A compound with the molecular formula C₁₀H₁₄gave the following ¹H NMR spectrum: doublet,$\delta$ 1.2 Singlet,$\delta$ 2.3 Septet,$\delta$ 2.8 Multiplet,$\delta$ 7.1 A possible structure for the compound is:

Question

Quizplus · Accepted Answer

From the NMR spectrum, we can deduce the following:
- There are 4 different types of protons in the molecule (4 peaks in NMR spectrum).
- The doublet at 1.2 ppm indicates a CH2 group adjacent to a CH3 group.
- The singlet at 2.3 ppm indicates a CH3 group.
- The septet at 2.8 ppm indicates a CH group adjacent to 2 CH3 groups.
- The multiplet at 7.1 ppm indicates an aromatic group.

Based on these observations, we can assign the different protons in the molecule to the following groups:
- CH2-CH3 group: 1.2 ppm (doublet)
- CH3 group: 2.3 ppm (singlet)
- CH-2CH3 group: 2.8 ppm (septet)
- Aromatic group: 7.1 ppm (multiplet)

Now, let's look at the possible structures:
- Structure I: Does not have an aromatic group.
- Structure II: Has an aromatic group, and all the proton assignments match the NMR spectrum.
- Structure III: Does not have an aromatic group.
- Structure IV: Has an aromatic group, but the proton assignments do not match the NMR spectrum. For example, the CH3 group is not a singlet in the spectrum.
- Structure V: Does not have an aromatic group.

Therefore, the best choice is Structure II.

A Compound with the Molecular Formula C10H14 gave the Following δ\deltaδ

A Compound with the Molecular Formula C₁₀H₁₄gave the Following $\delta$