daÃn xuaÁt halogen ancol – phenol - dayhoahoc.com · ths. lƯu huỲ nh vẠ n long...

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LUYEN THI AI HOC 2014

CHUYEN E HU C 2:

DAN XUAT HALOGEN ANCOL PHENOL

Khong tc gian v muon biet th khong gi m cho

Khong bc v khong hieu ro c th khong bay ve cho

Khong T

LU HNH NI B

2/2014

Lp BDKT va Luyen thi TN THPT, C-H

HOA HOC (0986.616.225)(0986.616.225)(0986.616.225)(0986.616.225)

www.hoahoc.edu.vnwww.hoahoc.edu.vnwww.hoahoc.edu.vnwww.hoahoc.edu.vn

CHUYEN E 2: DAN XUAT HALOGEN ANCOL PHENOL Website: www.hoahoc.edu.vn

ThS. LU HUNH VN LONG (Ging vin Trng H Th Du Mt- Bnh Dng) -1- CHUYN: Bi dng ki n thc Luy n thi TN THPT C & H mn HA HC

tm hiu v ng k hc, hy lin lc n ST: 0986.616.225 (T.Long). Email: [email protected]

GIAO KHOA

CU 1 (C 2010): Kh nng phn ng th nguyn t clo bng nhm OH ca cc cht c xp theo chiu tng dn t tri sang phi l:

A. anlyl clorua, phenyl clorua, propyl clorua B. anlyl clorua, propyl clorua, phenyl clorua C. phenyl clorua, anlyl clorua, propyl clorua D. phenyl clorua, propyl clorua, anlyl clorua

CU 2 (C 2011): Cho s chuyn ha: CH3CH2Cl +KCN X

+ o3+ H O , t C Y

Trong s trn, X v Y ln lt l

A. CH3CH2CN v CH3CH2OH B. CH3CH2NH2 v CH3CH2COOH C. CH3CH2CN v CH3CH2COOH D. CH3CH2CN v CH3CH2COOH CU 3 (H A 2012): Cho s chuyn ha:

CH3Cl+ KCNX

+3o

H Ot

Y

Cng thc cu to ca X, Y ln lt l:

A. CH3NH2, CH3COONH4. B. CH3CN, CH3CHO. C. CH3NH2, CH3COOH. D. CH3CN, CH3COOH CU 4 (H A 2009): Cho s chuyn ha:

CH3CH2Cl KCN X 30H Ot

+

Y

Cng thc cu to ca X, Y ln lt l:

A. CH3CH2CN, CH3CH2CHO. B. CH3CH2NH2, CH3CH2COOH. C. CH3CH2CN, CH3CH2COONH4. D. CH3CH2CN, CH3CH2COOH

CU 5 (H B 2008): Cho cc phn ng:

HBr + C2H5OH ot C C2H4 + HBr

C2H4 + Br2 C2H6 + Br2 (1:1)askt

S phn ng to C2H5Br l:

A. 4. B. 3. C. 2. D. 1 CU 6 (H B 2010): Pht biu no sau y ng? A. Khi un C2H5Br vi dung dch KOH ch thu c etilen B. Dung dch phenol lm phenolphtalein khng mu chuyn thnh mu hng C. Dy cc cht : C2H5Cl, C2H5Br, C2H5I c nhit si tng dn t tri sang phi D. un ancol etylic 1400C (xc tc H2SO4 c) thu c imetyl ete CU 7 (H A 2013): Trng hp no sau y khng xy ra phn ng ?

(a) = + 0t

2 2 2CH CH CH Cl H O

(b) + 3 2 2 2CH CH CH Cl H O

(c) ( ) + 0t cao,p cao6 5C H Cl NaOH ac ; vi (C6H5- l gc phenyl) (d) +

0t2 5C H Cl NaOH

A. (a) B. (c) C. (d) D. (b)




CU 8 (H B 2013): un si dung dch gm cht X v KOH c trong C2H5OH, thu c etilen. Cng thc ca X l

A. CH3COOH. B. CH3CHCl2. C. CH3CH2Cl. D. CH3COOCH=CH2. CU 9 (H B 2013): Cht no sau y trong phn t ch c lin kt n? A. Metyl fomat. B. Axit axetic . C. Anehit axetic . D. Ancol etylic . CU 10 (H A 2013): ng vi cng thc phn t C4H10O c bao nhiu ancol l ng phn cu to ca nhau?

A. 3 B. 5 C. 4 D. 2 CU 11(C 2012): S ancol bc I l ng phn cu to ca nhau c cng thc phn t C5H12O l A. 4. B. 1 C. 8. D. 3 CU 12 (C 2011): S ancol ng phn cu to ca nhau c cng thc phn t C5H12O, tc dng vi CuO un nng sinh ra xeton l:

A. 4 B. 2 C. 5 D. 3 CU 13 (C 2011): un si hn hp propyl bromua, kali hiroxit v etanol thu c sn phm hu c l

A. propin. B. propan-2-ol. C. propan. D. propen. CU 14 (H B 2007): Cho cc cht: etyl axetat, anilin, ancol (ru) etylic, axit acrylic,

phenol, phenylamoniclorua, ancol (ru) benzylic, p-crezol. Trong cc cht ny, s cht tc dng c vi dung dch NaOH l

A. 4. B. 6. C. 5. D. 3 CU 15 (H B 2007): Cc ng phn ng vi cng thc phn t C8H10O (u l dn xut ca

benzen) c tnh cht: tch nc thu c sn phm c th trng hp to polime, khng tc dng c vi NaOH. S lng ng phn ng vi cng thc phn t C8H10O, tho mn tnh cht trn l

A. 1. B. 4. C. 3. D. 2. CU 16 (C 2007): Cho cc cht c cng thc cu to nh sau: HOCH2-CH2OH (X); HOCH2-CH2-CH2OH (Y); HOCH2-CHOH-CH2OH (Z); CH3-CH2-O-CH2-CH3 (R); CH3-CHOH-CH2OH (T). Nhng cht tc dng c vi Cu(OH)2 to thnh dung dch mu xanh lam l

A. X, Y, R, T. B. X, Z, T. C. Z, R, T. D. X, Y, Z, T CU 17 (H A 2008): Dy gm cc cht c xp theo chiu nhit si tng dn t tri sang phi l:

A. CH3CHO, C2H5OH, C2H6, CH3COOH. B. C2H6, C2H5OH, CH3CHO, CH3COOH. C. C2H6, CH3CHO, C2H5OH, CH3COOH D. CH3COOH, C2H6, CH3CHO, C2H5OH.

CU 18 (H B 2008): Cho cc cht: ru (ancol) etylic, glixerin (glixerol), glucoz, imetyl ete v axit fomic. S cht tc dng c vi Cu(OH)2 l

A. 1. B. 3. C. 4. D. 2. CU 19 (C 2009): Dy gm cc cht u tc dng vi ancol etylic l: A. NaOH, K, MgO, HCOOH (xc tc). B. Na2CO3, CuO (t

o), CH3COOH (xc tc), (CH3CO)2O.

C. Ca, CuO (to), C6H5OH (phenol), HOCH2CH2OH. D. HBr (to), Na, CuO (to), CH3COOH (xc tc). CU 20 (H B 2009): Cho cc hp cht sau : (a) HOCH2-CH2OH (b) HOCH2-CH2-CH2OH

(c) HOCH2-CH(OH)-CH2OH (d) CH3-CH(OH)-CH2OH




(e) CH3-CH2OH (f) CH3-O-CH2CH3 Cc cht u tc dng c vi Na, Cu(OH)2 l

A. (c), (d), (f) B. (a), (b), (c) C. (a), (c), (d) D. (c), (d), (e) CU 21 (H B 2009): Cho s chuyn ho:

+ + oH SO ac,t HBr Mg,etekhan2 4Butan 2 ol X(anken) Y Z

Trong X, Y, Z l sn phm chnh. Cng thc ca Z l

A. CH3-CH(MgBr)-CH2-CH3 B. (CH3)3C-MgBr C. CH3-CH2-CH2-CH2-MgBr D. (CH3)2CH-CH2-MgBr CU 22 (H A 2010): Cho s chuyn ha:

C3H6 2dung dich Br X NaOH Y

0,CuO t Z 2 ,O xt T0

3 , ,CH OH t xt E (Este a chc). Tn gi ca Y l

A. propan-1,3-iol. B. propan-1,2-iol. C. propan-2-ol. D. glixerol.

CU 23 (H A 2012): Trong ancol X, oxi chim 26,667% v khi lng. un nng X vi H2SO4 c thu c anken Y. Phn t khi ca Y l

A. 42. B. 70. C. 28. D. 56.

HNG DN GII

un nng ancol X vi H2SO4 c thu c anken Y X l ancol no, n chc CnH2n+2O CnH2n

%O = XX

16 = 0,2667 M = 60

M X l C3H8O Y l C3H6 MY = 42

P N A

CU 24 (C 2007): C bao nhiu ru (ancol) bc 2, no, n chc, mch h l ng phn cu to ca nhau m phn t ca chng c phn trm khi lng cacbon bng 68,18%?

