Spectroscopy

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

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:

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

$\blacksquare$

• 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).

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.

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.

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.

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:

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

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).

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.

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.

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.

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.

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.

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:

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

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

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.

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.

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.

$\blacksquare$

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).

$\blacksquare$

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.

• 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.

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).

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).