Organic Spectroscopy: IR, ¹H NMR, and UV-Vis Spectroscopy
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Topic 10 — Key Facts for Kenyatta University (Kenya) Core concept: Spectroscopy techniques are used to identify the structure of organic molecules; IR spectroscopy identifies functional groups from their absorption of infrared light; ¹H NMR identifies hydrogen environments and their chemical shifts; UV-Vis identifies conjugated π-systems High-yield point: IR peaks: O–H and N–H (broad, 3200–3600 cm⁻¹), C≡N (sharp, 2240–2260 cm⁻¹), C≡C (sharp, 2100–2260 cm⁻¹), C=O (strong, 1700–1750 cm⁻¹), C=C (medium, 1620–1680 cm⁻¹); NMR chemical shifts: aldehyde H (9–10 ppm), benzylic H (2.3–2.8 ppm), alcohol O–H (variable 1–5 ppm, exchangeable with D₂O) ⚡ Exam tip: The splitting pattern in ¹H NMR (n+1 rule) comes from neighbouring non-equivalent hydrogens; be careful to identify which hydrogens are equivalent and which are neighbours
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Introduction to Organic Spectroscopy
Organic chemistry would be largely a matter of guesswork without spectroscopy. Modern spectroscopy techniques allow chemists to determine the exact structure of an organic molecule — the identity, number, and arrangement of atoms — without destroying the sample. The three main spectroscopic techniques for organic structure determination are Infrared (IR) Spectroscopy, Nuclear Magnetic Resonance (NMR) Spectroscopy, and Ultraviolet-Visible (UV-Vis) Spectroscopy.
Infrared (IR) Spectroscopy
Principle: Molecules absorb IR radiation at frequencies that correspond to the energies of their vibrational modes. When a molecule absorbs IR light, it transitions to a higher vibrational energy state. Different bonds (C–H, O–H, C=O, etc.) vibrate at characteristic frequencies.
IR Spectrum: Plotted as % Transmittance (or Absorbance) against wavenumber (cm⁻¹). 100% transmittance means no absorption; low transmittance means strong absorption.
Key IR Absorption Bands:
| Functional Group | Wavenumber (cm⁻¹) | Appearance |
|---|---|---|
| O–H (alcohol) | 3200–3600 | Broad, strong |
| O–H (carboxylic acid) | 2500–3300 | Very broad |
| N–H (amine) | 3300–3500 | Medium, sharp (2 peaks for 1° amine) |
| C≡N | 2240–2260 | Sharp |
| C≡C | 2100–2260 | Sharp |
| C=O (aldehyde/ketone) | 1700–1750 | Strong |
| C=O (ester) | 1735–1750 | Strong |
| C=O (amide) | 1630–1680 | Strong |
| C=O (acid) | 1710–1760 | Strong |
| C=C (alkene) | 1620–1680 | Medium |
| C=C (aromatic) | 1450–1600 | Medium (multiple peaks) |
| C–H (alkane) | 2850–2960 | Medium |
| C–H (alkene/aromatic) | 3000–3100 | Weak |
| C–O (alcohol/ether) | 1050–1150 | Strong |
| C–Cl | 700–800 | Medium |
| C–Br | 500–600 | Medium |
⚡ Exam Tip: A broad O–H peak spanning 2500–3300 cm⁻¹ immediately indicates a carboxylic acid. A broad peak at 3200–3600 cm⁻¹ alone could be an alcohol or water — look for accompanying C–O peak at 1050–1150 cm⁻¹ to confirm an alcohol.
¹H Nuclear Magnetic Resonance (NMR) Spectroscopy
Principle: Protons (¹H nuclei) spin like tiny magnets. In a strong magnetic field, proton spins can be aligned with or against the field. Radiofrequency radiation is applied to flip them from the lower energy state to the higher energy state. The exact frequency absorbed depends on the chemical environment of the proton.
