Ep 08 — Classic Textbook Exam Questions: Typical Molecular Spectrum Interpretation

Series: Encyclopedia of Infrared Spectroscopy: From Principle to Practice
Chapter: Part 1 · Introduction — The Code of Light
Target Audience: High school students, undergraduates, beginners in chemistry/materials/pharmacy
Prerequisites: Ep 04 (How to Read an Infrared Spectrum), Ep 05–07 (Characteristic Frequencies of Functional Groups and Fingerprint Region)
Reading Time: Approximately 28 minutes

Introduction: From "Seeing Peaks" to "Deducing Structures"

In previous episodes, we learned the basic concepts of infrared spectroscopy (Ep 04), characteristic absorption frequencies of functional groups (Ep 05–06), and fingerprint region discrimination rules (Ep 07). Now it's time to assemble these "parts" into a real "molecular inference machine."

The core of spectrum interpretation is not memorizing peak positions one by one, but establishing a thought chain of "characteristic peak → functional group → structural fragment → entire molecule" [1][2]. The starting point of this chain is the most prominent characteristic peak, and the endpoint is the complete structure of the molecule, with intermediate steps requiring interference elimination, cross-validation, and reasonable inference.

This episode first uses a complete five-step process to elaborate on three high-frequency exam molecules—benzoic acid, aniline, and acetamide; then supplements the plan with short cases for ethanol, acetone, ethyl acetate, polyethylene (PE), and polystyrene (PS), forming a spectrum reading ladder from small molecules to polymers [1][2][3].

🔗 During interpretation, you can refer to: carboxyl, amine, amide, carbonyl, hydroxyl, ester, alkyl C-H, aromatic ring.

I. Systematic Workflow for Spectrum Interpretation

Before diving into specific molecules, let's establish a general interpretation workflow [1][2]:

1.1 Five-Step Interpretation Method

Table 1: Five-step interpretation method (Data source: LibreTexts [1][2])

1.2 Common Pitfalls

The most common "traps" in exam questions are three [1][2][3]:

1. Confusing C=O and C=C: Both lie in 1600–1750 cm⁻¹, but C=O is usually stronger and sharper
2. Ignoring hydrogen bonding effects: Hydrogen bonding broadens O–H, N–H peaks and shifts them to lower wavenumbers
3. Misinterpreting water peaks: Peaks from KBr moisture absorption at 3400 (O–H) and 1640 (H–O–H) are easily mistaken as sample characteristics

With this workflow, we move to the first molecule.

II. Benzoic Acid (C₆H₅COOH) — Classic Case of Carboxylic Acid Dimer

2.1 Molecular Structure and Key Functional Groups

Benzoic acid (CAS 65-85-0) is the simplest aromatic carboxylic acid, with molecular formula C₇H₆O₂ and molecular weight 122.12 [4]. Its structure consists of a benzene ring directly connected to a carboxyl group (–COOH):

    COOH
     |
  /---\
 |     |       Benzoic acid: benzene ring + carboxyl group
  \---/

Key functional groups [3][5]:

• Carboxyl group –COOH: Contains C=O, C–O, and O–H bonds, dominating the spectrum
• Benzene ring: Monosubstituted aromatic ring, contributing aromatic C=C and C–H features
• Carboxylic acid dimer: In solid/liquid state, two carboxylic acid molecules form a dimer via hydrogen bonds, profoundly affecting the spectrum

2.2 Main Infrared Absorption Peaks

According to NIST Chemistry WebBook standard spectra [4], LibreTexts textbooks [1][2], and BenchChem application notes [3][5], the main infrared absorption peaks of benzoic acid are as follows:

Table 2: Main infrared absorption peaks of benzoic acid (Data sources: NIST [4], LibreTexts [1][2], BenchChem [3][5])

2.3 Standard Infrared Spectrum

📷 Figure 1: Standard infrared spectrum of benzoic acid (NIST)
Source: NIST Chemistry WebBook [4]
https://webbook.nist.gov/cgi/…

NIST provides spectra for different phases [4]:

• Solid (split mull): 2 cm⁻¹ resolution, Coblentz Society data
• Solution (2% CCl₄/CS₂): 2 cm⁻¹ resolution
• Vapor (160 °C): 2–4 cm⁻¹ resolution

This interpretation focuses on the solid-state spectrum, as it is the most commonly examined and practical phase.

