Elimination of `HBr`from 2 bromobutane results in the formation of .

Options

A. Predominantly 2-butyne
B. Predominantly 1-butene
C. Predominantly 2-butene
D. Equimolar mixture of 1 and 2-butene

Answer

To solve the question regarding the elimination of HBr from 2-bromobutane, we can follow these steps:

### Step 1: Identify the Structure of 2-Bromobutane
2-bromobutane has the following structure:
- It consists of a four-carbon chain with a bromine atom attached to the second carbon.
- The molecular formula is C4H9Br.

### Step 2: Understand the Elimination Reaction
The elimination of HBr from 2-bromobutane is a dehydrohalogenation reaction. This reaction typically occurs in the presence of a strong base, such as alcoholic KOH.

### Step 3: Mechanism of the Reaction
In the elimination reaction, the base (KOH) abstracts a proton (H) from a carbon adjacent to the carbon bearing the bromine atom (the β-carbon), while the bromine atom leaves as bromide ion (Br⁻). This results in the formation of a double bond between the two carbons.

### Step 4: Identify the Possible Products
From 2-bromobutane, two possible alkenes can form:
1. **Major Product (Zaitsev's Product)**: The more substituted alkene, which is more stable due to hyperconjugation and the inductive effect. This product is formed by removing a hydrogen from the β-carbon that has more hydrogen atoms.
- This product is 2-butene (CH3-CH=CH-CH3).

2. **Minor Product (Hofmann Product)**: The less substituted alkene, which is less stable. This product is formed by removing a hydrogen from the β-carbon that has fewer hydrogen atoms.
- This product is 1-butene (CH2=CH-CH2-CH3).

### Step 5: Determine the Major Product
In this case, the major product will be 2-butene, as it is more stable than 1-butene. The reaction predominantly favors the formation of the more substituted alkene due to Zaitsev's rule.

### Final Answer
The elimination of HBr from 2-bromobutane results in the formation of 2-butene (CH3-CH=CH-CH3).

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