Question

2. O alchina A contine 88,88% carbon si are o catena liniara cu un atom de carbon primar. Stabiliti formula moleculara a alchinei si completati ecuatiile reactiilor de mai jos: A + H 2 O(HgSO 4 /H 2 SO 4 )→ A + 2HCl(HgCl 2 /170C)→ A + 2Cl 2 (CCl4)→ A + [Ag(NH 3 ) 2 ]OH→ A + O 2 (aer) →

Answer 1

alchina = CnH2n-2, miu = 14n-2

100% ............................... 88,88% C

14n-2 ............................. 12n

=> n = 2/0,5 = 4 => C4H6, butina

=> HC≡C-CH2-CH3 => 1-butina

+ H2O -Hg²⁺/H⁺-> H3CC(O)CH2CH3, butanona

+ 2HCl -Hg/175oC-> H3CC(Cl2)CH2CH3, 2,2-diclorobutan

+ 2Cl2 --CCl4--> Cl2HCC(Cl2)CH2CH3, 1,1,2,2-tetraclorobutan

+  [Ag(NH3)2]OH -->  AgC≡C-CH2-CH3 + 2NH3 + H2O

+ 11/2O2 --> 4CO2 + 3H2O