2.4 Interpretation Thought Chain

Step ① Overview the entire spectrum [1][3]:
Scanning the full spectrum, the most striking feature is a very broad strong peak at 2500–3300 cm⁻¹ — the "trademark" of carboxylic acid O–H stretching. Next, a strong sharp peak at ~1690 cm⁻¹ corresponds to C=O stretching. Together, these two peaks almost confirm that the molecule contains a carboxyl group –COOH [1][2].

Step ② High-frequency region analysis [1][2]:

• 2500–3300 cm⁻¹ very broad peak → O–H stretching (hydrogen-bonded dimer). Note: Alcohol O–H usually appears at 3200–3600 cm⁻¹ and is narrower; carboxylic acid O–H extends significantly to lower wavenumbers, even covering the C–H peaks around 3000 cm⁻¹, a characteristic of hydrogen-bonded dimers [1][2][5]
• ~3070 cm⁻¹ weak peak → Aromatic =C–H stretching (>3000 cm⁻¹ indicates unsaturated C–H) [1]

Step ③ Double bond region analysis [3][5]:

• ~1690 cm⁻¹ strong peak → C=O stretching. Key judgment: 1690 cm⁻¹ is lower than typical saturated carboxylic acids (~1715 cm⁻¹), due to two reasons [2][5]:
  • Conjugation effect: The benzene ring conjugates with the C=O, reducing the C=O bond order
  • Hydrogen bonding effect: Dimer hydrogen bonding further weakens the C=O
• 1600, 1580, 1500 cm⁻¹ medium peaks → Aromatic C=C skeletal stretching, confirming the presence of a benzene ring [1]

Step ④ Fingerprint Region Analysis [3][6]:

• ~1290 cm⁻¹ → C–O stretching (carboxylic acid)
• ~920 cm⁻¹ → O–H out-of-plane bending, auxiliary diagnostic peak for carboxylic acid dimer [3][5]
• ~750 and ~700 cm⁻¹ double strong peaks → aromatic C–H out-of-plane bending, definitive evidence for monosubstituted benzene ring (ca. 770–730 + 710–690 cm⁻¹; see Ep 07) [6]

Step ⑤ Cross-Validation [1][3]:

• Carboxyl group (O–H + C=O + C–O) all three peaks present ✓
• Aromatic ring (C=C + C–H + monosubstitution pattern) evidence complete ✓
• Hydrogen-bonded dimer (very broad O–H + lowered C=O + 920 bending) self-consistent ✓

Conclusion: The molecule is benzoic acid (benzene ring + carboxyl group, monosubstituted) [1][3].

💡 Exam Tip: The C=O of benzoic acid is at ~1690 cm⁻¹, while that of benzaldehyde is at ~1705 cm⁻¹. The difference comes from the additional weakening by hydrogen bonding in the carboxylic acid dimer [2][5].

III. Aniline (C₆H₅NH₂) — Textbook Demonstration of the Primary Amine Doublet

3.1 Molecular Structure and Key Functional Groups

Aniline (CAS 62-53-3) is the simplest aromatic primary amine, with molecular formula C₆H₇N and molecular weight 93.13 [7]. Its structure consists of a benzene ring attached to an amino group (–NH₂):

    NH2
     |
  /---\
 |     |      Aniline: benzene ring + primary amino group
  \---/

Key functional groups [7][8][9]:

• Primary amino –NH₂: Contains two N–H bonds, producing characteristic "double peaks"
• Benzene ring: Monosubstituted aromatic ring
• C–N bond: Connects amino group to benzene ring; C–N frequency in aromatic amines is higher than in aliphatic amines

