icho39 dalythuyet chinh thuc

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HA HC: NGH THUT, KHOA HC V NHNG BT NG TH V

BI THI L THUYT

P N V THANG IM

NGY 20 THNG 7, 2007

MATXCVA, NGAOfficial English version


2/26

2

Bi1. Hiu ng ng hm i vi proton

1.1.1 Cu trc ca propandial v hai ng phn ca n

O=CHCH2CH=O 1 markH

O O

C C

H C H

H

OH

C HH C

C

1 mark

H O 1 markTi a 3 marks

1.1.2Nguyn thydro axit ca nhm CH2(dng enol th nguyn t hydro axit ca nhm OH).

1 markTnh axit ca nhm 2 c gy nn bi sn nh ca cacbanion lin hp vi hai nhm cacbonylCu tr li th nht l cu ng.

2 marksTi a 3 marks

1.2.1 Khong cch gia hai cc tiu trn ng cong nng lng l 0,06nm. l dng andehitH O

H O

khong cch gia hai proton dng ny l khng kh thi. Hiu ng ng hm ch xy ra duy nht

dng enol Z:H H

O O O O

C CH C H

H

C CH C H

H 1 mark cho mi cu trcTi a 2 marks

1.3.1Biu thc v th cho mt xc sut

[ ] )()()()()(21)0,()(222222 xxxxxxa LRLRL =++= 1 mark

Kh nng tm thy proton tp trung ch yu ging bn tri


3/26

3

2

L

-0,06 -0,04 -0,02 0,00 0,02 0,04 0,06

L Khong cchnm R

0.5 marks(b) Vo khong gia thi gian ta c

1 mark

th mt xc sut c dng i xng, proton dao ng gia hai ging:

(2+

2)/2

L R

-0,06 -0,04 -0,02 0,00 0,02 0,04 0,06

L Khong cch,nmR

0.5 marks

Xc sut tm thy proton tp trung ging bn phi:

2

R

-0,06 -0,04 -0,02 0,00 0,02 0,04 0,06

L Khong cch,nm R

0.5 marksTi a 4.5 marks

1.3.2 Xc sut tm thy proton ging bn tri l 1/2, do hm mt xc sut c dng i xng vc hai ging u ng nht.

2 marks

[ ])()(2

1

2, 222 xxx RL +=

[ ] )()()()()(2

1,)( 222222 xxxxxxc

RRLRL

=++=

1 mark


4/26

4

=

1.3.3 Thi gian di chuyn t mt ging sang mt ging khc l t= / .

t=3.14

4.851012 s. 2 marks6.481011

Vn tc proton:0.06 109

V= =124.85 1012

m/s. 2 marks

Ti a 4 marks

1.3.4 btnh vv tr ca proton xp x bng mt na khong cch gia hai cc tiu l 0.03 nm (cngchp nhn gi tr 0.06).

1 mark btnh cc tiu vvn tc c th nhn c t nguyn l bt nh:

So snh bt nh vn tc vi p s cu trn l 12m/s ta thy rng nh xc nh vn tc proton

trong qu trnh di chuyn gia cc ging th l khng kh thi. Nh vy ng hm proton l mt hintng thun ty lng t v khng thc gii thch bng l thuyt kinh in. Cu tr li thhai l chnh xc

2 marksTi a 6 marks

smxm

V /100010.03,0.

10.02,6

001,0.2

10.055,1

2 923

34

=

=

h

3 marks


5/26

5

a a a a

Bi2. Ha hc nano

2.1.1Nng lng Gibbs v hng scn bng ca phn ng (1)rG

o500 =G

of, 500(CoO,r) -G

of, 500(H2O,k)= -198,4 + 219,4 = 20,7kJ/mol

0 0 0

0

0.5 marksrG500 (1)

20700

K=e RT= e 8.3=14500 6.88103 0.5 marks1 mark maximum

2.1.2Nng lng Gibbs cho phn ng (1) vi tiu phn coban nano hnh cu c bn knh ra l

i vi cc tiu phn hnh cu c ra= 108, 109m tngng ta c

2Co-gasV(Co)

ra

o

= 210 v 2100 J/mol.

rG500(1, ra)bng 20.5 (a), v 18.6 (b) kJ/mol.Hng s cn bng c tnh t phng trnh sau

K(1,r)=7.22103; r =108m K(1,r)=11.4103; r =109m

Ti a 2 marks

rGo

500 (1, ra) = Go

khi, 500(CoO,r)+Go

500(H2,k) - Go

500(H2O,k) - Go

cu(Co) =

= Go 500(CoO,r)+Go

500(H2,k) - Go500(H2O,k) - =+

a

kCoo

r

CoVrCoG

)(2),(500

=rGo500 -a

kCo

rCoV )(2

molmM

CoVCo

Co /10.6,69,8

10.0,59)( 36

6

===

=RT

rGrK a

or

o),1(exp),1( 500


6/26

6

2.2.1Nng lng Gibbs tiu chun cho phn ng (1) bao gm c cc tiu phn nano ca coban l

rGo500(1)=20,7 kJ/mol.Vi cc tiu phn coban nano hnh cu c ra=2 nm,

Coban oxit rn c thc t hnh thnh khi nng lng Gibbs ca phn ng (1) m. Bt ng thc

i vi coban khi l

2r r r

v i vi coban nano hnh cu c ra= 2 nm:

r

o

rG500(1) = 20.7 kJ/mol. Vi tiu phn coban hnh cu c ra = 1 nm rGo500(1, r) J= 18.6 kJ/mol.

