Which of the following shows the correct order of decreasing basicity in gas phase?

Options

A. `(CH_(3))_(3)N gt (CH_(3))_(2)NH gt CH_(3)NH_(2) gt NH_(3)`
B. `(CH_(3))_(2)NH gt (CH_(3))_(3)N gt CH_(3)NH_(2) gt NH_(3)`
C. `(CH_(3))_(2)NH gt CH_(3)NH_(2) gt (CH_(3))_(3)N gt NH_(3)`
D. `(CH_(3))_(2)NH gt CH_(3)NH_(2) gt NH_(3) gt (CH_(3))_(3)N`

Answer

To determine the correct order of decreasing basicity in the gas phase for the given compounds, we will analyze the effect of the substituents on the nitrogen atom in each compound. Basicity is defined by the ability of a compound to donate an electron pair, and this ability can be influenced by the presence of electron-donating or electron-withdrawing groups.

### Step-by-Step Solution:

1. **Identify the Compounds**:
The compounds given are:
- A: \( \text{(CH}_3\text{)}_3\text{N} \) (Trimethylamine)
- B: \( \text{(CH}_3\text{)}_2\text{NH} \) (Dimethylamine)
- C: \( \text{CH}_3\text{NH}_2 \) (Methylamine)
- D: \( \text{NH}_3 \) (Ammonia)

2. **Evaluate Electron-Donating Effects**:
- **Trimethylamine (A)**: Has three methyl groups which are electron-donating through the +I (inductive) effect. This increases electron density on nitrogen, enhancing basicity.
- **Dimethylamine (B)**: Has two methyl groups, which also donate electrons, but less than trimethylamine.
- **Methylamine (C)**: Has one methyl group, providing some electron donation.
- **Ammonia (D)**: Has no alkyl groups, so it has no +I effect.

3. **Consider Steric Hindrance**:
- While trimethylamine has the highest electron density due to three methyl groups, the steric hindrance from these groups can impede the approach of protons, reducing its basicity compared to dimethylamine.
- Dimethylamine, with less steric hindrance, can more effectively donate its lone pair.

4. **Rank the Basicity**:
- **Most Basic**: Dimethylamine (B) - due to optimal balance of electron donation and minimal steric hindrance.
- **Next**: Trimethylamine (A) - although it has the highest electron density, steric hindrance reduces its basicity.
- **Next**: Methylamine (C) - has some basicity due to one methyl group.
- **Least Basic**: Ammonia (D) - has no alkyl groups to enhance basicity.

5. **Final Order**:
The order of decreasing basicity in the gas phase is:
- B (Dimethylamine) > A (Trimethylamine) > C (Methylamine) > D (Ammonia)

### Conclusion:
The correct order of decreasing basicity in the gas phase is B > A > C > D.