Question

Which of the following compounds exhibits two $^1$H-NMR signals and three $^{13}$C-NMR signals?

• A. 1,2,3,5-Tetramethylbenzene
• B. 1,4-Diethylbenzene
• C. 1,2,4,5-Tetramethylbenzene
• D. 1,2-Diethylbenzene

Hint:

When analyzing NMR spectra, consider the symmetry of the molecule. Symmetrical molecules tend to produce fewer signals.

Answer 1

The Correct Option is C

Step 1: Understanding NMR Signals.
In $^1$H-NMR, the number of distinct hydrogen atom types determines the number of signals. Similarly, in $^{13}$C-NMR, the number of distinct carbon atom types dictates the signal count.

Step 2: Analysis of Options.
- (1) 1,2,3,5-Tetramethylbenzene: This molecule possesses four unique methyl groups, thus generating multiple signals.
- (2) 1,4-Diethylbenzene: This compound features two distinct ethyl groups, resulting in multiple $^1$H-NMR and $^{13}$C-NMR signals.
- (3) 1,2,4,5-Tetramethylbenzene: This molecule exhibits two types of hydrogen atoms and three types of carbon atoms, leading to precisely two $^1$H-NMR signals and three $^{13}$C-NMR signals. This option is therefore correct.
- (4) 1,2-Diethylbenzene: Due to the varied placement of the ethyl groups, this compound produces more than two $^1$H-NMR signals.

Step 3: Conclusion.
Consequently, option (3), 1,2,4,5-Tetramethylbenzene, is the correct answer.

Final Answer: \[ \boxed{\text{(3) 1,2,4,5-Tetramethylbenzene}} \]