Problem 74 - A Complete Resource Book in Chemistry for JEE Main

Question

An organic compound ' \(\mathrm{X}\) ' with a molecular formula \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\) when reacts with excess of methyl magnesium chloride following by acidification gives an alcohol 'Y' only, when this alcohol ' \(Y\) ' is oxidized with \(\mathrm{NaOCl}\) followed by acidification, acidic acid is formed. Here the compound ' \(X\) ' is (a) Propyl methanoate (b) Isopropyl methanoate (c) Ethyl ethanoate (d) Both (a) and (b)

Step-by-Step Solution

Answer

Compound X is likely Ethyl ethanoate (c).

Step 1: Analyze the Reaction of X with Methyl Magnesium Chloride

The organic compound X (6} - Reacts with methyl magnesium chloride - Upon this reaction, Grignard reagent is formed which typically indicates ester hydrolysis and formation of tertiary/secondary alcohols. - Since it's specified to give alcohol 'Y', X may be an ester.

Step 2: Determine X as an Ester

Given that X has a structure allowing it to react with methyl magnesium chloride to give an alcohol Y, and X being suggested as an ester is logical. - With options provided suggesting different esters such as (a) Propyl methanoate and (b) Isopropyl methanoate or (c) Ethyl ethanoate, X is likely to align with these compounds being esters for this reaction type.

Step 3: Analyze for Alcohol Y after Ester Reaction

Compound X reacts with an excess of methyl magnesium chloride to form an alcohol Y. - Alcohol Y upon oxidation gives acetic acid implying alcohol Y is likely ethanol given the 2-carbon chain present in acetic acid. - Therefore, verify that compound X forms ethanol (via the reaction with methyl magnesium chloride) which implies precursor was probably an ethyl containing ester.

Step 4: Confirm the Synthesis Pathway to Determine X

Analyzing the molecular formula 6} - The expected conversion reaction: ester to secondary alcohol to carboxylic acid - Verify if given options align with forming ethanol to acetic acid implying ethyl chain functionality, primarily ethyl ethanoate.

Key Concepts

Grignard Reagent

In organic chemistry, the Grignard reagent is an essential tool for forming carbon-carbon bonds. It is formed when an organohalide, such as methyl magnesium chloride, reacts with magnesium metal in a dry ether solvent. This reaction produces a highly reactive organometallic compound that acts as a strong nucleophile, capable of attacking electrophilic centers like carbonyl groups.
When the Grignard reagent reacts with esters, it typically results in the formation of tertiary or secondary alcohols. This happens through a two-step mechanism:

• First, the Grignard reagent attacks the carbonyl carbon of an ester, forming an alkoxide intermediate.
• Subsequently, this intermediate is hydrolyzed upon the addition of acid, leading to the final alcohol product.

In the given exercise, the reaction of compound X with methyl magnesium chloride, forming alcohol Y, illustrates a classic Grignard reaction.
This implies that X is likely an ester compound that undergoes hydrolysis and reduction to yield an alcohol upon interaction with the Grignard reagent.

Ester Hydrolysis

Ester hydrolysis is a crucial reaction in organic chemistry, where an ester compound is converted into an alcohol and an acid or a salt. This reaction can occur under acidic or basic conditions, known as acid-catalyzed hydrolysis or saponification, respectively.
In the context of the exercise, compound X reacts with methyl magnesium chloride to create an alcohol. This suggests that X is an ester, which undergoes hydrolysis to yield alcohol Y. Here's how it generally works:

• The Grignard reagent initiates the attack on the carbonyl carbon of the ester.
• This forms a tetrahedral alkoxide intermediate, which then collapses to form a ketone or aldehyde.
• A second Grignard reagent molecule then attacks this aldehyde, eventually resulting in the formation of an alcohol after acidic workup.

This mechanism aligns with the behavior observed in ester hydrolysis via Grignard reagents in the exercise, where the process is simplified to yield only one type of alcohol.

Alcohol Oxidation

Alcohol oxidation is a vital transformation in organic chemistry, enabling the conversion of alcohols to aldehydes, ketones, or carboxylic acids. This reaction often employs oxidizing agents like chromium trioxide, sodium hypochlorite (NaOCl), or potassium permanganate.
In the exercise, alcohol Y undergoes oxidation using NaOCl to produce acetic acid, suggesting Y is likely ethanol. Sodium hypochlorite selectively oxidizes primary alcohols to carboxylic acids:

• For primary alcohols, the hydroxyl group is initially dehydrogenated into an aldehyde.
• The aldehyde is then further oxidized to yield the carboxylic acid.

This process is consistent with ethanol oxidation resulting in acetic acid because ethanol has two carbon atoms matching the carbon count of acetic acid.

Molecular Formula Analysis

Analyzing a molecular formula involves interpreting the atom count and their arrangements in a compound to deduce its possible structure. The molecular formula C_4H_8O_2 can suggest various compound structures.
To decipher the structure based on the exercise:

• Consider the functional groups that could inherently fit into the formula, such as esters (R-COOR).
• Given the formula, potential esters like propyl methanoate (propionate) or ethyl ethanoate logically fit the formula component setting.
• Cross-check the potential structures with the expected reaction outcomes, such as the production of ethanol and acetic acid, to determine any link between X and Y.

This exercise emphasizes recognizing esters under this formula context as logical candidates, especially for the conversion to other compounds through reactions.