jjai 9. Ta c6: n^„^ = n^,^^,o_ = 0,065 (mol).
Goi cong thufc chung ciia A, B la C^H^ÔN^ - n 0,065
=> n = — — — = ^ ^ , = ^ , 0 0,25nx 0,25.0,1
Phai CO 1 chat la CH2NH2-COOH (n = 2)
Goi cong thufc cua chat con lai la CnHm(C00H)x(NH2)y
2CH2NH2COOH + H2SO4 — > [NH3CH2C00H]2S04
(mol) a 0,5a
2C„H„(C00H),(NH2)y + yH2S04 ^ [C„H„,(COOH)x(NH3)y]2 ( 8 0 4 ) ^
(mol) b ^
H2SO, + 2NaOH —> Na2S04 + H2O
(mol) 0,006 0,012 ^> 0,5a + ^ + 0,006 = 0,112 x 0,5 ^ a + by = 0,1 (1) Do n A = a + b = 0,1 (2) Tii (1) va (2) => y = 1 +)Khichc) ~ X + Ba(0H)2: 2
2CH2NH2COOH + Ba(0H)2 > [CH2NH2COO]2Bax + 2H2O
0,5a (mol) 0,25a (mol) 0,25a (mol)
2C„H,„(C00H)^ NH2 + xBă0H)2 ^ [CnH,„(COO)^NH2]2Ba + 2XH2O
0,5b (mol) 0,25bx (mol) 0,25b (mol)
n,,^^,„„^ = 0,25a + 0,25bx = 0,03 hay a + bx = 0,12 (3) - a + bx 0,12
r=> X = = = 1,2 a + b 0,1
Theo de bai: moi chat chufa khong qua 2 nhom -COOH ma theo tren 1
chat CO chufa nhom -COOH (x' = 1) nen chat con lai c6 2 nhom
-COOH (x = 2)
Thay x vao (3) roi giai he phtfong trinh (2), (3), ta diioc: a = 0,08 (mol) va b = 0,02 (mol)
lat khac: n,,„ = 0,25 x 0,08(1 + 1) + 0,25 x 0,02(n + 2) = 0,065 ^ n = 3 m . = 285 X 0,25 x 0,08 + [(12 x 3 + m + 4 4 x 2 + 1 6 ) x 2
+ 137 x 2] X 0,25 X 0,02 = 8,58 => m = 5
V a y c6ng thuTc c a u t a o d a n g m a c h k h o n g n h ^ n h c u a A , B Ik: (A) : H 2 C — C O O H NH2 (B) : H O O C — C H 2 — C H - C H 2 — C O O H NH2 H O O C — C H — C H 2 — C H 2 — C O O H NH2
T h ^ n h p h a n p h a n t r S m k h o i lifcfng : HIA = 75 x 0,08 = 6 (gam) 6 X 100%
% A = = 67,11%
6 + 147 X 0,02
%B = 100% - 67,11% = 32,89%
Bai 10. a) H a i hap chat A , B ke tiep nhau t r o n g day dong dSng n e n :
M B = M A + 14 va c6 cung so nguyen tii oxi ( v i chufa cung n h o m chuTc) Goi z só nguyen ttf oxi t r o n g A va B
%0 t r o n g A = % 0 t r o n g B = 16z X 100 M A 16z X 100 + 14 = 53,33 = 43,24 53,33 43,24 + 14 M A = 60
A c6 cong thufc chung la CxHyOz, 3 a n (x, y, z) 1 phLfcng t r i n h :
M A = 12x + y + 16z = 60 Ta cho z = 1,2.. t i n h x, y Ta cho z = 1,2.. t i n h x, y +) Vcfi z = 2 =i> 12x + y + 16 = 60 => 12x + y = 44 X 2 3 4 y 2 0 8 a m Vay A la CgHgO va B Ik C4H10Ọ
iJng vdi 2 cong thufc phan tuf nay, ta c6 the c6 rifgu hoac ete nọ
RMu: C H 3 - C H 2 - C H 2 O H hoac C H g - C H O H - C H . s (cho A )
C H 3 - C H 2 - C H 2 - C H 2 O H hoac H 3 C — C H — C H 2 — C H 3 OH C H 3 H3C—C—CH3 (choB) OH 4 BOI OUflNG H O A H O C 12 Ete: C H 3- O- C H 2 - C H 3 (cho A ) C H 3 - O - C H 2 - C H 2 - C H 3 ; H 3 C — O — C M — C H 3 C H 3 C 2 H 5 - O - C 2 H 5 (cho B) +; Vdi z = 2 => 12x + y + 32 = 60 => 12x + y = 28 X 1 2 3 y 16 4 a m A la C2H4O2. B la C3H6O2.
