Xác định M:

\(\underbrace {RC{H_2}OH}_{_{14,72}}{\rm{ }} + {\rm{ }}CuO \to \underbrace {RCHO{\rm{ }} + {\rm{ }}RC{H_2}OH{\rm{ }}\left( {d} \right){\rm{ }} + {\rm{ }}{H_2}O}_{_{21,12}} + {\rm{ }}Cu\)

→ \({n_{RC{H_2}OH}} > n\left[ O \right] = \frac{{21,12 - 14,72}}{{16}} = 0,4 \to R < \frac{{14,72}}{{0,4}} - 31 = 5,8{\rm{ }}(R = H)\)

Ancol là CH_3­OH: 0,46 mol      

Xác định este:

\(\left\{ {\begin{array}{*{20}{c}}
{GlyNa:{\rm{ }}x}\\
{AlaNa:0,08}\\
{ValNa:{\rm{ }}y}
\end{array}} \right. \to \left\{ {\begin{array}{*{20}{c}}
{(2 + \frac{5}{4} - \frac{2}{2})x + 0,08(3 + \frac{7}{4} - \frac{2}{2}) + (5 + \frac{{11}}{4} - \frac{2}{2})y = 1,7625}\\
{BT{\rm{ }}Na:x{\rm{ }} + {\rm{ }}0,08{\rm{ }} + {\rm{ }}y{\rm{ }} = {\rm{ }}2.36,57/106}
\end{array}} \right.\)

⇒ x = 0,59; y = 0,02. Do n_M  = 0,46 mol nên

⇒ este: NH₂CH₂COOCH₃: 0,46 mol

Tìm công thức peptit và kết luận:

15,79 (g) YZ (C_nxH_2nx+2-nN_nO_n+1): Gly: 0,13; Ala: 0,08; Val: 0,02;

\(\left\{ {\begin{array}{*{20}{c}}
{a\left( {14nx{\rm{ }} + {\rm{ }}18{\rm{ }} + 29n} \right){\rm{ }} = {\rm{ }}15,79}\\
{BT{\rm{ }}C:{\rm{ }}anx{\rm{ }} = {\rm{ }}0,6}\\
{BT{\rm{ }}N:{\rm{ }}na{\rm{ }} = {\rm{ }}0,23}
\end{array}} \right. \to \left\{ {\begin{array}{*{20}{c}}
{a{\rm{ }} = {\rm{ }}0,04}\\
{n{\rm{ }} = {\rm{ }}5,75}
\end{array}} \right. \to \left\{ {\begin{array}{*{20}{c}}
{{A_5}:{\rm{ }}0,01}\\
{{A_6}:{\rm{ }}0,03}
\end{array}} \right.\)

\(\left\{ {\begin{array}{*{20}{c}}
{{A_5}\left( {Gl{y_a}Al{a_{3 - a}}Va{l_2}} \right){\rm{ }}0,01}\\
{{A_6}\left( {Gl{y_b}Al{a_{6 - b}}Va{l_2}} \right){\rm{ }}0,03}
\end{array}} \right. \to \left\{ {\begin{array}{*{20}{c}}
{0,01a{\rm{ }} + {\rm{ }}0,03b{\rm{ }} = {\rm{ }}0,13;a \le 3}\\
{Table:a{\rm{ }} = {\rm{ }}1;{\rm{ }}b{\rm{ }} = {\rm{ }}4}
\end{array}} \right.\)

⇒ Z(A₅)(GlyAla₂Val₂) 0,01 ⇒ %Z = 7,32%