# The H NMR analysis of the hydrogens in salicylic acid

I am currently working with an unknown and from the IR spectrum I was able to figure out that the compound I am working with is salicylic acid.

I am looking at the H NMR, but I am a bit confused on the deshielding, integration, and upsheilding of the hydrogens that are near the $$\ce{OH}$$ and $$\ce{COOH}$$ groups on the compound. Can someone explain this to me:

Attached is a picture

• The resolution of the picture is painfully low. – Karl Apr 30 '20 at 19:10
• Apologizes but this is the only picture I received... – Boo Apr 30 '20 at 22:47
• "Working with an unknown" means you got the IR and NMR spectra as a homework to identify, right? ;) – Karl Apr 30 '20 at 23:21

Since you have identified your unknown as salicylic acid (2-hydroxybenzoic acid), I'd help you understand how easily understand the substituents of aromatic nucleus induce shielding-deshielding effects. The following diagram illustrate the $$\mathrm{^1H}$$-$$\mathrm{NMR}$$ assignments of salicylic aromatic protons:
A lot of experts do not like explaining shielding-deshielding effects using electron densities around the sought atom. I agree with them, but a student of your capacity could easily related to that explanation. A good rule of thumb is o,p-directing electron donating groups (coursing positive mesomeric or resonance effect) usually shield the relevant protons at o,p-positions (inducing $$\delta^-$$-charge at the positions). Similarly, m-directing electron withdrawing groups (coursing negative mesomeric or resonance effect) usually deshield the relevant protons at o,p-positions (inducing $$\delta^+$$-charge at the positions). Accordingly, $$\ce{OH}$$-group shields $$\ce{H_a}$$ and $$\ce{H_c}$$ while $$\ce{COOH}$$-group deshields $$\ce{H_b}$$ and $$\ce{H_d}$$.
Thus, it is safe to say that the most up-field aromatic signals (sheilded) at $$\delta \ 6.97$$ (apparent $$triplet$$) and $$\delta \ 7.03$$ (apparent $$doublet$$) are due to $$\ce{H_a}$$ and $$\ce{H_c}$$. The assignments of $$\delta \ 6.97$$ is belong to $$\ce{H_c}$$ (two vicinal protons) and $$\delta \ 7.03$$ is belong to $$\ce{H_a}$$ (one vicinal proton) are solely due to the number of neighboring (vicinal) protons to each. Keep in mind that chemical shifts of these two signals can be moved to up- or down-field positions depending on the using locking solvent. For example, the locking solvent used to get present $$\mathrm{^1H}$$-$$\mathrm{NMR}$$ spectrum here is apparently $$\ce{CDCl3}$$ (based on solvent residue at $$\delta \ 7.28$$). However, two signals have been moved to more up-field positions (compared to that in $$\ce{CDCl3}$$) when the locking solvent is changed to $$d_6$$-$$\ce{DMSO}$$ (Ref.1). In that spectrum, two signals at $$\delta \ 6.92$$ ($$t,\: ^3\!\!J=\pu{7.4 Hz}$$) and $$\delta \ 6.96$$ ($$d,\: ^3\!\!J=\pu{7.5 Hz}$$) are assigned for $$\ce{H_c}$$ and $$\ce{H_a}$$, respectively.
Similarly, the most down-field aromatic signals (desheilded) at $$\delta \ 7.55$$ (apparent $$triplet \ of \ doublet$$) and $$\delta \ 7.96$$ (apparent $$doublet \ of \ doublet$$) are due to $$\ce{H_b}$$ and $$\ce{H_d}$$ (Note: The splitting of each signal with a small coupling constant is due to meta-or $$\omega$$-coupling generated by the presence of proton at meta to each; This phenomenon is generally called $$^4\!\!J$$-coupling). The assignments of $$\delta \ 7.55$$ is belong to $$\ce{H_b}$$ (two vicinal protons) and $$\delta \ 7.96$$ is belong to $$\ce{H_d}$$ (one vicinal proton) are also due to the number of neighboring (vicinal) protons to each. Similar to $$\ce{H_a}$$ and $$\ce{H_c}$$, these two signals have also been moved to more some what up-field positions (compared to that in $$\ce{CDCl3}$$) when the locking solvent is changed to $$d_6$$-$$\ce{DMSO}$$ (Ref.1). In that spectrum, two signals at $$\delta \ 7.52$$ ($$t,\: ^3\!\!J=\pu{7.7 Hz}$$) and $$\delta \ 7.82$$ ($$d,\: ^3\!\!J=\pu{7.7 Hz}$$) are assigned for $$\ce{H_b}$$ and $$\ce{H_d}$$, respectively.