Nuclear magnetic resonance (1H NMR) – In routine proton NMR spectra, how does scalar spin–spin coupling affect a given proton signal’s appearance (use the standard “n + 1” rule for nonequivalent vicinal neighbors)?

Difficulty: Easy

Correct Answer: Multiple peaks equal to the number of hydrogens on surrounding atoms, plus one

Explanation:


Introduction / Context:
Proton nuclear magnetic resonance (1H NMR) is a core analytical tool in organic chemistry. One of its most informative features is scalar spin–spin coupling, also called J-coupling, which splits a proton signal into a predictable multiplet. Understanding the “n + 1” rule allows quick inference of how many neighboring, nonequivalent protons are adjacent to a given proton, aiding structure elucidation and verification of synthetic products.


Given Data / Assumptions:

  • Consider first-order spectra where coupling is between nonequivalent vicinal (three-bond) protons.
  • Equivalent protons do not couple with each other.
  • Long-range couplings can occur but are typically weaker and often ignored in basic predictions.


Concept / Approach:
Scalar coupling splits a resonance into a multiplet with a number of lines defined by n + 1, where n is the count of equivalent neighboring hydrogens on adjacent carbons. For example, a methylene next to a methyl group (CH2–CH3) usually appears as a quartet (3 + 1) while the methyl appears as a triplet (2 + 1). The spacing between the lines is the coupling constant J, measured in hertz, and is characteristic of the geometric and electronic environment (e.g., larger for trans olefinic couplings than for cis or geminal in many cases).


Step-by-Step Solution:

Identify the proton of interest (Ha).Count nonequivalent neighboring hydrogens (n) on directly adjacent carbons.Predict multiplicity as n + 1 (singlet for n = 0, doublet for n = 1, triplet for n = 2, etc.).Recognize that integration gives relative proton count while splitting reveals neighborhood.


Verification / Alternative check:
Compare predicted patterns with experimental spectra of standard compounds (e.g., ethyl acetate), where the ethyl group provides classic triplet–quartet patterns. Matching calculated multiplets and J spacings validates the interpretation.


Why Other Options Are Wrong:

  • Two peaks: only true for n = 1; not general.
  • Equal to number of neighboring hydrogens (without +1): misses the rule by one.
  • Equal to number of neighboring carbons: coupling is through hydrogens, not the carbon count.
  • A single peak of reduced intensity: ignores J-coupling altogether.


Common Pitfalls:
Forgetting that chemically equivalent neighbors count as a single “set,” misreading second-order effects (overlap or complex patterns), and ignoring exchangeable protons (e.g., OH) that may show broadened singlets with suppressed coupling.


Final Answer:
Multiple peaks equal to the number of hydrogens on surrounding atoms, plus one

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