Problem 119 - A Complete Resource Book in Chemistry for JEE Main

Question

In the reaction \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}$$\stackrel{\mathrm{H}_{2} \mathrm{SO}_{4} \mathrm{HgSO}_{4}}{\longrightarrow}[\mathrm{A}]\) The compound [A] is (a) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{COOH}\) (b) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CHO}\) (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CO}-\mathrm{CH}_{3}\) (d) none of these

Step-by-Step Solution

Answer

The compound [A] is option (c): \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CO}-\mathrm{CH}_3\).

Step 1: Recognize the Reaction Type

The reaction provided is a hydration reaction involving an alkyne under acidic conditions with mercuric ion (\(\(\mathrm{Hg}^{2+}\)\)) as a catalyst.

Step 2: Understand Alkyne Hydration

In the presence of \(\mathrm{H}_2\mathrm{SO}_4\) and \(\mathrm{HgSO}_4\), alkynes undergo Markovnikov addition of water to yield a keto form (ketone or aldehyde) through an enol intermediate.

Step 3: Analyze the Alkyne Structure

The given alkyne is \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{C} \equiv \mathrm{CH}\). This is a terminal alkyne with one methyl (\(\mathrm{CH}_3\)) group attached to the end.

Step 4: Determine the Enol Form

The enol formed after hydration of \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{C} \equiv \mathrm{CH}\) is \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{C}(OH)=\mathrm{CH}_2\).

Step 5: Tautomerization to Keto Form

The enol tautomerizes to the more stable ketone form, \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{COCH}_3\), following Markovnikov's rule.

Step 6: Match the Product to Options

Going through the options, the keto product \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{COCH}_3\) matches option (c), which is \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CO}-\mathrm{CH}_3\).

Key Concepts

Markovnikov's Rule

Markovnikov’s Rule is a guiding principle in organic chemistry that helps predict the outcome of certain addition reactions. Let's break it down:

• What is Markovnikov's Rule? It predicts that in an addition reaction of an unsymmetrical reagent to an unsymmetrical alkene or alkyne, the more electronegative part of the reagent bonds with the more substituted carbon. This results in the formation of the more stable carbocation.
• Example in Alkyne Hydration: The reaction of the alkyne \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{C} \equiv \mathrm{CH}\) with water in the presence of acid and mercuric ions follows this rule. Water adds in such a way that the OH part attaches to the carbon that is more substituted, which often leads to greater stability of the intermediate.

Markovnikov’s Rule helps chemists predict the major product of an addition reaction, which in this case supports the formation of specific alcohol and subsequently, leads to the final ketone through further reactions.

Tautomerization

Tautomerization is a chemical reaction where an atom or group of atoms is transferred from one position to another within a molecule. When we talk about alkyne hydration, it often involves a specific type of tautomerization known as keto-enol tautomerism.

• What Happens During Tautomerization? In the hydration of an alkyne, we first get an unstable form called an enol (an alcohol with a double bond). Tautomerization rearranges the bonds resulting in a more stable ketone or aldehyde.
• Tautomerization in Our Reaction: After the initial Markovnikov addition of water, the enol form \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{C}(OH)=\mathrm{CH}_2\) is created. This enol converts to a ketone through tautomerization, making it a naturally occurring shift towards stability.

Tautomerization is essential in chemistry as it often determines the stability and reactivity of the resulting compounds. Simply put, enols often hide as the more stable ketones, which you'll usually find in the final product.

Ketone Formation

Forming a ketone from an alkyne follows a fascinating pathway dictated by Markovnikov’s Rule and followed by tautomerization. Ketones themselves are hugely significant in chemistry due to their distinct structure and reactivity.

• Formation Process: The hydration of a terminal alkyne like \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{C} \equiv \mathrm{CH}\) adds water across the triple bond, resulting first in an enol, which then tautomerizes into a ketone. This sequence is crucial for crafting a stable molecule.
• Ketone Characteristics: Ketones have the structure \(\mathrm{R}_2\mathrm{C}=\mathrm{O}\), where R can be any hydrocarbon chain. They are the result of the highest order of carbonyl reactivity, making them valuable intermediates in organic synthesis.

Their formation from alkynes demonstrates how seemingly simple transformations link into creating stable and functional organic compounds, crucial for further applications in chemistry.

Enol-Keto Conversion

The enol-keto conversion is a hallmark transformation in organic reactions involving alkynes, showcasing the dance between instability and stability in chemical structures.

• Understanding Enol Structures: Enols are versatile molecules that include a hydroxyl group (OH) directly bonded to a carbon-carbon double bond. In chemical terms, they're relatively unstable and inclined to isomerize into a keto form.
• The Keto Form: This isomerization leads to ketones or aldehydes, which are significantly more stable. The conversion occurs because enols are always on the lookout to become more stable. In our reaction, the enol formed initially is \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{C}(OH)=\mathrm{CH}_2\), which smoothly transitions into \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{COCH}_3\) during the process.

This conversion is not just a simple swapping of atomic positions; it represents a core principle of organic chemistry where molecules naturally shift to achieve stability and lower energy in their structures.