51110305 Chuyen de Amin Aminoaxit Protein

PHN DNG BI TP AMIN - AMINOAXIT 2011

1 Created by: Trn Vn Trung

AMIN DNG 1: TNH KHI LNG AMIN (MUI) TRONG PHN NG VI AXT HOC VI BROM * Cn nh 2 phng trnh: + Vi HCl: RNH2 + HCl RNH3Cl (1) R+16 R+52,5 Tng 36,5g

+Vi Brom: C6H5NH2 + 3Br2 C6H2Br3NH2 + 3HBr kt ta trng 93 333

X g Y g *PP: 1) p dng nh lut bo ton khi lng khi lng: mamin + maxit = mmui 2) Tnh mol ca cht bi cho ri t vo ptrnh suy ra s mol ca cht bi hi => tnh m 3) p dng nh lut tng gim khi lng C6H5NH2 + HCl C6H5NH3Cl 93 Tng 36,5

A g m = A.36,5

93 = B

mMui = A + B 3.1. Cho 9,3 gam anilin (C6H5NH2) tc dng va vi axit HCl. Khi lng mui thu c l A. 11,95 gam. B. 12,95 gam. C. 12,59 gam. D. 11,85 gam.

3.2. Cho 5,9 gam Propylamin (C3H7NH2) tc dng va vi axit HCl. Khi lng mui (C3H7NH3Cl) thu c l (Cho H = 1, C = 12, N = 14)

A. 8,15 gam. B. 9,65 gam. C. 8,10 gam. D. 9,55 gam.

3.3. Cho 4,5 gam etylamin (C2H5NH2) tc dng va vi axit HCl. Khi lng mui thu c l A. 7,65 gam. B. 8,15 gam. C. 8,10 gam. D. 0,85 gam.

3.4. Cho anilin tc dng va vi dung dch HCl thu c 38,85 gam mui. Khi lng anilin phn ng l: A. 18,6g B. 9,3g C. 37,2g D. 27,9g. 3.5. Th tch nc brom 3% (d = 1,3g/ml) cn dng iu ch 4,4 gam kt ta 2,4,6 tribrom anilin l A. 164,1ml. B. 49,23ml. C 146,1ml. D. 16,41ml. 3.6. Cho 11,25 gam C2H5NH2 tc dng vi 200 ml dung dch HCl x (M). Sau khi phn ng xong thu c dung dch c cha 22,2 gam cht tan. Gi tr ca x l

A. 1,3M B. 1,25M C. 1,36M D. 1,5M

3.7. Cho etyl amin tc dng 2000 ml dd HCl 0,3M. khi lng sn phm A. 48,3g. B. 48,9g. C. 94,8g. D. 84,9g.

3.8. Cho 7,75 metyl amin tc dng HCl khi lng sn phm l A. 11,7475. B. 16,785. C. 11,7495. D. 16,875.

3.9. Cho anilin tc dng 2000ml dd Br2 0,3M. Khi lng kt ta thu c l A.66.5g B.66g C.33g D.44g

3.10. Cho 9 gam C2H5NH2 tc dng va vi dung dch HCl, lng mui thu c l A. 0,85gam. B. 16,3 gam. C. 7,65gam. D. 8,10gam.

3.11. Cho 0,1 mol anilin (C6H5NH2) tc dng va vi axit HCl. Khi lng mui phenylamoniclorua ( C6H5NH3Cl) thu c l A. 25,900 gam . B. 6,475gam. C. 19,425gam. D. 12,950gam

3.12. Cho m gam Anilin tc dng ht vi dung dch Br2 thu c 9,9 gam kt ta. Gi tr m dng l A. 0,93 gam B. 2,79 gam C. 1,86 gam D. 3,72 gam

3.13. Th tch nc brom 5 % (d = 1,3g/ml) cn dng iu ch 3,96 gam kt ta 2,4,6 tribrom anilin l A. 164,1ml. B. 49,23ml. C 88,61 ml. D. s khc . 3.14. Cho 20g hn hpX gm ba amin no n chc l ng ng k tip nhau tc dng va vi dung dch HCl 1M, c cn dung dch thu c 31,68g mui. Xc nh th tch HCl dng ? A 16ml B 32ml C 160ml D 320ml

DNG 2: XC NH CU TO AMIN N CHC DA VO PHN NG CHY * i vi p t chy nh: + t CTTQ ca amin no n chc hoc amin n chc l: CxHyN

p dng CT: x

y = 2

22

CO

H O

n

n

+ Amin no n chc: 2 CnH2n+3N + (6n+3)/2 O2 2nCO2 + (2n+3)H2O+ N2



S mol amin =2

3

( nH2O n CO2) v 2

2 3

n

n = 2

2

CO

H O

n

n

n =?

