Animated Solution for Chemistry - Organic Chemistry: Which among the above compound/s does/do not form silver mirror when treated with Tollen's reagent?

Tollen's test is one of the most visually stunning and conceptually important reactions in organic chemistry. Tollen's reagent, which is an ammoniacal solution of silver nitrate ${\left[\text{Ag}({\text{NH}}_3{)}_2\right]}^{+}$, acts as a mild oxidizing agent. Its primary role is to selectively oxidize aldehydes into their corresponding carboxylic acids. During this process, the silver ions (${\text{Ag}}^{+}$) are reduced to elemental silver (${\text{Ag}}^0$), which deposits on the inner walls of the test tube, creating a beautiful silver mirror.

The golden rule of Tollen's test is simple: Aldehydes give a positive test, while ketones generally do not. However, as with all things in chemistry, the devil is in the details, and there are fascinating exceptions to this rule. Let's analyze the four compounds given in the problem to see how they behave.

The Straightforward Aldehyde

Compound (I) is $p$-hydroxybenzaldehyde. If you look closely at its structure, you will immediately spot the $-\text{CHO}$ group attached to the benzene ring.

Because it possesses a free, unhindered aldehyde group, it behaves exactly as expected. It will readily undergo oxidation by Tollen's reagent, yielding a positive silver mirror test. This one is a classic, textbook example.

The Deceptive Enol

Compound (II) is cyclohex-1-en-1-ol. At first glance, you might just see an $-\text{OH}$ group and a double bond. This specific arrangement is known as an enol (alkene + alcohol).

Enols are notoriously unstable and exist in a dynamic equilibrium with their keto forms through a process called tautomerization. In this case, cyclohex-1-en-1-ol rapidly tautomerizes into its much more stable keto counterpart, cyclohexanone.

Since cyclohexanone is a simple ketone, it lacks the easily oxidizable hydrogen atom that aldehydes possess. Therefore, it completely resists oxidation by Tollen's reagent. Compound (II) will not form a silver mirror.

The Special Case of $\alpha$-Hydroxy Ketones

Compound (III) is 2-hydroxycyclohexanone. This is an $\alpha$-hydroxy ketone, meaning it has a hydroxyl group on the carbon atom immediately adjacent to the carbonyl group.

Here is where the magic happens! While normal ketones fail Tollen's test, $\alpha$-hydroxy ketones are a famous exception. Under the alkaline conditions of Tollen's reagent, these molecules are highly susceptible to oxidation. They are easily oxidized into 1,2-diketones. Because they undergo this oxidation so readily, they successfully reduce the silver ions in the reagent.

Thus, despite being a ketone initially, compound (III) gives a positive silver mirror test.

The Hidden Aldehyde

Compound (IV) is a cyclic hemiacetal (specifically, tetrahydro-2H-pyran-2-ol). Hemiacetals are formed by the intramolecular reaction of an alcohol with an aldehyde.

In an aqueous solution, cyclic hemiacetals are not static; they exist in a dynamic equilibrium with their open-chain forms. When the ring of compound (IV) opens up, it reveals a free, reactive aldehyde group at one end of the chain.

Because this free aldehyde group is present in the equilibrium mixture, it will react with Tollen's reagent, driving the equilibrium forward until all the hemiacetal is consumed. Consequently, compound (IV) also gives a positive silver mirror test.

To summarize our findings: - Compound (I) is an aldehyde $\to$ Positive - Compound (II) tautomerizes to a simple ketone $\to$ Negative - Compound (III) is an $\alpha$-hydroxy ketone $\to$ Positive - Compound (IV) is a hemiacetal that opens to an aldehyde $\to$ Positive

The question specifically asks which compound does not form a silver mirror. Based on our rigorous analysis, only compound (II) fails the test. Therefore, the correct option is (c).