Spectroscopy

1. UV-Visible Spectroscopy

1.1 Beer-Lambert Law

Theorem 1 (Beer-Lambert Law):

$A = \varepsilon\,c\,l = \log_{10}\left(\frac{I_0}{I}\right)$

where $A$ is absorbance (unitless), $\varepsilon$ is the molar extinction coefficient (M$^{-1}$cm$^{-1}$), $c$ is concentration (M), and $l$ is path length (cm).

Transmittance: $T = I/I_0 = 10^{-A}$

1.2 Electronic Transitions

1.3 Effect of Conjugation

Theorem 2 (Conjugation Shift): Extending conjugation shifts $\lambda_{\max}$ to longer wavelengths (red shift, bathochromic shift) and increases $\varepsilon$.

For polyenes, the Woodward-Fieser rules estimate $\lambda_{\max}$:

$\lambda_{\max} = \lambda_{\text{base}} + n_{\text{double bonds}} \times 30 + \text{substituent corrections}$

Example 1: Predict $\lambda_{\max}$ for a diene with one exocyclic double bond and one alkyl substituent:

$\lambda_{\max} = 214 + 30 + 5 + 5 = 254 \text{ nm}$

$\blacksquare$

1.4 Applications

• Determining concentration (via Beer-Lambert).
• Monitoring reaction progress (kinetics).
• Protein quantification ($\lambda = 280$ nm for aromatic amino acids).
• Color and dyes (conjugated systems absorbing visible light).

2. Infrared Spectroscopy

2.1 Principles

IR spectroscopy measures absorption of infrared radiation corresponding to vibrational transitions. The frequency of absorption depends on:

$\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}$

where $k$ is the force constant (N/m) and $\mu = \frac{m_1 m_2}{m_1 + m_2}$ is the reduced mass.

Selection rule: Only transitions that change the dipole moment are IR-active.

2.2 Characteristic IR Absorptions

2.3 Fingerprint Region

Definition 1 (Fingerprint Region): The region 400–1500 cm$^{-1}$ contains complex patterns unique to each molecule (C–C, C–O, C–X stretches and bends). Useful for identification by comparison with reference spectra.

2.4 Key Functional Group Identification

Alcohols: Broad O–H stretch (3200–3600 cm$^{-1}$), C–O stretch (1000–1260 cm$^{-1}$).

Carboxylic acids: Very broad O–H (2500–3300 cm$^{-1}$), C=O (1710–1760 cm$^{-1}$), broad C–O.

Aldehydes: C=O (1720–1740 cm$^{-1}$), characteristic aldehyde C–H stretch (2720, 2820 cm$^{-1}$).

Ketones: Strong C=O (1705–1725 cm$^{-1}$).

Esters: C=O (1735–1750 cm$^{-1}$), two C–O stretches.

3. Nuclear Magnetic Resonance Spectroscopy

3.1 Basic Principles

Definition 2 (NMR): Nuclei with spin $I \neq 0$ (e.g., $^1\text{H}$, $I = 1/2$; $^{13}\text{C}$, $I = 1/2$) align with or against an external magnetic field $B_0$.

Resonance condition:

$\nu = \frac{\gamma}{2\pi}B_0$

where $\gamma$ is the gyromagnetic ratio. The frequency difference between two nuclei is expressed relative to a reference (TMS) as the chemical shift:

$\delta = \frac{\nu_{\text{sample}} - \nu_{\text{ref}}}{\nu_{\text{ref}}} \times 10^6 \text{ ppm}$

3.2 $^1\text{H}$ NMR: Chemical Shifts

3.3 Spin-Spin Splitting

Theorem 3 (n+1 Rule): A signal is split into $n + 1$ peaks by $n$ equivalent neighboring protons.

The splitting pattern follows Pascal”s triangle:

Coupling constant: $J$ (Hz) is the separation between adjacent peaks in a multiplet. $J$ is independent of $B_0$.

• Vicinal coupling: $^3J_{\text{HH}} \approx 6$–8 Hz (free rotation).
• Geminal coupling: $^2J_{\text{HH}} \approx 12$–16 Hz.
• $^3J_{\text{vic}}$ in alkenes: $cis$ ($J \approx 6$–12 Hz) vs $trans$ ($J \approx 12$–18 Hz).

3.4 Integration

The area under each signal is proportional to the number of protons giving that signal. Integration ratios reveal the relative numbers of different types of protons.

3.5 Complex Splitting

When non-equivalent neighboring protons have different coupling constants, more complex patterns arise:

• Doublet of doublets (dd): One proton coupled to two non-equivalent neighbors.
• Doublet of doublet of doublets (ddd): Three non-equivalent neighbors.
• AX, AB, and AA’XX’ systems: More complex analysis required for coupled aromatic rings.

