D.
2-metilpropena : CH2=C-CH3
CH3
Cl
CH2=C-CH3 + HCl --> CH3-C-CH3
CH3 CH3
2-cloro-2-metilpropan
(clorura de tertbutil)
a.
la probleme putem utiliza formulele restranse
560g n moli m g
C4H8 + HCl --> C4H9Cl
56 1 92,5
=> m = 560x92,5/56 = 925 g produs
b.
n = 560x1/56 = 10 moli HCl
c.
niu = V/Vm, Vm = 22,4 L/mol
=> V = niuxVm = 10x22,4 = 224 L HCl
d. vezi mai sus
E.
a. se cere natura at. de C
1= cuaternar , 2 = tertiar , 3= primar
b.
Fm. = C8H16, miu = 8x12+16x1 = 112 g/mol
raport de masa:
C: H = 96: 16 /:16 = 6: 1
c.
daca la 112g C8H16 ..................... 96g C
x g C8H16 ........................ 72g C
= 84g C8H16
d.
Fm. omolog inferior = C7H14
Fm. omolog superior = C9H18
e.
CH3
CH3-C-CH2-C=CH2
CH3 CH3
2,5,5-trimetil-1-pentena
f.
CH2=CH-CH2-CH2-CH2-CH2-CH2-CH3
1-octena
