MCAT® ExamChemical and Physical Foundations of Biological SystemsHard

A student is analyzing a ¹H NMR spectrum of an unknown organic compound. They observe a triplet at approximately 1.2 ppm and a quartet at approximately 3.5 ppm. These signals are characteristic of which of the following proton environments?

  1. AA benzene ring proton.
  2. BA methyl group adjacent to a carbonyl group.
  3. CA vinyl proton (C=CH₂).
  4. DAn isolated ethyl group (–CH₂CH₃).
Show answer & explanation

Correct answer: D. An isolated ethyl group (–CH₂CH₃).

The n+1 rule in NMR dictates that a signal will be split into n+1 peaks, where n is the number of equivalent protons on adjacent carbons. An ethyl group (–CH₂CH₃) consists of a methyl (CH₃) and a methylene (CH₂) group. The CH₃ protons are adjacent to two CH₂ protons, so they will appear as a triplet (2+1 = 3 peaks). The CH₂ protons are adjacent to three CH₃ protons, so they will appear as a quartet (3+1 = 4 peaks). The chemical shifts (1.2 ppm for CH₃, 3.5 ppm for CH₂ if attached to an electronegative atom like oxygen in an ether or alcohol) are also consistent with an ethyl group.

Why the other options are wrong

  • A. Benzene ring protons typically appear as complex multiplets in the 6.5-8.5 ppm range.
  • B. A methyl group adjacent to a carbonyl (CH₃-CO-) would typically be a singlet (no adjacent protons) or potentially a doublet if adjacent to a CH, and would have a chemical shift around 2.0-2.5 ppm.
  • C. Vinyl protons typically show complex splitting patterns and chemical shifts in the 4.5-6.0 ppm range.

¹H NMR Splitting (n+1 Rule)

In ¹H NMR spectroscopy, the 'n+1 rule' states that a proton signal will be split into n+1 peaks by n equivalent protons on adjacent carbon atoms.

  • Adjacent protons cause splitting (spin-spin coupling).
  • n = number of equivalent protons on adjacent carbons.
  • Common patterns: singlet (n=0), doublet (n=1), triplet (n=2), quartet (n=3).

Memory trick: Peaks Predict Protons' Partners.

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