Bài tập về chuỗi phản ứng este

C₆H₁₀O₂ có độ bất bão hòa k = (6.2 +2 -10)/2 = 2

X + NaOH → Y + Z

Y + HCl hoặc Z tác dụng với dd Br₂ đều thu được T => Y và Z có cùng số C trong phân tử và Z là andehit

=> CTCT của X là: CH₃CH₂COOCH=CHCH₃.

CH₃CH₂COOCH=CHCH₃. + NaOH → CH₃CH₂COONa (Y) + CH₃CH₂CH=O (Z)

CH₃CH₂COONa + HCl → CH₃CH₂COOH (T) + NaCl

CH₃CH₂CH=O + Br₂ + H₂O $\xrightarrow{{M{n^{2 + }}}}$ CH₃CH₂COOH (T) + 2HBr

$(1)\,2C{{H}_{4}}\xrightarrow{{{1500}^{0}}C}2{{C}_{2}}{{H}_{2}}+2{{H}_{2}}$

$(2)\,\,{{C}_{2}}{{H}_{2}}+{{H}_{2}}O\xrightarrow{HgS{{O}_{4}},{{H}_{2}}S{{O}_{4}}}C{{H}_{3}}CHO$

$(3)\,\,C{{H}_{3}}CHO+{{H}_{2}}\xrightarrow{Ni\,{{t}^{o}}}{{C}_{2}}{{H}_{5}}OH$

$(4)\,\,{{C}_{2}}{{H}_{5}}OH+{{O}_{2}}\text{ }\xrightarrow{men}\text{ }C{{H}_{3}}COOH\text{ }+\text{ }{{H}_{2}}O$

$(5)\,\,C{H_3}COOH + {C_2}{H_5}OH$ $C{H_3}COO{C_2}{H_5} + {H_2}O$

=> A là C₂H₂; C là C₂H₅OH.

$C{{H}_{4}}\xrightarrow{{{1500}^{0}}C}{{C}_{2}}{{H}_{2}}\xrightarrow{{{H}_{2}}O}C{{H}_{3}}CHO\xrightarrow{{{H}_{2}}}{{C}_{2}}{{H}_{5}}OH\xrightarrow{{{O}_{2}}}C{{H}_{3}}COOH\xrightarrow{CH\equiv CH}C{{H}_{3}}COOCH\text{ }=\text{ }C{{H}_{2}}$

PTHH:

(1) $C{{H}_{4}}\xrightarrow{{{1500}^{0}}C;\,\,lam\,\,lanh\,\,nhanh}{{C}_{2}}{{H}_{2}}+{{H}_{2}}$

(2) ${{C}_{2}}{{H}_{2}}+{{H}_{2}}O\xrightarrow{HgS{{O}_{4}},{{H}_{2}}S{{O}_{4}}}C{{H}_{3}}CHO$

(3) $C{{H}_{3}}CHO+{{H}_{2}}\xrightarrow{Ni\,{{t}^{o}}}{{C}_{2}}{{H}_{5}}OH$

(4) $C{{H}_{3}}CHO+{{H}_{2}}\xrightarrow{Ni\,{{t}^{o}}}{{C}_{2}}{{H}_{5}}OH$

(5) $C{{H}_{3}}COOH+{{C}_{2}}{{H}_{2}}\to C{{H}_{3}}COOCH\text{ }=\text{ }C{{H}_{2}}$

${{C}_{4}}{{H}_{10}}\xrightarrow{(1)}{{C}_{2}}{{H}_{4}}\xrightarrow{(2)}{{C}_{2}}{{H}_{5}}OH\xrightarrow{(3)}C{{H}_{3}}COOH\xrightarrow{(4)}C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}$

$(1)\,\,{{C}_{4}}{{H}_{10}}\text{ }\xrightarrow{cracking}\text{ }{{C}_{2}}{{H}_{4}}\text{ }+\text{ }{{C}_{2}}{{H}_{6~}}$

$(2)\,\,{{C}_{2}}{{H}_{4}}+{{H}_{2}}O\xrightarrow{{{H}_{2}}S{{O}_{4}}\,{{t}^{o}}}{{C}_{2}}{{H}_{5}}OH$

$(3){{C}_{2}}{{H}_{5}}OH+{{O}_{2}}\xrightarrow{{{t}^{0}}}C{{H}_{3}}COOH+{{H}_{2}}O$

=> A đúng C là muối CH₃COONH₄.

B đúng vì E là CH₃COOC₂H₅ được điều chế từ CH₃COOH và C₂H₅OH.

C sai vì CH₃COONH₄ không tráng gương

D đúng vì B là CH₃CHO.

