Tuyen Tap de Thi Hsg Hoa Hoc

8/12/2019 Tuyen Tap de Thi Hsg Hoa Hoc

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[Ti liu tng hp t cc thi HSG]
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SGIO DC V O TO

THNH PH NNG

HNG DN CHM THI HC SINH GII THNH PH

NM HC 2007 - 2008

MN: HA HC LP 1O

Thi gian: 150 pht (khng k thi gian giao )

Cu I (4 im)X v Y l cc nguyn t nhm A, u to hp cht vi hiro c dng RH (R l k hiu ca nguyn t X hoc Y).Gi A v B ln lt l hiroxit ng vi ha tr cao nht ca X v Y. Trong B, Y chim 35,323% khi lng. Trungha hon ton 50 gam dung d ch A 16,8% cn 150 mL dung dch B 1M.

1. Xc nh cc nguyn t X v Y.2. B l anion tng ng ca phn t B. (a) Hy cho bit (c cng thc minh ha) dng hnh hc ca B v B. (b)

So snh (c gii thch) di lin kt Y-O trong phn t B v B.3. Bit X c cu trc tinh th lp phng tm khi v mt sp xp tng i c nh ngha bng t l gia

th tch chim bi cc hnh cu trong t bo c sv th tch t bo c s. Hy tnh mt sp xp tng itrong tinh th ca X.

P N IM

1. Hp cht vi hiro c dng RH nn Y c th thuc nhm IA hoc VIIA.Trng hp : Nu Y thuc nhm IA th B c dng YOH

Ta c : 284,9Y

677,64

323,35

17

Y (loi do khng c nghim thch hp)

Trng hp 2 : Y thuc nhm VIIA th B c dng HYO4

Ta c : 5,35Y677,64

323,35

65

Y , vy Y l nguyn t clo (Cl).

B (HClO4) l mt axit, nn A l mt baz dng XOH

gam4,8gam50100

8,16mA

XOH + HClO4 XClO4 + H2O

mol15,0L/mol1L15,0nn4HClOA

mol15,0

gam4,8mol/gam17MX

MX = 39 gam/mol, vy X l nguyn t kali (K).

2. B l HClO4, B l ClO4-
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(a) Dng hnh hc :

Axit pecloric

(dng t din)

Ion peclorat

(dng t din u)

(b) Bc lin kt cng ln di lin kt cng nh, do vy :

3. Gi a l di cnh mng c sv R l bn knh nguyn t, ta c R4a3

S nguyn t c trong mt mng c sbng : 218

18

Vy %68a

R3

42

f3

3

v

Cu II (4 im)1. Cho bit s oxi ha ca mi nguyn t lu hunh (S) trong phn t axit thiosunfuric (H2S2O3) v ca mi

nguyn t cacbon trong phn t axit axetic (CH3COOH)2. Thm lng d dung dch KI (c pha h tinh bt) vo 5,00 mL dung dch K2Cr2O7 c nng a M trong

H2SO4, th dung dch thu c c mu xanh. Thm tip dung dch Na2S2O3 0,10 M vo cho n khi mu xanhbin mt th dng 15,00 mL dung dch ny. Vit cc phng trnh phn ng v tnh a. Bit sn phm oxi ha

S2O32- l S4O62-.3. Ha tan hon ton 9,06 gam mt mu hp kim Al-Mg (gi thit khng c tp cht no khc) bng dung dch

H2SO4 c, thu c 12,22 L kh SO2 (o 136,5oC; 1,1 atm) v 0,64 gam cht rn mu vng. Xc nh phn

trm khi lng mi kim loi trong mu hp kim trn.


4/65

P N IM

1. S oxi ha ca cc nguyn t S v C :

1.0

2. Phng trnh phn ng :6KI + K2Cr2O7 + 7H2SO4 3I2 + Cr2(SO4)3 + 4K2SO4 + 7H2O (1)

2Na2S2O3 + I2 2NaI + Na2S4O6 (2)

0,5

T (1) v (2) ta c : mol10.5,2L/mol1,0L015,06

1n

6

1n 4OSNaOCrK 322722

M05,0L10.5

mol10.5,2a

3

4

0,5

3. Gi x, y ln lt l s mol cc kim loi Mg v Al.

)mol(4,0

5,1273273

4,22

22,121,1n

2SO

; Cht rn l S, )mol(02,0

32

64,0nS

0

02.012,0

6

y

4

4,08,0

6

y3

30

y

x2

20

x

Se6S

Se2S

e3AlAl

e2MgMg

1,5

Ta c :

22,0y;13,0x

92,0y3x2

06,9y27x24

%44,34%100gam06,9

mol/gam24mol13,0m% Mg

v %56,65m% Al

0,5

Cu III (4 im)1. Nguyn t ca nguyn t X c tng s ht cc loi l 60, s ht mang in trong ht nhn bng s ht khng mang

in; Nguyn t ca nguyn t Y c 11 electron p; nguyn t nguyn t Z c 4 lp electron v 6 electron c thn.

a) Da trn cu hnh electron, cho bit v tr ca cc nguyn t trn trong bng h thng tun hon.

b) So snh (c gii thch) bn knh ca cc nguyn t v ion X, X2+ v Y-.2. V hnh m t cch tin hnh th nghim iu ch dung dch HCl bng nhng ha cht v dng cn gin c sn

trong phng th nghim sao cho an ton. Ghi r cc ch thch c n thit.


