Problem 82 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one of the following compounds will give a yellow precipitate on warming with a solution of iodine and aqueous alkali? (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}\) (c) CCC(C)O (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\)

Step-by-Step Solution

Answer

Option (c) will give a yellow precipitate.

Step 1: Understanding Iodoform Test

The compounds that give a yellow precipitate with iodine and aqueous alkali do so because they have a methyl ketone group \( (CH_3CO-) \) or secondary alcohols that can be oxidized to methyl ketones. This reaction is known as the iodoform test.

Step 2: Analyzing Option (a)

Option (a) \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3} \) is an ether. Ethers do not have a methyl ketone group and cannot oxidize to form one; hence it won't give the iodoform test.

Step 3: Analyzing Option (b)

Option (b) \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH} \) is a tertiary alcohol. Tertiary alcohols do not oxidize to form methyl ketones, so they will not give a yellow precipitate in the iodoform test.

Step 4: Analyzing Option (c)

Option (c) \( \mathrm{CCC(C)O} \) translates to \( \mathrm{CH_3CH_2CH(CH_3)OH} \) (butan-2-ol), a secondary alcohol. It can be oxidized to \( \mathrm{CH_3CH_2COCH_3} \) (methyl ketone), which will produce a yellow precipitate when warmed with iodine and aqueous alkali.

Step 5: Analyzing Option (d)

Option (d) \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} \) is a primary alcohol. Primary alcohols do not typically produce methyl ketones under oxidation, and thus will not give a yellow precipitate in the iodoform test.

Step 6: Concluding the Correct Compound

After analyzing each option, only \( \mathrm{CCC(C)O} \) has the right structure (secondary alcohol) that can be oxidized to a methyl ketone. Therefore, option (c) will give a yellow precipitate on warming with a solution of iodine and aqueous alkali.

Key Concepts

Methyl Ketone

A methyl ketone is a type of organic compound characterized by the presence of a carbonyl group (C=O) bonded directly to a methyl group (CH₃). This structure makes methyl ketones particularly reactive in certain chemical tests such as the iodoform test. The iodoform test is used to identify the presence of a methyl ketone or compounds that can be oxidized to a methyl ketone. These compounds, when warmed with iodine in the presence of an alkali, produce a yellow precipitate known as iodoform. This specific reaction is a classical example in organic chemistry that helps in the identification and study of ketones. It is essential to understand the structural requirements for a compound to display a positive iodoform test result.

Secondary Alcohol

Secondary alcohols are organic compounds where the hydroxyl group ( OH) is attached to a carbon atom that is also bonded to two other carbon atoms. A classic example is butan-2-ol. These types of alcohols are important in the iodoform test because they can be oxidized to form methyl ketones, which in turn react with iodine to produce yellow iodoform. This oxidation process involves the transformation of the secondary alcohol into a ketone, a process that highlights the versatility and importance of secondary alcohols in organic reactions. Understanding the role of secondary alcohols is crucial for deciphering reaction pathways in organic chemistry and predicting the outcomes of experiments like the iodoform test.

Oxidation Reaction

In organic chemistry, oxidation reactions involve the increase in the oxidation state of a molecule. Commonly, this is related to the addition of oxygen or the removal of hydrogen. Secondary alcohols are often oxidized to ketones, which is an example of an oxidation reaction. This transformation is essential in the iodoform test, where secondary alcohols are first oxidized to methyl ketones, enabling them to react with iodine and form a precipitate of iodoform. Such oxidation reactions are fundamental to understanding how different classifiers of alcohols behave under various chemical conditions. It is also a good demonstration of how oxidation reactions are pivotal in synthetic organic chemistry for functional group transformations.

Organic Chemistry Reactions

Organic chemistry reactions are processes that transform organic molecules into different compounds through chemical changes. These include rearrangements, combinations, decompositions, and many more. The iodoform test is a perfect example of a practical reaction used to identify specific functional groups within a compound. Such reactions are useful not only for identification but also for understanding the mechanisms by which these transformations occur. With the iodoform test, one can determine the presence of methyl ketones and secondary alcohols that can be oxidized into methyl ketones. This showcases the importance of reactions in organic chemistry, enabling chemists to infer structural features and reactivity of molecules through straightforward laboratory procedures. Each reaction tells a story about the molecular changes occurring and provides insight into the molecular architecture of the compounds involved.