Find the order of basic strength .(If R=Me) ?
`(I) R_(4)N^(+)OH^(-) " "(II) R_(3)N" "(III) R_(2)NH" "(IV)RNH_(2)`

Options

A. `I gt III gt IV gt II`
B. `IV gt III gt I gt II`
C. `II gt IV gt III gt I`
D. `II gt IV gt I gt III`

Answer

To determine the order of basic strength for the given compounds when R = Me (methyl), we need to analyze the basicity of each compound based on the structure and the presence of electron-donating or electron-withdrawing groups.

### Step-by-Step Solution:

1. **Identify the Compounds**:
- (I) \( R_4N^+OH^- \)
- (II) \( R_3N \)
- (III) \( R_2NH \)
- (IV) \( RNH_2 \)

2. **Understand Basic Strength**:
Basic strength is determined by the ability of a compound to donate a lone pair of electrons. The more electron density available on the nitrogen atom, the stronger the base.

3. **Analyze Each Compound**:
- **(I) \( R_4N^+OH^- \)**: This compound has a positively charged nitrogen and a hydroxide ion (OH⁻). The presence of OH⁻, which has a negative charge on an electronegative atom (oxygen), makes this compound very basic because it can easily donate the hydroxide ion.
- **(II) \( R_3N \)**: This is a tertiary amine. The three methyl groups (electron-donating groups) increase the electron density on nitrogen, making it a good base. However, steric hindrance due to the three bulky methyl groups can reduce its basicity compared to less hindered amines.
- **(III) \( R_2NH \)**: This is a secondary amine. It has two methyl groups donating electron density to nitrogen, which increases its basicity compared to the tertiary amine due to less steric hindrance.
- **(IV) \( RNH_2 \)**: This is a primary amine with one methyl group donating electron density. It is less sterically hindered than the tertiary amine and has a good ability to donate electrons, but it has the least electron-donating groups compared to the others.

4. **Rank the Basic Strength**:
- **Most Basic**: (I) \( R_4N^+OH^- \) - due to the presence of OH⁻.
- **Second Most Basic**: (III) \( R_2NH \) - due to two electron-donating methyl groups and less steric hindrance.
- **Third Most Basic**: (IV) \( RNH_2 \) - due to one electron-donating methyl group.
- **Least Basic**: (II) \( R_3N \) - despite having three methyl groups, steric hindrance reduces its basicity.

5. **Final Order of Basic Strength**:
The order of basic strength is:
\[ (I) > (III) > (IV) > (II) \]
This can be represented numerically as:
\[ 1 > 3 > 4 > 2 \]

### Conclusion:
Thus, the order of basic strength for the compounds when R = Me is \( 1, 3, 4, 2 \).