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VKC Notes · Spectroscopy · Civil Services P-II · IFoS P-II · Geo-Scientist P-III

13.2 Infrared Spectroscopy: Group Frequencies, Carbonyl Stretching and Hydrogen Bonding

★★★ 2021–2025: 24 parts, 163 marks (Civil Services 11 parts / 100 m · IFoS 5 / 30 m · Geo-Scientist 8 / 33 m). The favourite is “arrange these C=O (or C=C) bands and give the reason”: set in some form every year.

Syllabus lines. Civil Services and IFoS: vibrational spectra, “specific frequencies of functional groups in polyatomic molecules”. Geo-Scientist: infra-red spectrometry, “characteristic frequencies of organic molecules and interpretation of spectra”. Instruments are in 13.8; full structure problems in 13.5; diatomic theory (anharmonicity) in 13.7.

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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

13.2.1 Why a bond absorbs: Hooke's law ★★★

A bond behaves like a spring joining two masses. It absorbs IR light whose frequency equals its own vibration frequency, provided the vibration changes the dipole moment (selection rule). A molecule of N atoms has 3N − 6 vibrations (3N − 5 if linear): stretches (bond length changes) at higher ν̃, bends at lower ν̃.

(13.2.1)
m1m2spring constant k← stretch →stiffer bond (larger k): higher ν̃heavier atoms (larger μ): lower ν̃0.93C–H≈ 30006.00C–C≈ 11006.86C–O1050–13008.97C–Cl600–80010.43C–Br500–60010.96C–I≈ 500reduced mass μ / u (bars) and ν̃ / cm−1 (red)
Fig. 13.2.1 Two things set ν̃: the force constant k and the reduced mass μ.
Draw it in 60 s: two circles joined by a zigzag; write k on the spring. Beside it, six bars rising C–H → C–I with μ on top: the taller the bar, the lower ν̃.
  • C–H with k ≈ 500 N m⁻¹, μ = 0.93 u: ν̃ ≈ 3020 cm⁻¹; C=O with k ≈ 1200 N m⁻¹, μ = 6.86 u: ≈ 1720 cm⁻¹. Both match the observed bands, which is why the formula is worth writing in an answer.SN
  • Isotope test: replace H by D. k is unchanged, μ nearly doubles (0.93 → 1.71), so ν̃(C–D) ≈ ν̃(C–H)/√2 ≈ 2200 cm⁻¹. This is how a band is proved to be a C–H (or O–H) stretch.
  • Bond order sets k: C≡C ≈ 2150 > C=C ≈ 1650 > C–C ≈ 1100 cm⁻¹ (k roughly 3 : 2 : 1).SN
  • Single bonds to C, all with similar k, are ordered by μ: C–H > C–C > C–O > C–Cl > C–Br > C–I; from Cl to I the bond also weakens, so the gaps widen. (C–O sits above C–C in the spectrum, 1050–1300, because its k is larger.) Geo 2021 Q6(i) 2mSN
  • IR-inactive: a vibration with no dipole change gives no band: the C=C stretch of a symmetric tetrasubstituted alkene, N₂, O₂ and the symmetric stretch of CO₂. They show in the Raman spectrum instead.

