The major product (Y) in the following reaction is
`CH_3-overset(CH_3)overset(|)(CH)-C-=CHoverset(HgSO_4,H_2SO_4)rarrXunderset((ii)"Conc".H_2SO_4)overset((i)C_2H_5MgBr,H_2O)rarrY`

Options

A. `CH_3-overset(CH_3)overset(|)C=underset(C_2H_5)underset(|)C-CH_3`
B. `CH_3-overset(CH_3)overset(|)C=underset(CH_3)underset(|)C-CH_3`
C. `H_3C-overset(CH_2)overset(||)C=underset(C_2H_5)underset(|)(CH)-CH_3`
D. `CH_3-overset(CH_3)overset(|)(CH)=underset(CH_3)underset(|)C=CH-CH_3`

Answer

To solve the given reaction step-by-step, we will analyze the reaction sequence and identify the major product (Y).

### Step 1: Identify the starting compound
The starting compound is an alkyne, specifically 3-methyl-1-butyne, which can be represented as:
\[ \text{CH}_3-\text{C} \equiv \text{C}-\text{CH}_3 \]

### Step 2: Reaction with HgSO4 and H2SO4
When the alkyne is treated with mercuric sulfate (HgSO4) and sulfuric acid (H2SO4), it undergoes hydration to form a ketone. The reaction follows Markovnikov's rule, where the more substituted carbon gets the carbonyl group (C=O).

The hydration of 3-methyl-1-butyne results in:
\[ \text{CH}_3-\text{C}(\text{OH})-\text{C}(\text{CH}_3)=\text{C}(\text{O}) \]

This compound is 3-methyl-2-butanone (a ketone).

### Step 3: Identify the product (X)
The product after the first step (X) is:
\[ \text{CH}_3-\text{C}(\text{O})-\text{C}(\text{CH}_3)-\text{CH}_3 \]

### Step 4: Reaction with C2H5MgBr
Next, we react the ketone (X) with ethylmagnesium bromide (C2H5MgBr). This Grignard reagent acts as a nucleophile and attacks the carbonyl carbon of the ketone.

The reaction can be depicted as:
\[ \text{CH}_3-\text{C}(\text{O})-\text{C}(\text{CH}_3)-\text{CH}_3 + \text{C}_2\text{H}_5\text{MgBr} \rightarrow \text{CH}_3-\text{C}(\text{C}_2\text{H}_5)-\text{C}(\text{CH}_3)-\text{CH}_3 \]

### Step 5: Hydrolysis
After the Grignard reaction, hydrolysis occurs, replacing the magnesium bromide with a hydroxyl group (OH). The product after hydrolysis (let's call it Y) is:
\[ \text{CH}_3-\text{C}(\text{C}_2\text{H}_5)-\text{C}(\text{CH}_3)-\text{OH} \]

### Step 6: Elimination with Conc. H2SO4
The final step involves treating the alcohol with concentrated sulfuric acid (H2SO4), which promotes dehydration (elimination of water). This results in the formation of an alkene.

The dehydration of the alcohol leads to the formation of:
\[ \text{CH}_3-\text{C}=\text{C}(\text{CH}_3)-\text{C}_2\text{H}_5 \]

### Conclusion: Major Product (Y)
The major product (Y) after all the reactions is:
\[ \text{CH}_3-\text{C}=\text{C}(\text{CH}_3)-\text{C}_2\text{H}_5 \]