Animated Solution for Chemistry - Organic Chemistry: The correct statements(s) about of the following reaction sequence is(are)

The journey through this organic chemistry problem is like unravelling a beautifully crafted mystery. We are given a sequence of reactions starting from Cumene, and our goal is to identify the intermediates $P$, $Q$, $R$, and the final product $S$, while simultaneously testing their chemical properties. Let's dive into the molecular transformations!

The Birth of Phenol We begin with Cumene, which is isopropylbenzene. The first reaction involves treating cumene with oxygen ($O_2$) followed by acidic hydrolysis ($H_3{O}^{+}$).

This is the famous Cumene Hydroperoxide Rearrangement, a highly efficient industrial method for synthesizing phenol. The oxygen inserts into the benzylic C-H bond to form cumene hydroperoxide. Upon acid treatment, a beautiful rearrangement occurs, cleaving the molecule into two valuable products: Phenol and acetone.

Thus, our intermediate $P$ is Phenol.

The Reimer-Tiemann Reaction Next, Phenol ($P$) is treated with chloroform ($CHC{l}_3$) and sodium hydroxide ($NaOH$).

This is the classic Reimer-Tiemann reaction. The strong base reacts with chloroform to generate a highly reactive intermediate called dichlorocarbene ($:CC{l}_2$). This electrophile attacks the electron-rich phenoxide ring.

The reaction introduces an aldehyde ($-CHO$) group onto the aromatic ring. It yields two isomers: 1. Salicylaldehyde (ortho-hydroxybenzaldehyde), which is the major product due to the stabilizing effect of intramolecular hydrogen bonding. This is our product $Q$. 2. p-Hydroxybenzaldehyde, which is the minor product. This is our product $R$.

Analyzing Steam Volatility (Option A) Now, let's evaluate Option A, which claims that $R$ is steam volatile.

To understand steam volatility, we must look at hydrogen bonding. In molecule $Q$ (salicylaldehyde), the $-OH$ and $-CHO$ groups are adjacent to each other. They form an intramolecular hydrogen bond (a bond within the same molecule). This internal bonding prevents the molecules from strongly associating with one another, giving $Q$ a lower boiling point and making it steam volatile.

Conversely, in molecule $R$ (p-hydroxybenzaldehyde), the groups are too far apart for internal bonding. Instead, they form intermolecular hydrogen bonds with neighboring molecules, creating a massive, sticky network. This drastically raises its boiling point, making $R$ not steam volatile.

Therefore, Option A is incorrect.

The Ferric Chloride Test (Option B) Option B states that $Q$ gives a dark violet coloration with $1\%$ aqueous $FeC{l}_3$ solution.

The neutral Ferric Chloride ($FeC{l}_3$) test is a classic analytical test for the presence of a free phenolic $-OH$ group. Since $Q$ is salicylaldehyde, it possesses an intact phenolic hydroxyl group. When it reacts with $FeC{l}_3$, it forms a deeply colored violet coordination complex with the iron ion.

Thus, Option B is absolutely correct!

Williamson Ether Synthesis Let's move forward in the sequence to find $S$. Molecule $Q$ is treated with $NaOH$ and benzyl bromide ($PhC{H}_{2}Br$).

First, the strong base $NaOH$ deprotonates the acidic phenolic $-OH$ of $Q$, generating a nucleophilic phenoxide ion. Next, this phenoxide ion executes a textbook $S_{N}2$ attack on the electrophilic carbon of benzyl bromide, kicking out the bromide leaving group.

This is the Williamson Ether Synthesis. The resulting molecule, $S$, is 2-(benzyloxy)benzaldehyde. The crucial change here is that the free $-OH$ group has been converted into an ether linkage.

The 2,4-DNP Test (Option C) Option C claims that $S$ gives a yellow precipitate with 2,4-dinitrophenylhydrazine (2,4-DNP).

Brady's reagent, or 2,4-DNP, is the ultimate test for carbonyl groups (specifically aldehydes and ketones). If we look at the structure of $S$, the aldehyde ($-CHO$) group from the Reimer-Tiemann reaction is still perfectly intact!

When $S$ reacts with 2,4-DNP, it undergoes a condensation reaction to form a bright yellow-orange hydrazone precipitate.

Therefore, Option C is correct.

Re-evaluating the Ferric Chloride Test (Option D) Finally, Option D suggests that $S$ will also give a dark violet coloration with $FeC{l}_3$.

Recall our discussion from Step 4: the $FeC{l}_3$ test strictly requires a free phenolic $-OH$ group. In molecule $S$, that hydroxyl group was consumed during the Williamson ether synthesis to form a benzyl ether.

Since there is no longer a free phenol group, $S$ will not react with $FeC{l}_3$.

Therefore, Option D is incorrect.

The correct statements are indeed (B) and (C).