Question

Se considera sinteza...

Answer 1

C = a. mono.a si mono.c

=> H2N-CnH2n-COOH, 18,666% N

=> 100% ........................... 18,666% N

   14n+61 ......................... 14 g N => n = 1

=> C = H2N-CH2-COOH, glicina

a)

H2N-CH2-COOH <--NH3/-HCl-- Cl-CH2-COOH (B)

Cl-CH2-COOH <--Cl2(h.niu)/-HCl-- CH3-COOH (A)

b)

nH2N-CH2-COOH --H+-> H2N-CH2-CO(NH-CH2-CO)n-HN-CH2-COOH

miu.peptida = 132+57n

100% peptida ............... 22,222% N

132+57n ............................ 14n+28 g N

=> 132+57n = 4,5(14n+28)

=> 69 = 69n => n = 1 => tripeptida D = Glicil-Glicil-Glicina

c)

aH2N-CH2-COOH + bH2N-CH(CH3)-COOH --(-H2O)--> peptida

miu.peptida = 203 g/mol

75a + 89b - 18(a+b) = 203

daca a = 1 => b = 2 =>

verificam: 75x1+89x2-18x2 = 217 => NU

inversam raportul astfel

a = 2 si b = 1

verificam = 75x2+89-18x2 = 203 => DA

=> tripeptidele mixte:

Gli-Gli-Ala

Gli-Ala-Gli

Ala-Gli-Gli

d)

trei grupe petidice => tetrapeptida mixta

C = 45,833/12 = 3,82 l

H = 6,944/1 = 6,944   l

N = 19,444/14 = 1,388 l

O = 27,777/16 = 1,736 l: 1,388

=> Fb = (C2,75H5NO1,25)n

n = 4 => C11H20N4O5

=> miu = 288 g mol

75a + 89b - 18(a+b) = 288

daca a = 2 => b = 2,45 => NU

inversam raportul astfel

daca b = 2 => a = 2,56 => NU

daca b = 3 => a = 1

verificam = 75+89x3-18x3 = 288 => DA

=> A-A-A-G si tate celelalte variante....