Đáp án:

$CH\equiv C-CH_3+H_2\xrightarrow{{Pd, PbCO_3, t^o}} CH_2=CH-CH_3$ 

$CH\equiv C-CH_3+2H_2\xrightarrow{{Ni, t^o}} CH_3-CH_2-CH_3$ 

$CH\equiv C-CH_3+2Br_2\to CHBr_2-CBr_2-CH_3$ 

$CH\equiv C-CH_3+HBr\xrightarrow{{1:1}} CH_2=CBr-CH_3$ 

$CH\equiv C-CH_3+H_2O\xrightarrow{{HgSO_4, H_2SO_4, t^o}} CH_3-CO-CH_3$ 

$CH\equiv C-CH_3+AgNO_3+NH_3\to AgC\equiv C-CH_3+NH_4NO_3$