Spectroscopy

Spectroscopy for the MCAT

Part 1 of 4 — IR Spectroscopy: Reading the Bands

IR spectroscopy measures bond vibrations (stretching and bending). A vibration absorbs IR only if it changes the molecule's dipole moment. Two physical rules organize every frequency:

• Stiffer bonds vibrate faster: triple > double > single bond frequencies
• Lighter atoms vibrate faster: X-H stretches sit highest of all

Everything below about $1500\ \text{cm}^{-1}$ is the fingerprint region — a complex pattern used to match spectra, not to identify individual groups. Work only in the diagnostic region above it.

The Band Table (memorize these)

Sorting Out the C=O Neighborhood

Conjugation lowers a C=O frequency by roughly 20-30 $\text{cm}^{-1}$ (acetophenone ~1685 vs acetone ~1715) because resonance gives the C=O partial single-bond character. Ring strain raises it (small-ring lactones/ketones absorb higher).

The Interpretation Algorithm

1. Strong band near 1700? → carbonyl present; use its exact position + companions (broad O-H → acid; C-O stretch → ester; N-H → amide; 2720 → aldehyde)
2. Broad 3200-3550? → alcohol O-H (or N-H if spiked and less broad)
3. ~2250? → nitrile (or alkyne slightly lower and weaker)
4. Nothing but C-H? → hydrocarbon or ether (check 1000-1300 for C-O)

IR Interpretation 🎯

Key Takeaways — Part 1

• IR needs a dipole change; stiffer bonds and lighter atoms absorb at higher wavenumbers; below 1500 = fingerprint, don't interpret
• Anchor bands: broad O-H 3200-3550 (acid version smears 2500-3300), N-H spikes (1° two, 2° one), sp3 C-H below 3000 vs sp2 above, nitrile 2250, C=O ~1700-1750 strong
• Carbonyl ladder: acyl chloride 1800 > anhydride (two bands) > ester 1740 > aldehyde ~1725 > ketone 1715 ≈ acid 1710 > amide 1650-1690
• Conjugation lowers C=O ~20-30 $\text{cm}^{-1}$; ring strain raises it; aldehyde flag = weak 2720/2850 doublet
• Symmetric bonds (internal alkynes) can be IR-invisible

Worked Examples — IR Interpretation

Question: A compound C4H8O shows: sp3 C-H at 2900, strong sharp 1715, no bands above 3000, nothing at 2720. Identify it.

Solution:

1. Degrees of unsaturation: $(2 \times 4 + 2 - 8)/2 = 1$ → one ring OR one double bond
2. Strong 1715 → C=O accounts for the one degree of unsaturation
3. No broad O-H → not an alcohol/acid; no 2720/2850 pair → not an aldehyde
4. A C4 carbonyl that is not an aldehyde: 2-butanone (butanal would show the aldehyde C-H doublet)

MCAT Strategy: Always spend the degrees of unsaturation first — if C=O uses the only degree, there is no ring and no alkene, which collapses the candidate list fast.

Question: Two unknowns both show broad O-H stretches. Spectrum A: broad band centered near 3350, no carbonyl. Spectrum B: extremely broad absorption from 2500-3300 riding over the C-H peaks, strong 1705. Assign them.

Solution:

1. A: broad, rounded, 3200-3550, no C=O → alcohol
2. B: the O-H of a carboxylic acid is broadened dramatically by strong hydrogen-bonded dimer formation, dragging it down across the C-H region; paired with C=O near 1710 → carboxylic acid
3. The dimer's paired hydrogen bonds are why acid O-H looks so different from alcohol O-H

MCAT Strategy: Breadth + position of O-H tells you the class before you even look at the carbonyl region. Acid = the only group whose O-H swallows the C-H stretches.

Question: How do you distinguish propanamide from a mixture of a primary amine and a ketone using IR positions alone?

Solution:

1. Both show N-H stretches (~3300-3500) and a C=O
2. The KEY is the carbonyl position: an amide C=O is unusually LOW (1630-1690) because the nitrogen lone pair donates into the carbonyl (resonance → partial single bond)
3. A simple ketone sits near 1715 — outside the amide range
4. C=O at 1660 + N-H spikes → amide; C=O at 1715 + separate N-H → amine plus ketone

MCAT Strategy: The amide's low C=O frequency is the same resonance fact that makes peptide bonds planar and unreactive — one piece of physics, three exam contexts.

Question: List the diagnostic IR bands expected for methyl salicylate (an aromatic ring bearing an ester and a phenol O-H).

