Problem 141 - A Complete Resource Book in Chemistry for JEE Main

Question

To convert 2 -butanone into propionic acid (a) \(\mathrm{NaOH}+\mathrm{NaI} / \mathrm{H}^{+}\) (b) Fehling's test (c) \(\mathrm{I}_{2} / \mathrm{NaOH} / \mathrm{H}^{+}\) (d) Tollen's reagent

Step-by-Step Solution

Answer

Use option (c): \(\mathrm{I}_2 / \mathrm{NaOH} / \mathrm{H}^+\) for the Haloform reaction to convert 2-butanone to propionic acid.

Step 1: Analyze the Requirements

We need to convert 2-butanone, which is a ketone, into propionic acid, a carboxylic acid. The challenge is to remove one carbon from the 4-carbon ketone to form the 3-carbon carboxylic acid.

Step 2: Understanding the Haloform Reaction

The Haloform reaction is a reaction where methyl ketones are converted into carboxylic acids with one less carbon atom. The group that is cleaved off is transformed into a haloform (like chloroform, iodoform), making this a suitable conversion method.

Step 3: Evaluate the Options

Let's examine the options: - (a) - (b) is Fehling's test, which identifies aldehydes, not a fit for our ketone.- (c)  \(\mathrm{I}_{2} / \mathrm{NaOH} / \mathrm{H}^{+}\) is a classic reagent mix for the Haloform reaction, leading to carboxylic acids from methyl ketones.- (d) Tollen's reagent also detects aldehydes, again not suitable here.

Step 4: Selection and Conclusion

Option (c) employs the Haloform reaction, where 2-butanone reacts with \(\mathrm{I}_2\), \(\mathrm{NaOH}\), and \(\mathrm{H}^+\) to form propionic acid. This reagent combination cleaves the methyl group efficiently to create the desired carboxylic acid.

Key Concepts

Ketone Conversion

Ketone conversion is an essential chemical transformation where a ketone is transformed into another functional group. Ketones typically have the structure R-CO-R', where R and R' are carbon-containing groups. In our context, we need to convert a specific ketone, 2-butanone, into a carboxylic acid, propionic acid. This process involves changing the nature of the molecule by removing one of its carbon atoms. Understanding this transformation is crucial when dealing with chemical reactions that modify molecular structures. The conversion process often requires specific conditions, such as the presence of strong reagents and sometimes a catalyst to drive the reaction forward. For 2-butanone, the goal is to remove the terminal methyl group, which is achieved through a reaction that facilitates breaking this specific carbon-carbon bond. This is where the Haloform reaction plays a critical role, as it specializes in cleaving off a terminal carbon group from methyl ketones by transforming it into a stable byproduct like iodoform.

Carboxylic Acid Formation

Forming a carboxylic acid from a ketone involves a significant transformation in the functional group of the molecule. Carboxylic acids have the distinct structure R-COOH, where the key change from a ketone is the introduction of the -OH group along with the retention of the C=O bond.In the haloform reaction, the initial ketone compound first undergoes substitution where one or more hydrogen atoms are replaced by halogen atoms. The specific reagent mix of \(\mathrm{I}_{2}/\mathrm{NaOH}/\mathrm{H}^{+}\) facilitates this substitution, which primes the molecule for further reaction.

• The critical step is the formation of a carboxylate intermediate.
• This intermediate forms when the halogenated ketone undergoes cleavage, induced by the reagent mix.
• The subsequent acidification step converts this carboxylate into the final carboxylic acid.

Through this process, starting with substitution and finishing with cleavage and acid generation, the conversion to a carboxylic acid is completed.

2-butanone to Propionic Acid

The transition from 2-butanone to propionic acid is executed efficiently through a haloform reaction, which is a targeted approach to reduce the carbon chain by cleaving the methyl group at the end of the ketone structure. 2-butanone, which has the formula \(\mathrm{CH_3COC_2H_5}\), reacts with the reagents \(\mathrm{I}_{2}/\mathrm{NaOH}/\mathrm{H}^{+}\) to undergo this transformation.In this reaction:

• The methyl group (\(\mathrm{CH_3}\)) is halogenated, typically converting into a triiodomethyl group (\(\mathrm{C I_3}\)).
• Under alkaline conditions, this group is cleaved, resulting in the formation of triiodomethane (iodoform) and leaving behind a carboxylate ion.
• Upon acidification, the carboxylate ion is converted to propionic acid (\(\mathrm{CH_3CH_2COOH}\)).

Thus, the 2-butanone reduces by one carbon atom while transforming into the desired carboxylic acid product through this meticulous chemical path.