`CH_(3) CH (CH_(3))CH_(2)NH_(2) overset( NaNO_(2)//HCI ) to (A) overset( aq.KON ) 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, we will follow the steps outlined in the video transcript, detailing the reactions and transformations that occur.

### Step-by-Step Solution:

1. **Identify the Initial Compound**:
The starting compound is **CH₃-CH(CH₃)-CH₂-NH₂** (isobutylamine).

2. **Reaction with Sodium Nitrite and Hydrochloric Acid**:
When isobutylamine reacts with sodium nitrite (NaNO₂) in the presence of hydrochloric acid (HCl) at 0°C, it forms a diazonium salt. The reaction can be represented as:
\[ CH₃-CH(CH₃)-CH₂-NH₂ + NaNO₂ + HCl \rightarrow CH₃-CH(CH₃)-CH₂-N₂^+Cl^- \]
This product is an aliphatic diazonium salt, which is unstable.

3. **Decomposition of the Diazonium Salt**:
The diazonium salt is unstable and will decompose upon heating or even at room temperature, releasing nitrogen gas (N₂) and forming a carbocation:
\[ CH₃-CH(CH₃)-CH₂-N₂^+Cl^- \rightarrow CH₃-CH(CH₃)-CH₂^+ + N₂ \]

4. **Rearrangement of the Carbocation**:
The formed carbocation is unstable and can undergo rearrangement to form a more stable carbocation. In this case, a hydrogen atom from the adjacent carbon can shift to stabilize the carbocation:
\[ CH₃-CH(CH₃)-CH₂^+ \rightarrow CH₃-C^+(CH₃)-CH₃ \]
This results in a tertiary carbocation.

5. **Reaction with Aqueous Potassium Hydroxide (KOH)**:
The tertiary carbocation then reacts with aqueous KOH, where the hydroxide ion (OH⁻) attacks the carbocation:
\[ CH₃-C^+(CH₃)-CH₃ + OH^- \rightarrow CH₃-C(OH)(CH₃)-CH₃ \]
The final product is **2-methylpropan-2-ol** (tert-butyl alcohol).

### Final Answer:
The compound (B) is **2-methylpropan-2-ol**.

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