Solution:

1. Phenolic O-H: broad, 3200-3550 (intramolecular H-bonding to the ester carbonyl broadens/lowers it)
2. sp2 aromatic C-H just above 3000; sp3 C-H (the OCH3) just below 3000
3. Ester C=O near 1700-1735 (lowered from a normal ester by conjugation with the ring)
4. Strong C-O stretches 1100-1300; aromatic C=C ~1600

MCAT Strategy: Passage questions often run this in reverse — "which band confirms the product?" Answer with the band that CHANGES in the reaction (e.g., ester C=O appears, phenol O-H persists).

Spectroscopy for the MCAT

Part 2 of 4 — Proton NMR: Shift, Integration & Splitting

Proton ($^1H$) NMR reports on the hydrogens of a molecule. Nuclei in different electronic environments resonate at different frequencies, reported as chemical shift ($\delta$, in ppm) relative to TMS at 0 ppm. Three readouts answer three questions:

• Number of signals → how many chemically distinct H environments?
• Integration (area) → how many H's in each environment (a ratio, not an absolute count)?
• Splitting (multiplicity) → how many H's on the adjacent carbons?

Chemical Shift: Deshielding Moves You Downfield

Electron density shields a nucleus. Electronegative atoms and pi systems pull density away (deshielding), shifting signals downfield (higher ppm, drawn to the left).

Aromatic and aldehyde protons sit unusually far downfield because of anisotropy — the ring current (or C=O pi system) generates a local magnetic field that adds to the applied field at those positions. O-H and N-H shifts are variable (1-5 ppm typically), broadened by exchange, and often disappear on shaking with $D_2O$.

The n+1 Rule

A proton coupled to $n$ equivalent neighboring protons (on adjacent atoms, typically three bonds away) appears as $n+1$ peaks:

• 0 neighbors → singlet; 1 → doublet; 2 → triplet; 3 → quartet
• Equivalent protons do not split each other
• O-H and N-H usually do not split neighbors (fast exchange averages the coupling)

The classic ethyl pattern — a 2H quartet (3.5-4.5 if on O) plus a 3H triplet (~1.2) — flags $-OCH_2CH_3$ or $-CH_2CH_3$ instantly. An isopropyl group gives a 6H doublet plus a 1H septet.

Counting Environments with Symmetry

Symmetry-equivalent protons give ONE signal. Examples: acetone gives one singlet (6H); para-disubstituted benzenes with identical substituents give one aromatic signal; tert-butyl gives a 9H singlet. Fewer signals than you would naively expect is a symmetry clue.

Proton NMR Interpretation 🎯

Key Takeaways — Part 2

• Three readouts: signal count = environments; integration = H ratio; splitting = neighbors (n+1)
• Deshielding (electronegative atoms, pi anisotropy) moves signals downfield: alkyl 1, alpha-to-C=O 2.1-2.5, H-C-O 3-4.5, vinyl 4.5-6.5, aromatic 6.5-8.5, aldehyde 9-10, acid 10-12
• Ethyl = triplet + quartet; isopropyl = doublet + septet; tert-butyl = 9H singlet
• O-H/N-H: variable shift, broad, no reliable splitting, erased by $D_2O$ shake
• Symmetry merges signals — fewer signals than protons suggests a symmetric structure

Worked Examples — Proton NMR

Question: C4H8O2 shows: 2.0 ppm singlet (3H), 1.3 ppm triplet (3H), 4.1 ppm quartet (2H). Assign the structure.

Solution:

1. Degrees of unsaturation = $(2 \times 4 + 2 - 8)/2 = 1$ → likely a C=O with two oxygens → ester or acid; no 10-12 ppm signal rules out the acid
2. Triplet + quartet = ethyl; quartet at 4.1 → CH2 on oxygen → $-OCH_2CH_3$
3. 3H singlet at 2.0 = methyl alpha to C=O with no neighbors
4. Assemble: $CH_3CO$-$OCH_2CH_3$ — ethyl acetate

MCAT Strategy: Assign the most distinctive fragment first (the ethyl pattern), then place the leftover integration. The singlet's shift (2.0 vs 3.7) tells you which side of the ester oxygen a methyl sits on.

Question: Distinguish 1-propanol from 2-propanol by 1H NMR without using chemical shifts.

Solution:

1. 1-Propanol ($CH_3CH_2CH_2OH$): four environments — CH3 triplet, middle CH2 multiplet, O-CH2 triplet, plus O-H
2. 2-Propanol ($(CH_3)_2CHOH$): three environments — 6H doublet, 1H septet, O-H
3. The 6H doublet + septet (isopropyl signature) identifies 2-propanol immediately

MCAT Strategy: Symmetric branching compresses the spectrum. A large doublet integrating to 6H is nearly always $(CH_3)_2CH$-.