T l b nht cap(H2O)

p(H2)tng ng l 145.6 (a) v 87.7 (b).

p sut hydro l

p sut nh nht ca nc l1bar0.0015 = 1.5103bar

1.5103145.6 = 0.218 bar (a) v 1.510387.7 = 0.132 bar (b), ng vi coban khi v nano

H2O%(khi Co) = 21.8% H2O%(tiu phn nano c ra= 1*10-9m) = 13.2%.

Chng ta bit rng s c s hnh thnh coban oxitTi a 4 marks

2.2.2

i vi phn ng oxy ha t phtr a r

v

( 2

r

Pha bn tri bt ng thc s cng dng nu ra cng tng. mt thi im nht nh th btng thc si du nn phn ng oxy ha t pht s khng xy ra. Nh vy bo v tiu phncoban khi s oxy ha t gic th ch cn tng dn bn knh ra. Cu tr li (a) ng.

Ti a 2 marks

2.3.1Phng trnh thng p mol Gibbs ph thuc vo cc i lng ca CoO (lp ngoi)

1 mark

Gro

500(1,ra)=19,6kJ/mol

)(2

)1(),1( 500500 CoVr

GrGo

kCoo

ra

o

r

=

( )

+=

)(ln)1()1(

2

2500

OHp

HpRTGG orr 0(

)(ln)1(

2

2500

OHp

HpRTG or

( )

+=

)(ln),1(),1(

2

2500

OHp

HpRTrGrG a

o

rar 0(

)(ln)(

2)1(

2

2500

OHp

HpRTCoV

rG

a

kCoo

r

0)(

)(ln)(

2)1(),1(

2

2500

=

Hp

OHpRTCoV

rGrG

a

KCoo

rar

)(

)(ln)(

2)1(

2

2500

Hp

OHpRTCoV

rG

a

KCoo

r

Gocu(CoO,rb) = Gkhi(CoO)+ )(2

),()(2

CoOVr

rCoOGCoOVr b

kCoOo

b

kCoO +=


7/26

7

2.3.2Phng trnh thng p mol Gibbs ph thuc vo cc i lng ca Co (lp trong):

Phng trnh trong du ngoc kp cho bit p sut ni lp trong (xem hng dn).Ti a 5 marks

2.3.3Nng lng Gibbs tiu chun cho phn ng (1) vi tiu phn nano hai lp l

,

Ti a 2 marks

2.3.4.Di nhng iu kin ny ta c:

o o

Biu thc trong du ngoc n v phi dng

Ti a 3 marks

2.3.5.

Chiu t pht ca phn ng (1) xy ra khi rG(1,r0) 0, v

+i lng trong du ngoc n v tri l dng. V tri ca bt ng thc cng dng hn khi rogim xung. mt thi im nht nh th bt ng thc si du v phn ng oxy ha t pht skhng xy ra. bo v tiu phn nano khi vic b oxy ha trong trng hp ny th cn phi gim r0.p n ng l (b).

Ti a 2 marks

Gocu(Co,ra,rb) =Gkhi(Co)+V(Co)

+

a

CoCoO

b

kCoO

rr

22

= )(),( CoVrCoG o +

+

a

CoCoO

b

kCoO

rr

22

)(2

))()((2

)1(

)(2)(2

),(),(),(),(

),,(),(),(),(),,1(

22

22

CoVr

CoVCoOVr

G

rr

CoVCoOV

r

rCoGkOHGkHGrCoOG

rrCoGkOHGkHGrCoOGrrG

a

CoCoO

b

kCoOo

r

a

CoCoO

b

kCoO

b

kCoO

oooo

ba

o

cau

oo

b

o

cauba

o

r

+=

+

++=

+=

( )

+=

+==

)(2

3)(

2)1(

)(2

)()(2

)1(),1(),,1(

CoVCoOVr

G

CoVr

CoVCoOVr

GrGrrG

o

kCoOo

r

a

CoCoO

b

kCoOo

ro

o

rba

o

r

3610.56,6)(2

3)( mCoVCoOV =

rGo(1,ro) t l vi (1/ro). th (a) ng

2

2ln)(23)(2)1(

H

OH

o

kCoOo

rp

pRTCoVCoOV

rG

+


8/26

8

Bi3. Cc phn ng ha hc khng bn vng

3.1.1 Phn ng chungB + D P 1 mark

Phng trnh ng hc ca X

Ti a 2 marks

3.1.2 p dng nguyn l nng dng ta c

,

vi

[X] =k2[D]k1[B]

d[P] k2[D]2= 2 3 marks

dt k1[B]

Bc phn ng bng 2 ng vi cht D, 1 ng vi cht B; bc chung l 10.5 marks cho mi bc phn ngngTi a 4.5 marks

3.2.1 Trong h mth tc u ca phn ng l:

1) Nu [X]0> k2/k1, th d[X]/dt> 0 bt k thi im no v tc ca X tng tng ln:

