Viết CTCT có thể có của các chất sau: C4H10, C2H6O, C3H6O. Em cảm ơn nhiều ạ

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1. $$\begin{array}{|l|cr|} \hline \qquad \quad CTPT & CTCT\\ \hline &H\quad H \quad H \quad H\\ \qquad \,\,\,\,C_4H_{10}&H-\mathop{C}\limits_{|}^{|}-\mathop{C}\limits_{|}^{|}-\mathop{C}\limits_{|}^{|}-\mathop{C}\limits_{|}^{|}-H\\\quad \quad \,\,(butane)&H\quad H \quad H \quad H\\ \hline &H\quad H\quad H\\ \qquad \quad C_4H_{10}&H-\mathop{C}\limits_{|}^{|}-\mathop{C}\limits_{|}^{|}-\mathop{C}\limits_{|}^{|}-H\\ \,\,\,\quad (iso-butane)&H\quad \,\,\Big|\,\,\quad H\\ &H -\mathop{C}\limits_{|}-H\\ &\quad H\quad \\ \hline &H\quad\quad\quad H\\ \qquad \,\,\,\,C_2H_6O&H-\mathop{C}\limits_{|}^{|}-O-\mathop{C}\limits_{|}^{|}-H\\ \quad (dimethyl\,\,ether)&H\quad\quad\quad H\\ \hline &H\quad H\quad\quad\\ \qquad \,\,\, C_2H_6O&H-\mathop{C}\limits_{|}^{|}-\mathop{C}\limits_{|}^{|}-O-H\\ \qquad (ethanol)&H\quad H\quad\quad\\ \hline &H\quad \quad\quad H\\ \qquad \,\,\,C_3H_6O&H-\mathop{C}\limits_{|}^{|}-\mathop{C}\limits_{\Vert}-\mathop{C}\limits_{|}^{|}-H\\ \qquad (acetone) &H\quad O\quad H\\ \hline &H\quad H\quad \quad\\ \qquad \,\,\,C_3H_6O&H-\mathop{C}\limits_{|}^{|}-\mathop{C}\limits_{|}^{|}-\mathop{C}\limits_{|}=O\\ \quad\,\,\,\,\,(propanal)&H\quad H\quad H\,\,\\ \hline &H\quad \quad \quad\quad \quad\quad \\ \qquad \,\,\,\,C_3H_6O&H-\mathop{C}\limits_{|}^{|}-O-\mathop{C}\limits_{|}=\mathop{C}\limits_{|}-H\\ (methyl\,\,vinyl\,\, ether)&H\quad \quad \quad H\quad H\,\,\\ \hline &\quad \quad H\quad \quad\quad \quad \\ \qquad \,\,\,\,C_3H_6O&H- O-\mathop{C}\limits_{|}^{|}-\mathop{C}\limits_{|}=\mathop{C}\limits_{|}-H\\ \quad \,\,(allyl\,\,alcohol)&\quad\,\, H\,\quad H\,\quad H \\ \hline \end{array}$$

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2. Đáp án:

   Bạn tham khảo lời giải ở dưới nhé!!!

   Giải thích các bước giải:

   \(\begin{array}{l}
   {C_4}{H_{10}}\\
   C{H_3} – C{H_2} – C{H_2} – C{H_3}\\
   C{H_3} – CH(C{H_3}) – C{H_3}\\
   {C_2}{H_6}O\\
   C{H_3} – C{H_2} – OH\\
   C{H_3} – O – C{H_3}\\
   {C_3}{H_6}O\\
   C{H_3} – C{H_2} – CHO\\
   C{H_2} = CH – C{H_2} – OH\\
   C{H_3} – C(O) – C{H_3}\\
   C{H_2} = CH – O – C{H_3}
   \end{array}\)

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