V(Ji 1 n h o m chufc A , B la axit hay este
jixit: C H 3 C O O H (cho A ) , C H 3 - C H 2- C O O H (cho B)
Este: H C O O C H 3 (cho A ) , CH3COOCH3 hoac HCOOC2H5 (cho B) Vcfi 1 n h o m chufc A , B c6 tói da 2 nguyen tuf oxi
b ) 2 hgfp chat hufu ca tao r a td axit, tac d u n g diicfc vori N a O H , k h o n g tac dung vdri N a v a y 2 chat nay la este vay cong thufc cau tao la dung vdri N a v a y 2 chat nay la este vay cong thufc cau tao la
H C O O C H 3 CH3COOCH3 HCOOC2H5
Bai 11. A m i n o axit C tac dung vdri H2SO4 theo t i le m o l 2 C cho 1 m o l
H2SO4 vay C chi chufa 1 chufc a m i n . Gia suf C chufa n n h o m - C O O H C v d i H C l :
RN(COOH)„ + H C l > RNH(C00H)„C1
(mol) a a a Dung dich D t h u dugc chufa muoi RNH(COOH)nCl vdi so m o l
a = nnci = 0,5 x 0,05 = 0,025 (mol) So m o l N a O H dung de t r u n g hoa dung dich D
RNH(C00H)„C1 + (n + l ) N a O H > RN(COONa)n + N a C l + (n + 1)H20 nNaOH = (n + D a = (n + 1) x 0,025 = 2 x 0,05 = 0,1 (mol) 0,1 n + 1 = 0,025 Amino axit C chufa 3 chufc axit. Phan ufng dot chay C
= 4 r:i> n = 3 C,HyN(C00H)3 + X + y + 3 O 2 ' " I ' Q N G H O A H O C 12 '^-^ (x + 3)C02t + ^ ^ H 2 0 + ^ N 2 t 2 2 95
TM tich O 2 b^ng i the tich khong k h i n§n ta c6 5 V,^ ^V,,^ : V„ ^ „ = A M: 4 : 3 = 3 , 5 : 4 : 3 Vco, 2(x + 3) 4 o 3 , T T ^ = T - = 0 => y + 3 = - ( x + 3) V„^o y + 3 3 2 '^co, ^ X + 3 _ J _ 4 Thay y + 3 bang ^ ^ ± 3 ) 3 X + 3 4 8(x + 3) 4 (x + 3)3 3,5 8x + 3(x + 3) 3,5 8 7(x + 3) = l l x + 9 X = 3 2 ' 3
Vay cong thufc phan tuf cua C la C3H6N(COOH)3 C H 2 — C O O H hay N — C H 2 — C O O H
C H 2 — C O O H B a i 1 2 . a ) Phuang t r i n h hoa hoc: B a i 1 2 . a ) Phuang t r i n h hoa hoc:
( H 2 N ) , - C x H y-(C00H)^ + nHCl -> (CIH3N),-C^Hy-(COOH),„ (mol) 1 n (mol) 0,1 0 , l n 0,1 0 , l n = 0,1 => n = 1 Suy ra n: M(C I H3N , . , C . H, (C 0 0 H) „ , = ~ ~ = 183,5 (gam) M A = 183,5 - 35,5 - 1 = 147 (gam)
H2N-CxHy(C00H)m + mNaOH -> H g N- C J l / C O O N a ) ^ + mlịO
(gam) 22,05 28,65 (gam) 147 m
=> m = 191 (gam) => M m u o i chita Na - M A = 191 - 147 = 44 (gam).
9 6 B6I OI/SNG H6A HOC IZ
Cvt 1 nh6m - C O O H chuyen th^nh COONa t h i khoi lifgrng tSng: 67 - 45 = 22 (gam).
Suy ra trong phan tur A c6 2 nguyen tijf H diigc thay bang 2 nguyen tiJ Na: 44 = 2
Cong thufc phan tuf cua A: H2N-CxHy(COOH)2.
