Experimental Techniques and Structure Determination in Organic Chemistry
1. Polarimetry
A polarimeter passes monochromatic light (usually the sodium D line, 589 nm) through a polariser, the sample tube and a rotatable analyser; the angle through which the analyser must be turned to restore extinction is the observed rotation α, positive (dextrorotatory) if clockwise when looking towards the source. Because α depends on path length and concentration, the specific rotation [α]ᵀ_D = α/(l·c) is reported with l in dm and c in g/mL (or g/100 mL with a factor of 100), together with the temperature, wavelength and solvent, all of which change it. Comparing [α] with that of the pure enantiomer gives the optical purity, equal to the enantiomeric excess; chiral HPLC or GC gives the ee directly from peak areas. The variation of rotation with wavelength is optical rotatory dispersion, and the differential absorption of left and right circularly polarised light is circular dichroism, both used to assign absolute configuration.
2. Thin-layer, column, HPLC and GC chromatography
Chromatography separates compounds by their distribution between a stationary phase and a mobile phase. On silica or alumina, which are polar, separation is by adsorption: polar compounds (acids, alcohols, amines) are held more strongly and move more slowly. In TLC the retardation factor Rf = (distance moved by the spot)/(distance moved by the solvent front) is characteristic for a given plate and eluent; a more polar eluent raises every Rf. Spots are seen under UV (254 nm) or with iodine, permanganate, ninhydrin (amines, amino acids) or anisaldehyde stains; TLC monitors reactions and chooses the eluent for a column. Column (flash) chromatography scales the same separation up, eluting with solvents of increasing polarity along the eluotropic series hexane < toluene < dichloromethane < diethyl ether < ethyl acetate < acetone < methanol.
HPLC forces the mobile phase through a column of fine particles at high pressure. In normal-phase HPLC (silica, non-polar eluent) polar compounds elute last; in reversed-phase HPLC, the commonest, the stationary phase is non-polar (octadecylsilyl C18 silica) and the eluent aqueous acetonitrile or methanol, so polar compounds elute first. A chiral stationary phase separates enantiomers and gives the ee. Gas chromatography carries vaporised samples in an inert gas (He, N₂) through a capillary column coated with a liquid phase; compounds elute in order of volatility and polarity and are identified by retention time, and detected by flame ionisation (FID, for organics) or thermal conductivity; coupled to a mass spectrometer (GC–MS) it identifies each peak. Column efficiency is expressed as the number of theoretical plates N = 16(t_R/w)² (w the peak width at the base), and the separation of two peaks by the resolution R_s = 2(t_R2 − t_R1)/(w₁ + w₂), with R_s ≥ 1.5 meaning baseline separation.
3. UV–visible and IR spectroscopy in structure determination
UV–visible spectra identify chromophores. π → π* bands are strong (ε ≈ 10⁴) and n → π* bands of carbonyls weak (ε ≈ 10–100). Conjugation lowers the π → π* gap, moving λ_max to longer wavelength (a bathochromic shift) and raising ε: ethylene absorbs at 171 nm, butadiene at 217 nm, β-carotene near 450 nm. The Woodward–Fieser rules estimate λ_max. For dienes: base value 214 nm (heteroannular or acyclic) or 253 nm (homoannular), plus 5 nm for each alkyl substituent or ring residue and 5 for each exocyclic double bond, and 30 for each extra conjugated double bond. For α,β-unsaturated ketones (six-membered or acyclic): base 215 nm, plus 10 for an α-substituent, 12 for each β-substituent (alkyl or ring residue), 30 for extending conjugation, 5 for an exocyclic C=C and 39 for a homoannular diene component.
| Group | Wavenumber (cm⁻¹) | Appearance |
|---|---|---|
| O–H, alcohol (H-bonded) | 3200–3550 | strong, broad |
| O–H, carboxylic acid | 2500–3300 | very broad, over C–H |
| N–H | 3300–3500 | medium; two bands for NH₂ |
| C–H sp, sp², sp³ | ≈3300, 3000–3100, 2850–3000 | sharp |
| C–H of aldehyde | ≈2720 and 2820 | two weak bands |
| C≡N, C≡C | ≈2250; 2100–2260 | medium; C≡C weak if symmetric |
| C=O acid chloride, anhydride | ≈1800; ≈1820 and 1760 | strong |
| C=O ester, aldehyde, ketone | ≈1735–1750; ≈1725; ≈1715 | strong; conjugation lowers by 20–40 |
| C=O acid, amide | ≈1710; 1650–1690 | strong |
| C=C alkene, aromatic | 1640–1680; ≈1600 and 1500 | variable |
| NO₂ | ≈1550 and 1350 | two strong bands |
Ring strain raises the carbonyl frequency (cyclohexanone 1715, cyclopentanone 1745, cyclobutanone 1780 cm⁻¹) and conjugation lowers it (acetophenone 1690). The fingerprint region below 1500 cm⁻¹ identifies a known compound by comparison but is hard to assign.