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

HNG DN GII CTTQ ca ancol n no l CnH2n+2O (n 1)

5 1212n%C = *100 68,18 n = 5 (C H O)

12n+18=

C 3 CTCT ancol bc 2: CH3 CH2 CH2 CH CH3

OH

pentan-2-ol

CH3 CH2 CH CH2 CH3

pentan-3-ol

OH

CH3 CH

CH3

CH

OH

CH3

3-methylbutan-2-ol P N B

CU 25 (H A 2008): Khi phn tch thnh phn mt ru (ancol) n chc X th thu c kt qu: tng khi lng ca cacbon v hiro gp 3,625 ln khi lng oxi. S ng phn ru (ancol) ng vi cng thc phn t ca X l

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

HNG DN GII




t cng thc ru n chc CxHyO Theo bi: 12x + y = 3,625.16 = 58 CxHy = 58

x 1 2 3 4 y 46 34 22 10

KL loi loi loi nhn

CxHy l C4H10 ru l C4H10 O (no, n, h) c 4 ng phn P N B

TC DNG KL KI M

CU 26 (H A 2007): Cho 15,6 gam hn hp hai ancol (ru) n chc, k tip nhau trong dy ng ng tc dng ht vi 9,2 gam Na, thu c 24,5 gam cht rn. Hai ancol l

A. C3H5OH v C4H7OH. B. C2H5OH v C3H7OH.

C. C3H7OH v C4H9OH. D. CH3OH v C2H5OH.

HNG DN GII Theo L BTKL:

mhh ancol + mNa = mmui + mH2

mH2 = 15,6 + 9,2 24,5 = 0,3 (g)

2H

0,3n = 0,15( )2

mol=

Gi cng thc trung bnh 2 ancol l ROH

2

2

ROH H

ROH

1 ROH + Na RONa + H2

n = 2n = 0,3 (mol)

15,6M = 52 = R + 17 R = 35 0,3

=

Vy chn 2 ancol l C2H5OH v C3H7OH P N B

CU 27 (C 2010) : Cho 10 ml dung dch ancol etylic 460 phn ng ht vi kim li Na (d), thu c V lt kh H2 (ktc). Bit khi lng ring ca ancol etylic nguyn cht bng 0,8 g/ml. Gi tr ca V l:

A. 4,256 B. 0,896 C. 3,360 D. 2,128

HNG DN GII

C H OH C H OH 2 5 2 5

H O H O2 2

10.46V = 4,6(ml) m = 4,6.0,8 = 3,68 (g)100

V = 10 - 4,6 = 5,4 (ml) m = 5,4.1 = 5,4 (g)

=

Khi cho dung dch ancol tc dng vi Na s xy ra 2 phn ng to kh H2 2C2H5OH + 2Na 2C2H5ONa + H2 2H2O + 2Na 2NaOH + H2




H C H OH H O H 2 2 5 2 2

1 1 1 3,68 5,4n = n + n = ( ) 0,19( ol) V = 4,256 (lit)2 2 2 46 18

m+ =

P N A

OXI HA KHNG HON TON ANCOL

CU 28 (C 2010) : Oxi ho khng hon ton ancol isopropylic bng CuO nung nng, thu c cht hu c X. Tn gi ca X l

A. metyl phenyl xeton B. propanal C. metyl vinyl xeton D. imetyl xeton

CU 29 (C 2008): Oxi ho ancol n chc X bng CuO (un nng), sinh ra mt sn phm hu c duy nht l xeton Y (t khi hi ca Y so vi kh hiro bng 29). Cng thc cu to ca X l:

A. CH3-CHOH-CH3. B. CH3-CH2-CHOH-CH3.

C. CH3-CO-CH3. D. CH3-CH2-CH2-OH.

HNG DN GII Tm tt:

Ancol n chc X

CuO Xeton Y

2Y/Hd = 29

Oxi ha ancol n chc X to xeton xeton n chc: CnH2nO Ta c: M(xeton) = 14n + 16 = 29.2 = 58 n = 3 C3H6O

ancol tng ng l: CH3-CHOH-CH3 P N A

CU 30 (H B 2007): Cho m gam mt ancol (ru) no, n chc X qua bnh ng CuO (d), nung nng. Sau khi phn ng hon ton, khi lng cht rn trong bnh gim 0,32 gam. Hn hp hi thu c c t khi i vi hiro l 15,5. Gi tr ca m l:

A. 0,92. B. 0,32. C. 0,64. D. 0,46.

HNG DN GII CnH2n+2O + CuO CnH2nO + Cu + H2O

Khi lng cht rn gim l O trong CuO nn:

n 2n 2O CuO C H O H O0,32n = n = n = n = 0,02( )16

mol=

p dng L BTKL: mancol = 15,5*2*0,04 0,32 = 0,92 (g)

P N A Ch : Nu bi yu cu tm CTPT ca ancol:

+ Mancol = 2 60,92 46 C H O0,02

=

+ Hoc dng phng php ng cho:

CnH2nO ( 14n + 16)

H2O (18)31

13

14n-15

= 1 n = 2 C2H6O




CU 31 (C 2012): Cho m gam hn hp hoi X gm hai ancol (n chc, bc I, l ng ng k tip) phn ng vi CuO d, thu c hn hp hi Y gm nc v anehit. T khi hi ca Y so vi kh hiro bng 14,5. Cho ton b Y phn ng hon ton vi lng d dung dch AgNO3 trong NH3, thu c 97,2 gam Ag. Gi tr ca m l:

A. 14,0. B. 14,7. C. 10,1. D. 18,9.

HNG DN GII

Gi cng thc phn t trung bnh ca 2 ancol l n 2n+2C H Ovi s mol l x

n 2n+2C H O + CuO n 2nC H O + Cu + H2O

Vy hn hp Y gm n 2nC H O v H2O vi s mol bng nhau

p dng phng php ng cho: CnH2nO ( 14n + 16)

H2O (18)29

11

14n - 13

= 1 n = 1,71-

-

Do 2 ancol l ng ng k tip nn CTPT ca 2 ancol l CH3OH v C2H5OH

p dng phng php ng cho vi nguyn t cacbon:

Phn ng vi dung dch AgNO3/NH3.

CH3OH HCHO 4 Ag x x 4x

C2H5OH CH3CHO 2 Ag 2,5x 2,5x 5x

M: nAg = 4x + 5x = 0,9 x = 0,1 (mol)

m = 32.0,1 + 46.0,25 = 14,7 (g) P N B

CU 32 (H A 2008): Cho m gam hn hp X gm hai ru (ancol) no, n chc, k tip nhau trong dy ng ng tc dng vi CuO (d) nung nng, thu c mt hn hp rn Z v mt hn hp hi Y (c t khi hi so vi H2 l 13,75). Cho ton b Y phn ng vi mt lng d Ag2O (hoc AgNO3) trong dung dch NH3un nng, sinh ra 64,8 gam Ag. Gi tr ca m l

A. 7,8. B. 7,4 C. 9,2 D. 8,8

HNG DN GII

hhX(2ancol no,n chc) CuO hh ran Z + hh hi Y2Y/H

d = 13,75

+ Ag2O

64,8g Ag

m g

Gi cng thc phn t trung bnh ca 2 ancol l n 2n+2C H Ovi s mol l x

n 2n+2C H O + CuO n 2nC H O + Cu + H2O

x x x x x




Vy hn hp Y gm n 2nC H O (x mol) v H2O x (mol).

p dng phng php ng cho: (HS XEM THM PH NG PHP NG CHO)

Do 2 ancol l ng ng k tip nn CTPT ca 2 ancol l CH3OH v C2H5OH

Mt khc n = 1,5 l trung bnh cng ca 1(CH3OH) v 2(C2H5OH) s mol ca 2 ancol bng nhau

v bng 2x .

Phn ng vi dung dch AgNO3/NH3.

