In the following reaction
`CH_3-CH_2-CH_2-CH_3 underset"475 K"overset(H_2SO_4)to `

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

Explanation

To solve the problem, we need to analyze the reaction of butanol (CH3-CH2-CH2-CH3) in the presence of concentrated sulfuric acid (H2SO4) at a high temperature (475 K). The reaction involves dehydration, leading to the formation of an alkene.

### Step-by-Step Solution:

1. **Identify the Reactant**:
The reactant is butanol, which can be represented as CH3-CH2-CH2-CH3 (butane).

2. **Protonation of Alcohol**:
In the presence of concentrated sulfuric acid, the hydroxyl group (-OH) of butanol gets protonated to form a better leaving group (water, H2O). This step is crucial as it converts the alcohol into a more reactive species.
\[ CH3-CH2-CH2-CH2OH + H2SO4 \rightarrow CH3-CH2-CH2-CH2OH2^+ + HSO4^- \]

3. **Formation of Carbocation**:
The protonated alcohol (CH3-CH2-CH2-CH2OH2^+) loses a water molecule, resulting in the formation of a carbocation. The carbocation can form at the carbon where the -OH group was attached.
\[ CH3-CH2-CH2-CH2OH2^+ \rightarrow CH3-CH2-CH2^+ + H2O \]

4. **Carbocation Rearrangement**:
The formed carbocation can undergo rearrangement to form a more stable carbocation. In this case, the primary carbocation can rearrange to a more stable secondary carbocation.
\[ CH3-CH2-CH2^+ \rightarrow CH3-CH^+-CH2-CH3 \]

5. **Elimination of Beta Hydrogen**:
The next step involves the elimination of a beta hydrogen atom. There are two possible beta hydrogen atoms that can be eliminated, leading to two different alkenes:
- If the hydrogen from the first beta carbon is eliminated, 1-butene is formed.
- If the hydrogen from the second beta carbon is eliminated, 2-butene is formed.
\[ \text{Elimination of H from 1st beta carbon} \rightarrow CH3-CH2-CH=CH2 \quad (1-butene) \]
\[ \text{Elimination of H from 2nd beta carbon} \rightarrow CH3-CH=CH-CH3 \quad (2-butene) \]

6. **Determine the Major Product**:
According to Zaitsev's rule, the more substituted alkene is favored. In this case, 2-butene is more stable than 1-butene due to hyperconjugation and the stability of the double bond.
- 2-butene is the major product, while 1-butene is the minor product.

### Final Answer:
The predominant product of the reaction is **2-butene**.