Molecular Structure and Absorption Spectra

Spectroscopy lets you deduce a molecule's structure by measuring which photon energies it absorbs. The MCAT tests three windows—infrared (IR), ultraviolet-visible (UV-Vis), and nuclear magnetic resonance (NMR)—each probing a different structural feature.

Priority labels: Must know = cold; Know the logic = mechanism not names; Passage-level = recognize, don't memorize; Optional = skippable.

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The Big Picture: Why Different Energies Probe Different Features

Must know

A molecule's bonds vibrate, its electrons occupy orbitals, and its nuclei act like tiny magnets—each motion takes a different energy to perturb. When a photon's energy exactly matches the gap between two states, the molecule absorbs it; everything else passes through. That selectivity is what makes spectroscopy structurally informative.

The energy ordering: NMR (radiowave, nuclear spin) < IR (bond vibration) < UV-Vis (electronic transition). Photon energy is

$E = h\nu = \frac{hc}{\lambda}$

so higher frequency (shorter $\lambda$) = more energetic photon. That's why electronic transitions (UV-Vis) need shorter wavelengths than vibrations (IR), which need shorter wavelengths than nuclear-spin flips (NMR).

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Infrared Spectroscopy

Intramolecular Vibrations and Rotations

Must know

Model a bond as two masses on a spring. Its vibration frequency depends on spring stiffness (bond strength/order) and atomic masses (Hooke's Law):

$\tilde{\nu} \propto \sqrt{\frac{k}{\mu}}$

where $\tilde{\nu}$ is the wavenumber (cm$^{-1}$), $k$ the force constant (bond strength), and $\mu = \frac{m_1 m_2}{m_1 + m_2}$ the reduced mass.

Two takeaways:

1. Stronger/higher-order bonds vibrate at higher wavenumber: C≡C > C=C > C–C.
2. O–H and N–H absorb above C–H mainly because they have a higher force constant $k$ (stronger, more polar)—not reduced mass, since for any X–H bond $\mu \approx 1$ amu (dominated by H). The mass-driven case is C–H vs C–D: deuterium doubles $\mu$, lowering the wavenumber (basis of isotopic labeling).

Selection rule: a vibration is IR-active only if it changes the dipole moment. The oscillating dipole couples to the photon's electric field. Symmetric stretches with no dipole change ($\ce{N2}$, $\ce{O2}$, the symmetric stretch of $\ce{CO2}$) are IR-silent. Polar bonds (C=O, O–H, N–H) absorb strongly.

Vibration types: stretching (bond length changes; the most diagnostic peaks) and bending (bond angles change; lower energy, dominates the fingerprint region).

IR spectra plot transmittance (%) on y vs. wavenumber (cm$^{-1}$) on x, decreasing left (~4000) to right (~400). A peak pointing down = absorption.

Quick check: If you replace ethanol's –OH with –OD, does the O–D stretch appear at higher or lower wavenumber than O–H? Answer: Lower. Deuterium doubles the reduced mass $\mu$, and $\tilde{\nu} \propto 1/\sqrt{\mu}$ decreases.

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Recognizing Common Characteristic Group Absorptions and the Fingerprint Region

Must know

The spectrum has two zones.

Functional group region (4000–1500 cm$^{-1}$): diagnostic stretches. Know these:

Bond / Group	Region (cm$^{-1}$)	Key Feature
O–H (alcohol)	3200–3550	Broad (H-bonding)
O–H (carboxylic acid)	2500–3300	Very broad, overlaps C–H
N–H (amine/amide)	3300–3500	1° = two peaks, 2° = one
C–H (sp³ / sp²–sp)	<3000 / >3000	Splits at 3000 cm$^{-1}$
C≡N, C≡C	2100–2260	Sharp
C=O carbonyl	1700 (1630–1815)	Strong, sharp; position = class
C=C alkene	~1650	Moderate

Know the logic

Carbonyl position is the highest-yield IR pattern: the C=O wavenumber rises as C=O bond order rises. Electron donation into C=O (amide N lone pair) lowers it; electron withdrawal (Cl) raises it.

Compound Class	C=O Wavenumber
Amide	~1650 (lowest)
Carboxylic acid	~1710
Aldehyde / Ketone	~1715–1745
Ester	~1735–1750
Acid chloride / Anhydride	~1800 (highest)

Fingerprint region (400–1500 cm$^{-1}$): complex bends and C–X stretches, unique per molecule. Don't interpret peak-by-peak; its use is comparison—identical fingerprints = same compound.

Quick check: An IR shows a very broad absorption spanning 2500–3300 cm$^{-1}$ and a carbonyl at 1710 cm$^{-1}$. What group is present? Answer: Carboxylic acid—the broad H-bonded O–H plus a ~1710 C=O is the classic signature.