4. ¹H and ¹³C NMR in structure determination
| Proton type | δ |
|---|---|
| alkyl C–H (CH₃ < CH₂ < CH) | 0.9–1.8 |
| allylic, benzylic, α to C=O | 1.6–2.6 |
| H–C–X (X = O, N, halogen) | 2.5–4.5 |
| vinylic | 4.5–6.5 |
| aromatic | 6.5–8.5 |
| aldehyde CHO | 9–10 |
| carboxylic acid COOH | 10–13 (broad, exchanges with D₂O) |
| alcohol, amine O–H/N–H | 1–5, variable, exchanges with D₂O |
A ¹H spectrum gives four kinds of information: the number of signals (chemically distinct sets, found from symmetry — diastereotopic protons count separately), their chemical shifts, their integrals (relative numbers of protons) and their multiplicity. In a first-order spectrum a proton coupled equally to n neighbours shows n + 1 lines in binomial intensities: CH₃ next to CH₂ is a triplet, the CH₂ a quartet, an isopropyl CH a septet. When the couplings to two sets differ, the multiplicities multiply, (n + 1)(m + 1) lines — a CH₂ between a CH₃ and a CH₂ with different J values is a triplet of quartets, twelve lines. Typical coupling constants: vicinal sp³ about 7 Hz; alkene trans 12–18 Hz, cis 6–12 Hz, geminal 0–3 Hz; aromatic ortho 7–9, meta 1–3 Hz — so J distinguishes E from Z and the substitution pattern of a ring. Protons on O and N usually do not split their neighbours because they exchange rapidly, and they disappear on shaking with D₂O.
¹³C NMR is recorded proton-decoupled, so each distinct carbon gives one line; the count of lines reveals symmetry (toluene 5, p-xylene 3, p-dichlorobenzene 2). Shifts span about 220 ppm: sp³ carbons 0–80 (C–O and C–N 40–80), alkene and aromatic 100–160, nitrile 115–125, and carbonyls 160–220 (acids and esters 160–185, aldehydes and ketones 190–220). DEPT experiments sort carbons into CH₃, CH₂, CH and quaternary, and integrals are not reliable because relaxation and nuclear Overhauser enhancement differ from carbon to carbon.
5. Mass spectrometry, degrees of unsaturation and putting the data together
The first step with any formula is the degree of unsaturation (double-bond equivalents, rings plus π bonds): DBE = C − H/2 − X/2 + N/2 + 1, where X is the number of halogens; oxygen and sulfur do not enter. Benzene C₆H₆ has 4 (one ring, three C=C); four or more usually means an aromatic ring. In EI mass spectra the molecular ion M⁺• gives the molar mass (odd M means an odd number of nitrogens), isotope peaks reveal Cl (M + 2 at a third), Br (M + 2 equal to M) and S, and fragmentation follows the stability of the ions formed. Common patterns: α-cleavage next to a heteroatom (amines give iminium ions, m/z 30 for primary amines; ethers oxonium ions); loss of alkyl from ketones to acylium ions (m/z 43 for CH₃CO⁺); the tropylium ion, m/z 91, from benzylic compounds; loss of water (M − 18) from alcohols; and the McLafferty rearrangement, in which a carbonyl compound with a γ-hydrogen transfers it to oxygen through a six-membered transition state and loses an alkene, giving an enol radical cation — m/z 58 for 2-hexanone and other methyl ketones with a γ-hydrogen, m/z 44 for aldehydes such as butanal.
Key takeaways
- [α] = α/(l·c) with l in dm and c in g/mL at a stated T, λ and solvent; optical purity equals ee; chiral HPLC gives ee from peak areas.
- Rf = spot distance/front distance; polar compounds move less on silica and elute first in reversed-phase HPLC; N = 16(t_R/w)², R_s = 2Δt_R/(w₁ + w₂).
- Conjugation shifts λ_max to longer wavelength (Woodward–Fieser: diene 214 + 5 per substituent; enone 215 + 10 α + 12 β); IR C=O: acid chloride 1800 > ester 1740 > aldehyde 1725 > ketone 1715 > amide 1650–1690.