Sơ đồ hoàn chỉnh :

$C{{H}_{4}}~\xrightarrow{{{1500}^{0}}C}{{C}_{2}}{{H}_{2}}\xrightarrow[HgS{{O}_{4}},{{H}_{2}}S{{O}_{4}}]{+{{H}_{2}}O}C{{H}_{3}}CHO~\xrightarrow{+{{O}_{2}}}C{{H}_{3}}COOH$

${{C}_{2}}{{H}_{2}}\xrightarrow[Pd/PdC{{O}_{3}}]{{{H}_{2}},{{t}^{0}}}~{{C}_{2}}{{H}_{4}}\xrightarrow{+KMn{{O}_{4}}}~{{C}_{2}}{{H}_{4}}{{\left( OH \right)}_{2}}~\xrightarrow[{{H}_{2}}S{{O}_{4}},{{t}^{0}}]{+T}~{{\left( C{{H}_{3}}COO \right)}_{2}}{{C}_{2}}{{H}_{4}}\left[ E \right]$

Z, Q, E lần lượt là CH₃CHO, C₂H₄(OH)₂, (CH₃COO)₂C₂H₄

Từ phản ứng (3) => Z là C₂H₅COONa

=> T là C₂H₆.

Từ (2) => Y là C₂H₄(COONa)₂

=> X là CH₃CH=CHOOCC₂H₄COOC₆H₅ hoặc CH₃CH=CHOOCCH(CH₃)COOC₆H₅

$\begin{array}{l} {C_4}{H_6}{O_2}:{\rm{ }}HCOOCH{\rm{ }} = {\rm{ }}CH - C{H_3}\\ HCOOCH{\rm{ }} = {\rm{ }}CH - C{H_3} + NaOH \to HCOONa + {C_2}{H_5}CHO\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(X)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(Y)\\ HCOONa + {H_2}S{O_4}{\,_{loang}} \to HCOOH + N{a_2}S{O_4}\\ \,\,\,\,\,\,\,\,(X)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(Z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(T) \end{array}$

$\begin{array}{l} {C_4}{H_6}{O_2}:{\rm{ }}HCOOCH{\rm{ }} = {\rm{ }}CH - C{H_3}\\ HCOOCH{\rm{ }} = {\rm{ }}CH - C{H_3} + NaOH \to HCOONa + {C_2}{H_5}CHO\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(X)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(Y)\\ HCOONa + {H_2}S{O_4}{\,_{loang}} \to HCOOH + N{a_2}S{O_4}\\ \,\,\,\,\,\,\,\,(X)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(Z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(T) \end{array}$

$\begin{array}{l} {C_4}{H_6}{O_2}:{\rm{ }}HCOOCH{\rm{ }} = {\rm{ }}CH - C{H_3}\\ HCOOCH{\rm{ }} = {\rm{ }}CH - C{H_3} + NaOH \to HCOONa + {C_2}{H_5}CHO\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(X)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(Y)\\ HCOONa + {H_2}S{O_4}{\,_{loang}} \to HCOOH + N{a_2}S{O_4}\\ \,\,\,\,\,\,\,\,(X)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(Z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(T) \end{array}$

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

Công thức của E là :

+) HO – CH₂ – COO – CH₂ – COOH

Z: C₂H₅OH

Y: CH₃COONa

X: CH₃COOC₂H₅

(1) $C{H_3}COO{C_2}{H_5} + NaOH\xrightarrow{{{t^0}}}C{H_3}COON{\text{a}} + {C_2}{H_5}OH$

(2) CH₃COONa + HCl $\to$ CH₃COOH + NaCl

(3) ${C_2}{H_5}OH + {O_2}\xrightarrow{{en{\text{z}}im}}C{H_3}COOH + {H_2}O$

Y₁ là CH₃COOH; X có dạng là CH₃COOR

Y + O₂ → Y₁ nên Y có thể là CH₃CHO. Có 2 chất X thỏa mãn là CH₃COOCH=CH₂ và CH₃COOC₂H₅

$\begin{align}{{C}_{6}}{{H}_{4}}{{\left( COO \right)}_{2}}{{C}_{2}}{{H}_{4}}+\text{ }2NaOH\xrightarrow{{{H}_{2}}O,{{t}^{0}}}NaOOC{{C}_{6}}{{H}_{4}}COONa+\text{ }HOC{{H}_{2}}C{{H}_{2}}OH \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({{X}_{1}})\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({{X}_{2}}) \\ \end{align}$

$\begin{array}{l} NaOOC{C_6}{H_4}COONa + {\rm{ }}HCl{\rm{ }} \to HOOC{C_6}{H_4}COOH + {\rm{ }}2NaCl\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({X_3}) \end{array}$

$nHOOC{{C}_{6}}{{H}_{4}}COOH+\text{ }nHOC{{H}_{2}}C{{H}_{2}}OH\xrightarrow{{{t}^{0}}}poli\left( etilen-terephtalat \right)\text{ }+\text{ }2n{{H}_{2}}O$

