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I have no clue what the substituent may be.

The molar mass of the unsubstituted benzaldehyde is 105 g/mol (subtracting one H, where the substituent will be). So the substituent is around 30–33 g/mol. Also I'm having trouble identifying the molecular ion. Is it the really small 138 peak or the 137 peak?


The molecular ion $\ce{[M]^{\bullet +}}$ peak is likely at 136 m/z. The peak at 137 m/z is due to the 1.1% natural abundance of $\ce{^{13}C}$.

The peak at 135 m/z is the $\ce{[M-1]^{+}}$ peak typical of aldehydes:

$$\ce{ [R-C(=O)-H]^{\bullet +} -> [R-C+=O <-> R-C\equiv O+] +H ^{\bullet} }$$

Another typical fragmentation of aldehydes is the loss of the CHO for $\ce{[M-29]^{+}}$, which you see at 107 m/z.

$$\ce{ [R-C(=O)-H]^{\bullet +} -> R+ + [HC=O] ^{\bullet} }$$

Since the main contributor to the $\ce{[M+1]^{\bullet +}}$ peak is $^{13}\ce{C}$ we can estimate the number of carbon atoms ($C$) in the molecule from the ratio of intensities of the $\ce{[M]^{\bullet +}}$ to the $\ce{[M +1]^{\bullet +}}$ peak:

$$\frac{I_{M+1}}{I_{M}}=\frac{1.1}{98.9}C$$ $$C=\frac{98.9I_{M+1}}{1.1I_{M}}=89.9\frac{I_{M+1}}{I_{M}}$$

The intensities appear to be $I_M \approx75$ and $I_{M+1} \approx 7$. $$\therefore C=89.9\frac{7}{75}=8.4$$

Since the number of carbon atoms needs to be an integer, round down to $C=8$. We can now work backward to get a formula. Eight carbon atoms have a mass of $8*12=96$ g/mol, leaving 40 g/mol to be the rest of the formula. We know one is an oxygen (aldehyde) (16 g/mol), leaving 24 g/mol. The most number of hydrogen atoms that can be associated with eight carbon atoms is 18 (2n+2), so there has to be some other atom, maybe an oxygen or nitrogen?

I will leave rest to you:

  1. Maybe there is a second oxygen?
  2. How many hydrogens do you have?
  3. What is the final formula?
  4. How is it different from benzaldehyde?
  5. How could those extra atoms be placed on the ring? (there are many possibilities)
  6. Do you have any other evidence that suggests which of the possibilties from #5 is correct (IR, NMR, etc.)?
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  • $\begingroup$ Also 107-92 and 92-77 provide some big hints. $\endgroup$ – Lighthart Apr 8 '13 at 19:31

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