CH2 - CHO + 2AgNO3 + 3NH3 + H2O

CH3 - CH2 - COONH4 + 2NH4NO3 + 2Ag

CH3 - C - CH3 khng tham gia phn ng trng gng

O


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DNG 4:XC NH CNG THC PHN TV CNG THC CU TO CA CCAXIT HU C

V d

1:Khi ha hi 1,0g axit hu c n chc no (A) ta c mt th tch va ng bng thtch ca 0,535g oxi trong cng iu kin.

Cho mt lng d A tc dng vi 5,4g hn hp hai kim loi M v M thy sinh ra 0,45

mol hiro. T l smol M i vi M trong hn hp l 3:1. KLNT ca M bng 1/3 KLNT ca

M. Ha tr ca M l II, ha tr ca M l III. Este ca A i vi mt ru n chc no lu b

thy phn mt phn. trung ha hn hp sinh ra t 18,56g este ny phi dng 20ml dung dch

NaOH 0,50M v xphng ha lng este cn li phi dng thm 300ml dung dch NaOH ni

trn.

a) Xc nh KLPT v cng thc cu to ca axit.

b) Vit cc phng trnh phn ng xy ra.

c) Xc nh KLNT ca hai kim loi.

d) Xc nh cng thc phn t v vit cc cng thc cu to c th c ca este.

e) Vit cng thc cu to ca ru, bit rng khi oxi ha khng hon ton ru sinh ra

anehit tng ng, c mch nhnh.

Gii:

a) Cc kh (hi) trong cng iu kin c thtch nh nhau th cng c s mol bng nhau.

0,535g oxi ng vi0,535

32=

1

6mol oxi

Vy: 1g A ng vi1

60mol A. Suy ra KLPT ca A bng 60 .v.C

Bit A l axit no n chc, ta c:

CnH2n+1 - COOH = 60

V COOH = 45

Do CnH2n+1 = 60 - 45 = 15

Vy n = 1


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l axit axetic CH3COOH

b) Cc phng trnh phn ng:

Tc dng vi kim loi:

2CH3COOH + M M(CH3COOH)2 + H2 (1)

6CH3COOH + 2M 2M(CH3COOH)3 + 3H2 (2)

Thy phn este c ru CmH2m+1OH v axit axetic:

CH3COOCmH2m+1 + H2O CH3COOH + CmH2m+1OH (3)

Trung ha axit:

CH3COOH + NaOH + CH3COONa + H2O (4)X phng ha este:

CH3COOCmH2m+1 NaOH CH3COONa + CmH2m+1OH (5)

c) Xc nh KLPT ca kim loi:

Gi x, y l sgam M v M trong hn hp; M v 3M l KLPT ca chng. Ta c:

x + y = 5,4

x

M:

3

y

M= 3

Do : x = y =5, 4

2= 2,7g

Da theo (1) v (2) ta c:

2,7

M+

3.2,7

2.3M=

2,7

M+

2,7

2M= 0,45

Do : M = 9 (berili) v 3M = 27 (nhm)

d) Cng thc cu to ca este:

Theo (3) v (4) mi mol este b thy phn sinh ra mt mol axit tdo v do cn 1 mol

NaOH trung ha, smol dng trung ha bng mol este b thy phn:


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0,5.20

1000= 0,01 mol,

S mol este tham gia x phng ha

0,5.3001000

= 0,15 mol,

Smol este ban u (cha thy phn)

0,15 + 0,01 = 0,16 mol

Phn tlng ca este:

18,56.1

0,16= 116

Ta c th vit:

CH3COOCmH2m+1 = 116

59 + 12m + 2m + 1 = 116 m = 4

Cng thc este: CH3COOC4H9

Cc cng thc cu to c th c ca este:

CH3COO - CH2 - CH2 - CH2 - CH2

CH3COO - CH - CH2 - CH3

CH3

CH3COO - CH2 - CH(CH3)2, CH3COOC(CH3)3,

e) Cng thc cu to ca ru. Tng ng vi cc cu to trn ca este ta c cc cng

thc ca ru.

CH3 - CH2 - CH2 - CH2- OH,


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Trong ch c (CH3)2 CH - CH2- OH l c th boxi ha sinh ra anehit mch nhnh

nh nu trong bi.

V d 2:un hp cht A vi nc (dng axit v c lm xc tc) ta c axit hu c B v ruD, t khi hi ca B so vi nit l 2,57. Cho hi ru D i qua ng nung nng ng bt ng

lm xc tc th sinh ra hp cht E c khnng tham gia phn ng trng gng. t chy hon

ton 2,80g cht A phi dng ht 3,9l oxi (ktc), sn phm chy gm c kh cacbnic v hi nc

theo t l:

2COV = 2

3

2H OV

a) Cho bit A v E thuc nhng loi hp cht g?

b) Xc nh KLPT, cng thc phn t v cng thc cu to ca B.

c) Xc nh cng thc phn t ca D, bit rng D l ru 1 ln ru.

Gii:

a) Cht A l mt este v khi thy phn ta c mt axit hu c v ru.

Cht E l mt anehit v E to thnh khi un nng hi ru c Cu xc tc v v E c kh

nng phn ng trng gng.

b) KLPT ca B = 2,57.28 = 72 .v.C

V khi lng ca 2 nhm COOH = 2.45 = 90 nn B ch c thl axit n chc v cng

thc tng qut ca B l CnHmCOOH.

Suy ra gc CnHm = 72 - 45 = 27

V CnHm= 27 nn n < 3 v nguyn dng, do n ch c th bng 1 v 2

n m CnHm Nhn xt1 15 CH15 - V l v C khng c ha tr qu IV

2 3 C2H3 - ng, y l g c khng no ha tr I: CnH2n-1

Vy, cng thc ca B l C2H3COOH v cng thc cu to ca B l:


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2. V B l axit hai ln axit HOOC-COOH nn khi tc dng vi hn hp CH3OH vC2H5OH c th to thnh 5 este sauCH3OOCCOOH; CH3OOCCOOCH3;C2H5OOCCOOH; C2H5OOCCOOC2H5;CH3OOCCOOC2H5;

MT S BI TP TNG HP:

Bi 1: Cho hai axit cacbonxylic A v B. Nu cho hn hp A v B tc dng ht vi Na thu c

s mol H2 bng tng s mol ca A v B trong hn hp. Nu trn 20 gam dung dch axit A

23% vi 50 gam dung dch axit B 20,64% c dung dch D, trung ha hon ton D cn

200ml dung dch NaOH 1,1M.

a) Tm cng thc cu to ca A v B.

b) un nng hn hp A v B vi ru no X mch h, to ra hn hp cc este trong c

este E. E khng c khnng tc dng vi Na to ra H2. t chy hon ton V lt hi E cn 7,5V

O2 to ra 7V CO2v 5V hi H2O (thtch o cng iu kin nhit , p sut). Tm cng thc

cu to ca E v X.

Hng dn:

a) t cng thc: (A) R1(COOH)n (x mol)

(B) R2(COOH)m (x mol)

(A) R1(COOH)n + nNa R1(COONa)n +2

nH2 (1)

(B) R2(COOH)m + mNa R1(COONa)m +2

mH2 (2)

Theo : 0,5(nx + my) = 0,5(x + y) n = m = 1

(A), (B) u n chc.

mA =23.20

100= 4,6 (gam); mB =

20,64.50

100= 10,32 (gam)

(A) R1COOH + NaOH R1COONa + H2O (3)

(B) R2COOH + NaOH R2COONa + H2I (4)


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T (3, 4) v bi cho:

NNaOH = x + y = 0,2.1,1 = 2,2 (mol)

( )xM =( )

( )

x

x

m

n=

4,6 10,32

0,22

= 67,81

Suy ra: MA < ( )xM < MB

R1 + 45 < 67,81 < R2 + 45 R1 < 22,8

Vy gc (R1-) ch c thl (1) hay (15) ngha l (H-) hay (CH3-). Xt 2 trng hp:

* Nu (A) l HCOOH:

n(B) = 0,22 - 4, 646

= 0,12 (mol)

MB =10,32

0,12= 86

(B) l C3H5COOH (3 ng phn, c 1 ng phn cis-trans)

* Nu (A) l CH3COOH:

n(B) = 0,22 -4, 6

60

=

8, 6

60 (mol)

MB =10,32

8, 660

= 72

(B) l C2H3COOH (hay CH2 = CH - COOH)

b) Cng thc cu to E, X:

t cng thc phn t R l CxHyOz, theo phng trnh phn ng chy:

CxHyOz +24

zyx O2 xCO2 +

2

yH2O

1V 7,5V 7V 5V

T t l v th tch suy ra t l v s mol:


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CxHyOz + 7,5O2 7CO2 + 5H2O

Rt ra: x = 7; y = 10; z = 4

Cng thc phn t E: C7H10O4 (M = 158)

V E khng tc dng vi Na gii phng kh H2 nn E khng cha H linh ng trong -

COOH hoc -OH. Mt khc E c to ra tA, B l 2 axit n chc v ru no X. Nu E c

dng: (RCOO)nR; R, R l gc axit v ru.

E c 4 nguyn t O nn n = 2, cng thc phn t E l: (RCOO)2R.

Ta c: 2R + 88 + R = 158 2R + R = 70

Bin lun:

- Khi R = 1 (HCOOH): 2 x 1 + R = 70 R = 68

Cng thc phn t X: C5H8(OH)2l ru khng no (loi)

- Khi R = 41 (C3H5COOH): 82 + R = 70R < 0 (loi)

- Khi R = 42 (E to bi HCOOH v C3H5COOH)

(41 + 1) + R = 70 R = 28; Cng thc phn t X: C2H4(OH)2

- Khi R = 15 (CH3COOH) suy ra R = 40;

Cng thc phn t X: C3H4(OH)2 (loi)

- Khi R = 27 (C2H3COOH) suy ra R = 16 (loi)

- Khi R = 15 + 27 (E to bi CH3COOH v C2H3COOH)

R = 28; cng thc phn t X: C2H4(OH)2

Vy X l HO-CH2-CH2-OH ; E l HCOO-CH2-CH2-OOC-C3H5 hay CH3COO - CH2 -

CH2 - CH2 - OOC - C2H3.

Bi 2: C 2 dung dch axit hu c no, n chc A, B. Trn 1 lt A vi 3 lt B ta c 4 lt dung

dch D. trung ha 10 ml dung dch D cn 7,5 ml dung dch NaOH v to c 1,335 gam

mui.


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Trn 3 lt A vi 1 lt B ta c 4 lt dung dch E. trung ha 10 ml dung dch E cn

12,5 ml dung dch NaOH trn v to c 2,085 gam mui.

a) Xc nh cng thc phn t ca cc axit A, B.

b) Tnh nng mol ca dung dch NaOH.

Hng dn:

a) Cng thc phn t ca A, B.

Gi A: CnH2n+1COOH c nng mol x mol/l

B: CmH2m+1COOH c nng y mol/l

Gi nng mol ca NaOH z mol/l

Gi s 0 n < m

Phn ng trung ha:

CnH2n+1COOH + NaOH CnH2n+1COONa + H2O

CmH2m+1COOH + NaOH CmH2m+1COONa + H2O

Th nghim I:

10ml ddD10 1 2,5 ml ddA

4

10 2,5 7,5 ml ddB

x

T naxit = NNaOH2, 5 7, 5

1000

x y=

7, 5

1000

x x + 3y = 3z (I)

Th nghim II:

10ml ddE3

10 7,5 ml ddA

410 7,5 2,5 ml ddB

x

T naxit = NNaOH7, 5 2, 5

1000

x y=

12,5

1000

x 3x + y = 5z (II)

Gii hphng trnh (I) v (II) suy ra:


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x = 1,5z, y = 0,5z tc l: x = 3y

th nghim I:

mmui=2, 5

1000

x(14n + 68) +

7, 5

1000

y(14m + 68) = 1,335 ()

th nghim II:

mmui=7, 5

1000

x(14n + 68) +

2, 5

1000

y(14m + 68) = 2,085 ()

T () 7,5y[(14n + 68) + (14m + 68)] = 1335

y[(14n + 68) + (14m + 68)] = 178

y[14(n + m) = 136] = 178 ()

T () 22,5y(14n + 68) + 2,5 (14m + 68) = 2085

2,5y[9(14n + 68) + (14m + 68)] = 12085

y[14(9n + m) = 680] = 834 ()

Lp t s:( ')

( ')