[X]

t

2 marks

2) Nu [X]0< k2/k1, th d[X]/dt< 0 bt k thi im no, v nng ca cht X tnggim xung:

[ ][ ][ ] [ ][ ]XDkXBk

dt

Xd2

21 = 1 mark

[ ][ ][ ] [ ][ ]XDkXBk

dt

Pd2

21 ==

[ ][ ][ ] [ ] )( 21 kXkXB

dt

Xd=


9/26

9

[X]

t

2 marks

Ti a 4 marks

3.2.2 Trong mt h kn th tc u ca phn ng l:

Nh vy thi im u ca phn ng th [X] tng ln nhng n khng th tng mi v stn mt gi tr hng nh bi v phn ng th hai l bt thun nghch:

[X]

t

2 marks cho im cc i

1 mark cho ng tim cn

Ti a 3 marks

3.3.1 X C2H6O2, Y C2H4O, P C2H6O. Du chm ch O2v H2O.

C2H6+C2H6O2+O2 2C2H6O2C2H6O2+C2H4O 2C2H4O +H2OC2H6+C2H4O +H2O 2C2H6O

[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 0)( 212210

>===

kXkXBXDkXBkdtXd ooooooo

t


10/26

10

0.5 marks cho mi cht cha bit (X, Y, P, ba trng)

Ti a 3 marks

3.4.1 nhit cao nht c th xc nh c th tc phn ng bng nhau:

1 mark cho tnh ton2 marks cho p sng

Ti a 3 marks

K

A

AR

EET

RTEA

RTEA

AA

AA

354ln

expexp

1

2

1,2,

2,2

1,1

=

=

=


11/26

11

Bi4. Xc nh nc bng phng php chun Fischer

4.1.Phng trnh phn ng

I2+ SO2 + 2 H2O + 4 Py = 2 PyHI + (PyH)2SO4 1 mark(0.75 marks nu b qua s hnh thnh mui ca Py )

4.2.1. T bng vi:M(Na2C4H4O6.2H2O) = 230.05 2M(H2O) = 36.04

m(H2O) = 1.3520 36.04 / 230.05 = 0.2118 g = 211.8 mg

1 mark cho cng thc

T= 211.8 / 12.20 = 17.36 mg/mL

T= 17.36 mg/mL

kt qa (khng qu hai n v sau du phy)

0.25 marks cho p sng

Ti a 1.25 marks

4.2.2. T bng vi:

Tnh ton:

Th tch iot cn cho 10 mL CH3OH tinh khit = 2.2010.00 / 25.00 = 0.88mL

0.5 marks cho cng thc chun ng lng metanol tinh khit

T= 21.5370.01103 / (22.70 0.88) = 9.87 mg/mL

Chnh xc hn

10.00 mL dung dch cha (1000-21.5)10.00 / 1000 = 9.785 mL of metanol

Th tch iot cn cho 9,785 mL CH3OH tinh khit l = 2.209,785 / 25.00 =0.86 mL

T = 21.5370.01103 / (22.70 0.86) = 9.86 mg/mL

1 mark cho s chun nc, ch 0.5 marks n u t r i cho i lng0.88

T= 9.87 mg/mL 0.25 marks cho kt qa ng

Ti a 1.75 marks


12/26

12

4.2.3. T bng vi:

Tnh ton:

Cch 1.Cho rng 1 mL CH3OH chax mg H2O, vy 1 mL A cha

((1.000 0.006)x + 5.624) mg H2O.15.00T= 22.45(0.994x + 5.624) 1sttitration,10.00T= 25.00x + 10.79(0.994x + 5.624) 2ndtitration.

Vy,x = 1.13 mg/mL, T= 10.09 mg/mL (10.10 nu b qua i lng 0.994)

Cch 2.Cho rng y mL B c dng chun nc cha trong 1 mL of CH3OH.

Then T=

ln 2).

22.455.62415.00 22.450.994y

(chun ln1) = 10.795.62410.00 25.00y 10.79y

(chun

Vy,y = 0.1116 v T= 10.10 mg/mL

T= 10.09 mg/mL (10.10 nu b qua i lng 0.994)

2 marks cho cng thc ng (b hay khng b qua i lng 0.994 factor)v 0.25 marks cho kt qa ng (10.10 hay 10.09)

Ti a 2.25 marks

4.3.Phng trnh phn ng

CaO + SO2= CaSO32CaO + 2I2= CaI2+ Ca(OI)26CaO + 6I2= 5CaI2+ Ca(IO3)2

(Thay cho CaO, c th vit Ca(OH)2.)

