Identify Compounds Soluble in Aqueous Sodium Hydroxide Solution

To determine the number of compounds soluble in aqueous $\text{NaOH}$ at room temperature, we analyze the acidity and chemical nature of each compound given in the image.

Compounds that are acidic enough to react with aqueous $\text{NaOH}$ (a strong base) will form water-soluble ionic salts (such as sodium carboxylates or sodium phenolates) and thus dissolve in the solution. These typically include:

1. Carboxylic acids ($\text{R-COOH}$), which have $pK_a \approx 4 - 5$.
2. Phenols ($\text{Ar-OH}$), which have $pK_a \approx 9 - 10$.

Now, let's evaluate each of the 9 given compounds:

1. Compound (I) - Benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$): It is a neutral carbonyl compound and does not react with aqueous $\text{NaOH}$ to form a soluble salt. $\rightarrow$ Insoluble

2. Compound (II) - 1-Naphthol ($\alpha$-naphthol): It contains a phenolic $-\text{OH}$ group ($pK_a \approx 9.3$). It reacts with $\text{NaOH}$ to form water-soluble sodium 1-naphtholate. $\rightarrow$ Soluble

3. Compound (III) - 4-Hydroxypyridine / 4-Pyridone: This compound exists predominantly in its 4-pyridone keto-tautomeric form (amide-like structure) and is not classified as a standard acidic phenolic compound that forms a sodium salt soluble in aqueous $\text{NaOH}$. $\rightarrow$ Insoluble

4. Compound (IV) - $p$-Cresol (4-methylphenol): It contains a phenolic $-\text{OH}$ group ($pK_a \approx 10.2$). It reacts with $\text{NaOH}$ to form a soluble sodium salt. $\rightarrow$ Soluble

5. Compound (V) - Benzoic acid ($\text{C}_6\text{H}_5\text{COOH}$): It is a carboxylic acid ($pK_a \approx 4.2$). It reacts readily with $\text{NaOH}$ to form soluble sodium benzoate. $\rightarrow$ Soluble

6. Compound (VI) - $N,N$-Dimethylcyclohexylamine: It is a basic tertiary amine and does not react with bases like $\text{NaOH}$. $\rightarrow$ Insoluble

7. Compound (VII) - 1-Naphthoic acid: It is an aromatic carboxylic acid. It reacts with $\text{NaOH}$ to form a soluble sodium carboxylate salt. $\rightarrow$ Soluble

8. Compound (VIII) - 1,4-Di-tert-butylbenzene: It is a non-polar hydrocarbon and is insoluble in aqueous solutions. $\rightarrow$ Insoluble

9. Compound (IX) - 2-Naphthol ($\beta$-naphthol): It contains a phenolic $-\text{OH}$ group ($pK_a \approx 9.5$). It reacts with $\text{NaOH}$ to form water-soluble sodium 2-naphtholate. $\rightarrow$ Soluble

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Conclusion:

The compounds that are soluble in aqueous $\text{NaOH}$ are:

• (II) 1-Naphthol
• (IV) $p$-Cresol
• (V) Benzoic acid
• (VII) 1-Naphthoic acid
• (IX) 2-Naphthol

Total number of soluble compounds = $5$.

Thus, the correct option is A.