Đáp án:
${C_6}{H_{10}}{O_2}$
Giải thích các bước giải:
Gọi CTPT của este là ${C_x}{H_y}{O_2}$
${n_{KOH}} = \dfrac{{33,6.25\% }}{{56}} = 0,15mol$
$ \Rightarrow {n_{este}} = {n_{KOH}} = 0,15mol$
$ \Rightarrow {M_{este}} = 12x + y + 32 = \dfrac{{17,1}}{{0,15}} = 114 \Rightarrow 12x + y = 82$
$ \Rightarrow x = 6;y = 10$ là nghiệm duy nhất
⇒ CTPT của X là ${C_6}{H_{10}}{O_2}$
CTCT có thể có của X:
$HCOOCH = CH - {{\text{[}}C{H_2}{\text{]}}_2} - C{H_3};HCOOCH = C(C{H_3}) - C{H_2} - C{H_3}$
$HCOOC{H_2} - CH = CH - {C_2}{H_5};HCOOC{H_2} - CH = C{(C{H_3})_2}$
$HCOOCH = CH - CH{(C{H_3})_2};HCOOC(C{H_3}) = CH - {C_2}{H_5}$
$HCOOC{H_2} - C(C{H_3}) = CH - C{H_3};HCOOCH(C{H_3}) - CH = CH - C{H_3}$
$HCOO - {[C{H_2}{\text{]}}_3} - CH = C{H_2};HCOOCH(C{H_3}) - C{H_2} - CH = C{H_2}$
$HCOOC{H_2} - CH(C{H_3}) - CH = C{H_2};HCOOC{H_2} - C{H_2} - C(C{H_3}) = C{H_2}$
$\begin{gathered}
C{H_3}COOCH = CH - C{H_2} - C{H_3};C{H_3}COOC{H_2} - CH = CH - C{H_3} \hfill \\
C{H_3}COO - C(C{H_3}) = CH - C{H_3};C{H_3}COOCH = C{(C{H_3})_2} \hfill \\
C{H_3}COOC{H_2} - C(C{H_3}) = C{H_2};{C_2}{H_5}COOCH = C{H_2} - C{H_3} \hfill \\
{C_2}{H_5}COOC{H_2} - CH = C{H_2};{C_2}{H_5}COOCH(C{H_3}) = C{H_2} \hfill \\
\end{gathered} $
$C{H_2} = CH - COOC{H_2} - C{H_2} - C{H_3};C{H_2} = CH - C{H_2}COOCH{(C{H_3})_2}$
$\begin{gathered}
C{H_2} = CH - C{H_2} - COO{C_2}{H_5};C{H_3} - C{H_2} = CH - COO{C_2}{H_5} \hfill \\
C{H_2} = C(C{H_3}) - COO{C_2}{H_5};C{H_3} - C{H_2} - C{H_2} - COOCH = C{H_2} \hfill \\
{(C{H_3})_2}CH - COOCH = C{H_2};C{H_3} - C{H_2} - CH = CH - COOC{H_3} \hfill \\
{(C{H_3})_2}C = CH - COOC{H_3};C{H_3} - CH = CH - C{H_2} - COOC{H_3} \hfill \\
C{H_2} = CH - C{H_2} - C{H_2} - COOC{H_3};C{H_3} - CH = C(C{H_3}) - COOC{H_3} \hfill \\
\end{gathered} $
