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When we discussed oxidation of alcohols in Section \(30.5,\) we stated that the mechanism of oxidation of an aldehyde to carboxylic acid is analogous to that of the conversion of an alcohol to an aldehyde. Propose a mechanism for the oxidation of \(\mathrm{RCHO}\) to \(\mathrm{RCO}_{2} \mathrm{H}\) using \(\mathrm{KMnO}_{4}\) in acidic aqueous solution.

Short Answer

Expert verified
An aldehyde oxidizes to a carboxylic acid using KMnO4 in acid through geminal diol intermediate oxidation.

Step by step solution

01

Identify reactants and reagents

The reactant in this reaction is an aldehyde (\(\mathrm{RCHO}\)) and the reagent is potassium permanganate \((\mathrm{KMnO}_4)\) in acidic aqueous solution. The goal is to oxidize the aldehyde to a carboxylic acid (\(\mathrm{RCO}_2\mathrm{H}\)).
02

Formation of geminal diol

In acidic aqueous conditions, the aldehyde (\(\mathrm{RCHO}\)) is first converted to a geminal diol (a diol formed when water adds across the carbonyl group). This involves the nucleophilic attack of water on the carbonyl carbon, followed by protonation, yielding \(\text{RCH(OH)}_2\).
03

Oxidation by KMnO4

The geminal diol (\(\text{RCH(OH)}_2\)) is now ready to be oxidized by \(\mathrm{KMnO}_4\). In this step, manganese in \(\mathrm{KMnO}_4\) acts as an oxidizing agent and is reduced from +7 oxidation state to a lower oxidation state while oxidizing the diol to a carboxylic acid (\(\mathrm{RCO}_2\mathrm{H}\)).
04

Formation of carboxylic acid

In the oxidation process, the hydrogen atoms of the geminal diol are lost, forming \(\mathrm{RCO}_2\mathrm{H}\) and yielding \(\mathrm{MnO}_2\) as a side product in a lower oxidation state than \(\mathrm{KMnO}_4\). The carbonyl group is restored, and the substance is converted into a carboxylic acid.

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Key Concepts

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

Mechanism of oxidation
The mechanism of oxidation is an intriguing and essential process in organic chemistry. Understanding this mechanism helps us transform different organic compounds into more oxidized forms. Here we are specifically looking at how aldehydes, such as \( ext{RCHO}\), undergo oxidation to form carboxylic acids, such as \( ext{RCO}_2 ext{H}\). This transformation is analogous to the oxidation of alcohols into aldehydes, following a similar mechanistic pathway.First, it involves the creation of an intermediate species, in this case, a geminal diol. The carbonyl group of the aldehyde acts as an electrophile, being susceptible to attack by nucleophiles like water. This leads to the formation of the geminal diol through the addition of water. Subsequently, the intermediate geminal diol is oxidized, which effectively turns the intermediate diol into a new form—a carboxylic acid. The oxidation step involves an oxidizing agent, which in many reactions is \( ext{KMnO}_4\), a potent reagent that can further oxidize the carbon in the geminal diol, helping to achieve the final carboxylic acid structure.
Potassium permanganate
Potassium permanganate \((\text{KMnO}_4)\) is widely used in organic chemistry as a strong oxidizing agent. It is known for its versatility and efficiency, making it suitable for a range of oxidation reactions, including the oxidation of aldehydes to carboxylic acids.In the context of oxidizing an aldehyde to form a carboxylic acid, \( ext{KMnO}_4\) serves as an electron acceptor. Manganese in \( ext{KMnO}_4\) has a high oxidation state of +7, allowing it to efficiently transfer oxygen to the substrate and gain electrons. During the process, \( ext{KMnO}_4\) is reduced, typically to manganese dioxide \((\text{MnO}_2)\). This reduction is a clear indication that oxidation has occurred in the reaction, as \( ext{MnO}_2\) is visibly different due to its brown precipitate.The ability to control the reaction conditions, such as using an acidic medium, makes \( ext{KMnO}_4\) a suitable choice ensuring the complete oxidation of aldehydes. This precision allows for the efficient conversion to carboxylic acids, stressing the importance of \( ext{KMnO}_4\)'s role in oxidizing reactions.
Carboxylic acids
Carboxylic acids, like \( ext{RCO}_2 ext{H}\), are a fundamental functional group in organic chemistry, with applications spanning from polymer production to pharmaceuticals. They feature a carbonyl group attached to a hydroxyl group. This structure is typical and responsible for their characteristic chemical properties.Forming carboxylic acids from aldehydes signifies an increase in oxidation state of the molecule. The oxidation introduces another oxygen, breaking C-H bonds and creating C-O bonds, leading to the formation of the carboxylic acid group. The increased polarity due to the extra oxygen atoms also results in higher boiling points compared to aldehydes, due to stronger hydrogen bonding capabilities. This is a crucial characteristic helpful in distinguishing them.Carboxylic acids are not just important in synthesis but also occur naturally. They are the building blocks of numerous biological pathways, hence understanding their formation is beneficial for both industrial applications and biological science.
Reaction steps in chemistry
Breaking down complex reactions into simpler steps is key in understanding chemistry. Every reaction involves a series of distinct steps, often involving the formation and transformation of intermediate species. For the oxidation of an aldehyde to a carboxylic acid, each step supports the conversion process. First is the initiation of the reaction by identifying the reactants—an aldehyde and \( ext{KMnO}_4\)—and setting reaction conditions like acidity of the solution. This is pivotal, as it ensures the right pathway is pursued.Next is the formation of a geminal diol. This step is fundamental, as it temporarily transforms the aldehyde into a more oxidative-prone species, readying it for further transformation by the oxidizing agent.In the final steps, the reaction progresses through the loss of electrons by the substrate, helped by \( ext{KMnO}_4\). Each step seamlessly leads to the production of the desired carboxylic acid, demonstrating the importance of a methodical approach in chemistry.By analyzing these steps, chemists gain better predictive power and problem-solving ability, vital for both educational and professional chemical endeavors.

