In the following reactions `CH_3-overset(CH_3)overset(|)CH-underset(OH)underset(|)CH-CH_3overset(H^+//heat)rarrunderset([underset("product")"Major"])A+underset([underset("product")("Minor")])B`
(ii) `Aoverset("HBr, dark")underset("in absence of peroxide")rarrunderset([underset("product")"Mejor"])C+underset([underset("product")("Minor")])D`
The major products A and C are repectively

Options

A. `CH_2=overset(CH_3)overset(|)C-CH_2-CH_3" and" underset(Br)underset(|)CH_2-overset(CH_3)overset(|)CH-CH_2-CH_3`
B. `CH_3-overset(CH_3)overset(|)C=CH-CH_3 " and "CH_3-overset(CH_3)overset(|)underset(Br)underset(|)C-CH_2-CH_3`
C. `CH_3-overset(CH_3)overset(|)C=CH-CH_3 " and "CH_3-overset(CH_3)overset(|)CH-underset(Br)underset(|)CH-CH_3`
D. `CH_3-overset(CH_3)overset(|)C=CH-CH_3 " and "CH_3-underset(Br)underset(|)overset(CH_3)overset(|)C-CH_2-CH_3`

Explanation

To solve the given question step by step, we will analyze the reactions and determine the major products A and C.

### Step 1: Identify the starting material
The starting material is a compound with the structure:
```
CH3
|
CH3-CH-CH2-CH3
|
OH
```
This compound is a secondary alcohol.

### Step 2: Perform dehydration reaction
When the secondary alcohol is treated with H+ (acid) and heat, dehydration occurs. The hydroxyl group (OH) is protonated to form a better leaving group (H2O). The reaction proceeds as follows:
1. Protonation of the OH group leads to the formation of a good leaving group (H2O).
2. The water molecule leaves, forming a carbocation.

### Step 3: Carbocation formation and rearrangement
The initial carbocation formed is a secondary carbocation. However, secondary carbocations can rearrange to form more stable tertiary carbocations through a hydride shift:
- The hydrogen from the adjacent carbon shifts to the carbocation center, resulting in a more stable tertiary carbocation.

### Step 4: Formation of alkene (Product A)
From the tertiary carbocation, we can eliminate a hydrogen atom to form an alkene. The major product A will be formed by removing a hydrogen from the carbon adjacent to the carbocation:
- The major product A is:
```
CH3
|
CH3-C=CH-CH3
```
This is the more stable alkene due to the more substituted double bond.

### Step 5: Identify minor product B
The minor product B can be formed by removing a hydrogen from the original carbon that was not involved in the rearrangement. This leads to a less substituted double bond:
- The minor product B is:
```
CH2
|
CH3-CH=CH-CH3
```

### Step 6: Reaction of Product A with HBr
Next, we react product A with HBr in the dark (absence of peroxide). According to Markovnikov's rule, the H+ will add to the carbon with more hydrogens (the less substituted carbon), leading to the formation of a more stable carbocation:
- The major product C formed will be:
```
CH3
|
CH3-C(Br)-CH2-CH3
```
This is the major product C.

### Step 7: Identify minor product D
The minor product D can be formed by the addition of Br- to the less substituted carbon, leading to a less stable product:
- The minor product D is:
```
CH3
|
CH2-CH(Br)-CH3
```

### Conclusion
- Major product A: `CH3-CH=CH-CH3` (more substituted alkene)
- Major product C: `CH3-CH(Br)-CH2-CH3` (from the addition of HBr)