Cho X, Y, Z là ba peptit mạch hở (có số nguyên tử...

A

- Hướng dẫn giải

Phương pháp giải:

\(\begin{gathered}
\left\{ \begin{gathered}
X(5C) \hfill \\
Y(7C) \hfill \\
Z(11C) \hfill \\
T:{C_n}{H_{2n}}{O_2} \hfill \\
\end{gathered} \right.\xrightarrow{{quy\,doi}}\frac{1}{2}E\underbrace {\left\{ \begin{gathered}
C{H_2}:x \hfill \\
CONH:y \hfill \\
{H_2}O:z \hfill \\
{O_2}:t \hfill \\
\end{gathered} \right.}_{117,36(g)}\left| \begin{gathered}
P1\xrightarrow{{ + {O_2}:5,37}}\left\{ \begin{gathered}
C{O_2}:x + y(mol) \hfill \\
{H_2}O:x + 0,5y + z(mol) \hfill \\
{N_2} \hfill \\
\end{gathered} \right. \hfill \\
P2\xrightarrow{{ + NaOH:(y + t)\,mol}}{H_2}O:z\,mol + \,C{H_3}OH:t\,mol + G\left\{ \begin{gathered}
{C_n}{H_{2n}}{O_2}NNa \hfill \\
{C_m}{H_{2m - 1}}{O_2}Na \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} \right. \hfill \\
G\left\{ \begin{gathered}
{C_n}{H_{2n}}{O_2}NNa \hfill \\
{C_m}{H_{2m - 1}}{O_2}Na \hfill \\
\end{gathered} \right.\xrightarrow{{ + {O_2}}}\left\{ \begin{gathered}
N{a_2}C{O_3}:0,5y + 0,5t \hfill \\
C{O_2}:2,58(mol) \hfill \\
{H_2}O:2,8(mol) \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} \)

P1: BT “O”: y + z + 2t + 5,37.2 = 2(x + y) + x + 0,5y + z

=> 3x + 1,5y - 2t =  10,74 (1)

P2: n_NaOH = y + t => n_Na2CO3 = 0,5y + 0,5t

BT “C”: n_C(0,5E) – n_C(CH3OH) = n_Na2CO3 + n_CO2 hay x + y - t = 0,5y + 0,5t + 2,58

=> x + 0,5y – 1,5t = 2,58 (2)

BT “H”: n_H(0,5E) – n_H(CH3OH) = 2n_H2O => 2x + y + 2z + y + t - 2z – 4t = 2,8.2

=> 2x + 2y – 3t = 5,6 (3)

Phương trình về KL: 14x + 43y + 18z + 32t = 117,36 (4)

BTKL: 124,78 + 4(y+t) = 133,18 + 18z + 46t

 (1) (2) (3) (4) => x; y; z; t

Biện luận để tìm este

Giải chi tiết:

\(\begin{gathered}
\left\{ \begin{gathered}
X(5C) \hfill \\
Y(7C) \hfill \\
Z(11C) \hfill \\
T:{C_n}{H_{2n}}{O_2} \hfill \\
\end{gathered} \right.\xrightarrow{{quy\,doi}}\frac{1}{2}E\underbrace {\left\{ \begin{gathered}
C{H_2}:x \hfill \\
CONH:y \hfill \\
{H_2}O:z \hfill \\
{O_2}:t \hfill \\
\end{gathered} \right.}_{117,36(g)}\left| \begin{gathered}
P1\xrightarrow{{ + {O_2}:5,37}}\left\{ \begin{gathered}
C{O_2}:x + y(mol) \hfill \\
{H_2}O:x + 0,5y + z(mol) \hfill \\
{N_2} \hfill \\
\end{gathered} \right. \hfill \\
P2\xrightarrow{{ + NaOH:(y + t)\,mol}}{H_2}O:z\,mol + \,C{H_3}OH:t\,mol + G\left\{ \begin{gathered}
{C_n}{H_{2n}}{O_2}NNa \hfill \\
{C_m}{H_{2m - 1}}{O_2}Na \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} \right. \hfill \\
G\left\{ \begin{gathered}
{C_n}{H_{2n}}{O_2}NNa \hfill \\
{C_m}{H_{2m - 1}}{O_2}Na \hfill \\
\end{gathered} \right.\xrightarrow{{ + {O_2}}}\left\{ \begin{gathered}
N{a_2}C{O_3}:0,5y + 0,5t \hfill \\
C{O_2}:2,58(mol) \hfill \\
{H_2}O:2,8(mol) \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} \)

P1: BT “O”: y + z + 2t + 5,37.2 = 2(x + y) + x + 0,5y + z

=> 3x + 1,5y - 2t =  10,74 (1)

P2: n_NaOH = y + t => n_Na2CO3 = 0,5y + 0,5t

BT “C”: n_C(0,5E) – n_C(CH3OH) = n_Na2CO3 + n_CO2 hay x + y - t = 0,5y + 0,5t + 2,58

=> x + 0,5y – 1,5t = 2,58 (2)

BT “H”: n_H(0,5E) – n_H(CH3OH) = 2n_H2O => 2x + y + 2z + y + t - 2z – 4t = 2,8.2

=> 2x + 2y – 3t = 5,6 (3)

Phương trình về KL: 14x + 43y + 18z + 32t = 117,36 (4)

BTKL: 124,78 + 4(y+t) = 133,18 + 18z + 46t

 (1) (2) (3) (4) => x = 4,16; y = 0,44; z = 0,1; t = 1,2

\(\begin{gathered}
\left\{ \begin{gathered}
{n_{mat\,xich}} = 0,44 \hfill \\
{n_{peptit}} = 0,1 \hfill \\
{n_{este}} = 1,2 \hfill \\
{n_{C(hh)}} = 4,16 + 0,44 = 4,6 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
So\,mat\,xich\,trung\,binh = 4,4 \hfill \\
0,5 < nC(peptit) < 1,1 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
Gly - Ala \hfill \\
Gly - Val \hfill \\
Gl{y_4} - Ala \hfill \\
3,5 < nC({\text{es}}te) < 4,1 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
Gly - Ala \hfill \\
Gly - Val \hfill \\
Gl{y_4} - Ala \hfill \\
2,92 < C({\text{es}}te) < 3,42 \hfill \\
\end{gathered} \right. \hfill \\
\to Este:{C_3}{H_6}{O_2} \to nC({\text{pe}}ptit) = 4,6 - 1,2.3 = 1 \hfill \\
\to \left\{ \begin{gathered}
Z:Gl{y_4}Ala(m) \hfill \\
Y:GlyVal(n) \hfill \\
X:GlyAla(p) \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
m + n + p = 0,1 \hfill \\
11m + 7n + 5p = 1 \hfill \\
5m + 2n + 2p = 0,44 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
m = 0,08 \hfill \\
n = 0,01 \hfill \\
z = 0,01 \hfill \\
\end{gathered} \right. \hfill \\
\to \% {m_Y} = \frac{{0,01.(75 + 117 - 18)}}{{117,36}}.100\% = 1,48\% \hfill \\
\end{gathered} \)

Đáp án A