`CH_(3) CH (CH_(3))CH_(2)NH_(2) overset( NaNO_(2)//HCI ) to (A) overset( aq.KOH ) to (B) ` compound (B ) is :

Options

A. `CH_(3)CH(CH_(3))CH_2 OH`
B. `(CH_3)_2C(OH) C_2 H_5`
C. `(CH_3)_3COH`
D. None of these

Answer

To solve the question step by step, we will analyze the reactions involving the given amine and identify the products formed at each stage.

### Step 1: Identify the Given Amine
The starting compound is **CH₃CH(CH₃)CH₂NH₂**, which is a primary amine. The structure can be represented as follows:
- **CH₃** (methyl group)
- **CH(CH₃)** (isopropyl group)
- **CH₂** (methylene group)
- **NH₂** (amino group)

### Step 2: Reaction with NaNO₂/HCl
When the primary amine reacts with sodium nitrite (NaNO₂) in the presence of hydrochloric acid (HCl), it forms a diazonium salt. The reaction can be summarized as follows:
- **CH₃CH(CH₃)CH₂NH₂ + NaNO₂/HCl → CH₃CH(CH₃)CH₂N₂⁺Cl⁻ (A)**

The diazonium cation formed is **CH₃CH(CH₃)CH₂N₂⁺**. This compound is unstable and can decompose to form a carbocation.

### Step 3: Formation of Carbocation
The diazonium salt can lose nitrogen gas (N₂) to form a carbocation:
- **CH₃CH(CH₃)CH₂N₂⁺ → CH₃CH(CH₃)CH₂⁺ + N₂**

Here, the carbocation formed is **CH₃CH(CH₃)CH₂⁺**, which is a secondary carbocation.

### Step 4: Reaction with Aqueous KOH
The carbocation is then treated with aqueous KOH. The hydroxide ion (OH⁻) acts as a nucleophile and attacks the carbocation, leading to the formation of an alcohol:
- **CH₃CH(CH₃)CH₂⁺ + OH⁻ → CH₃CH(CH₃)CH₂OH (B)**

The product formed, **(B)**, is **CH₃CH(CH₃)CH₂OH**, which is 2-propanol (also known as isopropyl alcohol).

### Final Answer
Thus, the compound (B) is **2-propanol (isopropyl alcohol)**.

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