Question

Decreasing order of acidic strength of the following compounds is :

• A. $\left(z\right)>\left(y\right)>\left(x\right)$
• B. $\left(y\right)>\left(z\right)>\left(x\right)$
• C. $\left(z\right)>\left(x\right)>\left(y\right)$
• D. $\left(x\right)>\left(z\right)>\left(y\right)$

Answer 1

The correct option is C $\left(z\right)>\left(x\right)>\left(y\right)$

More the stability of conjugate base formed, more is the acidic strength of the compound :
For (x) :


After losing one $H^{+}$, it is observed that compound (x) has $6\pi$ electrons, hence obeys 4n+2 rule and becomes an aromatic compound.

For (y) :

After losing one $H^{+}$, it is observed that compound (y) has $8\pi$ electrons, hence obeys 4n rule and becomes an anti-aromatic compound.

For (z) :

After losing one $H^{+}$, it is observed that compound (z) has $6\pi$ electrons, hence obeys 4n+2 rule and becomes an aromatic compound.
Since, anti-aromatic compounds are the least stable, so (y) has the least acidic strength.
Between (x) and (z) :
Both are aromatic compounds, but in (z) , $-CN$ group attached shows -M effect and also -I which decrease the electron density on negative charge carbon, and thus stabilizes the conjugate base more as compared to (x), where no group is present.
So, acidic strength follows the order :
$\left(z\right)>\left(x\right)>\left(y\right)$