An organic compound A (C6H6O) gives dark green colouration with ferric chloride. On treatment with C

Q: Why does phenol give a green colour with FeCl3?

Phenols react with neutral ferric chloride to form a water-soluble complex, ferric phenoxide, which imparts a characteristic blue, violet, or green colour depending on the specific phenol.

Q: What is the electrophile in the Reimer-Tiemann reaction?

The electrophile is dichlorocarbene (:CCl2), which is generated in situ from chloroform and a strong base like KOH.

Q: Why does PCC stop the oxidation at the aldehyde stage?

PCC (Pyridinium chlorochromate) is a mild oxidizing agent that lacks the water necessary to form the hydrate intermediate, preventing further oxidation to a carboxylic acid.

Q: Can we use KMnO4 instead of PCC to get compound B?

No, KMnO4 is a strong oxidizing agent and would oxidize the primary alcohol all the way to a carboxylic acid (salicylic acid) instead of stopping at the aldehyde.

Q: Why does the aldehyde group attach at the ortho position in the Reimer-Tiemann reaction?

The phenoxide ion strongly activates the ortho and para positions. The ortho product is usually the major one due to the stabilization of the intermediate and the final product via intramolecular hydrogen bonding.

Question

An organic compound A ($C_6H_6O$) gives dark green colouration with ferric chloride. On treatment with $CHCl_3$ and KOH, followed by acidification gives compound B. Compound B can also be obtained from compound C on reaction with pyridinium chlorochromate (PCC). Identify A, B and C.

Sigma Percentile · Accepted Answer

### Decoding the Phenol Puzzle: Reimer-Tiemann and PCC Oxidation

Imagine you are a chemical detective, and you are handed a mystery compound, **A**, with the molecular formula $C_6H_6O$. The first clue is a classic one: it gives a dark green colouration when treated with neutral ferric chloride ($FeCl_3$). This is a hallmark test for phenols! The $Fe^{3+}$ ion forms a beautifully coloured, water-soluble complex with the phenoxide ion. So, right out of the gate, we know that **A is Phenol**.

### The Master Equation

Now, we take our phenol and treat it with chloroform ($CHCl_3$) and potassium hydroxide ($KOH$), followed by acidification. If you've been studying your name reactions, alarm bells should be ringing! This is the legendary **Reimer-Tiemann reaction**.

In this reaction, the strong base reacts with chloroform to generate a highly reactive intermediate called dichlorocarbene ($:CCl_2$). This electrophile attacks the electron-rich phenol ring, predominantly at the ortho position. After a series of steps and final acidification, an aldehyde group ($-CHO$) is installed. The resulting compound **B** is 2-hydroxybenzaldehyde, commonly known as **Salicylaldehyde**.

### Final Calculation

The final piece of the puzzle involves compound **C**. We are told that reacting **C** with Pyridinium Chlorochromate (PCC) also yields our compound **B** (Salicylaldehyde).

What does PCC do? It is a mild oxidizing agent that selectively oxidizes primary alcohols to aldehydes without taking them all the way to carboxylic acids. Since **B** is an aldehyde, **C** must be the corresponding primary alcohol. By simply replacing the $-CHO$ group in salicylaldehyde with a $-CH_2OH$ group, we find that **C** is **2-hydroxybenzyl alcohol**.

And there we have it! A beautiful sequence of logical deductions leading us straight to the correct answer.

Select Answer:

Visualized Solution

$A F e C l_{3} Dark Green Complex$

$A C H C l_{3}, K O H B$

$C P C C B$

$Final Conclusion$

The Sigma Insight: Alcohols, Phenols, Ethers

Decoding the Phenol Puzzle

The Master Equation

Final Calculation

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