CH3OH HCHO 4 Ag x/2 x/2 2x

C2H5OH CH3CHO 2 Ag x/2 x/2 x

M: nAg = 2x + x = 0,6 x = 0,2 (mol).

Do : m = 0,2.(14n+ 18) = 0,2.(14.1,5+18) = 7,8g HOC m = 32.0,1 + 46.0,1 = 7,8 (g) P N A

CU 33 (H B 2008): Oxi ho 1,2 gam CH3OH bng CuO nung nng, sau mt thi gian thu c hn hp sn phm X (gm HCHO, H2O v CH3OH d). Cho ton b X tc dng vi lng d Ag2O (hoc AgNO3) trong dung dch NH3, c 12,96 gam Ag. Hiu sut ca phn ng oxi ho CH3OH l:

A. 76,6%. B. 80,0%. C. 65,5%. D. 70,4%

HNG DN GII Tm tt:

1,2g CH3OHCuO

toChh X:

HCHO

H2O

CH3OH d

AgNO3/NH3 12,96g Ag

CH3OH + CuO HCHO + Cu + 2H2O (1)

Khi cho hn hp X tc dng vi dung dch AgNO3/NH3 d th ch c HCHO tc dng theo: HCHO 4Ag (2)

Theo (1) v (2) n(CH3OH p) = n(HCHO) = Ag1 n4

= 1 12,96.4 108

= 0,03 (mol)

n(CH3OH b) = 1,2 0,0375 (mol)32

= H = 0,03 .100% 80%

0,0375=

P N B

CU 34 (C 2010): Cho 4,6gam mt ancol no, n chc phn ng vi CuO nung nng, thu c 6,2 gam hn hp X gm anehit, nc v ancol d. Cho ton b lng hn hp X phn ng hon ton vi lng d dung dch AgNO3 trong NH3, un nng, thu c m gam Ag. Gi tr ca m l

A. 16,2 B. 43,2 C. 10,8 D. 21,6




HNG DN GII

RCH2OH + [O] ot C RCHO + H2O

Khi lng tng chnh l O: mO = 6,2 4,6 = 1,6 (g) n[O] = 0,1 (mol) S mol ancol p l 0,1 (mol) m sau phn ng li d ancol nn s mol ancol ban u > 0,1

(mol):

3

4,60,1 M < 46 CH OH

M>

3 3AgNO /NHCuO3

Ag

CH OH HCHO 4Ag

0,1 (mol) 0,4 (mol) m = 0,4.108 = 43,2 (g)

P N B

CU 35 (H A 2010): Oxi ho ht 2,2 gam hn hp hai ancol n chc thnh anehit cn va 4,8 gam CuO. Cho ton b lng anehit trn tc dng vi lng d dung dch AgNO3 trong NH3, thu c 23,76 gam Ag. Hai ancol l :

A. C2H5OH, C2H5CH2OH B. C2H5OH, C3H7CH2OH C. CH3OH, C2H5CH2OH D. CH3OH, C2H5OH

HNG DN GII

ancol CuO Ag 4,8 23,76n = n = 0,06( ); n = 0,22( )80 108

mol mol= =

Nhn xt: Ag

ancol

n 0,22 3,66 2n 0,06

= = > c HCHO hay c ancol CH3OH ban u Loi A v B

CH3OH HCHO 4Ag x x 4x

RCH2OH RCHO 2Ag y y 2y mol

Gii h phng trnh: 0,06 0,05

4 2 0,22 0,01x y x

x y y

+ = = + = =

RCH OH22,2 0,05.32M 60

0,01= = R = 29 (C2H5)

Vy 2 ancol ban u l CH3OH v C2H5CH2OH

P N C

CU 36 (H B 2009): Hn hp X gm hai ancol no, n chc, mch h, k tip nhau trong dy ng ng. Oxi ho hon ton 0,2 mol hn hp X c khi lng m gam bng CuO nhit thch hp, thu c hn hp sn phm hu c Y. Cho Y tc dng vi mt lng d dung dch AgNO3 trong NH3, thu c 54 gam Ag. Gi tr ca m l

A. 15,3 B. 8,5 C. 8,1 D. 13,5

HNG DN GII

Ag

hhY

n 0,5 2,5 2n 0,2

= = > trong hn hp Y c HCHO

Vy 2 ancol l CH3OH v C2H5OH vi s mol ln lt x v y

CH3OH HCHO 4Ag




x 4x

CH3CH2OH CH3CHO 2Ag y 2y

Ta c: x + y = 0,2 x =0,05

m = 32.0,05 + 46.0,15 = 8,5 (g)4x + 2y = 0,5 y = 0,15

P N B

CU 37 (H B 2012): Oxi ha 0,08 mol mt ancol n chc, thu c hn hp X gm mt axit cacboxylic, mt anehit, ancol d v nc. Ngng t ton b X ri chia lm hai phn bng nhau. Phn mt cho tc dng ht vi Na d, thu c 0,504 lt kh H2 (ktc). Phn hai cho phn ng trng bc hon ton thu c 9,72 gam Ag. Phn trm khi lng ancol b oxi ha l

A. 50,00% B. 62,50% C. 31,25% D. 40,00%

HNG DN GII

Gi: x (mol) RCH2OH to axit RCOOH

y (mol) RCH2OH to anehit RCHO

z (mol) RCH2OH d

RCH2OH 2O RCOOH + H2O

x x x RCH2OH 2

O RCHO + H2O y y y Cc cht tc dng vi Na: RCOOH (x); H2O(x + y) v RCH2OH d (z)

x + x + y + z = 2.0,0225 = 0,045 (1)

x + y + z = 0,04 (2)

Gi s RCHO khc HCHO nn: nAg = 2nRCHO 2y = 0,09 y = 0,045 (mol) V L Vy RCHO l HCHO. Do axit HCOOH cng trng gng:

2x + 4y = 0,09 (3)

Gii (1), (2), (3): x = 0,005; y = 0,02 v z = 0,015

ancol b oxi hoa 0,025% = .100% 62,5%0,04

=

P N B

CU 38 (H B 2013): Tin hnh ln men gim 460 ml ancol etylic 8o vi hiu sut bng 30%. Bit khi lng ring ca ancol etylic nguyn cht bng 0,8 g/ml v ca nc bng 1 g/ml. Nng phn trm ca axit axetic trong dung dch thu c l:

A. 2,51%. B. 2,47%. C. 3,76%. D. 7,99%.

HNG DN GII

CH3CH2OH + O2 len men giam CH3COOH + H2O

0,64 H 30%= 0,192




2 5 2 5 2 5C H OH C H OH C H OH460.8V 36,8 (ml) m 36,8.0.8 29,44 (g) n 0,64 (mol)100

= = = = =

2 2H O H OV 460 36,8 = 423,2 (ml) m 423,2 (g)= =

3CH COOHm 0,64.60.0,3 11,52 (g)= =

mdd sau p = 29,44 + 423,2 + 0,192.32 = 458,784 (gam)

3CH COOH11,52C% .100% 2,51 %458,784

= =

P N A

CU 39 (C 2013): Oxi ha m gam ancol n chc X, thu c hn hp Y gm axit cacboxylic, nc v ancol d. Chia Y lm hai phn bng nhau. Phn mt phn ng hon ton vi dung dch KHCO3 d, thu c 2,24 lt kh CO2 (ktc). Phn hai phn ng vi Na va , thu c 3,36 lt kh H2 (ktc) v 19 gam cht rn khan. Tn ca X l

A. metanol. B. etanol. C. propan-2-ol. D. propan-1-ol.

HNG DN GII 3

2

KHCO d2

O2 2 2Na (u)

2

Phan I: 2,24 (lit) CORCOOHm (g) ancol (X) RCH OH hh Y H O 3,36 (lit) H

Phan II:RCH OH d 19(g) ran

RCH2OH + O2ot C RCOOH + H2O

RCOOH + KHCO3 2 2 RCOOK + CO + H O

Phn 1: nRCOOH = n(ancol p) = 2H O

n = 0,1 (mol)

Phn 2: n(ancol d) = (0,15 0,1).2 = 0,1 (mol)

mrn = 2RCOONa RCH ONa NaOH

m m + m = 19+

0,1(R + 67) + 0,1(R + 53) + 0,1.40 = 19 R = 29 (C2H5) Vy anol l C2H5OH P N B

CU 40 : Oxi ho m gam etanol thu c hn hp X gm axetanehit, axit axetic, nc v etanol d. Cho ton b X tc dng vi dung dch NaHCO3 (d) thu c 0,56 lt kh CO2 ( ktc). Khi lng etanol b oxi ho to ra axit l

