Cho X, Y, Z là ba peptit đều mạch hở và Đốt cháy hoàn toàn a mol mỗi peptit

* Hướng dẫn giải

Đáp án D

X, Y, Z có dạng: $C_{nk}{H}_{2nk+2-k}{O}_{k+1}{N}_k$

$$\begin{gathered} C_{nk}{H}_{2nk+2-k}{O}_{k+1}{N}_k\stackrel{O_2}{\to }nkC{O}_2+\left(nk+1-0,5k\right)H_{2}O \\ a anka\left(nk+1-0,5k\right) \\ {n}_{C{O}_2}-n_{H_{2}O}=a\to ank-a\left(nk+1-0,5k\right)=a\to 0,5\text{a}k=2\text{a}\to k=4. \end{gathered}$$

Vậy X, Y, Z đều là tetrapeptit.

 $$\begin{gathered} 69,8E\left\{\begin{array}{l}{C}_2{H}_{3}NO:4x\\ C{H}_2:y\\ H_{2}O:x\end{array}\right.\stackrel{NaOH}{\to }101,04\left\{\begin{array}{l}{C}_2{H}_{4}N{O}_{2}Na:4\text{x}\\ C{H}_2:y\end{array}\right. \\ \to \left\{\begin{array}{l}57\text{x+14y+18x}=69,8\\ 97.4\text{x}+14.y=101,04\end{array}\right.\to \left\{\begin{array}{l}x=0,22\\ y=1,12\end{array}\right. \\ \to \left\{\begin{array}{l}Ala-Na:a\\ Val-Na:b\end{array}\right.\to \left\{\begin{array}{l}a+b=0,22.4\\ 111\text{a}+139b=11,04\end{array}\right.\to \left\{\begin{array}{l}a=0,76\\ b=0,12\end{array}\right. \end{gathered}$$Vì $n_{Val}<n_Z$ nên trong Z chỉ chứa Ala. Vậy Z: $Al{a}_4:0,16$

2 peptit còn lại: $\to \left\{\begin{array}{l}{n}_{X,Y}=0,22-0,16=0,06\\ m_{X,Y}=69,8-0,16.302=21,48\end{array}\right.\to \overline{M}=358$

$$\begin{gathered} \to Y:Al{a}_{3}Val\left(330\right) \\ X:\left\{\begin{array}{l}Al{a}_{2}Va{l}_2\left(358\right)\left(loai\right)\\ AlaVa{l}_3\left(386\right)\\ Va{l}_4\left(414\right)\end{array}\right. \end{gathered}$$

TH1: $\left\{\begin{array}{l}AlaVa{l}_3:z\\ Al{a}_{3}Val:t\end{array}\right.\to \left\{\begin{array}{l}z+t=0,06\\ 386\text{z}+330t=21,48\end{array}\right.\to \left\{\begin{array}{l}z=0,03\\ t=0,03\end{array}\right.$

 (loại)

TH2: $\left\{\begin{array}{l}Va{l}_4:z\\ Al{a}_{3}Val:t\end{array}\right.\to \left\{\begin{array}{l}z+t=0,06\\ 414\text{z}+330t=21,48\end{array}\right.\to \left\{\begin{array}{l}z=0,02\\ t=0,04\end{array}\right.$ (chọn)

$\to \%m_{Va{l}_4}=11,86.$