3.6 $^{13}\text{C}$ NMR

Definition 3 ($^{13}$C NMR): $^{13}\text{C}$ has $I = 1/2$ but only 1.1% natural abundance. Features:

• Broadband proton decoupling: all $^1$H–$^{13}$C couplings removed; one signal per unique carbon.
• DEPT (Distortionless Enhancement by Polarization Transfer): distinguishes CH, CH$_2$, CH$_3$, and quaternary carbons.
• Chemical shift range: 0–220 ppm.

3.7 2D NMR Techniques

Definition 4 (COSY — Correlation Spectroscopy): Reveals $^1$H–$^1$H coupling relationships. Cross-peaks indicate which protons are coupled.

Definition 5 (HSQC/HMQC): Shows $^1$H–$^{13}$C one-bond correlations. Each proton signal correlates to the carbon it is directly bonded to.

Definition 6 (HMBC): Shows $^1$H–$^{13}$C long-range correlations (2–3 bonds). Useful for connecting fragments through quaternary carbons.

Definition 7 (NOESY/ROESY): Nuclear Overhauser Effect spectroscopy; reveals spatial proximity ($< 5$ Å) rather than through-bond coupling. Essential for stereochemistry.

4. Mass Spectrometry

4.1 Principles

Definition 8 (Mass Spectrometry): Measures the mass-to-charge ratio ($m/z$) of ionized molecules.

The mass spectrum plots intensity vs $m/z$. The molecular ion peak ($\text{M}^+$ or $\text{M}^+\bullet$) gives the molecular weight.

Isotopic patterns: The natural abundance of isotopes creates characteristic patterns:

• $^{13}\text{C}$: 1.1% → $\text{M} + 1$ peak is $\sim 1.1\%$ of M per carbon.
• $^{35}\text{Cl}/^{37}\text{Cl}$: 3:1 ratio → $\text{M} + 2$ peak is $\sim 1/3$ of M for one Cl.
• $^{79}\text{Br}/^{81}\text{Br}$: 1:1 ratio → $\text{M} + 2$ peak is $\sim$ equal to M for one Br.

4.2 Ionization Methods

4.3 Fragmentation Patterns

McLafferty rearrangement: $\gamma$-hydrogen transfer to a carbonyl oxygen, followed by $\beta$-cleavage. Common in carbonyl compounds.

Alpha cleavage: Cleavage adjacent to a heteroatom or carbonyl:

$\text{R}–\text{C}(=\text{O})–\text{R}' \to \text{R}^+ + \cdot\text{C}(=\text{O})\text{R}'$

Common fragmentation:

4.4 High-Resolution MS

Definition 9 (High-Resolution MS): Determines exact mass to 4–6 decimal places, distinguishing between formulas with the same nominal mass:

$\text{C}_6\text{H}_{12}\text{O}_6: \text{exact mass} = 180.0634$ $\text{C}_8\text{H}_{12}\text{N}_2\text{O}_3: \text{exact mass} = 180.0899$

4.5 Nitrogen Rule

Theorem 4 (Nitrogen Rule): A molecule with an odd number of nitrogen atoms has an odd molecular weight; a molecule with an even number (or zero) of nitrogen atoms has an even molecular weight.

5. Combined Structure Elucidation

5.1 Strategy

Theorem 5 (Structure Elucidation Workflow):

1. Molecular formula from HRMS (exact mass) or elemental analysis. Calculate degree of unsaturation.
2. IR for functional groups (O–H, C=O, C≡N, C=C, etc.).
3. $^1$H NMR for proton environments (chemical shifts, integration, splitting).
4. $^{13}$C NMR for carbon framework (number of unique carbons, types).
5. 2D NMR (COSY, HSQC, HMBC) for connectivity.
6. MS fragmentation to confirm proposed structure.

5.2 Degree of Unsaturation

Theorem 6 (Degrees of Unsaturation): For a formula C$_c$H$_h$N$_n$O$_o$X$_x$ (X = halogen):

$\text{DoU} = c + 1 - \frac{h - n + x}{2}$

Each DoU corresponds to one double bond or ring.

Example 2: C$_8$H$_8$O$_2$: DoU $= 8 + 1 - 8/2 = 5$. Likely one benzene ring (4 DoU) plus one C=O.

$\blacksquare$

5.3 Worked Example

Unknown: Molecular ion at $m/z = 120$, HRMS = 120.0575, formula C$_7$H$_8$O.

DoU $= 7 + 1 - 8/2 = 4$.

