viet phuong trinh hinh học cua cac phan ung thuc hien so do chuyen hoa sau
Metan -> axetilen -> etilen- > poli etilen
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1.2 – di brom etan
viet phuong trinh hinh học cua cac phan ung thuc hien so do chuyen hoa sau
Metan -> axetilen -> etilen- > poli etilen
|
1.2 – di brom etan
Giải thích các bước giải:
$2CH_4\xrightarrow[\rm{lln}]{t^\circ}CH\equiv CH+3H_2$
$CH\equiv CH+H_2\xrightarrow{t^\circ, Pd}CH_2=CH_2$
$nCH_2=CH_2\xrightarrow{t^\circ,xt,p}(\kern-6pt-CH_2-CH_2-\kern-6pt)_n$
$CH_2=CH_2+Br_2\to CH_2Br-CH_2Br$
$(1)\quad \rm 2CH_4 \xrightarrow[làm\ lạnh\ nhanh]{1500^\circ} C_2H_2 + 3H_2$
$(2)\quad \rm C_2H_2 + H_2 \xrightarrow{t^\circ,\ Pd/PbCO_3} C_2H_4$
$(3)\quad \rm nCH_2=CH_2 \xrightarrow{t^\circ,\ xt,\ p} (\kern-6pt-CH_2-CH_2-\kern-6pt)_n$
$(4)\quad \rm CH_2=CH_2 + Br_2 \longrightarrow CH_2Br-CH_2Br$