Which one of the following does not lose water even in conc. \(\text{H}_{2}\text{SO}_{4}\)?
Show Hint
Dehydration reactions that lead to antiaromatic products (like cyclobutadiene, which has $4\pi$ electrons) are thermodynamically and kinetically highly unfavorable. Conversely, those leading to aromatic products (like the tropylium cation or benzene) occur with remarkable ease.
Step 1: Understanding the Concept:
The dehydration of alcohols in the presence of a strong acid like concentrated $H_2SO_4$ typically proceeds via an E1 (Elimination Unimolecular) mechanism.
This mechanism involves the protonation of the hydroxyl group to form a better leaving group ($H_2O$), followed by the loss of the water molecule to generate a carbocation intermediate.
The final step is the removal of a proton from an adjacent carbon to form a double bond.
The feasibility of this reaction is largely determined by the stability of the intermediate carbocation and the thermodynamic stability of the resulting alkene.
In cyclic conjugated systems, Hückel's Rule of aromaticity ($4n+2$ $\pi$-electrons) and antiaromaticity ($4n$ $\pi$-electrons) is the primary factor deciding the reaction's success. Step 2: Detailed Explanation:
Let's analyze the structures provided in the question:
- Compound 1 (Cyclopentanol): This secondary alcohol can undergo dehydration to form cyclopentene. Cyclopentene is a standard, non-aromatic, stable cyclic alkene. The reaction proceeds through a secondary carbocation and occurs readily.
- Compound 3 (Cycloheptatrienol): Dehydration of this compound leads to the formation of the cycloheptatrienyl cation, commonly known as the tropylium ion. This cation is planar, cyclic, and possesses $6\pi$ electrons ($n=1$ in $4n+2$), making it exceptionally stable due to its aromatic character. Thus, compound 3 dehydrates very easily.
- Compound 4 (Cyclohexa-1,3-dien-5-ol): The dehydration of this molecule results in the formation of a benzene ring. Since benzene is highly stable due to aromaticity, this reaction is extremely fast and energetically favorable.
- Compound 2 (Cyclobutenol derivative): If this molecule were to lose water, it would form a conjugated cyclic system with $4\pi$ electrons in a square-like ring, which is cyclobutadiene. According to Hückel's rule, a planar cyclic conjugated system with $4n$ $\pi$-electrons is antiaromatic. Antiaromatic compounds are highly unstable and possess significantly higher energy than their open-chain counterparts.
Because the formation of an antiaromatic product involves a very high activation energy barrier, the dehydration of compound 2 is practically impossible, even in concentrated sulfuric acid. Step 3: Final Answer:
Compound 2 does not lose water because the resulting product would be a highly unstable antiaromatic species (cyclobutadiene). The correct option is (B). Step 4: Final Verification:
All other compounds lead to stable non-aromatic or highly stable aromatic products, while compound 2 is prohibited by the stability rules of cyclic $\pi$-systems.