13.2.2 Regions of the spectrum ★★★

X–H stretchO–H, N–H, C–HtripleC≡N, C≡CdoubleC=O, C=C, C=NfingerprintC–O, C–X, bendsfree O–HH-bonded O–H=C–HC–HCHOC≡NC≡CC=OC=CC–OC–Cl / bends4000350030002500200015001000400ν̃ / cm−1 (scale halves below 2000)
Fig. 13.2.2 Four regions: read the left three for groups, use the fingerprint to match a known compound.
Draw it in 60 s: a line 4000 → 400 with 2500, 2000, 1500 marked; four boxes labelled X–H, triple, double, fingerprint; mark C=O at 1715 in red.
Bondν̃ / cm⁻¹Shape, intensity, note
O–H free (dilute, gas)3600–3650sharp
O–H hydrogen-bonded (alcohol, phenol)3200–3550broad, strong
O–H of COOH (dimer)2500–3300very broad, runs under the C–H bands
N–H (amine, amide)3300–3500NH₂ two bands, NH one band; weaker and sharper than O–H
≡C–H, =C–H, C–H (sp³)3300, 3010–3100, 2850–2960“above 3000 = unsaturated, below 3000 = saturated”
aldehyde C–H2820 and 2720Fermi doublet with the overtone of the C–H bend
C≡N, C≡C2210–2260, 2100–2260C≡N medium-sharp; internal C≡C weak or absent
C=O1650–1850strongest band in the spectrum (13.2.3)
C=C, arene ring1620–1680; 1600, 1500C=C weak when symmetric
NO₂1520 and 1350asymmetric and symmetric stretch, both strong
C–O (alcohol, ether, ester)1050–1300strong; aryl ether / phenol ≈ 1230–1260, 1° alcohol ≈ 1050
C–Cl, C–Br600–800, 500–600fingerprint region
  • Below 1500 cm⁻¹ the bands come from bends and coupled skeletal vibrations of the whole molecule: no two compounds match there, so it identifies a compound against a reference spectrum (the fingerprint), but single bands are hard to assign.SN
  • Intensity follows (dμ/dr)²: polar bonds (C=O, O–H, C–O) give strong bands, non-polar ones (C=C, C≡C) weak bands.

13.2.3 The carbonyl stretch and what moves it ★★★

anhydride (and 1760) 1820acid chloride 1800γ-lactone 1770vinyl / aryl ester 1760ester, δ-lactone 1742acid (H-bonded dimer) 1710amide (1650–1690) 16701785 cyclobutanone1745 cyclopentanone1725 aldehyde1715 ketone, cyclohexanone1685 α,β-unsaturated ketone1665 benzophenone1640 troponeacyl compounds C(=O)–Xaldehydes and ketones↑ −I of X,ring strain↓ conjugation,+M, H-bondingν(C=O) / cm−1
Fig. 13.2.3 The carbonyl ladder. Start every answer from saturated ketone = 1715 cm⁻¹ and move up or down.
Draw it in 60 s: a vertical line 1830 at the top, 1640 at the bottom; acyl compounds on the left (anhydride, acid chloride, ester, acid, amide), ketones on the right (cyclobutanone down to tropone).

Anything that makes C=O shorter and stiffer raises ν(C=O); anything that gives it single-bond character lowers it. Seven effects cover every question set:

  1. Inductive (−I) effect raises it. An electronegative X on the carbonyl carbon pulls σ density, the polar form C⁺–O⁻ is less favoured, C=O becomes more double: acid chloride 1800 > ester 1740 > ketone 1715. A monomeric acid (gas, dilute) absorbs near 1760, above a ketone (1720), for the same reason; the usual 1710 is the H-bonded dimer Geo 2024 Q10(c) 5m.SN
  2. Resonance (+M) lowers it. A lone pair on N fed into C=O gives the amide the lowest value, 1650–1690. In an ester, −I of O wins over its +M; in an amide, +M of N wins.SN
  3. Conjugation lowers it by 20–40 cm⁻¹. α,β-Unsaturated or aryl C=O: cyclohexenone 1685, acetophenone 1690, benzophenone 1665; tropone (aromatic tropylium-oxide form) near 1640. Order I < IV < V < II < III, tropone < benzophenone < enone < cyclohexanone < indan-2-one IFoS 2022 Q5(f) 5m Geo 2023 Q9(c) 5m. Citral (conjugated CHO, ≈ 1680, plus C=C 1635) and citronellal (isolated CHO, ≈ 1725) Geo 2022 Q6(h) 4m.SN
  4. Steric inhibition of resonance raises it back. 1-Acetylcyclohexene ≈ 1665; with a 2-methyl the acetyl twists out of the C=C plane, conjugation is lost, ν ≈ 1690 CSE 2022 Q8(a)(i) 5m.
  5. Ring strain raises it (13.2.4): cyclobutanone 1785, cyclopentanone 1745, cyclohexanone 1715; γ-lactone 1770, δ-lactone 1735 Geo 2021 Q6(d) 4m.SN
  6. Field effect (through space): chloroacetone shows two bands. The rotamer with Cl eclipsing O (dipoles side by side) absorbs at 1745; the gauche rotamer at 1725, close to acetone. IR is fast enough to see both conformers CSE 2021 Q8(c)(ii) 5m.
  7. Vibrational coupling splits two C=O on one O: anhydrides show an asymmetric (higher) and a symmetric stretch, butyric anhydride 1825 + 1750, while butyric acid (1725) and ethyl butyrate (1740) show one band each CSE 2023 Q6(a)(ii) 5m.SN
⟷ C=O gets single-bondcharacter: k falls,1715 → 1685 cm−1 ⟷ N lone pair (+M):amide C=O lowest1650–1690 cm−1 acid chloride 1800 ester 1740 ketone 1715−I of Cl, O pulls σ density from C: C=O shorter and stiffer; Cl gives almost no +M
Fig. 13.2.4 Resonance pushes C=O towards a single bond; −I pulls it towards a triple bond.
Draw it in 60 s: enone with two curly arrows (C=C → C–C, C=O → O) and its zwitterion; amide with the N lone pair arrow; then acid chloride 1800, ester 1740, ketone 1715 in a row.