- ¹H: count, shift, integral, multiplicity; n + 1 lines for equal coupling, (n + 1)(m + 1) for unequal; trans J 12–18 Hz, cis 6–12 Hz; ¹³C line count reveals symmetry.
- DBE = C − H/2 − X/2 + N/2 + 1; odd M means odd N; m/z 43 acylium, 91 tropylium, McLafferty 58 for methyl ketones with a γ-H.
Practice questions (20)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
What is the degree of unsaturation of C₈H₁₀O?
Numerical answer — type the value.
Show answer
Answer: 4
DBE = C − H/2 + 1 = 8 − 5 + 1 = 4; oxygen does not enter. Four degrees suggest a benzene ring, as in ethylphenols, methylbenzyl alcohols or phenetole. Counting O as reducing the value is a common slip.What is the degree of unsaturation of nitrobenzene, C₆H₅NO₂?
Numerical answer — type the value.
Show answer
Answer: 5
DBE = C − H/2 + N/2 + 1 = 6 − 2.5 + 0.5 + 1 = 5: four for the benzene ring and one for the N=O of the nitro group in its Lewis structure. Forgetting the N/2 term gives the non-integer 4.5, a signal that something was omitted.What is the degree of unsaturation of C₇H₇Cl?
Numerical answer — type the value.
Show answer
Answer: 4
A halogen counts like hydrogen: DBE = 7 − 7/2 − 1/2 + 1 = 4, consistent with a chlorotoluene or benzyl chloride. Ignoring the chlorine gives 4.5.In the first-order ¹H NMR spectrum of 2-bromopropane, (CH₃)₂CHBr, into how many lines is the CH signal split?
Numerical answer — type the value.
Show answer
Answer: 7
The methine proton has six equivalent neighbours on the two methyls, so n + 1 = 7 lines (a septet, intensities 1:6:15:20:15:6:1); the methyl protons appear as a doublet. Counting one methyl gives a quartet.In 1-chloropropane, CH₃CH₂CH₂Cl, the central CH₂ couples to the CH₃ and to the CH₂Cl protons with nearly equal J. How many lines does its signal show in a first-order spectrum?
Numerical answer — type the value.
Show answer
Answer: 6
With equal coupling all five neighbours (3 + 2) act as one set: n + 1 = 6, a sextet. Were the two couplings different, the pattern would be a triplet of quartets, (2 + 1)(3 + 1) = 12 lines.A CH₂ group is coupled to a CH₃ group with J = 7 Hz and to a vinylic CH with J = 12 Hz. In a first-order spectrum, how many lines does the CH₂ signal show?
Numerical answer — type the value.
Show answer
Answer: 8
Different couplings multiply: the vinylic CH (one proton) splits the signal into a doublet and the three methyl protons split each line into a quartet, (1 + 1)(3 + 1) = 8 lines, a doublet of quartets. Adding the neighbours as if the couplings were equal gives the wrong 5.How many signals appear in the proton-decoupled ¹³C NMR spectrum of toluene?
Numerical answer — type the value.
Show answer
Answer: 5
The mirror plane through CH₃ and C1 makes C2 = C6 and C3 = C5, leaving CH₃, C1 (ipso), C2/C6 (ortho), C3/C5 (meta) and C4 (para): 5 signals. Counting all seven carbons ignores symmetry; p-xylene, with more symmetry, shows 3.How many signals appear in the ¹H NMR spectrum of ethyl acetate, CH₃COOCH₂CH₃?
Numerical answer — type the value.
Show answer
Answer: 3
The acetyl CH₃ (singlet, δ 2.0), the OCH₂ (quartet, δ 4.1) and the ethyl CH₃ (triplet, δ 1.3) are three distinct environments, integrating 3:2:3. Counting lines instead of signals would give 1 + 4 + 3 = 8.A disubstituted alkene shows vinylic protons coupled with J = 16 Hz. The alkene is
Show answer
Answer: A — trans (E)
Vicinal alkene coupling depends on the dihedral angle: trans protons (180°) couple with 12–18 Hz, cis protons (0°) with 6–12 Hz, and the geminal protons of a 1,1-disubstituted alkene with only 0–3 Hz. So 16 Hz marks the E isomer.Which carbonyl compound absorbs at the highest IR frequency?
Show answer
Answer: A — Acetyl chloride
The electronegative chlorine withdraws electron density inductively and its lone pairs donate poorly, strengthening C=O (about 1800 cm⁻¹). Esters (about 1740) donate weakly through OR, ketones sit at 1715, and amides, with strong N lone-pair donation, fall to 1650–1690.Using the Woodward–Fieser rules (base value 214 nm for an acyclic diene, +5 nm per alkyl substituent), estimate λ_max of 2,3-dimethylbuta-1,3-diene, in nm.