A. 4,60 gam B. 1,15 gam C. 5,75 gam D. 2,30 gam

HNG DN GII

C2H5OH + O2 CH3COOH + H2O 0,025 0,025

CH3COOH + NaHCO3 CH3COONa + CO2 + H2O 0,025 0,025

mancol = 46.0,025 = 1,15 (g) P N B

CU 41: Hn hp X gm 1 ancol v 2 sn phm hp nc ca propen. T khi hi ca X so vi hiro bng 23. Cho m gam X i qua ng s ng CuO (d) nung nng. Sau khi cc phn ng xy ra hon ton, thu c hn hp Y gm 3 cht hu c v hi nc, khi lng ng s gim 3,2 gam. Cho Y tc




dng hon ton vi lng d dung dch AgNO3 trong NH3, to ra 48,6 gam Ag. Phn trm khi lng ca propan-1-ol trong X l

A. 65,2%. B. 16,3%. C. 48,9%. D. 83,7%.

HNG DN GII

Hai sn phm hp nc ca propen l C2H5CH2OH (propan-1-ol) v CH3CHOHCH3(propan-2-ol)

XM = 2.23 = 46 Trong X c CH3OH

Khi lng cht rn gim chnh l O ca CuO nO = nX = 2,016

2,3 = (mol)

Gi a l s mol ca 2 ancol C3H8O 32(0,2 a) + 60a = 46.0,2 a = 0,1

3CH OHn = 0,1 mol

nAg =48,6 0,45( )108

mol=

Do propan-2-ol b oxi ha to axeton nn khng tham gia phn ng trng gng nn: CH3OH HCHO 4Ag

0,1 0,1 0,4 mol C2H5CH2OH C2H5CHO 2Ag 0,025 mol 0,025 (0,45 0,4) mol

% m C2H5CH2OH = 0,025.60 .100 16,3%46.0,2

=

P N B

PHAN NG CHAY

CU 42 (C 2012): t chy hon ton hn hp X gm hai ancol no, hai chc, mch h cn va V1 lt kh O2, thu c V2 lt kh CO2 v a mol H2O. Cc kh u o iu kin tiu chun. Biu thc lin h gia cc gi tr V1, V2, a l:

A. V1 = 2V2 - 11,2a B. V1 = V2 +22,4a C. V1 = V2 - 22,4a D. V1 = 2V2 + 11,2

HNG DN GII

CnH2n+2O2 + O2 ot C CO2 + H2O

S mol hn hp X: nX = 2 2

2H O CO

Vn - n = a- (mol)22,4

Bo ton nguyn t Oxi: 2 2V VV1(a- ).2 .2 .2 + a.122,4 22,4 22,4

+ =

V1 = 2V2 - 11,2a P N A




CU 43 (H A 2009): Khi t chy hon ton m gam hn hp hai ancol no, n chc, mch h thu c V lt kh CO2 ( ktc) v a gam H2O. Biu thc lin h gia m, a v V l:

A. V

m a -5,6

= . B. Vm 2a11,2

= . C. Vm 2a22,4

= . D. Vm a5,6

= + .

HNG DN GII Cch 1: Tnh theo thnh phn nguyn t:

m = mC + mH + mO = V *2 V V*12 ( )*16 a-

22,4 18 18 22,4 5,6a a+ + =

Cch 2: Tnh theo cng thc:

2

2

COancol H O

mm = m -

11

22

COancol H O

mm = m -

11= V*44 V= a - a -

22,4*11 5,6=

P N A

CU 44 (H A 2010): t chy hon ton m gam hn hp 3 ancol n chc, thuc cng dy ng ng, thu c 3,808 lt kh CO2 (ktc) v 5,4 gam H2O. Gi tr ca m l

A. 4,72 B. 5,42 C. 7,42 D. 5,72

HNG DN GII

2COn = 0,17 mol ;

2H On = 0,3 mol

2H O

n > 2CO

n => ancol no n chc CnH2n+2O

nancol = 2H O

n - 2CO

n = 0,13 mol CO2ancol

n 0,17n = n 0,13

= = 1,307

mancol = (14.1,307 + 18)0,13 = 4,72 g P N A

CH :

2

2

COancol NO, n, h H O

mm = m -

11

CU 45 (C 2013): t chy hon ton 1 mol ancol no, mch h X cn va 3,5 mol O2. Cng thc phn t ca X l:

A. C3H8O3. B. C2H6O2. C. C2H6O. D. C3H8O2.

HNG DN GII

P N A

CU 46 (H A 2012): t chy hon ton mt lng ancol X to ra 0,4 mol CO2 v 0,5 mol H2O. X tc dng vi Cu(OH)2 to dung dch mu xanh lam. Oxi ha X bng CuO to hp cht hu c a chc Y. Nhn xt no sau y ng vi X?




A. Trong X c 3 nhm -CH3. B. Hirat ha but-2-en thu c X. C. Trong X c 2 nhm -OH lin kt vi hai nguyn t cacbon bc hai. D. X lm mt mu nc brom.

HNG DN GII

Ta c: 2 2CO H O

n = 0,4 < n = 0,5X l ancol no.

Khi s nguyn t cacbon trong X = 2 2

2 2

CO CO

X H O CO

n n 0,4 = = 4

n n n 0,5 0,4=

X tc dng vi Cu(OH)2 to dung dch mu xanh lam X c 2 nhm OH cnh nhau

Oxi ha X bng CuO to hp cht hu c a chc Y.

Vy X l: CH3-CHOH-CHOH-CH3 CH3-CO-CO-CH3 P N C

CU 47 (C 2008): t chy hon ton mt ru (ancol) a chc, mch h X, thu c H2O v CO2 vi t l s mol tng ng l 3:2. Cng thc phn t ca X l:

A. C2H6O2. B. C2H6O. C. C3H8O2. D. C4H10O2.

HNG DN GII Nhn xt:

Ancol a chc Loi B

2 2H O COn > n ancol no [Hoc cc em nhn vo 3 p n cn li u thy chng no]

CnH2n+2Ox nCO2 + (n+1)H2O

Ta c: n+1 3 n = 2 n 2

= Loi C v D

P N A CH :

Da vo t l S C : S H = 2 : 6 C2H6Oz Chn A

CU 48 (C 2007): t chy hon ton mt ru (ancol) X thu c CO2 v H2O c t l s mol tng ng l 3 : 4.Th tch kh oxi cn dng t chy X bng 1,5 ln th tch kh CO2 thu c ( cng iu kin). Cng thc phn t ca X l: A. C3H8O3. B. C3H4O. C. C3H8O2. D. C3H8O

HNG DN GII

x y z 2 2 2y yC H O + (x+ )O xCO + H O4 2 2

z

2

2

CO

H O

n 3 8 3n 4

2

xx y

y= = = (1)

Do cng iu kin nn: 2 2O CO

yn = 1,5n (x+ ) 1,5 2 24 2

zx x y z = = (2)

T (1) v (2) y = 8 x = 3 v z = 1 C3H8O

P N D




Ch :

T l 2

2

CO

H O

n 3n 4

= C3H8Oz

Th tch kh oxi cn dng bng 1,5 ln th tch kh CO2 Ancol NO, N, H

CU 49 (H A 2009): Cho hn hp X gm hai ancol a chc, mch h, thuc cng dy ng ng. t chy hon ton hn hp X, thu c CO2 v H2O c t l mol tng ng l 3 : 4. Hai ancol l

A. C2H4(OH)2 v C3H6(OH)2. B. C2H5OH v C4H9OH. C. C2H4(OH)2 v C4H8(OH)2. D. C3H5(OH)3 v C4H7(OH)3.

HNG DN GII

Hai ancol a chc Loi B

22

CO

H O

n 3n 4

= hai ancol no c cng thc trung bnh l C3H8Ox Loi A v D

P N C

CU 50 (C 2013): t chy hon ton mt lng ancol X cn va 8,96 lt kh O2 (ktc), thu c 6,72 lt kh CO2 (ktc) v 7,2 gam H2O. Bit X c kh nng phn ng vi Cu(OH)2. Tn ca X l