=

14( ) 136

14(9 ) 680

n m

n m

=

178

834

11676n + 11676m + 113424 = 22428n + 2492m + 121040

9184m - 10752n = 7617

143,5m - 168n = 119

n = 0 m = 0,82 (loi)

n = 1 m =2

Tip tc thay cc gi tr khc ca n u khng c m ph hp

Cng thc phn t ca A: CH3COOH

B: C2H5COOH

b) Nng mol/l ca dung dch NaOH

T y[14(n + m) + 136] = 178


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Thay n = 1, m = 2 y =178

14 3 136x = 1

T y = 0,5z z =1

0, 5= 2 mol/l

Vy nng mol/l ca dung dch NaOH l: 2 mol/l

Bi 3: C hai axit hu c no: (A) RCOOH; (B) R(COOH)m. Hn hp (X) cha x mol (A) v y

mol (B). trung ha (X) cn 500 ml dung dch NaOH 1M; cn nu t chy hon ton (X) th

thu c 11,2 lt CO2(ktc). Hn hp (Y) cha y mol (A) v x mol (B). trung ha (Y) cn

400 ml dung dch NaOH 1m. Bit x = y = 0,3 mol.

a) Xc nh CTPT ca cc axit v tnh % s mol ca mi axit trong hn hp (X).

b) Bit rng 1,26 gam tinh thaxit (B) R(COOH)m, 2H2O tc dng va vi 250 ml

dung dch KMnO4trong mi trng H2SO4 theo phn ng:

KMnO4 + (B) + H2SO4 K2SO4 + MnSO4 + SO2 + H2O

Tnh nng mol ca dung dch KMnO4

Hng dn:

a) Gi n l s nguyn t cacbon trung bnh ca 2 axit (A), (B) ta c:

n = 2CO

hh

n

n=

11,2

22,4

0, 3= 1,667 (A) phi l HCOOH

t cng thc (B) l CnH2n+2-m(COOH)mta c cc phng trnh phn ng:

HCOOH + NaOH HCOONa + H2O (1)

CnH2n+2-m(COOH)m + mNaOH CnH2n+2-m(COONa)m + mH2O (2)

HCOOH +1

2O2

ot

CO2 + H2O (3)


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CnH2n+2-m(COOH)m +3 1

2

n

ot

(n+m)CO2 + (n+1)H2O (4)

T (1,2,3,4) v bi cho:

Vi hn hp (X) nNaOH = x + my = 0,5 (I)

Vi hn hp (Y) nNaOH = mx + y = 0,4 (II)

2COn = x + (n + m)y =

11,2

22,4= 0,5 (III)

T (I, III): n = 0 m = 2. Vy trong (B) 2 nhm (-COOH) phi lin kt trc tip vi

nhau.

Cng thc cu to (B) l HOOC-COOH* Gi %n(A) = a(%) th %nB = (100 - a) (%)

n =5

3=

1. 2(100 )

100

a a

%n(A) = a = 33,33 (%)

V %n(B) = 100 - 33,33 = 66,67 (%)

b) Phng trnh phn ng:2KMnO4 + 5C2H2O4 + 3H2SO4 K2SO4 + 2MnSO4 + 10CO2+ 8H2O (5)

T (5) 2 2 4. 2.2C H O H O

n =1,26

90 36= 0,01 (mol)

Vy:4dd KMnO

MC = (2 / 5).0,01250

. 1000 = 0,016M

Bi 4: Cho 2,14g hn hp 2 hp cht hu c A, B tc dng ht vi xt ta c cht C v hn

hp mui natri ca 2 axit no, n chc, ng ng lin tip. Ton b C phn ng vi natri sinh ra

0,224 lt hir (tnh theo ktc). T khi hi ca C so vi khng kh bng 2. Cho D i qua ng Cu

t nng (xc tc) ta c cht D c th tham gia phn ng trng gng. A v B u lm mt

mu nc brm.

1. A, B, C, D thuc loi hp cht g?


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2. Xc nh cng thc phn t v cng thc cu to ca A, B, C.

3. Tnh % khi lng ca A v B trong hn hp ban u.

4. Hon thnh s bin ha:

C( )

ot

Cu D

2

2

O

Mn

E G

Hng dn:

1. Cht C tc dng vi Na gii phng H2, boxi ha thnh anehit. Vy C l ru (bc

nht). A, B tc dng vi NaOH to thnh ru v mui. Vy A, B phi l cc este. V A, B lm

mt mu nc brm m gc axit no, vy gc ru phi khng no. D c th tham gia phn ng

trng gng. Vy D l anehit.

2. KLPT ca ru M 29.2 = 58. Cng thc tng qut ca ru l ROH)n. Nu n = 2 ta c

R = 58 - 2.17 = 24 ngha l chcn 2 cacbon. Nh th cng thc ru l HO - C C - OH, ru

ny khng tn ti v khng boxi ha thnh anehit. Vy C phi l ru n chc R - OH hay

CxHy - OH. Ta c:

R = CxHy = 14x + y = 58 - 17 = 41

(C th bin lun hoc lp bng tm R)

x 1 2 3

y 29 17 5Kt lun Phi l Phi l C3H5

Vy cng thc ca ru l C3H5OH hay Ch2 = CH - CH2 - OH (khng th l CH3 - CH =

CH - OH v ru ny khng th boxi ha thnh anehit)

Gi cng thc ca A, B l CnH2n+1COOC3H5 v Cn+1H2n+3COOC3H5 ta c:

CnH2n+1COOC3H5 + NaOH CnH2n+1COONa + C3H5OH (1)

Cn+1H2n+3COOC3H5 + NaOH Cn+1H2n+3COONa + C3H5OH (2)

2C3H5OH + 2Na 2C3H5Ona + H2 (3)

Theo (1, 2, 3) s mol C = tng s mol A + B =

2 ln s mol H2 = 2.0,224

22,4= 0,02


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KLPTTB ca A v B l M=2,14

0,02= 107, do phi c:

MA < 107 < MB, ngha l:

14n + 86 < 107 < 14n + 100 hay 0,5 < n < 1,5 tc l n = 1 (v n phi nguyn, dng)

Cng thc ca A v B l:

CH3 - COO - CH2 - CH = CH2

V CH3 - CH2 - CO - CH2 - CH = CH2

(C th bin lun theo Mca gc axit = 22 (CH3-, C2H5-) hoc:

axitM = 67(CH3COOH, CH3 - CH2 - COOH).

3. Gi x, y l s mol ca A v B ta c:

x + y = 0,02 (a) trong 100 v 114 l KLPT ca A v B

100x + 114y = 2,14 (b)

Gii hphng trnh (a,b) ta c x = y = 0,01

Vy %A =0,01.100.100

2,14= 46,7%

%B = 100 - 46,7 = 53,3%

Bi 5:

Cho 50 ml dung dch A gm axit hu c RCOOH v mui kim loi kim ca axit tc dngvi 120 ml dung dch Ba(OH)2 0,125M, sau phn ng thu c dung dch B. trung haBa(OH)2d trong B cn cho them 3,75 gam dung dch HCl 14,6% sau c cn dung dch thuc 5,4325 gam mui khan. Mt khc, khi cho 50 ml dung dch A tc dng vi H2SO4d, unnng thu c 1,05 lt hi axit hu c trn (o 136,5oC, 1,12 atm).

a) Tnh nng mol ca cc cht trong Ab) Tm cng thc ca axit v ca muic) Tnh pH ca dung dch 0,1 mol/l ca axit tm thy trn, bit in li =1%

Hng dn:a) Tnh nng cc cht trong A, gm: axit RCOOH v mui RCOOM (M l kim loi

kim). S mol HCl cn trung ha Ba(OH)2 d trong dung dch B l:


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Ta c

2HCl + Ba(OH)2 BaCl2 + 2H2O (1)(mol) 0,015 0,0075 0,0075

Khi cho 50 ml dung dch A tc dng vi dung dch Ba(OH)2, ch c axit RCOOHtham gia phn ng:2RCOOH + Ba(OH)2 (RCOO)2Ba + 2H2O (2)Ta c:

( )

Vy CM(RCOOH) = 0,015 : 0,05 = 0,3MKhi cho 50 ml dung dch A tc dng vi H2SO4d, ch c RCOOM phn ng:RCOOM + H2SO4 RCOOH + MHSO4 (3)

Suy ra naxit (3) = 0,0350,015 = 0,02 molCM(RCOOM) = 0,02:0,05 = 0,4M

b) Lp cng thc phn t mui v axit:mmui khan = m(RCOO)2Ba + mRCOOM + mBaCl2 = 5,4325 gam

(2R+225)0,0075 + (208.0,0075) + (R + M + 44)0,02 = 5,4325

Nghim thch hp M=39 (Kali), R=15 (CH3)

Vy cng thc phn t ca axit: CH3COOH v ca mui: CH3COOKtc) Tnh pH ca dung dch CH3COOH phn li

Ta c

Theo phng trnh in li: CH3COOH CH3COO

- + H+[H+] = [CH3COOH]in li =10

-3 mol/l; pH = -lg[H+] = 3

Bi 6:

Oxi ha hn hp andehit fomic v andehit axetic hon ton thu c hn hp 2 axit tng ngc t khi hi so vi hn hp anehit ban u bng a. Hy tm khong bin thin ca a biton c nghim.

Hng dn:


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T (1) v (2) suy ra nHCOOH = nHCHO = x (mol)

v

Ta c

{

Vy khong bin thin ca x l 1,36


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T (1, 2, 3, 4)

Theo bi, ta c:

Gii ra: R = 15 (CH3). Vy cng thc ru v anehit l CH3CH2OH v CH3CHO

b)

Nn nu gi s:

Hn hp ch ton ru, s mol hn hp khi l: Hn hp ch gm ton anehit, s mol hn hp khi l:

Do :

c) Ta c: m = 400 gam x + y = 400.0,003 = 1,2 molV mhhban u = 46x + 44y = 53,2 gamGii ra: x = 0,2 mol, y = 1 molVy thnh phn % khi lng ban u:

PHN II: ESTE

Dng 1: Hon thnh s chuyn ha khi bit r cc cht trong s

Bi 1:


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33

4422

243322

CHCOOOOCCH

COOHHOOCCOONHNOOCHCHOHOCOHCHHOCH

CHOCHHOHCHOCHCOONaCHCHOCHHC

Gii:

CHOCHOHCHCH CHgSOO

380,

24

OHOCuCOONaCHNaOHOHCuCHOCHOt

22323 3)(2

324

,

3 CONaCHNaOHCOONaCHOtCuO

OHHCHOOCH CtNOO

2

,

24

CHOCHHOHCHOOH

2

)(

OHCHCHHOHCHOCHHOOtNi 22

,

22

OHCuCHOHOCCuOOHCHCHHO CtO

222 222

AgNONHCOONHNOOCHOHNHAgNOCHOHOC 44264 3444233

ClNHCOOHHOOCHClCOONHNOOCH 444 22

OHCHCOOOOCHCOHCHCOOHHOOCOtSOH

233

,

3 2242

Bi 2:

1.2.3. 2 2 2 2otCH CH CH OH CuO CH CH CHO Cu H O


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4. 22 2 212

MnCH CH CHO O CH CH COOH

5. 2 2 5 2 2 5 2CH CH COOH C H OH CH CH COOC H H O

6.

Gii:

23. 2 3HCOOCH CH NaOH HCOONa CH CHO

24.2

3 2 3

1

2

MnCH CHO O CH COOH

25. 3 3 2CH COOH CH CH CH COOCH CH

26.

27.

28.

Bi 3:


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Gii:

29.6 5 3 3 3 6 5 3C H OH CH C O C CH CH COOC H CH COOH

O O

30. 3 6 5 3 6 5CH COOC H NaOH CH COONa C H OH

31. 3 2 4 3 2 42CH COONa H SO CH COOH Na SO

32. 3 3 22 22 2CH COOH Ca OH CH COO Ca H O

33.

3 3 3 32

ot

CH COO Ca CH C CH CaCO

O

34.

35.

Bi 4:


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36. 4 10 4 6 26002o

Dehidro

CC H C H H

37. 2 2 2 2 2| |CH CH CH CH Br CH CH CH CH

Br Br

38.

39.

40.

41.

42.


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43.

Dng 2: Hon thnh s chuyn ha khi cha bit r cc cht trong s

Bi 1: Hon thnh chui bin ha sau:

A, B, D, F, ,G ,H l k hiu cc cht hu c cha bit, mi du hi l mt cht cn tm, mi mitn l mt phn ng.

Gii:

22

,

2 CHCHHCHCHOtNi (A)

OHCHCHOHCHCHCSOH O 23

180,

22242

OHBCHOCHOOHHCOtCu

23

,

252 2)(22


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OHCHCHHCHOCHOtNi

23,

23

CHOCHOHCHCH CHgSOO

380,

24 (B)

COOHCHOCHOCHOtMn

3

,

23

222

(D)

OHHCOOCCHOHHCCOOHCHOtSOH

2523

,

52342

COONaCHFOHHCNaOHHCOOCCH 352523 )(

CHClCHHClCHCHOtxt 2

, (G)

HClBCHOCHOHCHClCH OH

)(3)(

22

OHCHCHHBCHOCHOtNi

23

,

23 )(

25252 CHCHOHCCHCHOHHCxt

23

,2 CHClCHHClCHCHOtxt (H)

HClBCHOCHOHCHClCH OH 2)(3)(

223

AgCOOHCHOAgCHOCHNHAgNO

23/

2333

32,3 CHOCOCHCCHCOOHCHCHCHO

txt

Bi 2:

7. 2 5 2 5HCOOC H NaOH HCOONa C H OH

8. 2 4 2 42HCOONa H SO HCOOH Na SO


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9. 2HCOOH NaOH HCOONa H O

10. 2 5 2 2 5xt

HCOOC H H O HCOOH C H OH

11. 2 5 2 5 2HCOOH C H OH HCOOC H H O

12.

