Product formed in the following reaction is `CH_3CH_2CH(OH)CH_3overset(H_2SO_4)rarr`

Options

A. `CH_3CH=CHCH_3` predominates
B. `CH_2=CHCH_2CH_3` predominates
C. Both are formed in equal amounts
D. The amount of production depends on the nature of catalyst

Answer

To determine the product formed in the reaction of 2-degree alcohol (specifically, 2-butanol) in the presence of concentrated sulfuric acid (H₂SO₄), we can follow these steps:

### Step 1: Identify the Alcohol
The given alcohol is 2-butanol, which has the structure:
\[ \text{CH}_3\text{CH}_2\text{CHOHCH}_3 \]

### Step 2: Understand the Role of Concentrated H₂SO₄
Concentrated sulfuric acid acts as a dehydrating agent. It will facilitate the removal of a water molecule (H₂O) from the alcohol, leading to the formation of a carbocation.

### Step 3: Protonation of the Alcohol
When 2-butanol reacts with H₂SO₄, the hydroxyl group (-OH) of the alcohol gets protonated, forming a better leaving group (water):
\[ \text{CH}_3\text{CH}_2\text{C(OH)}\text{H}_3 + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{C(OH}_2^+)\text{H}_3 \]

### Step 4: Formation of Carbocation
The protonated alcohol (oxonium ion) loses a water molecule, resulting in the formation of a carbocation:
\[ \text{CH}_3\text{CH}_2\text{C(OH}_2^+)\text{H}_3 \rightarrow \text{CH}_3\text{CH}_2\text{C}^+\text{H}_3 + \text{H}_2\text{O} \]

### Step 5: Elimination to Form Alkenes
The carbocation can undergo elimination to form alkenes. In this case, two possible alkenes can be formed:
1. But-1-ene (from the elimination of a hydrogen from the terminal carbon)
2. But-2-ene (from the elimination of a hydrogen from the middle carbon)

### Step 6: Determine the Major Product
But-2-ene is more stable than but-1-ene due to hyperconjugation and the presence of more alkyl groups attached to the double bond. Therefore, but-2-ene will be the major product.

### Final Answer
The product formed in the reaction is **but-2-ene**.

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