What is the product of the following reactions?
`Me - overset(Me) overset(|)underset(Me ) underset(|) C-C=CHoverset({:((i)H_(2)SO_(4//HgSO_(4))), ((ii) NaBH_(4)):})underset({:((iii) Conc. H _(2) SO_(4)), ((iv) O _(3) ZnH_(2) O ):})to`

Options

A. only `CH_(3) CH_(2) CHO`
B. only `CH _(3) - overset(O) overset(|) C - CH _(3)`
C. only `CH _(3) CHO`
D. only `CH _(3) CH_(2) COOH`

Explanation

To solve the question regarding the product of the given reactions, we will break down the process step by step.

### Step 1: Identify the Starting Compound
The starting compound is a terminal alkyne, specifically 3-methyl-1-butyne, which can be represented as:
\[ \text{Me}-\text{C}\equiv\text{C}-\text{Me} \]

### Step 2: Reaction with HgSO4 and H2SO4
The first reaction involves treating the alkyne with dilute sulfuric acid (H2SO4) and mercuric sulfate (HgSO4). This reaction leads to the formation of an enol through electrophilic addition:
1. The triple bond reacts with Hg^2+ to form a vinyl cation.
2. Water (from the dilute acid) attacks the vinyl cation, leading to the formation of an enol.

The product after this step is:
\[ \text{Me}-\text{C}(\text{OH})=\text{C}(\text{Me})-\text{H} \]

### Step 3: Tautomerization to Ketone
The enol formed is not stable and undergoes tautomerization to form a ketone:
\[ \text{Me}-\text{C}(\text{O})=\text{C}(\text{Me})-\text{Me} \]
This product is 3-pentanone (or 3-methyl-2-butanone).

### Step 4: Reduction with NaBH4
Next, the ketone is reduced using sodium borohydride (NaBH4), which converts the ketone into an alcohol:
\[ \text{Me}-\text{C}(\text{OH})-\text{C}(\text{Me})-\text{Me} \]
This product is 3-pentanol.

### Step 5: Dehydration with Concentrated H2SO4
The alcohol undergoes dehydration when treated with concentrated sulfuric acid, leading to the formation of an alkene:
1. Protonation of the alcohol group occurs, forming a good leaving group.
2. Elimination of water leads to the formation of the alkene.

The product after dehydration is:
\[ \text{Me}-\text{C}=\text{C}(\text{Me})-\text{Me} \]
This is 3-methyl-2-pentene.

### Step 6: Ozonolysis
Finally, the alkene undergoes ozonolysis (O3 followed by Zn/H2O), which cleaves the double bond and forms two ketones:
\[ \text{Me}-\text{C}(\text{O})-\text{C}(\text{Me})-\text{O} \]
This results in the formation of two moles of acetone (CH3COCH3).

### Final Product
The final product of the reactions is two moles of acetone.

### Summary of Steps
1. Start with terminal alkyne.
2. React with HgSO4 and H2SO4 to form an enol.
3. Tautomerize the enol to form a ketone.
4. Reduce the ketone with NaBH4 to form an alcohol.
5. Dehydrate the alcohol with concentrated H2SO4 to form an alkene.
6. Perform ozonolysis to yield two moles of acetone.