1 mark cho BT K phn ng ng

Ti a 1 mark

4.4.1Phng trnh phn ng

Fe2(SO4)3+ 2HI = 2FeSO4+ I2+ H2SO4 1 mark

Fe2(SO4)3+ H2O + SO2+ CH3OH = 2FeSO4+ CH3OHSO3+ H2SO41 mark

(hay dng ionTi a 2 marks


13/26

13

4.4.2.Phng trnh phn ng

Fe2(SO4)3xH2O + (x 1)I2 +xSO2 + xCH3OH = 2FeSO4 + xCH3OHSO3 +H2SO4+ 2(x 1)HI 1 mark

4.4.3. Thnh phn ca hydrat tinh thl:Tnh ton:

M(Fe2(SO4)3xH2O) = 399.9 + 18.02x

m (g) =0.638718.02x

; 1 markH2O(399.9+18.02x)

mH2O(g) = 10.59(mL)15.46(mg/mL)0.001(g/mg)x

x1

1 mark

0.1637(399.9 + 18.02x) = 11.51x 11.51;

x = 8.994Cng thc: Fe2(SO4)3

.9 H2O x =

0.25 marks (cho p sng)

Ti a2.25 marks


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14

Bi5. Mt hn hp bn (tr chi trn tm trong ha hu c)

5.1.1 Cu trc ca sn phmDO

H3C O CH3

Etyl axetat, etyl etanoatBt k mt cu trc no khc hay vit dng bn khai trin (CH3COOC2H5) k c cc k hiucho cc gc t do hu cnh (Me, Et, Ac), hay vit tn IUPAC u c 5 marks

5.1.2Hp chtD thuc loi nhm chc no ? Tick vo ng.Lu ! Chchp nhn mt cu tr li ng. Nhiu cu tr li sdn n im 0 cho cu hi ny.

xeton ete axetal este ancol andehit glycol

Cu tr li ng duy nht l este 5 marks5.1.3Hiu sut to thnhD

Tnh ton:Cho rng phn ng t n cn bng m khng chng minh 1 markTr li c cu hi hiu sut thp hn 85% - 2 marks

Hiu sut n nh c tht c khi phn ng t n cn bng, v hng s cn bng lc nyc xem nh khng bnh hng bi nhit v bi thnh phn hn hp phn ng.

K=[AcOEt][H2O] =

(0.85)2=

[AcOH][EtOH] 0.15 1.15

Tnh ton hiu sut ng vi t l hn hp 1:1 cho kt qa 67%

Hiu sut = 67%Ti a 10 marks, nu hiu sut nm trong khong 671%5.2.1 Cu trc caA,B, v C.

OEt

OEt

OEt

CH3C(OEt)3

trietyl orthoaxetat, 1,1,1-trietoxyetan

OEt

HCCOEt

etoxyaxetylen, etynyletylete

COOEt

COOEt

CH2(COOEt)2

dietyl malonat

A B C

Mi cht u c cng thc cu to v cc k hiu, ng thng r rng c 10 marksTn h thng ng vi mi cu trc c 5 marks5.2.2 Vvo trng cc hp cht trung gian c hnh thnh trong qu trnh axit ha C, v baz

phn chtB.a) Axit malonic l hp cht trung gian khi thu phn dietyl malonat 5 marks

4,2


15/26

15

O O

COOEt H+/H2OCOOH t

COOEt COOH CO2

C

Nu trong trng l monoetyl malonat 2 marksTi a 5 marks

b) Thu phn etoxyaxetylen bt u bng bc cng ion hydroxi vo ni i cho dng enol kmbn vng ca etylaxetat, sau n ngay lp tc b chuyn ha

OEt OH/H2OOH O OH

/H2OCH3COO

+ C2H5OHOEt

B

OEt

Ch ra bt k dng xeto v enol ca etylaxetat 5 marksThy phn lin kt ete bn vng s cho ra hydroxyaxetilen, hay nhng cht no khc ng viqu trnh ny (xeten hay dixeten) l khng kh thi v khng c chp nhn - 0 marksTi a cho cu a) v b) 10 marks

5.3.1 Cu trc ca axit senexioicT duy nht axeton th qu trnh tng hp s qua giai on ngng t andol, dehydrat ha v phn ngiodofom3 marks

OH+

OH+ I2/OH

HO -H2O OH

Ch nu cu trc axit senexioic 4 marks, km theo s - Ti a 10 marks

5.3.2 Cu trc caE.Iodofom, triiodometan, CHI3 5 marks

CH3COOH + C2H5OH

2


16/26

16

Bi6.Silicat l mt thnh phn cbn trong v Tri t

6.1.1SiO3

2-+ 2CO2+ 2H2O = H2SiO3 (Gel silica axit) +2HCO3-hay

SiO2(OH)22-+2CO2+H2O=H2SiO3+2HCO3

-haySiO3

2-+ CO2 + H2O = H2SiO3 + CO32-

Ti a 3 marks2 marks nu vit axit silixic dng khc1 mark nu ch vit cacbonat hay bicacbonat m khng c axit silixic

6.1.2) proton ha ion orthosilicat dn n s hnh thnh nhm Si-OHSiO44-+ H2O = [SiO3(OH)]3-+ OH- haySiO4

4-+ H+= [SiO3(OH)]3- hay

[SiO2(OH)2]2-+ H+= [SiO(OH)3]

- Yes Nob) Hnh thnh anion hydrat ha [SiO4(H2O)2]

4-

SiO44-+ 2H2O = [SiO4(H2O)2]

4-

Yes No

c) Ngng t nhiu ion ortho-silicat dn n s hnh thnh lin kt Si-O-Si

2 SiO44-+ H2O = [O3Si-O-SiO3]6-+2 OH- hay2 SiO4

4-+ 2H+= [O3Si-O-SiO3]6-+ H2O hay

2SiO2(OH)22-+ 2O = [O-Si(OH)2-O-Si(OH)2-O]

