Bài tập về chuỗi phản ứng este

$C{H_4}\;\xrightarrow{{{{1500}^0}C}}{C_2}{H_2}\xrightarrow[{HgS{O_4},{H_2}S{O_4}}]{{ + {H_2}O}}C{H_3}CHO\;\xrightarrow{{ + {O_2}}}C{H_3}COOH$

${C_2}{H_2}\xrightarrow[{Pd/PdC{O_3}}]{{{H_2},{t^0}}}\;{C_2}{H_4}\xrightarrow{{ + KMn{O_4}}}\;{C_2}{H_4}{\left( {OH} \right)_2}\;\xrightarrow[{{H_2}S{O_4},{t^0}}]{{ + T}}\;{\left( {C{H_3}COO} \right)_2}{C_2}{H_4}\left[ E \right]$

${M_E} = 146$

(3) CH₃CHO + 2Cu(OH)₂ + NaOH $\xrightarrow{{{t}^{o}}}$ CH₃COONa + Cu₂O + 3H₂O

=> Z là CH₃COONa

 (4) CH₃COONa + NaOH $\xrightarrow{CuO,{{t}^{o}}}$ CH₄ + Na₂CO₃

=> T là CH₄.

(2) CH₂(COONa)₂ + 2NaOH $\xrightarrow{CaO,{{t}^{o}}}$ CH₄ + 2Na₂CO₃

=> Y là CH₂(COONa)₂

(1) X + 3NaOH $\xrightarrow{{{t}^{o}}}$ C₆H₅ONa + CH₂(COONa)₂ + CH₃CHO + H₂O

=> X là este 2 chức của axit CH₂(COOH)₂ với phenol và với gốc CH₂=CH-

=> X là CH₂=CHOOCCH₂COOC₆H₅

${C_3}{H_4}{O_2}:{\text{ }}HCOOCH{\text{ }} = {\text{ }}C{H_2}$

$HCOOCH{\text{ }} = {\text{ }}C{H_2} + NaOH\xrightarrow{{}}HCOONa + C{H_3}CHO$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(X)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(Y)$

$HCOONa + {H_2}S{O_4}{\,_{loang}}\xrightarrow{{}}HCOOH + N{a_2}S{O_4}$

$\,\,\,\,\,\,\,\,(X)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(Z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(T)$

Từ sơ đồ phản ứng, suy ra E có dạng HO – R – COO – R – COOH

E có 2 cấu tạo thỏa mãn là :

+) HO – CH₂ – CH₂ – COO – CH₂ – CH₂ – COOH

+) HO – CH(CH₃) – COO – CH (CH₃) – COOH 

${Z{\text{ }}:{\text{ }}C{H_3}CHO}$

${Y{\text{ }}:{\text{ }}C{H_3}COONa}$

${X{\text{ }}:{\text{ }}C{H_3}COOCH = C{H_2}\left( {{C_4}{H_6}{O_2}} \right)}$

$\left( 1 \right){\text{ }}C{H_3}COOCH = C{H_2}{\text{ }} + {\text{ }}NaOH\xrightarrow{{{t^0}}}\;C{H_3}COONa{\text{ }} + {\text{ }}C{H_3}CHO$

$\left( 2 \right){\text{ }}C{H_3}COONa{\text{ }} + {\text{ }}HCl\xrightarrow{{}}C{H_3}COOH{\text{ }} + {\text{ }}NaCl$

$\left( 3 \right)2C{H_3}CHO{\text{ }} + {\text{ }}{O_2}\xrightarrow{{xt, t^o}}2C{H_3}COOH{\text{ }} $

$HO{C_2}{H_4}COO{C_2}{H_4}COOH\,\,(X){\text{ }} + {\text{ }}2NaOH\xrightarrow{{{t^0}}}2HO - {C_2}{H_4} - COONa(Y){\text{ }} + {\text{ }}{H_2}O$

$HO - {C_2}{H_4} - COONa{\text{ }} + {\text{ }}HC{l_{loãng}}\xrightarrow{{}}{\text{ }}HO - {C_2}{H_4} - COOH{\text{ }} + {\text{ }}NaCl$

$HO - {C_2}{H_4} - COOH{\text{ }} + {\text{ }}2Na{\text{ }} \to \;NaO - {C_2}{H_4} - COONa{\text{ }} + {\text{ }}{H_2}$

$ =  > {\text{ }}{n_{H2}} = {\text{ }}{n_Z} = {\text{ }}0,1{\text{ }}mol$

${C_6}{H_4}{\left( {COO} \right)_2}{C_2}{H_4} + {\text{ }}2NaOH\xrightarrow{{{H_2}O,{t^0}}}NaOOC{C_6}{H_4}COONa + {\text{ }}HOC{H_2}C{H_2}OH$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({X_1})\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({X_2})$

$NaOOC{C_6}{H_4}COONa + {\text{ }}HCl{\text{ }}\xrightarrow{{}}{\text{ }}HOOC{C_6}{H_4}COOH + {\text{ }}2NaCl$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({X_3})$

$nHOOC{C_6}{H_4}COOH + {\text{ }}nHOC{H_2}C{H_2}OH\xrightarrow{{{t^0}}}poli\left( {etilen - terephtalat} \right){\text{ }} + {\text{ }}2n{H_2}O$

Các phản ứng xảy ra là:

CH₂ = C(CH₃)COOC₂H₅ + NaOH → CH₂ = C (CH₃)COONa (T) + C₂H₅OH

CH₂ = C (CH₃)COONa (T) + HCl → CH₂ = C (CH₃)COOH + NaCl

CH₂ = C(CH₃)COOH + CH₃OH $\overset {{H_2}S{O_4}} \leftrightarrows $  CH₂ = C (CH₃)COOCH₃ + H₂O

C₂H₅OH + O₂ → CH₃COOH (Y) + H₂O

CH₃COOH (Y) + NaOH → CH₃COONa (Z) + H₂O

CH₃COONa (Z) + NaOH → CH₄ + Na₂CO₃

$HCOOCH - CH\left( {C{H_3}} \right) - OOCH{\text{ }}\left( X \right){\text{ }} + {\text{ }}2NaOH{\text{ }} \to 2HCOONa{\text{ }}\left( {{X_1}} \right){\text{ }} + {\text{ }}HOC{H_2} - CH\left( {OH} \right) - C{H_3}\left( {{X_2}} \right)\;$

$HOC{H_2} - CH\left( {OH} \right) - C{H_3}\left( {{X_2}} \right){\text{ }} + {\text{ }}{O_2}\;\xrightarrow{{Cu,{t^0}}}OHC - CO - C{H_3}\left( {{X_3}} \right){\text{ }} + {\text{ }}2{H_2}O$

A. Đúng, X là este đa chức, có khả năng làm mất màu nước brom

B. Đúng, X₁ là HCOONa và ${M_{{X_1}}} = 68$

C. Đúng, HOCH₂-CH(OH)-CH₃ (X₂) là ancol hai chức liền kề có mạch C không phân nhánh.

D. Sai, OHC-CO-CH₃ (X₃) là hợp chất hữu cơ tạp chức.

X + NaOH tạo 2Z và Y (đều là 2 chất hữu cơ) => X là este

Oxi hóa 1 mol Y cần 2 mol CuO => Y có 2 nhóm OH => Y có ít nhất 2 Cacbon

Vậy X là : (HCOO)₂C₂H₄

=> Y : C₂H₄(OH)₂ => .T : (CHO)₂ có M_T = 58g