Xà phòng hóa 17,1(g) 1 este đơn chức X cần 33,6g dd KOH 25% đủ. Xác định CTCT có thể có của X Giúp mình với ạ 08/07/2021 Bởi Sarah Xà phòng hóa 17,1(g) 1 este đơn chức X cần 33,6g dd KOH 25% đủ. Xác định CTCT có thể có của X Giúp mình với ạ
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} $ Bình luận
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} $