Step 1: Understanding the Concept:
The \(pK_{a}\) value is a measure of the acidity of a compound. It is defined as the negative logarithm of the acid dissociation constant (\(K_{a}\)):
\[ pK_{a} = -\log_{10} K_{a} \]
A stronger acid dissociates more completely in water, resulting in a larger \(K_{a}\) value and, consequently, a lower \(pK_{a}\) value.
To find the compound with the lowest \(pK_{a}\), we must identify the strongest acid among the choices.
Step 2: Detailed Explanation:
Acidity is determined by the stability of the conjugate base. Let's compare the substances:
1. Ethanol (\(C_{2}H_{5}OH\)) and Propan-2-ol (\((CH_{3})_{2CHOH\)):}
These are aliphatic alcohols. Their conjugate bases are alkoxide ions. Alkyl groups are electron-donating (\(+I\) effect), which increases the negative charge density on the oxygen, destabilizing the ion.
Thus, alcohols are very weak acids. Propan-2-ol is even weaker than ethanol due to greater \(+I\) effect from branching.
2. Water (\(H_{2}O\)):
Water forms the hydroxide ion (\(OH^{-}\)). It lacks the destabilizing \(+I\) alkyl groups found in alcohols.
Water is generally slightly more acidic than simple aliphatic alcohols (excluding methanol). Its \(pK_{a}\) is \(15.7\).
3. Phenol (\(C_{6}H_{5}OH\)):
Phenol is far more acidic than water and alcohols.
When phenol loses a proton, it forms the phenoxide ion (\(C_{6}H_{5}O^{-}\)).
In the phenoxide ion, the negative charge on the oxygen is delocalized into the benzene ring via resonance. The pi-electron system of the ring spreads the charge to the ortho and para positions.
This resonance stabilization makes the phenoxide ion much more stable than hydroxide or alkoxide ions.
Stable conjugate base \(\implies\) Stronger acid \(\implies\) Lower \(pK_{a}\).
Phenol has a \(pK_{a}\) of approximately \(10\), which is much lower than water (\(15.7\)) or ethanol (\(\approx 16\)).
Step 3: Final Answer:
Phenol is the strongest acid among the given options, so it has the lowest \(pK_{a}\) value.
This corresponds to Option (C).