3.2 Major Infrared Absorption Peaks

Based on the NIST Chemistry WebBook standard spectrum [7], LibreTexts organic chemistry textbook [8], and OrgChemBoulder tutorial [9], the major IR absorption peaks of aniline are as follows:

Table 3: Major IR absorption peaks of aniline (data sources: NIST [7], LibreTexts [8], OrgChemBoulder [9])

3.3 Standard Infrared Spectrum

📷 Figure 2: Standard infrared spectrum of aniline (NIST)
Source: NIST Chemistry WebBook [7]
https://webbook.nist.gov/cgi/…

NIST provides spectra in both neat liquid and solution (CCl₄/CS₂) forms [7]. In the neat liquid spectrum, hydrogen bonding shifts the N–H peaks slightly to lower wavenumbers [8].

3.4 Interpretation Thought Chain

Step ① Overall Spectrum Survey [8][9]:
The most prominent features are two medium-intensity sharp peaks in the 3300–3500 cm⁻¹ region—this is the "doublet fingerprint" of primary amine N–H stretching. Unlike the very broad O–H peak of carboxylic acids, N–H peaks are noticeably narrower and weaker, which is key to distinguishing amines from alcohols/acids [8][9].

Step ② High-Frequency Region Analysis [8][9]:

• ~3430 and ~3350 cm⁻¹ doublet → N–H asymmetric and symmetric stretching. Doublet = primary amine (–NH₂); singlet = secondary amine (–NH–); no peak = tertiary amine (–N<) [8][9]
• Key discrimination: O–H also appears at 3200–3600 cm⁻¹ but is broader and stronger. N–H peaks are "narrow and weak", characteristic of amines [9]
• Weak shoulder at ~3200 cm⁻¹ → Fermi resonance between the overtone of N–H bending (~1620) and N–H stretching [8]
• ~3030 cm⁻¹ → Aromatic =C–H stretching

Step ③ Double Bond Region Analysis [8][9]:

• ~1620 cm⁻¹ → N–H bending (scissoring vibration). Caution: This peak is close to the C=O region and may be mistaken for a carbonyl! The difference: N–H bending is medium intensity and broader, while C=O is usually stronger and sharper [9]
• 1600, 1500 cm⁻¹ → Aromatic C=C skeleton stretching, confirming the presence of a benzene ring [1]

Step ④ Fingerprint Region Analysis [8][9]:

• ~1280 cm⁻¹ → C–N stretching. C–N in aromatic amines appears at 1250–1335 cm⁻¹, higher than in aliphatic amines (1000–1250 cm⁻¹) because the C–N bond in aromatic amines has partial double bond character [8][9]
• ~750/~700 cm⁻¹ double strong peaks → Aromatic C–H out-of-plane bending, evidence for monosubstituted benzene ring (ca. 770–730 + 710–690) [6]

Step ⑤ Cross-Validation [8][9]:

• Primary amine (N–H doublet + N–H bending + C–N) three features complete ✓
• Aromatic ring (C=C + C–H + monosubstitution pattern) evidence complete ✓
• No O–H or C=O carbonyl peaks → eliminates carboxylic acid and amide possibilities ✓

Conclusion: The molecule is aniline (benzene ring + primary amino group, monosubstituted) [8][9].

💡 Exam Tip: The N–H doublet of aniline (~3430/~3350) differs in position from that of acetamide (~3350/~3180)—the N–H frequency in aniline is higher because the lone pair on nitrogen in amides is delocalized into the C=O, weakening the N–H bond [2][8].