A. propan-1,3-iol. B. propan-1,2-iol. C. glixerol. D. etylen glicol.

HNG DN GII Theo L BTKL: mX = 0,03.44 + 7,2 0,4.32 = 7,6 (g)

nX = 2 2H O CO X

n - n = 0,1 (mol) M = 76 C3H8O2 Loi C, D

X c kh nng phn ng vi Cu(OH)2 Loi A P N B

CU 51 (C 2008): t chy hon ton hn hp M gm hai ru (ancol) X v Y l ng ng k tip ca nhau, thu c 0,3 mol CO2 v 0,425 mol H2O. Mt khc, cho 0,25 mol hn hp M tc dng vi Na (d), thu c cha n 0,15 mol H2. Cng thc phn t ca X, Y l:

A. C2H6O2, C3H8O2. B. C2H6O, CH4O.

C. C3H6O, C4H8O. D. C2H6O, C3H8O

HNG DN GII

hh M hai ancol kt

O2 0,3 mol CO2 + 0,425 mol H2O

0,25 mol M + Nad

< 0,15 mol

2 2H O CO

n > n 2 ancol no Loi C

Khi cho 0,25 mol hn hp M tc dng Na thu kh H2 < 0,15 mol 2 anol n chc Loi A t cng thc chung ca 2 ancol l CnH2n+2O

CnH2n+2O 2O nCO2 + (n+1)H2O

Lp t l: n+1 0,425 2,4n 0,3

n= = C2H6O v C3H8O




Hoc: 2CO

M

n 0,3C 2,4n 0,425 0,3

= = =

P N D

CU 52 (H A 2009): t chy hon ton 0,2 mol mt ancol X no, mch h, cn va 17,92 lt kh O2 ( ktc). Mt khc, nu cho 0,1 mol X tc dng va vi m gam Cu(OH)2 th to thnh dung dch c mu xanh lam. Gi tr ca m v tn gi ca X tng ng l:

A. 4,9 v propan-1,2-iol B. 9,8 v propan-1,2-iol C. 4,9 v glixerol. D. 4,9 v propan-1,3-iol

HNG DN GII

X ha tan Cu(OH)2 loi D S mol Cu(OH)2 =1/2s mol X = 0,05(mol) Khi lng Cu(OH)2 = 4,9(g) Loi B Th p n C: C3H8O3 + 3,5O2 3CO2 + 4H2O

0,2 0,7 (mol) # 0,8 (mol) Loi C

P N A

CU 53 (H B 2007): X l mt ancol (ru) no, mch h. t chy hon ton 0,05 mol X cn 5,6 gam oxi, thu c hi nc v 6,6 gam CO2. Cng thc ca X l

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

HNG DN GII

CnH2n+2Ox + O2 ot C CO2 + H2O

2CO

X

6,6n 44So C = = 3n 0,05

= Loi A

Do ancol no nn: 2 2 2X H O CO H O

n = n - n n = 0,05 + 0,15 = 0,2 (mol)

Bo ton nguyn t Oxi: S O = [(0,15 x 2 + 0,2) - 5,6 *232

]/ 0,05 = 3 C3H8O3

Hoc: Theo L BTKL mX + m(O2) = m(CO2) + m(H2O)

mX = 0,2.18 + 6,6 5,6 = 4,6 (g)

MX = 3 8 34,6 92 C H O0,05

=

P N C

CU 54 (C 2012): t chy hon ton m gam hn hp X gm hai ancol n chc, cng dy ng ng, thu c 15,68 lt kh CO2 (ktc) v 17,1 gam nc. Mt khc, thc hin phn ng este ha m gam X vi 15,6 gam axit axetic, thu c a gam este. Bit hiu sut phn ng este ha ca hai ancol u bng 60%. Gi tr ca a l

A. 15,48. B. 25,79. C. 24,80. D. 14,88.

HNG DN GII

2 2 3CO H O CH COOH15,6n = 0,7 (mol); n = 0,95 (mol); n = 0,26 (mol)60

=

2 2H O CO

n n> 2 ancol no, n, h




nancol = 0,95 0,7 = 0,25 < 3CH COOH

n Hiu sut tnh theo ancol

CH3COOH + ROH o

2 4H SO , t C 3CH COOR + H2O

Ta c: (ancol) 2,8 6,60,7C 2,8 C H OH hay ROH (R= 40,2)0,25

= =

Khi lng este to thnh: a = (59+40,2).0.25.0,6 =14,88 (g)

P N D

CU 55 (C 2011): t chy hon ton mt lng hn hp X gm 3 ancol thuc cng dy ng ng thu c 6,72 lt kh CO2 (ktc) v 9,90 gam H2O. Nu un nng cng lng hn hp X nh trn vi H2SO4 c nhit thch hp chuyn ht thnh ete th tng khi lng ete thu c l:

A. 6,45 gam B. 5,46 gam C. 7,40 gam D. 4,20 gam

HNG DN GII

2 2H O COn 0,55 (mol); n = 0,3 (mol)=

2

2 2 2 2

COH O CO ancol no H O CO

ancol

n 0,3n > n n n - n = 0,25 (mol) C 1,2n 0,25

= = = =

co ancol CH3OH nen thuoc ancol no, n h co cong thc chung: C1,2H4,4O (34,8) mX = 0,25.34,8 = 8,7 (g)

S : 2ancol 1ete + 1H2O

2

ancolH O

n 0,25n = = 0,125 (mol)2 2

=

Bo ton khi lng: m ancol = mete + 2H O

m

mete = 8,7 0,125.18 = 6,45 P N A

CU 56 (H A 2013): t chy hon ton hn hp X gm 0,07 mol mt ancol a chc v 0,03 mol mt ancol khng no, c mt lin kt i, mch h, thu c 0,23 mol kh CO2 v m gam H2O. Gi tr ca m l

A. 5,40 B. 2,34 C. 8,40 D. 2,70

HNG DN GII

Ancol a chc c s nguyn t cacbon 2 Ancol khng no, c mt lin kt i, mch h c s nguyn t cacbon 3

2CO

hh

nC 2,3

n= =

ancol a chc l C2H4(OH)2 (no, nh chc, h) Nhn xt: Do ancol khng no c 1C=C t cho

2 2CO H On = n nn:

2 2H O CO ancol non n = n

m 0,23 0,0718

= m = 5,4

P N A




Tm CTPT ancol khng no:

Bo ton nguyn t cacbon: 0,07.2 + 0,03.m = 0,23.1 m = 3 ( CH2 = CH CH2OH) CU 57 (H A 2013): Hn hp X gm ancol metylic, ancol etylic v glixerol. t chy hon ton m gam X, thu c 15,68 lt kh CO2 (ktc) v 18 gam H2O. Mt khc, 80 gam X ha tan c ti a 29,4 gam Cu(OH)2. Phn trm khi lng ca ancol etylic trong X l: A. 46% B. 16% C. 23% D. 8%

HNG DN GII

2

32O

2 52

3 5 3

2

CH OH15,68 (lit) CO

+ m (g) hh X C H OH18 (g) H O

C H (OH)80 (g) hh X + 29,4 (g) Cu(OH)

+

2C3H5(OH)3 + Cu(OH)2 (C3H7O3)2Cu + 2H2O 0,6 0,3 Gi a, b, c ln lt l s mol CH3OH, C2H5OH, C3H5(OH)3 trong 80 gam X

32a + 46b + 92.0,6 = 80 32a + 46b =24,8 (1) Gi s mol ca CH3OH, C2H5OH, C3H5(OH)3 trong m gam hn hp X l ka, kb, 0,6k.

