Cho chất hữu cơ E có công thức phân tử C9H8O4 và các sơ đồ phản ứng theo đúng tỉ lệ

* Hướng dẫn giải

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{A} \\ \mathrm{E}:\mathrm{HCOO}-{\mathrm{CH}}_2{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{OOCH}\text{ (o, m, p)} \\ \begin{array}{l}\mathrm{HCOO}-{\mathrm{CH}}_2{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{OOCH}+3\mathrm{NaOH}\underset{}{\overset{{\mathrm{t}}^{\mathrm{o}}}{\to }}2\mathrm{HCOONa}+{\mathrm{NaOC}}_6{\mathrm{H}}_4{\mathrm{CH}}_2\mathrm{OH}+{\mathrm{H}}_2\mathrm{O}\\ \text{ X Y}\end{array} \\ \to \mathrm{E}, \mathrm{X} \mathrm{tráng} \mathrm{bạc}, \mathrm{Y} \mathrm{tác} \mathrm{dụng} \mathrm{được} \mathrm{với} \mathrm{CO}2 \to \left(\mathrm{c}\right); \left(\mathrm{d}\right) \mathrm{đúng} \\ \begin{array}{l}2\mathrm{HCOONa}+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\underset{}{\overset{}{\to }}2\mathrm{HCOOH}+{\mathrm{Na}}_2{\mathrm{SO}}_4\\ \text{ X Z}\end{array} \\ \begin{array}{l}\text{HCOOH}+2{\mathrm{AgNO}}_3+\text{ 4}{\mathrm{NH}}_3+{\mathrm{H}}_2\mathrm{O}\underset{}{\overset{}{\to }} {\left({\mathrm{NH}}_4\right)}_2{\text{CO}}_3+2\mathrm{Ag}+2{\mathrm{NH}}_4{\mathrm{NO}}_3\\ \text{ Z T}\end{array} \\ \to \mathrm{T} \mathrm{là} \mathrm{hợp} \mathrm{chất} \mathrm{vô} \mathrm{cơ} \to \left(\mathrm{b}\right) \mathrm{sai} \\ \begin{array}{l}{\mathrm{NaOC}}_6{\mathrm{H}}_4{\mathrm{CH}}_2\mathrm{OH}+\mathrm{HCl}\underset{}{\overset{}{\to }}{\mathrm{HOC}}_6{\mathrm{H}}_4{\mathrm{CH}}_2\mathrm{OH}+\mathrm{NaCl}\\ \text{ Y F}\end{array} \\ \to \mathrm{F} \mathrm{là} \mathrm{hợp} \mathrm{chất} \mathrm{đa} \mathrm{chức} \to \left(\mathrm{e}\right) \mathrm{đúng}. \end{gathered}$$