Đáp án:

`↓`

Giải thích các bước giải:

`d.`

`-` `C_2H_2+H_2` $\xrightarrow[xt]{t^o, Pd}$ `C_2H_4`

`-` `nCH_2=CH2` $\xrightarrow[xt]{t^o, p}$ $(\kern-6pt- CH_2-CH_2 -\kern-6pt)_n$

`-` `C_2H_4+H_2O` $\xrightarrow[H_2SO_4]{t^o}$ `C_2H_5OH`

`-` `C_2H_5OH` $\xrightarrow[H_2SO_4]{t^o}$ `C_2H_4+H_2O`

`-` `3C_2H_4+2KMnO_4+4H_2O→2C_2H_4(OH)_2+2MnO_2+2KOH`

`-` `C_2H_4+Br_2→C_2H_4Br_2`

`-` `C_2H_4Br_2+Zn` $\xrightarrow[]{t^o}$ `C_2H_4+ZnBr_2`

`-` `C_2H_4+HCl` $\xrightarrow[]{as}$ `C_2H_5Cl`

$C_{2}H_{2}+H_{2}\xrightarrow[Pd]{t^o,xt}C_{2}H_{4}$

$nC_{2}H_{4}\xrightarrow[]{t^o,xt,p}(-CH_{2}-CH_{2}-)n$

$C_{2}H_{4}+H_{2}O\xrightarrow[]{t^o,H_{2}SO_{4}}C_{2}H_{5}OH$

$C_{2}H_{5}OH\xrightarrow[]{t^o,H_{2}SO_{4}}C_{2}H_{4}+H_{2}O$

$3C_{2}H_{4}+2KMnO_{4}+4H_{2}O→2C_{2}H_{4}(OH)_{2}+2MnO_{2}+2KOH$

$C_{2}H_{4}+Br_{2}→C_{2}H_{4}Br_{2}$

$C_{2}H_{4}Br_{2}+Zn\xrightarrow[]{t^o}ZnBr_{2}+C_{2}H_{4}$

$C_{2}H_{4}+HCl\xrightarrow[]{ánh sáng}C_{2}H_{5}Cl$