Animated Solution for Chemistry - Organic Chemistry: Silver mirror test is given by which one of the following compounds?

Imagine you are in a chemistry lab, holding a test tube containing an unknown clear liquid. Your task is to determine whether this liquid is an aldehyde or a ketone. How do you do it? Enter the legendary Silver Mirror Test, also known as Tollen's test.

This test is a beautiful demonstration of redox chemistry. Tollen's reagent is a mild oxidizing agent, essentially an ammoniacal solution of silver nitrate containing the diamminesilver(I) complex ion, $[Ag(N{H}_3{)}_2{]}^{+}$. The core principle is simple: aldehydes are relatively easy to oxidize because they have a hydrogen atom directly attached to the carbonyl carbon. Ketones, however, have two carbon groups attached to the carbonyl carbon, making them highly resistant to mild oxidation.

When an aldehyde is warmed with Tollen's reagent, it gets oxidized to a carboxylate ion. Simultaneously, the silver ions ($A{g}^{+}$) are reduced to elemental silver ($Ag$). This elemental silver precipitates out of the solution and coats the inner surface of the glass test tube, creating a stunning, reflective silver mirror.

Let's put our four candidates under the microscope to see who passes the test.

1. Acetaldehyde ($C{H}_{3}CHO$) If we look at the structure of acetaldehyde, we clearly see the defining feature of an aldehyde: the $-CHO$ group. Because it possesses that crucial, easily oxidizable hydrogen atom on the carbonyl carbon, it will readily react with Tollen's reagent.

2. Acetone ($C{H}_{3}COC{H}_3$) Next up is acetone. Notice its structure carefully. The carbonyl carbon is sandwiched right between two methyl groups. There is absolutely no hydrogen directly attached to it. This structural reality makes it a ketone. Since ketones resist mild oxidation, acetone will simply sit in the Tollen's reagent without reacting. No silver mirror for acetone.

3. Formaldehyde ($HCHO$) Moving on to formaldehyde, the simplest of all aldehydes. It not only has one, but technically two hydrogen atoms attached to the carbonyl carbon. It is an exceptionally reactive aldehyde. Without a doubt, formaldehyde will rapidly reduce the silver ions and produce a brilliant silver mirror.

4. Benzophenone ($PhCOPh$) Finally, let's examine benzophenone. Here, the carbonyl carbon is flanked by two bulky, electron-rich phenyl rings. It is a classic aromatic ketone. Just like acetone, it completely lacks the oxidizable hydrogen atom. Therefore, it will fail the silver mirror test.

The general chemical equation for this elegant transformation is:

Based on our structural analysis, only the aldehydes possess the chemical architecture required to drive this reaction forward. Therefore, both Acetaldehyde and Formaldehyde will successfully give the silver mirror test.