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Ketones can be oxidised by (A) Bayer's Villiger oxidation (B) Haloform reaction (C) Lucas regent (D) Oppenauer oxidation

Short Answer

Expert verified
Ketones can be oxidized by \( \textbf{(A) Baeyer-Villiger Oxidation} \). This oxidation reaction converts ketones to esters and cyclic ketones to lactones.

Step by step solution

01

Analyze Option A - Baeyer-Villiger Oxidation

Baeyer-Villiger Oxidation is an organic reaction that involves the oxidation of ketones to esters and cyclic ketones to lactones. This is typically performed using peroxyacids or hydrogen peroxide. Since this reaction oxidizes ketones, option A is correct.
02

Analyze Option B - Haloform Reaction

Haloform reaction is a reaction between haloform (such as chloroform, bromoform, or iodoform) and methyl ketones. It leads to the formation of a carboxylic acid and the corresponding haloform. Although it involves a reaction with ketones, it is not an oxidation of ketones. Therefore, option B is incorrect.
03

Analyze Option C - Lucas Reagent

Lucas reagent is a mixture of hydrochloric acid and zinc chloride, which is typically used as a reagent for the qualitative classification of alcohols. While it reacts with alcohols, it does not oxidize ketones. Thus, option C is incorrect.
04

Analyze Option D - Oppenauer Oxidation

Oppenauer oxidation is a reaction that selectively oxidizes secondary alcohols to ketones, using an aluminum alkoxide catalyst. Although it involves the formation of ketones, it does not oxidize ketones. So, option D is incorrect. #Conclusion# Based on the analysis, the correct option is: (A) Baeyer-Villiger Oxidation

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Key Concepts

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

Baeyer-Villiger Oxidation
The Baeyer-Villiger oxidation is a reaction used to transform ketones into esters and cyclic ketones into lactones. This process is primarily carried out using peroxyacids, such as mCPBA, or sometimes with hydrogen peroxide as the oxidizing agent. The mechanism of the reaction involves the migration of the more electron-rich alkyl group from the carbonyl carbon to the adjacent oxygen of the peroxyacid, resulting in the cleavage of the carbon-oxygen bond and the formation of an ester. Key points include:
  • Baeyer-Villiger oxidation specifically targets ketones and transforms them into more reactive and functional esters.
  • In cyclic ketones, this reaction results in the formation of lactones.
Understanding the selectivity and conditions required for this reaction is crucial in synthetic organic chemistry because it offers a method to extend carbon chains or introduce functional groups into molecules.
Haloform Reaction
The haloform reaction is a chemical process that involves the reaction of a methyl ketone with halogens like iodine, bromine, or chlorine. The product is a carboxylate ion and a haloform such as iodoform, chloroform, or bromoform. This reaction requires the presence of a methyl group directly attached to the carbonyl carbon, which distinguishes it from other reactions involving ketones. It proceeds via a halogenation of the methyl group followed by hydrolysis to yield the carboxylic acid.
  • Though the haloform reaction involves ketones, it does not oxidize them; rather, it converts them through a substitution mechanism.
  • This reaction is typically used for the qualitative test of a methyl ketone where a positive iodoform test indicates the presence of such a structure.
The haloform reaction can be a helpful analytical tool due to its specificity towards methyl ketones.
Lucas Reagent
Lucas reagent is a solution of zinc chloride in concentrated hydrochloric acid, mainly used to distinguish between primary, secondary, and tertiary alcohols based on their reactivity. It is not involved in the oxidation of ketones but works on alcohols to facilitate their transformation into alkyl chlorides. In a typical application:
  • Tertiary alcohols react almost immediately, while secondary alcohols take a few minutes to react, and primary alcohols react very slowly or not at all.
  • The turbidity or cloudiness formed during the reaction can be used to infer the class of alcohol.
This reagent does not alter ketones but is essential for classifying alcohols, and thus it is not applicable for oxidizing ketones.
Oppenauer Oxidation
The Oppenauer oxidation is an elegant method to selectively oxidize secondary alcohols into ketones using an aluminum alkoxide catalyst, typically in the presence of a nonpolar solvent like toluene. It is an example of reducing an alcohol without affecting other functionalities such as double bonds in a molecule. However, Oppenauer oxidation does not oxidize ketones, but rather creates them.
  • This reaction is the reverse of the Meerwein-Ponndorf-Verley reduction, which uses similar conditions to reduce ketones back to alcohols.
  • It demonstrates high chemoselectivity, allowing for the conversion of alcohols in multifunctional systems.
Understanding its mechanism can aid in designing synthesis routes where the selective oxidation of alcohols to ketones is required without affecting existing ketones.

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Most popular questions from this chapter

Statement - 1: When alkyne react with \(\mathrm{H}_{2} / \mathrm{Pd} / \mathrm{CaCO}_{3}-\mathrm{PbO}_{2}\) it will form alkene as a product. Statement - 2: \(\mathrm{CaCO}_{3}\) or \(\mathrm{BaSO}_{4}\) will act as a poison for \(\mathrm{Pd}\). (A) Both statements are correct and statement- 2 is correct explanation of statement-1. (B) Both statements are correct and statement- 2 is not correct explanation of statement-1. (C) Statement- 1 is correct but statement- 2 is a wrong statement. (D) Statement- 1 is wrong but statement- 2 is a correct statement.

Generally aldehydes are more reactive than ketones because of : (A) Less steric crowding (B) Hydrogen bonding (C) More electrophilic carbon (D) Both (A) and (C)

Choose the incorrect option(s). $$ \mathrm{CH}_{3} \mathrm{CN} \stackrel{' \mathrm{X}^{\prime}}{\longrightarrow}(\mathrm{P}) \stackrel{\mathrm{H}_{3} \mathrm{O}^{+}}{\longrightarrow}(\mathrm{Q}) $$ (A) If ' \(X\) ' is stannous chloride in presence of hydrochloric acid, then formation of \((Q)\) is known as 'Stephens reduction. (B) If \(\mathrm{X}\) is \(\mathrm{DIBAL}-\mathrm{H}\), then \(\mathrm{Q}\) is ethanal: (C) If \(\mathrm{X}\) is \(\mathrm{CH}_{3} \mathrm{MgI}\) then \((\mathrm{Q})\) gives (-ve) Tollen's as well as Fehling test. (D) If \(\mathrm{X}\) is \(\mathrm{LiAlH}_{4}\), then \((\mathrm{Q})\) can directly obtain by reacting \(\mathrm{CH}_{3} \mathrm{CN}\) with \(\mathrm{H}_{3} \mathrm{O}^{\oplus:}\)

In order to achieve the following conversion, O=C(C=CC1CCCCC1)OCC1CCC(C=CCO)CC1 which of the following reducing agents should be employed? (A) \(\mathrm{NaBH}_{4}\) (B) \(\mathrm{H}_{2}, \mathrm{Ni}\) (C) \(\mathrm{LiAlH}_{4}\) (D) $\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHO}\right]_{3} \mathrm{Al},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH}$

Which compound cannot reduce alkaline solution of iodine? (a) (b) (c) (d)

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