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3. Sc Cl2 vo dung dch KOH long thu c dung dch A, ha tan I2 vo dung dch KOH long thu c dungdch B (tin hnh nhit phng).a) Vit phng trnh ha hc xy ra v cho nhn xt.b) Nu hin tng v vit phng trnh ha hc xy ra khi cho ln lt cc dung dch hn hp HCl v FeCl2,

dung dch Br2, H2O2 vo dung dch A (khng c Cl2 d).

P N IM

1. a) Xc nh v tr da vo cu hnh electron:

X X X X X2Z N 60 ;Z N Z 20 , X l canxi (Ca), cu hnh electron ca 20Ca:[Ar] 4s2

Y c 11 electron p nn cu hnh ca Y l 1s22s22p63s23p5 hay [Ne] 3s2 3p5 Y l Cl

Theo gi thit th Z chnh l crom, cu hnh electron ca 24Cr: [Ar] 3d5

4s1

STT Chu k nguyn t Nhm nguyn t

Ca 20 4 IIA

Cl 17 3 VIIA

Cr 24 4 VIB

Trt t tng dn bn knh nguyn t: CaClCa RRR 2

b) Bn knh nguyn t t l vi thun vi s lp electron v t l nghch vi s n v in

tch ht nhn ca nguyn t . Bn knh ion Ca

2+

nh hn Cl

-

do c cng s lp electron (n= 3), nhng in tch ht nhn Ca2+ (Z = 20) ln hn Cl- (Z = 17). Bn knh nguyn t Ca lnnht do c s lp electron ln nht (n = 4).

2,0

2. iu ch HCl t dung dch H2SO4 m c v NaCl rn, hnh 5.5 trang 128 SGK Ha hc 10nng cao.

0,75

3. a) nhit thng:2KOH + Cl2 KCl + KClO + H2O

6KOH + 3I2 5KI + KIO3 + 3H2O

Gii thch: Trong mi trng kim tn ti cn bng: 3XO-X- + XO 3

Ion ClO-phn hy rt chm nhit thng v phn hy nhanh khi un nng, ion IO- phn

hy tt c cc nhit .

b) Cc phng trnh ha hc:

Ion ClO-c tnh oxi ha rt mnh, th hin trong cc phng trnh ha hc:

- Khi cho dung dch FeCl2 v HCl vo dung dch A: c kh vng lc thot ra v dung dch t

1,25


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khng mu chuyn sang mu vng nu:

2 FeCl2 + KClO + 2HCl 2 FeCl3 + Cl2 + H2O

- Khi cho dung dch Br2 vo dung dch A : dung dch brom mt mu :

Br2 + 5KClO + H2O 2HBrO3 + 5KCl

- Khi cho H2O2 vo dung dch A: c kh khng mu, khng mi thot ra:

H2O2 + KClO H2O + O2 + KCl

Cu IV (4 im)

1 Cho bit: nng lng lin kt ca cc lin kt H-H, O-O, O=O, H-O ln lt l 436, 142, 499, 460 ( kJ/mol).Hy vit phng trnh nhit ha hc ca phn ng gia kh hiro v kh oxi to ra hiropeoxit.

2 Cho phn ng: 2SO2 (k) + O2 (k) 2SO3 (k) H = - 198 kJa) tng hiu sut qu trnh tng hp SO3, ngi ta c th s dng bin php no lin quan n p sut, nhit

v cht xc tc? Gii thch?

b) Cho 10,51 mol kh SO2 v 37,17 mol khng kh (20% v th tch l O2 cn li l N2) c xc tc l V2O5. Thchin phn ng 4270C, 1 atm th phn ng t hiu sut 98%. Tnh hng s cn bng KC, KP ca phn ng 427

0C.

3 A l mt oxit ca st. Ly mt lng A chia lm 2 phn bng nhau. Phn I tc dng va vi a mol H2SO4trong dung dch H2SO4 long. Phn II tc dng va vi b mol H2SO4 trong dung dch H2SO4 c nng toSO2 (sn phm kh duy nht). Bit b = 1,25a, xc nh cng thc oxit st ban u.

P N IM

1. Hp= EH-H + EO=O - 2EO-H - EO-O

Hp = 436 + 499 -2.460142 = -127 (kJ)

H2(k) + O2(k) H2O2(k) H = - 127 kJ

1,0

- Gim nhit ca h phn ng, nhng khong 5000 C l thch hp v: nu gim thp qu thtc phn ng chm.

- Thi lin tc SO2 v khng kh c nn p sut cao vo l phn ng v phn ng xy ratheo chiu thun lm gim p sut ca h.

- Dng V2O5 lm xc tc phn ng mau chng t trng thi cn bng.

