IR Spectroscopy

Theory — IR Spectroscopy

1. What does IR measure?

Every molecule vibrates — bonds stretch and bend at characteristic frequencies. When infrared light passes through a sample, the molecule absorbs IR energy at frequencies that exactly match its vibrational frequencies. The transmitted light is missing those frequencies, and we see them as absorption peaks in the IR spectrum. Different bonds absorb at different frequencies; this is why IR is so useful for functional group identification.

The IR region of interest for organic chemistry is roughly 4000 cm⁻¹ (high frequency, short wavelength) to 400 cm⁻¹ (low frequency). The unit "wavenumber" (cm⁻¹) is the inverse of wavelength; higher wavenumber = higher energy. By convention, IR spectra are plotted with wavenumber DECREASING from left to right (4000 on the left, 400 on the right) and %transmittance on the y-axis (so peaks point DOWN, not up).

2. What determines a bond\'s vibrational frequency?

Two factors:

Bond strength. Stronger bonds vibrate at higher frequencies. Triple bonds (~2200 cm⁻¹) > double bonds (~1700 cm⁻¹) > single bonds (~1000 cm⁻¹).
Atomic masses. Lighter atoms vibrate faster (higher frequency). C-H bonds (light H, ~3000 cm⁻¹) vibrate faster than C-C (~1000) or C-Cl (~700).

The mathematical relationship is Hooke\'s law for diatomic vibration: ν = (1/2πc)√(k/μ), where k is the bond force constant (stiffness) and μ is the reduced mass of the atoms. This equation is mostly conceptual for our purposes; what matters is that you remember WHICH bonds appear in WHICH region.

3. The four diagnostic regions of an IR spectrum

A typical organic IR spectrum can be divided into four regions, each with a distinctive set of bonds:

Region (cm⁻¹)	What\'s vibrating	Diagnostic peaks
4000\u20132500	X-H stretches (X = O, N, C, S)	O-H broad ~3200-3550 (alcohols), O-H very broad ~2500-3300 (acids), N-H ~3300-3500, ≡C-H sharp ~3300, =C-H ~3030, sp³ C-H ~2850-2960
2500\u20132000	Triple bond stretches	C≡C ~2100-2260 (variable intensity), C≡N ~2210-2260 (sharp medium)
2000\u20131500	Double bond stretches	C=O ~1680-1800 (strong, the most diagnostic peak in organic IR), C=C alkene ~1640-1680 (often weak), C=C aromatic ~1450-1600 (often appears as a doublet)
1500\u2013400 (fingerprint)	Single-bond stretches and bending modes	C-O ~1000-1300, C-Cl ~600-800, C-N ~1180-1360, plus many compound-specific bends. This region is "the fingerprint" \u2014 it\'s unique to each compound but generally too cluttered to use peak-by-peak. Useful for identification by direct comparison with a reference spectrum.

Memorise these landmark peaks — 80% of IR diagnosis is based on them O-H broad: 3200\u20133550 cm⁻¹ (alcohols, water contamination)
O-H very broad: 2500\u20133300 cm⁻¹ (carboxylic acids \u2014 distinctive shape with C-H peaks underneath)
N-H: 3300\u20133500 cm⁻¹ (1° amines often show two peaks; 2° amines show one)
sp C-H: ~3300 cm⁻¹ (sharp; only terminal alkynes)
sp² =C-H: ~3030 cm⁻¹ (alkenes, aromatics)
sp³ C-H: 2850\u20132960 cm⁻¹ (almost everything organic)
C≡C: ~2100\u20132260 cm⁻¹ (weak)
C≡N: ~2210\u20132260 cm⁻¹ (sharp, medium intensity)
C=O: 1680\u20131800 cm⁻¹ (THE most diagnostic peak in organic IR)
C=C aromatic: 1450 + 1600 cm⁻¹ (doublet)
C-O: 1000\u20131300 cm⁻¹ (alcohols, ethers, esters)
NO₂: 1340 + 1520 cm⁻¹ (nitro groups; two strong peaks)
For most undergraduate IR work: identify the C=O region first, then check 4000\u20132500 for X-H, then check 2500\u20132000 for triple bonds.

