Question

Which of the following would undergo a haloform reaction when treated with \(\mathrm{Br}_{2}\) in the presence of NaOH: (a) pentan-3-one; (b) butanone; (c) propanone; (d) hexan-2-one; (e) hexan-3-one? Give the products in each case where you state the reaction will occur.

Short answer

Butanone (b) and propanone (c) undergo a haloform reaction, forming bromoform and sodium propanoate or sodium acetate, respectively.

Step-by-step solution

Identify Eligible Compounds for Haloform Reaction

The haloform reaction occurs with methyl ketones. These ketones have the structure RCOCH₃. For each compound, determine if it has this structure. Pentan-3-one (a), hexan-2-one (d), and hexan-3-one (e) do not have a methyl group attached directly to the carbonyl group, so they are not methyl ketones. Butanone (b) and propanone (c) are both methyl ketones.

Predict the Haloform Reaction Product

When methyl ketones undergo a haloform reaction with \( \mathrm{Br}_{2} \) in the presence of NaOH, the product is a carboxylate ion and a haloform (e.g., bromoform for \( \mathrm{Br}_{2} \)). For butanone (b), the reaction results in the formation of sodium propanoate and bromoform. For propanone (c), the reaction results in the formation of sodium acetate and bromoform.

Write Balanced Chemical Equations

For butanone reacting with \( \mathrm{Br}_{2} \) and NaOH:\[ \mathrm{CH}_3\mathrm{COCH}_2\mathrm{CH}_3 + 3\mathrm{Br}_2 + 4\mathrm{NaOH} \rightarrow \mathrm{CH}_3\mathrm{CH}_2\mathrm{COONa} + 3\mathrm{NaBr} + 3\mathrm{H}_2\mathrm{O} + \mathrm{CHBr}_3 \]For propanone reacting with \( \mathrm{Br}_{2} \) and NaOH:\[ \mathrm{CH}_3\mathrm{COCH}_3 + 3\mathrm{Br}_2 + 4\mathrm{NaOH} \rightarrow \mathrm{CH}_3\mathrm{COONa} + 3\mathrm{NaBr} + 3\mathrm{H}_2\mathrm{O} + \mathrm{CHBr}_3 \]

Key concepts

Methyl Ketones

Methyl ketones are a specific type of ketones that play a key role in the haloform reaction. The structure of a methyl ketone is characterized by the presence of a methyl group ( \( \mathrm{CH}_3 \) ) attached directly to the carbonyl group ( \( \mathrm{C}=\mathrm{O} \) ). This unique configuration allows methyl ketones to react distinctively compared to other ketone structures.

If we consider ketones like butanone and propanone, they both have the requisite methyl group next to the carbonyl. This feature makes them eligible candidates for the haloform reaction. In contrast, compounds such as pentan-3-one and hexan-3-one do not exhibit this critical structural attribute, thus excluding them from participating in the haloform reaction.

Understanding the structure of methyl ketones is crucial for predicting their reactivity and determining their suitability for specific chemical reactions.

Carboxylate Ion

The carboxylate ion is a fundamental product of the haloform reaction when it involves a methyl ketone treated with halogens like bromine ( \( \mathrm{Br}_2 \) ) in a basic environment. When a methyl ketone reacts, the transformation that occurs leads to the substitution of the methyl group for a carboxylate ion. This ion typically takes the form of either an acetate or propanoate, depending on the starting material.

During the haloform reaction, the methyl group adjacent to the carbonyl undergoes halogenation followed by hydrolysis. This process breaks down the structure, resulting in the formation of a carboxylate ion. For example, in the case of butanone as a reactant, the carboxylate ion produced is sodium propanoate. Meanwhile, propanone forms sodium acetate as its carboxylate product.

The presence of the carboxylate ion underscores the efficiency of the haloform reaction in converting methyl ketones into simpler organic molecules.

Bromoform

Bromoform ( \( \mathrm{CHBr}_3 \) ) is an important haloform produced when the haloform reaction is carried out using bromine ( \( \mathrm{Br}_2 \) ) as the halogen. This compound is characterized by its three bromine atoms bonded to a central carbon atom. Bromoform is thus a trihalomethane and one of the classic products in reactions involving methyl ketones.

The production of bromoform can be visualized in the haloform reaction, where a methyl ketone reacts under the influence of bromine and a basic medium like NaOH. This reaction not only produces a carboxylate ion but also results in the formation of bromoform as a byproduct. The creation of bromoform from the methyl group in the ketone's original structure marks an essential aspect of the chemical transformation.

Understanding how bromoform is formed and its characteristics helps in comprehending broader chemical processes involving the haloform reaction.

Chemical Reaction Equations

Chemical reaction equations serve as the blueprint for understanding the transformations taking place during the haloform reaction. These equations display the reactants, conditions, and products, maintaining a balanced view of the atoms involved.

For instance, the balance of reactants and products in the haloform reaction with butanone and propanone is critical. The equations for both compounds start with the addition of bromine ( \( \mathrm{Br}_2 \) ) and sodium hydroxide ( \( \mathrm{NaOH} \) ) to the respective methyl ketone. This leads to the creation of a carboxylate ion, along with bromoform, sodium bromide ( \( \mathrm{NaBr} \) ), and water ( \( \mathrm{H}_2\mathrm{O} \) ).

Writing and balancing these chemical equations is central for chemists to predict the outcome of reactions, helping them understand the stoichiometry and the conservation of mass during chemical processes.