0,75

2. nO2 b = 7,434 (mol), nN2 b = 29,736 (mol)

2SO2 (k) + O2 2SO3 (k) H = - 198 kJ

Ban u: 10,51 (mol) 7,434 (mol) 0

Lc phn ng: 10,3 (mol) 5,15 (mol) 10,3 (mol)

Lc CB: 0,21 (mol) 2,284 (mol) 10,3 (mol)

s mol hn hp TTCB = 0,21 + 2,284 + 10,3 + 29,736 = 42,53 (mol)

Pi = xi.P = xi.1 = xi

1,25


7/65

v

(R = 0,082, T = 427 + 273 = 700 0K,n = -1)

v

3. Gi FexOy l cng thc ca A

(1) 2FexOy + 2yH2SO4 xFe2(SO4)2y/x + 2y H2O

n ny

(2) 2FexOy + (6x-2y) H2SO4 xFe2(SO4)3 + (3x-2y) SO2 + ( 6x-2y) H2O

n n (3x-y)

Theo gi thit : n(3x-y) = 1,25 nyx 2,25 3

y 3 4 A l Fe3O4

1,0

Cu V ( 4 im)

1. T dung dch H2SO4 98% ( D= 1,84 g.mL-1) v dung dch HCl 5 M, trnh by ph ng php pha ch c

200 mL dung d ch hn hp H2SO4 1M v HCl 1M .

2. t chy hon ton a gam S ri cho sn phm sc qua 200 mL dung d ch NaOH b M thu c dung dch X.Chia X lm hai phn bng nhau. Phn 1 cho tc dng vi dung dch CaCl2 d thy xut hin c gam kt ta .Phn 2 tc dng vi dung dch nc vi d thy xut hin d gam kt ta . Bit d >c. Tm biu thc quan hgia a v b.

3. Cho cc dung dch ring bit mt nhn sau: Na2SO4, AlCl3, FeSO4, NaHSO4, FeCl3. Mt hc sinh cho rng nudng dung dch Na2S th c th phn bit cc dung dch trn ngay ln th u tin. Kt lun ca hc sinh c ng khng ? V sao?

P N IM

1.

*Phn tnh ton :

S mol H2SO4 cn ly = s mol HCl cn ly =

200 1

0,2(mol)1000

Khi lng dung dch H2SO4 cn ly : mdd =0,2 98 100

20(gam)98

Th tch dung dchH2SO4 cn ly = 20 : 1,84 = 10,87 (mL)

Th tch dung dch HCl cn ly = 0,2 : 5 = 0,04 (L) = 40 mL

* Cch tin hnh: Ly khong 100 - 120 mL nc cho vo bnh th tch 200 mL c chia vch.

1,0


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Cn 20 gam dung dch H2SO4 hoc ong 10,87 mL dung dch H2SO4, sau cho t t vobnh cha nc khuy u. i dung dch H2SO4 tht ngui, ong 40 mL dung dch HCl 5Mthm vo bnh, sau thm nc vo cho n vch 200 mL

2. Phng trnh :

(1) S + O2 SO2(2) SO2 + NaOH NaHSO3(3) SO2 + 2 NaOH Na2SO3 + H2OPhn I tc dng vi dung dch CaCl2 sinh kt ta, chng t dung dch X c cha Na2SO3, phnII tc dng vi dung dch Ca(OH)2 sinh nhiu kt ta hn chng t dung dch X c muiNaHSO3

(4) Na2SO3 + CaCl2 CaSO3 + 2NaCl(5) NaHSO3 + Ca(OH)2 CaSO3 + Na2SO3 + 2H2Ons = a/32 (mol) , nNaOH = 0,2 b ( mol)

Theo (2),(3), SO2 tc dng vi dung dch NaOH sinh 2 loi mui th :

2

NaOH NaOH

SO S

n n1 2

n n 1 (SCN)2 > I2. Vit phng trnh ha hc minh ha ktqu so snh trn .

2. Ti sao c s khc bit v gc lin kt ca Cl2O (1110) v OF2 (105

0) ?

3. tch brom c trong 1m3 nc bin di dng NaBr, ngi ta cho mt lng dung dch H2SO4 vo mtlng nc bin; tip n sc kh clo vo dung dch mi thu c; sau dng khng kh li cun hi brom

vo dung dch Na2CO3 ti khi bo ho brom. Cui cng cho H2SO4 vo dung dch bo ho brom, thu hibrom ri ho lng, c 28,05 ml Br2 c khi lng ring 3,12 g/ml 200C. Vit phng trnh ho hc ch yu

xy ra trong qu trnh v cho bit vai tr ca H2SO4. Tnh % khi lng ca brom trong nc bin bit khilng ring ca nc bin l 1,25 g/ml.

Cu IV: ( 2,0 im)

1. Cho hn hp Fe v FeS tc dng vi dung dch H2SO4 m c nng thu c dung dch X lm mt mu dungdch K2Cr2O7 trong H2SO4 long. Vit phng trnh ha hc ca cc qu trnh th nghim trn.

