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 signals corresponds to the number of distinct types of hydrogen atoms, and in $^{13}$C-NMR, it corresponds to the number of distinct carbon atoms.

Step 2: Analysis of Options.
- (1) 1,2,3,5-Tetramethylbenzene: This compound has four different methyl groups, leading to multiple signals.
- (2) 1,4-Diethylbenzene: This compound has two different ethyl groups, resulting in multiple $^1$H-NMR and $^{13}$C-NMR signals.
- (3) 1,2,4,5-Tetramethylbenzene: This compound has two types of hydrogen atoms and three types of carbon atoms, leading to exactly two $^1$H-NMR and three $^{13}$C-NMR signals. Hence, this is the correct option.
- (4) 1,2-Diethylbenzene: This compound results in more than two $^1$H-NMR signals due to the different positioning of the ethyl groups.

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

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