Problem 79 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

79\. Identify \((\mathrm{X}),(\mathrm{Y})\) and \((\mathrm{Z})\) in the following synthetic scheme and write their structures. \(\mathrm{Ba}^{*} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow(\mathrm{X})\) gas \(\left[\mathrm{C}^{*}\right.\) denotes \(\left.\mathrm{C}^{14}\right]\) $$ \begin{aligned} &\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Br} \frac{\text { (i) } \mathrm{Mg} / \text { ether }}{\text { (ii) } \mathrm{X} \text { (iii) } \mathrm{H}_{3} \mathrm{O}^{\prime}}(\mathrm{Y}) \stackrel{\mathrm{LiAlH}_{4}}{\longrightarrow}(\mathrm{Z}) \\ &\text { [2001-5 Marks] } \end{aligned} $$ Explain the formation of labelled formaldehyde \(\left(\mathrm{H}_{2} \mathrm{C}^{*} \mathrm{O}\right)\) as one of the products when compound (Z) is treated with HBr and subsequently ozonolysed. Mark the \(C^{*}\) carbon in the entire scheme.

Step-by-Step Solution

Answer

\((X) = \mathrm{C}^{*}{O}_2, (Y) = \mathrm{CH}_2=\mathrm{CH}-\mathrm{C}^{*}OH, (Z) = \mathrm{CH}_2=\mathrm{CH}-\mathrm{C}^{*}H_2OH\).

Step 1: Identify compound (X)

The first reaction is between radioactive barium carbonate \( \mathrm{Ba}^* \mathrm{CO}_3 \) and sulfuric acid \( \mathrm{H}_2 \mathrm{SO}_4 \). This reaction produces barium sulfate \( \mathrm{BaSO}_4 \) and carbon dioxide \( \mathrm{CO}_2 \), where the carbon atom is labeled \( \left(\mathrm{C}^{14}\right) \). Thus, \((X)\) is \( \mathrm{C}^{*}{O}_2 \).

Step 2: Understand the Grignard reaction

In the next reaction, \( \mathrm{CH}_2=\mathrm{CH}-\mathrm{Br} \) is treated with magnesium in ether to form a Grignard reagent \( \mathrm{CH}_2=\mathrm{CHMgBr} \). This will subsequently react with compound \((X)\), \( \mathrm{C}^{*}{O}_2 \), to form an intermediate \((Y)\) that is followed by hydrolysis with \( \mathrm{H}_3\mathrm{O}^+ \). The product \((Y)\) formed is \( \mathrm{CH}_2=\mathrm{CH}-\mathrm{C}^{*}OH \) (labeled acrylic acid).

Step 3: Reduction with LiAlH4

Compound \((Y)\) undergoes a reduction with lithium aluminum hydride \( \mathrm{LiAlH}_4 \). During the reduction, the carboxylic acid group \( \mathrm{-C}^{*}OH \) in \( \mathrm{CH}_2=\mathrm{CH}-\mathrm{C}^{*}OH \) is reduced to form an alcohol. Therefore, compound \((Z)\) is \( \mathrm{CH}_2=\mathrm{CH}-\mathrm{C}^{*}H_2OH \) (labeled allyl alcohol).

Step 4: Formation of labeled formaldehyde

When compound \((Z)\) is treated with hydrogen bromide \( \mathrm{HBr} \) and subsequently ozonolysed, it undergoes an addition reaction followed by ozonolysis. \((Z)\) forms \( \mathrm{CH}_2Br-\mathrm{CH}_2-\mathrm{C}^{*}H_2OH \) initially, then ozonolysis cleaves \( -\mathrm{CH}=\mathrm{CH}- \) to form the labeled formaldehyde \( \mathrm{H}_2\mathrm{C}^{*}O \), as the double bonds are broken creating carbonyl compounds.

Key Concepts

Barium Carbonate Reaction

When barium carbonate (\( \mathrm{BaCO}_3 \)) reacts with sulfuric acid (\( \mathrm{H}_2 \mathrm{SO}_4 \)), a series of notable chemical changes take place. This process involves the formation of barium sulfate (\( \mathrm{BaSO}_4 \)) and the release of carbon dioxide (\( \mathrm{CO}_2 \)). What is significant about this reaction is when barium carbonate is labeled, it becomes radioactive, especially if the carbon is tagged as carbon-14 (\( \mathrm{C}^{14} \)).
The released carbon dioxide will also contain the labeled carbon, making it \( \mathrm{C}^{*}{O}_2 \). This product of the reaction is not only a common demonstration of an acid-carbonate interaction but is pivotal in tracing and studying reaction mechanisms involving carbon labeled compounds. Understanding this process lays the foundation for further reactions that rely on isotopic labeling, which assists in meticulously mapping out carbon's journey through subsequent reactions.

Carbon-14 Labeling

Carbon-14 labeling is a powerful technique used in scientific studies to trace the path a specific carbon atom takes in a compound or reaction. Carbon-14 (\( \mathrm{C}^{14} \)) is a radioactive isotope of carbon, distinct due to its ability to be detected through its radioactive emissions.
At the core of this labeling method is the replacement of a regular carbon atom with \( \mathrm{C}^{14} \). When integrated into molecules, the carbon's path through chemical reactions can be precisely tracked, offering insights into mechanisms and pathways that are otherwise difficult to discern. In the given synthesis schema, the carbon atom in barium carbonate \( \mathrm{Ba}^{*} \mathrm{CO}_{3} \) undergoes a series of reactions, its presence marked and followed as it transitions from carbon dioxide into more complex molecules like acrylic acid and ultimately formaldehyde.

Alcohol Formation

Formation of alcohols from other functional groups is a critical transformation in organic chemistry. In this exercise, the focus is on the reduction of a carboxylic acid to form an alcohol. The starting compound, labeled acrylic acid (\( \mathrm{CH}_2=\mathrm{CH}-\mathrm{C}^{*}OH \)), undergoes a reduction via lithium aluminum hydride (\( \mathrm{LiAlH}_4 \)).
During reduction, the carboxylic group (\( -\mathrm{COOH} \)) loses its oxygen, and hydrogen atoms are added, converting it into an alcohol group (\( -\mathrm{CH}_2OH \)). This transformation results in labeled allyl alcohol (\( \mathrm{CH}_2=\mathrm{CH}-\mathrm{C}^{*}H_2OH \)). Understanding this reduction mechanism is essential because it helps convert functional groups, offering versatility in drug development and materials science that often requires such functional modifications.

Ozonolysis

Ozonolysis is a reaction used famously to break down alkenes into smaller compounds through the cleavage of carbon-carbon double bonds. This chemical process involves treating compound (\( \mathrm{Z} \)) with ozone, resulting in the disintegration of the double bonds present in the molecule.
The essential goal of ozonolysis is to sever these bonds, which consequently leads to the formation of carbonyl compounds such as aldehydes and ketones. In this exercise, the process focuses on breaking the structure (\( \mathrm{CH}_2=\mathrm{CH}- \)), leading to the formation of labeled formaldehyde (\( \mathrm{H}_2 \mathrm{C}^{*}O \)). Ozonolysis is a crucial step in understanding molecule degradation and analyzing the structural integrity of various compounds, making it indispensable in the synthetic pathway of complex molecules.