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Chapter 3: Problem 28

Which of the following species are involved in the reaction of phenol with chloroform and alkali to form salicylaldehyde?

Short Answer

Expert verified
Answer: The key species involved in the Reimer-Tiemann reaction are phenol, chloroform, and alkali. Phenol serves as the starting material, chloroform provides the dichlorocarbene that reacts with the phenol, and alkali (hydroxide ions) facilitates the formation of dichlorocarbene and the hydrolysis of the intermediate to form salicylaldehyde.

Step by step solution

01

Write the reaction equation

For the Reimer-Tiemann reaction, first, we should write the balanced chemical equation to see all the species involved. Phenol + Chloroform + Alkali → Salicylaldehyde + other products \[C_6H_5OH + CHCl_3 + OH^- \rightarrow C_7H_6O_2 + H_2O + 3Cl^-\]
02

Identify the species involved

Based on the balanced chemical equation for the Reimer-Tiemann reaction, the species involved are: 1. Phenol (\(C_6H_5OH\)) 2. Chloroform (\(CHCl_3\)) 3. Alkali (Hydroxide ions, \(OH^-\))
03

Describe the reaction mechanism

The Reimer-Tiemann reaction mechanism involves three major steps: 1. Formation of dichlorocarbene: The hydroxide ions (\(OH^-\)) from the alkali remove a hydrogen from chloroform (\(CHCl_3\)), creating dichlorocarbene (\(CCl_2\)). 2. Formation of an ortho intermediate: The dichlorocarbene then reacts with the phenol molecule by adding to the ortho position (next to the hydroxyl group) of the benzene ring, forming a dichloromethyl intermediate. 3. Hydrolysis: The intermediate undergoes hydrolysis, where water molecules replace the chlorines. This forms salicylaldehyde, as the final product. Hence, phenol, chloroform, and alkali are the key species involved in the Reimer-Tiemann reaction of phenol to form salicylaldehyde.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phenol
Phenol, represented by the chemical formula \(C_6H_5OH\), is the starting material in the Reimer-Tiemann reaction. This organic compound consists of a hydroxyl group (-OH) bonded to a benzene ring. Phenol is often used in the production of plastics, explosives, and as a precursor to many other chemical compounds.

In its role in the Reimer-Tiemann reaction, phenol acts as an electrophile, where its benzene ring undergoes electrophilic aromatic substitution to incorporate new functional groups, such as the formyl group (-CHO) to produce salicylaldehyde.
Chloroform
Chloroform, with the formula \(CHCl_3\), serves as the source of the dichlorocarbene \((:CCl_2)\) in the Reimer-Tiemann reaction. This volatile, colorless liquid has historically been used as an anesthetic, but due to health concerns, its applications have become limited in medicine.

During the reaction, chloroform reacts with a strong base like sodium hydroxide, resulting in the abstraction of a hydrogen atom and the formation of the reactive intermediate dichlorocarbene, which is integral to the reaction process. Chloroform's role is critical as it provides a carbon atom for the subsequent formation of the aldehyde group in salicylaldehyde.
Salicylaldehyde
Salicylaldehyde is an aromatic aldehyde compound characterized by the presence of a formyl group attached to the benzene ring of phenol at the ortho position. Salicylaldehyde has the chemical formula \(C_7H_6O_2\).

As the final product of the Reimer-Tiemann reaction, it's used as a precursor in the synthesis of other chemicals, including some pharmaceuticals and dyes. Its formation from phenol signifies a successful Reimer-Tiemann reaction. The name salicylaldehyde reflects its relationship to salicylic acid, as both share a similar structural motif and are commonly found in the synthesis of aspirin and other similar drugs.
Reaction Mechanism
The reaction mechanism of the Reimer-Tiemann reaction details the intricate steps of how phenol transforms into salicylaldehyde in the presence of chloroform and alkali. The mechanism is a prime example of electrophilic aromatic substitution:

- Firstly, the hydroxide ions \((OH^-)\) deprotonate chloroform, generating the highly-reactive dichlorocarbene intermediate.
- Dichlorocarbene then attacks the ortho position of phenol, forming an adduct.
- Finally, hydrolysis of this adduct leads to the replacement of chlorine atoms with a hydroxyl group and the unveiling of the formyl group that characterizes salicylaldehyde.
Organic Chemistry
Organic chemistry is the branch of chemical science focused on the structure, properties, and reactions of organic compounds—molecules containing carbon. Reactions like the Reimer-Tiemann reaction are a cornerstone in this field, as they demonstrate the transformation of simple organic molecules into more complex ones.

Understanding organic chemistry is vital for grasping how myriad compounds behave under different conditions and how they can be synthesized for various applications. It also provides a foundation for innovations in pharmaceuticals, agriculture, and materials science.

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Most popular questions from this chapter

Statement 1 \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{2}-\mathrm{CH}_{3}(\) ethyl benzene \()\) on reaction with \(\mathrm{NBS}\) gives mainly \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CHBr}-\mathrm{CH}_{3} .\) and Statement 2 The reaction with NBS proceeds by free radical mechanism.

The wrong statement relating to the influence of substituents on aromatic electrophilic substitution reactions is (a) The rate determining step in an electrophilic substitution is the one which leads to the formation of an arenium ion. (b) Electron releasing groups make the transition state more stable and the transition state resembles the arenium ion. (c) The arenium ion is a highly energetic intermediate and the step that leads to it is a highly exothermic step. (d) An electron withdrawing substituent makes the arenium ion less stable and the transition state leading to such an arenium ion is less stable.

\text { Starting from benzene how will you prepare m-nitrobenzoic acid? }

An organic compound \(\mathrm{X}\) having the molecular formula \(\mathrm{C}_{7} \mathrm{H}_{8} \mathrm{O}\) is soluble in \(\mathrm{NaOH}\). When treated with bromine water, \(X\) gives a compound \(Y\) having the molecular formula \(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{OBr}_{2}\). The oxidation of compound \(\mathrm{X}\) with alkaline \(\mathrm{KMnO}_{4}\) followed by acidification gives \(\mathrm{Z}\left(\mathrm{MF}: \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}\right)\) which is steam volatile. The compound \(\mathrm{X}\) is (a) anisole (b) benzyl alcohol (c) o-cresol (d) \(\mathrm{p}\) -cresol

Column I \(\quad\) Column II (a) Salicylic acid (p) characteristic colour with neutral \(\mathrm{FeCl}_{3}\) solution (b) \(\mathrm{p}\) -Hydroxy acetophenone (q) Kolbe's reaction (c) Quinol (r) Fries rearrangement (d) picric acid (s) Elbs persulphate oxidation (t) Decomposes \(\mathrm{NaHCO}_{3}\) solution
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