a)
..………………………………..........|÷0,68025
C: 40,81/12=3,4...................5
H: 4,76/1=4,76.....................7
Br: 54,42/80=0,68025.........1
f.bruta: C5H7Br
fiind monobromurat => f.bruta=f.moleculara
b)
NE=[2+5(4-2)+7(1-2)+1(1-2)]/2=(2+10-7-1)/2=4/2=2
c)
CH≡C-CH2-CH2-CH2-Br
Br-C≡C-CH2-CH2-CH3