`CH_(3)-CH_(2)-underset(Br)underset(|)(CH)-CH_(3) overset("alc.KOH")rarr`
`underset("Major")X" "," "CH_(3)-CH_(2) underset(oplus)underset(NMe_(3))underset(|)(CH)-CH_(3)`
`underset(Delta)overset("EtONa")rarr underset("Major")(Y)`
Product (X) and (Y) respectively is

Options

A. 1-butene, trans-2-butene
B. 1-butene, cis-2-butene
C. cis-2-butene, 1-butene
D. trans-2-butene, 1-butene

Answer

To solve the question, we need to identify the products X and Y formed from the given reactions. Let's break it down step by step.

### Step 1: Identify the Reactants
The starting compound is an alkyl halide:
\[ \text{CH}_3\text{CH}_2\text{C(Br)H}\text{CH}_3 \]
This compound has a bromine (Br) atom attached to a carbon chain.

### Step 2: Reaction with Alcoholic KOH
The alkyl halide reacts with alcoholic KOH, which is a strong base. This reaction typically leads to an elimination reaction (E2 mechanism) where the halogen (Br) and a hydrogen atom from an adjacent carbon are removed, forming a double bond.

### Step 3: Determine the Major Product (X)
To find the major product, we need to consider the possible elimination pathways:
1. Removing a hydrogen from the carbon adjacent to the carbon with Br (C1) leads to:
- **Product 1:** \[ \text{CH}_3\text{CH}_2\text{C}=\text{C}\text{H}\text{CH}_3 \] (but-2-ene)
2. Removing a hydrogen from the other adjacent carbon (C2) leads to:
- **Product 2:** \[ \text{CH}_3\text{CH}_2\text{C}=\text{C}\text{H}_2 \] (but-1-ene)

According to Zaitsev's rule, the major product will be the more substituted alkene. In this case, but-2-ene (specifically trans-2-butene) is more substituted than but-1-ene. Therefore, the major product (X) is:
\[ \text{X} = \text{trans-2-butene} \]

### Step 4: Identify the Second Reactant for Product Y
The second part of the question involves the product Y, which is formed from the compound:
\[ \text{CH}_3\text{CH}_2\text{C}(\text{NMe}_3)\text{H}\text{CH}_3 \]
This compound reacts with sodium ethoxide (a weak base).

### Step 5: Reaction with Sodium Ethoxide
In this case, sodium ethoxide will also lead to an elimination reaction, but since it is a weak base, it will favor the formation of the less substituted alkene (Hofmann product).

### Step 6: Determine the Major Product (Y)
For the elimination reaction with sodium ethoxide:
1. Removing a hydrogen from the carbon adjacent to the nitrogen (C1) leads to:
- **Product 1:** \[ \text{CH}_3\text{CH}_2\text{C}=\text{C}\text{H}\text{CH}_3 \] (but-2-ene)
2. Removing a hydrogen from the other adjacent carbon (C2) leads to:
- **Product 2:** \[ \text{CH}_3\text{CH}_2\text{C}=\text{C}\text{H}_2 \] (but-1-ene)

Since sodium ethoxide favors the formation of the less substituted alkene, the major product (Y) will be:
\[ \text{Y} = \text{but-1-ene} \]

### Final Answer
Thus, the products X and Y are:
- **X:** trans-2-butene
- **Y:** but-1-ene