Here is the mechanism of cyclic acetal format
ion

Hello, I am a bit confused with this, and would appreciate some help! I am trying to write the whole mechanism

Hello, I would appreciate some help with this question, thank you!

Hello, I would appreciate some help with this question, thank you!

Hello, I would appreciate some help with this question, thank you!

Hello, I would appreciate some help with this question, thank you!

Quote from: amimi on February 17, 2026, 12:44:43 AMHello! I have a few questions related to NMR that I would love clarification on, thank you so much!

There are 3 integrated signal groups visible:

one smaller signal near ~2.4 ppm nearly for 2H

one medium signal near ~2.1 ppm nearly for 3 H

one largest signal near ~1.0 ppm nearly for 9H
S the ratio is 2:3:9
With this molecule is A
Option c is type structure (CH₃COCH₂C(CH₃)3), which gives:

6H from the two equivalent tert-butyl groups → big signal near ~1.0 ppm

CH₃ next to C=O (3H) around ~2.1 ppm

CH₂ next to C=O (2H) around ~2.3–2.5 ppm

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)

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, 12:46:38 AMHello! I have a few questions related to NMR that I would love clarification on, thank you so much!

   

Q 13 -The arrow points to a vinylic carbon (terminal alkene CH₂).
Vinylic hydrogens typically appear around 4.6–6.5 ppm, and a terminal alkene =CH₂ is commonly near 4.8–5.0 ppm.
So 4.82 ppm is the best match.
Q 14-The arrow points to the CH₂ attached to –OH (an alcohol carbon, –CH₂OH), which is deshielded by oxygen and usually appears around 3.3–4.0 ppm.
So 3.50 ppm is the correct value.