+Amin khng no n chc c 1 lk i (CnH2n+1N)

2 CnH2n+1N + (6n+1)/2 O2 2nCO2 + (2n+1)H2O + N2

S mol amin = 2 ( nH2O n CO2) v 2

2 1

n

n = 2

2

CO

H O

n

n

+ Amin thm: 2CnH2n 5 N + (6n-5)/2 O2 2nCO2 + (2n-5)H2O +N2

3.2.1. t chy mt ng ng ca metylamin, ngi ta thy t l th tch cc kh v hi VCO2:VH2O sinh ra bng 2:3 Cng thc phn t ca amin l: A. C3H9N B. CH5N C. C2H7N D. C4H11N

3.2.2. t chy hon ton m (g) hn hp 3 amin thu c 3,36lt CO2; 1,12lt N2(ktc) v 5,4g H2O. Gi tr ca m l: A.3,6 B.3,8 C.4 D.3,1

3.2.3. t chy hon ton m gam mt amin n chc thu c 0,2 mol CO2 v 0,35 mol H2O. Cng thc phn t ca amin l A. C4H7N B. C2H7N C. C4H14N D. C2H5N

3.2.4. t chy hon ton mt amin n chc b 1 thu c CO2 v H2O theo t l 2

2

CO

H O=6

7. Vy CT amin l: A.

C3H7N B. C4H9N C. CH5N D. C2H7N

3.2.5. t chy hon ton mt amin n chc cha no thu c CO2 v H2O theo t l 2

2

CO

H O=8

9. Vy CT amin l:

A. C3H6N B. C4H9N C. C4H8N D. C3H7N

3.2.6. t chy hon ton mt amin n chc cha no c mt lin kt i mch cacbon ta thu c CO2 v H2O theo t l mol = 6:9. Vy cng thc phn t ca amin l cng thc no? A. CH3NH2 B. C4H9N C. C2H7N D. C3H9N

3.2.7. Khi t chy hon ton mt amin n chc X, thu c 8,4 lt kh CO2, 1,4 lt kh N2 (cc th tch kh o ktc) v

10,125 gam H2O. Cng thc phn t ca X l (cho H = 1, O = 16)

A. C2H7N. B. C4H9N. C. C3H7N. D. C3H9N.

3.2.8. t chy mt amin no n chc mch thng ta thu -c CO2 v H2O c t l mol 2 2

8 11CO H On : n : . CTCT ca

X l :A. (C2H5)2NH B. CH3(CH2)3NH2 C. CH3NHCH2CH2CH3 D. C 3

3.2.9. t chy mt amin n chc no (h) thu c t l s mol CO2 : H2O l 2 : 5. Amin cho c tn gi no di y? A. imetylamin. B. Metylamin. C. Trimetylamin. D. Izopropylamin 3.2.10. t chy hon ton 5,9 gam mt amin no h n chc X thu c 6,72 lt CO2, . Cng thc ca X l A.

C3H6O. B. C3H5NO3. C. C3H9N. D. C3H7NO2.

3.2.11. t chy hon ton 6,2 gam mt amin no h n chc, cn 10,08 lt O2 ktc. CTPT l A. C4H11N. B. CH5N. C. C3H9N. D. C5H13N.

3.2.12. t chy hon ton amin no n chc X, thu c 16,8 lt CO2 ; 2,8 lt N2 (ktc) v 20,25 g H2O. Cng thc phn t ca X l A. C4H9N. B. C3H7N. C. C2H7N. D. C3H9N.

3.2.14. Khi t chy hon ton mt amin n chc X, thu c 8,4 lt kh CO2 v 1,4 lt kh N2 v 10,125g H2O. Cng thc phn t l (cc kh o ktc) A. C3H5-NH2. B. C4H7-NH2. C. C3H7-NH2. D. C5H9-NH2.

