In the chemical reaction,
Aniline - (NaNO2 / HCl, 278 K) - A - (HBF4) - B
the compounds 'A' and

Q: Why do we need to maintain such a low temperature ($0-5^\circ\text{C}$) during diazotization?

Diazonium salts are highly unstable. If the temperature rises above $5^\circ\text{C}$, the diazonium salt will rapidly decompose by reacting with water in the solution to form phenol and release nitrogen gas.

Q: Why can't we just use direct fluorination to make fluorobenzene?

Direct fluorination of benzene using $F_2$ gas is extremely exothermic and violent. It is very difficult to control and usually results in polyfluorinated products or the complete cleavage of the carbon-carbon bonds in the benzene ring.

Q: What exactly is the intermediate formed before fluorobenzene in the second step?

When $HBF_4$ is added, an ion exchange occurs to form benzene diazonium fluoroborate ($C_6H_5N_2^+BF_4^-$). Unlike the chloride salt, this fluoroborate salt is relatively stable and precipitates out of the solution. It is then filtered, dried, and heated to yield fluorobenzene.

Q: Can we use $HF$ instead of $HBF_4$ to get fluorobenzene?

While theoretically possible, $HF$ is highly corrosive, toxic, and dangerous to handle. $HBF_4$ provides a much safer and more controlled way to form the fluoroborate salt, which decomposes cleanly upon heating.

Q: What happens if we warm the diazonium chloride solution without adding $HBF_4$?

If you warm the aqueous solution of benzene diazonium chloride, it will undergo hydrolysis. The $-N_2^+$ group will be replaced by an $-OH$ group from water, resulting in the formation of phenol.

Question

In the chemical reaction,

Options:

(A) nitrobenzene and fluorobenzene

(B) phenol and benzene

(C) benzene diazonium chloride and fluorobenzene

(D) nitrobenzene and chlorobenzene

Sigma Percentile · Accepted Answer

### The Challenge of Aromatic Fluorination

Introducing a fluorine atom directly onto a benzene ring is notoriously difficult. Direct fluorination using $F_2$ gas is highly exothermic, violent, and often leads to polyfluorination or even the complete destruction of the carbon skeleton. To achieve a clean, controlled fluorination, organic chemists rely on a brilliant two-step detour involving diazonium salts.

### Step 1: The Diazotization Gateway

The journey begins with **aniline** ($C_6H_5NH_2$), a primary aromatic amine. When aniline is treated with a mixture of sodium nitrite ($NaNO_2$) and hydrochloric acid ($HCl$) at a strictly controlled low temperature ($273-278	ext{ K}$ or $0-5^\circ	ext{C}$), a remarkable transformation occurs. The nitrous acid generated *in situ* reacts with the amino group, converting it into a highly reactive diazonium group ($-N_2^+$).

$$C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278	ext{ K}} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$$

This reaction is known as **diazotization**. The resulting product, **benzene diazonium chloride** (Compound A), is a versatile synthetic hub. The $-N_2^+$ group is an exceptional leaving group because it departs as stable nitrogen gas ($N_2$), driving the reaction forward thermodynamically.

### Step 2: The Balz-Schiemann Reaction

With our diazonium salt ready, we proceed to the second step. Benzene diazonium chloride is treated with fluoroboric acid ($HBF_4$). Initially, a simple ion exchange occurs, precipitating a stable, isolable salt called benzene diazonium fluoroborate ($C_6H_5N_2^+BF_4^-$).

$$C_6H_5N_2^+Cl^- + HBF_4 ightarrow C_6H_5N_2^+BF_4^- \downarrow + HCl$$

The magic happens when this dry precipitate is gently heated. It undergoes thermal decomposition in a controlled manner. The diazonium group leaves as nitrogen gas, and the fluoroborate ion donates a fluoride ion to the benzene ring, releasing boron trifluoride ($BF_3$) gas in the process.

$$C_6H_5N_2^+BF_4^- \xrightarrow{\Delta} C_6H_5F + N_2 \uparrow + BF_3 \uparrow$$

This elegant sequence is famously known as the **Balz-Schiemann reaction**. The final product is **fluorobenzene** (Compound B).

### Conclusion

By tracing the reaction pathway, we have successfully identified the intermediates and products. Compound A is benzene diazonium chloride, and Compound B is fluorobenzene. This perfectly aligns with the given options, making this a classic, high-yield concept for competitive exams.

Animated Solution for Chemistry - Organic Compounds Containing Nitrogen: In the chemical reaction,

Select Answer:

Visualized Solution

$Diazotization Reaction$

$Balz-Schiemann Reaction$

$Final Identification$

The Sigma Insight: Diazonium Salts

The Challenge of Aromatic Fluorination

Step 1

Step 2

Conclusion

Similar Questions

Toluene is nitrated and the resulting product is reduced with tin and hydrochloric acid. The product so obtained is diazotised and then heated with cuprous bromide. The reaction mixture so formed contains

Coupling of benzene diazonium chloride with 1-naphthol in alkaline medium will give

Consider the above reaction, compound $B$ is

What is the correct sequence of reagents used for converting nitrobenzene into m-dibromobenzene?

In the above chemical reaction, intermediate X and reagent / condition A are

In the chemical reactions given above A and B respectively are

The major product $Z$ obtained in the following reaction scheme is

In the reaction, The product E is