Problem 184 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Reaction Type

The process involves dehydrating alcohols to form alkenes using concentrated sulfuric acid ( ext{H}_{2} ext{SO}_{4}) as a catalyst. This indicates an elimination reaction, specifically an E1 mechanism for secondary and tertiary alcohols.

Step 2: Understand E1 Mechanism

The E1 mechanism (unimolecular elimination) typically involves two main steps: first, the formation of a carbocation intermediate, then the elimination of a leaving group to form a double bond (alkene), where a proton is removed to eliminate a molecule of water.

Step 3: Identify Initiation Step

In the dehydration process, the initiation step is the reaction between the alcohol and the concentrated sulfuric acid, where a proton ( ext{H}^+) from the acid protonates the alcohol, turning the hydroxyl group ( ext{OH}) into a better leaving group ( ext{H}_2 ext{O}). This step is crucial for the subsequent formation of a carbocation.

Step 4: Select the Correct Answer

Based on the mechanics of E1 elimination, the initiation step in the dehydration of alcohols is the  ext{protonation of alcohol molecule}. This aligns with choice (a).

Key Concepts

Dehydration of Alcohols

Dehydration of alcohols is a fundamental chemical reaction often encountered in organic chemistry. In essence, it involves the removal of a water molecule from an alcohol. This reaction typically uses a strong acid as a catalyst. For instance, concentrated sulfuric acid \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is commonly used.
The general goal of this reaction is to convert alcohols into alkenes. Here’s a simplified essence of the process:

• The alcohol molecule is heated in the presence of strong acid.
• A proton \(\mathrm{H}^+\) from the acid helps convert the hydroxyl group \(\mathrm{OH}\) into a good leaving group.
• After this, a molecule of water is detached, leading to the formation of a double bond, thereby creating an alkene.

Understanding this helps in grasping the larger picture of elimination reactions, where atoms or groups are removed to form double bonds in the molecules. The specific initiation step here involves protonation of the hydroxyl group, paving the way for further reaction steps.

E1 Mechanism

The E1 mechanism, or unimolecular elimination, is a common pathway for the dehydration of alcohols, particularly secondary and tertiary ones. It involves several key steps that take place once the alcohol is protonated.

• Firstly, the protonated alcohol transforms the \(\mathrm{OH}\) group into a good leaving group, forming water.
• Then, this leads to the creation of a carbocation intermediate, a positively charged ion.
• Finally, the removal of a proton from the adjacent carbon creates the double bond, generating the alkene.

It's important to note that E1 reactions are stepwise and depend on the stability of the formed carbocation. This mechanism is favored when the carbocation formed can be stabilized by the surrounding molecular structure. Stability is often greater in tertiary carbocations than in secondary, which is why E1 is more commonly observed with more substituted alcohols.

Carbocation Formation

Carbocation formation is a critical step in the E1 mechanism. During the dehydration of alcohols, once the alcohol is protonated, the \(\mathrm{OH}\) group leaves as a water molecule, creating a carbocation. Carbocations are reactive species with a positively charged carbon atom that lacks a full octet of electrons. Their stability influences the reaction pathway significantly. Here are some key points:

• Tertiary carbocations are the most stable, followed by secondary, and then primary.
• This stability derives from hyperconjugation and the inductive effect – where adjacent carbon atoms help distribute the positive charge.
• Once formed, this carbocation can quickly lead to the formation of an alkene through a further elimination step.

Understanding carbocation stability is crucial as it affects the rate at which the reaction proceeds. It also helps predict the major products in a reaction, as more stable carbocations lead to less energy-intensive, more favorable pathways towards alkene formation.