Step 1: Understanding the Concept:
Solubility in aqueous $NaOH$ depends on the acidity of the compound. $NaOH$ is a strong base; therefore, compounds that are significantly more acidic than water (such as phenols, carboxylic acids, and sulfonic acids) will react with $NaOH$ to form water-soluble salts.
Step 2: Key Formula or Approach:
1. Identify the functional groups: Carboxylic acids ($-COOH$), Phenols ($-OH$ on benzene), and Sulfonic acids ($-SO_3H$) are typically soluble.
2. Alcohols ($-OH$ on alkyl chains) and simple amines are generally not acidic enough to react with $NaOH$.
Step 3: Detailed Explanation:
1. In a standard set for this question (usually containing Phenol, Benzoic acid, o-Cresol, and $p$-Nitrophenol), these four would react with $NaOH$.
2. Phenol forms Sodium Phenoxide ($C_6H_5ONa$).
3. Benzoic acid forms Sodium Benzoate ($C_6H_5COONa$).
4. $p$-Nitrophenol is even more acidic than phenol due to the electron-withdrawing nitro group and dissolves readily.
5. If the list includes alcohols like Benzyl alcohol or Cyclohexanol, they will remain insoluble as they do not form stable salts with $NaOH$.
Step 4: Final Answer:
The total number of soluble compounds is 4.