3 4 3 3 22 22 2 2HCOOH Ag NH OH NH CO Ag NH H O

13. 4 3 3 2 2 32 2 2 2NH CO NaOH NH H O Na CO

14. 4 3 2 4 4 4 2 22 2NH CO H SO NH SO CO H O

Bi 3:

15.1500

4 2 2 2m lanh nhanh2 3

o C

lCH C H H

16.4

2 360 80o

HgSO

CCH CH H O CH CHO

17.

2

3 2 3

1

2

MnCH CHO O CH COOH

18. 3 3 2CH COOH CH CH CH COOCH CH

19. 3 2 3 3CH COOCH CH NaOH CH COONa CH CHO


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20. 3 2 4 3 2 42 2CH COONa H SO CH COOH Na SO

21.

3 3 3 4 3 22

2 2 3ot

CH CHO Ag NH OH CH COONH Ag NH H O

22.

3 4 2 4 3 4 422CH COONH H SO CH COOH NH SO

Dng 3: Xc nh c ch ca phn ng

Bi 1: Khi cho tert-butyl axetat v etyl axetat tc dng vi natri metylat trong dung mi metanol,u thu c metyl axetat, nhng phn ng ca etyl axetat chy nhanh gp 10 ln phn ngca tert-butyl axetat. Mt khc khi c mt lng nh hidroclorua th metanol nhanh chng phnng vi tert-butylaxxetat to tra axit axetic v tert-butyl metyl ete, trong khi methanol liphn ng rt chm vi etyl axetat to ra etanol v metyl axetat.

a) Vit phng trnh ha hc biu din c ch ca nhng phn ng trn.b) C th s dng ng v18O nh thno chng minh c c ch cc phn ngtrn?

Gii:

a) C ch ca cc phn ng vi natri metylat:

(*)

Khi gc R l gc tert-butyl (CH3)3C- th giai on u ca phn ng s xy ra chm

v gc tert-butyl c kch thc ln gy n ng khng gian mnh.Khi gc R l gc etyl C2H5- th s n ngkhng gian t hn rt nhiu nn phn ngca etyl axetat xy ra nhanh hn phn ng ca tert-butyl axetat.

C ch phn ng vi metanol khi c mt xc tc axit (HCl) c th theo mt trong 2 s sau:


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(1)

(2)

V cacbocation bc III bn v d to ra do vy khi R l gc tert-butyl (CH3)3C- th tora (CH3)3C

(+) phn ng xy ra nhanh, to ra axit v este tc l phn ng xy ra theo s (2). Khi R l gc etyl th vic to ra cation CH3CH2

(+) l khng thun li do phn ng xy ra theo c chs (1) to ra este mi v ancol.

b) chng minh c chno ng ta dng metanol chang v18O

(CH3-18O-H). Nu phn ng xy ra theo c ch (*) v (1) th ch c este to thnh mi

cha ng v18O. Ngc li nu theo c ch (2) th este to ra cha ng v18O.

Bi 2: Sildenafil (mt loi thuc tng lc) c tng hp theo s :


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1. Hy hon chnh dy phn ng trn, bit rng:

o Qu trnh chuyn sang G c to thnh axit sunfonic trung gian sau michuyn thnh sunfonyl clorua.

o N,N,-cacbonyliimiazol (CDI) l mt loi tc nhn dng hot ho axitcacboxylic cho phn ng th nucleophin ca nhm cacbonyl.

2. Vit c ch phn ng chuyn [I] thnh K.

Gii:


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2. C ch t [I] sang K

Bi 3: Vit cng thc cu trc cc dng enol ca dietylmalonat (1), Etylaxetoaxetat (2). Trongcc cu trc ca (2), cho bit dng no bn nht, dng no km bn. Gii thch?

Gii:

a. Cc cu trc


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b. Trong :

- Dng A t bn do ni i khng lin hp

- Dng B bn nhng khng c cng hng este

- Dng C bn nht do c ni i lin hp v cng hng este

Dng 4: Vit cng thc cu to, gi tn este khi bit cng thc phn t.


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Bi 1: Vit CTCT, gi tn cc ng phn n chc, mch hc th c ng vi CTPT C3H6O2

Gii:

C3H6O2c bt bo ha=1 v phn t c 2 nguyn t oxi

Nn c ng phn v este n chc no v axit cacboxylic n chc, no.

- ng phn este: HCOOC2H5 : etyl fomiatCH3COOCH3 : metyl axetat

- ng phn axit cacboxylic: CH3CH2COOH: axit propionicBi 2: Vit CTCTcc ng phn n chc, mch hc th c ng vi CTPT C4H6O2

Gii:

C3H6O2c bt bo ha=2 v phn t c 2 nguyn t oxi

Nn c ng phn veste n chc, khng no c mt ni oi gc v ng phn axitcacboxylic n chc, khng no c mt ni i gc.

- ng phn este: HCOOCH = CHCH3HCOOCH2CH = CH2

CH3COOCH = CH2

CH2 = CHCOOCH3

- ng phn axit cacboxylic: CH2 = CHCH2COOHCH3CH = CHCOOH

CH2 = C(CH3)COOH

Bi 3:Cho s phn ng:

CH3CH=CHCOOCH=CHCH3 dd Br2 (1:1) (B) dd NaOH (C) + (D)

a) Xc nh cng thc cu to ca cc sn phm trn v tn gi A, B, C, D.b) Cho bit loi c ch ca cc phn ng trn.c) Hy chr cc ng phn lp th ca A, B, C, D. C bao nhiu loi ng phn lpth?d) Khi A c cu hnh bn nht th cu trc ca B nh th no?e) Khi A c cu hnh bn nht. Tin hnh phn ng gia A vi D2/Ni, LiAlH4. Chobit cu trc cc sn phm to thnh.

Gii:

a)b)


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Tn gi: A: (1-propenyl) but-2-enoat.

B: (1,2-ibrom propyl) but-2-enoat.

C: natri but-2-enoat.

D: 2-hiroxi propanal.

c) A c 4 ng phn hnh hc v c 2 ni i C=C.B c 8 ng phn cu hnh, C c 2 ng phn cu hnh v D c 2 ng phn cu hnh.

d) K hiu gc:

e) A + D2 (Ni/to) => 4 ng phn cu hnh.

Dng 5: Xc nh este thng qua phn ng t chy.

Bi 1:t chy 1 este n chc no cn va 0,35 mol O2thu c 0,3 mol CO2. Xc nh CTEste bit rng khi tc dng vi xt thu c sn phm ckhnng trng gng. Xc nh tneste.

Gii:Ta c: n =3 => CTPT ca este : C3H6O2Khi tc dng xt thu c sn phm c khnng trng gng

=> CTPT l HCOOCH2-CH3Bi 2:un hp cht A vi H2O (c axit v c lm xc tc) c axit hu c B v ancol D. Tkhi hi ca B so vi nit l 2,57. Cho hi ancol D i qua ng un nng ng bt ng th sinhra hp cht E c khnng tham gia phn ng trng gng. t chy hon ton 2,8 g cht Aphi dng ht 3,92 lt O2(ktc). Sn phame chy gm c kh CO2v hi H2O theo t l s molbng 3:2.

a) Cho bit A v E thuc loi hp cht no?b) Xc nh cng thc cu to ca B.c) Xc nh cng thc cu to v gi tn ca A bit D l ancol n chc.

Gii:

a) un hp cht A vi H2O (c axit v c lm xc tc) c axit hu c v ancol nn A leste.Cho hi ancol D i qua ng un nng ng bt ng th sinh ra hp cht E c khnng

tham gia phn ng trng gng nn E l andehit.

b) mB = 28.2,57 = 72 (g/mol)V khi lng 2 nhm COOh l 90 nn phn t B chie c mt nhm COOH.


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t cng thc chung ca B l RCOOH.

MB = MR + 45 = 72 MR = 27 R l nhm C2H3 .

cng thc cu to ca B l: CH2 = CHCOOH

c) gmmoln OO 6,5175,0 22

Theo nh lut bo ton khi lng c:

gm

gmgm

gmgm

m

mC

gmmmm

O

HOH

CCO

OH

CO

OAOHCO

8,08,18,2

2,08,1

8,16,6

3

11

18.2

44.3:

)(4,86,58,2

2

2

2

2

222

t cng thc chung ca A l ZYX OHC

1:4:316

8,0:

1

2,0:

12

8,1:: zyx

Cng thc n gin nht ca A l C3H4O. Cng thc phn t ca A l C6H8O2 ( v D l ancoln chc nn A cng l este n chc).

Cng thc cu to cuae A: CH2 = CHCOOCH2CH = CH2

Bi 3: Cho hn hp X gm 2 este ca 2 axit k tip nhau trong dy ng ng l R1COOR vR2COOR. t chy hon ton 20,1g X cn 146,16 lt khng kh (ktc). Sn phm chy thu ccho qua bnh 1 ng H2SO4c v sau qua bnh 2 ng dung dch Ca(OH)2 d. Sau thnghim khi lng bnh 1 tng m gam v khi lng bnh 2 tng 41,6g. Mt khc, nu cho3,105g X tc dng va vi dung dch NaOH d thu c 2,529g hn hp mui.

a) Tnh m.b) Xc nh cng thc phn t ca 2 este.c) Tnh % khi lng mi este trong X.d) Xc nh cng thc cu to ca 2 este.e) Tnh khi lng mi mui sau phn ng x phng ha.

Gii:

a) nO2 = 146,16/22,4.20% = 1,305 ; nCO2 = 46,2/44 = 1,05.X + O2 => CO2 + H2O

p dng LBTKL ta c:


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mH2O = 20,1 + 1,305.32 - 46,2 = 15,66 (g) => nH2O = 15,66/18 = 0,87 (mol).

b) V 2 este l ca 2 axit ng ng k tip v cng ancol. cng thc trung bnh CxHyO2, trong x v y l s nguyn t C v H trung bnh.CxHyO2 + (x+y/4-1)O2 xCO2 + y/2H2O.

Gi a l tng s mol ca 2 este ta c: ax = nCO2 = 1,05 ; ay/2 = nH2O = 0,87 ;

a( x + y/4 -1) = nO2 = 1,305 => a = 0,18 ; x = 5,8 ; y = 9,7.

V s nguyn t H trong 2 este phi l s chn => cng thc phn t ca 2 este l C5H8O2 (M =100 vC) v C6H10O2(M = 114 vC).

c) C5H8O2 : 2 v C6H10O2 : (0,18) + O2to

nCO2= 5 + 6(0,18 ) = 1,05

= 0,03 % C5H8O2 = 14,93% v % C6H10O2 = 85,07%.

d) RCOOR + NaOH RCOONa + ROHnmui = neste = 0,18.3,015/20,1 = 0,027

Mmui =2,529/0,027 = 93,66 R = 93,66 67 = 26,66 (vC).

Gc R26,66 l C2H5- (M = 29)

Cng thc cu to 2 este l CH3COOC3H5 v C2H5COOC3H5.

e) V s mui t l vi s mol este ta c :nC2H5COOC3H5= 0,027.0,15/0,18 = 0,0225 mC2H5COOC3H5 = 2,16 (g)

m CH3COOC3H5 = 2,5292,16 = 0,369 (g).

Dng 6: Xc nh este thng qua phn ng thy phn.

Bi 1:t chy hon ton 1 mol este X thu c 3 mol CO2. Mt khc khi x phng ha honton 1 mol este thu c 8,2g mui Natri. Xc nh cng thc cu to ca esteS cacbon = = 3=> este n chc

Gii:V este n chc nn neste = nmui= 0,1(mol)Mmui = = 82

RCOONa=82R= 8267 =15

=>R l nhmCH3Vy CTCT ca X l: CH3COOCH3


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Bi 2:Thy phn hon ton mt este X c dX/H2 = 44 thu c mui Natri c khi lng bng41/44 meste. Xc nh CTCT ca este.

Gii:Ta c: dX/H2 = 44=>MX = 88 < 100

=>este X l este n chc => CT: CxHyO2=> neste = nmui => C4H8O2Ta c: mmui = 41/44 meste

=>Mmui = 82 MRCOONa = 82R = 15 => R l nhm -CH3

Vy CTCT ca X l CH3COOC2H5Bi 3:Mt este n chc X c dX/CH4= 6,25. Cho 20g X tc dng vi 300ml dd KOH 1M unnng. C cn dd thu c 28g cht rn. Xc nh CTCT ca X.

