Organic Chemistry · Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
Spectroscopy of Carboxylic Acid Derivatives
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In 30 seconds
Spectroscopy is how chemists identify carboxylic acid derivatives — and, just as important, how they tell the derivatives apart. Infrared (IR) spectroscopy is the workhorse: the carbonyl (C=O) stretching band appears in a characteristic window for each derivative, its position set by how strongly the attached heteroatom donates electrons into the π system. Nuclear magnetic resonance (NMR) details the alkyl groups and N–H/O–H protons; mass spectrometry (MS) gives the molecular mass and fragments.
The central idea of this topic: the more resonance donation the carbonyl receives from its neighbor (N > O > S > Cl), the weaker the C=O bond and the lower its stretching frequency. Amides absorb near 1650–1690 cm⁻¹, esters near 1735–1750 cm⁻¹, and acid chlorides near 1800 cm⁻¹. Learn the pattern, not just the numbers, and you can assign any unknown.
Why this matters
- Structure determination is a core skill in research, pharmaceutical quality control, forensics, and natural-product chemistry — unknowns are solved by combining IR, NMR, and MS.
- Isomer discrimination. Acid derivatives with the same molecular formula (esters vs. acids vs. amides) are distinguished quickly by IR.
- Exam relevance. "Given the IR/NMR/MS data, identify the compound" problems are standard on organic chemistry exams.
The college version
Core Concepts
IR: the C=O stretching frequency as a fingerprint
Typical carbonyl stretching frequencies (cm⁻¹):
| Derivative | C=O stretch (cm⁻¹) | Notes |
|---|---|---|
| acid chloride | ~1795–1815 | Cl withdraws electrons, raising the frequency |
| anhydride | ~1810–1820 and ~1750–1760 | two bands (symmetric and antisymmetric) |
| ester | ~1735–1750 | lactones higher (ring strain) |
| carboxylic acid | ~1710–1720 | plus a very broad O–H band, 2500–3300 cm⁻¹ |
| thioester | ~1690–1710 | less resonance stabilization than an O-ester |
| amide | ~1650–1690 | amide I band; lowest of the common derivatives |
Trend: more resonance donation from the heteroatom lowers the C=O frequency. Nitrogen's lone pair donates most strongly, so amides absorb lowest; chlorine withdraws rather than donates, so acid chlorides absorb highest. Ring strain (β-lactams, lactones) raises the frequency.
IR: N–H and O–H bands
- Primary amides (RCONH2) show TWO N–H stretching bands near 3300–3500 cm⁻¹ (symmetric and antisymmetric N–H stretches); secondary amides show one.
- Carboxylic acids show a very broad O–H band (2500–3300 cm⁻¹) from hydrogen bonding, overlapping the C–H region — a strong clue for an acid.
¹H NMR
- Methyl esters: OCH3 singlet near 3.6–3.8 ppm.
- Amides: N–CH3 singlet near 2.8–3.0 ppm; N–H protons appear 5–8 ppm as broad, exchangeable signals that vanish when D2O is added.
- α-Hydrogens next to the carbonyl appear 2.0–2.5 ppm; carboxylic acid O–H appears 10–12 ppm (broad).
¹³C NMR
Carbonyl carbons resonate from about 165 to 215 ppm, but the ranges overlap heavily: amides ~165–175, esters ~165–175, acids ~170–185, acid chlorides ~165–170, anhydrides ~165–170, thioesters ~190–200, and ketones/aldehydes ~195–215. Because of this overlap, ¹³C alone rarely identifies the derivative — pair it with IR.
Mass spectrometry
- The molecular ion (M⁺) gives the molar mass; the nitrogen rule Odd nominal molecular mass implies odd number of N Full entry → (an odd nominal mass implies an odd number of N) flags amides.
- McLafferty rearrangement MS fragmentation of carbonyl compounds with a γ-H Full entry →: esters and amides with a γ-hydrogen on the alkyl chain rearrange to a characteristic fragment — e.g., methyl butanoate (M⁺ 102) gives a strong peak at m/z 74.
A structure-elucidation workflow
- Compute the degrees of unsaturation from the formula.
- Use IR to identify the functional group (and any N–H/O–H).
- Use ¹H NMR integrations and splitting to map the alkyl groups.
- Count unique carbons in ¹³C NMR (watch for symmetry).
- Confirm with MS (M⁺ and fragments), then assemble the structure.
How It Works / Step-by-Step Process
Identifying an unknown acid derivative:
- Record the IR spectrum; note the C=O frequency and any N–H or O–H bands.
- Compare the C=O position against the derivative table to propose a class (amide ~1660, ester ~1740, acid ~1715 + broad O–H, etc.).
- Check ¹H NMR for the signature signals (OCH3 ~3.7, NCH3 ~2.9, N–H 5–8 broad).
- Use the molecular formula and MS to confirm the carbon count and functional group.
- Assemble and verify the candidate structure against ALL data.