Mf, „ = 147 - 90 - 16 = 41 (g) => 12x + y = 41 Bang bien luan:
X 1 2 3
y 29 (loai) 17 (loai) 5 (hoTp l i ) Cong thufc phan tuf cua A: H2N-C3H5(COOH)2 Cong thufc phan tuf cua A: H2N-C3H5(COOH)2
b) HOOC-CH2-CH2-CH(NH2)-COOH '
Bai 1 3 . a ) Goi amino axit I la: CnH2„(NH2)(COOH) amino axit I I la: C„,H2,„(NH2)(COOH)
Gpi n la so nguyen ttf cacbon trung binh trong goc hidrocacbon, cong thufc tLfOng dirong hai amino axit la:
C-H2- ( N H 2 ) ( C 0 0 H ) v(Ji n < n < m
Cac phan ufng: , C - n 2 - ( N H 2 ) ( C O O H ) + H C 1 ^ C-H2-(NH3Cl)(COOH) (1)
Dung dich A g6m C - H g - ( N H 3 C l ) ( C 0 0 H ) va C - H 2 - ( N H 2 ) ( C 0 0 H ) diT
hoac HCl dụ
+) Tac dung vdri K O H :
C-H2~(NH3Cl)(COOH) + 2 K 0 H - » C - H ^ - ( N H 2 ) ( C 0 0 K ) + KCl + 2H2O (2)
CnH2H (NH2)(C00H)^^, + KOH ^ C-H2- ( N H 2 ) ( C 0 0 K ) + H2O (3)
HCldu + K O H ^ KCl + H2O (3') Phan ufng chay: C - H g - ( N H 2 ) ( C 0 0 H ) ^ " + 3 +
(n + l) C 0 ;
Oo
+ ^ ^ H 2 0 + i N 2 (4)
2 ^ 2
San pham chay vao dung dich NaOH chi c6 nxidc vk CO2 bi hap thu nen
khoi li/orng CO2 v a khoi lifcfng HgO la 32,8 (gam).
+) Neu HCl het:
TO (1) ^ n c „ _ H , „ , H ^ e , ) ( c o o H ) = " H ^ ' = ^,11 x 2 = 0,22 (mol)
Txi (2) => H K O H = 2 X 0,22 = 0,44 (mol)
H K O H da dung = 0,14 X 3 = 0,42 mol < 0,44 (v6 li)
Vay H C l dạ
Goi só mol h6n hop hai amino axit la a mol
TCf (1) => H p „ = a (mol); n n c i = a (mol) H H C I dir = 0,22 - a (mol)
Tii (2) va (3) => H K O H: 2a + 0,22 - a = 0,42 ^ a = 0,2 (mol) TCf (4) =i> iicog = (n + l ) -0,2 (mol)
" H ^ O = ^ ^ " ^ ^ ) ' ^ ' ^ ^ ^ ^ ^ ^
Tii mco2 + m H^ o = 32,8 ^ (n + l).0,2.44 + (2n + 3) .0,1.18 = 32,8
=> 8,8n + 8,8 + 3,6n + 5,4 = 32,8 =^ n = 1,5 Vi n < n n e n n < 1,5 => n = 1
Vay amino axit I la: CH2(NH2)(COOH) (M = 75)
Phan tuf lugng ciia amino axit I I : M2 = 1,37 x 75 103 dvC
CraH2n.(NH2)(COOH) CO M = 103 ^ 14m + 45 + 16 = 103 => m = 3
Cong thufc cua amino axit I I la C3H6(NH2)(COOH) b) Goi X , y la só mol cua amino axit I va I I : x + y = 0,2
Só nguyen tuf C trung binh trong hai goc hidrocacbon CH2 va C3H6 la
n = ^L±_?y = 1,5 o X + 3y = 1,5.0,2 = 0,3 X + y
Theo de bai, ta c6 he phuong trinh: •x + y = 0,2
X + 3y = 0,3
Giai he phtfang t r i n h , ta du-gc: x = 0,15 va y = 0,05 0 15
Vay: % n a„ n n o a x i t i = - 7 - 7 7 - X 100% = 75%; % n a„ n „ o axit 11 = 25%
0, 2
CHl/dNG IV.
