C–H Stretches and Hybridisation

Introduction

Almost every organic molecule contains C–H bonds, so you might expect their IR bands to be useless for identification. In fact, the precise position of the C–H stretch reveals the hybridisation of the carbon atom it is attached to. A simple rule — look either side of 3000 cm⁻¹ — lets you tell an alkane from an alkene or arene at a glance, and spot the terminal hydrogen of an alkyne or an aldehyde.

Core explanation

The three types. C–H stretching wavenumbers depend on the hybridisation of the carbon:

C–H type Example Wavenumber / cm⁻¹ --- --- --- sp³ C–H alkanes, alkyl groups 2850–2960 sp² C–H alkenes, arenes 3000–3100 sp C–H terminal alkynes, ≡C–H about 3300 aldehyde C–H –CHO about 2720 and 2820

Why the trend? Hybrid orbitals with more s character hold electrons closer to the nucleus, because s orbitals are more penetrating than p orbitals. An sp orbital (50% s) forms a shorter, stronger C–H bond than an sp² orbital (33% s), which in turn is stronger than an sp³ bond (25% s). A stronger bond has a larger force constant and so vibrates at a higher wavenumber. The C–H bond dissociation energies follow the same order.

The 3000 cm⁻¹ rule. Almost all sp³ C–H stretches lie just below 3000 cm⁻¹, and almost all sp² C–H stretches lie just above it. Drawing an imaginary vertical line at 3000 cm⁻¹ therefore tells you immediately whether the molecule contains alkene or aromatic hydrogens. Hexane shows bands only below the line; hex-1-ene shows bands on both sides; benzene shows its C–H bands only above it.

Terminal alkynes. A ≡C–H bond absorbs as a strong, sharp band at about 3300 cm⁻¹. It lies in the same region as O–H and N–H, but its narrowness distinguishes it. It is usually accompanied by a weak C≡C band near 2100–2140 cm⁻¹.

Aldehydes. The hydrogen on a carbonyl carbon gives two weak but distinctive bands near 2720 and 2820 cm⁻¹. The lower one, at about 2720 cm⁻¹, stands on its own to the right of the ordinary C–H bands and is a useful clue for an aldehyde. The pair arises from interaction (Fermi resonance) between the C–H stretch and an overtone of a C–H bending vibration.

C–H bending. Alkyl groups also show bending bands near 1375 and 1450 cm⁻¹, but these occur in almost every organic compound and are rarely diagnostic.

Step-by-step reasoning

To use the C–H region:

1. Locate the 3000 cm⁻¹ line on the spectrum. 2. Bands only below it suggest a saturated compound (sp³ C–H only). 3. Bands above it indicate sp² C–H from an alkene or arene. 4. A sharp, strong band near 3300 cm⁻¹ suggests a terminal alkyne. 5. Weak bands near 2720 and 2820 cm⁻¹ with a C=O band indicate an aldehyde.

Visual explanation

Imagine a ruler laid vertically at 3000 cm⁻¹. For cyclohexane, all the spiky C–H dips hang to the right of the ruler. For methylbenzene, a small cluster of dips appears to the left of the ruler (aromatic C–H) and a larger cluster to the right (the CH₃ group).

Real-world analogy

Think of three springs made from the same wire but coiled more and more tightly. The tightest spring bounces fastest. Increasing s character works like tightening the coil: the C–H bond becomes stiffer, so it vibrates faster and absorbs at a higher wavenumber.

Real-world example

Polyethene and polystyrene can be distinguished in plastic recycling plants by IR sorting. Polyethene shows only sp³ C–H bands below 3000 cm⁻¹, while polystyrene's benzene rings add sp² C–H bands above 3000 cm⁻¹ and aromatic C=C bands near 1600 and 1500 cm⁻¹.

Why?

Why does more s character strengthen a bond? Electrons in s orbitals spend more time near the nucleus than electrons in p orbitals. A hybrid with more s character holds the bonding pair closer to the carbon nucleus, producing a shorter bond with stronger attraction between the nuclei and the shared electrons.

Common misconception

"Any band above 3000 cm⁻¹ must be O–H or N–H." Aromatic and alkene C–H bands sit between 3000 and 3100 cm⁻¹, and alkyne C–H near 3300 cm⁻¹. These are sharp and relatively narrow, unlike the broad O–H band.

Worked example

Question: Two isomers, C₆H₁₂, give IR spectra. Isomer P has C–H bands at 2850–2960 cm⁻¹ only. Isomer Q has bands at 2850–2960 cm⁻¹ and at 3080 cm⁻¹, plus a weak band at 1640 cm⁻¹. Identify the type of each.

Reasoning: C₆H₁₂ has one degree of unsaturation: a ring or a C=C bond. P has no sp² C–H and no C=C band, so the unsaturation must be a ring. Q has sp² C–H above 3000 cm⁻¹ and a C=C stretch at 1640 cm⁻¹.

Answer: P is a cycloalkane, such as cyclohexane; Q is an alkene, such as hex-1-ene.

Quick check

1. On which side of 3000 cm⁻¹ do the C–H stretches of an alkane appear, and why? Answer: Just below 3000 cm⁻¹, because sp³ C–H bonds have the lowest s character and are the weakest C–H bonds.

Exam focus

Remember the order sp > sp² > sp³ for C–H wavenumber and justify it with s character and bond strength. If asked to tell an alkane from an alkene, quote both the sp² C–H above 3000 cm⁻¹ and the C=C band near 1620–1680 cm⁻¹.

Advanced insight

The same s-character argument explains acidity: ethyne (pKa about 25) is far more acidic than ethene (about 44) or ethane (about 50). The carbanion formed by removing H⁺ holds its lone pair in an orbital with more s character, closer to the nucleus and better stabilised. IR and acidity both reflect the same underlying orbital effect.

Summary

C–H stretches depend on carbon hybridisation: sp³ at 2850–2960 cm⁻¹, sp² at 3000–3100 cm⁻¹ and sp near 3300 cm⁻¹. More s character gives shorter, stronger bonds and higher wavenumbers. The 3000 cm⁻¹ line separates saturated from unsaturated C–H, and aldehydes show a distinctive pair near 2720 and 2820 cm⁻¹.

Practice questions

1. Arrange sp, sp² and sp³ C–H bonds in order of increasing stretching wavenumber. Answer: sp³ (below 3000 cm⁻¹) < sp² (3000–3100 cm⁻¹) < sp (about 3300 cm⁻¹). 2. How could IR distinguish benzene from cyclohexane using only the C–H region? Answer: Benzene shows aromatic C–H bands just above 3000 cm⁻¹, whereas cyclohexane shows C–H bands only below 3000 cm⁻¹. 3. A spectrum shows a sharp, strong band at 3300 cm⁻¹ and a weak band at 2120 cm⁻¹. Suggest the functional group. Answer: A terminal alkyne: ≡C–H stretch at 3300 cm⁻¹ and C≡C stretch at 2120 cm⁻¹. 4. Which two weak bands help confirm an aldehyde, and what else must be present? Answer: Bands near 2720 and 2820 cm⁻¹ from the aldehyde C–H, together with a strong C=O band near 1720–1740 cm⁻¹.