Đáp án:
d, $CH_2=CH-CH_3+Br_2\to CH_2Br-CHBr-CH_3$
e, $ nCH_2=CH_2\xrightarrow{{t^o, p, xt}} (\kern-6pt-CH_2-CH_2-\kern-6pt)_n$
g, $n CH_2=CH-CH=CH_2 \xrightarrow{{t^o, p, xt}} (\kern-6pt-CH_2-CH=CH-CH_2-\kern-6pt)_n$
h, $CH_2=CH-CH=CH_2+Br_2\to \left[\begin{array}{I} CH_2Br-CHBr-CH=CH_2\quad(-80^oC)\\ CH_2Br-CH=CH-CH_2Br\quad(40^oC)\end{array}\right.$
i, $CH_2=CH-CH=CH_2+2H_2\xrightarrow{{t^o, Ni}} CH_3-CH_2-CH_2-CH_3$
k, $CH_2=CH-CH=CH_2+HBr\to \left[ \begin{array}{l} \left[ \begin{array}{l}CH_2Br-CH_2-CH=CH_2\quad(spp)\\ CH_3-CHBr-CH_2-CH_3\quad(spc)\end{array} \right. \quad(-80^oC)\\ CH_2Br-CH=CH-CH_3\quad(40^oC)\end{array} \right.$