IV. Acetamide (CH₃CONH₂) — The Best Teaching Molecule for Amide I/II Bands

4.1 Molecular Structure and Key Functional Groups

Acetamide (CAS 60-35-5) is one of the simplest amides, with molecular formula C₂H₅NO and molecular weight 59.07 [10]. Its structure consists of a methyl group attached to an amide group (–CONH₂):

  CH3-C(=O)-NH2      Acetamide: methyl + amide group

Key functional groups [10][11][12]:

• Amide group –CONH₂: C=O and –NH₂ are resonance-coupled, producing characteristic "amide I/II bands"
• Methyl –CH₃: C–H stretching and bending
• Primary amide: Contains two N–H bonds, giving a doublet

4.2 Major Infrared Absorption Peaks

Based on the NIST Chemistry WebBook standard spectrum [10], LibreTexts organic chemistry textbook [2], and BenchChem application notes [11][12], the major IR absorption peaks of acetamide are as follows:

Table 4: Main infrared absorption peaks of acetamide (data sources: NIST [10], LibreTexts [2], BenchChem [11][12])

4.3 Standard Infrared Spectrum

📷 Fig. 3: Standard infrared spectrum of acetamide (NIST)
Source: NIST Chemistry WebBook [10]
https://webbook.nist.gov/cgi/…

NIST provides the solid-state (split mull, fluorochlorocarbon oil + paraffin oil) spectrum [10].

4.4 Analytical Chain of Thought

Step ① Overall Spectrum Overview [2][11]:
The most prominent features are a strong sharp peak at ~1690 cm⁻¹ (amide I band) and a medium peak at ~1620 cm⁻¹ (amide II band), close but distinguishable. There is a doublet near 3300 cm⁻¹, attributed to N–H stretching [2].

Step ② High-Frequency Region Analysis [2][11][12]:

• Doublet at ~3350 and ~3180 cm⁻¹ → N–H asymmetric and symmetric stretching. Doublet = primary amide (–CONH₂) [11][12]
• Key comparison: The N–H doublet frequency of acetamide is lower than that of aniline (~3430/~3350), because the lone pair electrons on the N in the amide are delocalized to C=O, weakening the N–H bond [2][11]
• ~2960 cm⁻¹ → Methyl C–H stretching

Step ③ Double Bond Region Analysis [2][11][12]:

• Strong peak at ~1690 cm⁻¹ → Amide I band (C=O stretching). Key: Amide C=O frequency is lower than that of ketones/aldehydes (~1715), because [2][11]:
  • Resonance effect: Lone pair electrons on N delocalize to C=O, giving the C–O bond partial single-bond character
  • Hydrogen bonding effect: Hydrogen bonding in the solid state further lowers the frequency
• Medium peak at ~1620 cm⁻¹ → Amide II band (N–H bending coupled with C–N stretching). The amide II band is a decisive feature of amides; ketones/aldehydes lack this peak [2][11]

Step ④ Fingerprint Region Analysis [11][12]:

• ~1400 cm⁻¹ → C–N stretching
• ~600–800 cm⁻¹ → N–H out-of-plane wagging, characteristic of primary amides

Step ⑤ Cross-Validation [2][11]:

• All three characteristics of a primary amide (N–H doublet + amide I + amide II) are present ✓
• No aromatic C=C, C–H → excludes aromatic compounds ✓
• No broad O–H peak → excludes carboxylic acids ✓
• Methyl C–H present → consistent with CH₃– ✓

Conclusion: The molecule is acetamide (methyl + primary amide group) [2][11].

💡 Exam Tip: Both acetamide and benzoic acid have a strong peak at ~1690 cm⁻¹, but [2][5][11]:

• Benzoic acid: very broad O–H peak at 2500–3300 cm⁻¹ + aromatic C=C at 1600/1580/1500
• Acetamide: N–H doublet at 3350/3180 + amide II band at 1620 + no aromatic peaks
  This comparison is a frequent exam trap.