Gii:MX= 6,25 . 16 = 100(g) => CTPT l C5H8O2RCOOR + KOH RCOOK + ROH

nRCOOR = 20/100 = 02, (mol)nKOH = 0,3 (mol)=>KOH dnKOHd = 0,30,2 =0,1(mol)mcht rn = mmui + mKOH d = 28

0,2(R + 83) + 0,1.56 = 28R = 29 => R l nhm C2H5

Ta c: 29 + 12 + 16.2 + R = 100 => R = 27 (C2H3)Vy CTCT ca X l C2H5COOC2H3

Bi 4:t chy 1,6 g mt este E n chc c 3,52 g CO2 v 1,152 g H2O.

a) Tm cng thc phn t ca E.b) cho 10 g E tc dng vi lng d NaOH va , c cn dd sau phn ng c 14 g muikhan G. Cho G tc dng vi dd axit long thu c G1 khng phn nhnh. Tm cng thccu to ca E.c) X l mt ng phn ca E. X tc dng vi NaOH to ra mt ancol m khi t chy mtthtch hi ancol ny cn 3 th tch kh O2o cng iu kin. Xc nh cng thc cu toca X.

Gii:

a)Gi CTPT ca E l ZYX OHC

0(032,016

128,012.08,06,1

)(128,0064,0.2)(064,018

152,1

)(08,0)(08,044

52,3

2

2

moln

molnmoln

molnmoln

O

HOH

CCO

ta c t l: 2:8:5032,0:128,0:08,0:: zyx


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V E l este n chc nn phn t ch c hai nguyn t oxi

Nn CTPT ca E l 285 OHC

b)Gi CTTQ ca E l RCOOR.OHRRCOONaNaOHRCOOR ''

0,1 0,1 0,1 (mol)

nE = nNaOH = nRCOONa = 0,1 (mol) mNaOH = 4 g

mE + mNaOH = mmui + mROH mROH = 0

Suy ra : E l este mach vng c CTCT:

H2CCH2CH2CH2C = O

O

G1 l HOOC[CH2]3CH2OH

c) ancol sinh ra trong phn ng thy phn l C2H5OHVy CTCT ca X l : CH2 = CHCOOC2H5

Bi 5: Cho 2,76 gam cht hu c A cha C, H, O tc dng vi dd NaOH va , sau chngct kh th bay hi chc nc, phn cht rn khan cn li cha hai mui ca Natri chim khilng 4,44 gam. Nung nng hai mui ny trong oxi d, sau khi phn ng hon ton, ta thu c3,18 gam Na2CO3 ; 2,464 lt CO2(ktc) v 0,9 gam nc. Tm cng thc phn t, vit cng thccu to c th c ca A tha mn cc tnh cht trn bit rng cng thc n gin cng l cngthc phn t.

Gii:

Da trn d kin ca bi, ta c s phn ng:1,76g A + NaOH 4,44g mui + 4H2O

4,44g mui + O2 3,18g Na2CO3 + 2,464 lt CO2 + 0,9g H2O


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)(72,044,44,276,2

)(4,240.06,0

)(06,0106

18,3.22

2

32

gmmmm

gm

molnn

muiNaOHAOH

NaOH

CONaNaOH

mC trong A = mC trong CO2 + mC trong Na2CO368,136,032,.112.

106

18,312.

4,22

646,2

mH trong A = mH trong H2OmH trong NaOH

)(12,006,018,01.06,02.18

9,072,0g

mO trong A = mAmCmH = 2,761,680,12 = 0,96(g)

t A l ZYX OHC ta c:

3:6:716

96,0:

1

12,0:

12

68,1:: zyx

Vy cng thc n gin v cng thc phn t ca A u l C7H6O3

MA = 138

Smol A phn ng vi NaOH )(02,0138

76,2mol

S mol NaOH = 0,06

Vy c 1 mol A phn ng vi 3 mol NaOH, m trong phn t ch c nguyn t oxi, suy ra:

A c th c 3 nhmOH ( loi gi thit ny)

A c 1 nhm OH phenol v nhm este ca phenol.

V sau phn ng thu c hai mui vy ch c trng hp este ca phenol l tha mn.

Cng thc cu to ca A : HCOOC6H5OH

Bi6: Mt este quang hot A c cng thc C6H12O2. Thu phn hon ton 11,6g A trong dungdch kim. Sau khi thy phn em lc hn hp vi ete. Tch ly lp ete, lm kh ri chng ctui ete thu c mt cht lng quang hot c khi lng 7,4g. Dung dch di khng quanghot.

a) Xc nh cu to ca este .b) Vit c ch phn ng thy phn este trn.


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c) Tm cc ng phn cu to cha nhm chc axit v este ng vi cng thcC6H12O2.

Gii:

a) A tc dng vi dung dch NaOH cho cht lng tan trong ete v A n chc nn A l este

n chc v cht lng tan trong ete l ancol => A c dng: RCOOR nA = 11,6:116 = 0,1(mol).

RCOOR + NaOH => RCOONa + ROH

0,1 0,1

=> 0,1.(R+17) = 7,4 => R = 57 (C4H9)

ROH c tnh quang hot => A l:

CH3CH2-C*H(CH3)-O-C(O)-CH3

CH3CH2-C*H(CH3)-O-C(O)-CH3

NaOH CH3CH2-C*H(CH3)-OH.

b) C ch:

c) 26 ng phn.

Bi 7: Cho 100ml dung dch cha 2 este A, B n chc c cng nng 0,8M tc dng vi150ml dung dch NaOH 1M. Sn phm thu c gm 2 mui hu c C, D c khi lng l10,45g (t l MC:MD=41:65) v mt acol E c khi lng 2,9g. Ancol ny khng bn chuynthnh andehit. trung ha ht NaOH d sau phn ng cn dung 200ml dung dch HCl 0,2M.

Xc nh cng thc phn t v cng thc cu to ca A, B.

Gii:

nNaOH b = 0,15.1 = 0,15 (mol) ; nNaOH d = nHCl = 0,2.0,2 = 0,4 (mol).

NNaOH p/ = 0,15.0,04 = 0,11 > n2este = 0,18.0,1 = 0,08


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S c 1 este phn ng vi NaOH theo t l 1 : 2 v este ny c dng:

R-COO-Ar ( Ar l gc cha vng thm )

A l R1COOR ( a mol) v B l R2COORAr (b mol)

R1COOR + NaOH R1COONa + ROH

a a a a

R2COOR + 2NaOH R2COONa + RONa + H2O

b 2b b b

n2este = a + b = 0,08 ; nNaOH d= a + 2b = 0,11 a = 0,05 mol A ; b = 0,03 mol B

nE

= nA= 0,05 M

E= M

ROH= 2,9/0,05 = 58 (vC)

Do ROH khng bn RCHOR + 29 =58 R = 29 andehit l C2H5CHO to ra tancol khng bn CH3-CH=CHOH.

V phn ng x phng ha 2 este ch cho 2 mui trong 3 mui R1COONa, R2COONa vROna phi c 2 mui ging nhau R1 R2. Ta c:

M2 mui = MRCOONa + MROna = (a + b) + b

0,08(R + 67) + 0,03(R + 39) = 10,468R + 3R = 393 (1)

MC/MD = MRCOONa/MROna= (R + 67)/(R + 39) = 41/6541R -65R = 2756 (2)

T(1) v (2) R = 15 (-CH3) ; R = 91 (C6H4CH3)

A l este ca axit CH3COOH v ancol CH3-CH=CH-CH3.

B l este ca axit CH3COOH v phenol CH3C6H4OHcng thc cu to ca B l:

Dng 7: Ton v hng s cn bng v hiu sut phn ng este ha.


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Bi 1: Mt hp cht hu c X mch hch cha mt loi nhm chc c iu ch t axit noA v ru no B. Bit rng:

-a gam X thhi chim mt th tch bng th tch ca 6,4 gam oxi cng iu kinnhit v p sut.

- Khi t chy 1 mol ru B cn 2,5 mol oxi.- a gam X tc dng ht vi xt to ra c 32,8 gam mui.

a) Xc nh CTCT ca X.b) Cho 200 gam axit A tc dng vi 50 gam ru B ta thu c 87,6 gam X. Tnh hiu sutca phn ng.

Gii:

a) Gi CTPT ca ru no B l CnH2n+2-z(OH)z (z n)

Phn ng chy ca B:

OHnnCOOzn

OHHC ZZnn 22222 )1(2

13)(

1 mol2

13 zn mol

cho :2

13 zn =2,5

Suy ra : 34

43513

z

nznzn

z nguyn dng v z 2 ch c gi tr z = 2 ; n = 2 l ph hp.

Vy cng thc ca B l : C2H4(OH)2

S mol X ng vi a gam: )(2,032

4,6molnX

Gi X ch cha mt loi nhm chc v to taxit A v ru B nn X l este. Do B l ru haichc nn A c hai trng hp:

Trng hp 1: A l axit n chc no

Nu A l axit n chc no RCOOH vi R l gc hidrocacbon no.

Phn ng to X:

OHHCRCOOOHHCRCOOH 2422242 2)()(2


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Phn ng x phng ha X:

242422 )(2)( OHHCRCOONaNaOHHCRCOO

0,2 mol 0,4 mol

Phn tlng ca RCOONa : 824,0

8,32

)(158267 3 CHRR

Vy A l : (CH3COO)2C2H4

Cng thc cu to ca X: CH3COOCH2CH2COOCH3

Trng hp 2 : A l axit no hai chc

Phn ng to X:

OHHCCOOROHHCCOOHR 24222422 )()()(

X l hp cht mch kn (loi).

b) Hiu sut ca phn ng:

S mol axit A : )(33,360

200molnA

Smol ru B : )(806,062

50 molnB

Phn ng : OHHCCOOCHOHHCCOOHCH 242232423 2)()(2

V nA> nB nn hiu sut phn ng tnh theo B.

C 1 mol B to c: 146g X

Vy62

50mol B to c: )(14,117

62

50.146g

Nhng trong thc t chthu c 87,6 g este.

Vy hiu sut ca phn ng l :

%4,74%100.14,117

6,87H


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Bi 2: 25 oC, hng s cn bng ca phn ng:

OHHCOOCCHOHHCCOOHCH CtSOHO

2523

,

52342 l KC = 4

Cho nng ban u ca CH3COOH l 1 M. Tnh nng ca este to thnh khi cn bng, nu

nng ca C2H5OH l :

a) 1(M) b) 2 (M)

Gii:

OHHCOOCCHOHHCCOOHCHCtSOH O

2523

,

52342

Phn ng: x x x x

a) KhiOHHC

MC52

= 1M th [CH3COOH] = [C2H5OH] = 1x

Lc :3

24

)1(2

2

xx

x

Vy [CH3COOC2H5] =3

2

b) Khi OHHCMC 52 = 2MTng t : 85,04

)2)(1(

2

x

xx

x

Vy [CH3COOC2H5] = 0,85 M

Dng 8: Tnh khi lng este tphn ng este ha

Bi 1:Hn hp M gm 0,2 mol ancol no n chc mch hX, 0,3 mol axit cacboxylic c cng

s nguyn t cacbon. Nu t chy hon ton M th thu c 33,6 lt kh CO2(KC) v 25,2gnc. Mt khc nu nung nng M vi axit sunfuric c (H2SO4 ) thc hin phn ng este havi hiu sut 80% th khi lng este thu c l bao nhiu?

Gii:

Gi cng thc tng qut ca X: CnH2n+1OH, Y: CmHyCOOH v m+1=n ( do axit v ancol ccng s nguyn t cacbon)


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nCO2 = 1,5 mol , nH2O = 1,4 mol

Ta c : nc= 0,2n + 0,3m + 0,3 = 1,5 2n + 3m =12 n=2, m=3.

nH =2.1,4=0,2.8+0,3-y+0,3 y = 3 X: C3H7OH, Y: C2H3COOH

ptp: CH3CH2CH2OH + CH2CHCOOH CH2CHCOOCH2CH2CH3 + H2O

Khi lng este l m =0,8.0,2.114=18,24 g

Bi tp tng hp

Bi 1:Khi ha hi 1,0g axit hu c n chc no (A) ta c mt th tch va ng bng th tch

ca 0,535g oxi trong cng iu kin.