Chemical Shift (δ): Expressed in parts per million (ppm) relative to a standard (tetramethylsilane, TMS). Chemical shift indicates the electronic environment of a proton.
Reference: TMS is assigned a chemical shift of 0 ppm. Most organic protons appear between 0 and 12 ppm.
Characteristic ¹H Chemical Shifts:
| Proton Type | Chemical Shift (ppm) |
|---|---|
| TMS (reference) | 0.0 |
| Alkyl (R–CH₃) | 0.7–1.2 |
| Alkyl (R–CH₂–R) | 1.2–1.5 |
| Alkyl (R₃CH) | 1.5–2.0 |
| Allylic (C=C–C–H) | 1.6–2.8 |
| Benzylic (Ar–C–H) | 2.3–2.8 |
| Acetylenic (≡C–H) | 2.5–3.0 |
| Alcohol O–H | 1–5 (variable) |
| Aromatic (Ar–H) | 6.5–8.0 |
| Aldehyde (–CHO) | 9–10 |
| Carboxylic acid (–COOH) | 10–13 |
Integration and Splitting (n+1 Rule)
Integration: The area under each signal is proportional to the number of protons it represents. An NMR spectrum shows integration curves or numbers above peaks indicating relative proton counts.
Splitting (n+1 rule): A signal is split into n+1 peaks by n neighbouring non-equivalent protons:
| Neighbouring Hydrogens | Splitting Pattern | Peaks |
|---|---|---|
| 0 | Singlet (s) | 1 |
| 1 | Doublet (d) | 2 |
| 2 | Triplet (t) | 3 |
| 3 | Quartet (q) | 4 |
| 4 | Quintet | 5 |
Example: Ethanol (CH₃–CH₂–OH):
- CH₃ is attached to CH₂ → 2 neighbours → triplet
- CH₂ is attached to CH₃ (3 H) and OH → 4 neighbours → quartet
- OH proton has 0 neighbours → singlet
⚡ Exam Tip: Equivalent protons (protons in identical chemical environments) do not split each other. For example, the six protons of propane (CH₃–) are all equivalent and appear as a single singlet.
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Advanced NMR and UV-Vis Spectroscopy
Deuterium Exchange (D₂O)
Exchangeable protons (O–H, N–H, –COOH) can be replaced by deuterium (D) when D₂O is added to the NMR sample. After exchange:
- The signal for the exchangeable proton disappears
- This confirms the presence of O–H or N–H groups
- Useful for distinguishing OH signals from other signals in the 1–5 ppm region
Coupled and Decoupled NMR
Coupled NMR: Shows fine splitting from neighbouring protons — useful for determining connectivity.
Decoupled NMR (¹H broadband decoupled): All protons are simultaneously irradiated, removing all splitting. Each type of carbon appears as a single peak (singlet) showing only the number of equivalent carbons. This is the ¹³C NMR technique.
¹³C NMR
¹³C NMR detects carbon-13 nuclei rather than protons:
- ¹³C chemical shifts span 0–220 ppm
- No integration (too many carbons with similar response)
- Each carbon signal tells you its chemical environment
- No splitting from attached hydrogens (in broadband decoupled mode)
Characteristic ¹³C Chemical Shifts:
| Carbon Type | Chemical Shift (ppm) |
|---|---|
| Alkane C | 0–50 |
| Allylic C | 30–45 |
| Aromatic C | 100–150 |
| Alkyne C | 70–100 |
| Alkene C | 110–150 |
| Carbonyl C (ketone/aldehyde) | 195–220 |
| Carbonyl C (ester/acid) | 170–185 |
| Carbonyl C (amide) | 160–175 |
| Carboxylic acid C | 170–185 |
Spin-Spin Coupling in ¹³C NMR
DEPT (Distortionless Enhancement by Polarisation Transfer):
- DEPT-135: CH₃ and CH (positive), CH₂ (negative), quaternary C (absent)
- DEPT-90: Shows only CH carbons
- Used to determine the number of hydrogens attached to each carbon
UV-Visible Spectroscopy
Principle: Molecules absorb UV or visible light, causing electronic transitions. Most commonly, electrons in π orbitals are promoted from bonding (π) to antibonding (π*) orbitals.