Do ta c: k(2a + 3b + 4.0,6) = 2H O

n =1 (2)

k(a + 2b + 3.0,6) = 2CO

n = 0,7 (3)

Chia theo v (3) cho (2) c: a + 2b + 3.0,6 = 0,72a + 3b + 4.0,6

0,4a + 0,1b = 0,12 4a + b = 1,2 (4) Gii h (1),(4) c a = 0,2; b = 0,4

2 5C H OH0,4.46% .100% = 23 %80

=

P N C

CU 58 (H B 2013): Hn hp X gm ancol metylic, etylen glicol. Cho m gam X phn ng hon ton vi Na d, thu c 2,24 lt kh H2 (ktc). t chy hon ton m gam X, thu c a gam CO2. Gi tr ca a l

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

HNG DN GII

CH3OH Na 1

2 H2

x 0,5x C2H4(OH)2

Na H2 y y

2Hn = 0,5x + y = 0,1 (mol)

CH3OH 2O CO2

x x C2H4(OH)2 2

O 2CO2 y 2y

2COn = x +2y = 2.( 0,5x+y) = 0,2 (mol)

2COm = 0,2.44 = 8,8 (gam)

Nhn xt:

nC = nOH = 22H

n =2CO

n = 0,2 (mol) 2CO

m = 8,8 (gam)

P N C




CU 59 (H A 2012): Cho hn hp X gm ancol metylic, etylen glicol v glixerol. t chy hon ton m gam X thu c 6,72 lt kh CO2 (ktc). Cng m gam X trn cho tc dng vi Na d thu c ti a V lt kh H2 (ktc). Gi tr ca V l

A. 3,36 B. 11,20 C. 5,60 D. 6,72

HNG DN GII

CH3OH 2O CO2 C2H4(OH)2 2O 2CO2 C3H5(OH)3 2O 3CO2

CH3OH Na 1

2H2 C2H4(OH)2 Na H2 C3H5(OH)3 Na

32H2

Ta c: nOH = nC = 2CO

n = 0,3 (mol)

Mt khc: 2

OHH

nn2

= = 0,15 mol 2H

V = 3,36 (lit)

P N A

CU 60 (H B 2012): t chy hon ton m gam hn hp X gm hai ancol, thu c 13,44 lt kh CO2 (ktc) v 15,3 gam H2O. Mt khc, cho m gam X tc dng vi Na (d), thu c 4,48 lt kh H2 (ktc). Gi tr ca m l

A. 12,9 B. 15,3 C. 12,3 D. 16,9

HNG DN GII

Ancol tc dng vi Na th lun c: nO(ancol) = 2H

2n

Bo ton nguyn t:

m = mC + mH + mO = 13,44 15,3 4,48.12 .2 .2.16 15,3 (g)22,4 18 22,4

+ + =

P N B

CU 61 (H B 2010): t chy hon ton mt lng hn hp X gm 2 ancol (u no, a chc, mch h, c cng s nhm -OH) cn va V lt kh O2, thu c 11,2 lt kh CO2 va 12,6 gam H2O (cc th tch kh o ktc). Gi tr ca V l

A. 14,56 B. 15,68 C. 11,20 D. 4,48

HNG DN GII

ot Cx 2 2 2n 2n + 2

3n + 1 - xC H O + O nCO + (n + 1)H O2

nX = 2 2H O CO

n n = 0,7 0,5 = 0,2 (mol)

CO2X

n 0,5n = 2,5n 0,2

= = .

Nhn xt: mt ancol no, a chc, h c s nguyn t nh hn 2,5 th ancol phi l C2H6O2 (nh chc) theo ng iu kin bn ca ancol (s nhm OH s nguyn t C)

Bo ton nguyn t oxi:

O2

0,7 0,5.2 0,2.2V = .22,4 14,562

+ = (lt)

P N A




TCH NC

CU 62 (H B 2013): Tn gi ca anken (sn phm chnh) thu c khi un nng ancol c cng thc (CH3)2CHCH(OH)CH3 vi dung dch H2SO4 c l:

A. 3-metylbut-2-en. B. 2-metylbut-1-en. C. 2-metylbut-2-en. D. 3-metylbut-1-en.

CU 63 (H A 2007): Khi tch nc t mt cht X c cng thc phn t C4H10O to thnh ba anken l ng phn ca nhau (tnh c ng phn hnh hc). Cng thc cu to thu gn ca X l A. (CH3)3COH. B. CH3OCH2CH2CH3.

C. CH3CH(OH)CH2CH3. D. CH3CH(CH3)CH2OH.

HNG DN GII

Khi tch nc, X to 3 anken (k c ng phn hnh hc) X l ancol bc 2. Vy X l: CH3 CH

OH

CH2 CH3

P N C

CU 64 (C 2008): Khi un nng hn hp ru (ancol) gm CH3OH v C2H5OH (xc tc H2SO4

c, 140oC) th s ete thu c ti a l:

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

CU 65 (C 2013): Hn hp X gm hai ancol n chc, ng ng k tip. un nng 16,6 gam X vi H2SO4 c 140C, thu c 13,9 gam hn hp ete (khng c sn phm hu c no khc). Bit cc phn ng xy ra hon ton. Cng thc ca hai ancol trong X l

A. C3H5OH v C4H7OH. B. CH3OH v C2H5OH. C. C3H7OH v C4H9OH. D. C2H5OH v C3H7OH.

HNG DN GII o

2 4H SO , 140 C22ROH ROR + H O

Theo L BTKL: mancol = mete + 2H O

m 2H O

16,6 13,9n 0,15 (mol)18= =

nancol = 2.0,15 = 0,3 (mol)

ancol16,6M = 55,330,3

= 2 ancol l: C2H5OH v C3H7OH.




P N D

CU 66 (H A 2009): un nng hn hp hai ancol n chc, mch h vi H2SO4 c, thu c hn hp gm cc ete. Ly 7,2 gam mt trong cc ete em t chy hon ton, thu c 8,96 lt kh CO2 ( ktc) v 7,2 gam H2O. Hai ancol l

A. CH3OH v CH2=CH-CH2-OH. B. C2H5OH v CH2=CH-CH2-OH. C. CH3OH v C3H7OH. D. C2H5OH v CH3OH.

HNG DN GII Nhn xt: Hn hp 2 ancol s to ra 3 ete

t mt trong 3 ete bt k c: 2 2H O CO

n = n = 0,4 (mol) nn ete cha no cha 1 lin kt .

t CTTQ ca ete : CnH2nO

CnH2nO 2O nCO2 + nH2O

C: (14n + 16) g n (mol) bi: 7,2 g 0,4 (mol)

Lp t l: 4 814n + 16 n n = 4 C H O

7,2 0,4=

P N A

CU 67 (C 2007): Khi thc hin phn ng tch nc i vi ru (ancol) X, ch thu c mt anken duy nht. Oxi ho hon 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

HNG DN GII

2 2CO H O5,6 5,4n = 0,25 (mol); n = 0,3 (mol)22,4 18

= =

Ta c: 2 2H O COn > n

ancol no n chcancol tach nc tao anken

CnH2n+2O nCO2 + (n + 1) H2O

5 12n+1 0,3 n = 5 C H On 0,25

=

Khi tch nc ancol thu c 1 anken duy nht ancol i xng hoc ancol bc 1 (tr dng c bit)

CH3CH2CH2CH2CH2OH

pentan-1-ol

CH3 CH2 CH CH2

CH3

OH

2-methylbutan-1-ol

CH3 CH CH2 CH2 OH

CH3

3-methylbutan-1-ol

CH3 CH2 CH CH2 CH3

OH

pentan-3-ol P N B

CU 68 (H B 2008): un nng hn hp gm hai ru (ancol) n chc, mch h, k tip nhau

trong dy ng ng vi H2SO4 c 140oC. Sau khi cc phn ng kt thc, thu c 6 gam hn

hp gm ba ete v 1,8 gam nc. Cng thc phn t ca hai ru trn l




A. CH3OH v C2H5OH. B. C2H5OH v C3H7OH.

C. C3H5OH v C4H7OH. D. C3H7OH v C4H9OH.

HNG DN GII

hh 2 ancol X,Y(n chc, kt)

H2SO4 ac

140oC6g hh 3 ete + 1,8g H2O

Gi cng thc chung ca 2 ancol l ROH :

2ROH o

2 4H SO ,140 C ROR + H2O

Nhn thy: n(ete) = n(nc) = 1,8 0,1 (mol)18

= ete 6M 60 2R 16 R 220,1

= = = + = .

chn CH3OH v C2H5OH. P N A

CU 69 (H B 2008): un nng mt ru (ancol) n chc X vi dung dch H2SO4 c trong iu kin nhit thch hp sinh ra cht hu c Y, t khi hi ca X so vi Y l 1,6428. Cng thc phn t ca X l

A. C3H8O. B. C2H6O. C. CH4O. D. C4H8O

HNG DN GII Tm tt:

ancol XH2SO4 ac

toCY X/Yd = 1,6428biet

Khi un ancol X trong H2SO4 c nhit thch hp to hp cht hu c Y m

X/Yd 1,6428 1= > phn ng to anken ancol X n, no, h CnH2n+2O CnH2n + H2O

X/Y 2 614n + 18d 1,6428 n = 2 X: C H O

14n= =

P N B Ch :

Ancol n, no (X) o

2 4dH SO ,t C Sn phm hu c (Y): Nu dY/X < 1 Y l anken Nu dY/X > 1 Y l ete

CU 70 (H A 2010): Tch nc hn hp gm ancol etylic v ancol Y ch to ra 2 anken. t chy cng s mol mi ancol th lng nc sinh ra t ancol ny bng 5/3 ln lng nc sinh ra t ancol kia. Ancol Y l

A. CH3-CH2-CH(OH)-CH3. B. CH3-CH2-CH2-CH2-OH. C. CH3-CH2-CH2-OH. D. CH3-CH(OH)-CH3.

HNG DN GII

Hn hp gm ancol etylic v ancol Y tch nc ch to ra 2 anken trong ancol etylic to ra 1 anken nn ancol Y ch to 1 anken Loi A (v ancol Y tch nc to 2 anken ng phn cu to).