Cho mt lng d A tc dng vi 5,4g hn hp hai kim loi M v M thy sinh ra 0,45mol hiro. T l smol M i vi M trong hn hp l 3:1. KLNT ca M bng 1/3 KLNT caM. Ha tr ca M l II, ha tr ca M l III. Este ca A i vi mt ru n chc no lu bthy phn mt phn. trung ha hn hp sinh ra t 18,56g este ny phi dng 20ml dung dchNaOH 0,50M v x phng ha lng este cn li phi dng thm 300ml dung dch NaOH nitrn.

a) Xc nh KLPT v cng thc cu to ca axit.

b) Vit cc phng trnh phn ng xy ra.

c) Xc nh KLNT ca hai kim loi.

d) Xc nh cng thc phn t v vit cc cng thc cu to c th c ca este.

e) Vit cng thc cu to ca ru, bit rng khi oxi ha khng hon ton ru sinh raanehit tng ng, c mch nhnh.

Gii:

a) Cc kh (hi) trong cng iu kin c thtch nh nhau th cng c s mol bng nhau.

0,535g oxi ng vi0,535

32=

1

6mol oxi

Vy: 1g A ng vi1

60mol A. Suy ra KLPT ca A bng 60 .v.C


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Bit A l axit no n chc, ta c:

CnH2n+1 - COOH = 60

V COOH = 45

Do CnH2n+1 = 60 - 45 = 15

Vy n = 1

l axit axetic CH3COOH

b) Cc phng trnh phn ng:

Tc dng vi kim loi:

2CH3COOH + M M(CH3COOH)2 + H2 (1)

6CH3COOH + 2M 2M(CH3COOH)3 + 3H2 (2)

Thy phn este c ru CmH2m+1OH v axit axetic:

CH3COOCmH2m+1 + H2O CH3COOH + CmH2m+1OH (3)

Trung ha axit:

CH3COOH + NaOH + CH3COONa + H2O (4)

X phng ha este:

CH3COOCmH2m+1 NaOH CH3COONa + CmH2m+1OH (5)

c) Xc nh KLPT ca kim loi:

Gi x, y l sgam M v M trong hn hp; M v 3M l KLPT ca chng. Ta c:

x + y = 5,4

x

M:

3

y

M= 3

Do : x = y =5, 4

2= 2,7g

Da theo (1) v (2) ta c:

2,7

M+

3.2,7

2.3M=

2,7

M+

2,7

2M= 0,45


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Do : M = 9 (berili) v 3M = 27 (nhm)

d) Cng thc cu to ca este:

Theo (3) v (4) mi mol este b thy phn sinh ra mt mol axit tdo v do cn 1 mol

NaOH trung ha, smol dng trung ha bng mol este b thy phn:0,5.20

1000= 0,01 mol,

S mol este tham gia x phng ha

0,5.300

1000= 0,15 mol,

Smol este ban u (cha thy phn)

0,15 + 0,01 = 0,16 mol

Phn tlng ca este:

18,56.1

0,16= 116

Ta c th vit:

CH3COOCmH2m+1 = 116

59 + 12m + 2m + 1 = 116 m = 4

Cng thc este: CH3COOC4H9

Cc cng thc cu to c th c ca este:

CH3COO - CH2 - CH2 - CH2 - CH2

CH3COO - CH - CH2 - CH3

CH3

CH3COO - CH2 - CH(CH3)2, CH3COOC(CH3)3,

e) Cng thc cu to ca ru. Tng ng vi cc cu to trn ca este ta c cc cngthc ca ru.

CH3 - CH2 - CH2 - CH2- OH,


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Trong ch c (CH3)2 CH - CH2- OH l c th boxi ha sinh ra anehit mch nhnhnh nu trong bi.

CC DNG BI TP TRC NGHIM ESTE

I. NG PHN, DANH PHP, CNG THC TNG QUT.

Cu 1: Nhn nh no sau y khngng?

A.Tn este RCOOR gm: tn gc hirocacbon R + tn anion gc axit (ui at).

B.Khi thay nguyn t H nhmCOOH ca axit cacboxylic bng gc hirocacbon th c este.

C.Phn ng thuphn este trong mi trng kim l phn ng 1 chiu v gi l phn ng x phng ho.

D.Este c nhit si thp hn so vi axit v ancol c cng s nguyn t C v este c khi lng phn t nhhn.

Cu 2: Trong s mi lin h gia hirocacbon v dn xut cha oxi, ankan c t trung tm v

A.ankan tng i tr v mt ho hc.

B.ankan c th tch H2 to thnh cc hirocacbon khng no v cng O2 sinh ra dn xut cha oxi.

C. ngnh cng nghip ho cht ly du m lm nn tng. T ankan trong du m ngi ta sn xut ra cc

hirocacbon khc v cc loi dn xut ca hirocacbon.

D. l do khc.

Cu 3:Cng thc tng qut ca este to bi axit n chc no mch hv ancol n chc no mch hc dng.

A.CnH2n+2O2( n 2) B.CnH2nO2(n 2) C. CnH2nO2 ( n 3) D.CnH2n-2O2( n 4)

Cu 4: Cho glixerol phn ng vi hn hp axit bo gm C17H35COOH v C15H31COOH, s loi trieste ti a c tora l

A.3. B.4.C.5. D. 6.

Cu 5:Khi un nng glixerol vi hn hp 3 axit bo C17H35COOH, C17H33COOH, C17H31COOH thu c chtbo khc nhau. S CTCT c th c l bao nhiu?

A.21 B.18C.16 D.19


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Cu 6:Sng phn este ng vi cng thc phn t C4H8O2 l:

A.5 B.2C.4 D.6

Cu 7: Sng phn este ng vi cng thc phn t C5H10O2 l:

A.10 B.9C.7 D.5

Cu 8. Hy cho bit c bao nhiu cht hu c n chc c cng thc phn t l C3H6O2?

A.4 B.2C.3 D.5

Cu 9:Glixerol C3H5(OH)3 c khnng to ra 3 ln este (trieste). Nu un nngglixerol vi hn hp axit R'COOH v

R''COOH (c H2SO4 c xc tc) th thu c ti a l bao nhiueste?

A.2 B.6C.4 D.8

Cu 10. t chy hon ton mt lng hn hp hai este cho sn phm chy qua bnh ng P2O5d thy khi lngbnh tng thm 6,21g, sau cho qua dd Ca(OH)2d thu c 34,5g kt ta. Cc este ni trn thuc loi:

A.No n chc B.Khng no n chc C.No a chcD.Khng no a chc.

II. TNH CHT.

* Tnh cht.

Cu 1:Pht biu ng l:

A.Phn ng gia axit v ancol c mt H2SO4c l phn ng mt chiu.

B.Tt c cc este phn ng vi dung dch kim lun thu c sn phm cui cng l mui v ancol.

C.Khi thu phn cht bo lun thu c C2H4(OH)2.

D.Phn ng thuphn este trong mi trng axit l phn ng thun nghch

Cu 2: Nhn nhkhngng l

A.CH3CH2COOCH = CH2cng dy ng ng vi CH2 = CHCOOCH3.

B.CH3CH2COOCH = CH2 tc dng vi dung dch NaOH thu c anehit v mui.

C.CH3CH2COOCH = CH2 tc dng vi dung dch Br2.

D.CH3CH2COOCH = CH2 c th trng hp to polime.
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Cu 3: Chn sn phm chnh cho phn ng sau:

C2H5COOCH3 4LiAlH A + B

A, B l:

A.C2H5OH, CH3COOH B.C3H7OH, CH3OH C.C3H7OH, HCOOHD.C2H5OH, CH3COOH

Cu 4. Axit Fomic khng tc dng vi cc cht no trong cc cht sau

A.CH3OH B.NaClC.C6H5NH2 D.Cu(OH)2 (xt OH

-, to)

Cu 5: Cho cc cht: etyl axetat, etanol, axit acrylic, phenol, phenylamoni clorua, phenyl axetat. Trong cc cht ny, scht tc dng c vi dung dch NaOH l

A.3. B.4.

C. 5. D.6.

Cu 6: Cho cc cht: axit propionic (X); axit axetic (Y); ancol etylic (Z) v metyl axetat (T). Dy gm cc cht csp xp theo chiu tng dn nhit si l

A.T, Z, Y, X. B.Z, T, Y, X. C. T, X, Y, Z.D.Y, T, X, Z.

Cu 7: Cho tt ccc ng phn n chc, mch h, c cng thc phn t C2H4O2 ln lt tc dng vi: Na, NaOH,NaHCO3. S phn ng xy ra l

A.2.

B.3.C. 4. D.5.

Cu 8: C2H4O2c 3 ng phn mch h. Cho cc ng phn tc dng vi: NaOH, Na, AgNO3/NH3 th sphngtrnh phn ng xy ra l

A.3. B.4.C.5. D. 6.

Cu 9:C4H6O2c bao nhiu ng phn mch hphn ng c vi dung dch NaOH?

A.5 ng phn. B.6 ng phn. C. 7

ng phn. D.8 ng phn.

Cu 10: S hp cht n chc, ng phn cu to ca nhau c cng cng thc phn t C4H8O2, u tc dng vi dungdch NaOH

A.3 B.4C.5 D.6


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Cu 11: Hp cht X khng no mch hc cng thc phn t C5H8O2 khi tham gia phn ng x phng ha thu c 1anehit v 1 mui ca axit hu c. C bao nhiu cng thc cu to ph hp vi X ( khng kng phn cis,tran )?

A.2 B.3 C. 4D.5

Cu 12:Este A n chc, mch h, c t khi hi so vi metan bng 6,25 v khi tham gia phn ng x phng ho tora mt anehit v mt mui ca axit hu c. C bao nhiu cng thc cu to ph hp vi A?

A.3. B.4.C.2. D.5.

Cu 13: Thy phn 1 mol este cho 2 mui v nc . CTCT ca este c dng: (R l gc hirocacbon trong cacbon mang ha tr l cacbon no)

A.RCOOR B.RCOOCH=CHR C.RCOOC6H5 D.C6H5COOR

Cu 14. Este X c cng thc phn t l C5H10O2. un nng X vi NaOH thu c mui Y v ancol Z trong MY y) thnh hai phn bng nhau.

- Phn 1: Cho tc dng vi Na d thu c 5,6 lt H2 (ktc).- Phn 2: un nng vi H2SO4c ti phn ng hon ton c 8,8 gam este.

Gi tr ca x v y l

A.x = 0,4; y = 0,1. B.x = 0,8; y = 0,2. C.x = 0,3; y = 0,2.D.x = 0,5; y = 0,4.

Cu 6: Cho 2,72 gam CH3COOC6H5 vo 500 ml dung dch NaOH 0,1M. C cn dung dch sau phn ng thu c sgam cht rn l

A.1,64g B.3,96gC.2,84g D.4,36g

Cu 7: Cho 4,48 gam hn hp gm CH3COOC2H5 v CH3COOC6H5 (c t l mol l 1:1) tc dng vi 800 ml dungdch NaOH 0,1 M thu c dung dch X. C cn dung dch X th khi lng cht rn thu c l

A.5,6 gam B.3,28 gam C.6,4 gamD.4,88 gam


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Cu 8: Khi thc hin phn ng este ha 1 mol CH3COOH v 1 mol C2H5OH, lng este ln nht thu c l 2/3 mol.t hiu sut cc i l 90% (tnh theo axit). Khi tin hnh este ha 1 mol CH3COOH cn s mol C2H5OH l (bitcc phn ng este ho thc hin cng nhit )

A.2,115. B.2,925. C.2,412. D.0,456.

Cu 9: Hn hp X gm axit fomic v axit axetic (t l mol 1:1). Ly 5,3 gam hn hp X tc dng vi 5,75 gam ancoletylic (c xc tc H2SO4c) thu c m gam hn hp este (hiu sut ca cc phn ng este ho u bng 80%). Gitr ca m l

A.10,125. B.6,48. C. 8,10.D.16,20.

Cu 10: Tnh khi lng este metyl metacrylat thu c khi un nng 215 gam axit metacrylat vi 100 gam ancolmetylic. Gi thit phn ng este ho t hiu sut 60%.

A.125 gam B.175 gamC.150 gam D.200 gam

Cu 11:un nng 6,0 gam CH3COOH vi 6,0 gam C2H5OH ( c H2SO4 lm xc tc, hiu sut phn ng este hobng 50%). Khi lng este to thnh l:

A.6,0 gam B.4,4 gamC.8,8 gam D.5,2

Cu 12: Cho bit hng s cn bng ca phn ng este ho:

CH3COOH + C2H5OH CH3COOC2H5 + H2O K = 4

Nu cho hn hp cng s mol axit v ancol tc dng vi nhau th khi phn ng t n trng thi cn bng th % ancolv axit b este ho l

A.50%. B.66,7%. C.33,3%. D.65%.

Cu 13: Cho cn bng sau: CH3COOH + C2H5OH CH3COOC2H5 + H2O K = 4

Khi cho 1 mol axit tc dng vi 1,6 mol ancol, khi ht n trng thi cn bng th hiu sut ca phn ng l

A.66,67%. B.33,33%. C. 80%.

D.50%.

