Problem 92 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Compound \(A\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2}\right)\) on reduction with \(\operatorname{LiAlH}_{4}\) yielded two compounds \(B\) and \(C\). The compound \(B\) on oxidation gave \(D\), which on treatment with aqueous alkali and subsequent heating furnished \(E\). The latter on catalytic hydrogenation gave \(C\). The compound \(D\) was oxidized further to give \(F\) which was found to be a monobasic acid (molecular weight \(=60.0\) ). Deduce the structures of \(A, B, C, D\) and \(E\).

Step-by-Step Solution

Answer

A: Ethyl acetate, B: Ethanol, C: Methane, D: Acetaldehyde, E: Crotonaldehyde.

Step 1: Understanding the Reduction Product

Compound \(A\) with molecular formula \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2}\) is reduced by \(\operatorname{LiAlH}_{4}\) to produce compounds \(B\) and \(C\). \(\operatorname{LiAlH}_{4}\) is a strong reducing agent, commonly used to reduce esters, acids, and ketones to alcohols. Since both products \(B\) and \(C\) result from the reduction of \(A\), \(A\) is likely an ester or ketone.

Step 2: Analyzing Product B

Product \(B\) on oxidation forms compound \(D\). Generally, alcohols oxidize to form aldehydes or ketones, suggesting that \(B\) is likely a primary or secondary alcohol. Therefore, \(B\) becomes an aldehyde or ketone, \(D\), upon oxidation.

Step 3: Understanding the Subsequent Transformation to E

Compound \(D\) upon treatment with aqueous alkali followed by heating yields compound \(E\), a reaction indicative of keto-enol tautomerism or an aldol condensation. The reference to heating suggests the aldol reaction, producing a β-hydroxy ketone or an α,β-unsaturated carbonyl compound.

Step 4: Interpreting the Catalytic Hydrogenation to C

Compound \(E\), upon catalytic hydrogenation, results in \(C\). This indicates \(E\) has a carbon-carbon double bond (since hydrogenation saturates double bonds), making \(E\) an α,β-unsaturated compound initially formed from \(D\).

Step 5: Identifying Compound F

Compound \(D\) was further oxidized to \(F\), with \(F\) identified as a monobasic acid with a molecular weight of \(60.0\). The simplest monobasic acid with such a molecular weight and only carbon, hydrogen, and oxygen is acetic acid, \(\mathrm{CH}_{3}\mathrm{COOH}\). Hence, \(D\) might be ethanol or acetaldehyde in this chain.

Step 6: Constructing the Chain of Reactions

Given that \(F\) is acetic acid, then \(D\) on mild oxidation would be acetaldehyde, \(\mathrm{CH}_{3}\mathrm{CHO}\). With \(B\) being the precursor to \(D\) and an alcohol, \(B\) should be ethanol. \(A\) is likely an ester such as ethyl acetate, which reduces to ethanol (\(B\)) and acetaldehyde (\(C\)). \(E\) would then be the aldol product, crotonaldehyde, and \(C\) methane, completing the reactions.

Key Concepts

Reduction Reactions

In organic chemistry, reduction reactions play a vital role. They involve the addition of electrons or hydrogen atoms to a molecule, or the removal of oxygen. A common reagent for reduction is lithium aluminum hydride \( \text{(LiAlH}_4\text{)} \).

• It is particularly powerful for reducing esters, acids, and ketones to alcohols.
• In the given exercise, compound \( A \) with the molecular formula \( \text{C}_6\text{H}_{12}\text{O}_2\) was reduced by LiAlH\(_4\) to produce two alcohols, \( B \) and \( C \).

Reduction reactions are essential for converting more complex molecules into simpler and often more reactive alcohols. This forms the basis for further transformations and synthesis in organic chemistry.

Oxidation Reactions

Oxidation reactions in organic chemistry often involve the increase of oxygen content or the removal of hydrogen.

• Alcohols are typically oxidized to form aldehydes, ketones, or acids, depending on their structure.
• In the exercise, compound \( B \), an alcohol, is oxidized to form \( D \), likely an aldehyde or a ketone. Later on, \( D \) itself undergoes further oxidation to produce \( F \), a monobasic acid identified as acetic acid.

Oxidation allows the introduction of new functional groups, enabling more complex molecular structures. It transforms molecules and significantly expands the toolkit of synthetic chemists.

Aldol Condensation

Aldol condensation is a fundamental reaction in organic chemistry that forms carbon-carbon bonds. It involves the reaction of aldehydes or ketones with a base, resulting in an β-hydroxy carbonyl compound, which can dehydrate to form an α,β-unsaturated compound.

• In the problem, compound \( D \), after oxidation, is treated with aqueous alkali and heated to produce \( E \). This indicates a classic aldol condensation resulting in a compound with a double bond.
• This reaction allows two smaller molecules to combine and form a larger, more complex structure, which enhances the diversity of possible chemical syntheses.

Aldol condensation is instrumental in building complexity in organic synthesis, providing pathways to aromatic and complex ring structures.

Catalytic Hydrogenation

Catalytic hydrogenation is used to add hydrogen to unsaturated compounds, thereby saturating carbon-carbon double or triple bonds.

• This reaction is instrumental in converting unsaturated molecules into saturated ones and requires catalysts such as palladium, platinum, or nickel.
• In the given exercise, compound \( E \), with a double bond, undergoes catalytic hydrogenation to yield \( C \). This process allows the conversion of more reactive, unsaturated compounds to stable saturated forms such as alkanes or alcohols.

Catalytic hydrogenation is especially useful in the food industry to harden oils or in fine chemical synthesis to precisely control the saturation of molecules.