Quy đổi ankadien gồm $C_3H_4$ (x mol), $CH_2$(y mol)
$\Rightarrow 40x+14y=5,4$ $(1)$
$n_{CO_2}=\dfrac{8,96}{22,4}=0,4(mol)$
Bảo toàn $C$: $3n_{C_3H_4}+n_{CH_2}=n_{CO_2}$
$\Rightarrow 3x+y=0,4$ $(2)$
$(1)(2)\Rightarrow x=y=0,1$
Số nhóm $CH_2$: $\dfrac{y}{x}=1$
$\to$ ankadien có $3+1=4C$
CTPT: $C_4H_6$
CTCT:
$CH_2=C=CH-CH_3$ (buta-1,2-đien)
$CH_2=CH-CH=CH_2$ (buta-1,3-đien)
Dạng liên hợp của $C_4H_6$ là buta-1,3 đien
$CH_2=CH-CH=CH_2+H_2\buildrel{{Pd/PbCO_3, t^o}}\over\longrightarrow CH_3-CH_2-CH=CH_2$
$CH_2=CH-CH=CH_2+Br_2\buildrel{{-80^oC}}\over\longrightarrow CH_2Br-CHBr-CH=CH_2$
$CH_2=CH-CH=CH_2+Br_2\buildrel{{40^oC}}\over\longrightarrow CH_2Br-CH=CH-CH_2Br$
$CH_2=CH-CH=CH_2+HCl\buildrel{{-80^oC}}\over\longrightarrow CH_2-CHCl-CH=CH_2$ (spc)
$CH_2=CH-CH=CH_2+ HCl\buildrel{{40^oC}}\over\longrightarrow CH_2-CH=CH-CH_3$
$CH_2=CH-CH=CH_2+HOH\buildrel{{-80^oC}}\over\longrightarrow CH_2-CHOH-CH=CH_2$ (spc)
$CH_2=CH-CH=CH_2+ HOH\buildrel{{40^oC}}\over\longrightarrow CH_2OH-CH=CH-CH_3$