2-+ 2 OH- Yes NoTi a 9 marks2 marks cho mi phn ng ng1 mark khi nh tick vo ng

6.2.1 n= 6 (cho rng s oxy ha ca silic l (+4) v oxy l (-2), hay lu n cu trc v intch ca ion orthosilicat l (-4))Ti a 2 marksTri 1 mark nu tnh nhm

6.2.2 Si3O9 3 [SiO4] 3 O, do c 3 nguyn t oxy chung cho cc t din k nhauTi a 2 marksTri 1 marks nu tnh nhm

6.2.3

Ti a 3 marks

6.2.4Tnh ton:m=4 (cho rng s oxy ha ca silic l (+4) v oxy l (-2), hay lu n cu trc v in tch ca ionorthosilicat l (-4))Si4O10 4[SiO4] 6O, do cu trc ca t din lc ny l SiO2.5, ph hp kh i 1 nguyn t Othuc v t din ny v ba nguyn t oxy cn l i chung g ia ha i t din (h sng gp l =3/2). iu ny l kh thi nu t din c coi nh phng v lin kt vinhau thng qua cc nh chung ca t din


17/26

17

Ti a 10 marks2 marks cho xc nh in tch ng3 marks cho xc nh ng cc nguyn t oxy to cu5 marks for the correct structureTr 1 mark nu ch v t 6 n 15 t din m vn th hin ng cc lin ktTr 3 marks nu ch v t hn 6 t din (do khng thy r cc lp a din)Tr 4 marks nu v lin kt qua cc nh chung m dng 3DTr 4 marks nu lin kt vi nhau qua cc nh chung nhng trn 1 ng thng (1D)0 mark nu v mt cu trc bt k no khc

6.3.1 = 4

Ti a 5 marksTr 1 mark nu tnh nhmTr 2 marks nu vit nhm biu thc lin h tnh [H+] qua Ka

I

Tr 2 marks nu c sai st trong nh ngha tnh pH (v d dng ln thay cho lg)Tr 3 marks nu vit sai phn ng thy phn

6.3.2uSO4+ Na2SiO3+ 2H2O = Cu(OH)2 + H2SiO3 + Na2SO4hay 2uSO4+ Na2SiO3+ 2H2O = (CuOH)2SO4 + H2SiO3 + Na2SO4Cc phn ng ny (xt ring t s hnh thnh ng silicat) c thc suy lun t kin cho rng

phn ng ny m t s thy phn ln nhau (t khuch i). N n t phn trc ca bi tp: pH ca fLGL s ln hn 7 (xem cu 6.2), v pH ca dung dch ng sunfat s b hn 7 (xem 6.3.1).

Ti a 3 marks2 marks nu h s cn bng sai1 mark nu ch cho thy mt trong hai kt ta (Cu(OH)2 hay H2SiO3)


18/26

18

E1

Bi7. Chng tch mlm dy thnh ng mch v cc hp cht trung

gian trong qa trnh sinh tng hp cholesterol

7.1.1 2-4 xc tc cho mt loi phn ng (hoc cc phn ng tng t).Phn ng duynht xut hin ba ln trong mt dy l phn ng monophotphoryl ha (tt c cc loi

phn ng cn li u khng ng vi cht ban u hay sn phm cui cng). iu nycng c xc nhn bi s c mt ca on mch pyrophotphat trong IPP v s gii phng ccsn phm v c(bao gm cc photphat v c) trong qu trnh t phn hu ca 1.

l mt axit monocacboxylic c to thnh t ba nguyn t: cacbon, hydro v oxy. Nkhng cha lu hunh c tm thy trong CoA hay photpho c tm thy trong cc hpcht trung gian ca qu trnh bin i t HMG-CoA thnh IPP hay hin din trong CoA. Nhvy, 1 xc tc cho phn ng loi CoA t HMG-CoA khng qua thu phn. Do nc khngtham gia phn ng nn CoA gii phng buc phi tham gia vo mt phn ng khc lin quan nnhm cacboxyl c este ha trong HMG-CoA. Qua trnh duy nht kh thi y chnh l qutrnh kh 4 electron to thnh nhm hydroxyl. 1 khng th xc tc cho phn ng dehydrat hado c hot tnh quang hc (s loi nc s lm mt i trung tm bt i duy nht). Phn ngdecacboxyl ha cng b loi tr do l mt axit nn phi tn ti nhm cacboxyk trong phn t. Oxy

ha nhm hydroxyl bc ba trong HMG-CoA s lm cho cch oxy ha - trnn khng th thc hinc. Mt iu hiu nhin na l nhm cacboxyl bao gm c s hnh thnh lin kt thioeste hin dintrong on mch cha nhm hydroxyl trong IPP. Nh vy:

1 4, 5

3 6

Ti a 12 marksE1: 9 marks nu c hai kt qa u ng.4 marks cha ra mt kt qa ng4 marks nu a ra hai kt qa ng v mt kt qa sai

0 mark nu a ra mt kt qa ng cn cc kt qa cn li u sai0 mark nu a ra nhiu hn ba kt qa.3: 3 marks nu a ra duy nht kt qa ng. Tt c cc trng hp khc khng cho im no

7.1.2 Da vo cc phn ng c xc tc bi 1 v cu hnh ca trung tm bt i trong HMG-CoAth cu trc ca cht l:

HOOC

HO

(R) OH

, axit mevalonic

Lu rng cu hnh tuyt i trung tm bt i thay i do c s thay i v hn cp cc nhmth trong qu trnh trao i cht t HMG-CoA thnh axit mevalonic.