Common Confusions
| Do not confuse | With | Difference |
|---|---|---|
| "C=O is always near 1700 cm⁻¹" | Derivative-dependent positions | The C=O range spans ~1650–1820 cm⁻¹; amides and acid chlorides differ by ~150 cm⁻¹. |
| Amide C=O lower than ketone | A weak or absent carbonyl | The amide C=O is strong; it is simply lower in frequency because of N resonance donation. |
| One N–H band | Two N–H bands | Primary amides show two (symmetric/antisymmetric); secondary amides show one. |
| ¹³C carbonyl shift alone | IR data | ¹³C carbonyl ranges overlap heavily; IR is the reliable derivative discriminator. |
| McLafferty fragment in any ester | Only with a γ-H | Methyl propanoate (no γ-H) gives no McLafferty peak; methyl butanoate (γ-H present) does. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Spectrometers are like detectives' tools. IR tells you what kind of clip holds the molecule together — each clip has its own "voice": amide clips sing low (near 1660), ester clips mid (near 1740), acid-chloride clips high (near 1800). NMR counts the different neighborhoods of hydrogen atoms, and the mass spectrometer weighs the molecule and its broken pieces. Together they name a molecule without seeing it.
Worked example
Example 1: Converting a wavenumber to a wavelength
An ester's C=O stretch appears at ν̃ = 1735 cm-1. Wavelength and wavenumber are related by:
λ= 1ν̃
Substituting:
λ= 11735 cm-1 = 5.76 × 10-4 cm
Convert to micrometers (1 cm = 10⁴ μm):
λ= 5.76 × 10-4 cm × 104 μm1 cm = 5.76 μm
An amide C=O at 1665 cm⁻¹ gives λ = 1/1665 cm⁻¹ = 6.01 μm — a longer wavelength, consistent with the weaker amide carbonyl.
Example 2: Identifying an unknown ester
An unknown has formula C5H10O2, a strong IR band at 1740 cm⁻¹, ¹H NMR signals at 0.94 (t, 3H), 1.65 (m, 2H), 2.28 (t, 2H), and 3.67 (s, 3H), and MS: M⁺ = 102 with a strong peak at m/z 74.
Step 1 — degrees of unsaturation:
DBE = C - H2 + N2 + 1 = 5 - 102 + 0 + 1 = 1
One degree of unsaturation fits a single carbonyl, no ring or alkene.
Step 2 — IR 1740 cm⁻¹ with no broad O–H and no N–H bands: an ester, not an acid or amide.
Step 3 — ¹H NMR: the 3H singlet at 3.67 ppm is the classic OCH3 of a methyl ester; triplet 0.94 (3H), multiplet 1.65 (2H), and triplet 2.28 (2H) describe a CH3CH2CH2– chain (the 2.28 triplet is the CH2 next to C=O).
Step 4 — MS: M⁺ = 102 matches C5H10O2; the m/z 74 peak is the McLafferty fragment, which requires a γ-hydrogen — present on the butyl chain.
Answer: methyl butanoate, CH3CH2CH2COOCH3.
Key takeaways
- C=O IR (cm⁻¹, typical): acid chloride ~1800 > anhydride ~1820/1760 > ester ~1740 > acid ~1715 > thioester ~1700 > amide ~1660.
- Two C=O bands = anhydride; two N–H bands = primary amide; very broad O–H (2500–3300) = carboxylic acid.
- ¹H NMR: OCH3 ~3.6–3.8 (esters); NCH3 ~2.8–3.0 (amides); N–H 5–8 ppm broad, D2O-exchangeable; acid O–H 10–12 ppm.
- ¹³C carbonyls overlap heavily (165–215 ppm) — use IR, not ¹³C alone, to assign the derivative.
- McLafferty rearrangement requires a γ-hydrogen (n-propyl or longer chain).
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
Order acid chloride, ester, amide, and acid by decreasing C=O stretching frequency.
Show answer
Acid chloride (~1800) > ester (~1740) > acid (~1715) > amide (~1660). (Anhydrides show two bands, ~1820 and ~1760.)
Why does the amide C=O absorb at lower frequency than the ester C=O?
Show answer
The nitrogen lone pair donates into the carbonyl π system more effectively than the ester oxygen's lone pair, weakening the C=O bond and lowering its stretching frequency.
What IR features distinguish a carboxylic acid from a primary amide?
Show answer
The acid shows a very broad O–H band (2500–3300 cm⁻¹); the primary amide shows two N–H bands (~3300–3500 cm⁻¹) and no broad O–H.
Compute the degrees of unsaturation for C5H10O2.
Show answer
DBE = 5 − 10/2 + 0 + 1 = 1 — one carbonyl, no ring or alkene.
What structural requirement must a methyl ester meet to show a McLafferty rearrangement in its mass spectrum?
Show answer
A γ-hydrogen on the alkyl chain (e.g., an n-propyl or longer group), which migrates in the rearrangement; methyl butanoate shows m/z 74, methyl propanoate does not.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- wavenumber (cm⁻¹)
- Frequency unit for IR light, proportional to energy
- C=O stretching band
- IR absorption from the carbonyl bond vibration
- amide I band
- The strong C=O stretch of amides (~1650–1690 cm⁻¹)
- McLafferty rearrangement
- MS fragmentation of carbonyl compounds with a γ-H
- degrees of unsaturation (DBE)
- Number of rings plus π bonds from the formula
- D2O exchange
- Adding heavy water removes N–H/O–H NMR signals
- nitrogen rule
- Odd nominal molecular mass implies odd number of N
Sources & references
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