Arrange the basic strength of the following compounds.
`(i) CH_(2) = CHCH_(2)NH_(2)`
`(ii) CH_(3)CH_(2)CH_(2)NH_(2)`
`(iii) CH -= C - CH_(2)NH_(2)`

Answer

To determine the basic strength of the given compounds, we need to analyze the availability of the lone pair of electrons on the nitrogen atom in each amine. The basic strength of amines is influenced by the hybridization of the carbon atoms adjacent to the nitrogen atom, as well as the overall electron density around the nitrogen.

### Step-by-Step Solution:

1. **Identify the Compounds**:
- (i) CH₂=CHCH₂NH₂ (Allyl amine)
- (ii) CH₃CH₂CH₂NH₂ (Propyl amine)
- (iii) CH≡C-CH₂NH₂ (Propyne amine)

2. **Analyze the Hybridization**:
- In compound (i), the nitrogen is bonded to a carbon that is sp² hybridized (due to the double bond).
- In compound (ii), the nitrogen is bonded to a carbon that is sp³ hybridized.
- In compound (iii), the nitrogen is bonded to a carbon that is sp hybridized (due to the triple bond).

3. **Consider the Electronegativity and Electron Density**:
- The sp hybridized carbon (in compound iii) is more electronegative than sp² and sp³ hybridized carbons. This means it will pull electron density away from the nitrogen, making the lone pair less available for protonation.
- The sp² hybridized carbon (in compound i) will also attract electron density, but not as strongly as the sp carbon.
- The sp³ hybridized carbon (in compound ii) does not withdraw electron density as effectively, allowing the lone pair on nitrogen to be more available for protonation.

4. **Rank the Basic Strength**:
- Compound (ii) will have the highest basic strength because the lone pair on nitrogen is most available for protonation.
- Compound (i) will have intermediate basic strength due to the presence of the sp² carbon.
- Compound (iii) will have the lowest basic strength due to the sp carbon withdrawing electron density from nitrogen.

### Final Order of Basic Strength:
**(ii) > (i) > (iii)**