2. A l hp cht ca lu hunh, tan rt tt trong nc to dung dch X cha mt cht tan. Ha tan 25,8 gam A vonc thu c 200 ml dung dch X ( D = 1,15 g/ml). Chia dung dch X lm hai phn bng nhau. Phn 1 trung ha

va vi 100 ml dung dch NaOH 10% (D = 1,2 g/ml). Phn 2 tc dng vi dung dch BaCl2 d thu c mtkt ta trng khng tan trong axit. Xc nh cng thc phn t v vit cng thc cu to ca A.

Cu V: (2,0 im)

1. Hng s cn bng KC 250C ca phn ng Cl2 (k) 2Cl (k) l 1,4.10

-18. Tnh nng Cl 250C lc cn

bng bit nng ca Cl2 lc cn bng l 0,896 M. T cho bit 250C, clo tn ti ch yu dng phn t

hay nguyn t ?

2. Cho bit cn bng 2FeCl2 (aq) + Cl2 (aq) FeCl3 (aq) s chuyn di theo chiu no khi sc mt lng kh H2S

thch hp vo dung dch? Cho V36,1E,V77,0ECl2

Cl0

FeFe

02

2

3

.

3. Phn ng chuyn ho mt loi khng sinh trong c th ngi nhit 370C c hng s tc bng 4,2.10-5

(s-1

). Vic iu tr bng loi khng sinh trn ch c kt qu nu hm lng khng sinh lun lun ln hn 2,00mg trn 1,00 kg trng lng c th. Mt bnh nhn nng 58 kg ung mi ln mt vin thuc cha 300 mgkhng sinh .a) Hi bc ca phn ng chuyn ho?b) Khong thi gian gia 2 ln ung thuc k tip l bao lu?c) Khi bnh nhn st n 38,50C th khong cch gia 2 ln ung thuc thay i nh th no? Bit nng

lng hot ho ca phn ng bng93,322 kJ.mol1.

----------------HT ---------------

Ch : Hcsinh c sdng H thng tun hon cc nguyn tha hc v my tnh c nhn n gin theo quynh ca B Gio dc v o to. Gim th khng gii thch g thm.


11/65

S GIO DC V O TO

THNH PH NNG

HNG DN CHM K THI HC SINH GII THNH PH

NM HC 2004 - 2005

MN: HA HC LP 11


1.

2.

3. 35,6pK 1a 33,10pK 2a

P N IM

1.

N

H HH

N

F FF

-

- -

2.

1

1

1

2 P

1

4P

P

PK

2

2

ON

2

NO

ON

2

NO

P

42

2

42

2

17,01)2,0(1

)2,0(4P

1

4K

2

2

2

2

P

%)6,54(546,017,010,01

42

2

mol75,092

69n


12/65

)1(9225,020

300082,0)1(75,0P

42ON

845,120

300082,0.75,0.2P

2NO

17,0)1(9225,0

)845,1(K2

P

%)27,19(1927,0

3. 01,005,02,0n,mol01,04,22

224,0n NaOHCO2

3,833,1035,62

1)pKpK(

2

1pH 21

2.

3.

P N IM

1.

2.

mol035,01175,05

1n

5

1n 2

4 Fe)3(MnO


13/65

mol115,0035,075,02,0n)2,1(MnO 4

0625,0y

025,0x

115,0y5

6

x5

810y96x160

%60%10010

960625,0m% SCu

1.

2.

P N IM

1.

2.


14/65

01,0ba

36292414,102,0

b28a44

02,04,22

448,0ba

mol3,0)bd(nmol22,0)pu(nHH

04,0y

02,0x

18,0y3x3

2,2y27x56

%9,50%1002,2

5602,0m% Fe

%1,49m% Al

mol34,040

6,13nmol3,0n4n3nn

)bd(OHAlFeH)4,3,2,1(OH 33

g6,116001,0m 32OFe


15/65

mol03,078

34,2n

3)OH(Al

mol07,003,004,0nnn3)OH(AlOHH

L14,05,0

07,0V

mol01,003,004,0n )7()OH(Al 3

mol11,003,004,004,0n3nnn )7()OH(AlAlOOHH 32 L22,0

5,0

11,0V

P N IM

1. kcal4,368H kcal2,337H

kcal32,68H

kcal1,37)32,68()4,368()2,337(H

2.

CH3 CH CH C

H

CH3

C2H5 (4-metylhex-2-en)


16/65

C

C

CH3H

C2H

5

C

H

HCH3

C

C

HCH3

C2H

5

C

C

CH3H

C2H

5

C

CH3

HH

C

C

HCH3

C2H

5

H

H

CCH3

H

CH3

CH

(E)(R) (E)(S)(Z)(R) (Z)(S)

CH3 CH CH CH

CH3

C2H5

CH3 CH CH CH

CH3

C2H5

Br Br

CH3 CH CH CH

CH3

C2H5

OH Br

CH3 CH CH CH

CH3

C2H5

Cl Br

CH

O

H CH

O

C

O

H

andehit fomic andehit oxalic

P N IM

1.

