PHƯƠNG PHÁP GIẢI BÀI TẬP PHẦN DẪN XUẤT HALOGEN

8/8/2019 PHNG PHP GII BI TP PHN DN XUT HALOGEN

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Dnh cho hc sinh yu thch mn Ho hc

PHNG PHP GII BI TP PHN DN XUT HALOGEN

- ANCOL - PHENOL

A. LU CCH TR LI TRC NGHIM

Phn ng tch nc ca ancol

To anken:

Sn phm chnh c xc nh theo quy tc Zaixep.

Quy tc Zaixep: Nhm -OH u tin tch ra cng vi H cacbon bc cao hn bn cnh to thnh linkt i C=C mang nhiu nhm ankyl hn.

+ To ete:

(Vi n loi ancol s to ra2

)1( +nnloi ete, trong c n loi ete i xng)

c bit: Ring vi etanol c kh nng tch nc to but-1,3- ien:

Phn ng oxi ha:

Oxi ha khng hon ton:

Ancol bc I b oxi ha thnh anehit:

Ancol bc II b oxi ha thnh xeton:

Ancol bc III khng b oxi ha

Oxi ha hon ton: CnH2n+1OH + 2

3n

O2

o

t

nCO2 + (n+1)H2O(Sn phm chnh c xc nh theo quy tc Maccpnhicp)

(sn phm chnh)

Nhn bit ancol

- Phn bit cc ancol c bc khc nhau

* un nng vi CuO (hoc t nng trn si dy ng)

Ancol bc I b oxi ha thnh anehit (nhn bit sn phm to thnh bng phn ng trng bc). Ancol

bc II b oxi ha thnh xeton (sn phm to thnh khng tham gia phn ng trng bc). Ancol bc IIIkhng b oxi ha trong iu kin trn.

Tp ch Ho hc v ng dng 1

H2SO4 ,c

170o

+ H2OCnH2n+1OH CnH2n

C2H5OH

H2SO4 ,c

140o

+ + H2OC2H5OH C2H5OC2H5

2C2H5OH + 2H2OAl2O3, ZnO

450oCH2=CH- CH=CH2 + 2H2

H2ORCH2OH + O2Cu

toRCHO +

H2SO4 ,c

140o

ROH + + H2OR'OH ROR'

R C

O

R CH R'

OH

R'O2+Cu

to+ H2O

CH2 CH CH3 + HOHH2SO4,l

CH3 CH

OH

CH3


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* Cng c th phn bit cc bc ca ancol bng thuc th Luca l hn hp ca HCl m c v ZnCl2

Ancol bc III s phn ng ngay lp tc to ra dn xut clo khng tan trong nc.

Ancol bc II phn ng chm hn, thng phi ch t pht mi to ra dn xut clo.

Ancol bc mt khng cho dn xut clo nhit phng.

Ch : Phenol khng tc dng trc tip vi axit hu c nh ancol. Mun iu ch este ca phenolphi dng clorua axit hoc anhirit axit v mt in tch dng nhm C=O ln hn axit v phnng c thc hin trong mi trng kim

V d

C6H5OH + (CH3CO)2O CH3COOC6H5 + CH3COOH

Mt s lu khi gii bi tp

1. ru: l s ml ru nguyn cht c trong 100 ml dung dch ru

V d: Trong 100 ml ru 960 c cha 96 ml ru nguyn cht

2. Trong phn ng ete ha ancol n chc cn lu

Vi n loi ancol s to ra2

)1( +nnloi ete, trong c n loi ete i xng

S mol H2O to ra = tng mol ete =1

2tng mol cc ancol tham gia phn ng

Nu cc ete to ra c s mol bng nhau th cc ancol tham gia phn ng ete ha c s mol nh nhau

3. CTPT chung ca ancol

- Ancol no n chc : CnH2n+1OH

- Ancol no a chc, mch h : CnH2n+2-a(OH)a (ancol bn nu n a)

- Ancol khng no ch bn khi -OH lin kt vi C c lin kt n. Nu -OH lin kt vi C khng no(ca lin kt i, ba) th ancol khng bn v b chuyn ha ngay thnh anehit hoc xeton

- Trong ancol no, a chc mi nhm -OH ch lin kt trn mi cacbon. Nu nhiu nhm -OH cng linkt trn mt nguyn t cacbon th phn t ancol t tch nc to thnh anehit, xeton hoc axit.

4. CTTQ ca phenol n chc, gc hirocacbon lin kt vi nhn benzen l gc hirocacbon no :CnH2n-7OH (n 6)

5. Phn bit phenol v ru

Phenol c th tan trong dung dch kim to thnh dung dch trong sut.

