Question

La oxidarea unui amestec echimolar de doua hidrocarburi aromatice mononucleare omoloage A si B se obtine acelasi acid caboxilic C. Daca acelasi amestec este supus bromurarii, se formeaza un amestec de monobromurati ce contin 44,95 % Br.
Compusii A, B si C sunt :
A. o-xilen, p-xilen, respectiv acid ftalic;
B. m-xilen, cumen ( izopropilbenzen ), respectiv acid benzoic;
C. toluen, etilbenzen, respectiv acid benzoic;
D. etilbenzen, cumen, respectiv acid benzoic;
E. o-xilen, etilbenzen, respectiv acid ftalic.

Answer 1

A   : n moli  CxH(2x-6) + Br2 ⇒ CxH(2x-7)Br+HBr    nBr = n     m1Br = 80n
                  M A = 14x-6 = M
B  : n moli   C(x+1)H(2x-4) + Br2 ⇒ C(x+1)H(2x-5)Br    m2Br = 80n
                  M B = 14x +8 = M + 14
160n·100/[n(14x - 7 + 80 + 14x +7 + 80)] = 44,95
16000/(28x + 160) = 44,95      28x + 160 = 356  x = 7   
A = toluen     B = etilbenzen
raspuns : C