Từ chất X (C5H8O4) mạch hở, không

Chọn đáp án D

Giải thích:

$\begin{array}{l}+\left\{\begin{array}{l}{\pi }_X=2;O_X=4\\ X+2NaOH\stackrel{}{\to }{H}_{2}O\end{array}\right.\Rightarrow \left\{\begin{array}{l}Xco1chucaxit(-COOH)\\ Xco1chuceste(-COO-)\end{array}\right.\\ +\left\{\begin{array}{l}\left(2\right)\Rightarrow Zcon\mathrm{hom}-COONa\Rightarrow Tconhom-COOH\\ \left(3\right)T+H_{2}S{O}_{4dac}taoQlammatmaudungdichB{r}_2\end{array}\right.\Rightarrow trongTcon\mathrm{hoùm}OHdathuchienphanungtachnuoctaoanken\end{array}$

$\begin{array}{l}Tcodang\mathrm{...}-C{H}_2-C{H}_{2}OHhoac\mathrm{...}-CH\left(OH\right)C{H}_3\\ \Rightarrow Xla\left[\begin{array}{l}HOOC-C{H}_2-C{H}_{2}OOCC{H}_3\\ HOOC-CH\left(OOCC{H}_3\right)C{H}_3\end{array}\right.\Rightarrow \left\{\begin{array}{l}YlaC{H}_{3}COONa\\ ZlaNaOOC-C{H}_2-C{H}_2-OHhoacNaOOC-CH\left(OH\right)-C{H}_3\\ TlaHOOC-C{H}_2-C{H}_2-OHhoacHOOC-CH\left(OH\right)-C{H}_3\\ QlaHOOC-CH=C{H}_2\end{array}\right.\end{array}$