Multiple Bonds and the Triple-Bond Region

Introduction

Between about 2000 and 2300 cm⁻¹ the IR spectra of most organic compounds are almost blank. This quiet stretch is the triple-bond region , and any band appearing there immediately draws attention: it usually means a nitrile or an alkyne. A little further right, near 1600–1680 cm⁻¹, carbon–carbon double bonds make their presence known, although often only faintly. This page explains where multiple bonds absorb and why their intensities vary so much.

Core explanation

Bond order and wavenumber. Bond strength increases with bond order: C≡C is stronger than C=C, which is stronger than C–C. With similar atomic masses, the stronger bond has the larger force constant and absorbs at the higher wavenumber. Approximate stretching positions are C–C about 1000–1200 cm⁻¹, C=C about 1620–1680 cm⁻¹ and C≡C about 2100–2260 cm⁻¹.

Nitriles (C≡N). The C≡N stretch appears at 2220–2260 cm⁻¹ as a sharp band of medium intensity. Because the bond is polar, the band is reliably visible, and since very little else absorbs in this window it is highly diagnostic. Conjugation with a benzene ring lowers it slightly, to about 2220–2230 cm⁻¹ in benzonitrile.

Alkynes (C≡C). A terminal alkyne, R–C≡C–H, gives a weak but sharp C≡C band at about 2100–2140 cm⁻¹, together with the strong, sharp ≡C–H band near 3300 cm⁻¹. An internal alkyne, R–C≡C–R′, absorbs at about 2190–2260 cm⁻¹, but the band is often very weak. In a perfectly symmetrical alkyne such as but-2-yne, stretching the C≡C does not change the dipole moment at all, so the vibration is IR inactive and no band appears.

Alkenes (C=C). Non-conjugated alkenes absorb at 1620–1680 cm⁻¹. The band is typically weak to medium, because C=C is almost non-polar. Terminal alkenes (RCH=CH₂) give clearer bands than highly substituted, nearly symmetrical ones. Supporting evidence comes from sp² C–H stretches just above 3000 cm⁻¹ and strong out-of-plane C–H bending bands between about 650 and 1000 cm⁻¹.

Aromatic rings. Benzene rings do not have isolated C=C bonds; instead they show several ring-stretching bands, typically near 1600 and 1500 cm⁻¹ (and often 1450 cm⁻¹). Together with aromatic C–H above 3000 cm⁻¹, these mark an arene.

Other features in the region. Carbon dioxide from the air absorbs near 2350 cm⁻¹ and can appear in poorly corrected spectra; it should not be mistaken for a triple bond.

Step-by-step reasoning

To investigate multiple bonds:

1. Scan 2000–2300 cm⁻¹ for any band. 2. A sharp, medium band near 2220–2260 cm⁻¹ suggests C≡N. 3. A weak band near 2100–2260 cm⁻¹ suggests C≡C; check for ≡C–H at 3300 cm⁻¹. 4. Scan 1620–1680 cm⁻¹ for a weak C=C band, and 3000–3100 cm⁻¹ for sp² C–H. 5. Bands near 1600 and 1500 cm⁻¹ with aromatic C–H suggest a benzene ring.

Visual explanation

Picture the spectrum as a long road. The stretch from 2000 to 2300 cm⁻¹ is an empty desert; a nitrile appears as a single tall signpost in the middle of it. Further along, near 1650 cm⁻¹, a C=C band is a small roadside marker easily missed next to the towering C=O landmark.

Real-world analogy

A perfectly balanced seesaw with identical twins at each end can bounce without tipping its centre of balance. A symmetrical C≡C bond is like this: it vibrates, but its "electrical balance" never shifts, so infrared radiation has nothing to push against and no band appears.

Real-world example

Acrylic fibres used in clothing and carpets are made from polymers containing many –C≡N groups. Textile analysts confirm acrylic fibres by the sharp nitrile band near 2240 cm⁻¹, which distinguishes them from polyester (strong C=O near 1715 cm⁻¹) and nylon (amide bands).

Why?

Why do triple bonds absorb in a region almost free of other bands? Few common bonds are stiff enough to vibrate there: X–H stretches are higher because hydrogen is so light, and double-bond stretches are lower because they are weaker. Only triple bonds, and a few cumulated systems, fall in between.

Common misconception

"No band near 2150 cm⁻¹ proves there is no C≡C bond." A symmetrical or nearly symmetrical internal alkyne may give no visible band at all. Absence of a band cannot rule out a symmetrical multiple bond; other techniques such as Raman or NMR are needed.

Worked example

Question: Compound X, C₄H₅N, shows a sharp band at 2250 cm⁻¹, a weak band at 1645 cm⁻¹ and C–H bands both above and below 3000 cm⁻¹. Identify its functional groups and suggest a structure.

Reasoning: The sharp 2250 cm⁻¹ band indicates C≡N. The weak 1645 cm⁻¹ band plus sp² C–H above 3000 cm⁻¹ indicate C=C. sp³ C–H below 3000 cm⁻¹ indicates a CH₂ or CH₃ group.

Answer: X contains a nitrile and an alkene; a possible structure is but-3-enenitrile, CH₂=CH–CH₂–C≡N.

Quick check

1. Why does but-2-yne show no C≡C stretching band in its IR spectrum? Answer: The molecule is symmetrical, so stretching the C≡C bond causes no change in dipole moment and the vibration is IR inactive.

Exam focus

Learn the C≡N range (2220–2260 cm⁻¹) and the C=C range (1620–1680 cm⁻¹). When asked to distinguish an alkene from an alkane, give both the C=C band and sp² C–H above 3000 cm⁻¹. Be ready to explain weak bands in terms of a small change in dipole moment.

Advanced insight

Raman spectroscopy complements IR: it detects vibrations that change a bond's polarisability rather than its dipole. Symmetrical C=C and C≡C stretches, weak or silent in IR, are often strong in Raman spectra. Using both techniques together gives a fuller picture of molecular vibrations.

Summary

Stronger bonds absorb at higher wavenumbers, so triple bonds (2100–2260 cm⁻¹) absorb above double bonds (1620–1680 cm⁻¹). Nitriles give a sharp, medium band at 2220–2260 cm⁻¹. C≡C and C=C bands are often weak because these bonds are nearly non-polar, and symmetrical ones may be IR inactive. Arenes show ring bands near 1600 and 1500 cm⁻¹.

Practice questions

1. State the range in which a nitrile C≡N stretch absorbs and describe the band. Answer: 2220–2260 cm⁻¹, a sharp band of medium intensity. 2. Explain why a C≡C bond absorbs at a higher wavenumber than a C=C bond. Answer: The triple bond is stronger, with a larger force constant, and the atoms have the same masses, so it vibrates at a higher frequency. 3. How could IR distinguish hex-1-ene from cyclohexane? Answer: Hex-1-ene shows a C=C band near 1640 cm⁻¹ and sp² C–H bands above 3000 cm⁻¹; cyclohexane shows neither. 4. Suggest why the C=C band of ethene itself is absent from its IR spectrum. Answer: Ethene is symmetrical, so the C=C stretch produces no change in dipole moment and is IR inactive.