Chromophores: The part of a molecule responsible for UV-Vis absorption is called a chromophore. Isolated double bonds absorb at ~170 nm (vacuum UV, not measured). Conjugated double bonds absorb at longer wavelengths:
| Chromophore | λmax (nm) | ε (extinction coefficient) |
|---|---|---|
| Isolated C=C | ~170 | ~10,000 |
| Conjugated C=C (diene) | ~220 | ~20,000 |
| Conjugated triene | ~260 | ~35,000 |
| α,β-unsaturated carbonyl | ~220–250 (π→π*) | high |
| Benzene | ~254 | ~200 |
| Carvone (α,β-unsat. ketone) | ~300 | moderate |
Woodward-Fieser Rules: Used to calculate λmax for conjugated dienes and enones based on:
- Base value
- Increments for each alkyl substituent or ring residue
- Increment for each double bond extension
- Increment for each exocyclic double bond
- Increment for each auxiliary chromophore
⚡ Exam Tip: Conjugation lowers the energy gap between π and π* orbitals, causing absorption at longer wavelengths. Benzene absorbs at 254 nm (UV). β-Carotene (11 conjugated double bonds) absorbs at ~450 nm (visible, blue light — hence its orange colour).
Structure Determination: Putting It All Together
A typical problem requires combining IR, NMR, and UV-Vis data to identify an unknown compound:
Step 1 — Molecular Formula: If provided (e.g., C₈H₈O), calculate the degree of unsaturation (DBE): DBE = (2C + 2 + N – H – X)/2 For C₈H₈O: DBE = (2×8 + 2 – 8)/2 = 5 → suggests a benzene ring (4 DBE) plus one more double bond
Step 2 — IR: Identify functional groups from characteristic absorption bands.
Step 3 — NMR: Determine:
- Number of signals = number of distinct proton environments
- Chemical shift of each signal → type of proton
- Integration → relative number of each type of proton
- Splitting pattern → number of neighbouring protons
Step 4 — UV-Vis (if applicable): Confirm presence of conjugation
Worked Example: Unknown Compound C₃H₆O
IR:
- Strong peak at 1715 cm⁻¹ → C=O (aldehyde or ketone)
- No broad O–H band → not alcohol or carboxylic acid
- C–H stretch at 2850–2960 cm⁻¹ and 2720 cm⁻¹ (aldehyde C–H stretch)
¹H NMR:
- Signal at 9.7 ppm (1H, triplet) → aldehyde proton (–CHO)
- Signal at 2.3 ppm (2H, quartet) → CH₂ adjacent to CHO
- Signal at 1.1 ppm (3H, triplet) → CH₃ adjacent to CH₂
Conclusion:
- CH₃–CH₂–CHO = Propanal (propionaldehyde)
⚡ Exam Tip: Always use molecular formula → DBE → functional group possibilities → IR confirmation → NMR detailed structure. This systematic approach is essential for solving structure elucidation problems in the Kenyatta University organic chemistry examination.
Common Spectroscopic Pitfalls to Avoid
- O–H vs N–H: Both appear in 3200–3600 cm⁻¹. O–H is typically broader; N–H shows two distinct peaks for primary amines.
- Aldehyde C–H stretch: Appears as two weak bands at ~2720 and 2820 cm⁻¹ — look for this to confirm an aldehyde even if the C=O peak is ambiguous.
- Exchangeable protons: O–H and N–H protons appear at variable chemical shifts and may not split. Add D₂O to confirm.
- Equivalent protons: Protons on the same carbon in identical environments (like the three CH₃ protons of a methyl group) are equivalent and give a single signal.
- Alkene H chemical shifts: Vinyl protons appear at 4.5–6.5 ppm, not to be confused with aromatic protons (6.5–8 ppm).
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