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 \(\mathrm{C}_{9} \mathrm{H}_{14} \mathrm{O}\). '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

The compounds are propanone and mesityl oxide, answer (b).

Step 1: Understanding the Iodoform Test

The iodoform test is used to identify methyl ketones or secondary alcohols with at least one methyl group at the alpha position. Compound 'A' undergoes the iodoform test, suggesting it has the structure of a methyl ketone or a secondary alcohol with an appropriate group.

Step 2: Analyzing the Molecular Formula of 'A'

The molecular formula of 'A' is \( C_3H_6O \). Possible candidates include propanone (acetone), propanal (aldehyde), or isopropanol (alcohol). Given that 'A' undergoes the iodoform test, propanone is the most likely candidate, as it has a methyl ketone structure.

Step 3: Interpreting Reaction with HCl

The compound 'A' gives 'B', with a molecular formula \( C_9H_{14}O \), when saturated with HCl. This suggests a condensation reaction, possibly aldol condensation, in which acetone, as 'A', might condense to form a compound 'B' with the given formula.

Step 4: Identifying Compound 'B'

Upon reaction, acetone can undergo a self-condensation reaction in the presence of HCl to form mesityl oxide (2,6-dimethyl-2,5-heptadien-4-one) or mesitylene (1,3,5-trimethylbenzene). However, based on the given options and molecular formula \( C_9H_{14}O \), 'B' is likely to be mesityl oxide.

Key Concepts

Iodoform Test

The Iodoform Test is a well-known laboratory technique used to detect the presence of methyl ketones. These are compounds where a carbonyl group (\(\text{C}=\text{O}\)) is directly attached to a methyl group (\(-\text{CH}_3\)). During this test, the methyl ketone is treated with iodine and a base, such as sodium hydroxide. If the test is positive, a yellow precipitate of iodoform (\(\text{CHI}_3\)) forms. This distinct yellow solid acts as a clear indicator. Secondary alcohols that can be oxidized to methyl ketones also give a positive result in this test.
This is why the presence of a positive iodoform test points to compounds like propanone (acetone), which is common in the given exercise scenario. A student can remember this test as a 'reveal' for the hidden structures of certain alcohols and ketones, making it an essential tool in organic chemistry labs for identifying particular molecular features.
When considering compound 'A' in this exercise, the positive iodoform test strongly suggests it is propanone, as propanone meets the criteria of having the necessary alpha-methyl group attached to the carbonyl.

Molecular Formula Analysis

Molecular formula analysis is a technique used to deduce the structure of an organic compound from its molecular formula. The molecular formula provides the count of each type of atom within the compound. For compound 'A', this formula is \(\text{C}_3\text{H}_6\text{O}\), giving clues about its structure.
In analyzing such a formula, it’s crucial to consider all possible structural isomers. These structural configurations vary despite having the same formula, pointing to possibilities like propanone (acetone), propanal, or isopropanol. The key task here is to match the molecular insights from the formula with chemical behavior, such as the positive iodoform test, to narrow it down to the correct structure.
In this exercise, because 'A' undergoes the iodoform test, propanone emerges as the likely candidate. Its structure fits the observations from both its molecular formula and its chemical reactivity.

Condensation Reaction

Condensation reactions are significant in organic chemistry for their role in forming bigger molecules by combining smaller ones, with water often being a byproduct. In this exercise, compound 'A' reacts to form compound 'B' through a condensation mechanism.
Specifically, acetone, acting as 'A', may undergo an aldol condensation. This type of reaction involves the formation of a new carbon-carbon bond, ideal for synthesizing larger molecules. Under acidic conditions, such as the saturation of \(\text{HCl}\), acetone molecules react with each other. This reaction leads to the formation of larger molecules, assembling to create mesityl oxide, which has a molecular formula of \(\text{C}_9\text{H}_{14}\text{O}\).
This example highlights the versatility and importance of condensation reactions, especially in synthetic organic chemistry, where forming complex structures is key. Students should remember that by controlling conditions, we can predict and promote specific reaction pathways, just as in forming mesityl oxide from acetone.

Methyl Ketone Identification

Methyl ketone identification is a crucial practice for determining organic compound structures. Methyl ketones are easily identifiable due to the carbonyl group bonded to a methyl group. They are pivotal in many organic reactions and often serve as substrates in different chemical tests such as the iodoform test.
The process of confirming a methyl ketone involves understanding the molecular formula and reactivity characteristics. For exercise "A", involving a compound \(\text{C}_3\text{H}_6\text{O}\) that tests positive for the iodoform reaction, methyl ketone identification is accomplished by determining which structure aligns with these criteria.
In realistic scenarios, this involves evaluating possible isomers and their reaction types. This reinforces learning about ketone properties and the types of reactions such compounds undergo, thereby making the task of identification clearer and more structured. The identification intricately links analytical thinking with practical experimentation, crucial for mastery in organic chemistry.