Consider the given compounds :
(a) `CH_(3)-CH_(2)-NH_(2)` (b) `CH_(3)-CH=NH`
(c ) `CH_(3)-C=N` (d) `C_(2)H_(5)-NH-C_(2)H_(5)`
Arrange basicity of these compounds in decreasing order :

Options

A. `4 gt 1 gt 2 gt 3`
B. `1 gt 2 gt 3 gt 4`
C. `1 gt 4 gt 2 gt 3`
D. `4 gt 1 gt 3 gt 2`

Answer

To determine the basicity of the given compounds, we need to analyze the availability of lone pairs of electrons on the nitrogen atom in each compound. Basicity is influenced by the hybridization of the nitrogen atom and the electron-donating effects of alkyl groups.

### Step-by-Step Solution:

1. **Identify the Compounds:**
- (a) `CH₃-CH₂-NH₂` (Ethylamine)
- (b) `CH₃-CH=NH` (Vinylamine)
- (c) `CH₃-C≡N` (Methyl cyanide)
- (d) `C₂H₅-NH-C₂H₅` (Diethylamine)

2. **Analyze the Hybridization of Nitrogen:**
- In compound (a) `CH₃-CH₂-NH₂`, the nitrogen is **sp³** hybridized.
- In compound (b) `CH₃-CH=NH`, the nitrogen is **sp²** hybridized.
- In compound (c) `CH₃-C≡N`, the nitrogen is **sp** hybridized.
- In compound (d) `C₂H₅-NH-C₂H₅`, the nitrogen is also **sp³** hybridized.

3. **Consider the Electronegativity and Basicity:**
- The order of electronegativity based on hybridization is: **sp > sp² > sp³**.
- Higher electronegativity means less availability of lone pairs for donation, thus lower basicity.
- Therefore, **sp hybridized nitrogen** will be the least basic, followed by **sp²**, and then **sp³** will be the most basic.

4. **Evaluate the Electron-Donating Effects:**
- Alkyl groups (like ethyl in compound d) have a +I (inductive) effect, which increases electron density on nitrogen, enhancing basicity.
- Compound (d) has two ethyl groups, which significantly increases the electron density on nitrogen.

5. **Rank the Compounds Based on Basicity:**
- **Compound (d)** `C₂H₅-NH-C₂H₅` (Diethylamine) - Most basic due to sp³ hybridization and +I effect from two ethyl groups.
- **Compound (a)** `CH₃-CH₂-NH₂` (Ethylamine) - Next basic due to sp³ hybridization and one ethyl group.
- **Compound (b)** `CH₃-CH=NH` (Vinylamine) - Less basic due to sp² hybridization.
- **Compound (c)** `CH₃-C≡N` (Methyl cyanide) - Least basic due to sp hybridization.

### Final Basicity Order:
**(d) > (a) > (b) > (c)**