Solved PYQ CSE 2024 Q8(a) 15m

Photolysis of a bicyclic enone I in ethanol gives II, III, IV; the mixture shows ν(C=O) 1787, 1740, 1715, 1685 cm⁻¹. Assign.

  1. 1787: II, a cyclobutanone (four-membered ring, C–CO–C ≈ 90°: strain raises ν).
  2. 1740: IV, an ester whose C=C is β,γ (not conjugated): normal saturated ester.
  3. 1715: III, an α,β-unsaturated ester (conjugation takes 25 off 1740).
  4. 1685: I, the unchanged conjugated cyclohexenone (1715 − 30).SN
  • Lactones and esters in pairs: an enol (vinyl) lactone or ester absorbs higher (≈ 1760–1800) than its saturated partner because the ring O lone pair is busy with the C=C and cannot feed C=O; an α,β-unsaturated ester absorbs lower (≈ 1720) CSE 2021 Q6(a)(ii) 5m. Furan-2(3H)-one ≈ 1800 > γ-butyrolactone 1770 > δ-valerolactone 1735 Geo 2021 Q6(d).
  • Two amides with one N–H each (acetanilide, PhNHCOCH₃, and N-methylbenzamide, PhCONHCH₃): both show one N–H near 3300 and amide I near 1660. In N-methylbenzamide the C=O is conjugated with the ring (lower, ≈ 1640); in acetanilide the N lone pair is shared with the ring, so less goes to C=O (higher, ≈ 1665) IFoS 2024 Q5(f) 5m.

13.2.4 Ring size: C=O and C=C ★★★

cyclobutanone 1785 cyclopentanone 1745 cyclohexanone 1715 1780 1678 1657 1651exocyclic C=C: smaller ring, higher ν(C=C) cyclobutene 1566 cyclopentene 1611 cyclohexene 1646endocyclic C=C: smaller ring, lower ν(C=C)
Fig. 13.2.5 Small rings raise an exocyclic C=O or C=C, and lower an endocyclic C=C.
Draw it in 60 s: three ring ketones with their values; under them four methylenecycloalkanes (1780 → 1651); last row three cycloalkenes (1566 → 1646).
  • Exocyclic C=O and C=C rise as the ring shrinks. The small ring angle (60°, 90°) forces the ring bonds to use orbitals with more p character, so the exocyclic σ bond gets more s character: shorter, stiffer, larger k. The C=C stretch also couples with the ring C–C stretches, pushing it up further. Methylenecyclopropane ≈ 1780 > methylenecyclobutane ≈ 1678 > methylenecyclopentane 1657 > methylenecyclohexane 1651 CSE 2023 Q5(b) 10m Geo 2024 Q6(j) 4m.SN
  • Endocyclic C=C falls as the ring shrinks: cyclohexene 1646 > cyclopentene 1611 > cyclobutene 1566 (the C=C stretch couples with the strained ring bonds and its effective k drops). Cyclopropene (≈ 1640) is the exception that breaks the trend Geo 2022 Q9(c) 5m.SN
  • Ketone vs ring: cyclobutanone 1785, cyclopentanone 1745, cyclohexanone 1715, cycloheptanone 1700; indan-2-one (a cyclopentanone) ≈ 1750 Geo 2023 Q9(c). Cyclobutanone vs cyclohexanone Geo 2025 Q6(h) 4m.

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