Animated Solution for Chemistry - Organic Chemistry: Consider the following reaction : The product ‘P’ gives positive ceric ammonium nitrate test. This is because of the presence of which of these $\text{—OH}$ group(s)?

The journey to solving this problem begins with a careful examination of the complex, steroid-like reactant molecule. At first glance, it might look intimidating, but the key is to break it down into its functional groups.

We are presented with four distinct hydroxyl ($-OH$) groups, labeled $\stackrel{◯}{a}$, $\stackrel{◯}{b}$, $\stackrel{◯}{c}$, and $\stackrel{◯}{d}$. Let's classify each one based on the carbon atom it's attached to.

Group $\stackrel{◯}{a}$ is located at the end of an alkyl side chain. The carbon it's attached to is bonded to only one other carbon, making it a primary ($1^{\circ }$) alcohol.

Group $\stackrel{◯}{b}$ is situated on the cyclopentane ring. The carbon holding this $-OH$ group is also bonded to two ring carbons and a methyl ($-C{H}_3$) group. Being bonded to three carbons makes this a tertiary ($3^{\circ }$) alcohol.

Group $\stackrel{◯}{c}$ is on a cyclohexane ring. The carbon it's attached to is bonded to two other ring carbons, classifying it as a secondary ($2^{\circ }$) alcohol.

Finally, group $\stackrel{◯}{d}$ is directly attached to an aromatic benzene ring. This specific arrangement is known as a phenol.

The problem states that this molecule is treated with chromic anhydride ($Cr{O}_3$). Chromic anhydride is a robust and aggressive oxidizing agent. To predict the product 'P', we must recall the standard oxidation rules for alcohols.

When treated with a strong oxidizing agent like $Cr{O}_3$: - Primary alcohols are oxidized all the way to carboxylic acids ($-COOH$). - Secondary alcohols are oxidized to ketones ($>C=O$). - Tertiary alcohols lack a hydrogen atom on the carbinol carbon and are therefore resistant to oxidation under normal conditions. - Phenols can sometimes undergo complex oxidations, but in standard textbook problems like this, they are generally considered to remain intact unless specified otherwise.

Now, let's apply these rules to our specific molecule to deduce the structure of product 'P'.

The primary alcohol, group $\stackrel{◯}{a}$, will be oxidized to a carboxylic acid.

The secondary alcohol, group $\stackrel{◯}{c}$, will be oxidized to a ketone.

The tertiary alcohol, group $\stackrel{◯}{b}$, and the phenol, group $\stackrel{◯}{d}$, will remain completely unchanged.

The final piece of the puzzle lies in the Ceric Ammonium Nitrate (CAN) test. The problem tells us that product 'P' gives a positive CAN test.

The CAN test is a classic qualitative chemical test used specifically to detect the presence of aliphatic alcohols. When an aliphatic alcohol reacts with the CAN reagent, it forms a characteristic red-colored complex. It is crucial to remember that phenols do not produce this specific red color; they typically yield brown or black precipitates.

Let's look at our newly formed product 'P'. The primary and secondary alcohols have been destroyed by oxidation. The only aliphatic alcohol that survived the chromic anhydride treatment is the tertiary alcohol, group $\stackrel{◯}{b}$.

Therefore, the positive ceric ammonium nitrate test observed for product 'P' is entirely due to the presence of the unoxidized tertiary alcohol, group $\stackrel{◯}{b}$.

This leads us to the final conclusion that the correct option is (c).