V. Cross-Comparison of Three Molecules—Exam Traps and Differentiation Strategies

5.1 Comparison Table of Characteristic Peaks

Table 5: Comparison of characteristic peaks for three molecules

5.2 Analysis of Common Traps

Trap 1: Confusion between C=O and C=C [1][2]

• C=O (~1690–1715) and aromatic C=C (~1600) are close
• Differentiation: C=O is stronger and sharper; aromatic C=C usually appears as a group (1600+1580+1500)

Trap 2: Confusion between N–H and O–H [8][9]

• Both appear in the 3300–3500 cm⁻¹ region
• Differentiation: O–H is broader and stronger (especially carboxylic acid O–H covering 2500–3300); N–H is narrower and weaker

Trap 3: N–H doublet of aniline vs. N–H doublet of acetamide [2][8][11]

• Aniline: ~3430/~3350 (higher frequency)
• Acetamide: ~3350/~3180 (lower frequency, resonance weakens N–H)
• Auxiliary differentiation: acetamide has amide I at ~1690 + amide II at ~1620; aniline has aromatic C=C

Trap 4: Water peak interference [1]

• KBr hygroscopicity: 3400 (O–H) + 1640 (H–O–H)
• Identification: measure pure KBr background; if the same peaks appear, they are water peaks

VI. Short Cases: Ethanol, Acetone, Ethyl Acetate, PE, PS

The complete five-step procedure has been demonstrated with the three molecules above. Below, short cases using “key peaks → conclusion” supplement common textbook examples (peak positions are typical values; actual measurements may shift by several wavenumbers) [1][2].

6.1 Ethanol (CH₃CH₂OH)

• Broad strong peak at 3200–3600 cm⁻¹ → Hydrogen-bonded O–H (hydroxyl)
• ~2970/2930/2870 → Alkyl C–H (alkyl C-H)
• ~1050–1100 → C–O stretching
• No strong sharp peak at ~1700 → excludes ketones/aldehydes/esters/carboxylic acids
• Conclusion: Saturated alcohol

6.2 Acetone (CH₃COCH₃)

• Strong sharp peak at ~1715 cm⁻¹ → Ketone C=O (carbonyl)
• ~1360 vicinity → CH₃ bending often enhanced
• No broad O–H peak, no ester C–O doublet pattern → distinguishes from carboxylic acids/esters
• Conclusion: Simple ketone

6.3 Ethyl Acetate (CH₃COOCH₂CH₃)

• Strong peak at ~1740 cm⁻¹ → Ester C=O (usually higher than ketones; ester)
• ~1240 and ~1050 region → C–O–C related absorptions
• No very broad O–H at 2500–3300 → distinguishes from carboxylic acids
• Conclusion: Saturated ester

6.4 Polyethylene (PE)

• ~2915/2848 → Dominant CH₂ stretching; almost no aromatic, no C=O
• ~1465, ~720 → CH₂ bending and long-chain rocking (see Ep 07)
• Conclusion: Aliphatic polyolefin backbone (alkyl C-H)

6.5 Polystyrene (PS)

• >3000 weak–medium + 1600/1490 region → Aromatic ring (aromatic)
• ~750/~700 → Out-of-plane bending of monosubstituted benzene
• Alkyl C–H still present, but distinct from the “pure aliphatic chain” of PE
• Conclusion: Vinyl polymer containing benzene ring; PS film is often used for wavenumber calibration (see Ep 50), here only as an identification fingerprint

Summary of This Episode

| Aniline | ~3430/~3350 N–H doublet + ~1620 N–H bending + ~750/~700 monosubstitution |
| Acetamide | ~3350/~3180 N–H doublet + ~1690 amide I + ~1620 amide II |
| Carboxylic acid O–H feature | Very broad (2500–3300), hydrogen-bonded dimer, covers C–H region |
| Primary amine N–H doublet | Asymmetric + symmetric stretching, doublet = primary amine |
| Amide I/II bands | I=C=O stretching (~1690); II=N–H bending + C–N stretching (~1620) |
| Monosubstituted benzene ring | Around 770–730 + 710–690 cm⁻¹ (often noted ~750/~700) |
| Hydrogen bonding effect | Broadens O–H, N–H and shifts to lower wavenumber |
| Resonance effect | Amide C=O lower than ketone/aldehyde; aromatic amine C–N higher than aliphatic amine |
| Ethanol/Acetone/Ethyl acetate | Broad O–H + no C=O / strong C=O ~1715 / ester C=O ~1740 + C–O |
| PE / PS | Almost only alkyl C–H / aromatic ring + monosubstituted fingerprint |

