`CH_(3)-overset(O)overset(||)C-CH_(2)-CH_(3)+CH_(3)MgBrrarr X overset(H_(3)O^(+))rarr Yoverset(H_(2)SO_(4))underset(170^(@)C)rarr Z.`
What is Z ?

Options

A. `CH_(3)-overset(OH)overset("| ")underset(CH_(3))underset("| ")"C "-CH_(2)-CH_(3)`
B. `CH_(3)-underset(CH_(3))underset("| ")"C "=CH-CH_(3)`
C. `CH_(3)-overset(C_(2)H_(5))overset("| ")underset(CH_(3))underset("| ")"C "-O-overset(C_(2)H_(5))overset("| ")underset(CH_(3))underset("| ")"C "-CH_(3)`
D. `CH_(2)=underset(CH_(3))underset("| ")"C "-CH_(2)-CH_(3)`

Explanation

To solve the given reaction sequence, we will break it down step by step, identifying the products at each stage.

### Step 1: Formation of X
We start with the compound:
\[ CH_3C(=O)CH_2CH_3 + CH_3MgBr \]

This is a reaction between an aldehyde (or ketone) and a Grignard reagent (CH₃MgBr). The Grignard reagent acts as a nucleophile and attacks the electrophilic carbon of the carbonyl group.

- **Mechanism**: The nucleophilic CH₃⁻ from the Grignard reagent attacks the carbonyl carbon (C=O), leading to the formation of an alkoxide intermediate.

The reaction can be represented as:
\[ CH_3C(=O)CH_2CH_3 + CH_3MgBr \rightarrow CH_3C(OH)(CH_2CH_3)CH_3MgBr \]

This intermediate is denoted as X:
\[ X = CH_3C(OH)(CH_2CH_3)CH_3MgBr \]

### Step 2: Formation of Y
Next, we treat X with dilute acid (H₃O⁺) to protonate the alkoxide and convert it into an alcohol.

- **Reaction**:
\[ CH_3C(OH)(CH_2CH_3)CH_3MgBr + H_3O^+ \rightarrow CH_3C(OH)(CH_2CH_3)CH_3 + MgBr^+ \]

This gives us the alcohol Y:
\[ Y = CH_3C(OH)(CH_2CH_3)CH_3 \]

### Step 3: Formation of Z
Now, we treat Y with concentrated sulfuric acid (H₂SO₄) at 170°C. This will lead to dehydration, where water is eliminated to form an alkene.

- **Mechanism**: The alcohol undergoes dehydration to form a carbocation, which can then rearrange or eliminate to form the double bond.

The reaction can be represented as:
\[ CH_3C(OH)(CH_2CH_3) \xrightarrow{H_2SO_4, 170°C} CH_3C^+(CH_2CH_3) + H_2O \]

From the carbocation, two possible alkenes can form:
1. \( CH_3C(CH_2)CH_2 \) (1-butene)
2. \( CH_3C(CH_3)CH \) (2-butene)

The more stable alkene will be the one with more alkyl substituents (more alpha hydrogens), which is 2-butene.

Thus, the final product Z is:
\[ Z = CH_3C(CH_3)=CH_2 \] (2-butene)

### Final Answer:
Z is 2-butene.

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