Dehydration of cyclopentyl carbinol with conc.`H_(2)SO_(4)` forms

Options

A. Cyclopentene
B. Cyclohexene
C. Cyclohexane
D. none of these

Answer

To solve the problem of what product is formed when cyclopentyl carbinol undergoes dehydration with concentrated sulfuric acid, we can follow these steps:

### Step 1: Identify the Reactant
The reactant is cyclopentyl carbinol, which can be represented as:
\[ \text{C}_5\text{H}_{10}\text{O} \]
This structure consists of a cyclopentane ring with a hydroxyl (OH) group attached to a carbon atom.

### Step 2: Protonation of the Alcohol
When cyclopentyl carbinol is treated with concentrated sulfuric acid (H₂SO₄), the first step is the protonation of the hydroxyl group. The sulfuric acid donates a proton (H⁺) to the oxygen of the hydroxyl group, making it a better leaving group:
\[ \text{C}_5\text{H}_{10}\text{OH} + \text{H}^+ \rightarrow \text{C}_5\text{H}_{10}\text{OH}_2^+ \]

### Step 3: Formation of Carbocation
After protonation, water (H₂O) can leave, leading to the formation of a carbocation:
\[ \text{C}_5\text{H}_{10}\text{OH}_2^+ \rightarrow \text{C}_5\text{H}_{10}^+ + \text{H}_2\text{O} \]
This results in a cyclopentyl carbocation.

### Step 4: Ring Expansion
Cyclopentyl carbocation is less stable due to ring strain. According to Baeyer strain theory, five-membered rings are less stable than six-membered rings. Therefore, the carbocation can undergo ring expansion to form a more stable six-membered ring (cyclohexane):
\[ \text{C}_5\text{H}_{10}^+ \rightarrow \text{C}_6\text{H}_{12}^+ \]
This results in the formation of a cyclohexyl carbocation.

### Step 5: Deprotonation
The final step involves the loss of a proton from the cyclohexyl carbocation, leading to the formation of cyclohexene:
\[ \text{C}_6\text{H}_{12}^+ \rightarrow \text{C}_6\text{H}_{10} + \text{H}^+ \]

### Final Product
The final product of the dehydration of cyclopentyl carbinol with concentrated sulfuric acid is cyclohexene.

### Summary
Thus, the product formed is:
\[ \text{Cyclohexene} \]

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