8y

6x2x2y80yx12 )3(


17/65

CH2 CH CH CH CH CH2

H2C O OHC

HC O O

HC

HC O O CH2

A (hexa-1,3,5-trien)

CH2 CH CH CH CH CH2Br2

CH2 CH CH CH CH CH2

CH2 CH CH CH CH CH2

Br

Br

CH2 CH CH CH CH CH2Br

CH2 CH CH CH CH CH2

CH2 CH CH CH CH CH2

Br

Br

CH2 CH CH CH CH CH2Br

Br

Br

Br

(X) 5,6-dibromhexa-1,3-dien; (Y) 3,6-dibromhexa-1,4-dien;

(X)

(Y)

(Z)

(Z) 1,6-dibromhexa-2,4-dien

2. 6

1. 2. 3.


18/65

P N IM

1. mol08,04,22

584,3

2

1n C,B,A

mol01,008,0100

5,12nn ankin

g82,14001,022,2mn2nHC mol065,0201,01606,13n )2(Br2

065,0

1

82,1

n14

mol005,0065,001,008,0n2n2nHC

2O2

1n3

015,0197

955,2nn

32 BaCOCO 3n

015,0

n

005,0

1

2.

C3H8 C3H7Cl CH3CH=CH2 CH3CHBr-CH2Br

CH3 C CH

CH3CH2CHBr2

CH3CBr2CH3

Cl2,as KOH/ROH

KOH/ROH

Br2

HBr

HBr

peoxit

3. + 2KMnO4CH3 C CH CH3 C C

O O

+ 2MnO2 + KOHOK


19/65

S 4

S GIO DC V O TO

THNH PH NNG


NM HC 2005 - 2006

MN: HA HC LP 11


1.

2.

P N IM

1.

2.


20/65

0V

2CO

HHCOCO nnn2 323

mol01,0n2

1n

HCO2

mol015,0n2CO

L336,04,22015,0V4,2201,0L224,0 2CO


21/65

2

1

2

1

2

1

2

1

1.

2.

P N IM


22/65

)n6m2(n32M2

4,2m

M

8,4

3,2,1m,nm2n6

mn64M

mol075,064

8,4nCu

NaOHNO nmol3,0075,022n 2

1.

2.

3.

4. 5.


23/65

P N IM

1.

CH3 CH CH3+ Cl2

- HCl

CH3 CH CH2 Cl

CH3 C CH3

Cl

1-clo-2-metylpropan

2-clo-2-metylpropan

CH3

CH3

CH3

(9x1,0)(9x1,0) + (1x5,0)

= 64,3%

(1x5,0)(9x1,0) + (1x5,0)

= 35,7%

CH3 CH CH3+ Br2

- HCl

CH3 CH CH2 Br

CH3 C CH3

Br

1-brom-2-metylpropan

2-brom-2-metylpropan

CH3

CH3

CH3

(9x1,0)(9x1,0) + (1x1600)

= 0,56%

(1x1600)(9x1,0) + (1x1600)

= 99,44%

2.

H+

-H+

3. 32

128.22

0

2

188.22

22

148.22


24/65

CH3 C C

H H

HC

CH3

C* C CH3 2Z-4-metylhept-2-en-5-in(A)

CH3CH2CH2CH(CH3)CH 2CH2CH3 4-metylheptan(B)

CH3 C C

H H

HC

CH3

C C CH3 2Z,5Z-4-metylhepta-2,5-dien(C)

H H

1.

CH3 C

O

CH2 CH2 CH

CH2COOH

C O

H3C

(A)

2.


25/65

P N IM

1.

O + O=C=O

OO

OH

limonen

OH

OH OH

OH

4.

42SOH

HO - NO2 + H2SO4 H - O - NO2 + HSO4

H

(+) (-)

H - O - NO2 + H2SO4

H

(+)

H3O(+)

+ HSO4(-)

+ NO2

(+)


26/65

+ NO2(+)

H NO2

+

NO2

+ H(+)

gam512,9123

7815m )1(HC 66

%88

mL/g8,0mL5,13

g512,9H

P N IM

1.


27/65

mol1,0yx

85,242

y197

2

yx100

15,02

yx

mol2,0yxn

2CO

mol14,018

44.2,032,11ng32,11mmm OHCOOH 222

14y,10x14,0.2

y

2,0

x

02,0

1

4

C

CH3

CH3

CH3

(t-butylbenzen)

2.

+

C(CH3)3H

+ H(+)

+ (CH3)3C+

C(CH3)3

3.

C

CH3

CH3

CH3

O2N


28/65

S GIO DC V O TO

THNH PH NNG

HNG DN CHM THI HC SINH GII THNH PH

NM HC 2006 - 2007

MN: HA HC LP 11


1.

2.

(a) (b)

P N IM

1. M08,0L125,0

L.mol200,0L050,0C

1o

ClNH4

M06,0L125,0

L.mol100,0L075,0C

1oNaOH

5

3

4b 10.8,1

x06,0

x)x02,0(

]NH[

]OH][NH[K

M10.4,5

02,0

06,010.8,1x 55

73,9)]10.4,5lg([14pH 5

0,50


29/65

1,00

2.

10,25

mol10.037,1L.mol100,0L01037,0ax 31

mol10.037,1L.mol010,0L02074,05x 31

mol10.037,1

gam5,03


30/65

1.