Numerical answer — type the value.
Show answer
Answer: 224
214 + 2 × 5 = 224 nm (observed 226 nm). Only substituents on the diene carbons count, here the two methyls on C2 and C3; there are no ring residues or exocyclic double bonds. Using the homoannular base (253) applies only when both double bonds lie in one ring.Estimate λ_max of 3-methylcyclohex-2-en-1-one by the Woodward–Fieser rules for enones (base 215 nm; α-substituent +10; each β-substituent or ring residue +12), in nm.
Numerical answer — type the value.
Show answer
Answer: 239
The β-carbon (C3) carries the methyl group and the ring residue C4, two β-substituents: 215 + 2 × 12 = 239 nm (observed about 235 nm). The α-carbon C2 bears only H, so nothing is added for it; counting the ring bond to C1 as an α-substituent is the usual error.On a TLC plate the solvent front has moved 7.5 cm and a spot 3.0 cm from the origin. What is the Rf value, to two decimal places?
Numerical answer — type the value.
Show answer
Answer: 0.4
Rf = 3.0/7.5 = 0.40. Rf is always below 1; inverting the ratio gives 2.5. A more polar eluent would raise it, and a more polar compound on silica would lower it.A GC peak has a retention time of 5.0 min and a base width of 0.25 min. What is the number of theoretical plates of the column?
Numerical answer — type the value.
Show answer
Answer: 6400
N = 16(t_R/w)² = 16 × (5.0/0.25)² = 16 × 20² = 6400. With the width at half height the formula would be 5.54(t_R/w½)²; using 16 with a half-height width overstates N fourfold.Two peaks elute at 5.0 and 6.0 min, each with a base width of 0.40 min. What is their resolution R_s, to one decimal place?
Numerical answer — type the value.
Show answer
Answer: 2.5
R_s = 2(t_R2 − t_R1)/(w₁ + w₂) = 2(1.0)/(0.80) = 2.5, well above the 1.5 needed for baseline separation. Dividing by one width instead of their sum doubles the result to 5.0.Which statements about chromatography are correct?
Show answer
Answer: A — In reversed-phase HPLC on C18 silica, more polar compounds elute earlier; B — On a silica TLC plate, a carboxylic acid usually has a lower Rf than a hydrocarbon; C — A chiral stationary phase can separate enantiomers
A non-polar C18 phase retains non-polar solutes, so polar ones leave first; polar silica holds the acid strongly; a chiral phase forms diastereomeric interactions with the two enantiomers. GC needs samples that vaporise without decomposing, which proteins cannot — they are analysed by HPLC or electrophoresis.Chiral HPLC of a product shows peaks for the two enantiomers in the area ratio 95:5. What is the enantiomeric excess, in per cent?
Numerical answer — type the value.
Show answer
Answer: 90
ee = (95 − 5)/(95 + 5) × 100 = 90%. The HPLC measures ee directly, without needing the specific rotation of the pure enantiomer; quoting 95 gives the percentage of the major enantiomer.Which quantity does NOT affect the specific rotation of a compound measured in a polarimeter?
Show answer
Answer: A — The length of the sample tube, once the observed rotation is divided by it
Specific rotation is defined by dividing the observed rotation by l and c, so the tube length is normalised away. Wavelength (optical rotatory dispersion), temperature and solvent all change [α], which is why they are always quoted with it.In the EI mass spectrum of 2-hexanone, what is the m/z of the fragment ion produced by the McLafferty rearrangement?
Numerical answer — type the value.
Show answer
Answer: 58
The γ-hydrogen (on C5) transfers to the carbonyl oxygen through a six-membered transition state and the C3–C4 bond breaks, releasing propene (42) and leaving the enol radical cation CH₂=C(OH)CH₃⁺•, m/z 58 (100 − 42). The acylium ion CH₃CO⁺ from α-cleavage appears at m/z 43.A compound shows its molecular ion at m/z 121. What does the nitrogen rule indicate?
Show answer
Answer: A — It contains an odd number of nitrogen atoms
Nitrogen has an even mass (14) but an odd valence, so a molecule with an odd number of N atoms has an odd molar mass, and one with zero or an even number has an even mass. An odd M⁺ at 121 (for example C₈H₁₁N) therefore implies one, three … nitrogens. Chlorine is recognised by its M + 2 peak, not by an odd mass.