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Ultraviolet-Visible (UV-Vis) Spectroscopy

$\pi$-Electron and Nonbonding Electron Transitions

Must know

UV-Vis promotes valence electrons between molecular orbitals. Three relevant types:

• $\sigma$ electrons (single bonds): very high energy (far UV, not relevant).
• $\pi$ electrons (double/triple bonds, aromatics): undergo $\pi \to \pi^*$.
• n electrons (lone pairs on O, N, S, halogens): undergo $n \to \pi^*$.

Know the logic

$\pi \to \pi^*$ is allowed = intense band; $n \to \pi^*$ is symmetry-forbidden = weak band at longer wavelength.

A chromophore is the part responsible for absorption (conjugated $\pi$ system or C=O). An auxochrome is a substituent (–OH, –NH$_2$) that shifts $\lambda_{max}$ without absorbing strongly itself.

Quick check: A carbonyl shows a weak absorption at 310 nm. $\pi \to \pi^*$ or $n \to \pi^*$? Answer: $n \to \pi^*$—weak and at longer wavelength, the symmetry-forbidden lone-pair promotion.

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Conjugated Systems and the Effect of Structural Changes

Must know

The central idea: conjugation lowers the HOMO–LUMO gap, shifting absorption to longer wavelength. Each added conjugated double bond gives a bathochromic (red) shift; losing conjugation gives a hypsochromic (blue) shift. (Particle-in-a-box intuition: a longer "box" has closer-spaced levels—no quantitative formula needed.)

The qualitative trend (you do not need Woodward–Fieser increments): isolated C=C absorbs deep in UV (~180 nm); a conjugated diene at ~220–250 nm; extended polyenes reach into the visible; benzene ~250–280 nm.

Color: $\beta$-carotene's 11 conjugated double bonds absorb blue-violet (~450–480 nm), so we see the complementary color, orange (why carrots are orange). For any colored molecule, you see what is not absorbed.

Passage-level

Complementary color pairs (recognize, don't memorize the full table): violet↔yellow, blue↔orange, green↔red, yellow↔violet, red↔blue-green.

Know the logic

Indicators: pH indicators change conjugation with protonation state. Phenolphthalein is colorless in acid (lactone form, broken conjugation) and pink in base ($\ce{OH-}$ opens the lactone to a fully conjugated trianion). The structural change extends the $\pi$ system, shifting $\lambda_{max}$ into the visible. General rule: extending conjugation red-shifts, disrupting it blue-shifts.

Beer-Lambert Law (know the relationship; quantitative work is Passage-level):

$A = \varepsilon \ell c$

$A$ = absorbance, $\varepsilon$ = molar absorptivity (L mol$^{-1}$ cm$^{-1}$), $\ell$ = path length (cm), $c$ = concentration (mol L$^{-1}$). Absorbance and transmittance:

$A = -\log T = -\log\left(\frac{I}{I_0}\right)$

Worked Example: A dye in a 1.00 cm cuvette gives $A = 0.60$, with $\varepsilon = 6000$ L mol$^{-1}$ cm$^{-1}$. Find $c$.

$c = \frac{A}{\varepsilon \ell} = \frac{0.60}{6000 \times 1.00} = 1.0 \times 10^{-4}\ \text{M}$

Sanity check: $A = 0.60 \Rightarrow T \approx 0.25$, so ~75% absorbed—reasonable for a tinted solution.

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NMR Spectroscopy

Protons in a Magnetic Field and Equivalent Protons

Must know

A $^1$H nucleus (spin 1/2) can align with (lower energy) or against a magnetic field; the gap is in the radiowave region and scales with field strength $B_0$. RF energy matching the gap makes protons resonate (flip spin). The exact frequency depends on the local electronic environment: surrounding electrons partly shield each proton, so

$B_{\text{eff}} = B_0(1 - \sigma)$

More shielding → lower frequency → upfield (lower chemical shift $\delta$); less shielding (near electronegative atoms) → downfield (higher $\delta$).

$\delta$ is in ppm relative to TMS ($\delta = 0$). All organic protons appear downfield of TMS.

Canonical shift ranges:

Proton Environment	$\delta$ (ppm)
Alkyl (CH$_3$, CH$_2$)	0.9–1.5
α to C=O	2.0–2.7
Next to N/O (–OCH$_3$)	3.0–4.0
Vinyl (C=C–H)	4.5–6.5
Aromatic (Ar–H)	6.5–8.5
Aldehyde (–CHO)	9–10
Carboxylic acid (–COOH)	10–12

Aromatic protons are especially downfield because the benzene ring current deshields the H atoms.

