Question

An organic compound of molecular formula \(\mathrm{C}_{7} \mathrm{H}_{10}\) on reaction with \(\mathrm{O}_{3}\), Zn gives Acetic acid as one of the major product. Compound is: (A) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) (B) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\) (C) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) (D) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\)

Short answer

The correct structure of the given organic compound is option C: $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{3}$.

Step-by-step solution

Option A: CH3-CH=CH-CH2-C≡C-CH3

Perform the ozonolysis and reduction reaction: Ozone breaks the multiple bonds of the compound and forms ozonides, which are then reduced to carbonyl compounds by Zn. We will have the following products: Acetone (from the double bond): \(CH_3-CO-CH_3\) Acid anhydride (from the triple bond): \(CH_3-C(O)-O-C(O)-CH_3\) Acetic acid is not one of the major products in this reaction. So, option A is incorrect.

Option B: CH3-CH=CH-CH=CH-CH=CH2

Perform the ozonolysis and reduction reaction: Ozone breaks the multiple bonds of the compound and forms ozonides, which are then reduced to carbonyl compounds by Zn. We will have the following products: 3 molecules of Acetaldehyde (from the three double bonds): \(3 \times CH_3-CO-H\) Acetic acid is not one of the major products in this reaction. So, option B is incorrect.

Option C: CH2=CH-CH2-C≡C-CH2-CH3

Perform the ozonolysis and reduction reaction: Ozone breaks the multiple bonds of the compound and forms ozonides, which are then reduced to carbonyl compounds by Zn. We will have the following products: Acetaldehyde (from the double bond): \(CH_3-CO-H\) Acetic acid (from the triple bond): \(CH_3-CO-OH\) Since acetic acid is one of the major products in this reaction, option C is the correct answer.

Option D: CH2=CH-CH2-CH2-CH=CH-CH3

For completeness, perform the ozonolysis and reduction reaction: Ozone breaks the multiple bonds of the compound and forms ozonides, which are then reduced to carbonyl compounds by Zn. We will have the following products: Acetaldehyde (from the first double bond): \(CH_3-CO-H\) Propanal (from the second double bond): \(CH_3-CH_2-CO-H\) Acetic acid is not one of the major products in this reaction. So, option D is incorrect. To conclude, the structure of the given organic compound is option C: $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{3}$.

Key concepts

Organic Compound Structure Identification

Understanding the structure of organic compounds is fundamental to organic chemistry. The process often involves deducing the arrangement of atoms based on certain chemical reactions and their products. As illustrated in the exercise, structure identification can be narrowed down using the molecular formula and certain reactions such as ozonolysis.

In the given problem, each option represents a different structure with varying types and positions of bonds. When a molecule undergoes ozonolysis, the reaction specifically targets multiple bonds, such as double or triple bonds, to form carbonyl compounds. The formation of acetic acid as a product is a crucial piece of information, as it indicates that the molecule must have a particular set-up of bonds that would yield acetic acid upon ozonolysis and subsequent reduction. By methodically analyzing each option and considering the products of ozonolysis, the correct structure is identified. This approach teaches a systematic way to narrow down structural possibilities, enhancing problem-solving skills for JEE Main and Advanced exams.

Carbonyl Compound Formation

Carbonyl compounds are characterized by a carbon atom double-bonded to an oxygen atom. They are commonly formed through ozonolysis, which splits carbon-carbon multiple bonds (double or triple bonds) in alkenes or alkynes.

Ozonolysis cleaves the bonds, and the reaction with zinc reduces the intermediate ozonides to form carbonyl compounds such as aldehydes or ketones. For instance, in the exercise, ozonolysis of a double bond leads to acetaldehyde, while cleavage of a triple bond gives acetic acid. Knowing how these products arise from different bond types is crucial for predicting the outcome of reactions and is an important aspect ofJEE Main and Advanced syllabus, helping students to solve complex organic chemistry problems.

JEE Main and Advanced Organic Chemistry

The JEE Main and Advanced exams are critical gateways for students aspiring for engineering courses in India, and organic chemistry is an integral part of these tests. It encompasses a wide range of topics, including reaction mechanisms, compound identification, and synthesis.

The understanding of reactions like ozonolysis and the ability to infer structural information from reaction products form core skills tested in these exams. Deep knowledge of functional groups, reaction pathways, and structural analysis can help students not only in answering multiple choice questions but also in tackling the comprehensive problem solving required for advanced levels. Additionally, exercises similar to the one discussed guarantee a better grasp of organic principles and enhance students' readiness for the competitive environment of JEE exams.