Ti a 12 marks

HOOCHO

(S)

CoAS

OHOOC

HO

(R) OH

8 marks cho cng thc ng4 marks cho lp thng (ch cho im khi cng thc ng v xc nh cu hnh tuyt i trung tm

bt i l R, tt c nhng trng hp khc khng cho im)Khng trim no nu lp th sai hay khng xc nh cu hnh tuyt i


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7.2.1 Phng trnh phn ng ozon phn

Ti a 5 marks

O

- -

4.5 marks cho sn phm ng (1.5 cho mi sn phm), sai sn phm khng b trim0.5 mark cho h s t lng ng

7.2.2 Phn t DAP ch cha mt nguyn t cacbon c tham gia vo s to thnh lin kt qua qu trnh sinh tng hp cht Y. Bt chp cch cc phn t t hp thnh Y nh th no th vic ozon

phn mnh ny u cho sn phm l dimetyl xeton (axeton) (Xem phn ng ozon phn DAP cu7.2.1). Nh vy axeton chnh l Y1, do n cha ba nguyn t cacbon (Y2 v Y3 cha ln lt 5 v4 nguyn t cacbon). Ch n t l cc sn phm ozon phn th ta s c c biu thc tnh s cacbontrong Y nh sau:

nY(C)= 2*nY1(C)+4* nY2(C)+ nY3(C)= 2*3 + 4*5 + 4 = 30Y l phn t mch h, nh vy on mch DAP ch c th c t m thy m t u.Y ch c hai u (cn ti thiu ba u to thnh mt phn t c mch nhnh). Do vic ozon phn

mt phn t Y to thnh hai phn t axeton nn Yphi cha 30 nguyn t cacbon xc nh s nguyn t hydro th phi xc nh s lin kt i trong Y. S hnh thnh m il i n k t i s lm m t 2 nguyn t hydro t rong sn phm ghp mch so vi tngs cc nguyn t c t rong cc ch t ban u. T l ca Y trong sn phm ca phn ngozon phn l 1:7 (2+4+1) ng vi 6 lin kt i trong cht Y. Nh vy khi s dng cng thc chungcho ankan chng ta c:

n(H)= 2*nY(C)+22*nc=c=30*2 + 2 - 6*2 = 50Cng thc phn t cht Y (squalen) 3050.

S nguyn t cacbon 30Tnh ton:nY(C)= 2*nY1(C)+4*nY2(C)+ nY3(C)= 2*3 + 4*5 + 4 = 30

S nguyn t hydro 50Tnh ton:

n(H)= 2*nY(C)+22*nc=c=30*2 + 2 - 6*2 = 50Cng thc phn tY 3050Ti a 12 marks8 marks cho vic xc nh ng cng thc phn t4 marks cho vic kt lun ng cng thc phn t

7.2.3 IPP v DAP l cc ng phn cha 5 nguyn t cacbon. Do tt c cc nguyn t cacbonca hai cht ny u c trong cht Y nn c th tnh c s lng phn t IPP v DAP cn thit tng hp cht Y:

n(IPP&DAP)= nY(C)/5=30/5=6S phn t DAP cn thit l 2 v c xc nh t trc. Nh vy cn c thm 4 phnt IP P

n(IPP&DAP)= nY(C)/5=30/5=6

S phn t DAP cn thit 2 S phn t IPP cn thit 4Ti a 7 marks3.5 marks cho tnh ton ng tng s phn t DAP v IPP3.5 marks cho vic tnh ton s lng mi phn t DAP v IPP


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7.2.4 Tt c cc hng kt hp m khng lm thay i b khung cacbon c ch ra di y(khng v cc n v pyrophotphat). Hai nhm sn phm sinh ra do s khc nhau ca ccnguyn t cacbon tham gia ghp mch c phn ra bng ng nt t. an mchIPP s ni vi on mch DAP sao cho vic ozon phn sn phm ny s to Y2 cha 5nguyn t cacbon. Ch duy nht mt ng phn ph hp nu khng v chi tit lp th, cn nu vchi tit lp th s c hai ng phn

No

No

* * *+ +

No No

Yes

(E)O

O-P O-

OO P

OO-

hay

-OO

O-

(Z)

P PO

OOO-

Ti a 8.5 marksng phn pha trn l geranyl pyrophotphat

8.5 marks cho cng thc ngKhng trim no v mt ha lp th, bt k cng thc chnh xc no cng c chp nhn2.5 marks nu ozon phn sn phm sinh ra axeton nhng khng to c sn phm cha5 nguyn t cacbon2.5 marks nuozon phn sn phm to thnh hp cht cha 5 nguyn t cacbon nhng khng sinh raaxeton0 mark nu thuc v cc trng hp khc