2.

3.

(a)

(b)

P N IM

1.

2.

H N

H

H

H O

H

H N

H

H

... ... ......

3.

32 NHOHNOH4H6H4H

kJ908)kJ46(4[)]kJ242(6[)kJ904(H


31/65

1.

2.

(a)

(b) C180,SOH o

42

(c) o

42 t,SOH

3.

P N IM

(a)

CH3 CH CH2 + HCl

CHCH3 CH3

Cl

CH2CH3 CH2 Cl

CH3 CH CH2H

+ CH3 CH CH3

CH3 CH2 CH2

Cl-

CHCH3 CH3

Cl

(X)

(Y)


32/65

(b)

CH3 CH2H2SO4

CH CH3

OH

CH3 CH CH CH3 + H2O

CH2 CH CH2 CH3 + H2O

CH3 CH2 CH CH3

OH

CH3 CH CH CH3

CH2 CH CH2 CH3

H+

CH3 CH2 CH CH3+OH2

-H2O

(X)

(Y)

(c)

+ HONO2

H2SO4

+ H2O

CH3

CH3

NO2

+ H2O

CH3

NO2

+NO2

+NO2

-H+CH3 CH3

CH3 CH3

-H+

CH3 CH3

H

NO2

H NO2

NO2

NO2

H

CH3

H

CH3


33/65

H

CH3

H

CH3

1.

(a) (b)

2. (a) (b) (c)

P N IM

1.

mol02,0mol/gam170

gam4,3)ankin(n mol04,0)ankin(n2n

2Br

mol03,0L/mol15,0L2,0


34/65

mol01,0mol01,0mol/L4,22

L672,0

2.

mol045,0n2)OH(Ca

mol02,0n3CaCO

mol2,0mol/gam16

gam2,3

2

2

2

CHCH3 CCH3

CH CH CH3 (3-metylhexa-2,4-dien)


35/65

CH3C C

CH3

C C

CH3

HH

H

H3C

C C

CH3

C C

H

CH3H

H

H

C C

CH3

C C

CH3

HH

CH3

H

C C

CH3

C C

H

CH3H

CH3

cis-cis cis-trans trans-cis trans-trans

CHCH3 C

CH3

CH CH CH3+ Br2

- Br-

CHCH3 C

CH3

HC CHBr CH3

HCCH3 C

CH3

CH CHBr

HCCH3 C

CH3

CH CH CH3

CH3

Br

+ Br-

CH3CH=C(CH3)-CHBr-CHBr-CH3

CH3-CHBr-C(CH3)=CH-CHBr-CH3

CH3-CHBr-CBr(CH3)-CH=CH-CH3

1.

(a) (b)

2. (a) (b) (c) (d)

3. (a) (b)

P N IM


36/65

NO2

+ HONO2

H2SO4

+ Cl2

Fe

NO2

Cl

(a)

Cl

+ HONO2

H2SO4

+ Cl2

Fe

Cl

(b)+H2SO4

SO3H

Cl

SO3H

t

Cl

NO2

NO2

CH3

+ CH3Cl

AlCl3

+ Br2

Fe

COOH

(c)+ KMnO4

COOH

Br

CH3

+ Br2

Fe

+CH3Cl

AlCl3

CH3

(d)+H2SO4

SO3H

CH3

SO3H

t

CH3

Br

Br+ KMnO4

COOH

Br

7

(naphtalen) (tetralin) (decalin)


37/65

+ HONO2H2SO4

+ H2O

NO2

+ Br2Fe

+ HBr

Fe


38/65

S GIO DC V O TO

THNH PH NNG


NM HC 2004 - 2005

MN: HA HC LP 12 - BNG A


Cu I (3 im)1.

2.

3.

p n im

1. T31 1823 10.810.02,62

18

10

ktn

nln

o

2/1t

693,0k

T31

40

3,12

693,0

6 e10.35,5n

T31

2.

H3N

H3N NH3

NH3

Cl

ClH3N

H3N Cl

NH3

Cl

NH3trans- cis-

2+ 2+


39/65

Cl

Cl

Cl

Cl

trans- cis-

enen

en

en

cis-en

en

F- F- F- F- F- F-

CN- CN- CN- CN-CN- CN-

3. t

t

t

C0o

Cu II (4 im)1.

2.


40/65

V771,0EoFe/Fe 23

V51,1Eo

Mn/H,nOM 24

3.

p n im

1. ]NH[

]NH[lgpKpH

4

3a

]NH[

]NH[lg)76,414(5,9

4

3

8,1

]NH[

]NH[

4

3

M1,0NH4 M18,0M1,08,1NH3

mol05,0)L5,0()L.mol1,0(n 1NH4

mol09,0)L5,0()L.mol18,0(n 1NH3

L03,0)L.mol0,3(

)mol09,0(V

1ddNaOH

g49,7)mol/g5,53()mol14,0(m ClNH4

2. V671,015,0

003,0

lg059,0771,0]Fe[

]Fe[

lg059,0EE 2

3o

Fe/FeFe 23

V52,1005,0

)1)(02,0(lg

5

059,051,1

]Mn[

]H[]MnO[lg

5

059,0EE

8

2

84o

Mn/nOMMn 24

63,62059,0

)771,051,1.(5

059,0

En

101010K


41/65

V79,005,0

103,0lg059,0771,0

]Fe[

]Fe[lg059,0EEE

2

3o

Fe/FeFeMn 23

3.

