Bottles containing C6H5I and C6H5CH2I lost their original labels. They were labelled A and B for tes

Q: Why doesn't iodobenzene react with NaOH even when boiled?

In iodobenzene, the lone pairs on the iodine atom are in conjugation with the pi-electron system of the benzene ring. This resonance gives the carbon-iodine bond a partial double bond character, making it significantly stronger and shorter than a typical single bond. Therefore, it is very difficult to break under normal nucleophilic substitution conditions.

Q: Why does benzyl iodide react so easily?

Benzyl iodide is an alkyl halide where the iodine is attached to an $sp^3$ hybridized carbon. The carbon-iodine bond is a standard single bond. Moreover, the resulting benzyl carbocation (if it goes via $S_N1$) or the transition state (if $S_N2$) is highly stabilized by resonance with the adjacent benzene ring, making the substitution reaction very favorable.

Q: What exactly is the yellow precipitate?

The yellow precipitate is silver iodide ($\text{AgI}$). It forms when free silver ions ($\text{Ag}^+$) from the silver nitrate reagent combine with free iodide ions ($\text{I}^-$) in the solution.

Q: Why must we add dilute $\text{HNO}_3$ before adding $\text{AgNO}_3$?

The initial reaction was carried out in a basic medium (NaOH). If we add $\text{AgNO}_3$ directly to a basic solution, the silver ions will react with the hydroxide ions to form a brown precipitate of silver oxide ($\text{Ag}_2\text{O}$), which would interfere with our observation. Nitric acid neutralizes the excess base, preventing this side reaction.

Q: Could we use $\text{HCl}$ or $\text{H}_2\text{SO}_4$ instead of $\text{HNO}_3$ to acidify the solution?

No, we cannot use $\text{HCl}$ because it contains chloride ions, which would react with silver nitrate to form a white precipitate of $\text{AgCl}$, confusing the results. $\text{H}_2\text{SO}_4$ is also generally avoided as silver sulfate is only sparingly soluble and might precipitate. Nitric acid is ideal because all nitrates are soluble.

Question

Bottles containing $	ext{C}_6	ext{H}_5	ext{I}$ and $	ext{C}_6	ext{H}_5	ext{CH}_2	ext{I}$ lost their original labels. They were labelled $A$ and $B$ for testing. $A$ and $B$ were separately taken in test tubes and boiled with NaOH solution. The end solution in each tube was made acidic with dilute $	ext{HNO}_3$ and then some $	ext{AgNO}_3$ solution was added. Substance $B$ gave a yellow precipitate. Which one of the following statements is true for this experiment?

Options:

(A) $A$ was $	ext{C}_6	ext{H}_5	ext{I}$

(B) $A$ was $	ext{C}_6	ext{H}_5	ext{CH}_2	ext{I}$

(C) $B$ was $	ext{C}_6	ext{H}_5	ext{I}$

(D) Addition of $	ext{HNO}_3$ was unnecessary

Sigma Percentile · Accepted Answer

The Mystery of the Missing Labels: Aryl vs. Alkyl Halides

Imagine walking into a chemistry lab and finding two bottles with missing labels. You know one contains **iodobenzene** ($	ext{C}_6	ext{H}_5	ext{I}$) and the other contains **benzyl iodide** ($	ext{C}_6	ext{H}_5	ext{CH}_2	ext{I}$). How do you tell them apart? This classic problem tests your understanding of the fundamental differences in reactivity between aryl halides and alkyl halides.

Let's break down the elegant chemical detective work used to solve this mystery.

### The Setup: Two Unlabeled Bottles

We start by labeling our mystery bottles as $A$ and $B$. To figure out their true identities, we subject both to a chemical interrogation. The first step is to boil samples from both bottles with an aqueous solution of sodium hydroxide ($	ext{NaOH}$).

Sodium hydroxide is a strong base that dissociates to give hydroxide ions ($	ext{OH}^-$). These hydroxide ions are excellent nucleophiles, always on the hunt for an electron-deficient carbon atom to attack. We are essentially trying to force a **nucleophilic substitution reaction** to kick out the iodine atom as an iodide ion ($	ext{I}^-$).

### The Boiling NaOH Test

Let's analyze how our two suspects react to this interrogation.

First, consider **iodobenzene** ($	ext{C}_6	ext{H}_5	ext{I}$). In this molecule, the iodine atom is directly attached to the aromatic benzene ring. The lone pairs of electrons on the iodine atom are in conjugation with the delocalized pi-electron system of the ring. This resonance phenomenon imparts a **partial double bond character** to the carbon-iodine bond.

Because of this partial double bond character, the $	ext{C}-	ext{I}$ bond in iodobenzene is incredibly strong and short. It stubbornly refuses to break under normal conditions. Therefore, boiling iodobenzene with $	ext{NaOH}$ yields absolutely **no reaction**. Crucially, no free iodide ions ($	ext{I}^-$) are released into the solution.

