Consider the following reaction sequence, the correct option(s) is(are) (A) P = H2/Pd, ethanol R = N

Q: Why can't we just oxidize the methyl group first using KMnO4?

If you oxidize p-nitrotoluene first, you get p-nitrobenzoic acid. While this is a valid reaction, it doesn't match the intermediate $\mathbf{Q}$ which then undergoes diazotization. Carboxylic acids don't form diazonium salts!

Q: What is the difference between using H3PO2 and Ethanol for deamination?

Both are effective mild reducing agents used to replace a diazonium group with a hydrogen atom. $\text{H}_3\text{PO}_2$ gets oxidized to $\text{H}_3\text{PO}_3$, while ethanol gets oxidized to acetaldehyde. In the context of this problem, they are interchangeable for step $\mathbf{U}_1$.

Q: Why does water convert the diazonium salt to a phenol instead of just dissolving it?

While diazonium salts are soluble in cold water, warming the solution provides enough energy for the excellent leaving group ($-\text{N}_2^\oplus$) to depart as nitrogen gas. The resulting highly unstable phenyl cation is immediately attacked by water, forming a phenol.

Q: Could we use LiAlH4 instead of Sn/HCl for step P?

No! $\text{LiAlH}_4$ is a powerful reducing agent, but when it reacts with nitroarenes, it typically produces azo compounds (like azobenzene) rather than primary amines. $\text{Sn/HCl}$ or catalytic hydrogenation ($\text{H}_2\text{/Pd}$) are the correct choices for cleanly getting the amine.

Q: Why is KOH used alongside KMnO4 in the final oxidation step?

Alkaline $\text{KMnO}_4$ is a standard, highly effective reagent for side-chain oxidation of alkylbenzenes. The alkaline medium helps in the oxidative cleavage, initially forming the potassium salt of the carboxylic acid, which is then acidified to yield the free benzoic acid.

Question

Consider the following reaction sequence,

the correct option(s) is(are)

Options:

(A) P = $	ext{H}_2	ext{/Pd}$, ethanol \quad R = $	ext{NaNO}_2	ext{/HCl}$ \quad U = 1. $	ext{H}_3	ext{PO}_2$, 2. $	ext{KMnO}_4	ext{ - KOH}$, heat

(B) P = Sn/HCl \quad R = $	ext{HNO}_2$

(C) U = 1. $	ext{CH}_3	ext{CH}_2	ext{OH}$, 2. $	ext{KMnO}_4	ext{ - KOH}$, heat

(D) R = $	ext{H}_2	ext{/Pd}$, ethanol

Sigma Percentile · Accepted Answer

### The Beauty of Sequential Synthesis

Organic chemistry is often like a highly strategic game of chess. Every move must be calculated, and the order of operations is absolutely critical. In this problem, we are presented with a fascinating reaction sequence starting from **p-nitrotoluene**. Our mission is to decode the identity of the reagents and intermediates as we navigate through a series of transformations.

Let's begin by carefully examining our starting material. We have a benzene ring adorned with two functional groups sitting opposite to each other: a methyl group ($-	ext{CH}_3$) and a nitro group ($-	ext{NO}_2$). Our goal is to trace how these groups transform through a series of carefully chosen reagents.

### Step 1: The Selective Reduction

The first transformation involves reagent $\mathbf{P}$, which converts our starting material into product $\mathbf{Q}$. If we peek at the options provided, $\mathbf{P}$ is suggested to be either catalytic hydrogenation ($	ext{H}_2	ext{/Pd}$) or a metal-acid reduction ($	ext{Sn/HCl}$).

Both of these are classic, powerful reducing agents. What is their primary target here? They selectively reduce the nitro group into a primary amine group ($-	ext{NH}_2$), leaving the robust methyl group completely untouched.

Therefore, the product $\mathbf{Q}$ must be **p-toluidine**.

### Step 2: The Magic of Diazotization

Moving forward, our newly formed primary aromatic amine $\mathbf{Q}$ reacts with reagent $\mathbf{R}$ to form intermediate $\mathbf{S}$. The options suggest $\mathbf{R}$ is nitrous acid ($	ext{HNO}_2$), typically generated in situ from sodium nitrite and hydrochloric acid ($	ext{NaNO}_2	ext{/HCl}$) at ice-cold temperatures ($0-5^\circ	ext{C}$).

This is a textbook **diazotization** reaction! The amine group is converted into a highly reactive diazonium salt.

Consequently, $\mathbf{S}$ is identified as **p-toluenediazonium chloride**.

### Step 3: Hydrolysis to Phenol

Next, let's observe what happens when we treat our diazonium salt $\mathbf{S}$ with warm water ($	ext{H}_2	ext{O}$). The diazonium group ($-	ext{N}_2^\oplus$) is a phenomenal leaving group because it departs as highly stable nitrogen gas.

When warmed with water, the diazonium group leaves, and a hydroxyl group ($-	ext{OH}$) takes its place on the aromatic ring via a nucleophilic aromatic substitution mechanism.

This hydrolysis yields product $\mathbf{T}$, which is **p-cresol**.

### Step 4: The Grand Finale - Deamination and Oxidation

Now for the most fascinating part of the sequence. We need to convert the diazonium salt $\mathbf{S}$ all the way to **benzoic acid** using reagent $\mathbf{U}$. Notice the two massive structural changes here: the diazonium group has completely vanished, and the methyl group has been oxidized to a carboxylic acid.

This transformation cannot happen in a single step! It requires a strategic two-step sequence:

1.  **Deamination:** First, we must remove the diazonium group. We achieve this using a mild reducing agent like hypophosphorous acid ($	ext{H}_3	ext{PO}_2$) or ethanol ($	ext{CH}_3	ext{CH}_2	ext{OH}$). These reagents replace the diazonium group with a hydrogen atom, effectively giving us **toluene**.
2.  **Side-Chain Oxidation:** Then, we hit the toluene with a powerful oxidizing agent like alkaline potassium permanganate with heat ($	ext{KMnO}_4	ext{/KOH}, \Delta$). This aggressively oxidizes the methyl side chain directly into a carboxyl group ($-	ext{COOH}$), yielding the final product, **benzoic acid**.

### Conclusion

Evaluating the given options based on our derived sequence:
-   **Option (A)** perfectly matches our reagents for $\mathbf{P}$, $\mathbf{R}$, and the two-step process for $\mathbf{U}$.
-   **Option (B)** correctly identifies $\mathbf{P}$, $\mathbf{R}$, and the structure of the diazonium salt $\mathbf{S}$.
-   **Option (C)** correctly identifies $\mathbf{S}$, $\mathbf{T}$, and an alternative valid two-step sequence for $\mathbf{U}$ using ethanol.
-   **Option (D)** incorrectly claims $\mathbf{Q}$ is p-nitrobenzoic acid.

Thus, the correct options are **(A), (B), and (C)**.

Animated Solution for Chemistry - Organic Chemistry: Consider the following reaction sequence, the correct option(s) is(are)

Select Answer:

Visualized Solution

$A na l y z in g t h e R e a c t an t$

$S t e p P : R e d u c t i o n o f N i t r o G r o u p$

$S t e p R : D ia z o t i z a t i o n$

$S t e p H_{2} O : H y d r o l y s i s$

$S t e p U : D e amina t i o n an d O x i d a t i o n$

$E v a l u a t in g t h e O pt i o n s$

$T h e W a y F or w a r d$

The Sigma Insight: Amines

The Beauty of Sequential Synthesis

Step 1

Step 2

Step 3

Step 4

Conclusion

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