Phenol to kt ta trng (2,4,6-tribromphenol) vi dung dch nc brom.

B. BI TRC NGHIM C LI GII

1. Khi thc hin phn ng tch nc i vi ru (ancol) X, ch thu c mt anken duy nht. Oxi hohon ton mt lng cht X thu c 5,6 lt CO2 ( ktc) v 5,4 gam nc. C bao nhiu cng thc cu to

ph hp vi X?

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

(Trch thi tuyn sinh C khi A- nm 2007)

p n B

Hng dn

V loi nc thu c anken nn X l ancol no, n chct CTPT ca X l CnH2n+2O



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t chy 1 mol X thu c n mol CO2 v (n+1) mol H2O

Theo : S mol CO2 l 5,6 : 22,4 = 0,25 mol

S mol H2O l 5,4 : 18 = 0,3 mol

Ta c n = 5

Vy CTPT l C5H12OX c 4 CTCT ph hp l CH3-CH2-CH2-CH2-CH2OH

CH3-CH(CH3)-CH2-CH2OH

CH3-CH2-CHOH-CH2-CH3

CH3-CH2-CH(CH3)-CH2OH

2. Hp cht hu c X (phn t c vng benzen) c cng thc phn t l C7H8O2, tc dng c vi Na vvi NaOH. Bit rng khi cho X tc dng vi Na d, s mol H2 thu c bng s mol X tham gia phn ngv X ch tc dng c vi NaOH theo t l s mol 1:1. Cng thc cu to thu gn ca X l

A. C6H5CH(OH)2. B. HOC6H4CH2OH.

C. CH3C6H3(OH)2. D. CH3OC6H4OH.(Trch thi tuyn sinh C khi A- nm 2007)

p n B

Hng dn

- X tc dng vi Na d, s mol H2 thu c bng s mol X tham gia phn ng.

X c 2 nhm -OH. X ch tc dng c vi NaOH theo t l s mol 1:1 X c 1 nhm OH nhtrc tip vi vng benzen (phenol), v 1 nhm OH nh nhnh (ru)

3. Cho hn hp hai anken ng ng k tip nhau tc dng vi nc (c H2SO4 lm xc tc) thu c hnhp Z gm hai ru (ancol) X v Y. t chy hon ton 1,06 gam hn hp Z sau hp th ton b sn

phm chy vo 2 lt dung dch NaOH 0,1M thu c dung dch T trong nng ca NaOH bng0,05M. Cng thc cu to thu gn ca X v Y l (Cho: H = 1; C = 12; O = 16; th tch dung dch thay ikhng ng k).

A. C2H5OH v C3H7OH. B. C3H7OH v C4H9OH.

C. C2H5OH v C4H9OH. D. C4H9OH v C5H11OH.


p n A

Hng dn

Theo X, Y l sn phm cng nc vo anken nn X, Y l ancol no n chc

t CT chung ca X, Y l 2 2n nC H O+

2 2n nC H O+ +3

2

nO2 n CO2 + (n +1) H2O

S mol NaOH cn d: 0,05. 2 = 0,1 mol CO2 b hp th hon ton theo phn ng CO2 + 2 NaOH Na2CO3 + H2O

S mol NaOH tham gia phn ng l 2. 0,1 - 0,1 = 0,1 mol

s mol CO2 = 0,05 mol

Ta c s mol hai ancol = 1,06/ (14n +18) m nCO2 = n . nancol



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nn1,06.

18 14.

n

n+= 0,05 n = 2,5

Do hai anken l ng ng k tip nn X, Y cng l ng ng k tip

Cng thc ca X, Y l C2H5OH v C3H7OH

4. C bao nhiu ru (ancol) bc 2, no, n chc, mch h l ng phn cu to ca nhau m phn t cachng c phn trm khi lng cacbon bng 68,18%?

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


p n B

Hng dn

CTPT ca ru (ancol) no, n chc: CnH2n+2O.

