Problem 47 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An alkene (A) \(\mathrm{C}_{16} \mathrm{H}_{16}\) on ozonolysis gives only one product (B) \(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}\). Compund (B) on reaction with \(\mathrm{NaOH} / \mathrm{I}_{2}\) yields sodium benzoate. Compound (B) reacts with \(\mathrm{KOH} / \mathrm{NH}_{2} \mathrm{NH}_{2}\) yielding a hydrocarbon (C) \(\mathrm{C}_{8} \mathrm{H}_{10}\). Write the structures of compounds (B) and (C). Based on this information, two isomeric structures can be proposed for alkene (A). Write their structures and identify the isomer which on catalytic hydrogenation \(\left(\mathrm{H}_{2} / \mathrm{Pd}-\mathrm{C}\right)\) gives a racemic mixture.

Step-by-Step Solution

Answer

Compound B is benzaldehyde; C is ethyl benzene. Alkene A is cis-stilbene for racemic mixture.

Step 1: Determine the Structure of Compound B

The ozonolysis of alkene A (\( \mathrm{C}_{16} \mathrm{H}_{16} \)) results in a single product B (\( \mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O} \)). Since it gives sodium benzoate upon reaction with \( \mathrm{NaOH} / \mathrm{I}_{2} \), compound B likely contains a benzaldehyde group, as benzaldehyde can undergo this reaction to form sodium benzoate.

Step 2: Identify Possible Isomeric Structures for Compound A

Compound A can have two double bonds since it is an alkene with the formula \( \mathrm{C}_{16} \mathrm{H}_{16} \). It should be constructed to generate two identical benzaldehyde molecules upon ozonolysis. Hence, compound A is likely stilbene, which can be either cis-stilbene or trans-stilbene.

Step 3: Reactions of Compound B to Find Compound C

When compound B reacts with \( \mathrm{KOH} / \mathrm{NH}_{2} \mathrm{NH}_{2} \), it undergoes a reduction reaction, likely the Wolff-Kishner reduction, converting benzaldehyde to toluene (\( \mathrm{C}_{7} \mathrm{H}_{8} \)) to give compound C, which has the formula \( \mathrm{C}_{8} \mathrm{H}_{10} \). This indicates compound C contains an additional methyl group, or ethyl benzene.

Step 4: Determine the Racemic Mixture

On catalytic hydrogenation, cis-stilbene would convert to a racemic mixture of 1,2-diphenylethane, because the two phenyl groups will be added on the same side of what was the plane of the double bond, leading to chiral centers. Trans-stilbene would not lead to such a mixture as the groups are added across from each other.

Key Concepts

Understanding Ozonolysis

Ozonolysis is a crucial reaction in organic chemistry, especially for alkenes. In this process, ozone (O₃) reacts with a double bond to cleave it, forming carbonyl compounds. This mechanism includes the following steps:

• The ozone molecule adds to the carbon-carbon double bond, creating an ozonide intermediate.
• The unstable ozonide breaks down, often in the presence of a reductive or oxidative workup, resulting in carbonyl compounds like aldehydes and ketones.

In the exercise, compound (A) undergoes ozonolysis to give a single product (B), a carbonyl compound with the formula C₈H₈O. Given that product B forms sodium benzoate upon treatment with NaOH/I₂, it's most likely a benzaldehyde structure. This reaction is used frequently to determine the position of double bonds in larger molecules by analyzing the resultant smaller fragments.

Exploring Stilbene Isomers

Stilbene represents a class of compounds with the general formula C₁₄H₁₂, characterized by a central carbon-carbon double bond flanked by two phenyl groups. There are two geometric isomers of stilbene:

• Cis-stilbene: The phenyl groups are on the same side of the double bond, making this isomer less stable due to steric hindrance.
• Trans-stilbene: The phenyl groups are on opposite sides, resulting in a more stable configuration.

In the provided exercise, the ozonolysis of the alkene A forms polybenzaldehyde compounds denoting the structure of an isomer of stilbene. Cis-stilbene, as specified, results in a racemic mixture upon catalytic hydrogenation (H₂/Pd-C), creating a symmetry that allows for chiral centers. This racemic mixture feature is noteworthy for illustrating the distinctive properties between the isomers due to their molecular geometry differences.

Wolff-Kishner Reduction Explained

The Wolff-Kishner reduction is a valuable organic reaction employed to systematically convert carbonyl groups, like aldehydes and ketones, into hydrocarbons. This process involves:

• Treatment with hydrazine (NH₂NH₂) forming an intermediate hydrazone.
• The base, typically KOH, facilitates the removal of nitrogen gas, translating the carbonyl to a methylene group.

In the textbook exercise, compound B undergoes Wolff-Kishner reduction to transform into a hydrocarbon (C), C₈H₁₀. This hydrocarbon resembles ethyl benzene, marking a significant step in organic synthesis where functional groups are selectively removed to form simpler hydrocarbons. As such, understanding this reaction sequence not only reinforces concepts of molecular transformations but illustrates real-world applications in organic product formation.