Question

The major product R obtained in the following reaction is/are : $$ \mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2} \frac{\mathrm{O}_{3}}{\underset{\left(\mathrm{l}, \mathrm{H}_{2}\right)}{\mathrm{Z}} \mathrm{O}} \mathrm{P}+\mathrm{Q} \frac{\mathrm{H}_{2} / \mathrm{Pd}}{\mathrm{B}_{2} \mathrm{SO}_{4}}{\longrightarrow} \mathrm{R} $$ (A) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{OH}\) (B) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{O}\) (C) \(\mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{OH}\) (D) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH}\)

Short answer

The major product R in the given reaction is (D) CH3-CH2-CH2-OH.

Step-by-step solution

Ozonolysis of the reactant

For the first step, we have the reactant HC≡C-CH=CH2, undergoing ozonolysis using O3 (ozone) and Zn/H2O as reducing agents. Ozonolysis is the cleavage of alkenes and alkynes to form carbonyl compounds (aldehydes or ketones). The alkyne will break at its triple bond, and the alkene will break at its double bond. Thus, the product formation can be summarized as \( HC \equiv C - CH = CH_{2} \xrightarrow{O_{3}/Zn} P + Q\). For the alkyne part, HC≡C becomes HC=O (formaldehyde) due to the cleavage. For the alkene part, CH=CH2 becomes CH2=O (acetaldehyde). So, P = Formaldehyde (HCHO) and Q = Acetaldehyde (CH3CHO).

Hydrogenation

During the second step, we have the reaction of P and Q with H2/Pd(B2SO4) as the catalyst. H2/Pd is a hydrogenation catalyst that can reduce the carbonyl group to form alcohols. Reaction: P + Q → R HCHO + CH3CHO → R Formaldehyde is reduced to methanol (CH3OH), and acetaldehyde is reduced to ethanol (CH3CH2OH). Finally, the two alcohols will combine to form the final product.

The Major Product

Based on the above steps, the \(Major Product (R) = CH_{3}OH + CH_{3}CH_{2}OH = CH_{3} CH_{2} CH_{2} OH\). Thus, the correct answer is: (D) CH3-CH2-CH2-OH

Key concepts

Ozonolysis in Organic Reactions

Organic chemistry JEE advanced questions often explore the fascinating process of ozonolysis. This reaction involves the cleavage of carbon-carbon double or triple bonds in alkenes or alkynes when treated with ozone, a powerful oxidizing agent.

Ozonolysis is commonly used to determine the position of a double bond in an unknown alkene or to cleave a double bond in order to synthesize smaller, functional compounds. During the reaction, ozone reacts with the double bond to form an unstable ozonide which is then reduced to yield carbonyl compounds. If an alkyne with a terminal triple bond like in the given exercise HC≡C-CH=CH2 undergoes ozonolysis, we would expect the formation of separate carbonyl compounds corresponding to the cleaved parts.

In the step-by-step solution provided, we saw that the alkyne part, HC≡C, yielded formaldehyde (HCHO), while the alkene part, CH=CH2, produced acetaldehyde (CH3CHO). This demonstrates the utility of ozonolysis in breaking down complex molecules into simpler molecules, which can then be further manipulated through organic reactions.

Understanding the basics of ozonolysis through such exercises can be crucial for students preparing for competitive exams like the JEE Advanced, as it lays the foundation for more complex synthetic pathways and analysis of organic molecular structures.

Alkyne and Alkene Reactions

Alkynes and alkenes are hydrocarbons with triple and double bonds respectively. These unsaturated bonds are reactive sites that make these molecules versatile in organic synthesis. Reactions of alkynes and alkenes are essential parts of organic chemistry molecules often react through addition reactions, where other atoms or molecules attach to the carbon atoms involved in the double or triple bond.

For instance, in the case of our JEE advanced exercise, the reactant molecule HC≡C-CH=CH2 features both an alkyne and an alkene. During ozonolysis, as discussed earlier, these unsaturated bonds are cleaved to form their respective carbonyl compounds.

In a broader scope, reactions involving alkynes might lead to similar or different products compared to alkenes due to the involvement of additional pi bonds. Alkynes, for example, can undergo hydrohalogenation, halogenation, hydroboration-oxidation, and hydrogenation, just like alkenes, but the process might require different conditions or may lead to different products due to the extra pi bond.

For students, understanding these reactions is crucial not only for solving problems but also for designing new synthetic routes in the lab. These principles form the building blocks for a wide range of organic synthetic strategies and are regularly tested in competitive exams like the JEE Advanced.

Hydrogenation of Carbonyl Compounds

Hydrogenation, the addition of hydrogen to a molecule, plays a significant role in transforming carbonyl compounds into alcohols. This reaction is a type of reduction, wherein the carbonyl group (C=O) gets converted to a hydroxyl group (OH).

In the context of the JEE Advanced problem given, after the initial ozonolysis, the produced carbonyl compounds, formaldehyde (HCHO) and acetaldehyde (CH3CHO), were subjected to hydrogenation. Using a palladium catalyst supported by Barium sulfate (Pd/B2SO4) and hydrogen gas (H2), these compounds were reduced to their corresponding alcohols: methanol (CH3OH) and ethanol (CH3CH2OH), respectively.

Hydrogenation typically requires a metal catalyst such as palladium, platinum, or nickel to facilitate the addition of hydrogen atoms to the carbon atom of the carbonyl group. The catalyst provides a surface where the reaction can take place more readily. This step is essential in synthesizing alcohols from carbonyl compounds, which has wide applications in industries ranging from pharmaceuticals to perfumeries.

Students examining this reaction should bear in mind the reagents and conditions needed, as various catalysts and hydrogen sources can influence the rate and outcome of the reduction. Such reactions are a testament to the intricacy and precision necessary for successful transformations in organic chemistry, and a clear understanding of these can significantly assist those aiming for high scores in exams like the JEE Advanced.