`(CH_3)_3CCloverset(NaCl)rarrAoverset(dil.H_2SO_4)rarrB`Compound B is

Options

A. `(CH)_3COOOH`
B. `(CH_3)_3COH`
C. `(CH_3)_3COC(CH_3)_3`
D. All the three

Explanation

To solve the problem, we need to analyze the reactions step by step.

### Step 1: Identify the starting compound
The starting compound is `(CH₃)₃CCl`, which is tert-butyl chloride (tert-butyl group with a chlorine atom).

### Step 2: Reaction with NaCl
When tert-butyl chloride reacts with NaCl, we can expect a nucleophilic substitution reaction. However, since tert-butyl chloride is a tertiary halide, it is more likely to undergo an elimination reaction rather than substitution.

In the presence of NaCl, the chloride ion (Cl⁻) can leave, leading to the formation of a tert-butyl cation `(CH₃)₃C⁺`.

### Step 3: Elimination Reaction
The tert-butyl cation is very stable and can undergo an elimination reaction. In the presence of a strong base, such as NaCN, it can lose a proton (H⁺) to form an alkene.

The elimination of H⁺ from the tert-butyl cation results in the formation of isobutylene (2-methylpropene):
\[ (CH₃)₃C⁺ \rightarrow CH₂=C(CH₃)₂ \]

### Step 4: Reaction with Dilute H₂SO₄
Next, the alkene (isobutylene) reacts with dilute sulfuric acid (H₂SO₄). In this reaction, the alkene undergoes hydration. The double bond in isobutylene will react with H⁺ from H₂SO₄, leading to the formation of a carbocation.

The carbocation formed can then react with water (H₂O) to form an alcohol. The water molecule acts as a nucleophile and attacks the carbocation, resulting in the formation of tert-butyl alcohol:
\[ CH₂=C(CH₃)₂ + H₂O \rightarrow (CH₃)₃COH \]

### Step 5: Final Product
The final product, compound B, is tert-butyl alcohol, which can be represented as:
\[ (CH₃)₃COH \]

### Conclusion
Thus, the compound B formed after the reactions is tert-butyl alcohol.

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