Problem 83 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An ester \(A\left(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\right)\), on treatment with excess methyl magnesium chloride followed by acidification, gives an alcohol \(B\) as the sole organic product. Alcohol \(B\), on oxidation with \(\mathrm{NaOCl}\) followed by acidification, gives acetic acid. Deduce the structures of \(A\) and \(B\). Show the reactions involved.

Step-by-Step Solution

Answer

Ester A is methyl acetate; Alcohol B is isopropyl alcohol.

Step 1: Identify the Functional Groups in Ester A

Ester A is given by the molecular formula \(\mathrm{C}_4\mathrm{H}_8\mathrm{O}_2\). Given the molecular structure, ester A can be presumed to have a structural formula of an ester group which is represented by the formula \(\mathrm{RCOOR'}\). Here, \(\mathrm{R}\) and \(\mathrm{R'}\) may represent alkyl groups. Begin by identifying these alkyl groups so that the molecular formula is satisfied.

Step 2: Reaction with Methyl Magnesium Chloride

When ester A is treated with methyl magnesium chloride \((\mathrm{CH}_3\mathrm{MgCl})\), it undergoes a Grignard reaction that typically breaks the ester into two parts. This reaction results in the formation of an alcohol B from the carbonyl functioning group with the addition of a methyl group \((\mathrm{CH}_3)\) to each side.

Step 3: Predicting Structure for Alcohol B

The information states that alcohol B is the only organic product. Since B gives acetic acid on oxidation, it must be \(\mathrm{C}_3\mathrm{H}_7\mathrm{OH}\), typically propanol types. Therefore, it suggests that after the addition from methyl magnesium chloride, \(\mathrm{B}\) must be isopropyl alcohol (\(\mathrm{C}_3\mathrm{H}_8\mathrm{O}\)).

Step 4: Determine Structure of Ester A

Before the Grignard reaction, ester A must have a structured allowing a single organic product like isopropyl alcohol to form. Therefore, it is likely to be methyl acetate, \((\mathrm{CH}_3\mathrm{COOCH}_3)\), as it would form isopropyl alcohol after Grignard addition and acidification.

Step 5: Verify with Reaction

Start with the reaction:\[\text{Methyl acetate} : \mathrm{CH}_3\mathrm{COOCH}_3 + \mathrm{CH}_3\mathrm{MgCl} \rightarrow \mathrm{CH}_3\mathrm{CHOHCH}_3 \]With the Grignard reaction forming isopropyl alcohol \((\mathrm{CH}_3\mathrm{CHOH}\mathrm{CH}_3)\). This confirms the prediction that Ester A is Methyl Acetate and Alcohol B is Isopropyl Alcohol.

Step 6: Oxidation of B to Acetic Acid

During oxidation of isopropyl alcohol \((\mathrm{CH}_3\mathrm{CHOH}\mathrm{CH}_3)\), it is converted to acetone \((\mathrm{CH}_3\mathrm{COCH}_3)\), which can further generate acetic acid \((\mathrm{CH}_3\mathrm{COOH})\) under specific conditions like reaction with \(\mathrm{NaOCl}\) followed by acidification.

Key Concepts

Ester Hydrolysis

Ester hydrolysis is a crucial reaction in organic chemistry where esters are broken down into alcohols and acids. This process can be acid-catalyzed, where a proton from the acid activates the ester’s carbonyl carbon, making it more susceptible to nucleophilic attack.
In our exercise, the ester undergoes a Grignard reaction, introducing a methyl group, rather than a typical hydrolysis. However, understanding hydrolysis is essential because after the Grignard reaction, the structure adjusts as if going through hydrolysis resulting in alcohol formation.
Here is the general reaction for acid hydrolysis: \[\text{RCOOR'} + \text{H}_2\text{O} \rightarrow \text{RCOOH} + \text{R'OH}\]Despite differing conditions in the exercise, the concept remains that chemical structures change and adapt, similar to hydrolysis, when nucleophilic agents are involved. The nuances of this reaction highlight the transformational nature of chemical compounds under various conditions.

Oxidation Reactions

Oxidation reactions involve an increase in the oxidation state of a molecule. This typically means adding oxygen or removing hydrogen, transforming the organic compounds into more oxidized states.
In the given exercise, alcohol B (isopropyl alcohol) undergoes oxidation. Using sodium hypochlorite (NaOCl) as the oxidizing agent, the alcohol group (\(-OH\)) in isopropyl alcohol is converted to a ketone group in acetone. This is a classic oxidation, transforming an alcohol to a ketone: \[\text{RCHOH} \rightarrow \text{RCOOH}\]Further acidification leads acetone to continue reacting, ultimately forming acetic acid, illustrating how oxidation journeys substances through multiple transformations, each one increasing oxygen or reducing hydrogen in the molecule.

Chemical Structure Elucidation

Chemical structure elucidation is the process of determining the structures of unknown compounds. This can involve identifying molecular formations, like functional groups that define how structures react chemically.
Our exercise provides molecular formulas for ester A and alcohol B, asking us to infer their structures based on reactions described. The approach involves piecing together chemical changes through reactions like the Grignard reaction and oxidation.

• Using a known molecular formula (C4H8O2 for ester, like methyl acetate).
• Executing a Grignard reaction using CH3MgCl creating isopropyl alcohol (B).
• Observing the product, oxidation transforming alcohol to an organic acid, assists in deducing intermediate compounds.

These steps detail how the scientific methodology aids in deciphering chemical structures, emphasizing how reactions explain hidden features of compounds.