Sơ đồ chuyển hoá:

$\rm CaC_2 \xrightarrow{(1)} C_2H_2\xrightarrow{(2)} C_2H_4\xrightarrow{(3)} C_2H_5OH \xrightarrow{(4)} CH_3COOH\xrightarrow{(5)} CH_3COOC_2H_5$

$(1)\quad \rm CaC_2 + 2H_2O \longrightarrow C_2H_2 + Ca(OH)_2$

$(2)\quad \rm C_2H_2 + H_2 \xrightarrow{t^\circ,\ Pd/PbCO_3} C_2H_4$

$(3)\quad \rm C_2H_4 + H_2O \xrightarrow{t^\circ,\ H_2SO_4} C_2H_5OH$

$(4)\quad \rm C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$

$(5)\quad \rm CH_3COOH + C_2H_5OH \mathop{\rightleftharpoons}\limits^{t^\circ}_{H_2SO_4} CH_3COOC_2H_5 + H_2O$

Đáp án:

Giải thích các bước giải:

$CaC_{2}+2H_{2}O→C_{2}H_{2}+Ca(OH)_{2}$

$C_{2}H_{2}+H_{2}→C_{2}H_{4}$

$C_{2}H_{4}+H_{2}O\overset{t^o}→C_{2}H_{5}OH$

$5C_{2}H_{5}OH+4KMnO_{4}+6H_{2}SO_{4}→5CH_{3}COOH+2K_{2}SO_{4}+4MnSO_{4}+11H_{2}O$

$CH_{3}COOH+C_{2}H_{5}OH→CH_{3}COOC_{2}H_{5}+H_{2}O$