4. Reading the C=O region carefully

The carbonyl (C=O) is THE most diagnostic peak in IR. Different carbonyl-containing compounds absorb at slightly different positions, and you can often distinguish them:

Compound class	C=O position (cm⁻¹)	Clue
Aliphatic aldehyde	~1720-1740	Plus C-H of CHO at ~2720 + ~2820 (Fermi doublet)
Aliphatic ketone	~1705-1720	C=O peak only; no special CH
Aliphatic carboxylic acid	~1700-1725	Plus the very broad O-H 2500-3300
Aliphatic ester	~1735-1750	C=O slightly higher than ketone; plus strong C-O 1100-1300
Aromatic carbonyl	~1680-1700	Conjugation lowers C=O frequency by ~20-30 cm⁻¹
Amide	~1640-1690	The lowest carbonyl; conjugation with N lone pair

5. Real vs simulated spectra in this lab

The instrument bench in this lab uses a mix of real reference spectra (annotated as REAL) and simulated spectra (annotated as SIMULATED). Real spectra are computer renderings of published IR data from the SDBS database, NIST Webbook, or Sigma-Aldrich; simulated spectra are constructed from known peak positions extrapolated to compounds where reference data is less accessible. Each spectrum in the lab is labeled at the top so you always know which you\'re looking at.

Why both? Real data teaches you what real spectra actually look like (slightly noisy, broad peaks, fingerprint clutter). Simulated data lets us cover compounds across the full curriculum even where high-quality reference spectra aren\'t freely available. Both modes show the same diagnostic peaks at the same positions; the difference is mainly in the level of background noise.

6. The systematic IR analysis approach

Here\'s the approach that works for most undergraduate IR puzzles:

Look at C=O region (1680-1800 cm⁻¹) first. If there\'s a strong peak there, the compound has a carbonyl. Then narrow it down: aldehyde, ketone, acid, ester, amide.
Check 3200-3600 region for O-H/N-H. Broad O-H (alcohol). Very broad O-H + C-H underneath (acid). Pair of peaks (1° amine). Single peak (2° amine).
Check 2500-2300 for triple bonds. C≡C (often weak). C≡N (sharp medium).
Check ~3030 cm⁻¹ for =C-H. Indicates alkene or aromatic. Combine with the doublet at 1450/1600 for aromatic.
Use the fingerprint region only for direct comparison with a reference spectrum, not for individual peak interpretation.

7. What IR can\'t tell you

IR is excellent at functional group identification but limited in other ways:

IR doesn\'t tell you the carbon skeleton or the molecular weight (use mass spec for that).
IR can\'t distinguish stereoisomers (need NMR or chiroptical methods).
IR doesn\'t give detailed connectivity (which atoms bond to which) \u2014 only what types of bonds are present.
For full structural identification, IR is one piece of the puzzle alongside NMR, mass spec, and UV/Vis.

Instructions

This lab\'s Simulation section has four parts. Complete them in order.

Section I — Diagnostic Peak Library. Eight peak-identification cases. For each: identify the wavenumber position; identify what bond is vibrating; identify the compound class.

Section II — IR Instrument Bench. Six unknown samples. For each: select the bottle, click "Inject & Scan", view the spectrum, identify the functional group(s), and verify using the peak labels (revealed after you answer).

Section III — Spectrum Interpretation. Eight harder puzzles: distinguishing similar compounds (alcohol vs amine, ketone vs aldehyde), interpreting multi-peak spectra, recognising contamination peaks.

Section IV — SDS & Microscale Sample Prep. Read SDS extracts for four IR materials (KBr salt plate, NaCl plate, IR-grade chloroform, CCl₄ or KBr disk; 16 questions). Then six microscale tests for IR sample preparation (KBr pellet, neat liquid film, ATR diamond, dry sample technique).

Prepare your lab notebook. Use the Example Report as your template.

Prerequisite: Familiarity with functional group structures (Functional Group Tests lab) and the major compound classes (Alcohols, Aldehydes & Ketones, Carboxylic Acids, Esters, Amines, etc.). The Spectroscopy quartet is most useful AFTER you\'ve completed those other labs.

Simulation

Four interactive parts. Use the ↺ Reset Simulation button at any time to clear all answers and start over.

