Problem 127 - A Complete Resource Book in Chemistry for JEE Main

Question

A compound 'A' has a molecular formula \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{OH}\). It reduces Fehling's solution and on oxidation, gives a monocarboxylic acid ' \(\mathrm{B}\) '. 'A' can be obtained by the action of chlorine on ethyl alcohol. 'A' is (a) chloroform (b) chloral (c) methyl chloride (d) monochloroacetic acid

Step-by-Step Solution

Answer

The compound 'A' is chloral.

Step 1: Analyze the Molecular Formula

The molecular formula given is \( \mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{OH} \). We need to analyze which compound matches this structure.

Step 2: Identify Reducing Ability with Fehling's Solution

Fehling's solution is reduced by aldehydes, so compound 'A' must contain an aldehyde group.

Step 3: Determine Oxidation Product

On oxidation, compound 'A' gives a monocarboxylic acid 'B'. This further suggests the presence of an aldehyde group, which can be oxidized to a carboxylic acid.

Step 4: Consider Production Route from Ethyl Alcohol

The problem states that 'A' can be formed by the action of chlorine on ethyl alcohol (\(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \)), suggesting a substitution reaction leading to a chlorinated aldehyde.

Step 5: Identify Compound 'A'

Based on the molecular formula \( \mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{OH} \), the reducing ability, oxidation behavior, and production method, 'A' is a chlorinated aldehyde. Therefore, 'A' is chloral (\( \mathrm{CCl}_3\mathrm{CHO} \)).

Key Concepts

Molecular Formula Analysis

Molecular formula analysis is an integral part of organic chemistry. It involves breaking down a given molecular formula to better understand the structure and properties of a compound. For example, the molecular formula \( \mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{OH} \) describes a compound with two carbon atoms, three chlorine atoms, one oxygen atom, and one hydrogen atom. By examining each component:

• Carbon (C): Forming the backbone of organic compounds.
• Chlorine (Cl): Typically indicates a halogenated compound.
• Oxygen (O) & Hydrogen (H): Suggests the presence of a functional group like alcohol, aldehyde, or carboxylic acid.

Such analysis provides clues about possible functional groups and behaviors. Recognizing that 'A' reduces Fehling's solution and oxidizes to a monocarboxylic acid, we deduce 'A' must have an aldehyde group. This makes formula analysis crucial for identifying possible structure and reactivity.

Fehling's Solution

Fehling's solution is a chemical reagent used to differentiate between water-soluble aldehyde and ketone functional groups. It contains copper(II) ions and is deep blue in color. This solution is particularly useful for identifying aldehydes, which reduce Fehling's solution to form a red precipitate of copper(I) oxide. When a compound like the one in our exercise reduces Fehling's solution, it confirms the presence of an aldehyde group, distinguishing it from ketones, which usually do not react. Understanding the reaction with Fehling's solution helps in confirming the structural component of an aldehyde in compound 'A'. This property was key to determine the reducing nature of ‘A’ and thus helped narrow down its identity.

Oxidation Reactions

Oxidation reactions are crucial in organic chemistry to transform functional groups to higher oxidation states. When an aldehyde is subjected to oxidation, it typically forms a carboxylic acid. In our case, compound 'A' (chloral) oxidizes to form a monocarboxylic acid 'B'. This process involves:

• Acceptance of Oxygen: Addition of an oxygen or the increase of any electronegative element bonded to carbon.
• Hydrogen Removal: Loss of hydrogen atoms from the molecule.

This specific transformation from aldehyde to carboxylic acid highlights both the role of aldehydes in oxidation reactions and the predictable pathways such reactions follow. Understanding these transformations is fundamental as they are common in organic synthesis.

Alcohol Chlorination

Alcohol chlorination involves replacing hydrogen atoms in an alcohol with chlorine atoms, usually through a substitution reaction. In the context of the exercise, ethyl alcohol (\( \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \)) is exposed to chlorine, resulting in chlorination to form compound 'A'.This reaction typically occurs in stages:

• Initial Chlorination: Adds chlorine atoms one at a time.
• Formation of Chlorinated Alcohol or Aldehyde: Leads to a compound like chloral (\( \mathrm{CCl}_3\mathrm{CHO} \)).

Such reactions are pivotal in forming versatile compounds with a range of uses in synthetic chemistry. The process not only highlights the reactivity of alcohols in the presence of halogens but also underscores the predictability and control possible in organic transformations.