MCAT® ExamChemical and Physical Foundations of Biological SystemsMedium
An organic chemist is synthesizing a compound and wants to determine its structure using ¹H NMR spectroscopy. The spectrum shows a signal at approximately 9.5 ppm, integrating for one proton, and appearing as a doublet. Which of the following functional groups is most likely present in the compound?
- AMethyl ketone
- BAldehyde
- CAlcohol
- DCarboxylic acid
Show answer & explanationAnswer & explanation
Correct answer: B. Aldehyde
A signal at approximately 9.5 ppm is characteristic of an aldehyde proton (R-CHO). The integration of one proton confirms a single aldehyde proton. The doublet splitting suggests that the aldehyde proton is coupled to one adjacent proton (n+1 rule, where n=1 for a doublet). This is consistent with an aldehyde where the carbon adjacent to the carbonyl carbon has one hydrogen (e.g., R-CH(H)-CHO).
Why the other options are wrong
- A. Methyl ketones (R-CO-CH3) would show a signal for methyl protons around 2.1-2.4 ppm, not 9.5 ppm.
- C. Alcohol protons (R-OH) typically appear in a broad range (1-5 ppm, sometimes higher for hydrogen bonding), but are almost always broad singlets due to rapid exchange, and not typically at 9.5 ppm.
- D. Carboxylic acid protons (R-COOH) typically appear much further downfield, usually between 10-13 ppm, and are often broad singlets (due to exchange).
¹H NMR Chemical Shifts
¹H NMR chemical shifts indicate the electronic environment of protons, helping to identify functional groups.
- High ppm (downfield) = deshielded protons (near electronegative atoms or π systems).
- Low ppm (upfield) = shielded protons (away from electronegative atoms).
- Characteristic ranges: Aldehyde (~9-10 ppm), Carboxylic acid (~10-13 ppm), Alcohol (~1-5 ppm, variable), Alkyl (~0.9-2.0 ppm).
Memory trick: NMR: Downfield means Deshielded, Doublet means one neighbor.