Y tch nc to anken Y l ancol no, n chc, h:




C2H5OH 2O 3H2O

CnH2n + 1OH 2O (n + 1)H2O

t chy cng s mol mi ancol th khi lng nc sinh ra t l s s mol:

4 9n+1 5 n = 4 C H OH3 3

= Loi C v D

P N B

CU 71 (H B 2010): t chy han tan m gam hn hp X gm ba ancol (n chc, thuc cng dy ng ng), thu c 8,96 lt kh CO2 (ktc) v 11,7 gam H2O. Mt khc, nu un nng m gam X vi H2SO4 c th tng khi lng ete ti a thu c l:

A. 7,85 gam. B. 7,40 gam. C. 6,50 gam. D. 5,60 gam.

HNG DN GII

2COn = 0,4 mol ;

2H On = 0,65 mol

2CO

n < 2H O

n ancol no n chc: n 2n + 1

C H OH

nX = 2CO

n 2H O

n = 0,25 mol

S mol ete = S mol H2O = s mol ancol = 0,125 mol

Bo ton khi lng: mete = mancol 2H O

m = 0,25(14.1,6 + 18) 0,125.18 = 7,85 (g)

m = 0,125(28.1,6 +18) = 7,85g

P N A

CU 72 (H B 2011): Chia hn hp gm hai ancol n chc X v Y (phn t khi ca X nh hn ca Y) l ng ng k tip thnh hai phn bng nhau:

- t chy hon ton phn 1 thu c 5,6 lt CO2 (ktc) v 6,3 gam H2O.

- un nng phn 2 vi H2SO4 c 1400C to thnh 1,25 gam hn hp ba ete. Ha hi hon ton

hn hp ba ete trn, thu c th tch ca 0,42 gam N2 (trong cng iu kin nhit , p sut).

Hiu sut ca phn ng to ete ca X, Y ln lt l:

A. 30% v 30% B. 25% v 35% C. 40% v 20% D. 20% v 40%

HNG DN GII

Phn 1 : 2 2CO H O

n = 0,25 (mol) < n = 0,35 (mol) 2 ancol no, n chc, h c s mol l 0,1 (mol)

t cng thc chung 2 ancol: n 2n 1C H OH+

2COhh

nn 2,5

n= = Hai ancol l C2H5OH v C3H7OH c s mol bng nhau v bng 0,05 (mol)

Phn 2: nete = 2H O

n = 0,4228

= 0,015 (mol) n2 ancol p = 2nete = 0,03 (mol)

n 2ancol d = 0,07 (mol) Khi lng 2ancol p = 1,25 + 0,015.18 = 1,52 (g)

Gi a, b ln lt l s mol CH3OH v C2H5OH phn ng:

46a +60b = 1,52 a 0,02a + b = 0,03 b 0,01

= =

Hiu sut este ha ca X v Y l 40% v 20%

P N C




GIO KHOA

CU 1 (H A 2013): Phenol phn ng c vi dung dch no sau y ? A. NaCl. B. KOH. C. NaHCO3. D. HCl. CU 2 (C 2013): Dung dch phenol (C6H5OH) khng phn ng c vi cht no sau y? A. NaOH. B. NaCl. C. Br2. D. Na . CU 3 (C 2007): Trong cng nghip, axeton c iu ch t A. xiclopropan. B. propan-1-ol. C. propan-2-ol. D. cumen. CU 4 (H B 2008): nh hng ca nhm OH- n gc C6H5

- trong phn t phenol th hin qua phn ng gia phenol vi

A. dung dch NaOH. B. Na kim loi. C. nc Br2. D. H2 (Ni, nung nng).

CU 5 (C 2008): Cho dy cc cht: phenol, anilin, phenylamoni clorua, natri phenolat, etanol. S cht trong dy phn ng c vi NaOH (trong dung dch) l

A. 3. B. 2. C. 1. D. 4 CU 6 (H A 2009): Hp cht hu c X tc dng c vi dung dch NaOH v dung dch brom nhng khng tc dng vi dung dch NaHCO3. Tn gi ca X l

A. metyl axetat. B. axit acrylic. C. anilin. D. phenol.

CU 7 (H B 2007): S cht ng vi cng thc phn t C7H8O (l dn xut ca benzen) u tc dng c vi dung dch NaOH l

A. 2. B. 4. C. 3. D. 1 CU 8 (C 2013): S ng phn cha vng benzen, c cng thc phn t C7H8O, phn ng c vi Na l

A. 3. B. 5. C. 4. D. 2. CU 9 (H B 2012): C bao nhiu cht cha vng benzene c cng cng thc phn t C7H8O? A. 3 B. 5 C. 6 D. 4 CU 10 (C 2011): S hp cht ng phn cu to ca nhau c cng thc phn t C8H10O, trong phn t c vng benzen, tc dng c vi Na, khng tc dng c vi NaOH l

A. 4. B. 6. C. 7. D. 5. CU 11 (H A 2011): Cho dy cc cht: stiren, ancol benzylic, anilin, toluen, phenol (C6H5OH). S cht trong dy c kh nng lm mt mu nc brom l:

A. 5. B. 4. C. 3. D. 2. CU 12 (H B 2013): Cho cc pht biu sau: (a) Cc cht CH3NH2, C2H5OH, NaHCO3 u c kh nng phn ng vi HCOOH.

(b) Phn ng th brom vo vng benzen ca phenol (C6H5OH) d hn ca benzen.

(c) Oxi ha khng hon ton etilen l phng php hin i sn xut anehit axetic .

(d) Phenol (C6H5OH) tan t trong etanol.

Trong cc pht biu trn, s pht biu ng l A. 4. B. 2. C. 3. D. 1. CU 13 (H A 2010): Trong s cc pht biu sau v phenol (C6H5OH): (1) Phenol tan t trong nc nhng tan nhiu trong dung dch HCl




(2) Phenol c tnh axit, dung dch phenol khng lm i mu qu tm

(3) Phenol dng sn xut keo dn, cht dit nm mc

(4) Phenol tham gia phn ng th brom v th nitro d hn benzen

Cc pht biu ng l A. (1), (2), (4) B. (2), (3), (4) C. (1), (2), (3) D. (1), (3), (4) CU 14 (H A 2011): Cho cc pht biu sau v phenol (C6H5OH):

(a) Phenol tan nhiu trong nc lnh.

(b) Phenol c tnh axit nhng dung dch phenol trong nc khng lm i mu qu tm.

(c) Phenol c dng sn xut phm nhum, cht dit nm mc.

(d) Nguyn t H ca vng benzen trong phenol d b thay th hn nguyn t H trong benzen.

(e) Cho nc brom vo dung dch phenol thy xut hin kt ta.

S pht biu ng l A. 4. B. 2. C. 5. D. 3.

CU 15 (H B 2007): Dy gm cc cht u phn ng vi phenol l: A. dung dch NaCl, dung dch NaOH, kim loi Na. B. nc brom, anhirit axetic, dung dch NaOH. C. nc brom, axit axetic, dung dch NaOH.

D. nc brom, anehit axetic, dung dch NaOH CU 16 (H A 2007): Pht biu khng ng l: A. Axit axetic phn ng vi dung dch NaOH, ly dung dch mui va to ra cho tc dng vi kh CO2 li thu c axit axetic.