POLIME VA VAT LIEU POLIME
Ạ KIEN THLfC CAN NHCJ
ị DAI ClfUNG VE POLIME
a) Phan vtng phan cdt mgch polime:
- Polime CO nhom chiíc trong mach de bi thuy phan Thi du: 4-CH2-CH--
n
3()0"(:
; i HjC^CH
b) Phan vfng giS nguyen mgch polime
-HjC—CH=:C—CHp- I + ntiCl CI I "H2C CH2 C CH2 CH,
Poliisopren poliisopren hidroclo hoa
c) Phan ihng tang mgch polime
Khi CO dieu kien thich hap (nhiet do, chat xiic tdc, ...) cac mgch polime CO the nói vdi nhau qua cau thanh mgch dai han hogc thanh mgng liidị
o n -CH,
CH,
-CH2-OH OH
^- VAr LIEU POLIME a) Chat dco:
+) Khai niem ve chat deo va vat lieu compozit
Chat deo Id nhilng vat lieu polime c6 tinh deọ
Vat lieu compozit la vat lieu hdn hap gom it nhdt hai thanh phdn phan tan vao nhau ma khong tan vao nhaụ
+} Mot so polime dung lam chat deo
> Polietilen (PE): (-CH2-CH,-),,
> Polifvinyl clorua) (PVC): (-CH2-CHCI-),,
CH,
1
> Poli(metyl metacrylat): HjC—C C O O C H 3 _
> Polyphenol - fomandehit) (PPF)
Poli(phenol-fomandehit) c6 ba dang: nhiia novolac, nhiia rezol va nhúa rezit.
b) Ta
+) Khdi niem: Ta la nhitng vat lieu polime hinh sai đi va inanh vdi do
ben nhdt dinh.
+) Mot so logi tcf tong hap thiT&ng gap
> Ta nilon - 6,6
' nH2N-[CH2]a-NH2 + nHOOC-[CH2]4-COOH —> -> [-NH-[CH2]6-NHCO-[CH2]4-CO-)n + 2nH20
poli[hexametylen adipamit) hay nilon-6,6 > Td nitron (hay olon)
H 2 C = C H —
CN CN
c) Cao su
+) Khai niem: Cao su la logi vat lieu polime c6 tinh dan hoị +) Phan logi
> Cao su thien nhien: lay ti( mu cay cao sụ
> Cao su tong hap: dieu che túcac ankadien bang phan itng trtmg hdp.
• Cao su buna
Na, p, t° ,
nCH2=CH-CH=CH2
buta-l,3~dien «:H2 -CH^ CH-CH2 -)u polibuta -1,3-dien polibuta -1,3-dien • Cao su buna-S: dong triing hap cua hutãl,3-dien vdi stiren
• Cao su bima-N: dong triing hap cua butãl,3-dien vdi acrilonitrin.
d) Keo dan tong help
+) Khdi niem: Keo dan la logi vgt lieu c6 khd ndng ket dinh hai
manh vgt lieu rdn giong hogc khdc nhau ma khong lam bien doi ban chat cua cac vgt lieu ditac ket dinh.
+) M o f so logi keo dan tong hap thong dung
P Nhúa va sdm > Keo dan epoxi
> Keo dan ure-fomandehit:
nH2N-CO-NH2 + nCH2=0 xt, t" -> (-HN-CO-NH-CH2-)n + nH20
5. BAI TAP AP D U N G
pai 1- Ti^ ^h.di dau l a bot g6 va cac chat v6 co can t h i e t , neu h a i so
do p h a n uTng dieu ché t h a n h 2 l o a i horp chat cao p h a n tuT theo p h a n
urng t r u n g hop va t r u n g ngiAig. V i e t cac phiTcfng t r i n h h o a h o c .
Bai 2. V i e t phiJcfng t r i n h hoa h o c cua p h a n ufng d o n g t r u n g h g p tao t h a n h cac polime t i f cac monome s a u :
a) V i n y l clorua v d i v i n y l axetat.
b ) B u t a - l , 3 - d i e n vdri s t i r e n .
c) A x i t m e t a c r i l i c vdfi b u t a - 1 , 3 - d i e n .
Bai 3..Cho cac c h a t : OsN-CCHgle-NOa va BrCCHalsBr.
a) V i e t phifong t r i n h hoa h o c c u a p h a n ufng tao t h a n h tcf n i l o n - 6 , 6 . b ) H a y cho b i e t d a c d i e m c u a loai tcf t r e n . b ) H a y cho b i e t d a c d i e m c u a loai tcf t r e n .
