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: Analysis of ¹H NMR signals with intensity ratio 1:2:3:

(A) (CH₃)₃B:N(CH₃)₃:

• B-CH₃: 9H (singlet)
• N-CH₃: 9H (singlet)
• Ratio: 1:1 ✗

(B) (CH₃CH₂)₃B:N(CH₂CH₃)₃:

• B-CH₂: 6H (quartet)
• B-CH₃: 9H (triplet)
• N-CH₂: 6H (quartet)
• N-CH₃: 9H (triplet)
• Four signals, not three ✗

(C) H₃B:N(CH₂CH₃)₃:

• B-H₃: 3H (quartet due to coupling with ¹¹B, appears as one signal)
• N-CH₂: 6H (quartet)
• N-CH₃: 9H (triplet)
• Three signals with ratio 3:6:9 = 1:2:3 ✓

(D) (CH₃CH₂)₃B:NH₃:

• B-CH₂: 6H (quartet)
• B-CH₃: 9H (triplet)
• N-H₃: 3H (broad singlet)
• Three signals with ratio 6:9:3 = 2:3:1 (or 3:6:9 rearranged)
• Can be written as 1:2:3 depending on assignment ✓

Answer: (C) and (D) are correct.