Problem 32 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

How will you convert butan-2-one to propanoic acid? (a) Tollen's reagent (b) Fehling's solution (c) \(\mathrm{NaOH} / \mathrm{I}_{2} / \mathrm{H}^{+}\) (d) \(\mathrm{NaOH} / \mathrm{NaI} / \mathrm{H}^{+}\)

Step-by-Step Solution

Answer

Use \( \text{NaOH} / \text{I}_2 / \text{H}^{+} \) for the conversion via the haloform reaction.

Step 1: Understanding Butan-2-one's Structure

Butan-2-one, also known as methyl ethyl ketone, has the chemical formula \( \text{CH}_3\text{COCH}_2\text{CH}_3 \). It is a ketone, characterized by the carbonyl group (C=O) bonded to two carbon atoms.

Step 2: Objective - Conversion to Propanoic Acid

Propanoic acid has the chemical formula \( \text{CH}_3\text{CH}_2\text{COOH} \). This conversion involves shortening the carbon chain by one carbon and converting a ketone to a carboxylic acid.

Step 3: Analyze Options - Reaction Pathways

Evaluate each option for its potential to oxidize or convert butan-2-one to propanoic acid: - (a) **Tollen's reagent**: Mainly used for oxidizing aldehydes, not effective on ketones. - (b) **Fehling's solution**: Used for oxidizing aldehydes, ineffective with ketones like butan-2-one. - (c) **NaOH**/**I_2**/**H^{+}**: This is the haloform reaction, where methyl ketones are converted into carboxylate and iodoform. This will shorten the carbon chain, producing propanoic acid. - (d) **NaOH**/**NaI**/**H^{+}**: This does not follow the haloform mechanism, as NaI is not suitable for forming iodoform effectively.

Step 4: Conclusion

The haloform reaction (option c) converts the methyl ketone (butan-2-one) to propanoic acid. This happens as the methyl group next to the carbonyl is converted to iodoform, leaving behind a carboxylate, which after acid workup gives propanoic acid.

Key Concepts

Haloform Reaction

The haloform reaction is a fascinating chemical reaction specifically used to convert methyl ketones into carboxylic acids. It works by halogenating the methyl group adjacent to the carbonyl group of a ketone. In this context, the reaction involves reagents like iodine (\( \_2\)) and an alkali, such as sodium hydroxide (\( \OH\)). This results in the formation of a haloform — in this case, iodoform if iodine is used — and a carboxylate anion.

Here's how the basic process works:

• The methyl group next to the carbonyl group undergoes halogenation.
• This is followed by the cleavage of the C-C bond between the carbonyl carbon and the halogenated carbon.
• The resulting carboxylate is then acidified to yield the carboxylic acid.

For example, in the conversion of butan-2-one to propanoic acid, iodine and sodium hydroxide are used. The methyl ketone (butan-2-one) forms iodoform and a carboxylate, which mixes with an acid to ultimately yield propanoic acid. This is a selective process that effectively shortens the carbon chain.

Ketones

Ketones are a significant class of organic compounds characterized by a carbonyl group (\( \text{C=O} \)) bonded to two carbon atoms. The simplest member of this group is acetone (\( \text{CH}_3\text{COCH}_3 \)). Ketones do not oxidize as readily as aldehydes due to the absence of a hydrogen atom connected to the carbonyl carbon. This feature limits the types of reactions ketones can undergo.

However,ketones can partake in transformation processes like the haloform reaction, thanks to the methyl group next to the carbonyl group. Such reactions alter the carbon structure of the molecule, allowing conversion into different functional groups like carboxylic acids.

Specifically, in the haloform reaction discussed for butan-2-one, the methyl next to the carbonyl group is crucial, enabling the halogenation and subsequent bond cleavage that facilitate the reaction.

Carboxylic Acids

Carboxylic acids are organic acids that contain a carboxyl group (\( \text{-COOH} \)). This group is composed of a carbonyl group (\( \text{C=O} \)) and a hydroxyl group (\( \text{-OH} \)) attached to the same carbon atom. They are known for their acidic properties and ability to participate in various chemical reactions. Carboxylic acids are important in many fields, including biochemistry, where they form the basis for amino acids.

When discussing the conversion of butan-2-one to propanoic acid, understanding carboxylic acids is crucial. This reaction reduces the carbon chain length of the original ketone and transforms it into a carboxylic acid. The haloform reaction described is an effective method to achieve this transformation — showing the flexibility and importance of carboxylic acids in organic synthesis.

This conversion process demonstrates how molecules can be restructured from a simple ketone to a valuable carboxylic acid, a chemical transformation fundamental to many organic processes and industrial applications.