Problem 137 - A Complete Resource Book in Chemistry for JEE Main

Question

The compound which gives the most stable carbonium ion on dehydration is (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{OH}\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (d) \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3}\)

Step-by-Step Solution

Answer

(b) \((\mathrm{CH}_{3})_{3} \mathrm{COH}\) forms the most stable carbonium ion on dehydration.

Step 1: Understanding Dehydration and Carbonium Ions

Dehydration of an alcohol involves the removal of a water molecule, leading to the formation of a carbocation (carbonium ion). These ions are more stable when they are tertiary, as more alkyl groups can donate electrons to stabilize the positive charge.

Step 2: Analyzing Option (a)

The structure \((\mathrm{CH}_{3})_{2} \mathrm{CHCH}_{2} \mathrm{OH}\) corresponds to a secondary alcohol. Dehydration would yield a secondary carbocation, which is less stable than a tertiary carbocation.

Step 3: Analyzing Option (b)

The structure \( (\mathrm{CH}_{3})_{3} \mathrm{COH} \) is a tertiary alcohol. Upon dehydration, it will form a tertiary carbocation, which is highly stable due to the presence of three electron-donating methyl groups.

Step 4: Analyzing Option (c)

The structure \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} \) is a primary alcohol. Dehydration would lead to a primary carbocation, which is the least stable due to minimal electron-donating group support.

Step 5: Analyzing Option (d)

The structure \( \mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3} \) is a secondary alcohol, which would form a secondary carbocation upon dehydration. Though more stable than a primary carbocation, it is less stable than a tertiary carbocation.

Step 6: Determining the Most Stable Carbonium Ion

Based on the stability of carbocations, tertiary carbocations are the most stable. Option (b) \( (\mathrm{CH}_{3})_{3} \mathrm{COH} \) forms a tertiary carbocation and thus is the most stable.

Key Concepts

Tertiary Carbocations

Carbocations are key intermediates in many organic reactions, especially in dehydration processes involving alcohols. A tertiary carbocation forms when a positively charged carbon atom is attached to three alkyl groups. This configuration is extremely stable due to several factors.
The more alkyl groups attached to the positively charged carbon, the better they can donate electrons through hyperconjugation and inductive effects. This effectively disperses the positive charge among the neighboring carbon atoms, thus stabilizing the carbocation.

• Tertiary carbocations have three electron-donating groups.
• Such groups are essential in stabilizing the positive charge.
• Tertiary structures are more stable than secondary or primary ones.

So, when evaluating options like (a) (\((\text{CH}_3)_2 \text{CHCH}_2 \text{OH}\)) and (b) (\((\text{CH}_3)_3 \text{COH}\)), recognizing the formation of tertiary carbocations directly determines the stability of the intermediates.

Carbonium Ion Stability

Stability in carbocations or carbonium ions differs greatly depending on their structure. Electron-donating effects from adjacent alkyl groups can dictate how stable these ions are once formed during a reaction like dehydration of alcohols.
The relative stability can generally be summarized as tertiary > secondary > primary. In essence, a tertiary carbonium ion is the most stable due to:

• Hyperconjugation: Overlapping of \(\sigma\) bonding orbitals with the empty p orbital on the carbon atom helps stabilize the positive charge.
• Inductive Effect: The electron-releasing effect from alkyl groups also aids in stabilizing the positive charge on the carbonium ion.

Being aware of this stability order helps predict reaction outcomes, especially in organic chemistry where carbocation intermediates play a crucial role.

Alcohols Reaction Mechanism

Let's see how the reaction mechanism unfolds for alcohols undergoing dehydration. Dehydration is the elimination of a water molecule, typically accomplished under acidic conditions. In this process, the alcohol converts to a carbocation before it typically rearranges into a more stable structure, if possible.

• Step 1: Protonation of the hydroxyl group \(\text{OH}\) in the alcohol, making it a better leaving group (water \(\text{H}_2\text{O}\)).
• Step 2: Formation of carbocation by the departure of water. This is a slow, rate-determining step.
• Step 3: Carbocation rearrangement might occur if a more stable carbocation can form due to migration of hydride or alkyl groups.

This mechanism helps understand why tertiary alcohols, leading to tertiary carbocations, dehydrate more readily compared to secondary or primary alcohols. Tertiary carbocations are less likely to undergo loss, making them favorable intermediates.