`(CH_3)_2 CH CH (CH_3) OHoverset(Acid)underset(-H_2O)rarr X`
The major product obtained in this reaction is

Options

A. `(CH_3)_2 CHCH=CH_2`
B. `(CH_3)_2C=CH-CH_3`
C. 1:1 mixture of (1) and (2)
D. None of the above

Answer

To solve the problem of determining the major product obtained from the dehydration of (CH₃)₂CHCH(CH₃)OH in the presence of acid, we can follow these steps:

### Step 1: Identify the Reactant
The reactant is (CH₃)₂CHCH(CH₃)OH, which is 2-methyl-2-butanol. This alcohol will undergo dehydration in the presence of an acid.

**Hint:** Look for the structure of the alcohol and identify the functional groups.

### Step 2: Protonation of the Alcohol
In the presence of acid (H⁺), the hydroxyl (-OH) group of the alcohol gets protonated, forming a better leaving group (water). The oxygen of the -OH group uses its lone pair to bond with the proton (H⁺), resulting in a positively charged oxonium ion.

**Hint:** Remember that protonation increases the leaving ability of the -OH group.

### Step 3: Formation of Carbocation
Once the water molecule leaves, a carbocation is formed. In this case, the carbocation formed is a secondary carbocation because the positively charged carbon is attached to two other carbon atoms.

**Hint:** Identify the type of carbocation formed (primary, secondary, tertiary) as it influences stability.

### Step 4: Carbocation Rearrangement
The secondary carbocation can undergo rearrangement to form a more stable tertiary carbocation through a 1,2-hydride shift. A hydrogen atom from an adjacent carbon shifts to the carbocation, stabilizing it.

**Hint:** Look for possible rearrangements that can lead to a more stable carbocation.

### Step 5: Elimination to Form Alkene
After the formation of the more stable carbocation, a proton (H⁺) is removed from an adjacent carbon atom, leading to the formation of a double bond. This results in the formation of an alkene.

**Hint:** Identify the location of the double bond and ensure it follows Zaitsev's rule, which states that the more substituted alkene is favored.

### Step 6: Determine the Major Product
The major product will be the more substituted alkene, which is formed by the elimination of a proton from the adjacent carbon to the carbocation. The resulting product is 2-methyl-2-butene.

**Hint:** Compare the possible products and choose the one with the most alkyl substituents on the double bond.

### Final Answer
The major product obtained from the dehydration of (CH₃)₂CHCH(CH₃)OH in the presence of acid is **2-methyl-2-butene**.

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