Eight peak-identification cases. For each: (a) wavenumber range; (b) which bond is vibrating; (c) the compound class indicated.

Score: 0 / 24 (3 questions × 8 cases)

Six unknown samples. Select a sample bottle, click Inject & Scan, then identify the functional group from the spectrum. Peak labels appear after you answer.

Score: 0 / 6

Eight harder puzzles: distinguishing similar functional groups, interpreting multi-peak spectra, recognising contamination, predicting peaks of new compounds.

Score: 0 / 8

Round 1 — SDS interpretation

Four common IR sample preparation materials. Each has 4 questions.

SDS score: 0 / 16

Round 2 — Microscale sample prep

Six sample preparation scenarios. Identify the appropriate method.

Microscale score: 0 / 6

Team Questions

Discuss with your team before answering.

Question 1 — The most diagnostic peak. Which single peak position in an IR spectrum is the MOST informative for organic compounds, and why?

Question 2 — Acid vs alcohol. Both alcohols and carboxylic acids have an O-H stretch in the 2500-3550 region. How do you distinguish them by IR?

Question 3 — Aldehyde signature. Aldehydes have a C=O around 1720-1740, similar to ketones. What additional IR feature distinguishes an aldehyde from a ketone?

Question 4 — Conjugation effect. What happens to the C=O peak position when the carbonyl is in conjugation with a C=C or aromatic ring? Which direction does it shift?

Question 5 — Limitations. Name two important things IR spectroscopy CANNOT tell you about a molecule.

Question 6 — Sample contamination. A student\'s "neat liquid" IR spectrum shows a broad peak at 3300-3500 that they didn\'t expect. What is the most likely explanation, and how can they fix it?

Example Lab Notebook Entry

Use the format below as a template.

IR Spectroscopy — Lab Notebook Entry

Submitted by: [Student Name]

Course: Organic Chemistry I · Section: 201-A · Date: May 6, 2026

Objective

To learn the diagnostic peak positions of common organic functional groups in IR spectroscopy; to interpret IR spectra of unknown samples and identify their functional groups; to distinguish similar functional groups (alcohol vs amine, ketone vs aldehyde, ester vs ketone) by careful peak analysis; to understand the appropriate sample preparation method for IR (KBr pellet, neat liquid, ATR) and the SDS profile of common IR materials.

Diagnostic peak library (Section I)

Peak (cm⁻¹)	Bond / vibration	Compound class indicator
3200-3550 (broad)	O-H stretch (free + H-bonded)	Alcohol
2500-3300 (very broad)	O-H stretch (carboxylic acid, dimer)	Carboxylic acid (with C=O ~1710)
3300-3500 (one or two peaks)	N-H stretch (1° amine: 2 peaks; 2°: 1 peak)	Amine
~3300 (sharp)	sp C-H (terminal alkyne)	Terminal alkyne
~3030	sp² =C-H	Alkene or aromatic
2850-2960	sp³ C-H	Aliphatic (almost everything)
2210-2260	C≡N	Nitrile
1680-1800 (strong)	C=O	Carbonyl class (aldehyde 1720-1740, ketone 1705-1720, acid 1700-1725, ester 1735-1750, amide 1640-1690)

Instrument bench results (Section II)

Sample	Compound	Key peaks observed	Functional group identified
1	Ethanol	Broad 3300, sp³ C-H 2950, C-O 1050	Alcohol
2	Acetone	C-H 2900, strong C=O 1715	Ketone
3	1-Hexene	=C-H 3080, sp³ C-H 2950, C=C 1640	Alkene
4	1-Pentyne	Sharp ≡C-H 3300, C≡C 2120	Terminal alkyne
5	Benzoic acid	Very broad O-H 2500-3300, C-H underneath, C=O 1690, aromatic doublet 1450/1600	Aromatic carboxylic acid
6	Butyronitrile	sp³ C-H 2950, sharp C≡N 2250	Nitrile

Microscale sample prep results (Section IV)