Now, let's look at **benzyl iodide** ($	ext{C}_6	ext{H}_5	ext{CH}_2	ext{I}$). Here, the iodine atom is attached to an $sp^3$ hybridized carbon atom, which is then attached to the benzene ring. This makes it a typical alkyl halide. The carbon-iodine bond is a standard single bond.

When boiled with $	ext{NaOH}$, the hydroxide ion easily attacks this $sp^3$ carbon, displacing the iodide ion. Benzyl iodide readily undergoes nucleophilic substitution to form benzyl alcohol ($	ext{C}_6	ext{H}_5	ext{CH}_2	ext{OH}$), and most importantly, it releases **free iodide ions** ($	ext{I}^-$) into the solution.

### The Silver Nitrate Confirmation

After the boiling $	ext{NaOH}$ treatment, we have two test tubes. One contains unreacted iodobenzene, and the other contains benzyl alcohol along with free iodide ions. To visually confirm which is which, we use the classic silver nitrate test.

First, we acidify the solutions with dilute nitric acid ($	ext{HNO}_3$). Then, we add a few drops of silver nitrate ($	ext{AgNO}_3$) solution. Silver ions ($	ext{Ag}^+$) have a tremendous affinity for halide ions. If they encounter free iodide ions, they instantly combine to form a bright **yellow precipitate** of silver iodide ($	ext{AgI}$).

$$ 	ext{Ag}^+_{(aq)} + 	ext{I}^-_{(aq)} ightarrow 	ext{AgI}_{(s)} \downarrow 	ext{ (Yellow precipitate)} $$

Since the iodobenzene tube has no free iodide ions, it remains clear. However, the benzyl iodide tube, teeming with iodide ions, immediately produces the beautiful yellow precipitate.

### The Final Verdict

The problem states that substance $B$ gave the yellow precipitate. This is our smoking gun! It implies that substance $B$ must be the one that released iodide ions during the $	ext{NaOH}$ treatment.

Therefore, substance $B$ is definitively **benzyl iodide** ($	ext{C}_6	ext{H}_5	ext{CH}_2	ext{I}$). By process of elimination, substance $A$ must be **iodobenzene** ($	ext{C}_6	ext{H}_5	ext{I}$).

Looking at our options, the correct statement is that $A$ was $	ext{C}_6	ext{H}_5	ext{I}$.

### The Crucial Role of Nitric Acid

You might be wondering about option (d), which suggests that the addition of $	ext{HNO}_3$ was unnecessary. This is a dangerous trap!

Why did we add dilute nitric acid before the silver nitrate? Remember that our initial reaction was carried out in a strongly basic medium ($	ext{NaOH}$). If we were to add $	ext{AgNO}_3$ directly to this basic solution, the silver ions would react with the abundant hydroxide ions to form a brown precipitate of silver oxide ($	ext{Ag}_2	ext{O}$).

$$ 2	ext{Ag}^+_{(aq)} + 2	ext{OH}^-_{(aq)} ightarrow 	ext{Ag}_2	ext{O}_{(s)} \downarrow + 	ext{H}_2	ext{O}_{(l)} $$

This brown precipitate would completely mask the yellow silver iodide precipitate, ruining our test. The dilute nitric acid is absolutely essential to **neutralize the excess base**, ensuring that the silver ions only react with the halide ions. Always respect the role of every reagent in a qualitative analysis!

Select Answer:

Visualized Solution

$The Setup$

$The Reagents$

$Analyzing Iodobenzene$

$Analyzing Benzyl Iodide$

$The Silver Nitrate Test$

$Conclusion$

$Role of Nitric Acid$

The Sigma Insight: Haloalkanes & Haloarenes

The Setup

The Boiling NaOH Test

The Silver Nitrate Confirmation

The Final Verdict

The Crucial Role of Nitric Acid

Similar Questions

Compound (A), $C_{8} H_{9} Br$ gives a white precipitate when warmed with alcoholic $AgNO_{3}$ . Oxidation of (A) gives an acid (B), $C_{8} H_{6} O_{4}$ . (B) easily forms anyhydride on heating. Identify the compound (A).

In the following sequence of reactions, $C_{3} H_{6} H^{+} / H_{2} O A K I O dil. KOH B + C$ The compounds $B$ and $C$ respectively are

Which of the following compounds will produce a precipitate with $AgNO_{3}$ ?

Identify $A$ and $B$ in the chemical reaction.

The compounds A and B in the following reaction are, respectively

Which of the following compounds will form the precipitate with aq. $A g N O_{3}$ solution most readily?

The product $A$ and $B$ formed in above reactions are

The major product obtained in the following reaction is

The product of the reaction given below is

Which one of the following is likely to give a precipitate with $A g N O_{3}$ solution?

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Reagents

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Iodobenzene

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Benzyl Iodide

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Silver Nitrate Test

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Role of Nitric Acid