Which of the following is a stronger base? Give reason to justify your answer.
`{:(CH_(2)=CH-NH_(2)" "CH_(2)=N-CH_(3)),(" "I" "II):}`

Answer

To determine which of the two compounds is a stronger base, we need to analyze the structures and the availability of the lone pair of electrons on the nitrogen atom in each compound.

### Step 1: Identify the Compounds
- Compound I: \( CH_2=CH-NH_2 \) (Vinylamine)
- Compound II: \( CH_2=N-CH_3 \) (Imine)

### Step 2: Understand Basicity
Basicity is the ability of a compound to donate a lone pair of electrons. The more available the lone pair is, the stronger the base.

### Step 3: Analyze Compound I (Vinylamine)
- In vinylamine, the nitrogen atom has a lone pair of electrons. However, due to resonance, this lone pair can participate in resonance with the adjacent double bond.
- The resonance structure can be represented as:
\[ CH_2=CH-NH_2 \leftrightarrow CH_2=CH^+-NH_2^- \]
- This resonance delocalizes the lone pair on nitrogen, making it less available for bonding with protons (H⁺). Thus, the basicity of compound I is decreased.

### Step 4: Analyze Compound II (Imine)
- In the imine (compound II), the nitrogen atom is bonded to a carbon atom with a double bond. The lone pair on nitrogen is not involved in resonance with any double bond.
- Therefore, the lone pair on nitrogen is fully available to accept protons. This makes compound II a stronger base compared to compound I.

### Step 5: Conclusion
Based on the analysis, we conclude that:
- **Compound II \( (CH_2=N-CH_3) \) is a stronger base than Compound I \( (CH_2=CH-NH_2) \)** because the lone pair on nitrogen in compound II is more available for protonation, while in compound I, the lone pair is delocalized due to resonance.

### Final Answer
**Compound II \( (CH_2=N-CH_3) \) is the stronger base.**

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