Cu 14: un nng hn hp X gm 1 mol ancol etylic v 1 mol axit axetic (c 0,1 mol H2SO4c lm xc tc), khiphn ng t n trng thi cn bng c hn hp Y trong c 0,667 mol etyl axetat. Hng s cn bng KC caphn ng l

A.KC = 2. B.KC = 3. C. KC =4. D.KC = 5.


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Cu 15: un 12 gam axit axetic vi 1 lung d ancol etylic ( c H2SO4c lm xc tc). n khi phn ng dng lithu c 11 gam este. Hiu sut ca phn ng este ho l bao nhiu?

A.70% B.75%C.62,5% D.50%

Cu 16:Khi un nng 25,8 gam hn hp ancol etylic v axit axetic c H2SO4c lm xc tc thu c 14,08 gam este.Nu t chy hon ton lng hn hp thu c 23,4 ml H2O. Tm thnh phn trm mi cht trong hn hp u vhiu sut ca phn ng este ho.

A.53,5% C2H5OH; 46,5% CH3COOH v hiu sut 80% B. 55,3% C2H5OH; 44,7% CH3COOH vhiu sut 80%

C.60,0% C2H5OH; 40,0% CH3COOH v hiu sut 75% D. 45,0% C2H5OH; 55,0% CH3COOH vhiu sut 60%

Cu 17:un 12 gam axit axetic vi 13,8 gam etanol ( c H2SO4c lm xc tc) n khi phn ng t ti trng thi

cn bng, thu c 11 gam este. Hiu sut ca phn ng este ho l:

A.55% B.50%C.62,5% D.75%

Cu 18: Bit rng phn ng este ho CH3COOH + C2H5OH CH3COOC2H5 + H2O

C hng s cn bng K = 4, tnh % Ancol etylic b este ho nu bt u vi [C2H5OH] = 1 M, [CH3COOH] = 2 M.

A.80% B.68%C.75% D.84,5%

IV. XC NH CU TO ESTE KHI BIT CTPT.

* Este thng thng.

Cu 1:un este E ( C6H12O2) vi dung dch NaOH ta c 1 acol A khng b oxi ho bi CuO.E c tn l:

A. isopropyl propionat B. isopropyl axetat C.butyl axetatD. tert-butyl axetat.

Cu 2: Este X ( C4H8O2) thomn cc iu kin:

X HOH ,2 Y1 + Y2 Y1 xtO ,2 Y2

X c tn l:

A. Isopropyl fomiat B.propyl fomiat C. Metylpropionat D.Etyl axetat.

Cu 3: Hai cht hu c X, Y c cng cng thc phn t C3H4O2. X phn ng vi NaHCO3 v phn ng trng hp, Yphn ng vi NaOH nhng khng phn ng vi Na. Cng thc cu to ca X, Y ln lt l


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A.C2H5COOH, CH3COOCH3. B.C2H5COOH, CH2 = CHCOOCH3.

C.CH2 = CHCOOH, HCOOCH = CH2. D. CH2 = CH CH-

2COOH, HCOOCH = CH2.

Cu 4: Thu phn este c cng thc phn t C4H8O2 (xc tc H+), thu c 2 sn phm hu c X v Y. T X c thiu ch trc tip ra Y. Vy cht X l

A.metanol. B.Etyl axetat. C. Axit axetic.D.Etanol.

Cu 5: Thu phn este C4H6O2 (X) bng dung dch NaOH chthu c 1 mui duy nht. Cng thc cu to ca X l

A.CH3COOCH = CH2. B.HCOOCH2CH = CH2. D.

CH3CH = CHCOOH.

Cu 6:X c cng thc phn t C3H4O2. Khi cho X phn ng vi dung dchNaOH thu c 1 sn phm duy nht. Xcnh cng thc cu to ca X ?

A.CH2=CH-COOH. B.HCOOCH=CH2. C.

H3CHC C O

O .D.tt cu ng.

Cu 7: Hp cht X c cng thc phn t CnH2nO2 khng tc dng vi Na, khi un nng X vi axit v c c 2 cht

Y1v Y2. Bit Y2 b oxi ho cho metanal cn Y1 tham gia phn ng trng gng. Vy gi tr ca n l

A.1. B.2.C.3. D.4.

Cu 8: Cht X c cng thc phn t C7H6O3(M = 138). Bit 27,6 gam X tc dng va vi 600 ml dung dchNaOH 1M. Cng thc cu to ca X l

A. (HO)2C6H3CHO. B.HOC6H4CHO. C. (HO)3C6H2CH3.D.HCOOC6H4OH.

Cu 9: Cho 10,4 gam este X (cng thc phn t: C4H8O3) tc dng va vi 100 ml dung dch NaOH 1M c 9,8

gam mui khan. Cng thc cu to ca X l

A.HCOOCH2CH2CHO. B.CH3COOCH2CH2OH. C.HOCH2COOC2H5.D.CH3CH(OH)COOCH3.

Cu 10: Khi thu phn mt este c cng thc C4H8O2ta c axit X v ancol Y. Oxi ho Y vi K2Cr2O7 trong H2SO4ta c li X. Este c cng thc cu to no sau y?

O

C = O

C. (CH2)3


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A.CH3COOC2H5 B.HCOOC3H7 C. C2H5COOCH3D.Khng xc nh c.

* Este v sn phm c phn ng trng gng.

Cu 1: Thu phn este C2H5COOCH=CH2trong mi trng axit to thnh nhng sn phm g?

A.C2H5COOH, CH2=CH-OH B.C2H5COOH, HCHO C.C2H5COOH, CH3CHO D. C2H5COOH,CH3CH2OH

Cu 2. Este X c CTCP C4H6O2.Bit X thuphn trong mi trng kim to ra mui v anhit.. Cng thc cu toca X l.

A.CH3COOCH= CH2 B.HCOOCH2- CH= CH2 C.HCOOCH2- CH= CH2 D.CH3COOCH2CH3

Cu 3: Thu phn este C4H6O2trong mi trng axit thu c hn hp 2 cht u tham gia phn ng trng gng.Cng thc cu to ca este l

A.HCOOCH2CH = CH2 B.HCOOC(CH3) = CH2 C.CH2 = CHCOOCH3D.HCOOCH = CHCH3

Cu 4: Cho cht X tc dng vi 1 lng va dung dch NaOH, sau c cn dung dch thu c cht rn Y vcht hu c Z. Cho Z tc dng vi AgNO3/NH3c cht hu c T. Cho cht T tc dng vi NaOH li thu c chtY. Cht X c th l

A.HCOOCH = CH2. B.HCOOCH3. C. CH3COOCH =CHCH3. D.CH3COOCH = CH2.

Cu 5: Cht X c cng thc phn t C4H6O3, X c cc tnh cht ho hc sau:- Tc dng vi H2 (Ni, t0), Na, AgNO3/NH3.- Tc dng vi NaOH thu c mui v anehit n chc.Cng thc cu to ca X l

A.HCOOCH2CH2CHO. B.OHC-CH2CH2-COOH. C. HCOOCH(OH)-CH=CH2. D.CH3-CO-CH2-COOH.

Cu 6: Cho cht X c cng thc phn t C4H6O2 bit:

2 4 2 4

X + NaOH Y + Z

Y + H SO Na SO + T

Z v T u c khnng tham gia phn ng trng bc.

Cng thc phn t ca X l

A.CH3COOCH = CH2. B.HCOOCH2CH = CH2. C.HCOOC(CH3) = CH2.D.HCOOCH = CHCH3.


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Cu 7: Este C4H8O2 tham gia phn ng trng bc c th c tn sau:

A.Etyl fomiat B.propyl fomiat C.isopropyl fomiat D.B, C u ng

Cu 8: un este E (C4H6O2) vi HCl thu c sn phm c khnng c phn ng trng gng. E c tn l:

A.Vinyl axetat B.propenyl axetat C.Alyl fomiatD.CA, B, C u ng.

Cu 9: Mt cht hu c A c CTPT C3H6O2 tha mn: A tc dng c dd NaOH un nng v ddAgNO3/NH3,t

0.Vy A c CTCT l:

A.C2H5COOH B.CH3-COO- CH3 C. H-COO- C2H5 D.HOC-CH2-CH2OH

* Este thm.

Cu 1: Cho este X (C8H8O2) tc dng vi dung dch NaOH thu c hn hp mui u c khi lng phn t lnhn 70 vc. Cng thc cu to ca X l

A.HCOOC6H4CH3. B.CH3COOC6H5. C.C6H5COOCH3.D.HCOOCH2C6H5.

Cu 2: Hp cht thm X thuc loi este c cng thc phn t C8H8O2. X khng thiu ch t phn ng ca axit vancol tng ng v khng tham gia phn ng trng gng. Cng thc cu to ca X l

A.C6H5COOCH3. B.CH3COOC6H5. C.HCOOCH2C6H5.D.HCOOC6H4CH3.

Cu 3. Hai este A v B l dn xut ca benzen c CTPT l: C9H8O2. A v B u cng hp vi Br2 theo t l mol 1:1. Atc dng vi NaOH cho mt mui v mt anehit, B tc dng vi NaOH cho 2 mui v nc . Cc mui c khilng phn t ln hn khi lng phn t ca CH3COONa . CTCT ca A v B c th l:

A.HOOC-C6H4-CH=CH2 v CH2=CH-COO-C6H5 B. C6H5COOCH=CH2v C6H5-CH=CH-COOH

C.HCOO-C6H4-CH=CH2v HCOO-CH=CH-C6H5 D.C6H5COOCH=CH2 vCH2=CH-COO-C6H5

Cu 4: Hai este A, B l dn xut ca benzen c CTPT l C9H8O2. A v B u cng hp Br2 theo t l 1:1. A tc dng

vi NaOH to 1 mui v 1 anehit. B tc dng vi NaOH cho 2 mui v H2O. A, B c CTCT ln lt l:

A.C6H5COOCH=CH2, CH2=CH-COOC6H5 B.CH2=CH-COOC6H5,C6H5COOCH=CH2

C.HCOOCH=CH-C6H5, C6H5COOCH=CH2 D.C6H5COOCH=CH2,HCOOCH=CH-C6H5


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Cu 5: Este X l dn xut ca benzen c cng thc C9H8O2. X tc dng c vi dung dch Br2 theo t l 1:1. Khi choX tc dng vi dung dch NaOH un nng thu c 1 mui v mt anehit. Mui thu c c khi lng phn t lnhn 82. Cng thc cu to thu gn ca X l

A.C6H5COOCH=CH2. B.HCOOC6H4CH=CH2. C.HCOOCH=CHC6H5.D.HCOOC(C

6H

5)=CH

2.

Cu 6: Este X l dn xut ca benzen c cng thc C9H8O2. X tc dng c vi dung dch Br2 theo t l 1:1. Khi choX tc dng vi dung dch NaOH un nng thu c 1 mui v 1 xeton. Cng thc cu to thu gn ca X l

A.C6H5COOCH=CH2. B.HCOOC6H4CH=CH2. C.C6H5COOCH=CHCH3. D.HCOOC(C6H5)=CH2.

Cu 7: Este X l dn xut ca benzen c cng thc C9H8O2. X tc dng c vi dung dch Br2 theo t l 1:1. Khi choX tc dng vi dung dch NaOH un nng, d thu c 2 mui v nc . Cc mui u c khi lng phn t lnhn 82. Cng thc cu to thu gn ca X l

A.HCOOC6H4CH=CH2. B.CH2=CHCOOC6H5. C.CH2=CHCOOC6H4CH3.D.C2H5COOC6H5.

Cu 8: Este X c cng thc C9H8O2 tc dng vi dung dch NaOH to thnh 2 mui v nc . Nung nng 1 trong 2mui vi vi ti xt thu c etilen. X l

A.phenyl axetat. B.phenyl propionat. C.phenyl acrylat. D.benzyl axetat.

Cu 9:Mt hn hp X gm 2 este X, Y c cng cng thc phn t C8H8O2v u cha vng benzen. X phng hoht 0,2 mol X, ta cn 0,3 lit dung dch NaOH 1M thu c 3 mui.Tnh khi lng mi mui.

A.8,2 gam CH3COONa; 14,4 gam C6H5COONa; 11,6 gam C6H5ONa

B.4,1 gam CH3COONa; 14,4 gam C6H5COONa; 11,6 gam C6H5ONa

C.8,2 gam CH3COONa; 7,2 gam C6H5COONa; 5,8 gam C6H5ONa

D.4,1 gam CH3COONa; 14,4 gam C6H5COONa; 17,4 gam C6H5ONa

* Este a chc.

Cu 1: Este X c cng thc phn t C7H12O4, khi cho 16 gam X tc dng va vi 200 gam dung dch NaOH 4%th thu c 1 ancol A v 17,8 gam hn hp hai mui. Cng thc cu to ca X l

A.CH3COO(CH2)2OOCC2H5. B.HCOO(CH2)3OOCC2H5. C.HCOO(CH2)3OOCCH3.D.CH3COO(CH2)3OOCCH3.

