The products (A) and (B) are respectively.
(i) `CH_3-underset(Br)underset(|)(CH)-underset(CH_3)underset(|)(CH)-CH_3overset(C_2H_5ONa)to(A)`
(ii) `CH_2=overset(CH_3)overset(|)(C)-CH_2CH_3underset("Peroxide")overset(HBr)to(B)`

Options

A. `CH_3-underset(Br)underset(|)CH-underset(CH_3)underset(|)CH-CH_3overset(C_2H_5ONa)rarr(A)`
B. `CH_2=overset(CH_3)overset(|)C-CH_2CH_3overset(Hbr)underset("Peroxide")rarr(B)`
C. `CH_3CH_2-overset(CH_3)overset(|)"CH"-CH_3,CH_3(CH_2)_2CH_3`
D. `CH_3CH_2-underset(OC_2H_5)underset(|)C(CH_3),underset(Br)underset(|)CH_2-underset(CH_3)underset(|)"CH"-CH_2-CH_2`

Answer

To solve the question, we will analyze the two reactions step by step to determine the products A and B.

### Step 1: Analyze the first reaction
The first reaction is:
\[ \text{CH}_3\text{Br} + \text{C}_2\text{H}_5\text{O}^- \text{(sodium ethoxide)} \]

1. **Identify the structure**: The starting compound is a bromoalkane with two methyl groups on adjacent carbons.
2. **Determine the reaction type**: Sodium ethoxide is a strong base, which suggests that an elimination reaction (E2 mechanism) will occur rather than a substitution reaction.
3. **Identify the beta hydrogens**: The alpha carbon (the one attached to the bromine) has two beta carbons, each with one hydrogen atom that can be eliminated.
4. **Perform elimination**:
- If the hydrogen on the left beta carbon is removed, we will form a double bond between the alpha carbon and the adjacent carbon, leading to the product:
\[ \text{CH}_3\text{C}(\text{CH}_3)=\text{C}(\text{CH}_3)\text{CH}_2 \]
- If the hydrogen on the right beta carbon is removed, we will form a different alkene:
\[ \text{CH}_3\text{C}(\text{CH}_3)=\text{C}(\text{CH}_3)\text{CH}_2 \]
5. **Determine the major product**: The more stable alkene (with more alkyl substituents) will be the major product. The product with the double bond between the two central carbons is more stable due to hyperconjugation and steric effects.

Thus, the product A is:
\[ \text{A} = \text{CH}_3\text{C}(\text{CH}_3)=\text{C}(\text{CH}_3)\text{CH}_2 \]

### Step 2: Analyze the second reaction
The second reaction is:
\[ \text{CH}_2=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_3 + \text{HBr (in peroxide)} \]

1. **Identify the structure**: The starting compound is an alkene with a double bond between two carbons.
2. **Determine the reaction type**: The presence of peroxide indicates that the reaction will follow the anti-Markovnikov addition mechanism.
3. **Apply anti-Markovnikov's rule**: In this case, the H from HBr will add to the carbon with fewer hydrogen atoms, while Br will add to the carbon with more hydrogen atoms.
4. **Perform the addition**:
- H will add to the terminal carbon (the one with fewer hydrogens), and Br will add to the internal carbon.
5. **Write the final product**: The resulting product will be:
\[ \text{B} = \text{CH}_2\text{Br}-\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_3 \]

### Final Products
- Product A:
\[ \text{A} = \text{CH}_3\text{C}(\text{CH}_3)=\text{C}(\text{CH}_3)\text{CH}_2 \]
- Product B:
\[ \text{B} = \text{CH}_2\text{Br}-\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_3 \]