Equivalent protons: protons interconvertible by molecular symmetry are chemically equivalent and give one signal (e.g., the 3 H of CH$_3$Cl; the 6 H of benzene). The number of non-equivalent environments = number of signals.

Worked Example: How many $^1$H signals does ethyl acetate ($\ce{CH3COOCH2CH3}$) show? Three environments—acetyl CH$_3$ (~2.0), –OCH$_2$– (~4.1), ethyl CH$_3$ (~1.2). Answer: 3 signals. The –OCH$_2$– is most downfield (bonded to O).

Quick check: How many $^1$H signals does para-xylene show? Answer: Two—the four aromatic H's are equivalent by symmetry (one signal ~7 ppm) and the two methyls are equivalent (~2.3 ppm).

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Spin-Spin Splitting

Must know

The n + 1 rule: neighboring non-equivalent protons split each other's signal through bonds (vicinal, 2–3 bonds). A proton with $n$ adjacent non-equivalent protons gives $n + 1$ lines, with intensities from Pascal's triangle:

n	Pattern	Intensities
0	singlet	1
1	doublet	1:1
2	triplet	1:2:1
3	quartet	1:3:3:1

The line spacing is the coupling constant $J$ (~6–8 Hz vicinal). Equivalent protons do NOT split each other (a CH$_3$'s own H's don't split each other).

Worked Example — Ethanol ($\ce{CH3CH2OH}$): CH$_3$ (next to 2 H) → triplet; CH$_2$ (next to 3 H) → quartet; OH → broad singlet (fast exchange suppresses coupling in protic solvents). This triplet + quartet is the canonical ethyl signature.

Integration ∝ number of protons giving a signal, so it gives the ratio of proton types—key for structure elucidation.

Quick check: In CH$_3$CHO, the CH$_3$ is a doublet and the CHO is a quartet. Why? Answer: CH$_3$ (3 H) is adjacent to 1 CHO proton → n+1 = doublet; CHO (1 H) is adjacent to 3 CH$_3$ protons → n+1 = quartet. Each splits the other.

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Common Confusions & Tricks

1. IR transmittance vs. absorbance: y-axis is transmittance, so peaks point downward; a dip means absorption.

2. Carbonyl wavenumber order: "Amides Are Always Low"—amide (~1650) < acid (~1710) < ketone/aldehyde (~1720) < ester (~1740). Ester is always higher than ketone.

3. Broad O–H vs. sharp N–H: both near 3300, but O–H (esp. carboxylic acid) is very broad from H-bonding. Primary amines show two N–H peaks, secondary one, tertiary none—a way to distinguish amine types.

4. UV-Vis: you see the complement. $\beta$-Carotene absorbs blue-violet, so it looks orange, NOT blue. Always flip to the complementary color.

5. $\pi \to \pi^*$ vs. $n \to \pi^*$: the $n \to \pi^*$ band is always weaker and at longer wavelength. Of two carbonyl UV bands, the longer/weaker one is $n \to \pi^*$.

6. NMR: equivalent protons don't split each other. Benzene's six H's give a singlet; a CH$_3$'s H's split only neighboring non-equivalent H's.

7. NMR: n+1 assumes equivalent neighbors. With non-equivalent neighbors in different environments you get complex multiplets; the MCAT tests only the clean cases.

8. Chemical shift direction: higher ppm = downfield = less shielded. Electronegative groups deshield → downfield. Don't reverse it.

9. IR vs. NMR roles: IR identifies what functional groups are present; NMR tells you how many and where (connectivity). Complementary tools.

10. Conjugation and color: more conjugation → smaller gap → longer $\lambda$ absorbed (not shorter/bluer—a common reversal).

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Key Equations

Equation	Variables & Usage
$E = h\nu = hc/\lambda$	Photon energy; absorption when $E$ matches a quantized transition.
$\tilde{\nu} \propto \sqrt{k / \mu}$	Wavenumber; $k$ = force constant, $\mu = m_1m_2/(m_1+m_2)$. Predicts IR shift with bond order or mass.
$A = \varepsilon \ell c$	Beer-Lambert; $\varepsilon$ = molar absorptivity, $\ell$ = path length, $c$ = concentration.
$A = -\log T = -\log(I/I_0)$	Absorbance vs. transmittance. $A=1 \Rightarrow T=10\%$; $A=2 \Rightarrow T=1\%$.
$\delta\ (\text{ppm}) = \dfrac{\nu_\text{sample} - \nu_\text{TMS}}{\nu_\text{spectrometer}} \times 10^6$	Chemical shift, referenced to TMS. Higher $\delta$ = downfield.
Splitting $= n + 1$	Lines in a multiplet = adjacent non-equivalent protons + 1.
$B_\text{eff} = B_0(1 - \sigma)$	Effective field on a proton; greater $\sigma$ → upfield.