Đáp án:
$3.$
$a, CH_{3} - CH_{2} - CH_{3} + Cl_{2} \xrightarrow{\text{$1:1 , as $}} $ \(\left[ \begin{array}{l}CH_{2}Cl -CH_{2} - CH_{3} \\CH_{3} -CHCl - CH_{3}\end{array} \right.\) $+HCl$
$b, CH_{3} - CH_{2} - CH_{3} \xrightarrow{\text{$ t^{o},xt,p$}} CH_{3} - CH = CH_{2} + H_{2}$
$c, 2C_{6}H_{14} + 19O_{2} \xrightarrow{\text{$ t^{o}$}}12CO_{2} + 14H_{2}O$
