Question

The boron adduct(s), which show(s) three signals in \( ^1 \text{H} \) NMR spectrum with the intensity ratio 1:2:3 is(are)

• A. \( (\text{CH}_3)_3\text{B}:\text{N}(\text{CH}_3)_3 \)
• B. \( (\text{CH}_3\text{CH}_2)_3\text{B}:\text{N}(\text{CH}_2\text{CH}_3)_3 \)
• C. \( \text{H}_3\text{B}:\text{N}(\text{CH}_2\text{CH}_3)_3 \)
• D. \( (\text{CH}_3)_3\text{B}:\text{NH}_3 \)

Hint:

In \( ^1 \text{H} \) NMR, the intensity of signals reflects the number of protons in each distinct environment. Look for the intensity ratio to deduce the correct environment assignment.

Answer 1

The Correct Option is C, D

Step 1: Understanding the NMR spectrum.
The \( ^1 \text{H} \) NMR spectrum shows the relative intensity of signals based on the number of protons in different environments. The ratio of 1:2:3 indicates a set of three different proton environments, with the number of protons in each environment being in the ratio 1:2:3.

Step 2: Analyzing the options.
- Option (A) is correct as it shows three distinct environments: the three methyl groups of \( (\text{CH}_3)_3\text{B} \), the methyl group attached to nitrogen, and the nitrogen-bound protons. This leads to the 1:2:3 intensity ratio.
- Other options (B), (C), and (D) do not show the correct proton environments for the given intensity ratio.

Step 3: Conclusion.
The correct boron adduct is (A).