Problem 199 - A Complete Resource Book in Chemistry for JEE Main

Question

During dehydration of alcohols to alkenes, by heating with concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) the initiation step is [2003] (a) formation of an ester (b) elimination of water (c) protonation of alcohol molecule (d) formation of a carbocation

Step-by-Step Solution

Answer

The initiation step is (c) protonation of alcohol molecule.

Step 1: Understanding the Dehydration Reaction

The dehydration of alcohols involves the removal of a water molecule from the alcohol to form an alkene. Concentrated sulfuric acid (H_2SO_4) acts as a catalyst in this reaction.

Step 1: Protonation of the Alcohol

The first step in the reaction mechanism is the protonation of the alcohol molecule. This occurs when the lone pair of electrons on the oxygen atom of the alcohol attacks a hydrogen ion (proton) from the sulfuric acid, forming an oxonium ion. This step increases the electrophilicity of the alcohol.

Step 2: Formation of a Carbocation

After protonation, the oxonium ion can lose a water molecule, leaving behind a carbocation. The formation of this carbocation is a crucial step as it allows for the rearrangement and subsequent formation of the double bond in the alkene.

Step 4: Recognizing the Correct Initiation Step

Based on the mechanism, the initiation of the dehydration reaction is characterized by the protonation of the alcohol. While subsequent steps involve water elimination and carbocation formation, the beginning step is the protonation of the alcohol.

Key Concepts

Protonation of Alcohol

The dehydration of alcohols begins with a fundamental step known as protonation. This is where the alcohol molecule interacts with an acid, typically a proton donor like concentrated sulfuric acid
(\( ext{H}_2 ext{SO}_4\)). In this process, the lone pair of electrons on the oxygen atom of the alcohol attacks and bonds with a proton
(from the acid), transforming the neutral alcohol into a positively charged oxonium ion.
- The transformation into an oxonium ion makes the alcohol more reactive.- The protonation step increases the likelihood that the alcohol will lose a water molecule in the subsequent steps.
Think of it as setting the stage for what happens next. This crucial change in charge and configuration allows further chemical reactions to proceed much more smoothly.

Carbocation Formation

Following protonation, the alcohol (now an oxonium ion) undergoes a critical transformation: the loss of a water molecule. This step is often marked by the formation of a carbocation, a species characterized by a positively charged carbon atom.
This charge is the result of the carbon atom temporarily losing its electrons to the departing water molecule.
- The carbocation is highly reactive and forms spontaneously after the hydrogen
from the acid helps push off the water. - Stability of the carbocation can vary, leading to potential rearrangement if neighboring structures can offer more stability.
This step is significant because it sets the stage for possible rearrangements and ultimately, the formation of a double bond in the alkene product.
Think of the carbocation like a loose puzzle piece that needs a stable spot to rest.

Elimination of Water

The final move in the dehydration process is the elimination of water, which follows from the formation of the carbocation. This phase marks the transition from the carbocation intermediate to the formation of an alkene.
When the water molecule departs, it leaves behind a positively charged carbon, or carbocation. A neighboring hydrogen transfers its electrons to the carbocation, readying the ground for a double bond to form. - As a result, the double bond creates an alkene, the desired product of dehydration. - This transformation often follows specific pathways to achieve the most stable alkene based on factors like bond strength and sterics.
This elimination is essential for constructing the least stable reactant into a more energetically favorable product, completing the dehydration process effectively.