\(\begin{array}{l}
2)\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\\
3)\\
a)\\
nC{O_2} = \dfrac{{6,6}}{{44}} = 0,15\,mol = > nC = 0,15\,mol\\
n{H_2}O = \dfrac{{2,7}}{{18}} = 0,15\,mol = > nH = 0,3\,mol\\
mC + mH = 0,15 \times 12 + 0,3 = 2,1g\\
= > A:C,H\\
b)\\
nC:nH = 0,15:0,3 = 1:2\\
= > CTDGN:C{H_2}\\
MA = 42g/mol\\
= > 14n = 42 = > n = 3\\
= > CTPT:{C_3}{H_6}\\
c)\\
C{H_2} = CH - C{H_3}
\end{array}\)