Đáp án: $6$ đồng phân
Giải thích các bước giải:
Đặt CTTQ anken: $C_nH_{2n}$
BTKL: $m_{H_2}=3,6-3,5=0,1g$
$\to n_A=n_{H_2}=\dfrac{0,1}{2}=0,05(mol)$
$\to M_A=\dfrac{3,5}{0,05}=70=14n$
$\to n=5\quad (C_5H_{10}$)
CTCT:
$CH_2=CH-CH_2-CH_2-CH_3$
$CH_3-CH=CH-CH_2-CH_3$ (cis-trans)
$CH_2=C(CH_3)-CH_2-CH_3$
$CH_3-C(CH_3)=CH-CH_3$
$CH_3-CH(CH_3)-CH=CH_2$