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Which one of the following sets of reactants is used in Reimer-Tiemann reaction? (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}+\mathrm{CO}_{2}+\mathrm{KOH}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{KOH}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{CHCl}_{3}+\mathrm{KOH}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}^{2}+\mathrm{CHCl}_{3}+\mathrm{KOH}\)

Short Answer

Expert verified
Option (d) is correct: \( \mathrm{C_6H_5OH} + \mathrm{CHCl_3} + \mathrm{KOH} \).

Step by step solution

01

Understand the Reimer-Tiemann Reaction

The Reimer-Tiemann reaction is a chemical reaction used to ortho-formylate phenols into salicylaldehyde using chloroform (\( \mathrm{CHCl_3} \)) and a base such as \( \mathrm{KOH} \). It specifically involves phenol (\( \mathrm{C_6H_5OH} \)) as one of the reactants.
02

Identify the Reactants in Each Option

Examine each set of reactants:(a) \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} + \mathrm{CO}_{2} + \mathrm{KOH} \)(b) \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} + \mathrm{CH}_{3} \mathrm{Cl} + \mathrm{KOH} \)(c) \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} + \mathrm{CHCl}_{3} + \mathrm{KOH} \)(d) \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} + \mathrm{CHCl}_{3} + \mathrm{KOH} \)
03

Match Reactants with Reaction Requirements

For the Reimer-Tiemann reaction, we need \( \mathrm{C_6H_5OH} \), \( \mathrm{CHCl_3} \), and \( \mathrm{KOH} \). Check the options:- Option a: Incorrect, as it contains \( \mathrm{CO_2} \) instead of \( \mathrm{CHCl_3} \).- Option b: Incorrect, as neither phenol nor \( \mathrm{CHCl_3} \) is present.- Option c: Incorrect, as it uses aniline \( \mathrm{C_6H_5NH_2} \) instead of phenol, and it lacks \( \mathrm{CHCl_3} \).- Option d: Correct, as it contains all needed reactants: \( \mathrm{C_6H_5OH} \), \( \mathrm{CHCl_3} \), and \( \mathrm{KOH} \).

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Key Concepts

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

ortho-formylation
Ortho-formylation is a distinctive chemical process requiring the addition of an -CHO group to the ortho position of an aromatic ring. This specific placement adjacent to a functional group like -OH is crucial for subsequent reactions. An intriguing aspect of ortho-formylation is how it is primarily facilitated in phenolic compounds, especially phenols, leading directly to useful products. The Reimer-Tiemann reaction is particularly ingenious because it uses chloroform and a base to achieve this task seamlessly. With chloroform acting as a source of the formyl group, the presence of a strong base such as potassium hydroxide helps in generating the dichlorocarbene intermediate. This step is a fascinating part of the reaction, as it allows the formation of the formyl group at the ortho position. Finally, the outcome is crucial for synthesizing various aromatic aldehydes, like salicylaldehyde, illustrating the utility of ortho-formylation.
phenols
Phenols are widely known as aromatic compounds that bear a hydroxyl group (-OH) directly bonded to a benzene ring. This unique structure endows them with distinctive properties and reactivity.
Phenols are more acidic than alcohols due to the stabilization of their anion by resonance, which contributes to their diverse applications in chemical synthesis.
  • They serve as precursors for numerous chemical reactions, including the Reimer-Tiemann reaction.
  • Phenols are the primary substrates in the ortho-formylation process using the Reimer-Tiemann method.
  • They play a pivotal role in the production of industrial chemicals, dyes, and pharmaceuticals.
The ability of phenols to undergo electrophilic substitution reactions with ease is particularly significant in this context. Their reactivity provides an excellent opportunity to introduce various substituents into the aromatic ring, which, when harnessed properly, leads to valuable chemical transformations. This underlines the importance of phenols not just as individual compounds, but as integral players in complex chemical synthesis.
salicylaldehyde synthesis
Salicylaldehyde is a notable chemical compound primarily synthesized via the Reimer-Tiemann reaction. This process elegantly transforms phenol into salicylaldehyde by ortho-formylating the aromatic ring. The synthesis of salicylaldehyde is significant due to the compound's myriad applications, including:
  • Usage in the synthesis of pharmaceuticals, where it acts as a precursor to other active pharmaceutical ingredients.
  • In the design and production of flavoring agents due to its pleasant almond fragrance.
  • Its role in the production of complex organic compounds and fine chemicals.
The direct conversion of phenol to salicylaldehyde through this reaction underscores its efficiency and effectiveness.
The Reimer-Tiemann reaction not only serves as a simple method but also provides a high yield of the desired aldehyde. Moreover, understanding the role of each reagent involved, such as chloroform and caustic potassium hydroxide, becomes critical in recognizing how these elements facilitate the transformation efficiently.

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Most popular questions from this chapter

An organic compound 'A' reacts with methyl magnesium iodide, to form an addition product which on hydrolysis forms the compound 'B'. Compound 'B' gives blue colour salt in Victor Meyer's test. The compounds 'A' and 'B' are respectively (a) acetaldehyde and isopropyl alcohol (b) acetone and isopropyl alcohol (c) acetaldehyde and tertiary butyl alcohol (d) acetaldehyde and ethyl alcohol

The molecule which the highest boiling point is (a) \(\mathrm{CH}_{3}-\mathrm{CHCl}-\mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3}-\mathrm{CHOH}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) (d) \(\mathrm{CH}_{3}-\mathrm{CHOH}-\mathrm{CH}_{3}\)

Vinyl carbinol is (a) \(\mathrm{HO}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}_{2}\) (b) \(\mathrm{CH}_{3}-\mathrm{C}\left(\mathrm{CH}_{2} \mathrm{OH}\right)=\mathrm{CH}_{2}\) (c) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{OH}\) (d) \(\mathrm{CH}_{3} \mathrm{C}(\mathrm{OH})=\mathrm{CH}_{2}\)

1-propanol and 2 -propanol can be best distinguished by (a) oxidation with \(\mathrm{KMnO}_{4}\) followed by reaction with Fehling solution (b) oxidation with acidic dichromate followed by reaction with Fehling solution (c) oxidation by heating with copper followed by reaction with Fehling solution (d) oxidation with concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) followed by reaction

Which of the following compounds on reaction with \(\mathrm{CH}_{3} \mathrm{MgBr}\) will give a tertiary alcohol? CC(C)(C)C 11 (a) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\)

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