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.

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



[/quote]
Ha and Hb are vinylic hydrogens across a C=C double bond, and from the drawing they are trans to each other.
trans (H–C=C–H) → 11–18 Hz

cis → 5–10 Hz

geminal (same carbon) → 0–3 Hz

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!

In aromatic rings:

ortho coupling (adjacent H's) is the largest → typically ~6–9 Hz

meta coupling is smaller → typically ~1–3 Hz

para coupling is very small (often ~0–1 Hz)
From the structure, Hc and Hb are adjacent (ortho), so J(Hc–Hb) is larger than J(Ha–Hc) (meta).

17 -Downfield menas more deshielded or less electron cloud on the proton or it is close to some more electronegative atom.
Here H on sp2 carbon is most deshielded as in aromatic ring.Check the chemical shiift values-
D is a vinylic proton (attached to a C=C), which typically appears around 5–6.5 ppm.

The others are more upfield:

A = ester OCH₃ (~3.7 ppm)

B = CH₂ alpha to carbonyl (~2.1–2.5 ppm)

C = allylic/alkyl proton (~1.5–2.5 ppm range depending exact position)

Quote from: amimi on February 17, 2026, 12:47:33 AMHello! I have a few questions related to NMR that I would love clarification on, thank you so much!

Most deshielded will have highest chemical shift value and will appear more downfield.Least deshielded which is E (sp3C-H) are more uplfieldwith lowest chemical shift value.
So the order is
E<D<A<C<B
E = simple alkyl CH₃ (most upfield, ~0.9 ppm)

D = CH₂ next to C=O (deshielded, ~2.1–2.5 ppm)

A = OCH₃ (attached to O, ~3.2–3.8 ppm)

C = vinylic proton in conjugated system (enone), ~6–7 ppm

B = aromatic protons (usually ~7–8 ppm, furthest downfield here)

 triplet

The arrow points to a CH₂ carbon (two hydrogens on that carbon).

Its adjacent carbons are:

one CH on the left → 1 neighboring H

one CH on the right → 1 neighboring H

So total neighboring hydrogens = 2

Using the n + 1 rule: 𝑛+1=2+1=3

So the signal is a triplet.
Splitting is done by the magentic field of neighbouring(adjacent) carbons hydrogens.