Question: An unknown C8H8O has signals at 9.9 (1H, singlet), 7.8 and 7.3 (2H each, doublets), and 2.4 (3H, singlet). Identify it.

Solution:

1. 9.9 singlet = aldehyde H (no neighboring H's — the carbonyl carbon's neighbor is the ring)
2. Two 2H aromatic doublets = the classic para-disubstituted benzene pattern (two pairs of equivalent H's splitting each other)
3. 2.4 singlet (3H) = methyl on the ring (benzylic shift)
4. Structure: 4-methylbenzaldehyde (p-tolualdehyde)

MCAT Strategy: A pair of aromatic doublets ("AA'BB' pattern") = para substitution. Ortho/meta patterns are messier multiplets — the MCAT uses the clean para case.

Question: A symmetric ether shows only a triplet integrating to 3 and a quartet integrating to 2. The molecular formula is C4H10O. Reconcile the integration with 10 protons.

Solution:

1. Integration gives ratios: 3:2 could be 3H:2H, 6H:4H, etc.
2. C4H10O with an ethyl pattern and total 10 H → 6H:4H → two equivalent ethyl groups
3. Structure: diethyl ether, $CH_3CH_2$-$O$-$CH_2CH_3$ — symmetry halves the apparent proton count

MCAT Strategy: Always scale integrations to the molecular formula before assigning. A "3H" triplet in a 10-proton molecule may really be 6H of two identical groups.

Spectroscopy for the MCAT

Part 3 of 4 — Carbon-13 NMR, UV-Vis & Mass Spectrometry

Three more tools round out the MCAT's spectroscopy kit. Each answers a different structural question.

Carbon-13 NMR: Counting Carbon Environments

$^{13}C$ is only ~1.1% abundant, so carbon-carbon splitting is not observed and routine spectra are proton-decoupled: every distinct carbon environment gives one singlet. Use it to count environments and to spot carbonyls.

Two workhorse facts: a peak beyond ~190 means aldehyde or ketone (acid-derivative carbonyls stop near 185), and a molecule with fewer $^{13}C$ signals than carbons has symmetry.

UV-Vis: Conjugation Detector

UV-Vis absorption promotes pi electrons to antibonding orbitals ($\pi \rightarrow \pi^*$). The key trend: more conjugation → smaller HOMO-LUMO gap → longer wavelength absorbed ($E = hc/\lambda$). Isolated double bonds absorb below ~200 nm (invisible on the instrument); extended conjugation walks the absorption into the near-UV and, eventually, the visible (beta-carotene's 11 conjugated C=C absorb blue light, so it looks orange). Biochemistry hooks: proteins absorb at 280 nm (aromatic Trp/Tyr), nucleic acids at 260 nm, and NADH absorbs at 340 nm while $NAD^+$ does not — the standard way to follow redox enzyme kinetics.

Mass Spectrometry: Weighing the Pieces

The sample is ionized (losing one electron gives the molecular ion $M^{+}$, a radical cation), fragments, and ions are sorted by mass-to-charge ratio ($m/z$).

• $M^{+}$ gives the molecular weight; the tallest peak is the base peak (most stable ion, set to 100%)
• M+2 isotope flags: Cl gives M+2 about one-third of M (ratio 3:1, from $^{35}Cl/^{37}Cl$); Br gives M+2 roughly equal to M (1:1)
• Nitrogen rule: an odd $M^{+}$ mass means an odd number of nitrogens
• Fragments form the most stable cations: benzylic/allylic and tertiary carbocations, and acylium ions ($RCO^{+}$, e.g. $m/z = 43$ for $CH_3CO^{+}$)
• Common neutral losses: M-15 (methyl), M-18 (water, typical of alcohols), M-29 (CHO or ethyl)

Which Tool for Which Question?

Carbon NMR, UV-Vis & MS 🎯

Key Takeaways — Part 3

• $^{13}C$ NMR: one singlet per unique carbon; >190 ppm = aldehyde/ketone, 160-185 = acid derivative; fewer signals than carbons = symmetry
• UV-Vis: more conjugation → smaller HOMO-LUMO gap → longer wavelength; 260 nm nucleic acids, 280 nm proteins, 340 nm NADH (not $NAD^+$)
• MS: $M^{+}$ = molecular weight; M+2 flags — Cl 3:1, Br 1:1; odd mass = odd N count
• Fragments follow cation stability: benzylic/allylic/tertiary and acylium; losses of 15 (CH3), 18 (H2O), 29 (CHO/C2H5)
• Match the tool to the question: IR = groups, 1H NMR = H arrangement, 13C = C environments, UV = conjugation, MS = mass/halogens

Worked Examples — 13C, UV-Vis & MS

Question: A compound C8H10 shows only three $^{13}C$ signals: 21, 126, and 137 ppm. Propose a structure.

