Which of the following reaction(s) will not give p-aminoazobenzene?
(a) A only
(b) C only
(c) B only

Q: Why doesn't $\text{NaBH}_4$ reduce nitrobenzene?

$\text{NaBH}_4$ is a source of hydride ions ($H^-$), which are nucleophiles. The nitro group has electron-rich oxygen atoms and lacks a suitable electrophilic center for hydride attack. Therefore, mild hydride donors like $\text{NaBH}_4$ are not used for this specific conversion. We need dissolving metal reduction (like Sn/HCl) or catalytic hydrogenation.

Q: What is the active electrophile in the diazotization reaction?

The active electrophile is the nitrosonium ion ($\text{NO}^+$). It is generated in situ by the reaction of sodium nitrite ($\text{NaNO}_2$) with a strong acid like $\text{HCl}$.

Q: Why does the coupling reaction with aniline occur at the para position?

The $-\text{NH}_2$ group is strongly activating and ortho/para directing. However, the diazonium ion is a relatively bulky electrophile. Steric hindrance makes the ortho position less favorable, so the attack predominantly occurs at the less hindered para position.

Q: Why is a slightly acidic medium used for coupling with aniline?

In a slightly acidic medium, a significant portion of aniline exists as the free base ($\text{Ph-NH}_2$), which is highly nucleophilic and reactive towards the diazonium ion. If the medium is too acidic, aniline is completely protonated to the anilinium ion ($\text{Ph-NH}_3^+$), which is strongly deactivating and will not undergo coupling.

Q: What happens if we use phenol instead of aniline for coupling?

If phenol is used, the coupling reaction is typically carried out in a slightly alkaline medium (pH 9-10). This converts phenol into the phenoxide ion, which is even more strongly activating than phenol itself, facilitating the electrophilic attack by the weak diazonium electrophile.

Question

Which of the following reaction(s) will not give $p$-aminoazobenzene?

Options:

(A) A only

(B) C only

(C) B only

(D) A and B

Sigma Percentile · Accepted Answer

### The Goal: Synthesizing a Dye

In the fascinating world of organic chemistry, synthesizing vibrant azo dyes is a classic and visually rewarding process. Our objective in this problem is to evaluate three distinct reaction sequences and identify the imposter—the sequence that fails to produce the brilliant yellow-orange dye known as **$p$-aminoazobenzene**.

To successfully synthesize an azo dye, we generally need two key components: a **diazonium salt** (the electrophile) and a highly activated aromatic ring (the nucleophile) to undergo a coupling reaction. Let's dissect each pathway to see if it meets these criteria.

### Analyzing Reaction A: The Classic Route

Reaction A begins with nitrobenzene ($	ext{Ph-NO}_2$). The first step employs tin and hydrochloric acid ($	ext{Sn/HCl}$). This is a robust dissolving metal reduction. The metal transfers electrons to the electron-deficient nitro group in the acidic medium, effectively stripping away the oxygen atoms and replacing them with hydrogen. The result? A pristine molecule of aniline ($	ext{Ph-NH}_2$).

Next, the aniline is treated with nitrous acid ($	ext{HNO}_2$) at low temperatures ($0-5^\circ	ext{C}$). This is the legendary **diazotization** reaction. The nitrous acid generates a highly reactive nitrosonium ion ($	ext{NO}^+$), which attacks the amine, ultimately transforming it into a benzene diazonium salt ($	ext{Ph-N}_2^+$).

Finally, a fresh batch of aniline is introduced. The diazonium ion, acting as a weak electrophile, seeks out an electron-rich target. The $-	ext{NH}_2$ group on the new aniline molecule strongly activates the ring, pushing electron density to the ortho and para positions. Due to the bulky nature of the diazonium group, steric hindrance blocks the ortho position, directing the attack almost exclusively to the para position. The result is a successful coupling, yielding our target: **$p$-aminoazobenzene**. Reaction A is a success!

### Analyzing Reaction C: A Shortcut

Reaction C is essentially a streamlined version of Reaction A. Instead of starting with nitrobenzene and reducing it, we begin directly with aniline.

The first step is identical: diazotization with $	ext{HNO}_2$ to form the benzene diazonium salt.

In the second step, aniline is added along with a catalytic amount of $	ext{HCl}$. The slightly acidic medium is crucial here. It ensures that the aniline remains largely as the free, nucleophilic base rather than being completely protonated to the unreactive anilinium ion. The coupling proceeds smoothly, just as in Reaction A, attacking the para position and forming **$p$-aminoazobenzene**. Reaction C is also a success!

### Analyzing Reaction B: The Trap

Now, let's examine Reaction B. Like Reaction A, it starts with nitrobenzene. However, the first reagent is sodium borohydride ($	ext{NaBH}_4$).

Here lies the critical trap! Sodium borohydride is a mild hydride donor. It is fantastic for reducing polarized double bonds like aldehydes and ketones. However, the nitro group is a different beast. Its oxygen atoms are electron-rich, and there is no suitable electrophilic center for the nucleophilic hydride ion ($H^-$) to attack. Consequently, **$	ext{NaBH}_4$ is completely unreactive towards nitro groups**.

Because the very first step fails to produce aniline, the entire sequence collapses. Adding $	ext{NaOH}$ and aniline later on accomplishes nothing because the essential diazonium electrophile was never generated.

### The Final Verdict

Reactions A and C successfully navigate the path of reduction, diazotization, and coupling to produce the desired azo dye. Reaction B, however, stumbles at the starting line due to an inappropriate choice of reducing agent. Therefore, the only reaction that will **not** give $p$-aminoazobenzene is Reaction B.

Animated Solution for Chemistry - Organic Compounds Containing Nitrogen: Which of the following reaction(s) will not give $p$ -aminoazobenzene?

Select Answer:

Visualized Solution

$O bj ec t i v e : Identify the non-yielding reaction$

$R e a c t i o n A : S t e p 1$

$R e a c t i o n A : S t e p 2$

$R e a c t i o n A : S t e p 3$

$R e a c t i o n C : S t e p 1$

$R e a c t i o n C : S t e p 2$

$R e a c t i o n B : S t e p 1$

$R e a c t i o n B : C o n c l u s i o n$

$F ina l A n s w er$

The Sigma Insight: Amines

The Goal

Analyzing Reaction A

Analyzing Reaction C

Analyzing Reaction B

The Final Verdict

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Animated Solution for Chemistry - Organic Compounds Containing Nitrogen: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which of the following reaction(s) will not give p-aminoazobenzene?