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!



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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.

R    CH3OH(g)+      HCl(g)⇌       CH3Cl(g)+  H2O(g)       

Initial  0.250 atm   0.600 atm             0   0
Change     -x   -x                  +x   +x
Equilibrium 0.250−x   0.600−x                  x   x


   Kp=P_CH3OHP_HCl /P_CH3ClP_H2O

x2 /(0.250−x) (0.600−x) =4.7×103
        x=0.249962 atm
P(HCl,eq)=0.600-x=0.600-0.249962=0.350"atm"

You can do this question without RICE table also.
Since K_pis very large, reaction goes almost to completion (methanol is limiting), so x≈0.250.
Here temperature and volume are same for both gases so pressures are directly proportional to moles.
HCl is in excess left unused.
Finally, methanol is zero, and convert pressure of methanol to HCl pressure used taking them as mole ratio.
0.250 atm of methanol reacts with 0.250 atm of HCl .Therefore,
final partial pressure of HCl left: 0.600- 0.250 = 0.350 atm

Hi Mam (This is Mahee)!

Attached is the question I made mistakes on. (The second screenshot is the question)  

NaF is completely soluble in water, so it dissociates to give Na⁺(aq) and F⁻(aq). The fluoride ion, F⁻, is the conjugate base of the weak acid HF, so it behaves as a Brønsted–Lowry base (proton acceptor) in water. Water, in this reaction, acts as the Brønsted–Lowry acid (proton donor).

Hydrolysis equation (definition-based):

F-(aq) + H2O(l) <--> HF(aq) + OH-(aq)

F⁻ accepts H⁺ → forms HF

H₂O donates H⁺ → forms OH⁻

Because HF is weak, the equilibrium lies mostly to the left, but enough OH⁻ forms to make the solution basic.

Hi Mam!

I am confused about how to do this problem.