Sample type	Recommended prep	Notes
Liquid (low viscosity)	Neat thin film between NaCl plates	One drop sandwiched, scan immediately
Solid (powder)	KBr pellet (1-2 mg sample : 100 mg KBr)	Grind, press at 8-10 ton in evacuated die
Solid (small amount)	ATR diamond crystal	No prep needed; press sample onto crystal; quick scan
Air-sensitive	Sealed cell, fast scan, dry N₂ purge	Avoid moisture; OH peak 3300 is contamination indicator
Aqueous solution	NOT IR-friendly without modification	Water has strong absorption everywhere; use ATR with thin film, or D₂O substitution
Crystalline organic	KBr pellet OR Nujol mull	Mull in liquid paraffin if pellet pressing fails

Discussion

IR spectroscopy detects molecular vibrations: every bond has a characteristic stretching or bending frequency in the IR region. The position depends on bond strength (stronger \u2192 higher frequency) and atomic mass (lighter \u2192 higher frequency). Triple bonds appear above 2000 cm⁻¹, double bonds in the 1500-2000 region, and single-bond stretches in the fingerprint region below 1500.

The carbonyl C=O is the single most diagnostic peak in organic IR. Its strong, distinctive absorption between 1680-1800 cm⁻¹ immediately tells you a carbonyl is present, and its precise position helps narrow down which kind: aldehyde 1720-1740, ketone 1705-1720, ester 1735-1750, amide 1640-1690. Conjugation with C=C or aromatic ring lowers the C=O frequency by 20-30 cm⁻¹.

The 3000-3500 region tells the second-most-important story. Broad O-H (alcohols) is centred ~3300; very broad O-H of carboxylic acids extends from 2500 right through the C-H region. N-H of primary amines appears as TWO peaks (asymmetric and symmetric N-H stretches); secondary amines show ONE peak; tertiary amines show no N-H stretch (no N-H bond). Identifying these features distinguishes the major hydrogen-bonded functional groups.

The instrument bench in Section II demonstrated this on six real samples. Ethanol shows the alcohol signature (broad O-H, C-O); acetone shows the ketone signature (sharp C=O alone); 1-pentyne shows the terminal alkyne signature (sharp =C-H AND C≡C); benzoic acid shows the carboxylic acid signature (very broad O-H, C=O, aromatic doublet); butyronitrile shows the nitrile signature (sharp C≡N at 2250). Each sample\'s spectrum is annotated with peak labels to make the diagnostic features explicit.

Section III\'s harder puzzles practiced distinguishing similar compounds: alcohol vs amine (broad O-H around 3300 vs N-H pair around 3400); aldehyde vs ketone (Fermi doublet at 2720+2820 vs single C=O); ester vs ketone (C=O position 1735-1750 vs 1705-1720, plus strong C-O 1100-1300 in ester). With practice, these distinctions become straightforward.

For sample preparation, the choice of method depends on the sample. Neat liquid film between NaCl plates is the fastest method for clean liquids. KBr pellet (sample ground in dry KBr, pressed into a transparent disc) is standard for solids. ATR (attenuated total reflectance) on a diamond crystal requires no preparation and is the modern method of choice in most labs. NaCl and KBr plates are themselves IR-transparent, but they react with water; CaF₂ plates are used when aqueous samples are unavoidable.

Conclusion

IR spectroscopy is the most accessible technique for functional group identification in organic chemistry. The systematic approach \u2014 check C=O region first, then 3000-3600 for X-H, then 2000-2500 for triple bonds, then sp² C-H ~3030 for alkene/aromatic \u2014 reliably identifies most functional groups with practice. IR is one of three techniques every chemistry student should master; combined with NMR and mass spectrometry, the three techniques together give a near-complete structural picture of an unknown compound. The next labs in this Spectroscopy quartet (Mass Spectrometry, NMR, UV/Vis) build on the same logic but probe different molecular properties.

References

1. Pavia, D. L.; Lampman, G. M.; Kriz, G. S.; Vyvyan, J. R. Introduction to Spectroscopy, 5th ed., Cengage, 2015, Ch 2.
2. Silverstein, R. M.; Webster, F. X.; Kiemle, D. J. Spectrometric Identification of Organic Compounds, 8th ed., Wiley, 2014, Ch 2.
3. SDBS — Spectral Database for Organic Compounds, AIST Japan: https://sdbs.db.aist.go.jp.
4. NIST Chemistry WebBook: https://webbook.nist.gov/chemistry.
5. Sigma-Aldrich SDS for benzoic acid (CAS 65-85-0), butyronitrile (CAS 109-74-0), KBr (CAS 7758-02-3), accessed online March 2026.

