Problem 114 - A Complete Resource Book in Chemistry for JEE Main

Question

An organic compound 'A' has the molecular formula \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O} .\) It undergoes iodoform test. When saturated with HCl it gives 'B' of molecular formula \(\mathrm{C}_{9} \mathrm{H}_{14} \mathrm{O}\). \(' \mathrm{~A}\) ' and ' \(\mathrm{B}\) ' respectively are (a) Propanone and 2,6 -dimethyl-2, 5-heptadien-4-one (b) propanone and mesitylene oxide (c) propanal and mesitylene (d) propanol and mesityl oxide

Step-by-Step Solution

Answer

(b) Propanone and mesitylene oxide

Step 1: Analyze the Molecular Formula of Compound 'A'

The molecular formula is \(\mathrm{C}_{3}\mathrm{H}_{6}\mathrm{O}\). This suggests it could be a ketone or an aldehyde.

Step 2: Determine the Functional Group from the Iodoform Test

The iodoform test is positive for methyl ketones or ethanol. Since compound 'A' goes through the iodoform test, it is likely a methyl ketone, which supports that 'A' is propanone (acetone), \(\mathrm{CH}_3\mathrm{COCH}_3\).

Step 3: Understand the Reaction with HCl

When 'A' is saturated with HCl, it forms 'B'. Since 'B' has the formula \(\mathrm{C}_{9}\mathrm{H}_{14}\mathrm{O}\) and knowing 'A' is propanone, this suggests a condensation reaction like aldol or similar polymerization might have occurred to form compound 'B'.

Step 4: Evaluate the Options

Given (a) to (d), it's known that propanone (acetone) could form mesityl oxide through self-condensation under acidic conditions. Mesityl oxide has the molecular formula \(\mathrm{C}_{9}\mathrm{H}_{14}\mathrm{O}\), which matches our results.

Key Concepts

Iodoform Test

The iodoform test is a chemical reaction used to identify specific functional groups in organic compounds. It is particularly useful for detecting methyl ketones, which have the structural feature of a methyl group attached to the carbonyl carbon (\[\text{-C(=O)CH}_3\]). In the presence of iodine and a base, such as sodium hydroxide, these compounds undergo a reaction to form iodoform (a yellow precipitate, \(\text{CHI}_3\)). This reaction is very specific because few other structures will give a positive iodoform test, primarily methyl alcohols or specific secondary alcohols that form methyl ketones upon oxidation.
Some important points about the Iodoform Test:

• The presence of a yellow precipitate confirms a positive test.
• Methyl ketones, like acetone, easily form iodoform due to their structure.
• The reaction requires iodine (I2) and a base, typically NaOH.

In our exercise, compound 'A', with the formula \(\text{C}_3\text{H}_6\text{O}\), gives a positive iodoform test indicating it likely contains a methyl ketone, confirming its identity as propanone (acetone).

Molecular Formula Analysis

Analyzing the molecular formula of an organic compound helps in deducing its possible structures. The molecular formula provides the types and numbers of atoms in a molecule but not how they are arranged. For example, the formula \(\text{C}_3\text{H}_6\text{O}\) could correspond to several structures, including ketones and aldehydes, like propanone or propanal.
Key aspects of molecular formula analysis include:

• Counting the degree of unsaturation to determine possible rings or multiple bonds.
• Identifying functional groups that match the given number of atoms.
• Considering isomers and tautomer forms that fit the formula.

In the given exercise, the molecular formula suggested 'A' as \(\text{C}_3\text{H}_6\text{O}\), which aligns with propanone—a typical methyl ketone known to respond to the iodoform test.

Condensation Reactions

Condensation reactions are a fundamental process in organic chemistry where two molecules combine, releasing a small molecule, typically water or an alcohol. In our context, the focus is on the aldol condensation, which is a reaction between enolizable carbonyl compounds that forms β-hydroxy ketones or aldehydes, followed by dehydration to yield α,β-unsaturated carbonyl compounds.
Essential details:

• Aldol condensation involves enol or enolate ions reacting with carbonyl groups.
• It requires a base or acid to proceed, often forming larger, complex structures.
• It's common in forming conjugated systems with double bonds.

In this exercise, propanone undergoes self-condensation when treated with HCl, leading to the formation of a larger compound, compound 'B'—likely mesityl oxide, fitting the molecular formula \(\text{C}_9\text{H}_{14}\text{O}\).

Methyl Ketones

Methyl ketones are a class of ketones characterized by a methyl group bonded directly to the carbonyl group. A common example is acetone, which is the simplest methyl ketone. These compounds are significant in organic chemistry due to their reactivity, especially in the iodoform test and various synthesis reactions.
Attributes of methyl ketones include:

• A carbonyl group (\(\text{C=O}\)) bonded to a methyl group (\(\text{CH}_3}\)).
• Tendency to participate in halogenation reactions, like the iodoform test.
• Key roles in condensation reactions, forming larger constructs.

In the context of the exercise, the identification of compound 'A' as propanone—a methyl ketone, allows it to participate in iodoform testing and subsequent condensation reactions, linking it directly to the formation of compound 'B.'