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Organic Chemistry · Amines and Heterocycles

Spectroscopy of Amines

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6 key termsSourcesReport an error
IR frequencies (N–H 3300–3500 cm⁻¹, N–H bend 1580–1650 cm⁻¹, C–N 1020–1250 cm⁻¹), NMR shifts (N–CH₃ ≈ 30 ppm ¹³C, α-protons 2.2–3.1 ppm ¹H), and exact atomic masses (¹H 1.00783, ¹²C 12.000, ¹⁴N 14.00307) are standard literature values. General laboratory principles only; no specific experimental procedures are given.
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On this page 9 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Check yourself
  8. Study tools
  9. Sources & references

In 30 seconds

Spectroscopy is how chemists learn a molecule's structure without breaking it apart. For amines, four techniques work together, and each answers a different question:

  • Infrared (IR) spectroscopy: "is there an N–H bond, and how many?" The number of N–H stretching bands reveals whether the amine is primary, secondary, or tertiary.
  • ¹H NMR spectroscopy: "where are the hydrogens?" The N–H is a broad, exchangeable signal; α-protons appear downfield of ordinary alkanes.
  • ¹³C NMR spectroscopy: "what is the carbon skeleton?" Carbons attached to nitrogen appear at 30–60 ppm.
  • Mass spectrometry (MS): "what is the molecular mass and formula?" The flags nitrogen, and gives a diagnostic fragment ion.

The classic exam task — "given a formula and spectra, deduce the structure" — is these four questions asked in order.

Why this matters

Most pharmaceutical drugs are amines; spectroscopy confirms that the right amine was made, detects impurities, and identifies unknowns in forensics and metabolomics. In industry, a new drug candidate's structure is verified with exactly these techniques; in research, unknown natural products are solved the same way. IR alone can tell a chemist whether a reaction "worked" (the N–H bands appear or disappear), and the odd-nitrogen rule instantly tests a formula for nitrogen. These are everyday, transferable skills — and a reliable source of exam problems.

The college version

Core Concepts

Infrared: count the N–H bands

The N–H stretch appears around 3300–3500 cm⁻¹, and its multiplicity reports the amine class:

  • Primary amine (RNH2): two bands near 3350 and 3280 cm⁻¹ (asymmetric and symmetric stretches).
  • Secondary amine (R2NH): one band near 3300 cm⁻¹.
  • Tertiary amine (R3N): no N–H stretch at all.

Supporting signals: the N–H bend near 1580–1650 cm⁻¹ and the C–N stretch near 1020–1250 cm⁻¹. Do not confuse N–H with O–H: O–H is very broad and intense (hydrogen bonding); N–H is narrower, and only amines show the two-vs-one band pattern.

¹H NMR: the exchangeable N–H and the α-protons

Three features identify an amine:

  • The N–H proton: a broad singlet whose chemical shift wanders from ~0.5 to 5 ppm depending on concentration, solvent, and temperature. It is exchangeable: shaking the sample with D₂O replaces N–H with N–D, and the signal disappears — a positive test for an N–H (or O–H) proton.
  • Protons on α-carbons (adjacent to nitrogen) are deshielded: N–CH3 ≈ 2.2–2.3 ppm, N–CH2 ≈ 2.4–2.6, N–CH ≈ 2.6–3.1. These are the most reliable landmarks.
  • Coupling to N–H is often absent: fast exchange means N–H usually neither splits nor is split — an ethyl group on nitrogen still looks like a clean triplet + quartet.

¹³C NMR: carbons next to nitrogen

Nitrogen deshields attached carbons by ~20–40 ppm relative to alkanes: N–CH3 ≈ 30 ppm, N–CH2 ≈ 44, N–CH ≈ 50. A carbon at 40–60 ppm with no attached oxygen strongly suggests C–N.

Mass spectrometry: the odd-nitrogen rule and α-cleavage

Two rules do most of the work:

  • Odd-nitrogen rule: a molecule with an odd number of nitrogen atoms has an odd nominal (integer) molecular mass. Nitrogen's most abundant isotope has nominal mass 14 but forms three bonds, so each nitrogen flips the mass's parity.
  • α-Cleavage: the bond between the α-carbon and nitrogen breaks, giving a resonance-stabilized . For a primary amine the fragment is CH2=NH2+, m/z 30, often the base peak; each carbon added to the chain shifts the fragment by 14 (e.g., CH3CH=NH2+ at m/z 44).

Together with the molecular ion, these fragments let you reconstruct the alkyl groups on the nitrogen.

How It Works / Step-by-Step Process

To solve an unknown amine structure:

  1. Get the formula. Compute DOU: (2C + 2 + N - H - X)/2. Zero means a saturated, acyclic amine.
  2. Apply the odd-nitrogen rule to confirm the nitrogen count; record the molecular ion mass from MS.
  3. Read the IR: two N–H bands = primary, one = secondary, none = tertiary.
  4. Read the ¹H NMR: find α-protons (2.2–3.1 ppm), count them by integration, and check whether an N–H signal disappears on D₂O shake.
  5. Check MS fragments: m/z 30 (or 44, 58…) sizes the alkyl groups lost from nitrogen.

