Question

The decreasing order of acidic characters of the following is:
I. $CH\equiv CH$
II.
III.

• A. I > II > III
• B. II > I > III
• C. III > II > I
• D. I > III > II

Answer 1

The correct option is B II > I > III

Conjugate base of I, II and III:
I. $\begin{array}{c}HC\equiv C^{\ominus }\\ \left(\begin{array}{c}\text{Negative charge on}\\ \text{sp C atom}\end{array}\right)\end{array}$
sp carbons have 50% s character. More the s- character, more will be the electronegativity. Hence it will stabilise the negative charge on carbon.

(II) In this conjugate base, it has $6\pi$ electrons which will obey the 4n+2 rule. Thus it is aromatic in nature. Hence it stabilise the conjugate base more than others.
(III)

In this conjugate base, it has $4\pi$ electrons which will obey 4n $\pi$ electron rule. Hence it is anti-aromatic and it destabilise the conjugate base more.

Stability of conjugate base $\propto$ acidic character
Hence, the decreasing acidic character follows II > I > III