Thought Questions

1. An unknown compound spectrum shows a very broad strong peak at 2500–3300 cm⁻¹, a strong sharp peak at ~1690 cm⁻¹, and double strong peaks at ~750/~700 cm⁻¹. Deduce the molecule and explain the basis.
2. How to distinguish primary, secondary, and tertiary amines using only the 3300–3500 cm⁻¹ region?
3. Why is the N–H doublet of acetamide (~3350/~3180) at lower frequency than that of aniline (~3430/~3350)? Explain from an electronic structure perspective.
4. A spectrum shows peaks at ~1690 and ~1620 cm⁻¹. How to determine if it is an amide or other carbonyl compound?
5. Benzoic acid's C=O (~1690) is lower than that of saturated carboxylic acids (~1715). What are the reasons?
6. In a KBr pellet spectrum, peaks at 3400 and 1640 cm⁻¹ appear. Are they necessarily sample characteristics? How to distinguish?
7. Design a procedure to distinguish benzoic acid, aniline, and acetamide (three white solids) using infrared spectroscopy.

References

Spectrum Interpretation Methods

[1] LibreTexts. "13.3: Interpreting Infrared Spectra." Cañada College CHEM 231.
https://chem.libretexts.org/C…

[2] LibreTexts. "21.10: Spectroscopy of Carboxylic Acid Derivatives." Organic Chemistry (Morsch et al.).
https://chem.libretexts.org/B…

Benzoic Acid

[3] BenchChem. "Core Principles of Infrared Spectroscopy of Benzoic Acid Derivatives." May 2026.
https://pdf.benchchem.com/29/…

[4] NIST Chemistry WebBook. Benzoic acid (CAS 65-85-0). IR Spectrum.
https://webbook.nist.gov/cgi/…

[5] LibreTexts. "9.9: Spectroscopy of Carboxylic Acids and Nitriles." Shasta College Organic Chemistry II.
https://chem.libretexts.org/C…

Aniline

[6] LibreTexts. "15.7: Spectroscopy of Aromatic Compounds."
https://chem.libretexts.org/@…

[7] NIST Chemistry WebBook. Aniline (CAS 62-53-3). IR Spectrum.
https://webbook.nist.gov/cgi/…

[8] LibreTexts. "25.5 Spectroscopic Properties." Amines.
https://chem.libretexts.org/@…

[9] OrgChemBoulder (University of Colorado Boulder). "IR Spectroscopy Tutorial: Amines."
https://www.orgchemboulder.co…

Acetamide

[10] NIST Chemistry WebBook. Acetamide (CAS 60-35-5). IR Spectrum.
https://webbook.nist.gov/cgi/…

[11] BenchChem. "Spectroscopic Comparison of Acetamide and Its Isomers." April 2026.
https://pdf.benchchem.com/32/…

[12] BenchChem. "A Senior Application Scientist's Guide to Primary Amide Functional Group Validation Using IR Spectroscopy." May 2026.
https://pdf.benchchem.com/246…

Database Resources

[SDBS] AIST, Japan. "Spectral Database for Organic Compounds (SDBS)."
https://sdbs.db.aist.go.jp/sd…

[NIST] NIST Chemistry WebBook.

https://webbook.nist.gov/chem…

[ftir.fun] ftir.fun FTIR Functional Group and Peak Position Database.
https://ftir.fun ; Examples: carboxyl, amine, amide, ester, aromatic ring

Next Episode Preview: Ep 09 — IR vs Raman: Both Vibrational Spectroscopy, How Are They Different?
We will compare the principles, selection rules, and complementarity of infrared absorption and Raman scattering, explain why some information is only provided by IR and some only by Raman, and how to choose the appropriate technique for different samples.

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