Common Confusions

Do Not ConfuseWithDifference
N–H stretchO–H stretchBoth near 3300 cm⁻¹, but O–H is very broad and intense; N–H is narrower, and primary amines show two bands
"The N–H proton always splits neighbors"Ordinary couplingFast exchange usually kills N–H coupling — expect broad singlets and clean α-protons
"Tertiary amines show an N–H band"1°/2° amine spectraTertiary amines have no N–H bond and no N–H stretch
¹³C shift of N–C vs O–CBoth deshielded carbonsO–CH₃ ≈ 50 and O–CH₂ ≈ 60 ppm sit further downfield than N–CH₃ (≈ 30) and N–CH₂ (≈ 44)
m/z 30 vs m/z 44 fragmentsAny primary aminem/z 30 (CH2=NH2+) comes from any primary amine; m/z 44 implies an ethyl substituent
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Think of spectroscopy as giving a molecule a fingerprint test. IR counts how many "arms" (N–H bonds) the nitrogen is holding — two arms means primary, one means secondary, none means tertiary. NMR reads the molecule's "name tag," showing where its hydrogen and carbon atoms sit. Mass spectrometry weighs the molecule and snaps it in half to show what pieces it was built from — an odd weight means a nitrogen is hiding inside.

Worked example

Example 1: Degrees of unsaturation and the odd-nitrogen rule for C₃H₉N

Compute DOU:

DOU = 2C + 2 + N - H - X2 = 2(3) + 2 + 1 - 9 - 02 = 02 = 0

The molecule is saturated and acyclic — it must be a propylamine isomer. Nominal mass:

M = 3(12) + 9(1) + 1(14) = 59

Odd, as the rule requires for one nitrogen. IR shows two N–H bands (primary); ¹H NMR has a triplet at 0.9 ppm (3H), a sextet at 1.5 ppm (2H), a triplet at 2.6 ppm (2H, N–CH₂), and a broad 2H signal that vanishes with D₂O (NH₂). Structure: propan-1-amine; α-cleavage gives CH2=NH2+ at m/z 30.

Example 2: IR and MS distinguish three isomers of C₄H₁₁N

The isomers butan-1-amine, diethylamine, and N,N-dimethylethanamine share the formula C4H11N. IR decides: primary shows two N–H stretches, secondary one, tertiary none. MS agrees: α-cleavage gives m/z 30 for the primary amine, m/z 44 (loss of ethyl) for the secondary, and m/z 58 (loss of methyl) for the tertiary.

Example 3: Exact mass of a drug-like amine

Ethylamine, C2H7N, with exact atomic masses C = 12.000, H = 1.00783, N = 14.00307:

M = 2(12.000) + 7(1.00783) + 1(14.00307) = 24.000 + 7.0548 + 14.0031 = 45.058

Nominal mass is 45 (odd). Exact mass distinguishes C2H7N (45.058) from CHO2 (45.000) — same nominal mass, different formula, resolved only by exact measurement.

Key takeaways

  • IR: primary = two N–H bands (~3350, ~3280 cm⁻¹); secondary = one (~3300); tertiary = none.
  • ¹H NMR: N–H is broad, 0.5–5 ppm, disappears on D₂O shake; α-protons at 2.2–3.1 ppm are the landmarks.
  • ¹³C NMR: N–CH₃ ≈ 30, N–CH₂ ≈ 44, N–CH ≈ 50 ppm.
  • MS: odd nominal molecular mass ⇔ odd number of nitrogens; α-cleavage gives CH2=NH2+ at m/z 30 (primary amines).
  • Degrees of unsaturation: DOU = (2C + 2 + N - H - X)/2 — nitrogen adds one to the numerator.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

  1. How many N–H IR bands do primary, secondary, and tertiary amines each show?

    Show answer

    Primary: two bands (~3350 and ~3280 cm⁻¹); secondary: one (~3300 cm⁻¹); tertiary: none.

  2. What happens to the N–H proton signal when a sample is shaken with D₂O, and why?

    Show answer

    The N–H proton exchanges with deuterium from D₂O, becoming N–D; the ¹H signal disappears, confirming an exchangeable N–H (or O–H) proton.

  3. Apply the odd-nitrogen rule to C₅H₁₁N: is the nominal mass odd or even, and what does that imply?

    Show answer

    Nominal mass = 5(12) + 11(1) + 14 = 85, odd — consistent with exactly one nitrogen.

  4. What fragment ion appears at m/z 30 for a primary amine, and how is it formed?

    Show answer

    The iminium ion CH2=NH2+ (m/z 30), formed by α-cleavage: the C–N bond breaks and the charge is stabilized by resonance on nitrogen.

  5. Compute the DOU for C₄H₁₁N and interpret the result.

    Show answer

    DOU = [2(4) + 2 + 1 − 11]/2 = 0: saturated and acyclic — an open-chain amine.

Keep learning

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

IR stretch
A bond vibration absorbing infrared light at a characteristic frequency
α-proton
A hydrogen on the carbon directly attached to nitrogen
odd-nitrogen rule
Odd number of N ⇒ odd nominal molecular mass
α-cleavage
Fragmentation of the C–N bond next to the nitrogen
iminium ion
A C=N⁺ fragment such as CH2=NH2+
proton
A hydrogen nucleus, H+, with no electrons.

Sources & references

  1. openstax.org — Organic Chemistry

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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