Bai 4. Goi t e n cac p h a n ufng va v i e t phtfomg t r i n h hoa h o c cua p h a n ufng
polime h o a cac monome sau:
a) C H 3- C H = CH2; •
b ) CH2=CC1-CH=CH2; c) CH2= C(CH3) - C H=CH2; c) CH2= C(CH3) - C H=CH2;
d ) C H 2 O H - C H 2 O H va m-C6H4(COOH)2 (axit isophtalic);
e ) NH2[CH2]ioCOOH.
Bai 5. Thifc h i e n p h a n ufng clo hoa p o l i ( v i n y l clorua) t h u di/gc m o t l o a i tcf c l o r i n chufa 6 6 , 7 % clọ Xac d i n h só m a t x i c h t r u n g b i n h cua p o l i ( v i n y l clorua) da tac d u n g vdri m o t p h a n tuf clọ
Bai 6. De san x u a t p o l i m e c l o r i n ngifdi t a clo hoa P V C b a n g clọ San p h a m t h u dtfoc chufa 66,7% clọ T i n h x e m t r u n g b i n h m 6 i p h a n ttf clo tac
d u n g vofi m a y m a t x i c h - C H 2- C H C I - t r o n g p h a n tuf PVC, gia t h i e t r k n g he so polime hoa n k h o n g t h a y doi sau p h a n ufng. V i e t cong thufc cau tao m o t doan p h a n tuf polime t r e n .
HifdNG DAN GIAI
fi^i 1. T h a n h p h a n c h i n h cua b o t go l a xenlulozcf.
+) Tuf xenlulozof tao t h a n h cao su buna theo so do chuyen d o i sau:
+H O
(CeHioOs) ^^^-^ CetlnOe sll^c ^ C2H5OH
xt > C H 2= C H - C H= C H 2 ^ ^ ^ ^ — > ( - C H 2- C H = C H- C H 2- ) n
+) TiX xenlulozcf tao thanh nhira phenolfomandehit theo so do phan iftig sau;
> (CeHioOsX. > CeHisOe CH3COOH > C H s C O O N a
enzim
30 .35OC > C2H5OH men giam
NaOH voi to N a O H CH3C1 >CH30H > H C H O fCl, > T H 1500°c p „ +c,600"c „ -^^'2 V n u ni
^ l&mlanhuhanh ^ ^2^2 > CgHe bot F e ^ CgHsCl +NaOH
nCH20 + n C s H s O H '^k^—^
Bai 2. a) V i n y l clorua vdri vinyl axetat: nCH2=CHCl + n H 3 C - C H OCOCH3 > CeHgOH + nH20 xt, t", p b) B u t a - l , 3 - d i e n vdi stiren: n C H 2 = C H - C H = C H 2 + nH2C=CH H 2 C — C H - C H o - C H CI O C O C H 3 xt, t", p ^ -H2C-CH=CH-CH2-CH2-CH - c) A x i t metacrilic vdri b u t a - l, 3- d i e n n C H 2 = C H - C H = C H 2 + n H 2 C = C H- C 0 0 H P - ^ CH-: COOH Bai 3. - H 2 C - C H = C H - C H 2 - C H - C H 2 CH. a) 02N-[CH2]6-N02 + 1 2[ H ] Fe+HC! -> H 2 N - [ C H 2 ] 6 - N H 2 + 4H2O B r- [ C H 2] 6 - B r + 2 N a O H — H O- C H 2 - [ C H 2] 4- C H 2- O H + 2 N a B r
H O - C H 2 [ C H 2] 4- C H 2 O H + 2CuO OHC[CH2]4CHO + 2Cu + 2H2O
-> H O O C - [ C H 2] 4- C O O H OHC[CH2]4CHO + O2
n H O O C - [ C H 2] 4- C O O H + nH2N-[CH2]6NH2 x t , t "
102 [ - C O- [ C H 2] 4 - C O - N H- [ C H 2] 6 - N H - ] „ +Bdl D U S N G H O A HOC 12 2 n H 2 0
J,) Dac diem cau tao cua loai to nilon-6,6:
_ Mach thang khong phan nhanh.
I Mach g6m n mat xich: - C O - C C H a k - C O - N H - L C H s l e - N H -, Trong m 6 i mat xich c6 2 nhom: - C O - N H - ,