Practice Questions

Work through each before peeking at the hint.

Practice 1 — Reading wavenumber

An IR spectrum shows: broad peak 3200-3500, peaks 2900-3000, strong peak at 1715, peaks 1100-1300. What functional groups are likely present, and what compound class might this be?

Hint: Broad 3200-3500 = O-H of alcohol. 2900-3000 = sp³ C-H. 1715 = C=O of ketone or carboxylic acid (need to check 2500-3300 for very broad acid O-H). 1100-1300 = C-O. Combined: looks like a carboxylic acid (broad O-H + C=O 1715 + C-O 1300). The slightly lower wavenumber of the O-H (3200-3500) suggests this could also be an alcohol with separate C=O group, but most likely — given C=O at 1715 \u2014 a carboxylic acid like acetic acid, propanoic acid, etc.

Practice 2 — Distinguishing

Two unknown solids have similar molecular formulas. One has IR peaks at 3380, 3300 (two peaks both medium), 1620 (medium); the other has peaks at 3350 (broad), 1715 (strong). What are they?

Hint: First sample: TWO peaks at 3380 and 3300 = primary amine (1° amines show two N-H stretches: asymmetric and symmetric). The 1620 is N-H bending. So this is a primary amine \u2014 RNH₂. Second sample: ONE broad peak at 3350 + strong C=O at 1715 = the C=O is a ketone or aldehyde or acid. The 3350 broad is O-H of alcohol. Combined: hydroxy ketone or hydroxy acid (or actually, the combination suggests a ketone with separate alcohol elsewhere on the molecule). The most diagnostic distinction: TWO peaks at 3300-3400 = primary amine; ONE broad peak at 3300-3550 = alcohol.

Practice 3 — Carbonyl identification

An IR spectrum shows a strong peak at 1735 cm⁻¹ and another strong peak at 1230 cm⁻¹. What carbonyl-containing compound class is this most likely to be?

Hint: C=O at 1735 is in the ester range (1735-1750). The strong peak at 1230 is C-O stretch \u2014 esters always have a strong C-O around 1100-1300. The combination C=O 1735 + strong C-O 1230 strongly suggests an ester (R-COO-R\'). Compare: ketones have C=O ~1715 (lower) and no strong C-O peak; carboxylic acids have C=O 1700-1725 plus the very broad O-H 2500-3300 (which the question doesn\'t mention, ruling out acid).

Practice 4 — Aromatic identification

A liquid sample shows IR peaks at 3030, 1605, 1500, 750, 700. What functional group does this suggest, and what specific feature indicates it?

Hint: 3030 = sp² =C-H stretches. 1605 + 1500 = the AROMATIC C=C doublet (the most diagnostic feature of aromatic compounds in IR \u2014 a pair of medium peaks at ~1500 + 1600). 750 + 700 = aromatic C-H out-of-plane bending pattern (specifically, this 750+700 combination is characteristic of MONO-substituted benzene like toluene, ethylbenzene). So this is mono-substituted benzene. Different substitution patterns give different out-of-plane bending: ortho ~750 (single peak); meta ~780 + 690; para ~810. The OOP bending pattern is diagnostic for substitution.

Practice 5 — Aldehyde signature

A liquid shows C=O at 1725 and TWO unusual peaks at 2720 and 2820 cm⁻¹ (medium intensity, in the C-H region but well below the typical sp³ C-H range). What does this indicate?

Hint: The two peaks at 2720 and 2820 are the characteristic FERMI DOUBLET of an aldehyde C-H. The aldehyde\'s -CHO has a unique C-H stretch around 2700-2900 that splits into two peaks via Fermi resonance with an overtone of the C-H bend. C=O at 1725 is consistent with aldehyde. So this is an aldehyde (R-CHO), like benzaldehyde, propanal, or hexanal. The Fermi doublet is the diagnostic feature that distinguishes aldehydes from ketones (which lack it because they have no C-H attached to the C=O).