Cu 2: Mt hp cht hu c X c cng thc phn t C8H14O4 . Khi X tc dng hon ton vi dd NaOH to ra haiancol A v B c s nguyn t cacbon gp i nhau . Khi un nng ln lt A , B vi H2SO4c 170

0C th A to ramt olefin duy nht , B to ra 3 olefin ng phn . X c cng thc cu to l.


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A.C2H5OOCCOOCH2- CH2- CH2- CH3 B.CH3OOC- (CH2)3-COO- CH2- CH3

C.C2H5OOCCOOCH(CH3) - CH2- CH3 D.C2H5OOCCOOCH2-CH(CH3)- CH3

V. XC NH CTPT, CTCT 1 ESTE.

* Este n chc.

Cu 1: Mt este n chc no c 48,65 % C trong phn t th sng phn este l:

A.1 B.2C.3 D.4

Cu 2:t chy hon ton 7,5 gam este X ta thu c 11 gam CO2 v 4,5 gam H2O.Nu X n chc th X c cng

thc phn t l:

A.C3H6O2 B.C4H8O2C.C5H10O2 D.C2H4O2

Cu 3: t chy hon ton 1,46 gam cht hu c A gm C, H, O th thu c 1,344 lit CO2(ktc) v 0,9 gam H2O.Cng thc no di y c th l cng thc ng .

A. (COOC2H5)2 B.CH3COOH C.CH3COOCH3 D.HOOC-C6H4-COOH

Cu 4: Thu phn mt este trong mi trng kim ta c ancol etylic m khi lng ancol bng 62% khi lngphn t este. Cng thc este c th l cng thc no di y?

A.HCOOCH3 B.HCOOC2H5C.CH3COOC2H5 D.C2H5COOC2H5

Cu 5:Thu phn mt este trong mi trng kim thu c mt mui natri c khi lng 41/37 khi lng este.Bitkhi lm bay hi 7,4 gam este th thtch hi ca n ng th tch ca 3,2 gam O2cng iu kin.Cng thc cu toca este c th l cng thc no di y?

A.HCOOCH3 B.HCOOC2H5C.CH3COOCH3 D.CH3COOC2H5

Cu 6:Khi t chy hon ton 4,4 gam cht hu c X n chc thu c sn phm chy ch gm 4,48 lit CO2 (ktc) v 3,6 gam H2O. Nu cho 4,4 gam cht X tc dng vi dung dch NaOH va n khi phn ng hon ton, thuc 4,8 gam mui ca axit hu c Y v cht hu c Z.Tn ca X l:

A.Etyl propionat B.Metyl propionatC. isopropyl axetat D.etyl axetat


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Cu 7: Este X khng no, mch h, c t khi hi so vi oxi bng 3,125 v khi tham gia phn ng x phng ho to ramt anehit v mt mui ca axit hu c. C bao nhiu cng thc ph hp vi X?

A.2 B.3C.4 D.5

Cu 8: Cho cht X tc dng vi mt lng va dung dchNaOH, sau c cn dung dch thu c cht rn Y vcht hu c Z. Cho Z tc dng vi dung dch AgNO3 trong NH3thu c cht hu c T. Cho T tc dng vi dungdch NaOH li thu c cht Y. Cht X c th l

A.HCOOCH=CH2 B.CH3COOCH=CH2 C.HCOOCH3 D.CH3COOCH=CH-CH3

Cu 9: X l mt este no n chc, c t khi hi so vi CH4 l 5,5. Nu em un 2,2 gam este X vi dd NaOH d, thuc 2,05 gam mui. Cng thc cu to thu gn ca X l:

A.HCOOCH2CH2CH3 B.HCOOCH(CH3)2 C.C2H5COOCH3D.CH3COOC2H5

Cu 10: A c cng thc phn t trng vi cng thc n gin. Khi phn tch A thu c kt qu: 50% C, 5,56% H,44,44%O theo khi lng. Khi thu phn A bng dung dch H2SO4long thu c 2 sn phm u tham gia phn ngtrng bc. Cng thc cu to ca A l

A.HCOO-CH=CH-CH3. B.HCOO-CH=CH2. C.(HCOO)2C2H4. D.CH2=CH-CHO.

Cu 11: Cho 13,2 g este n chc no E tc dng ht vi 150 ml dung dch NaOH 1M thu c 12,3 g mui . Xcnh E.

A.HCOOCH3 B.CH3-COOC2H5 C.HCOOC2H5D.CH3COOCH3

Cu 12: Thy phn 1 este n chc no E bng dung dch NaOH thu c mui khan c khi lng phn t bng24/29 khi lng phn t E.T khi hi ca E i vi khng kh bng 4. Cng thc cu to.

A.C2H5COOCH3. B.C2H5COOC3H7 C. C3H7COOCH3D.Kt qu khc

Cu 13: X l este ca mt axit hu c n chc v ancol n chc. thu phn hon ton 6,6g cht X, ngi ta

dng 34,10ml dung dch NaOH 10% c D = 1,1g/ml. Lng NaOH ny d 25% so vi lng NaOH cn dng chophn ng. X c cng thc cu to no sau y?

A.HCOOC3H7 vCH3COOC2H5 B.HCOOC3H7 C.CH3COOC2H5D.C2H5COOCH3

Cu 14: x phng ho 17,4g mt este no n chc cn dng 300ml dung dch NaOH 0,5M. Este c cng thcphn t l


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A.C3H6O2 B.C5H10O2C.C4H8O2 D.Kt qu khc

Cu 15: 12,9g mt este n chc, mch htc dng ht vi 150ml dung dch KOH 1M. Sau phn ng thu c mtmui vanehit. Cng thc cu to ca este l cng thc no sau y?

A.HCOOCH=CH-CH3 B.CH3COOCH=CH2 C.C2H5COOCH=CH2 D.A v B ng.

Cu 16: Mt este n chc c thnh phn khi lng mC:mO = 9:8 .Cho este trn tc dng vi mt lng dung dchNaOH va thu c mt mui c khi lng bng 41/37 khi lng este. Cng thc cu to este l:

A.HCOOCH=CH2 B.HCOOC=CH-CH3 C.HCOOC2H5D.CH3COOCH3

Cu 17: t chy 3g mt este Y ta thu c 2,24lt kh CO2(ktc) v 1,8g H2O. Y c cng thc cu to no sau y?

A.HCOOCH3 B.CH3COOCH3C.CH2=CHCOOCH3 D.A, B, C u sai

Cu 18: Este X to bi ancol no n chc v axit cacboxylic khng no (c 1 lin kt i) n chc. t chy m mol Xthu c 22,4 lt CO2(ktc) v 9g H2O .Gi tr ca m l bao nhiu trong cc scho di y?

A.1 mol B.2 molC.3 mol D.Kt qu khc

Cu 19: t chy hon ton 0,1mol este X thu c 0,3mol CO2 v 0,3 mol H2O. Nu cho 0,1mol X tc dng ht viNaOH th thu c 8,2g mui. X l cng thc cu to no sau y:

A.CH3COOCH3 B.HCOOCH3C.CH3COOC2H5 D.HCOOC2H5

Cu 20: un nng 1,1g este no n chc M vi dung dch KOH d, ngi ta thu c 1,4g mui. T khi ca M sovi kh CO2 l 2. M c cng thc cu to no sau y?

A.C2H5COOCH3 B.CH3COOC2H5 C.HCOOC3H7 D.CH3COOC2H5

Cu 21:t chy hon ton 0,2 mol este X ri dn sn phm chy vo dung dch Ca(OH)2d thu c 40g kt ta. Xc cng thc phn t l:

A.HCOOC2H5 B.CH3COOCH3C.HCOOCH3 D.Khng xc nh c.

Cu 22:Khi t chy hon ton este no n chc th s mol CO2 sinh ra bng s mol O2 phn ng. Tn gi caeste l: A.etyl axetat B.metyl axetat

C.metyl fomiat D.propyl axetat


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Cu 23: Este n chc X c t khi hi so vi CH4 l 6,25.Cho 20 gam X tc dng vi 300 ml dung dch KOH1M(un nng). C cn dung dch c sau phn ng thu c 28 gam cht rn khan. Cng thc cu to ca X l:

A.CH2=CH-CH2COOCH3 B.CH2=CH-COOCH2CH3 C.CH3COOCH=CH-CH3D.CH

3-CH

2COOCH=CH

2

Cu 24. Cht X l mt hp cht n chc mch h, tc dng c vi dd NaOH c khi lng phn t l 88 dvc. Khicho 4,4g X tc dng va vi dd NaOH, c cn dung dich sau phn ng c 4,1g cht rn. X l cht no trong cccht sau:

A.Axit Butanoic B.Metyl PropionatC.Etyl Axetat D. Isopropyl Fomiat .

Cu 25: Mt cht hu c X mch hc khi lng phn tl 60 vC tha mn iu kin sau:

X khng tc dng vi Na, X tc dng vi d2 NaOH, v X phn ng vi Ag2O.NH3. Vy X l cht no trong cc

cht sau:

A.CH3COOH B.HCOOCH3 C.C3H7OH D.HOCH2CHO

Cu 26. Mt este X c to ra bi mt axit no n chc v ancol no n chc c dX/CO2=2. Cng thc phn t caX l:

A.C2H402 B.C3H602C.C4H602 D.C4H802

Cu 27.Cho 4,2g este n chc no E tc dng ht vi dd NaOH ta thu c 4,76g mui natri. Vy cng thc cu to

ca E c th l:

A.CH3COOCH3 B.C2H5COOCH3 C.CH3COOC2H5 D.HCOOC2H5

Cu 28: Hp cht hu c X n chc cha (C, H, O) khng tc dng vi Na nhng tc dng vi dung dch NaOHtheo t l mol 1 : 1 hoc 1 : 2. Khi t chy 1 mol X thu c 7 mol CO2. Cng thc cu to ca X l

A.C2H5COOC4H9. B.HCOOC6H5. C. C6H5COOH.D.C3H7COOC3H7.

Cu 29: Khi t chy hon ton 4,4 gam hp cht hu c X n chc thu c sn phm chy ch gm 4,48 lt CO2

(ktc) v 3,6 gam nc. Nu cho 4,4 gam hp cht X tc dng vi dung dch NaOH va n khi phn ng honton, thu c 4,8 gam mui ca axit hu c Y v cht hu c Z. Tn ca X l

D.etyl propionat. B.Metyl propionat. C.Isopropyl axetat. D.Etyl axetat.

Cu 30: X l mt este no n chc mch h, t khi hi i vi CH4 l 5,5. Nu un nng 2,2 gam este X vi dungdch NaOH (d), thu c 2,05 gam mui. Cng thc cu to ca X l


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A.HCOOCH2CH2CH3. B.C2H5COOCH3. C.CH3COOC2H5. D.HCOOCH(CH3)2.

Cu 31: Hai este n chc X v Y l ng phn ca nhau. Khi ho hi 1,85 gam X, thu c hi ng bng th tchhi ca 0,7 gam N2(o cng iu kin). Cng thc cu to thu gn ca X v Y l

A.HCOOC2H5 v CH3COOCH3.B.C2H5COOCH3 v HCOOCH(CH3)2.

C.C2H3COOC2H5 v C2H5COOC2H3. D.HCOOCH2CH2CH3 v CH3COOC2H5.

Cu 32: t chy 1,6 gam mt este X n chc thu c 3,52 gam CO2 v 1,152 gam H2O. Cho 10 gam X tc dngvi dung dch NaOH (va ), c cn dung dch sau phn ng thu c 14 gam mui khan Y. Cho Y tc dng vi axitv c long thu c Z khng phn nhnh. Cng thc cu to ca Z l

A.CH3(CH2)3COOH. B.CH2 = CH(CH2)2COOH. C.

HO(CH2)4COOH. D.HO(CH2)4OH.Cu 33: X l mt este khng no (cha 1 lin kt i C = C) n chc, mch h. t chy hon ton 4,3 gam X cnva 7,2 gam O2. X c ti a bao nhiu cng thc cu to?

A.3. B.4.C.5. D.6.

Cu 34:Hn hp A gm 2 este n chc no, ng phn. Khi trn 0,1 mol hn hp A vi O2 va ri t chy thuc 0,6 mol sn phm gm CO2v hi nc. Cng thc phn t2 este l

A.C4H8O2. B.C5H10O2. C.

C3H6O2. D.C3H8O2.

Cu 35: Mt este to bi axit n chc v ancol n chc c t khi hi so vi CO2 bng 2. Khi un nng este ny vidung dch NaOH to ra lng mui c khi lng ln hn lng este phn ng . Este l

A.Metyl axetat. B.Propyl axetat. C. Metylpropionat. D.Etyl axetat.