3.2.15. Phn tch nh lng hp cht hu c X ta thy t l khi lng gia 4 nguyn t C, H, O, N l mC : mH : mO : mN = 4,8 : 1 : 6,4 : 2,8. T khi hi ca X so vi He bng 18,75. Cng thc phn t ca X l A. C2H5O2N. B. C3H7O2N. C. C4H10O4N2. D. C2H8O2N2 3.2.16. t chy hon ton a mol amin no, n chc thu c 13,2g CO2 v 8,1g H2O.Gi tr ca a l:

A.0,05 B.0,1 C.0,07 D.0,2

3.2.17. t chy hon ton amin no, n chc bc mt ,mch h thu c t l mol CO2 v H2O l 4:7 .Tn gi ca amin l:

A.etyl amin B. imetyl amin C.etyl metyl amin D.propyl amin



3.2.18. t chy hon ton 10,4 gam hai amin no, n chc, ng ng lin tip nhau, thu c 11,2 lt kh CO2 (ktc). Hai amin c cng thc phn t l A. CH4N v C2H7N B. C2H5N v C3H9N C. C2H7N v C3H7N D. C2H7N v C3H9N

DNG 3: XC NH CU TO AMIN N CHC DA VO PHN NG TO MUI

* Phn ng to mui: t CTTQ ca amin no n chc CnH2n+1NH2 hoc amin n chc l: R-NH2 V n chc nn ta lun c nHCl = nNH2 + Vi HCl: RNH2 + HCl RNH3Cl (1)

p dng CT: 2RNH

M =2RNH

HCl

m

n= a => MR =a -16 => CTPT

(hoc M RN = m

RN

n

HCl

= A M R = A - 14 => CTPT)

3.3.1. trung ha 25 gam dung dch ca mt amin n chc X nng 12,4% cn dng 100ml dung dch HCl 1M. Cng thc phn t ca X l (Cho H = 1; C = 12; N = 14) A. C3H7N. B. C2H7N. C. C3H5N. D. CH5N.

3.3.2. trung ha 20 gam dung dch ca mt amin n chc X nng 22,5% cn dng 100ml dung dch HCl 1M. Cng thc phn t ca X l (Cho H = 1; C = 12; N = 14) A. C2H7N B. CH5N C. C3H5N D. C3H7N

3.3.3. Trung ha 3,1 gam mt amin n chc cn 100 ml dung dch HCl 1M. Cng thc phn t ca X l A. C2H5N B. CH5N C. C3H9N D. C3H7N

3.3.4.Cho 0,4 mol amin no, n chc X tc dng vi dd HCl va thu c 32,6g mui.CTPT ca X l: A.CH3NH2 B.C2H5NH2 C.C3H7NH2 D.C4H9NH2

3.2.5. Trung ha 11,8 gam mt amin n chc cn 200 ml dung dch HCl 1M. Cng thc phn t ca X l A. C2H5N B. CH5N C. C3H9N D. C3H7N

3.3.6. Cho 10 gam amin n chc X phn ng hon ton vi HCl (d-), thu -c 15 gam mui. S ng phn cu to ca

X l A. 5. B. 8. C. 7. D. 4.

3.4.1. Cho 17,4 gam hn hp 2 amin n chc bc I c t khi so vi khng kh bng 2. Tc dng vi dung dch FeCl3 d thu c kt ta, em nung kt ta n khi lng khng i c m gam cht rn. Gi tr ca m l A. 16,0 gam B. 10,7 gam C. 24,0 gam D. 8,0 gam

3.4.2. Cho hn hp 2 amin n chc bc I c t khi hi so vi hiro l 19 (bit c mt amin c s mol bng 0,15) tc dng vi dung dch FeCl3 d thu c kt ta A. em nung A n khi lng khng i thu c 8 gam cht rn. Cng thc ca 2 amin l

A. CH3NH2 v C2H5NH2 B. CH3NH2 v C2H3NH2 C. C2H5NH2 v C2H3NH2 D. CH3NH2 v CH3NHCH3

3.4.3. Cho m gam hn hp hai amin n chc bc I c t khi hi so vi hiro l 30 tc dng vi FeCl2 d thu c kt ta X. ly kt ta nung trong khng kh n khi lng khng i c 18,0 gam cht rn. Vy gi tr ca m l

A. 30,0 gam B. 15,0 gam C. 40,5 gam D. 27,0 gam

AMINOAXIT DNG 1: TNH KHI LNG CA AMINO AXIT TRONG PHN NG VI AXT HOC BAZ * Phi vit c 2 phng trnh c dng: + Vi axt HCl:

H2N R COOH + HCl ClH3N R COOH R + 61 R+ 97,5 tng 36,5

Dng 4: AMIN TAC DUNG VI DUNG DCH MUOI (tng t amoniac)

RNH2 + H2O + FeCl3 Fe(OH)3 + RNH3+

Cl-



A g m = A.36,5

R+61 = B

mMui = A + B + Vi baz NaOH: H2N R COOH+ NaOH H2N R COONa+ H2O

R + 61 R+ 83 tng 22

A g m = A.22

R+61 = B

mMui = A + B

**PP: - p dng phng php tng gim khi lng - p dng inh lut bo ton khi lng - t s mol vo ptrnh ri tnh ging pp ca cc bi trnh by trn.

