Determining structure from C-NMR spectrum, is my answer correct?

Question

Determine the structure of the compound whose spectral data is given below and explain your reasoning (5 marks for structure, 15 marks for reasoning). The peak at 7.27 ppm does not belong to the compound. What is the source of this peak? (2.5 marks) Looking at the full spectrum you will see small peaks at about 4.2 and 3.7 ppm. What is the origin of these peaks? (2.5 marks)

Formula: $\ce{C9H10O3}$

IR: small peaks observed between 3010 and 2853. The largest peak is at 1693. All peaks are in $\pu{cm-1}.$

$\ce{^{13}C}$ NMR (DEPT-135 result in brackets): δ 190 (+), 153 (no peak), 149 (no peak), 131 (no peak), 127 (+), 111 (+), 109 (+), 58 (+), 57 (+) ppm.

$\ce{^1H}$ NMR (300 MHz. See expansions on next page):

I mainly looked at the $\ce{^{13}C}$-$\mathrm{NMR}$ aspect of this question. Given that there is a peak at $\pu{190ppm}$, it is probably an aldehyde with a $\ce{CH}$ (which explains the +DEPT sign). The molecular formula also showed a degree of unsaturation of 5, which could be explained by a benzene ring and a double bond (which links up nicely with the carbonyl group). But in order to come up with no peaks in an aromatic ring I added some methyl groups. I eventually came up with the structure 2,3-dimethylbenzaldehyde. The +signs indicated $\ce{CH}$ or $\ce{CH3}$, the 5 $\ce{CH3/CH}$ groups could be explained by the methyl groups and the rest of the carbons in the benzene ring. Do you think this structure is correct?

Answer 1

It is not always a good idea to focus only on $\ce{^{13}C}$-$\mathrm{NMR}$ when you are doing structure elucidation on unknown compound using spectral data. Yes, $\ce{^{13}C}$-$\mathrm{NMR}$ gives you valuable info, so does FTIR to some extent. Yet, $\ce{^{1}H}$-$\mathrm{NMR}$ gives the most info about the compound as I depicted in following diagram:

All your analysis is correct except for the final conclusion on the compound. It is not 2,3-dimethoxybenzaldehyde as you concluded. But it is 3,4-dimethoxybenzaldehyde. How do I know that? Look at the diagram and analyze the coupling pattern (disregard the small but significant meta-coupling, which give close to $\pu{2 ppm}$ here).

If the compound is 2,3-dimethoxybenzaldehyde, the $\ce{^{1}H}$-$\mathrm{NMR}$ spectrum of it should displays a doublet, a triplet (or dd), and a doublet in aromatic region (~$\pu{6.7-8.5 ppm}$), in that order going from upfield to downfield. However, what given $\ce{^{1}H}$-$\mathrm{NMR}$ exhibit is a doublet, a singlet, and a doublet in the aromatic region (disregarding the meta-coupling). The singlet indicates the presence of isolated aromatic ring proton, and two separate doubelets also indicate the presence of two isolated but neighboring aromatic ring protons. Since three substitution groups are one formaldehylde (-$\ce{CHO}$) and two methoxy groups (-$\ce{OCH3}$), there are three possibilities: The compound would be 2,5-dimethoxybenzaldehyde, or 2,4-dimethoxybenzaldehyde, or 3,4-dimethoxybenzaldehyde:

The possibility of been 2,4-dimethoxybenzaldehyde can easily be ruled out, because it would show the upfield shift of the chemical shift of the singlet corresponding to the proton on position 3 of the ring (been ortho to two -$\ce{OCH3}$ groups), even shielded than the doublet at the $\pu{6.99 ppm}$. This is justifiable conclusion since the resonance structures of 2,4-dimethoxybenzaldehyde (using only 2-methoxy group) indicate that the ortho ($\ce{C}$-3) and para ($\ce{C}$-5) positions will be more shielded than meta positions. Also, with respect to 4-methoxy group, the ortho ($\ce{C}$-3 and $\ce{C}$-5) positions will be even more shielded than corresponding ortho positions in 3,4-dimethoxybenzaldehyde.

However, distingushing 3,4-dimethoxybenzaldehyde from 2,5-dimethoxybenzaldehyde is tricky. Both would have a doublet at $\pu{6.99 ppm}$ (for the proton at $\ce{C}$-5 for the former and the proton at $\ce{C}$-3 for the latter) since substitution effects are the same for both. Same argument is valid for the singlet (with meta-coupling) at $\pu{7.43 ppm}$ (for the proton at $\ce{C}$-2 for 3,4-dimethoxybenzaldehyde and the proton at $\ce{C}$-6 for 2,5-dimethoxybenzaldehyde) since, again, the substitution effects are identical for both. Yet, chemical shift for the proton at $\ce{C}$-6 position of 3,4-dimethoxybenzaldehyde and that of the proton at $\ce{C}$-4 position of 2,5-dimethoxybenzaldehyde would not be identical. That is because ortho-effect of a substitution group is somewhat larger than para-effect of the same group in general (there are few examples you may find here), regardless of its sheilding or deshielding effect. Thus, the chemical shift for the proton at $\ce{C}$-4 position of 2,5-dimethoxybenzaldehyde should be more shielded than that for the proton at $\ce{C}$-6 position of 3,4-dimethoxybenzaldehyde. My reasoning:

1. The proton at $\ce{C}$-4 position of 2,5-dimethoxybenzaldehyde should be deshielded by para-$\ce{CHO}$ group (blue shift), while shielded by ortho-$\ce{OCH3}$ group (larger red shift).
2. The proton at $\ce{C}$-6 position of 3,4-dimethoxybenzaldehyde should be deshielded by ortho-$\ce{CHO}$ group (larger blue shift), while shielded by para-$\ce{OCH3}$ group (red shift).
3. The both protons at $\ce{C}$-6 position of 2,5-dimethoxybenzaldehyde and at $\ce{C}$-2 position of 3,4-dimethoxybenzaldehyde are deshielded by ortho-$\ce{CHO}$ group (larger blue shift), while shielded by ortho-$\ce{OCH3}$ group (larger red shift).
4. Comparition of reason 1 & 3 clearly shows these two chemical shifts should be switched (instead of singlet followed by doublet, it should be doublet followed by singlet).

Thus correct resonance order would be given by 3,4-dimethoxybenzaldehyde.