Cu 36: Cho 1,76 gam mt este no, n chc phn ng va ht vi 40 ml dung dch NaOH 0,5M thu c cht X vcht Y. t chy hon ton 1,2 gam cht Y c 2,64 gam CO2 v 1,44 gam H2O. Cng thc cu to ca este l

A.HCOOCH2CH2CH3. B.CH3COOC2H5. C.

C2H5COOCH3. D.CH3COOCH(CH3)2.

Cu 37: un nng hp cht X vi H2O (xc tc H+) c axit hu c Y v ancol Z n chc. Cho hi Z i qua ng

ng CuO, t0c hp cht T c khnng tham gia phn ng trng bc. t chy hon ton 2,8 gam X phi dng ht

3,92 lt oxi (ktc), c kh CO2v hi nc theo t l th tch:2 2CO H O

V : V 3 : 2 . Bit2

YN

d 2,57 . Cng

thc cu to ca X l


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A.CH2= CHCOOC3H7. B.CH2= CHCOOCH2CH = CH2. C.C2H5COOCH = CH2. D.CH2=CHCH2COOCH=CH2.

Cu 38: X l este ca mt axit cacboxylic n chc v ancol etylic. Thy phn hon ton 7,4 gam X ngi ta dng125 ml dung dch NaOH 1M. Lng NaOH d 25% so vi l thuyt (lng cn thit). Cng thc cu to ca X l

A.HCOOC2H5. B.CH3COOC2H5. C.C2H5COOC2H5. D.HCOOCH3.

Cu 39: t chy hon ton 0,2 mol este n chc X ri cho sn phm chy ln lt qua bnh 1 ng 100 gam dungdch H2SO496,48%; bnh 2 ng dung dch KOH d. Sau th nghim thy nng H2SO4bnh 1 gim cn 87,08%;bnh 2 c 82,8 gam mui. Cng thc phn t ca X l

A.C2H4O2. B.C3H6O2. C.C4H8O2. D.C3H4O2.

Cu 40: Cho ancol X tc dng vi axit Y c este E. Lm bay hi 8,6 gam E c thtch hi bng th tch ca 3,2

gam kh oxi (o cng iu kin), bit MY > MX. Cng thc cu to ca E l :

A.HCOOCH2CH = CH2. B.CH3COOCH = CH2. C.CH2 = CHCOOCH3.D.HCOOCH = CHCH3.

Cu 41: Mt este n chc X c phn t khi l 88 vC. Cho 17,6 gam X tc dng vi 300 ml dung dch NaOH 1M.Khi phn ng xy ra hon ton, c cn dung dch sau phn ng thu c 23,2 gam cht rn khan. Cng thc cu toca X l

A.HCOOCH2CH2CH3. B.HCOOC3H7. C.CH3CH2COOCH3. D.CH3COOCH2CH3.

Cu 42: t chy hon ton 4,44 gam cht hu c X n chc (cha C, H, O). Cho ton b sn phm chy hp thhon ton vo bnh ng dung dch Ca(OH)2d thy khi lng bnh tng 11,16 gam ng thi thu c 18 gam ktta. Ly m1 gam X cho tc dng vi dung dch NaOH (va ), c cn dung dch sau phn ng c m2gam chtrnkhan. Bit m2 < m1. Cng thc cu to ca X l

A.HCOOC2H5. B.CH3COOCH3. C. C2H5COOH.D.CH2 = CHCOOCH3.

Cu 43: Hn hp M gm mt axit X n chc, mt ancol Y n chc v mt este to ra t X v Y. Khi cho 25,2gam hn hp M tc dng va vi 100 ml dung dch NaOH 2M c 13,6 gam mui khan. Nu un nng Y viH2SO4c th thu c cht hu c Y1 c t khi hi so vi Y bng 1,7 (coi hiu sut t 100%). Cng thc cu to

ca este l

A.HCOOCH2CH2CH3. B.CH3COOC3H7. C.HCOOCH(CH3)2. D.HCOOC2H4CH3 hoc HCOOCH(CH3)2.

* Este a chc.


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Cu 1: Cho 21,8 gam cht hu c X mch hch cha mt loi nhm chc tc dng vi 1 lt dung dch NaOH 0,5Mthu c 24,6 gam mui v 0,1 mol ancol. Lng NaOH d c trung ho va ht bi 0,5 lt dung dch HCl 0,4M.Cng thc cu to ca X l

A. (HCOO)3C3H5. B. (CH3COO)2C2H4. C.(CH

3COO)

3C

3H

5. D.C

3H

5(COOCH

3)3.

Cu 2: Thu phn hon ton 444 gam mt lipit thu c 46 gam glixerol v hai loi axit bo. Hai loi axit bo l

A.C15H31COOH v C17H35COOH.B.C17H33COOH v C15H31COOH.

C.C17H31COOH v C17H33COOH.D.C17H33COOH v C17H35COOH.

Cu 3: thu phn 0,01 mol este to bi mt ancol a chc v mt axit cacboxylic n chc cn dng 1,2 gamNaOH. Mt khc thy phn 6,35 gam este cn 3 gam NaOH, sau phn ng thu c 7,05 gam mui. Cng thc

cu to ca este l

A. (CH3COO)3C3H5. B. (CH2 = CHCOO)3C3H5. C. (CH2 = CHCOO)2C2H4.D. (C3H5COO)3C3H5.

Cu 4: iu ch mt este X, dng lm thuc chng mui gi tt l DEP ngi ta cho axit Y tc dng vi mt lngd ancol Z. Mun trung ho dung dch cha 1,66 gam Y cn 100 ml dung dch NaOH 0,2M. Trong dung dch ancol Z

94% (theo khi lng) t l s mol2Z H O

n : n 86 :14 . Bit 100 < MY< 200. Cng thc cu to ca X l

A.CH2 = CHCOOCH3. B.C6H5COOC2H5. C.C6H4(COOC2H5)2. D. (C2H5COO)2C6H4.

Cu 5.un nng 0,1 mol X vi lng va dd NaOH thu c 13,4g mui ca axit hu c a chc B v 9,2g ancoln chc C.Cho ancol C bay hi 1270C v 600 mmHg s chim th tch 8,32 lt.Cng thc phn t ca cht X l:

A.

CH COOCH3

COOCH3

COOCH3 B.

CH2

CH2

COOCH3

COOCH3

C. O

OOC

2H5

OC2H5

D. O

OOCH

3

OCH3

Cu 6:X l este ca glixerol v axit hu c Y. t chy hon ton 0,1 mol X ri hp th tt c sn phm chy vodung dch Ca(OH)2d thu c 60g kt ta. X c cng thc cu to l:

A.(HCOO)3C3H5 B. (CH3COO)3C3H5C. (C17H35COO)3C3H5 D. (C17H33COO)3C3H5


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Cu 7:X l este mch hdo axit no A v ancol no B to ra. Khi cho 0,2 mol X phn ng vi NaOH thu c 32,8gam mui. t chy 1 mol B cn dng 2,5 mol O2. Cng thc cu to ca X l

A.(CH3COO)2C2H4. B.(HCOO)2C2H4.C.(C2H5COO)2C2H4. D.(CH3COO)3C3H5.

VI. XC NH CTPT, CTCT HN HP ESTE.

Cu 1: X phng ha hon ton 9,7 gam hn hp hai este n chc X, Y cn 150 ml dung dch KOH 1M. Sau phnng c cn dung dch thu c hn hp hai ancol ng ng k tip v mt mui duy nht. Cng thc cu to thu gnca X, Y ln lt l

A.HCOOCH3, HCOOC2H5.B.C2H5COOCH3, C2H5COOC2H5.

C.CH3COOCH3, CH3COOC2H5D.C2H3COOCH3, C2H3COOC2H5.

Cu 2: Mt hn hp X gm 2 este A, B n chc ng ng lin tip, khi b x phng ho cho ra 2 mui caaxitcacboxylic v 1 ancol. Th tch dung dch NaOH 1M cn dng x phng ho este ny l 0,3 lit. Xc nh CTCTv s mol mi este trong hn hp X. Bit rng khi lng mX=23,6 gam v trong 2 axit A, B khng c axit no chophn ng trng gng.

A.0,1 mol CH3COOCH3 v 0,2 mol CH3COOC2H5 B. 0,2 mol CH3COOCH3v 0,1 mol CH3COOC2H5

C.0,2 mol CH3COOCH3 v 0,1 mol C2H5COOCH3 D. 0,2 mol HCOOCH3 v0,2 mol CH3COOC2H5

Cu 3: Cho 35,2 gam hn hp gm 2 este no n chc l ng phn ca nhau c t khi hi i vi H2 bng 44 tcdng vi 2 lit dung dch NaOH 0,4 M, ri c cn dung dch va thu c, ta c 44,6 gam cht rn B. Cng thc ca2 este l:

A.HCOOC2H5 v CH3COOCH3B.C2H5COOCH3 v CH3COOC2H5

C.HCOOC3H7 v CH3COOC2H5D.HCOOC3H7 v CH3COOCH3

Cu 4: t chy hon ton 3,7g hn hp 2 este ng phn X v Y ta thu c 3,36 lt kh CO2(ktc) v 2,7g H2O. X

v Y c cng thc cu to l:A.CH2=CHCOOCH3 v HCOOCH2CH=CH2 B. CH3COOCH3 v

HCOOC2H5

C.CH2=CHCOOC2H5v C2H5COOCH=CH2 D.Kt qu khc.


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Cu 5: x phng ho hon ton 2,22g hn hp 2 este ng phn X v Y cn dng ht 30ml dung dch KOH 1M.Khi t chy hon ton hn hp 2 este th thu c kh CO2v hi nc c th tch bng nhau v o cng iukin. Cng thc phn t ca X, Y l:

A.CH3COOCH3 v HCOOC2H5B.C

2H

5COOCH

3v CH

3COOC

2H

5

C.C3H7COOCH3 v CH3COOC3H7 D. Ktqu khc.

Cu 6: Hai este n chc X v Y l ng phn ca nhau. Khi ho hi 1,85 gam X, thu c thtch hi ng bng thtch ca 0,7 gam N2( o cng iu kin). Cng thc cu to thu gn ca X, Y l:

A.C2H5COOCH3 v HCOOCH(CH3)2 B.HCOOC2H5 v CH3COOCH3

C.C2H3COOC2H5 v C2H5COOC2H3 D.

HCOOCH2CH2CH3 v CH3COOC2H5

Cu 7: Lm bay hi 5,98 gam hn hp 2 este ca axit axetic v 2 ancol ng ng k tip ca ancol metylic. N chimthtch 1,344 lit (ktc). Cng thc cu to ca 2 este l:

A.HCOOC2H5 v HCOOC3H7B.CH3COOCH3 v CH3COOC2H5

C.CH3COOC2H5 v CH3COOC3H7 D.CH3COOCH3 v CH3COOC2H5

Cu 8: X l hn hp 2 este ng phn c to thnh t mt ancol n chc, mch cacbon khng phn nhnh vi axit

n chc. T khi hi ca X so vi hiro bng 44. Cng thc phn t ca X l:

A.C3H6O2 B.C4H8O2C.C5H10O2 D.C6H12O2

Cu 9: t chy hon ton mt lng hn hp hai este n chc no, mch hcn 3,976 lt oxi (o diu kin tiuchun) thu c 6,38 g CO2. Cho lng este ny tc dng va vi KOH thu c hn hp hai ru k tip v 3,92g mui ca mt axit hu c. Cng thc cu to ca hai cht hu c trong hn hp u l:

A.HCOOCH3 v C2H5COOCH3B.CH3COOC2H5 v C3H7OH

C.CH3COOCH3 V CH3COOC2H5 D. CH3COOCH3V CH3COOC2H5

Cu 10: Hn hp Y gm hai este n chc mch hl ng phn ca nhau. Cho m gam hn hp Y tc dng va vi 100ml dung dch NaOH 0,5M, thu c mt mui ca mt axit cacboxylic v hn hp hai ru. Mt khc tchy hon ton m gam hn hp Y cn dng 5,6 lt O2v thu c 4,48 lt CO2 (cc thtch kh o iu kin tiuchun). Cng thc cu to ca 2 este trong hn hp Y l:


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A.CH3COOCH3 v HCOOC2H5 B.C2H5COOCH3 v HCOOC3H

C.CH3COOCH3 v CH3COOC2H5 D.HCOO-CH(CH3)CH3

VII. ESTE V CC HP CHT KHC.

Cu 1: Mt hn hp X gm 2 cht hu c n chc. Cho X phn ng va vi 500ml dung dch KOH 1M. Sauphn ng thu c hn hp Y gm 2 mui ca hai axit cacboxylic v mt ancol. Cho ton blng ancol thu c trn tc dng vi Na d, sinh ra 3,36 lit H2( ktc). Hn hp X gm:

A.mt axit v mt este B.mt este v mt ancol C.hai esteD.mt axit v mt ancol

bài tập Axit-Este

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