Question

When acetone reacts with chloroform in the presence of alkali like \(\mathrm{NaOH}\), the product formed is (a) Chloropionium (b) acetic acid (c) Chloretone (d) mesityloxide

Short answer

Answer: Chloretone (CH2=COCHCl).

Step-by-step solution

Identify the reactants and their functional groups

In this reaction, we have the following reactants: 1. Acetone: a ketone with the structure CH3-CO-CH3 2. Chloroform: a "haloform" with the structure CHCl3 3. NaOH: a base The functional groups are the carbonyl group (C=O) in acetone and the haloform group in chloroform.

Determine the reaction mechanism

Given the reactants and the presence of a base, this reaction is an example of the haloform reaction. In the haloform reaction, a haloform (like chloroform) undergoes a nucleophilic attack on the carbonyl group of a ketone (like acetone) in the presence of a base (like NaOH). The mechanism proceeds through deprotonation, formation of an enolate ion, and a nucleophilic substitution reaction.

Write down the reaction steps

Here is the step-by-step breakdown of the reaction: 1. Deprotonation: The NaOH base abstracts a proton from acetone, forming water and an enolate ion. \[\mathrm{CH}_{3}\mathrm{COCH}_{3} + \mathrm{NaOH} \rightarrow \mathrm{CH}_{2}\mathrm{COCH}_{3}^{-} + \mathrm{H}_{2}\mathrm{O} + \mathrm{Na}^{+}\] 2. Nucleophilic substitution: The enolate ion formed in the first step attacks the haloform (chloroform) at the carbon center, displacing a chloride ion and forming an intermediate product. \[\mathrm{CH}_{2}\mathrm{COCH}_{3}^{-} + \mathrm{CHCl}_{3} \rightarrow \mathrm{CH}_{2}\mathrm{COCH}_{2}\mathrm{Cl} + \mathrm{Cl}^{-}\] 3. Proton abstraction: Another molecule of NaOH deprotonates the intermediate product, leading to the formation of the final product and regeneration of the enolate ion. \[\mathrm{CH}_{2}\mathrm{COCH}_{2}\mathrm{Cl} + \mathrm{NaOH} \rightarrow \mathrm{CH}_{2}\mathrm{COCHCl} + \mathrm{H}_{2}\mathrm{O} + \mathrm{Na}^{+}\]

Identify the correct product

The final product formed in the reaction is CH2=COCHCl, also known as chloretone. Therefore, the correct answer is (c) Chloretone.

Key concepts

Haloform Reaction

The haloform reaction is a fascinating chemical process where a molecule containing a methyl ketone group reacts with a source of halogen, such as chloroform, in the presence of a base like sodium hydroxide (NaOH). This reaction typically results in the formation of a haloform (a trihalomethane), such as chloroform, and a carboxylate ion.

The reaction begins with the base deprotonating the methyl ketone to form an enolate ion, which is crucial for the next step. Following this, the newly formed enolate ion adds to the halogen molecule (like CHCl₃) through a nucleophilic attack. Subsequent steps involve multiple halogenations and another deprotonation, which culminate in cleavage of the carbon skeleton and release of the haloform along with the formation of a carboxylate ion.

It's significant to note, in our example with acetone and chloroform, the reaction product is chloretone rather than a classic haloform and carboxylate, since chloretone contains the chloride substituent from the final nucleophilic substitution reaction.

Nucleophilic Substitution

The concept of nucleophilic substitution is central to understanding a myriad of organic reactions, including the haloform reaction. It involves the substitution of a leaving group by a nucleophile. In simpler terms, a more electron-rich atom or molecule (the nucleophile) attacks an electron-deficient center (often a carbon atom within a larger molecule), effectively exchanging places with a group that departs (the leaving group).

In the context of our example, the enolate ion created during the haloform reaction acts as the nucleophile. During the reaction, it attacks the carbon in chloroform, which is bonded to halogen atoms. These halogens are good leaving groups because they are electronegative and can stabilize the negative charge after departure. The result of this nucleophilic attack is the formation of a new bond and the release of a halide ion, establishing the groundwork for the eventual product, chloretone.

Enolate Ion

An enolate ion is a negatively charged and highly reactive intermediate that plays a pivotal role in several organic reactions, including aldol condensations and, relevant here, the haloform reaction. Enolate ions form when a base abstracts a proton from an alpha-carbon next to a carbonyl group (found in aldehydes and ketones), resulting in a resonance-stabilized anion.

The formation of the enolate is key to the haloform reaction as it enables nucleophilic attack on the haloform. When acetone reacts with chloroform in an alkaline environment, an enolate ion is formed from the acetone. The NaOH in our example removes a proton from a methyl group of acetone, resulting in the enolate, which then acts as the nucleophile to advance the reaction. Understanding the enolate's role is crucial for grasping not just the haloform reaction but a host of other important mechanisms in organic chemistry.