Question

Which amongst the following be most readily dehydrated under acidic conditions?

• A.
• B.
• C.
• D.

Answer 1

The Correct Option is C

To determine which compound is most readily dehydrated under acidic conditions, we need to consider the stability of the carbocation intermediate formed during the dehydration process. Dehydration under acidic conditions typically involves the following steps:

1. Protonation of the hydroxyl group to form water, a better leaving group. 
2. Loss of water to form a carbocation intermediate.
3. Rearrangement if necessary to form a more stable carbocation.
4. Elimination of a proton to form a double bond, resulting in an alkene.

Among the options provided, the compound that forms the most stable carbocation will dehydrate most readily. The stability of carbocations follows the order: tertiary > secondary > primary, due to hyperconjugation and inductive effects.

Let's examine the given options:

Option	Carbocation Stability
	Secondary carbocation
	Primary carbocation
	Tertiary carbocation
	Secondary carbocation with resonance stability

Option 3, which forms a tertiary carbocation, is the most stable. Therefore, it will be dehydrated most readily under acidic conditions.

Answer 2

Because of the existence of conjugation in the resulting product the correct answer is option (C):

Correct Answer: Option 3

This question tests your understanding of the acid-catalyzed dehydration of alcohols, a key concept in organic chemistry. Dehydration of alcohols typically proceeds via an E1 (unimolecular elimination) mechanism under acidic conditions, especially when the formation of a stable carbocation intermediate is possible.

Let’s examine the core principle: In acidic medium, an alcohol loses water to form a carbocation, and the ease of dehydration depends largely on the stability of the resulting carbocation. More stable carbocations (like tertiary > secondary > primary) favor faster dehydration.

Option 1: This compound contains a nitro group (NO₂) at the β-position. The OH group is on a secondary carbon, but the adjacent nitro group is electron-withdrawing, which destabilizes the carbocation intermediate. Hence, this compound will not dehydrate readily.

Option 2: Similar to option 1, it contains a nitro group adjacent to the OH-bearing carbon. Although the carbon is more substituted (potential for tertiary carbocation), the strong electron-withdrawing nature of NO₂ again reduces carbocation stability.

Option 3: This is the correct option. It contains a tertiary alcohol group (the OH is bonded to a carbon that is itself bonded to three other carbon atoms). Upon dehydration under acidic conditions, this compound will form a highly stable tertiary carbocation, which greatly facilitates the elimination of water and formation of an alkene. Tertiary carbocations are significantly more stable than secondary or primary ones due to hyperconjugation and inductive effects.

Option 4: This compound also has a nitro group, and the OH groups are on less substituted carbons. The presence of NO₂ again destabilizes the intermediate, and the molecule lacks the structural environment for tertiary carbocation stabilization.

Conclusion: Among all the choices, Option 3 forms a stable tertiary carbocation upon dehydration, making it the compound that will be most readily dehydrated under acidic conditions. Understanding the role of carbocation stability and electron-donating or withdrawing groups is essential for mastering elimination reactions and reaction mechanisms in organic chemistry