Animated Solution for Chemistry - Organic Chemistry: An organic compound X showing the following solubility profile is

Imagine you are a chemical detective, and you've just been handed a vial containing a mysterious organic compound, labeled simply as 'Compound X'. Your mission, should you choose to accept it, is to identify this compound using nothing but its solubility profile. This is a classic problem in practical organic chemistry, where we use the acid-base properties of functional groups to unmask their identity.

Our first clue is that Compound X is insoluble in water. This tells us it's likely a relatively large organic molecule, as small polar molecules tend to dissolve in water. Next, we see it's insoluble in $5\%$ $\text{HCl}$. Hydrochloric acid is a strong acid, and it typically dissolves basic organic compounds, like amines, by protonating them to form water-soluble salts. Since X doesn't dissolve, we can confidently say it's not a base.

The plot thickens when we test it with $10\%$ $\text{NaOH}$. Ah! It dissolves! Sodium hydroxide is a strong base, which means our Compound X must be acidic. It reacts with the $\text{NaOH}$ to form a water-soluble sodium salt.

But wait, there's a twist. When we test it with $10\%$ ${\text{NaHCO}}_3$, a weak base, it remains insoluble. This is the crucial piece of evidence. It means Compound X is a weak acid—strong enough to react with $\text{NaOH}$, but not strong enough to react with ${\text{NaHCO}}_3$. Specifically, it must be a weaker acid than carbonic acid (${\text{H}}_2{\text{CO}}_3$), because a weaker acid cannot displace a stronger acid from its salt.

Now, let's bring in our suspects: $o$-toluidine, oleic acid, $m$-cresol, and benzamide.

First up is $o$-toluidine. It contains an amino group ($-{\text{NH}}_2$) attached to the benzene ring. This makes it basic. It would easily dissolve in $5\%$ $\text{HCl}$, which contradicts our profile. So, $o$-toluidine is out.

Next, we have oleic acid. As the name suggests, it has a carboxylic acid group ($-\text{COOH}$). Carboxylic acids are relatively strong organic acids. They will dissolve in both strong bases like $\text{NaOH}$ and weak bases like ${\text{NaHCO}}_3$. Since X is insoluble in sodium bicarbonate, oleic acid is also incorrect.

Let's look at benzamide. The amide group ($-{\text{CONH}}_2$) is essentially neutral due to resonance stabilization. It won't react with dilute acids or dilute bases at room temperature. So, it would be insoluble in both $\text{HCl}$ and $\text{NaOH}$. But our compound dissolves in $\text{NaOH}$. So, benzamide is not the answer.

Finally, let's examine $m$-cresol. It's a phenol derivative, meaning it has a hydroxyl group ($\text{-OH}$) directly attached to the benzene ring. Phenols are weakly acidic. They are acidic enough to react with a strong base like $\text{NaOH}$ to form a water-soluble sodium phenoxide salt.

However, they are weaker acids than carbonic acid, so they cannot react with sodium bicarbonate. This perfectly matches our solubility profile! By understanding the subtle differences in acidity between carboxylic acids and phenols, we've successfully cracked the case. The mystery compound is indeed $m$-cresol.