Methyl methacrylate có công thức cấu tạo là CH2=C(CH3)COOCH3.

Đáp án đúng là:

a. Đúng. Do có liên kết đôi $\text{C}=\text{C}$ trong gốc hydrocarbon nên methyl methacrylate có khả năng cộng bromine, làm mất màu nước bromine ở nhiệt độ thường.

b. Đúng. $M_{\text{C}{\text{H}}_2=\text{C}\left(\text{C}{\text{H}}_3\right)\text{C}\text{O}\text{O}\text{C}{\text{H}}_3}=100\text{g}\text{/}\text{m}\text{o}\text{l}\Rightarrow n_{\text{e}\text{s}\text{t}\text{e}\text{r}}=\frac{15}{100}=0,15\text{m}\text{o}\text{l}$.

$n_{\text{N}\text{a}\text{O}\text{H}}=0,1\times 2=0,2\text{m}\text{o}\text{l}$. Phương trình phản ứng:

$\text{C}{\text{H}}_2=\text{C}\left(\text{C}{\text{H}}_3\right)\text{C}\text{O}\text{O}\text{C}{\text{H}}_3+\text{N}\text{a}\text{O}\text{H}\stackrel{t{}^{\circ }}{\to }\text{C}{\text{H}}_2=\text{C}\left(\text{C}{\text{H}}_3\right)\text{C}\text{O}\text{O}\text{N}\text{a}+\text{C}{\text{H}}_3\text{O}\text{H}$

        0,1                            0,2 mol

So sánh tỉ lệ có NaOH dư, ester phản ứng hết.

Muối thu được là sodium methacrylate: $\text{C}{\text{H}}_2=\text{C}\left(\text{C}{\text{H}}_3\right)\text{C}\text{O}\text{O}\text{N}\text{a}$ ($M=108\text{g}\text{/}\text{m}\text{o}\text{l}$).

$m_{\text{m}\text{u}\text{o}\text{i}}=0,15\times 108=16,2\text{g}\text{a}\text{m}$.

c. Đúng.

d. Sai. $\text{C}{\text{H}}_2=\text{C}\left(\text{C}{\text{H}}_3\right)\text{C}\text{O}\text{O}\text{C}{\text{H}}_3+{\text{H}}_2\text{O}\stackrel{{\text{H}}^{+},t^{\circ }}{\rightleftharpoons }\text{C}{\text{H}}_2=\text{C}\left(\text{C}{\text{H}}_3\right)\text{C}\text{O}\text{O}\text{H}+\text{C}{\text{H}}_3\text{O}\text{H}$

Dung dịch thu được gồm methacrylic acid; methyl alcohol (methanol); ester và nước.