Hello! I have a few questions related to NMR that I would love clarification on, thank you so much!
 

Quote from: amimi on February 17, 2026, 12:45:51 AMHello! I have a few questions related to NMR that I would love clarification on, thank you so much!

Signals confirms para disubstituted aromatic ring with two methyl groups on it.Only molecule 2 has two methyl groups and it is also para substituted.
The spectrum shows only 4 sets of protons:

2H + 2H in aromatic region (~6.8–7.5 ppm)
→ this is the classic pattern for a para-disubstituted benzene ring (two equivalent pairs of aromatic H's).

3H + 3H around ~2–2.6 ppm
→ two different methyl groups, both somewhat deshielded:

one Ar–CH₃ (tolyl methyl, ~2.2–2.4 ppm)

one CO–CH₃ (methyl next to carbonyl, ~2.4–2.7 ppm)

This matches p-methylacetophenone (option 2).

Quote from: amimi on February 17, 2026, 12:45:51 AMHello! I have a few questions related to NMR that I would love clarification on, thank you so much!
 

Para symmetrical disubtituted aromatic ring with two methyls on it.
The spectrum shows only 2 signals:

~7 ppm, integration = 4H → aromatic protons

~2.2–2.4 ppm, integration = 6H → two equivalent methyl groups attached to a benzene ring (Ar–CH₃)

This means:

The molecule has high symmetry

All 4 aromatic H are equivalent as two overlapping equivalent sets (appearing together here)

Both CH₃ groups are equivalent → total 6H

That is exactly para-xylene (1,4-dimethylbenzene).(option 3)