Which one shows the best formation of adduct ?

Options

A. `O(CH_(3))_(2)+BF_(3)`
B. `O(SiH_(3))_(2)+BF_(3)`
C. `O(SiH_(3))_(2)+BBr_(3)`
D. `O(CH_(3))_(2)+BBr_(3)`

Answer

To determine which compound shows the best formation of adducts, we need to analyze the electron-donating and electron-accepting capabilities of the given options. An adduct is formed when a Lewis base (electron donor) donates a lone pair of electrons to a Lewis acid (electron acceptor).

### Step-by-Step Solution:

1. **Identify the Electron Donors and Acceptors:**
- Look at the compounds provided in the options. Identify which species have lone pairs (potential electron donors) and which are electron deficient (potential electron acceptors).

2. **Analyze the Electron Deficiency:**
- For a good adduct formation, the electron acceptor should be highly electron deficient. For example, boron compounds like BF3 and BBr3 are known to be electron deficient because boron has only six electrons in its valence shell.

3. **Evaluate the Electron Donors:**
- Among the options, check which species can donate electrons effectively. Oxygen (O) is a strong donor due to its lone pair. Methyl groups (CH3) can also donate electrons, but they are generally weaker donors compared to oxygen.

4. **Compare the Electron Acceptors:**
- Between BF3 and BBr3, we need to determine which is more electron deficient. BF3 has a smaller atomic radius and a higher electronegativity compared to BBr3, making BF3 a stronger Lewis acid. However, BBr3 has larger bromine atoms which can lead to a more significant electron deficiency due to the larger size and lower electronegativity of bromine compared to fluorine.

5. **Determine the Best Combination:**
- The best adduct formation will occur when a strong electron donor (like O) interacts with a highly electron-deficient acceptor (like BBr3). Therefore, we can conclude that BBr3 will be the better electron acceptor compared to BF3.

6. **Final Conclusion:**
- Based on the analysis, the best formation of adducts will occur with the combination of an effective electron donor (like O) and a highly electron-deficient acceptor (like BBr3). Thus, the answer to the question is **Option D: BBr3**.