Answer
See the explanation
Work Step by Step
To explain the acidities of the given compounds, we can consider the role of resonance in stabilizing the conjugate base formed after the loss of the acidic hydrogen.
a. $CH_{3}-CH=C-CH_{3}$
In this compound, there is no resonance stabilization possible in the conjugate base formed after the loss of the acidic hydrogen. The conjugate base would be a simple carbanion, which is not particularly stable. Therefore, this compound would be a relatively weak acid.
c. $C_{6}H_{5}-OH^{\ast}$
In this compound, the phenol group ($C_{6}H_{5}-OH$) exhibits resonance stabilization in the conjugate base formed after the loss of the acidic hydrogen. The negative charge can be delocalized throughout the aromatic ring, resulting in a more stable conjugate base. This resonance stabilization makes the phenol group a relatively stronger acid compared to the compound in part (a).
The presence of resonance in the conjugate base is a key factor in explaining the relative acidities of these compounds. Resonance stabilization of the conjugate base leads to increased acidity, as the more stable the conjugate base, the more readily the proton is released from the parent compound.
