30.
A = C2H2, B = C6H6
a = C2H4, b = Cl-CH2-CH2-Cl
c = C6H5-CH2-CH3
d = C6H5-CH(Cl)-CH3
e = C6H5-CH(C6H5)-CH3
f = C6H5-CH2-CH2-H5C6
g = C6H5-CH(Cl)-CH2-H5C6
h = C6H5-CH=CH-H5C6
i = C6H5-COOH
31.
hidrocarbura aromatica cu N.E = 7 => naftalina sau derivat al naftalinei
CnH2n-12
100% ............................... 93,75% C
14n-12 .............................. 12n C
=> n = 10 => C10H8 = naftalina simpla
a = C10H7-CH2CH3 in alfa
b = C10H7-COOH in alfa
c = anhidrida ftalica
d = C10H7-SO3H in alfa
e si f = vezi foto
