The products for the following reactions are
(i) `CH_(3)-underset(H)underset(|)overset(Br)overset(|)(C)-CH_(2)-CH_(3)+"alc. KOH"to X`
(ii) `CH_(3)-underset(CH_(3))underset(|)(C)H-CH=CH_(2) overset(O_(3))to Y+Z`

Options

A. `X=(CH_(3))_(2)C=CH_(2),Y=CH_(3)CH_(2)CHO,Z=CH_(3)CH_(2)CHO`
B. `X=CH_(2)=CH_(2),Y=CH_(3)CHO,Z=CH_(3)COOH`
C. `X=CH_(3)-CH=CH-CH_(3)`
`Y=CH_(3)-overset(CH_(3))overset(|)(C)H-CHO,Z=HCHO`
D. `X=CH_(3)-CH=C(CH_(3))_(2),Y=HCHO,Z=CH_(3)CHO`

Explanation

To solve the reactions given in the question, we will analyze each reaction step by step.

### Reaction (i):
**Given Reaction:**
\[ \text{CH}_3-\text{C}(\text{Br})-\text{CH}_2-\text{CH}_3 + \text{alc. KOH} \rightarrow X \]

**Step 1: Identify the Reactants**
- The reactant is 1-bromo-2-butane (CH3-CHBr-CH2-CH3) and alcoholic KOH.

**Step 2: Understand the Role of Alcoholic KOH**
- Alcoholic KOH is a strong base and will facilitate elimination reactions (dehydrohalogenation) rather than substitution.

**Step 3: Formation of Carbocation**
- The base (OH-) will abstract a proton (H+) from the carbon adjacent to the carbon bearing the bromine (Br). This leads to the formation of a carbocation.
- The more stable carbocation (secondary) will form at the carbon adjacent to the bromine.

**Step 4: Elimination Reaction**
- The bromine (Br) will leave, and a double bond will form between the carbons, resulting in the formation of an alkene.

**Step 5: Write the Product**
- The product formed is 2-butene (CH3-CH=CH-CH3).
- Therefore, \( X = \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3 \).

### Reaction (ii):
**Given Reaction:**
\[ \text{CH}_3-\text{C}(\text{CH}_3)-\text{CH}=\text{CH}_2 + \text{O}_3 \rightarrow Y + Z \]

**Step 1: Identify the Reactants**
- The reactant is 3-methyl-1-butene (CH3-CH(CH3)-CH=CH2) and ozone (O3).

**Step 2: Understand Ozonolysis**
- Ozonolysis involves the cleavage of double bonds in the presence of ozone, resulting in the formation of carbonyl compounds.

**Step 3: Break the Double Bond**
- The double bond between the carbons will break, and oxygen will be added to the resulting fragments.

**Step 4: Write the Products**
- The ozonolysis of 3-methyl-1-butene will yield two products:
- 2-methylpropanal (from the carbon chain with the methyl group).
- Formaldehyde (HCHO) from the other fragment.

Thus, the products are:
- \( Y = \text{2-methylpropanal} \)
- \( Z = \text{formaldehyde (HCHO)} \)

### Final Answers:
- \( X = \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3 \) (2-butene)
- \( Y = \text{2-methylpropanal} \)
- \( Z = \text{formaldehyde (HCHO)} \)