7.2.5 T phn ng ghp mch hnh 2 th ta thy rng Y4 cha 15 nguyn t cacbon hay cha 1on mch DAP v hai on mch IPP, cc phn c gn vo lin tip vi nhau. Mt lu quan trng lY3 khng uc tm thy trong hai on mch hydrocacbon sinh ra tY4, do Y3 l ktqa ca phn ng ozon phn Y vi t l 1 : 1. Nh vy geranyl photphat l hp cht trunggian trong qu trnh tng hp Y (tt c cc lin kt i u c cu hnh trans). Gn mt on mchIPP tip theo vo geranyl photphat dn n s to thnh sn phm m khi ozon phn s cho 1 phntY1 v 2 phn tY2. Nh vy cu trc ca Y4 vi cc chi tit lp th nh sau:

(E) (E)O

O-P O-O

O P

OO-

Y4, farnesyl pyrophotphat


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Kt hp hai on mch hydrocacbon ca Y4 v lu rng ni i gia chng s mt i th ta thu ccng thc ca cht Y:

(E) (E)(E) (E)

Y, squalenTi a 16 marks

9 marks cho farnesyl pyrophotphat (6.5 marks cho cng thc ng v 2.5 marks cho lp thng)7 marks cho squalen (5 marks cho cng thc ng v 2 marks cho lp thng)Tr 2.5 marks nu khng loi b ni i trong squalen


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Bi 8. Phng php ATRP cho php tng hp cc polymermi

8.1.1 Phng trnh tc cho cc bc phn ng scp ca ATRPs: hot ha (vact), phn hot ha(vdeact), pht trin mch (vp) v tt mch (vt)

vact= kact[R-Hal][CuHal(Ligand)k] 2 marksvdeact= kdeact[R

][CuHal2(Ligand)k] 2 marksvp= kp[R][M] 2 marksvt= 2kt[R

]2 2 marks (khng tr nu thiu s 2)8 marks maximum

8.1.2 So snh tc cc bc phn ng scp ca ATRPDo tt c cc mch u pht trin vi tc bng nhau nn qu trnh vn polymer ha vnlun tip din. Cc gc t do polymer ha ch tn ti nu nng ca cc gc t do hot haqu b ngn cn bc chuyn mch v tt mch. Nh vy:

vdeact>> vactS phn ct cc gc t do phi xy ra vi tc rt nh nn cn bng dch chuyn v pha cc tiu

phn khng hot ng

vdeact >> vpBc pht trin mch phi lun chm hn bc phn hot ha to s pht trin mch vi tc bng nhau.

vdeact>> vtBc tt mch khng xy ra khi s phn t polymer khng t n mt s lng mch nht nh - bngs phn t tham gia giai on khi mo.

vdeact>> vact 3 marksvdeact>> vp 3 marksvdeact>> vt 3 marks

Ti a 9 marks

8.2.1 Tnh khi lngs (m) ca polymer nhn c.Cch th nht[M] = [M]0 exp(kP[R]t) hayn(MMA) =n0 (MMA) exp(kP[R]t)

Lng monomer MMA cn li sau 1400s l1 mark

31.0 exp(1616 1.76 107 1400) = 20.8 mmol. 2 marksLng monomer tiu th trong qu trnh polymer ha: 31-20.8=10.2 mmol 1 markLng polymer nhn c l m =n(MMA) M(MMA) = (10.2 /1000) 100.1=1.03 g 1 markCch th hai[M] = [M]0 exp(kP[R]t) orn(MMA) =n0 (MMA) exp(kP[R]t)

Lng monomer MMA cn li sau 1400s l1 mark

n(MMA)=n0(MMA)(1exp(kp

3 marks

[R]t))=31.0(116161.76107 1400)=10.2 mmol

Lng polymer nhn c l m =n(MMA) M(MMA) = (10.2 /1000) 100.1=1.03 g 1 markCch th ba

[M][M]0

mark

=e0.398 = 0.672 1

[ ][ ]

398,01400.10.76,1.1616ln 7 ===

Rtk

M

MP

o

1 mark


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P

[M][M]0

=n(MMA)

n0 (MMA)

n(MMA) = 0.672 n0(MMA) = 20.8 mmol 1

markLng monomer tiu th trong qu trnh polymer ha 31-20.8=10.2 mmol 1 markLng polymer nhn c l m =n(MMA) M(MMA) = (10.2 /1000) 100.1=1.03 g 1 mark

m = 1.03 g Ti a 5 marks

8.2.2 Tnh polymer ha (DP) ca polymer nhn c.S lng mch ang pht trin bng s phn t TsCl (0.12 mmol) 2 marks bc th nht tiu th ht 7,3mmol MMA (0.73/100.1).Tng lng monomerthi im ban u bc th hai l 23.7 + 23.7 = 47.4 mmol.

