Question

Iodoform can be prepared from all except (a) 3 - methyl \(-2\)-butanone (b) Isopropyl alcohol (c) Ethyl methyl ketone (d) Isobutyl alcohol

Short answer

The answer is (d) isobutyl alcohol.

Step-by-step solution

Identify the Reaction

The question is about preparing iodoform, which is typically formed in an iodoform test. The iodoform test is a chemical reaction to identify methyl ketones or ethanol. It involves the reaction with iodine and a base.

Understand the Criteria for Iodoform Test

The compounds that give a positive iodoform test have either a methyl ketone group (CH3CO-) or secondary alcohols that can be oxidized to methyl ketones, like ethanol and isopropyl alcohol.

Analyze Each Option

- (a) 3-Methyl-2-butanone: Contains a methyl group next to the ketone, fits CH3CO- - (b) Isopropyl alcohol: It's a secondary alcohol that oxidizes to acetone, fits the requirement. - (c) Ethyl methyl ketone: Contains CH3COCH2CH3, giving a positive test. - (d) Isobutyl alcohol: A primary alcohol, not oxidized to a methyl ketone, and won't give iodoform.

Conclude Based on Analysis

Since isobutyl alcohol (option d) is a primary alcohol and does not form a methyl ketone upon oxidation, it does not meet the criteria for the iodoform test.

Key concepts

Methyl Ketones

Methyl ketones are a category of organic compounds that contain the functional group -COCH3. This group is what makes these compounds reactive in a distinct manner. Methyl ketones play a crucial role in the iodoform test, which is a common method in organic chemistry used to identify them or their derivatives. The presence of the methyl group (CH₃) adjacent to the carbonyl group (C=O) in methyl ketones is key for reactivity. This configuration allows methyl ketones to easily undergo halogenation reactions. When iodine (\(I_2\) or triiodide, I₃⁻) is used in the presence of a base like hydroxide (OH⁻), these ketones form a yellow precipitate known as iodoform (CHI₃). This identity is specific for compounds possessing the -COCH₃ group.

It's important to recognize that not all ketones will give a positive iodoform reaction. Only those that contain the methyl group directly attached to the carbonyl carbon do so. This feature ensures that methyl ketones can be distinctly identified and separated from other carbonyl-containing compounds.

Iodoform Test

The iodoform test is a qualitative chemical test used to detect the presence of methyl ketones or secondary alcohols oxidizable to them. This test relies on a reaction that occurs when iodine is mixed with the compound in the presence of a strong base, usually sodium hydroxide (NaOH). During the test, iodine reacts with the methyl ketone or the secondary alcohol, forming a yellow precipitate of iodoform.

• Appearance: The precipitate is a yellow solid, often identified by its distinct smell.
• Scope: The test shows a positive result not only for methyl ketones but also for secondary alcohols that can be oxidized to such ketones, like ethanol and isopropyl alcohol.

Most organic chemistry students can perform this test easily because it serves as a clear indicator, making it a staple in laboratory experiments.

However, not all alcohols respond positively. For example, primary alcohols like isobutyl alcohol do not oxidize to methyl ketones and thus will not yield a yellow precipitate, allowing us to differentiate among types based on their oxidation potential.

Chemical Reactions

Chemical reactions are fundamental in understanding the transformation of substances. In the context of the iodoform reaction, chemical processes enable the identification of specific functional groups, like methyl ketones. When considering the iodoform test, chemical reactions occur in two main stages:

• Oxidation: Compounds like secondary alcohols are first oxidized to methyl ketones. This transformation is crucial because only methyl ketones can further react with iodine.
• Substitution: Iodine reacts with the methyl ketone to replace hydrogen atoms, leading to the formation of triiodomethane (iodoform, CHI₃), which precipitates out.

These reactions are guided by the principles of organic chemistry, where the reactivity of carbonyl groups plays a central role. The reactions demonstrate how molecular structure and environmental conditions, such as the presence of oxidizing agents and bases, dictate the pathway and products of chemical processes.

Understanding these reactions not only aids in practical laboratory work but also enhances comprehension of how molecular interactions result in observable phenomena like the yellow precipitate of the iodoform test.