Question

Yellow crystal of \(\mathrm{CHI}_{3}\) is obtained using iodoform test by (A) \(\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{OH}\) (B) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{O}\) (C) \(\mathrm{CH}_{3}-\mathrm{C}-\mathrm{CH}_{2}-\mathrm{C}-\mathrm{CH}_{3}\) (D) All of these

Short answer

The correct answer is (D) All of these, as both (A) CH3CH2OH and (B) CH3CH=O will form yellow crystals of CHI3 using the iodoform test, while (C) CH3CCH2CCH3 will not.

Step-by-step solution

Analyze compound (A)

CH3CH2OH (ethanol) can undergo the iodoform test due to the presence of a CH3CH2OH group. The test will produce a positive result, forming a yellow precipitate of CHI3.

Analyze compound (B)

CH3CH=O (acetone) can also undergo the iodoform test as it contains a CH3CO- group (acetone is a methyl ketone). The test will result in a positive outcome, forming a yellow precipitate of CHI3.

Analyze compound (C)

CH3CCH2CCH3 (diethyl ether) does not contain a CH3CO- group or ethanol group, so it cannot give a positive iodoform test. Therefore, this compound will not form a yellow precipitate of CHI3. #Conclusion# Based on the analysis of the compounds, (A) CH3CH2OH and (B) CH3CH=O will result in the formation of yellow crystals of CHI3 using the iodoform test, while (C) CH3CCH2CCH3 will not. The correct answer is: (D) All of these

Key concepts

Understanding Organic Chemistry

Organic chemistry is a vast and fascinating branch of chemistry that encompasses the study of carbon-containing compounds and their properties, reactions, and applications. Not only does it cover compounds with various functional groups, but it also includes an understanding of reaction mechanisms and chemical synthesis. This is particularly relevant for students preparing for competitive exams like the Joint Entrance Examination (JEE), which is a gateway to prestigious engineering institutes in India.

For JEE aspirants, mastering organic chemistry involves learning about various reactions, including nucleophilic substitution, elimination, addition, and more specialized reactions such as the haloform reaction. Fluency in identifying functional groups and predicting their behavior in reactions is crucial. Understanding the foundational theories, such as hybridization, resonance, and electronic effects, enhances the ability to tackle complex problems, which ultimately reflects in a student's performance in the organic chemistry section of JEE.

Strategizing for JEE Organic Chemistry

Competitive exams like the JEE require applicants to not only remember organic chemistry concepts but also apply them quickly and efficiently. To excel in the JEE organic chemistry section, one must hone problem-solving skills through consistent practice and application of concepts. Strategies that can benefit students include:

• Creating a strong foundation by thoroughly understanding the basic principles of organic reactions and mechanisms.
• Practicing reaction mechanisms until they become intuitive, thus speeding up the problem-solving process.
• Approaching multi-step synthesis problems by working backwards from the final product to the reactants.
• Keeping handy summarised notes of reaction types, reagents, conditions, and exceptions, which are crucial for last-minute revisions.

By integrating these strategies, students can approach JEE organic chemistry problems with confidence, whether they are aiming for a top rank or simply looking to clear the cutoff.

Demystifying the Haloform Reaction

The haloform reaction is a type of organic reaction where a methyl ketone (or similar structure, such as ethanol) reacts in the presence of halogens and a base to form a haloform. The iodoform test is a practical application of this reaction and is used to identify the presence of a methyl ketone or ethanol in a compound.

In the iodoform test, iodine and a base (commonly sodium hydroxide) react with the compound in question. If the compound contains a methyl ketone group, characterized by CH3CO-, or an ethanol group (CH3CH2OH), a yellow precipitate of iodoform (CHI3) is formed. This can be observed as yellow crystals. This test, therefore, provides a visual confirmation of the functional groups present and is a valuable tool for organic chemists.

The mechanism involves several steps, starting with the halogenation of the methyl group adjacent to the carbonyl or alcohol group, followed by the formation and cleavage of a trihalomethyl intermediate. This reaction is not limited to only iodine but can also be performed with chlorine and bromine, creating chloroform and bromoform, respectively. Understanding the haloform reaction is crucial as it not only serves a diagnostic purpose but is also representative of how halogens behave in organic compounds.