When `1`pentyne `(A)` is treated with `4N` alcoholic `KOH` at `175^(@)C`, it is slowly converted into an equilibrium mixture of `1.3%` of `1`pentyne `(A), 95.2% 2`-pentyne `(B)` and `3.5%` of `1,2`-pentandiene `(C )`. The equilibrium was maintained at `175^(@)C`. Calculate `DeltaG^(Theta)` for the following equilibria:
`B hArr A, DeltaG^(Theta)underset(1) = ?`
`B hArr C, DeltaG^(Theta)underset(2) =?`
From the calculated value of `DeltaG^(Theta)underset(1)`and `DeltaG^(Theta)underset(2)`, indicate the order of stability of `A,B` and `C`. Write a reasonable reaction mechanism showing all intermediates leading to `A,B` and `C`.

Explanation

`{:("Pentyne"-1hArr,"Pentyne"-2+,1.2-"pentadiene",,),((A),(B),(C),,),(t_(Eq)1.3,95.2,3.5,,):}`
`K_(Eq) = ([B][C])/([A]) = (95.2xx3.5)/(1.3) = 256.31`
`B hArr A`
`K_(1) = ([A])/([B]) = ([C])/(K_(Eq)) = (3.5)/(256.31) = 0.013`
`DeltaG^(Theta)underset(1).=-2.303 RT log_(10)K_(1)`
`= -2.303 xx 8.314 xx 448 log 0.013`
` = 1617J = 16.178 kJ`
for `B hArr C`
`K_(2) =([C])/([B]) = (K_(Eq)[A])/([B]^(2)) = (256.31xx1.3)/((9.5.2)^(2)) = 0.037`
`DeltaG^(Theta)underset(2). =- 2.303 Rt log_(10)K_(2)`
`=- 2.303 xx 8.314 xx 448 log 0.037`
` = 12282 J = 12.282 kJ`
Stability will lie in the order
`B gt C gt A`