Thủy phân hoàn toàn peptit X và Y trong môi trường bazơ theo phương trinh phản ứng sau:X + 6NaOH → 6A + H2O; Y + 5NaOH → 5B + H

* Hướng dẫn giải

E: C_nxH_2nx+2-nN_nO_n+1: a mol

\(\left\{ {\begin{array}{*{20}{c}}
{a\left( {14nx{\rm{ }} + {\rm{ }}29n{\rm{ }} + {\rm{ }}18} \right){\rm{ }} = {\rm{ }}21,45}\\
{a\left( {62nx{\rm{ }}--{\rm{ }}9n{\rm{ }} + {\rm{ }}18} \right){\rm{ }} = {\rm{ }}47,79}\\
{an\left( {1,5x{\rm{ }}--{\rm{ }}0,75} \right){\rm{ }} = {\rm{ }}0,9675}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{nxa{\rm{ }} = {\rm{ }}0,81}\\
{na{\rm{ }} = {\rm{ }}0,33}\\
{a{\rm{ }} = {\rm{ }}0,03}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{x{\rm{ }} = {\rm{ }}27/11}\\
{{n_X} = {\rm{ }}0,015}\\
{{n_Y} = {\rm{ }}0,015}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{{A_{\left( {2C} \right)}}:{\rm{ }}0,015.6{\rm{ }} = {\rm{ }}0,09}\\
{B:{\rm{ }}0,015.5{\rm{ }} = {\rm{ }}0,075}
\end{array}} \right.\)

\(\frac{{0,09.2 + 0,075.{C_B}}}{{0,09 + 0,075}} = \frac{{27}}{{11}}\)

→ C_B = 3 →  \(\left\{ {\begin{array}{*{20}{c}}
{A:{\rm{ }}{C_2}{H_5}N{O_2}\left( {75} \right)}\\
{B:{\rm{ }}{C_3}{H_7}N{O_2}\left( {89} \right)}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{{M_X} = {\rm{ }}360}\\
{{M_X} = {\rm{ }}360}
\end{array}} \right.\)