Choose the correct order for the energy barrier to rotation around the B-N bond:
[`R=CH_(3)` is all above cases]

Options

A. `H_(2)B-NR_(2)ltBH(NR_(2))_(2)ltB(NR_(2))_(3)`
B. `B(NR_(2))_(3)ltBH(NR_(2))_(2)ltBH_(2)-NR_(2)`
C. `BH(NR_(2))_(2)gtB(NR_(2))_(3)gtBh_(2)-NR_(2)`
D. `BH_(2)-NR_(2)gtB(NR_(2))_(3)gtBH-(NR_(2))_(2)`

Answer

To determine the correct order for the energy barrier to rotation around the B-N bond in the given compounds, we need to analyze the steric hindrance caused by the substituents attached to the boron atom. The compounds can be represented as follows:

1. **Compound 1**: H2B-N(CH3)2 (two methyl groups attached to nitrogen)
2. **Compound 2**: H-B(N(CH3)2)2 (two bulky groups on nitrogen)
3. **Compound 3**: B(N(CH3)2)3 (three bulky groups on nitrogen)

### Step-by-Step Solution:

1. **Identify the Compounds**:
- The first compound has two methyl groups attached to nitrogen and two hydrogen atoms attached to boron (H2B-N(CH3)2).
- The second compound has two bulky N(CH3)2 groups attached to boron (H-B(N(CH3)2)2).
- The third compound has three bulky N(CH3)2 groups attached to boron (B(N(CH3)2)3).

2. **Assess Steric Hindrance**:
- The energy barrier to rotation around the B-N bond is influenced by steric hindrance. The more bulky groups attached to the boron, the higher the steric hindrance and, consequently, the higher the energy barrier to rotation.
- Compound 3 (B(N(CH3)2)3) has three bulky groups, leading to the highest steric hindrance and the highest energy barrier.
- Compound 2 (H-B(N(CH3)2)2) has two bulky groups, resulting in a lower energy barrier than Compound 3 but higher than Compound 1.
- Compound 1 (H2B-N(CH3)2) has only two bulky groups and two hydrogens, leading to the least steric hindrance and the lowest energy barrier.

3. **Determine the Order**:
- Based on the analysis of steric hindrance:
- Highest energy barrier: Compound 3 (B(N(CH3)2)3)
- Intermediate energy barrier: Compound 2 (H-B(N(CH3)2)2)
- Lowest energy barrier: Compound 1 (H2B-N(CH3)2)

### Final Order:
The correct order for the energy barrier to rotation around the B-N bond is:
**B(N(CH3)2)3 > H-B(N(CH3)2)2 > H2B-N(CH3)2**