21212

22

10K.K]SH[

]S[]H[

M10.1,1003,0

1,010

17

2

21

M010

1,010 4

29

21


42/65

Cu III (3 im)1. G

(a) (b)

2.

H oH oS P X 3.

p n im

1.

141411No

mol.J10.8)10ln()K298)(mol.K.J314,8(KlnRTG

])OH][Hln([RTGG o 14431114

mol.J10.4)1010ln()K298)(mol.K.J314,8(mol.J10.8G

0G

148121114 mol.J10.2,3)10.210ln()K298)(mol.K.J314,8(mol.J10.8G 0G

2.

21

o

1

2

T

1

T

1

R

H

K

Kln

313

1

298

1

314,8

H

10.8,1

10.45,3ln

o

3

3


43/65

11

2

22o

omol.K.J6,175

T

KlnRTHS

13

1CP atm726,0298.314,8

10.8,1

)RT(KK

726,0PKK Px

3. t

A

ktA

1

A

1

o

2

1t

2

AA 0

0002/1

Ak

1

A

1

2

A

1

k

1t

s257)M272,0)(sM0143,0(

1

Ak

1t

11

o2

1

Cu IV (3 im)1.

H

4422 HgSO,SOH,OH

2.

p n im

1.

H

4422 HgSO,SOH,OH

4KMnO

2.


44/65

C

O O

C

OHHO

C

O O

C

O-

HO

C

O O

C

OO 2-

Cu V (3 im)

1. 2.

3.

OH

HO C5H11-n

Olivetol

OH

C5H11-n

1-3,4-trans-tetrahidrocannabinol (I)O

OH

-tecpineol

p n im

1. 3

CH3 C

CH3

OH O C

OH

CH2 CH2 C OH

CH3

O C

OH

C OH

OH

CH3 C

CH3

CH CH2 CH2 C

CH3

CH CH O


45/65


46/65

2.

p n im

1. 1300)117115165132131(nM

Val Orn Leu Phe

Pro

ValOrnLeuPhe

Pro

2.

OMe

HOOH

CH2OH

O

OMe

MeOOMe

CH2OH

O

CHO

MeOOMe

CH2OH

OH

COOH

MeOOMe

COOH

OMe

CHOCHO

CH2OH

O

(Q)


47/65

S GIO DC V O TO

THNH PH NNG

P N K THI HC SINH GII THNH PH

NM HC 2004 - 2005

MN: HA HC LP 12 - BNG B


Cu VII (3 im)

4.

23XO

(a)

(b)

5.

p n im

1.

23SO 2

3*SOS 232OS

23*SOS 3

*SOBaS

3

*SOBaS

O

S

S

OO

2-*


48/65

HO

HO

SH

SH

S

O

OOOO

d ng ch V d ng tam gi c phng dng t din

2.

Cu VIII (4 im)

4.

5.

6.

p n im

1. 2

1

ot

2

1

2. 3y

x

67,6

33,93

y12

x56


49/65

t

t

3. g07,1)mol.g5,53()L.mol2,0()L1,0(m 11ClNH4

4

3b

NH

NHlg)pK14(pH

1

3

L.mol2,0

NHlg

mol115,0)L1,0()L.mol15,1(n1

NH3

L10.8,7L.mol8,14

mol115,0V

3

1ddNH3

Cu IX (3 im)

p n im

3


50/65

3

3

3

3

Al

5

N

2

Mg

5

N

3

Al

5

N

2

Mg

5

N

5

N

3

4

3

4


51/65

Cu X (3 im)

3.

4.

)1( )2( )3(

)1( )2( )3( )4(

p n im

1.

men

t,CuO 2

2 Mn,O

t,SOH 42 4KMnO

t,CaO,NaOH

o600400,NO

22 Mn,O

2. )1( )2( )3(


52/65

)1( )2( )3(

)4(

Cu XI (4 im)

4. 7:36m:m HC

Ni,H2

CuO H,KMnO

4

5.

6.

p n im

7:31

7:

12

36n:n HC


53/65

Cl OH OH

O

n-C6H14Al2O3/Cr2O3 Cl2, as H2O

425o

H2, Ni

CuO, t KMnO4, H+

HOOC(CH2)4COOHH2N(CH2)6NH2

T nilon-6,6

0

t

0

t

CH3 CCH3

CH3

CHOH

CH3 CH3 CCH3

CH3

CHOH2

CH3 CH3 CCH3

CH3

CH CH3

CH3 C

CH3

CH CH3

CH3

H+

- H2O

-H+

CH3 C

CH3

C CH3

CH3


54/65

CH2OH CH2OH2H

+

- H2O - H+

Cu XII (3 im)

p n im

mol975,0n2O mol8,0n

2CO

mol65,018

448,032975,07,15n OH2

mol15,02

2975,028,065,0n Y

33,515,0

8,0C

66,8215,0

65,0H

66,5015,0

6,7M ROH


55/65

Cu 1

Po21084

Cu 1

1.

eV6,781

3,026,132

n

Z6,132E

2

2*

2**He

eV4,541

26,13

n

Z6,13E

2

2

2*

He

eV2,24)eV6,78()eV4,54(EEI *He*

He1

2.

