N.E. = (8x2+2-8)/2 = 5
=> un compus aromatic + 1 dubla legatura
daca poate reactiona doar cu un 2 moli Na si 1 mol NaOH => un acid aromatic cu grupare OH alcool si deci putem avea
HO-CH2-CH2-C6H4-COOH
HO-CH2-CH2-C6H4-COOH + 2PCl5 -->
Cl-CH2-CH2-C6H4-COCl (B) + 2PCl3 + 2HCl
HO-CH2-CH2-C6H4-COOH + 2[O] --> HOOC-CH2-C6H4-COOH (C) + H2O
Cl-CH2-CH2-C6H4-COCl + 2H2O --KOH-->
HO-CH2-CH2-C6H4-COOH + 2HCl