Solution:

1. C8H10 has $(16 + 2 - 10)/2 = 4$ degrees of unsaturation → benzene ring likely
2. Only 3 signals for 8 carbons → high symmetry
3. Para-xylene: two equivalent methyls (21), two equivalent substituted ring carbons (137), four equivalent ring C-H carbons (126) — exactly three environments
4. Ortho- or meta-xylene would show more ring environments (4 signals each)

MCAT Strategy: For aromatic isomer problems, count expected 13C signals for each candidate first — para isomers are always the most symmetric.

Question: Unknown A: $M^{+}$ = 92, M+2 = 94 at ~33% of M. Unknown B: $M^{+}$ = 136, M+2 = 138 at ~98% of M. Assign the halogens.

Solution:

1. A: M+2 about one-third of M → the 3:1 signature of one chlorine
2. B: M+2 roughly equal to M → the 1:1 signature of one bromine
3. Check masses: A = 92 fits $C_4H_9Cl$ (57 for $C_4H_9$ + 35 for $^{35}Cl$); B = 136 fits $C_4H_9Br$ (57 + 79)

MCAT Strategy: Read the M+2 ratio BEFORE doing any arithmetic on the mass — it instantly fixes the halogen and simplifies the formula search.

Question: An alcohol's mass spectrum shows a weak peak at m/z 74 and a strong peak at m/z 56. Interpret.

Solution:

1. 74 - 56 = 18 → loss of water, the signature fragmentation of alcohols
2. $M^{+}$ = 74 fits C4H10O (butanol isomer); dehydration in the spectrometer gives the alkene radical cation at 56
3. Alcohols often show weak or missing molecular ions — the M-18 peak is the tell

MCAT Strategy: Memorize the neutral-loss table (15, 18, 29). A prominent M-18 says "alcohol" as surely as a broad IR band at 3350.

Question: Lactate dehydrogenase converts pyruvate + NADH to lactate + $NAD^+$. A passage reports absorbance at 340 nm falling over time. What is being measured, and what would a flat trace mean?

Solution:

1. NADH absorbs at 340 nm; $NAD^+$ does not
2. Falling A340 = NADH consumed → the reaction is running in the direction written
3. A flat trace = no NADH turnover → enzyme inhibited, denatured, or missing substrate
4. Rate of absorbance change is proportional to reaction velocity (Beer's law: $A = \epsilon l c$, linear in concentration)

MCAT Strategy: Any passage pairing a dehydrogenase with a spectrophotometer is a 340 nm question. Link the SIGN of the absorbance change to the reaction direction before touching the answer choices.

Spectroscopy for the MCAT

Part 4 of 4 — Structure Elucidation: MCAT Integration

Real MCAT questions rarely hand you one clean spectrum. They combine fragments of IR, NMR, and MS data inside a passage and ask you to identify a product, confirm a reaction happened, or choose which technique would settle an ambiguity. This part builds the workflow.

The Five-Step Elucidation Algorithm

1. Molecular formula first. From MS ($M^{+}$) or the passage. Compute degrees of unsaturation: $DoU = (2C + 2 + N - H - X)/2$ (oxygen is ignored). DoU = 4 or more suggests a benzene ring.
2. IR for functional groups. Carbonyl? Broad O-H? N-H spikes? Nitrile? This partitions the formula into fragments.
3. $^1H$ NMR for the carbon skeleton. Scale integrations to the formula, assign signature patterns (ethyl, isopropyl, para-aromatic), and use shifts to place fragments next to O, N, or C=O.
4. $^{13}C$ / symmetry check. Signal count vs carbon count catches symmetric candidates.
5. Assemble and verify. Every atom of the formula, every degree of unsaturation, and every spectral feature must be accounted for. One contradiction kills a candidate.

