ms = 200 g, c% = 4%
din c% => md = msxc%/100
= 200x4/100 = 8 g Br2 initial
ms2 = ms-a, a = masa de Br2 consumata in urma reactiei de aditie
md2 = md-a
=> c2% = md2x100/ms2
=> 2,04 = (8-a)x100/(200-a)
=> a = 392/102,04 = 3,84 g Br2 consumat
notam alchena = CnH2n
1,75 g 3,84 g
CnH2n + Br2 --CCl4--> CnH2nBr2
14n 160
=> n = 5
=> alchena = C5H10, pentena
alchena care contine in molecula un atom de acrbon cuaternar este:
H2C=C-CH2-CH3
CH3
sau
H3C-C=CH-CH3
CH3