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Most popular questions from this chapter

Draw the structures of (a) pentan-l-ol; (b) heptan- \(3-01\); (c) 2 -methylpentan-2-ol; (d) propane- 1,2,3 -triol.

Suggest reasons for the following trend in \(\mathrm{p} K_{\mathrm{a}}\) values: EtOH, p \(K_{\mathrm{a}}=16.0 ; \mathrm{Me}_{3} \mathrm{COH}, \mathrm{p} K_{\mathrm{a}}=18.0\) \\[ \mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{OH}, \mathrm{p} K_{\mathrm{a}}=12.4 ;\left(\mathrm{CF}_{3}\right)_{3} \mathrm{COH}, \mathrm{p} K_{\mathrm{a}}=5.4 \\]

Suggest starting materials for the preparation of the following alcohols using a Grignard reagent as one precursor. In each case, state how you might make the Grignard reagent: (a) 2 -methylpropan- 2 \(\mathrm{ol}\) (b) heptan-I-ol; (c) pentan-2-ol.

What spectroscopic technique(s) would you choose to distinguish between (a) propan-1-ol and propan-2-ol; (b) diethyl ether and butan-1-ol; (c) cyclohexanol and hexan-1-ol?

Suggest explanations for the following. (a) The \(^{1} \mathrm{H}\) NMR spectrum of \(\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{OH}\) contains a quartet \((J 9 \mathrm{Hz})\) at \(\delta+3.9 \mathrm{ppm}\) in addition to the signal assigned to the OH proton. (b) The addition of \(\mathrm{D}_{2} \mathrm{O}\left(\mathrm{D}=^{2} \mathrm{H}\right)\) to hexanol causes the disappearance of the signal assigned to the OH proton. (c) Whereas alcohols exhibit relatively high boiling points and enthalpies of vaporization, the same is not true of thiols, \(\mathrm{RSH}\), e.g. propan-1-ol, bp \(=370.2 \mathrm{K}\) \(\Delta_{\mathrm{vap}} H(\mathrm{bp})=41.4 \mathrm{kJ} \mathrm{mol}^{-1} ;\) propane-l-thiol \(\mathrm{bp}=340.8 \mathrm{K}, \Delta_{\mathrm{vap}} H(\mathrm{bp})=29.5 \mathrm{kJ} \mathrm{mol}^{-1}\)

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