X₁ có nhiệt độ nóng chảy cao hơn X₃ => A sai

X₁ có tổng số nguyên tử H bằng  4 => B sai

X₂ có 2 nhóm OH => X₂ hòa tan được Cu(OH)₂ => C sai

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  CH₂ = C (CH₃)COOCH₃ + H₂O

C₂H₅OH + CuO → CH₃CHO+ Cu + H₂O

$HCOOC{H_2} - C{H_2} - OOCH{\rm{ }}\left( X \right){\rm{ }} + {\rm{ }}2NaOH{\rm{ }} \to 2HCOONa{\rm{ }}\left( {{X_1}} \right){\rm{ }} + {\rm{ }}HOC{H_2} - C{H_2}(OH)\left( {{X_2}} \right)\;$

$HOC{{H}_{2}}-C{{H}_{2}}OH\left( {{X}_{2}} \right)\text{ }+\text{ }{{O}_{2}}~\xrightarrow{Cu,{{t}^{0}}}OHC-CHO\left( {{X}_{3}} \right)\text{ }+\text{ }2{{H}_{2}}O$

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

B.Sai, X₁ là HCOONa là muối của axit đơn chức.

C. Đúng, X₂ là HOCH₂-CH₂-OH .

D. Sai, OHC-CHO (X₃) là hợp chất hữu cơ đa chức

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

Oxi hóa 1 mol Z cần 1 mol CuO => Z là ancol no đơn chức

Vậy X là : CH₃OOCH₂COOCH₃

=> Z : CH₃OH => .T : HCHO có M_T = 30 đvc

A – Từ ankan không điều chế được este

C – Từ C₂H₆O₂  (C₂H₄(OH)₂) không điều chế được C₂H₅OH

D - Từ C₂H₅Cl  không điều chế được CH₃COOH

${C_2}{H_2}\xrightarrow{{(1)}}C{H_3}CHO\xrightarrow{{(2)}}{C_2}{H_5}OH\xrightarrow{{(3)}}C{H_3}COOH\xrightarrow{{(4)}}C{H_3}COO{C_2}{H_5}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(A)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(B)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(D)$

$(1)\,\,{C_2}{H_2} + {H_2}O\xrightarrow{{HgS{O_4},{H_2}S{O_4}}}C{H_3}CHO$

$(2)\,\,C{H_3}CHO + {H_2}\xrightarrow{{Ni\,{t^o}}}{C_2}{H_5}OH$

$(3)\,\,{C_2}{H_5}OH + {O_2}{\text{ }}\xrightarrow{{men}}{\text{ }}C{H_3}COOH{\text{ }} + {\text{ }}{H_2}O$

$(4)\,\,C{H_3}COOH + {C_2}{H_5}OH\overset {{H_2}S{O_4}} \leftrightarrows C{H_3}COO{C_2}{H_5} + {H_2}O$

$C{H_4}\xrightarrow{{{{1500}^0}C}}{C_2}{H_2}\xrightarrow{{{H_2}O}}C{H_3}CHO\xrightarrow{{{H_2}}}{C_2}{H_5}OH\xrightarrow{{{O_2}}}C{H_3}COOH\xrightarrow{{{C_2}{H_2}}}C{H_3}COOCH{\text{ }} = {\text{ }}C{H_2}$

$C{H_4}\xrightarrow{{{{1500}^0}C;\,\,lam\,\,lanh\,\,nhanh}}{C_2}{H_2} + {H_2}$

${C_2}{H_2} + {H_2}O\xrightarrow{{HgS{O_4},{H_2}S{O_4}}}C{H_3}CHO$

$C{H_3}CHO + {H_2}\xrightarrow{{Ni\,{t^o}}}{C_2}{H_5}OH$

${C_2}{H_5}OH + {O_2}{\text{ }}\xrightarrow{{men}}{\text{ }}C{H_3}COOH{\text{ }} + {\text{ }}{H_2}O$

$C{H_3}COOH + {C_2}{H_2} \to C{H_3}COOCH{\text{ }} = {\text{ }}C{H_2}$

${C_4}{H_{10}}\xrightarrow{{(1)}}{C_2}{H_4}\xrightarrow{{(2)}}{C_2}{H_5}OH\xrightarrow{{(3)}}C{H_3}COO{C_2}{H_5}$

$(1)\,\,{C_4}{H_{10}}{\text{ }}\xrightarrow{{cracking}}{\text{ }}{C_2}{H_4}{\text{ }} + {\text{ }}{C_2}{H_{6\;}}$

$(2)\,\,{C_2}{H_4} + {H_2}O\xrightarrow{{{H_2}S{O_4}\,{t^o}}}{C_2}{H_5}OH$

$(3)\,\,{C_2}{H_5}OH + C{H_3}COOH\overset {{H_2}S{O_4}} \leftrightarrows C{H_3}COO{C_2}{H_5} + {H_2}O$