Question

When 1-pentyne $\left(A\right)$ is treated with $4N$ alcoholic $KOH$ at $175$, it is converted slowly into an equilibrium mixture of 1.3% 1-pentyne $\left(A\right)$ 95.2% 2-pentyne $\left(B\right)$ and 3.5% of 1, 2-pentadiene $\left(C\right)$. The equilibrium was maintained at $175$. Calculate $\Delta G$ for the following equilibria.
$B\rightleftharpoons A$; $\Delta G_1=?$
$B\rightleftharpoons C$; $\Delta G_2=?$
From the calculate value of $\Delta G_1$ and $\Delta G_2$ indicate the order of stability of $A,B$ and $C$. Write a reasonable reaction mechanism showing all international leading to $A,B$ and $C$.

Answer 1

Pentyne-1 $\rightleftharpoons$ Pentyne-2 + Pentadiene-1-2
$\left(A\right)\left(B\right)\left(C\right)$
At equilibrium $1.395.23.5$
$K_{equilibrium}=\frac{\left[B\right]\left[C\right]}{\left[A\right]}=\frac{95.2\times 3.5}{1.3}=256.31$ ...(i)

Now for, $B\rightleftharpoons A$
$K_1=\frac{\left[A\right]}{\left[B\right]}$ ...(ii)

then from equations (i) and (ii)
$K_1=\frac{\left[C\right]}{K_{equilibrium}}=\frac{3.5}{256.31}=0.013$

$\therefore \Delta G=-2.303RT{\log }_{10}K$

$=-2.303\times 8.314\times 448\times \log 0.013$
$=16178J=16.178kJ$

Stability order for $A$ and $B$ is $B>A$

Similarly for $B\rightleftharpoons C$

$K_2=\frac{\left[C\right]}{\left[B\right]}=\frac{K_{equilibrium}\left[A\right]}{{\left[B\right]}^2}=\frac{256.31\times 1.3}{95.2\times 95.2}=0.037$

$\Delta G=-2.303RT{\log }_{10}K$

$=-2.303\times 8.314\times 448\times {\log }_{10}0.037$
$=12282J=12.282kJ$

Thus, stability order for $B$ and $C$ is $B>C$

The total stability order is $B>C>A$

The reaction is:
$C{H}_3\cdot C{H}_{2}C{H}_{2}C\equiv CH\stackrel{alc.KOH}{\to }[C{H}_{3}C{H}_{2}CH=C=C{H}_2]⟶[C{H}_{3}C{H}_{2}C\equiv C-C{H}_3]$