B. Phenol phn ng vi dung dch NaOH, ly mui va to ra cho tc dng vi dung dch HCl li thu c phenol.

C. Anilin phn ng vi dung dch HCl, ly mui va to ra cho tc dng vi dung dch NaOH li thu c anilin.

D. Dung dch natri phenolat phn ng vi kh CO2, ly kt ta va to ra cho tc dng vi dung dch NaOH li thu c natri phenolat.

CU 17 (H B 2007): Cho cc cht sau: phenol, etanol, axit axetic, natri phenolat, natri hiroxit. S cp cht tc dng c vi nhau l

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

CU 18 (C 2007): Hp cht hu c X (phn t c vng benzen) c cng thc phn t l C7H8O2, tc dng c vi Na v vi NaOH. Bit rng khi cho X tc dng vi Na d, s mol H2 thu c bng s mol X tham gia phn ng v 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

HNG DN GII Cht X C7H8O2 ( c vng benzen) c cc tnh cht:

+ X tc dng vi Na v NaOH c nhm OH gn trc tip vi vng benzen

+ X tc dng vi Na cho s mol H2 bng s mol X X c 2 nhm -OH

+ X tc dng vi NaOH theo t l 1: 1 X c 1 nhm OH gn trc tip vi vng benzen

Kt lun X l: HOC6H4CH2OH

P N B




CU 19 (H A 2009): Cho dy chuyn ha sau:

Phenol X+Phenyl axetat 0 (du)NaOHt+ Y (hp cht thm)

Hai cht X,Y trong s trn ln lt l:

A. anhirit axetic, phenol. B. anhirit axetic, natri phenolat C. axit axetic, natri phenolat. D. axit axetic, phenol

CU 20 (H B 2012): Cho phenol (C6H5OH) ln lt tc dng vi (CH3CO)2O v cc dung dch: NaOH, HCl, Br2, HNO3, CH3COOH. S trng hp xy ra phn ng l

A. 3 B. 4 C. 2 D. 1 CU 21 (H B 2012): Ha tan cht X vo nc thu c dung dch trong sut, ri thm tip dung dch cht Y th thu c cht Z (lm vn c dung dch). Cc cht X, Y, Z ln lt l:

A. phenol, natri hiroxit, natri phenolat B. natri phenolat, axit clohiric, phenol C. phenylamoni clorua, axit clohiric, anilin D. anilin, axit clohiric, phenylamoni clorua

CU 22 (H B 2007): Khi t 0,1 mol mt cht X (dn xut ca benzen), khi lng CO2 thu c nh hn 35,2 gam. Bit rng, 1 mol X ch tc dng c vi 1 mol NaOH. Cng thc cu to thu gn ca X l

A. C2H5C6H4OH. B. HOCH2C6H4COOH.

C. HOC6H4CH2OH. D. C6H4(OH)2.

HNG DN GII

+ Ta c: 2CO

35,2 0,8n < = 0,8(mol) So C < 844 0,1

= ( m s C 6) Loi A, B

+ 1 mol X ch tc dng vi 1 mol NaOH X c 1 nhm OHphenol Loi D P N C

CU 23 (H B 2009): Cho X l hp cht thm; a mol X phn ng va ht vi a lt dung dch NaOH 1M. Mt khc nu cho a mol X phn ng vi Na (d) th sau phn ng thu c 22,4a lt kh H2 ( ktc). Cng thc cu to thu gn ca X l

A. CH3-C6H3(OH)2. B. HO-C6H4-COOCH3. C. HO-CH2-C6H4-OH. D. HO-C6H4-COOH.

HNG DN GII

nX = nNaOH Loi A, D nX =

2Hn Loi B

P N C

CU 24 (H B 2010): Cho 13,74 gam 2,4,6-trinitrophenol vo bnh kn ri nung nng nhit cao. Sau khi phn ng xy ra hon ton, thu c x mol hn hp kh gm: CO2, CO, N2 v H2. Gi tr ca x l

A. 0,60. B. 0,36. C. 0,54. D. 0,45.

HNG DN GII

hay C6H3N3O7




2C6H3N3O7 ot C 2CO2 + 3H2 + 3N2 + 10CO

0,06 mol 0,06 0,09 0,09 0,3 mol

x = 0,06 + 0,09 + 0,09 + 0,3 = 0,54 mol

P N C CU 25 (H B 2010): Cho cc cht : (1) axit picric; (2) cumen; (3) xiclohexanol; (4) 1,2-ihiroxi-4-metylbenzen; (5) 4-metylphenol; (6) -naphtol. Cc cht thuc loi phenol l: A. (1), (3), (5), (6) B. (1), (2), (4), (6) C. (1), (2), (4), (5) D. (1), (4), (5), (6) CU 26 (C 2010): Pht biu ng l A. Phenol phn ng c vi dung dch NaHCO3 B. Phenol phn ng c vi nc brom C. Vinyl axetat phn ng vi dung dch NaOH sinh ra ancol etylic D. Thu phn benzyl clorua thu c phenol CU 27 (C 2009) : Cho cc cht HCl (X); C2H5OH (Y); CH3COOH (Z); C6H5OH (phenol) (T). Dy gm cc cht c sp xp theo tnh axit tng dn (t tri sang phi) l :

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

BI TP

CU 28 (C 2012): Cho dung dch cha m gam hn hp gm phenol (C6H5OH) v axit axetic tc dng va vi nc brom, thu c dung dch X v 33,1 gam kt ta 2,4,6-tribromphenol. Trung ha hon ton X cn va 500 ml dung dch NaOH 1M. Gi tr ca m l

A. 21,4 B. 24,8 C. 33,4 D. 39,4

HNG DN GII

Ch c phenol tc dng vi nc brom: C6H5OH + 3Br2 C6H2OHBr3 + 3HBr 0,1 0,1 mol 0,3 (mol) nphenol = n = 0,1 (mol)

Hn hp X gm HBr v CH3COOH tc dng vi NaOH HBr + NaOH NaBr + H2O 0,3 0,3 CH3COOH + NaOH CH3COONa + H2O 0,2 (mol) (0,5 0,3) mol

Khi lng hn hp ban u: m = 0,2.60 + 0,1.94 = 21,4g

P N A

CU 29 (C 2011): Cho m gam hn hp X gm phenol v etanol phn ng hon ton vi natri (d), thu c 2,24 lt kh H2 (ktc). Mt khc, phn ng hon ton vi m gam X cn 100 ml dung dch NaOH 1M. Gi tr ca m l:

A. 7,0 B. 14,0 C. 10,5 D. 21,0

HNG DN GII




C6H5OH + Na C6H5ONa + H2 C2H5OH + Na C2H5ONa + H2 C6H5OH + NaOH C6H5ONa + H2O

2phenol + ancol Hn = 2n = 2.0,1 = 0,2 v nphenol = nNaOH = 0,1 (mol)

nancol = 0,1 (mol) mX = 0,1.94 + 0,1.46 = 14 (g) P N B

CU 30 (H A 2012): Cho dy cc hp cht thm: p-HO-CH2-C6H4-OH, p-HO-C6H4-COOC2H5, p-HO-C6H4-COOH, p-HCOO-C6H4-OH, p-CH3O-C6H4-OH. C bao nhiu cht trong dy tha mn ng thi 2 iu kin sau?

(a) Ch tc dng vi NaOH theo t l mol 1 : 1.

(b) Tc dng c vi Na (d) to ra s mol H2 bng s mol cht phn ng.

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

HNG DN GII

p-HO-CH2-C6H4-OH + NaOH p-HO-CH2-C6H4-ONa + H2O p-HO-CH2-C6H4-OH + 2Na p-NaO-CH2-C6H4-ONa + H2 P N C

CU 31 (H A 2011): Hp cht hu c X cha vng benzen c cng thc phn t trng vi cng thc n gin nht. Trong X, t l khi lng cc nguyn t l mC : mH : mO = 21 : 2 : 8. Bit khi X phn ng hon ton vi Na th thu c s mol kh hir bng s mol ca X phn ng. X c bao nhiu ng phn (cha vng benzen) tha mn cc tnh cht trn ?

A. 9. B. 3. C. 7. D. 10.

HNG DN GII t CTTQ ca X: CxHyOz

Lp t l: x : y : z = 21 2 8: :12 1 16

= 7 : 8 : 2 X l C7H8O2

nX =2H

n trong X c 2 nguyn t H linh ng

+ Nu X c 1 chc OH ancol v 1 chc OH phenol th c 3 CTCT:

+ Nu X c 2 chc OH phenol th c 6 CTCT:

P N A




CU 32 (H A 2010): Axeton c iu ch bng cch oxi ho cumen nh oxi, sau thu phn trong dung dch H2SO4 long. thu c 145 gam axeton th lng cumen cn dng (gi s hiu sut qu trnh iu ch t 75%) l

A. 300 gam B. 500 gam C. 400 gam D. 600 gam

HNG DN GII

C6H5C3H7 CH3COCH3 + C6H5OH 120g 58g

120.145 100. 400 g

58 80= 145g

P N C




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