IR: 3350 cm$^{-1}$ (broad, O–H), 1600, 1500 cm$^{-1}$ (aromatic C=C), no C=O.

$^1$H NMR: $\delta$ 7.1 (2H, d, $J = 8$ Hz), 6.7 (2H, d, $J = 8$ Hz), 4.5 (1H, br s), 2.2 (3H, s).

Interpretation: Para-disubstituted benzene (AA’BB’ pattern, 4H). Methyl group ($\delta$ 2.2). OH ($\delta$ 4.5, broad). Structure: 4-methylphenol ($p$-cresol).

$\blacksquare$

Common Pitfalls

1. Wrong chemical shift assignments. O–H and N–H protons are highly variable and can appear anywhere from $\delta$ 0.5 to 13 ppm depending on concentration and hydrogen bonding. Fix: Always consider exchangeable protons as variable; use D$_2$O shake to identify them.
2. Ignoring coupling constants for stereochemistry. $J$ values distinguish cis ($\sim 6$–12 Hz) from trans ($\sim 12$–18 Hz) alkenes and axial-axial ($\sim 8$–12 Hz) from other couplings in cyclohexanes. Fix: Always extract $J$ values from spectra.
3. Misinterpreting IR broadness. A broad O–H stretch indicates hydrogen bonding (alcohol or acid); a sharp O–H indicates free OH (dilute solution). Fix: Run spectra at different concentrations to distinguish.
4. Wrong molecular ion in EI-MS. EI is a hard ionization; the molecular ion may be absent for fragile molecules. Fix: Use CI, ESI, or soft ionization to find M$^+$.
5. Confusing mass spectral isotope patterns. One Br gives M and M+2 in 1:1 ratio; two Br gives M:M+2:M+4 in 1:2:1. Fix: Calculate expected ratios and compare.
6. Overlooking NOE for stereochemistry. NOE shows proximity, not bonding. An NOE between protons on different fragments confirms their spatial relationship. Fix: Use NOESY/ROESY when stereochemistry is needed.
7. Wrong degree of unsaturation calculation. For halogens, add their count to H; for nitrogen, subtract. Fix: Use DoU $= c + 1 - (h + x - n)/2$ where $x$ = halogens.

Summary

• UV-Vis: $\lambda_{\max}$ depends on conjugation; Beer-Lambert law for quantitation.
• IR: Functional group identification; fingerprint region for comparison; O–H, C=O, C≡N are most diagnostic.
• $^1$H NMR: Chemical shift (environment), splitting ($n + 1$ rule), integration (number of protons), coupling constant $J$ (geometry).
• $^{13}$C NMR: Carbon framework; DEPT distinguishes CH$_n$ types; chemical shifts 0–220 ppm.
• 2D NMR: COSY (H–H connectivity), HSQC (one-bond H–C), HMBC (long-range H–C), NOESY (spatial).
• MS: Molecular weight, formula (HRMS), fragmentation patterns, isotopic patterns.
• Structure elucidation: Combine all techniques systematically with DoU calculation.

Worked Examples

Example 1: Interpreting an IR Spectrum

Problem: A compound with molecular formula C4H8O shows a strong IR absorption at 1715 cm^-1 and no absorption above 3000 cm^-1 except at 2850-2950 cm^-1. What functional group is present? Solution: The absorption at 1715 cm^-1 indicates a carbonyl (C=O) stretch. The absence of broad O-H absorption above 3000 cm^-1 rules out a carboxylic acid. The absence of N-H stretch rules out an amide. The sharp, moderate-wavenumber carbonyl is consistent with an aldehyde (no O-H, but C-H stretch around 2720 cm^-1 should be checked). Given C4H8O and one C=O: likely butanal (CH3CH2CH2CHO) or butanone (CH3CH2COCH3).

Example 2: Degree of Unsaturation Calculation

Problem: A compound with molecular formula C6H6O has the following spectroscopic data: IR: 3300 (broad), 1600, 1500 cm^-1. 1H NMR: 7.2 (5H, multiplet), 4.5 (1H, singlet), 2.0 (1H, singlet, exchanges with D2O). Identify the compound. Solution: DoU = 2C + 2 + N - H/2 - X/2 = 12 + 2 - 3 = 4. IR at 3300 (broad) indicates O-H (phenol or alcohol). 1600, 1500 cm^-1: aromatic ring (DoU >= 4). 1H NMR: 5H multiplet at 7.2 ppm = monosubstituted benzene. 1H singlet at 4.5 ppm = CH attached to O. 1H singlet at 2.0 ppm exchanging with D2O = OH. Compound: phenol (C6H5OH).

Cross-References