3.1.1. Khi cho 3,75 gam axit amino axetic ( NH2CH2COOH) tc dng ht vi dung dch NaOH, khi lng mui to thnh l A. 4,5gam. B. 9,7gam. C. 4,85gam. D. 10gam.

3.1.2. Cho 7,5 gam axit aminoaxetic (H2N-CH2-COOH) phn ng ht vi dung dch HCl. Sau phn ng, khi lng mui thu c l (Cho H = 1, C = 12, O = 16, Cl = 35, 5)


3.1.3. Cho 8,9 gam alanin ( CH3CH(NH2)COOH) phn ng ht vi dung dch NaOH. Khi lng mui thu c l. A. 11,2gam. B. 31,9gam. C. 11,1gam. D. 30,9 gam.

3.1.4. Cho 7,5 gam axit aminoaxetic (H2N-CH2-COOH) phn ng ht vi dung dch HCl. Sau phn ng, khi lng mui thu c l (Cho H = 1, C = 12, O = 16, Cl = 35, 5)


3.1.5. Cho m gam alanin phn ng ht vi dung dch NaOH. Sau phn ng, khi lng mui thu c 11,1 gam. Gi tr m dng l (Cho H = 1, C = 12, O = 16, Na = 23)


3.1.6. Cho 7,5 gam axit aminoaxetic (H2N-CH2-COOH) phn ng ht vi dung dch NaOH. Sau phn ng, khi lng mui thu c l (Cho H = 1, C = 12, O = 16, Na = 23)


3.1.7. Khi cho 3,75 gam axit amino axetic ( NH2CH2COOH) tc dng ht vi dung dch NaOH, khi lng mui to thnh l A. 4,5gam. B. 9,7gam. C. 4,85gam. D. 10gam.

DNG 2: TM CTCT CA AMINO AXT DA VO P TO MUI

*T/ dng vi NaOH:. Ptpu: (H2N)a R (COOH)b + bNaOH (H2N)a R (COONa)b + bH2O

mina

NaOH

n

n = b = s nhm chc axit ( COOH)

* T/d vi HCl Ptpu: (H2N)a R (COOH)b + aHCl (ClH3N)a R (COOH)b

mina

HCl

n

n = a = s nhm chc baz (NH2)

3.2.1. Mt - amino axit X ch cha 1 nhm amino v 1 nhm cacboxyl. Cho 10,68 gam X tc dng vi HCl d thu c 15,06 gam mui. Tn gi ca X l

A. axit glutamic. B. valin. C. alanin. D. Glixin

3.2.2. 0,1 mol aminoaxit X phn ng va vi 0,2 mol HCl hoc 0,1 mol NaOH. Cng thc ca X c dng l. A. (H2N)2R(COOH)2. B. H2NRCOOH. C. H2NR(COOH)2. D. (H2N)2RCOOH

3.2.3. X l mt - amioaxit no ch cha 1 nhm -NH2 v 1 nhm -COOH. Cho 23,4 gam X tc dng vi HCl d thu c 30,7 gam mui. Cng thc cu to thu gn ca X l cng thc no? A. CH3-CH(NH2)-COOH B. H2N-CH2-COOH

C.H2N-CH2CH2-COOH D.CH2=C(CH3)CH(NH2)COOH

3.2.4. X l mt aminoaxit no ch cha 1 nhm - NH2 v 1 nhm COOH. Cho 0,89 gam X tc dng vi HCl va to r a 1,255 gam mui. Cng thc cu to ca X l cng thc no sau y? A. H2N- CH2-COOH B. CH3- CH(NH2)-COOH.C. CH3-CH(NH2)-CH2-COOH. D. C3H7-CH(NH2)-COOH

3.2.5. X l mt - amioaxit no ch cha 1 nhm -NH2 v 1 nhm -COOH. Cho 15,1 gam X tc dng vi HCl d thu c 18,75 gam mui. Cng thc cu to ca X l cng thc no? A. C6H5- CH(NH2)-COOH B. CH3- CH(NH2)-COOH

C. CH3-CH(NH2)-CH2-COOH D. C3H7CH(NH2)CH2COOH



3.2.6. X l mt amino axit. Khi cho 0,01 mol X tc dng vi HCl th dng ht 80 ml dung dch HCl 0,125 M v thu -c 1,835 g mui khan. Cn khi cho 0,01 mol X tc dng vi dung dch NaOH th cn 25 gam dung dch NaOH 3,2%. Cng thc no sau y l ca X ?