Practice 6 — Predicting peaks

Predict 3-4 key IR peaks you would expect to see for: (a) acetic acid (CH₃COOH); (b) acetonitrile (CH₃CN); (c) p-cresol (4-methylphenol).

Hint: (a) Acetic acid: sp³ C-H 2950, very broad O-H 2500-3300 (overlapping the C-H), C=O 1715, C-O 1280. (b) Acetonitrile: sp³ C-H 2950, sharp medium C≡N 2255, C-N 1040. (c) p-cresol: sp³ C-H 2920, broad O-H ~3250, sp² =C-H 3030, aromatic doublet 1500 + 1600, p-substitution OOP bend ~810.

Practice 7 — Contamination

A student\'s IR spectrum of a "pure ester" shows: expected ester C=O at 1735, expected C-O at 1230, but ALSO an unexpected very broad peak at 3300-3500. What is the most likely explanation, and how would you fix it?

Hint: The very broad peak at 3300-3500 is most likely WATER CONTAMINATION (H₂O has two strong stretches in this region). Alternative: hydrolysed ester (so a small amount of acid + alcohol present). Fix: dry the sample (anhydrous Na₂SO₄ or MgSO₄ before scanning); use freshly distilled solvent; avoid moisture by capping vials and the IR sample cell quickly; if KBr pellet, ensure KBr is dry (heat at 110°C overnight) or use ATR to avoid the issue. Water absorption is the #1 contamination problem in IR spectroscopy.

Practice 8 — Conjugation effect

Compare the expected C=O frequency of: (a) cyclohexanone (saturated cyclic ketone); (b) 2-cyclohexen-1-one (conjugated cyclic ketone); (c) methyl vinyl ketone, CH₂=CH-CO-CH₃ (conjugated open-chain ketone). Predict the order of decreasing C=O frequency.

Hint: Conjugation with C=C lowers C=O frequency by 20-30 cm⁻¹ (because the C=O bond order is partially reduced by sharing density with the conjugated π system). So: cyclohexanone (no conjugation) ~1715; methyl vinyl ketone (conjugated) ~1680; 2-cyclohexenone (conjugated, in ring) ~1680. Order of decreasing frequency: cyclohexanone > 2-cyclohexenone ≈ methyl vinyl ketone (the conjugated ones are about the same; both lower than non-conjugated). General rule: each additional conjugation lowers C=O by ~20-30 cm⁻¹; aromatic carbonyls (like acetophenone) show this clearly.

Practice 9 — Nitro vs nitrile

Both nitro compounds (R-NO₂) and nitriles (R-C≡N) contain a C-N bond and might be confused. How do their IR signatures differ, and what positions distinguish them?

Hint: Nitrile (R-C≡N): ONE sharp medium peak at 2210-2260 cm⁻¹ (the C≡N stretch). Located in the triple-bond region. Nitro (R-NO₂): TWO STRONG peaks at ~1340 and ~1520 cm⁻¹ (asymmetric and symmetric N=O stretches). Located in the double-bond region. Completely different positions and pattern. Plus the nitro\'s two peaks are STRONG and characteristic; the nitrile\'s single peak is MEDIUM and sharp. Easy to tell apart on a spectrum.

Practice 10 — Big picture

An organic chemistry student isolates a clear liquid product after a synthesis. They run an IR and observe: sp³ C-H 2920, NO peak in 1680-1800 region, NO broad O-H or N-H peaks, peaks at 1450 + 1505 (medium), peak at 740. Beyond identifying that this is "an aromatic with sp³ carbons", what specific kind of substituted benzene might this be?

Hint: sp³ C-H + aromatic doublet (1450/1505 + 740 = mono-substituted benzene by OOP bending) + no functional group in 1680-1800 = an alkylbenzene. The 740 (single OOP peak) suggests ortho-disubstitution actually, OR the 740 alone (without 700) suggests a particular pattern. With the 1450/1505 in the doublet, plus the absence of any other functional group, this looks like an alkylbenzene like toluene, ethylbenzene, xylene, etc. The absence of a broad O-H or C=O rules out phenol or aromatic acid/ester. So: alkylbenzene with no other functional groups. The full identification (which alkylbenzene specifically) needs the molecular weight from mass spec or H NMR for chain length.