NH

H

H..

H

N

HH


56/65

ClF

F

F.. ..

F Cl

F

F

Xe

FF

FF .

.

..

F

XeF F

F

3.

00507,085,6100

100ln

14

1kkt

m

mln

t

0

137

00507,0

693,0

k

2lnt 2/1

Po21084 454

90100

100ln

00507,0

1t

Cu 2

V51,1EoMn/MnO 24

V5355,0Eo I/I3

Cu 2

1. M010,045

1503,0C -

3 OCOCH

M100,0

45

3015,0C -OHCO


57/65

25,11HCOO'b

24,11OCOCHb

10CK10CK -3

]OHCO[)K(]OCOCH[K1

Kh

-1'a

-3

1a

w

9

175,3276,4

14

1 10.96,210.1010.101

10h

o-

3976,4

76,4

1-

3 ]OCOCH[010,010.96,210

10010,0]OCOCH[

o-

975,3

75,3

1- ]OHCO[10,0

10.96,210

1010,0]OHCO[

2.


58/65

01,0

]H[05,0lg

5

059,051,1

]Mn[

]H][MnO[lg

5

059,0EE

8

2

84o

Mn/MnOp 24

V574,0)1,0(

02,0lg

2

059,05355,0

]I[

]I[lg

2

059,0EE

33

3o

I3/It 3

E 5

059,0

ChK

hK

hC

h

a

2

a

2

M334,0C

4HSO

3.

24,9NH

10K4


59/65

M2015,0C10)2,0.(1,0

10.)2,0C( 24,112

82

Cu 3

U H

Cu 3

1.

12p12p12VV TTRCnTTRCnTTCnQU

J74,1013815,27315,27315,1

314,81,2928

100

TTP

P

RCn 111

2p

J05,1419415,27315,2731

5,11,29

28

100TT

P

PCnTTCnH 11

1

2p12p


60/65

11

1

2p12pp TT

V

VCnTTCnHQ

J09,2838815,27315,273.21,2928

100

J60,811015,273.314,8.28

100TT

V

VnRTTnRVPW 11

1

212

J,20278811028388WQU

2.

CBr)CH(OHkdt

CBr)CH(dv 332

33

CBr)CH(k

dt

CBr)CH(dv 33

33

t

AAk

1o1

1o1

A

Aln

t

1k

1o1

1o1

AA

AA

t

1k


61/65

3. 2

n

2

3HN

2NH

3HN

2NH

PRT

VK

RT

V

nn

n

PP

PK

22

3

22

3

constnn

nK 3

HN

2NH

n

22

3

2

nPRT

VKK

Cu 4

1.

H

NH

HCH3

H3CH

OHHO

O

OH

2.


62/65

Cu 4

H

NH

HCH3

H3CH

OHHO

O

OH

E

S

R

**

O NH

NNH N

H

Cu 5

1.


63/65

2.

Cu 5

1.

SOCl2 EtO(CH2)3CdCl

N

COOH

N

COCl

N

C

O

CH2

CH2

CH2

EtO

G

Amin bc 2

NH3, H2

Ni

HBr

IH

N

CHCH2

CH2

CH2

EtONH2

N

CHCH2

CH2

CH2

BrNH2

N

*

N

H

H

(+)-Nicotin

Al kyl ha

CH3I, OH-

N

*

N

H3C

H

2. 2LiNH

ClCHCHBrCH 222

Mg


64/65

H,CHOHCn 2110

(CH2)3C

H

C

n-C5H11

H

CH

OH

C10H21-nH2Lindlar

(CH2)3C

H

Cn-C5H11

H

CO

C10H21-nCrO3

(Z)-6-henicosen-11-on

Cu 6

1.

2.

Cu 6

Melibioz

Raffinoz

O

HO

OH

O

HO

CH2OH

OHO

OH

OHO

CH2

CH2OH

HOH2C O

H

HO

HO

Galactoz

Saccaroz

FructozGlucoz


65/65

Melibioz

6-O-(-D-Galactopyranosyl)-D-glucopyra noz(-anomer)

O

OH

OH

O

HO

CH2OH

OH O

OHHO

CH2

HO

N2H4

Gly-Ala

+

Phtalhydrazua

Kali phtalimit

+H3NCH2C NHCHCOO

-

CH3

ONH

NH

O

O

NCH2C

O

O

NHCHCOO-

O

CH3

NCH2COCl

O

O

Ala SOCl2NCH2COOH

O

O

H+

H2O

NCH2COOEt

O

O

BrCH2COOEtNK

O

O

Tuyen Tap de Thi Hsg Hoa Hoc

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