A. C7H12-(NH)-COOH B. C3H6-(NH)-COOH C. NH2-C3H5-(COOH) D. (NH2)2-C3H5-COOH

3.2.7. Cho 4,41g mt aminoaxit X tc dng vi dung dch NaOH d- cho ra 5,73g mui. Mt khc cng l-ng X nh- trn

nu cho tc dng vi dung dch HCl d- thu -c 5,505 g mui clorua. X CTCT ca X.

A. HOOC-CH2CH2CH(NH2)COOH B. CH3CH(NH2)COOH C. HOOCCH2CH(NH2)CH2COOH D.C A,B

3.2.8. 0,1 mol aminoaxit A phn ng va vi 0,2 mol HCl, sn phm to thnh p va ht vi 0,3 mol NaOH. Gi tr n, m ln lt l: A. (H2N)2R(COOH)2. B. H2NRCOOH. C. H2NR(COOH)2. D. (H2N)2RCOOH

3.2.9. Cho X l mt amino axit. Khi cho 0,01mol X tc dng vi HCl th dng ht 80ml dung dch HCl 0,125M v thu c 1,835gam mui khan . Cn khi cho 0,01mol X tc dng vi dung dch NaOH th cn dng 25gam NaOH 3,2%. Cng thc cu to ca X l A. NH2C3H6COOH B. ClNH3C3H3(COOH)2

C. NH2C3H5(COOH)2. D. (NH2)2C3H5COOH

3.2.10. Cho 14,7 gam mt amino axit X (c 1 nhm NH2) tc dng vi NaOH d thu c 19,1 gam mui. Mt khc cng lng amino axit trn phn ng vi HCl d to 18,35 gam mui. Cng thc cu to ca X c th l A. NH2-CH2-COOH B. CH3-CH(NH2)COOH

C. NH2-(CH2)6 -COOH D. HOOC-CH2-CH2-CH(NH2)COOH

3.2.11. X l mt amin aamino axit cha 1 nhmCOOH v 1 nhm NH2. Cho 8,9g X tc dng vi dung dch HCl. Sau c cn dung dch th thu c 12,55g mui khan. Cng thc ng ca X l

3.2.12. Amino axit X cha 1 nhmCOOH v 2 nhm NH2. Cho 0,1 mol X tc dng ht vi 270ml dung dch NaOH 0,5M c cn thu c 15,4g cht rn. Cng thc phn t c th c ca X l A. C4H10N2O2 B. C5H12N2O2 C. C5H10NO2 D. C3H9NO4

3.2.13. X l mt a amino axit cha 1 nhm COOH v 1 nhm NH2. Cho 8,9g X tc dng vi 200ml dung dch HCl 1M. Thu c dung dch Y. phn ng ht vi cc cht trong dung dch Y cn dng 300ml dung dch NaOH 1M. Cng thc ng ca X l

DNG 3: T CHY AMINOAXIT

t CTTQ CxHyOzNt

x: y: z :t = : : :12 1 16 14

C O NHm m mm

= nCO2 : 2.nH2O : 16

Om : 2.nN2

Hay x: y: z :t = % % % %

: : :12 1 16 14

C H O N

3.3.1. Mt amino axit (X) c cng thc tng qut NH2RCOOH. t chy hon ton a mol X thu -c 6,729 (l) CO2 (ktc) v 6,75 g H2O. CTCT ca X l :

A. CH2NH2COOH B. CH2NH2CH2COOH C. CH3CH(NH2)COOH D. C B v C

3.3.2. Este X c to bi ancol metylic v - amino axit A. T khi hi ca X so vi H2 l 51,5. Amino axit A l

A. Axit - aminocaproic B. Alanin C. Glyxin D. Axit glutamic

3.3.3. Hp cht X c 40,45%C, 7,86%H, 15,73%N v cn li l oxi. Khi lng mol phn t ca X nh hn 100 gam. Bit X tc dng c vi hiro nguyn t. Cng thc cu to ca X l

A. CH3CH(NH2)COOH B. CH3-CH2-CH2-CH2NO2

C. H2NCH2CH2COOH D. CH3-CH2-CH2-NO2

A. CH2 COOH B. CH3 CH2 CH COOH

C. CH3 CH COOH D.

NH2 NH2

COOH CH2

NH2

CH2

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51110305 Chuyen de Amin Aminoaxit Protein