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

The compound (b) \\(CH_3)_3COH\\) gives the most stable carbonium ion on dehydration.

Step 1: Understanding Dehydration Reaction

In a dehydration reaction, an alcohol loses a water molecule to form an alkene. The process involves the formation of a carbonium ion (carbocation) intermediate. The stability of this carbocation is crucial for determining the final product distribution.

Step 2: Analyzing Stability of Carbocations

Carbocation stability is influenced by the number of alkyl groups attached to the positively charged carbon. Generally, tertiary (3°) carbocations are more stable than secondary (2°) and primary (1°) carbocations due to greater hyperconjugation and inductive effects.

Step 3: Evaluating Each Option

(a) \((CH_3)_2CHCH_2OH\) forms a secondary carbocation upon dehydration. (b) \((CH_3)_3COH\) forms a tertiary carbocation, \((CH_3)_3C^+\), which is very stable. (c) \((CH_3CH_2CH_2CH_2OH)\) forms a primary carbocation, which is the least stable. (d) \(CH_3CHOHCH_2CH_3\) forms a secondary carbocation.

Step 4: Conclusion Based on Stability

The most stable carbocation from the options given is the tertiary carbocation formed from compound (b). Recall that stability increases from primary to tertiary carbocations. Therefore, the dehydration of \((CH_3)_3COH\) will give the most stable carbonium ion.

Key Concepts

Dehydration Reaction

A dehydration reaction is a vital process in organic chemistry. This type of reaction involves the removal of a water molecule from an alcohol to form an alkene. The process occurs under acidic conditions, usually with the help of a strong acid like sulfuric acid.

Here's how it works:

• The alcohol molecule is protonated by the acid, making the -OH group a better leaving group, which then leaves as water.
• This generates a carbocation intermediate, a positively charged carbon atom.
• The most stable carbocation will then release a proton to form the alkene, which is the dehydration product.

Understanding how a dehydration reaction leads to different products based on carbocation stability is crucial. This guides organic chemists in predicting reaction outcomes, ensuring that the most stable alkene is formed during the process.

Carbocation Intermediate

In a dehydration reaction, the formation of a carbocation intermediate is a key step. A carbocation is an ion with a positively charged carbon atom.

The stability of carbocations depends heavily on the nature of the carbon atom:

• A primary carbocation has a single alkyl group attached.
• A secondary carbocation has two alkyl groups attached.
• A tertiary carbocation has three alkyl groups attached, making it the most stable due to optimal hyperconjugation and inductive effects.

During a dehydration reaction, the alcohol's hydroxyl group leaves, resulting in a carbocation. The formation of a stable carbocation allows the dehydration process to proceed smoothly, directing the reaction towards forming stable products.

Tertiary Carbocation Stability

Tertiary carbocations are considered the pinnacle of stability when it comes to carbocation species. This stability plays a critical role in reactions like dehydration.

Why are tertiary carbocations stable?

• Hyperconjugation: Involves the delocalization of electrons from adjacent C-H bonds to the positively charged carbon, elevating stability.
• Inductive effect: Electron-donating alkyl groups help reduce the positive charge on the carbon, enhancing stability further.
• Tertiary carbocations have three alkyl groups, providing the utmost combination of hyperconjugation and inductive stabilization.

Since stability is paramount for the progression of the dehydration reaction, being able to identify the formation of a tertiary carbocation can help predict where the most stable product will arise. This is why in dehydration reactions involving multiple alcohols, those that form tertiary carbocations are typically more reactive and efficient.