$\begin{gathered}- A\xrightarrow{{NaOH}}\left\langle \begin{gathered}B\xrightarrow{{NaO{H_{(r)}}}}K(d{\,_{K/{O_2}}} = 0,5) \hfill \\C\xrightarrow{{Na}}2,24lit\,\,{H_2} \hfill \\ \end{gathered} \right.\hfill\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C\xrightarrow{{CuO,{t^0}}}D\,\,({k^0}\,\,\,trang\,\,\,guong) \hfill \\ \end{gathered} $

$\begin{gathered}{n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1\,\,mol \hfill \\{n_{NaOH}} = 0,1.3 = 0,3\,\,mol \hfill \\ \end{gathered} $

- C + Na →H₂ => C là ancol. Oxi hóa C ra D không tráng gương => C là ancol bậc 2

=> A là este đơn chức. B là muối Na

M_K= 32 . 0,5 =16 => K: CH₄=> Gốc R trong K là –CH₃.

$A:C{H_3}COOR'$

$C{H_3}COOR'\xrightarrow{{ + NaOH}}\left\{ \begin{gathered}C{H_3}COONa \hfill \\R'OH\xrightarrow{{ + Na}}R'ONa + {H_2} \hfill \\ \end{gathered} \right.$

${n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1mol = > {n_{ancol}} = 0,2mol.$

${n_{{\text{es}}te}} = {n_{ancol}} = 0,2\,\,mol$

${n_{NaOH}} = 0,1.3 = 0,3\,\,mol$

n_NaOH>n_este => NaOH dư, este hết

M_este=102

=>M_R’= 43(-C₃H₇: gốc ancol bậc 2)

=> Este: CH₃COOCH(CH₃)₂