Confirming a Reaction Happened

Passages love "which spectral change confirms the product?" Answer with the band or signal that changes:

Ranking Techniques for a Given Ambiguity

• Distinguishing constitutional isomers with different H patterns → $^1H$ NMR
• Distinguishing functional group classes (ester vs ketone) → IR
• Distinguishing halogenated candidates or different MW → MS
• Enantiomers: IR, NMR, UV, and MS are ALL identical — only optical rotation (polarimetry) or a chiral environment distinguishes them
• Diastereomers: physically distinct, so NMR CAN distinguish them (different shifts/couplings)

Passage Traps

• Integration scaling — ratios must be rescaled to the true formula (symmetry!)
• Exchangeable protons — O-H/N-H wander, broaden, and vanish with $D_2O$; do not force them into the n+1 count
• Conjugation shifts — a C=O at 1680 can still be a ketone if conjugated; check for the aromatic/alkene evidence before calling it an amide
• Missing molecular ions — alcohols and branched substrates fragment readily; the highest visible m/z is not always $M^{+}$

Structure Elucidation Strategy 🎯

Key Takeaways — Part 4

• Workflow: formula + DoU → IR groups → 1H NMR skeleton → 13C/symmetry check → assemble and verify every feature
• Confirm reactions with the features that CHANGE (O-H disappearing, C=O appearing, shift moves)
• Technique choice: isomers with same formula → NMR; group classes → IR; mass/halogen → MS; conjugation/redox assay → UV-Vis
• Enantiomers are spectroscopically identical (need polarimetry or a chiral environment); diastereomers are NMR-distinguishable
• Traps: rescale integrations, exclude exchangeable H from n+1, allow conjugation shifts, distrust the highest m/z as M+ for alcohols

Worked Examples — Full Elucidation

Question: $M^{+}$ = 150. IR: 1745, 1225, no O-H. 1H NMR: 7.3 (5H, multiplet), 5.1 (2H, singlet), 2.1 (3H, singlet). Identify the compound.

Solution:

1. C9H10O2: $DoU = (18 + 2 - 10)/2 = 5$ → ring (4) + one C=O
2. IR: ester (1745 + strong C-O, no O-H) — uses the fifth degree
3. NMR: 5H multiplet = monosubstituted phenyl; 2H singlet at 5.1 = deshielded benzylic CH2 on oxygen; 3H singlet at 2.1 = acetyl methyl
4. Assemble: benzyl acetate, $CH_3CO$-$O$-$CH_2C_6H_5$

MCAT Strategy: The 2H singlet near 5 ppm is the giveaway: only a CH2 flanked by BOTH oxygen and an aromatic ring gets that far downfield with no splitting.

Question: A passage claims NaBH4 reduced 4-tert-butylcyclohexanone to the alcohol. Rank the usefulness of (i) IR, (ii) optical rotation, (iii) MS for confirming the reduction.

Solution:

1. IR: excellent — 1715 vanishes, broad 3350 appears. Direct and diagnostic.
2. MS: helpful — MW rises by 2 (154 → 156), a small but real shift
3. Optical rotation: useless — both compounds are achiral (the molecule has a mirror plane), so rotation is zero before and after
4. Ranking: IR > MS > polarimetry

MCAT Strategy: Before reaching for stereochemical tools, check whether the substrate is even chiral. Passage writers plant achiral substrates to punish reflexive "measure the rotation" answers.

Question: An unknown C8H8O shows C=O at 1685, aromatic signals at 7.4-7.9 (5H), and a 3H singlet at 2.6 ppm. A student concludes "amide, because the C=O is below 1700." Correct the reasoning.

Solution:

1. The formula contains no nitrogen — an amide is impossible. Formula constraints outrank band positions.
2. DoU = 5: benzene ring + one C=O
3. 1685 is a KETONE C=O lowered ~30 wavenumbers by conjugation with the ring
4. 3H singlet at 2.6 (alpha to C=O, benzylic-adjacent shift) completes acetophenone

MCAT Strategy: Conjugation-lowered ketones (1680-1690) overlap the amide window. Resolve the ambiguity with the formula (any N?) and with N-H stretches (amides that have N-H show it; acetophenone shows none).

Question: C10H14, $M^{+}$ = 134, base peak at m/z 91. NMR: 7.2 (5H, multiplet), 2.5 (2H, triplet), 1.6 (2H, multiplet), 1.3 (2H, multiplet), 0.9 (3H, triplet). What does m/z 91 tell you, and what is the compound?

Solution:

1. 134 - 91 = 43 → loss of $C_3H_7$ (propyl radical)
2. m/z 91 is the famous tropylium/benzyl cation ($C_7H_7^{+}$) — the signature of a benzylic cleavage
3. NMR: monosubstituted ring + an unbranched four-carbon chain (triplet CH3, benzylic CH2 at 2.5)
4. Compound: n-butylbenzene; cleavage at the benzylic bond gives the stabilized $C_7H_7^{+}$

MCAT Strategy: m/z 91 in any aromatic mass spectrum = benzylic cleavage. Fragmentation follows cation stability, so the strongest peaks mark the most stable cations — the same hierarchy you learned for SN1.