Đáp án:

$\rm k_{C_8H_{12}O_5}=\dfrac{2×8+2-12}{2}=3$

Vậy X có: $\rm 2\pi_{C\ =\ O}$ và $\rm 1\pi_{C\ =\ C}$.

Công thức thỏa mãn:

$\begin{array}{l}{\rm\kern1,95cm CH_3COO}\\{\rm CH_2=CH-COO}\\{\rm\kern3,17cm HO}\\\end{array} \left\rangle \begin{array}{l}{}\\{}\\\end{array} \right.\!\!\rm C_3H_5$

$\begin{array}{l}{\rm CH_2=CH-COO}\\{\rm\kern1,95cm CH_3COO}\\{\rm\kern3,17cm HO}\\\end{array} \left\rangle \begin{array}{l}{}\\{}\\\end{array} \right.\!\!\rm C_3H_5$

$\begin{array}{l}{\rm CH_2=CH-COO}\\{\rm\kern3,17cm HO}\\{\rm\kern1,95cm CH_3COO}\\\end{array} \left\rangle \begin{array}{l}{}\\{}\\\end{array} \right.\!\!\rm C_3H_5$

Phương trình phản ứng (chọn 1 trong 3 đồng phân cấu tạo):

$\begin{array}{l}{\rm\kern1,95cm CH_3COO}\\{\rm CH_2=CH-COO}\\{\rm\kern3,17cm HO}\\\end{array} \left\rangle \begin{array}{l}{}\\{}\\\end{array} \right.\!\!\rm C_3H_5+2NaOH\xrightarrow{t^°}CH_2COONa+CH_2=CH-COONa+C_3H_5(OH)_3$