Giải thích các bước giải:

$(C_6H_{10}O_5)_n+nH_2O\xrightarrow{H^+,\ t^o}nC_6H_{12}O_6$

$C_6H_{12}O_6\xrightarrow[enzim]{30-35^oC}2C_2H_5OH+2CO_2$

$C_2H_5OH\xrightarrow[H_2SO_{4\ (đ)}]{180^oC}C_2H_4↑+H_2O$

$C_2H_4+Br_2\to C_2H_4Br_2$

$C_2H_4Br_2+2NaOH\xrightarrow{C_2H_5OH\ (khan)}C_2H_2+2NaBr+2H_2O$

$CH_3COOH+HC≡CH\xrightarrow{Hg^{2+}}CH_3COOC=CH_2$

$CH_3COOH+C_2H_4\xrightarrow{O_2,\ Pd}CH_3COOCH=CH_2+H_2$

$nCH_3COOCH=CH_2\xrightarrow{t^o,\ xt,\ p}(\kern-6pt- CH_2-CH(OOCCH_3) -\kern-6pt)_n$

$(\kern-6pt- CH_2-CH(OOCCH_3) -\kern-6pt)_n+nNaOH\to (\kern-6pt- CH(OH)-CH_2 -\kern-6pt)_n+nCH_3COONa$

$A:\ C_2H_4$

$B:\ C_2H_4Br_2$

$C:\ C_2H_2$

$D:\ CH_3COOCH=CH_2$

$E:\ (\kern-6pt- CH_2-CH(OOCCH_3) -\kern-6pt)_n$

$F:\ (\kern-6pt- CH(OH)-CH_2 -\kern-6pt)_n$

Đáp án + Giải thích các bước giải:

$-(-C_6H_{10}O_5-)-_n + nH_2O \xrightarrow{H^+} nC_6H_{12}O_6$

$C_6H_{12}O_6 \xrightarrow[enzim, H^+]{t^o} 2C_2H_5OH + 2CO_2 \uparrow$

$C_2H_5OH + O_2 \xrightarrow{\text{men giấm}} CH_3COOH + H_2O$

$C_2H_5OH \xrightarrow[axit]{t^o} C_2H_4 \uparrow + H_2O$

$C_2H_4 + Br_2 \rightarrow C_2H_4Br_2$

$C_2H_4Br_2 + 2NaOH \rightarrow C_2H_4(OH)_2 + 2NaBr$

$C_2H_4(OH)_2 + 2CH_3COOH \xrightarrow{HgSO_4/H_2SO_4} (CH_3COO)_2C_2H_4 + 2H_2O$

$2C_2H_4 + O_2 + 4CH_3COOH \xrightarrow{Pd} 2(CH_3COO)_2C_2H_4 + 2H_2O$

$n(CH_3COO)_2C_2H_4 \xrightarrow{\text{trùng hợp}} 2-(-CH_3COOCH_2-)-_n$

$-(-CH_3COOCH_2-)-_n + NaOH \rightarrow -(-C_6H_9OCH_2COONa-)-_n$ (em chưa cân bằng được ._.)

Vậy:

$A:C_2H_4$

$B:C_2H_4Br_2$

$C:C_2H_4(OH)_2$

$D: (CH_3COO)C_2H_4$

$E: CH_3COOCH_2$

$F: -(-C_6H_9OCH_2COONa-)-_n$