%mC =12

.10014 18

n

n += 68,18% n = 5. C5H12O

C 3 ng phn ru bc 2: CH3-CH2-CH2-CH(OH)-CH3

CH3-CH2 -CH(OH) -CH2-CH3

CH3-CH(CH3) -CH(OH) -CH3

5. Cho mt hn hp hi metanol v etanol i qua ng cha CuO nung nng, khng c khng kh. Cc snphm kh v hi sinh ra c dn i qua nhng bnh cha ring r H2SO4 c v KOH. Sau th nghim,thyng ng CuO gim 80 gam, bnh ng H2SO4 tng 54 gam. Khi lng etanol tham gia phn ng l

A. 46 gam B. 15,33 gam C. 23 gam D. 14,67 gam

p n B

Hng dn iu kin trn (CuO nung ), CuO s cung cp oxi oxi ha hon ton to CO2 v H2O

Gi x, y ln lt l s mol CH3OH v C2H5OH

CH3OH + 3 CuOo

t CO2 + 2 H2O + 3 Cu

x mol 3x mol x mol 2x mol

C2H5OH + 6 CuOo

t 2 CO2 + 3 H2O + 6 Cu

y mol 6y mol 2y mol 3y mol

S mol oxi dng: 3x + 6y = 80 : 16 = 5 mol

S mol H2O sinh ra : 2x + 3y = 54 : 18 = 3 molGii ra ta c x = 1 mol, y = 1/3 mol

Khi lng etanol l 46. 1/3 = 15,33 gam

6. T mt tn khoai cha 20% tinh bt, bng phng php ln men ngi ta iu ch c 100 lt ruetylic tuyt i c khi lng ring l 0,8 g/ml. Hiu sut ca qu trnh phn ng l

A. 100 % B. 70% C. 80% D. 75%

p nB

Hng dn

S qu trnh iu ch

(C6H10O5)n + nH2O m e n nC6H12O6 (1)



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C6H12O6 m e n 2C2H5OH + 2CO2 (2)

Khi lng tinh bt : 620

.10100

= 2. 105 gam

T phng trnh (1) v (2) ta c khi lng ru etylic thu c l

5

2.10 . .2.46162 nn= 113580. 24 g

Hiu sut ca qu trnh sn xut l

100.0,8.1000.100

113580,24= 70%

7. un mt ancol X vi hn hp (ly d) KBr v H2SO4 c thu c 12,3 gam cht hu c Y. Hiu sutphn ng t 60%. Cht Y cha 29,27% C, 5,69% H v 65,04% mt nguyn t khc. Hi ca 12,3 gam Yni trn chim mt th tch bng th tch ca 2,8 gam nit trong cng iu kin. Bit khi oxi ha ancol X

bi CuO thu c mt anehit. Cng thc cu to thu gn ca X v Y l

A. CH3-CH2OH v CH3CH2BrB. CH3-CH2-CH2OH v CH3CH2CH2Br

C. CH3-CH2-CH2OH v CH3-CHBr-CH3

D. CH2=CH-CH2OH v CH2=CHCH2Br

p n B

Hng dn

S mol ca 12,3 gam Y bng s mol ca 2,8 gam nit tc l 2,8 : 28 = 0,1 mol

Do MB = 12,3 : 0,1 = 123

Y l dn xut cha brom. t cng thc phn t ca Y l CxHyBrz

Ta c: x : y : z =29,27 5,69 65,04

: :12 1 80

= 3 : 7 : 1

(C3H7Br)n = 123 suy ra n = 1. Cng thc phn t ca Y l C3H7Br cn cng thc ca X l C3H7OH

V khi oxi ha X thu c anehit nn X l ancol bc 1

Vy cng thc cu to ca X, Y l CH3-CH2-CH2OH v CH3CH2CH2Br

8. X l mt ancol no, mch h. t chy hon ton 0,05 mol X cn 5,6 gam oxi, thu c hi nc v 6,6gam CO2. Cng thc ca X l

A. C2H4(OH)2 B. C3H6(OH)2 C. C3H5(OH)3 D. C3H7OH

(Trch thi tuyn sinh H 2007, khi B)p n C

Hng dn

2

5,60,175

32On = = mol;

2

6,61,5

44COn = = mol

Phn ng chy: 2 2 2 2 23 1

( 1)2n n x

n xC H O O nCO n H O

+

+ + + +

0,05 mol 0,175 mol 1,5 mol

n = 3;

3 1

3,52

n x+ = x= 3.



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9. Khi t 0,1 mol cht X (dn xut ca benzen), khi lng CO2 thu c nh hn 35,2 gam. Bit rng 1mol X ch tc dng c vi 1 mol NaOH. Cng thc cu to thu gn ca X l

A. HOCH2C6H4COOH B. C6H4(OH)2

C. C2H5C6H4OH D. C6H4(CH3)OH

(Trch thi tuyn sinh H 2007, khi B)

p n D

Hng dn

2

35,20,8

44CO

n = = mol;

t chy CxHyOzxCO2 x

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