2 marksDo cc monomer c hot tnh nh nhau nn n s tham gia polymer ha vi tc bng nhau.Lng monomer tiu th trong bc phn ng th hai ln=n0(1exp(k [R]t))=47.4(1exp(16161.7610

71295))=14.6 mmol. 4 marks

Tng cng 7.3+14.6=21.9 mmol monomerc polymer ha qua hai bc phn ng 2 marksDP=21.9/0.12=182.5 1 mark

DP = 182-183 (chp nhn tt c cc cu tr li trong khon ny) Ti a 11 marks

8.2.3 Cu trc ca polymer nhn c.Sn phm polymer ha l mt poylmer khi ng trng hp do n thu c bi s trng hp lin tiphai mch polymer khc nhau.Khi n v th nht ch gm cc n v MMA. polymer ha DP l 7.3/0.12=60.861 n vmonomer.Khi n v th hai nhn c bng sng trng hp hai monomer thnh phn vi tc bngnhau. Nh vy n l mt polymer vi cc monomerc sp xp ngu nhin. S lng on mchA v B khi polymer th hai bng nhau do nng ca chng trong hn h p phnng thi im ban u bc th hai bng nhau. polymer ha ca khi polymerth hai l 183-61 = 122 n v monomer (121 cng l kt qa ng nu s dng kt qa DP

cu 8.2.2 l 182).Ts-A61-block-(A-stat-B)61-Cl or Ts-A61-block-(A61-stat-B61)-ClTi a 14 marks4 marks cho khi polymerng trng hp vi khi A v ng khi AB4 marks cho vic ch ra cc n v monomerc sp xp ngu nhin khi polymer th hai1 mark cho vic xc nh tng cc on mch A v B khi polymer th hai bng nhau2 marks cho vic tnh chnh xc DP mi khi1 mark cho vic ch ra chnh xc cc nhm cui mch

8.3.1 X nh cc tn hiu 1H NMR ng vi cc cu trc nh cho trong phiu tr li* CH2 *

O CH2a, b, g 3x1.5 marks

*

H H

H H

H

c 2 marks

*

H H

H H

*

d 2 marks


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24

*

*

H Cl

e 4 marks

*

Cl

H H

f 4 marks

Ti a 16.5 marks

8.3.2Xc nh phn mol ca cc n v C v D v khi lng phn tca P1 v P2.Tn hiu ca vn a b v g l 40.2, nh vy cng cho 1 proton l 40.2/4/58=0.173 cho mi phca polymerng trng h p Cng ca vn a l 13.0, ng vi 13.0/0.173=75 proton. Lu rng mi vng styren c5 proton vng thm, polymer ha ca khi styren l 75/5=15. 2 marksPhn mol ca n v styren l P1 bng 15/(15+58) = 20.5% 1 markCng ca vn a d l 10.4, ng vi 10.4/0.173=60 proton. Do mi n v monomer ca

p-clometylstyren (PCS) c 4 proton, polymer ha ca PCS l 60/4=15. 2 marksPhn mol ca D l 15/(15+58) = 20.5% 1 markM(P1) = 15.03+58x44.05+72.06+15x104.15+35.45 = 4240 2 marks

M(P2) = 15.03+58 x44.05+72.06+15x152.62+35.45 = 4967 2 marksM(P1) = 4240 M(P2) = 4967n(C) = 20.5% n(D) = 20.5%Ti a 12 marks

8.3.3 Tt c cc kh nng khi mo c thxy ra trong qu trnh tng hp P1 v P2Ti a 10 marks

(+2)

y k hiu R s dng cho on mch khi mo kch thc ln vi s c mt ca mt hay vi on

mch styren.

P2: (1.5+2+3) marks


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2 k

2

+

k k

OH3C

O

CH3O58

Cl

Cu(+)Cl(Ligand)kO

H3C

O

O CH58

CH3

+ Cu(+2)Cl (Ligand)

R

Cl

+ Cu(+)

Cl(Ligand)

R

CH

+ Cu(+2)Cl (Ligand)

Cl Cl

R R

Cl Cl

y k hiu R s dng cho on mch khi mo kch thc ln vi s c mt ca mt hay vi phnt p-clometylstyren.

8.3.4 Cu trc ca P1 v mt trong snhng cu trc c thc ca P2P1 l mt polymer khi ca PEO v PS. Khi PS cha 15 n v .

P2 l mt polymer khi c hnh thnh t khi PEO v mch nhnh styren.Cng ca vn a f l 2.75, nn 2.75/0.173=15.9, ng vi 16 proton hay 8 nhm clometyl.d) Nu trong phn t P2 khng c nhnh th n s cha 15 nhm clometyl. Mi nhnh s lm gims lng nhm i 1. Vy P2 c 15-8 = 7 nhnh. C mi cu trc ca 7 nhnh ny s chnh xcnu nh mi n v monomer lin kt vi t hn ba n v monomer khc

1 RC C C

C CC C C C

C C C CC

C Cl

+Cu(+)Cl(Ligand)k + Cu(2+)Cl(Ligand)k

Cl CH2.


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D

2Cl

Cl

DCl

D ClCl

Cl D

D D ClD

ClR D

Cl

D

Cl

D D

Cl D Cl

DCl

D

D

ClCl

Cl

Ti a 13.5 marks2 marks cho P17.5 marks cho cu trc ng hon chnh ca P24 marks cho cu trc ca P2 vi s nhnh khc 0 (mc d khng ng vi p n)Tr 4 marks nu c mt n v lin kt vi nhiu hn 3 n v monomer khc

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