Organic Chemistry · Alcohols and Phenols
Spectroscopy of Alcohols and Phenols
On this page 9 sections
In 30 seconds
Spectroscopy is how chemists "see" functional groups without breaking the molecule apart. Alcohols and phenols leave distinctive fingerprints in four techniques: infrared (IR) spectroscopy, ¹H NMR Counts and characterizes protons by magnetic environment Full entry →, ¹³C NMR Counts unique carbons and reports their environments Full entry →, and mass spectrometry Ionizes and weighs molecules/fragments Full entry → (MS).
The single most diagnostic feature is the O–H group. In IR, an alcohol or phenol shows a broad O–H stretch between about 3200 and 3600 cm⁻¹ (broad because of hydrogen bonding Attraction between an O–H and a lone pair Full entry →); in ¹H NMR, the O–H proton is an exchangeable, concentration-dependent signal that vanishes when you shake the sample with D₂O. In ¹³C NMR, the carbon bearing the –OH appears downfield (50–90 ppm for alcohols; ~155 ppm for the aromatic ipso carbon of a phenol). In MS, alcohols give characteristic fragmentations (loss of water; α-cleavage Bond break next to the C–O bond in MS Full entry → ions like m/z 31 for primary alcohols), while phenols give a strong molecular ion.
Why this matters
Structure determination is the daily work of organic chemistry — identifying an unknown alcohol, distinguishing a 1° from a 3° alcohol, or proving that a synthesized phenol is pure. In pharmaceutical analysis, IR and NMR confirm that a drug candidate has the intended –OH groups (many drugs are phenols or alcohols). For exams, spectroscopy questions are guaranteed: you will be handed an IR spectrum, a ¹H NMR spectrum, or a mass spectrum and asked to assign the peaks, count the protons, or pick the correct isomer. Knowing the alcohol/phenol signatures cold is high-yield.
The college version
Core Concepts
IR spectroscopy: the O–H stretch
- O–H stretch: broad band 3200–3600 cm⁻¹. Broadness comes from hydrogen bonding: molecules in many different H-bonded arrangements absorb over a range of energies. A free (non-H-bonded) O–H, seen only in dilute solution or gas phase, is a sharp peak near 3600 cm⁻¹.
- C–O stretch: 1050–1200 cm⁻¹ for alcohols — and the exact position reveals substitution level: primary ~1050–1085, secondary ~1085–1125, tertiary ~1125–1200 cm⁻¹. Phenols show C–O near 1230–1260 cm⁻¹ plus aromatic C–C stretches at 1450–1600 cm⁻¹ and aromatic C–H just above 3000 cm⁻¹.
- Distinguish from amines: N–H gives sharper bands (two for –NH₂), and C–N is below 1250 cm⁻¹.
¹H NMR: finding the O–H proton
- The O–H proton resonates over a wide, variable range: roughly 0.5–5 ppm for alcohols and 4–7 ppm for phenols, depending on concentration, temperature, solvent, and hydrogen bonding. It is usually a broad singlet that does not couple to neighboring C–H protons (exchange is fast).
- D₂O shake Adding deuterium oxide to exchange labile H Full entry →: add D₂O, and the O–H signal disappears (the proton exchanges: ROH + D₂O → ROD + HOD). This is the definitive test that a signal is an O–H (or N–H).
- The carbon-bound protons on –CH(OH)– appear at 3.2–4.0 ppm (deshielded by oxygen). Phenols show aromatic protons at 6.5–8 ppm.
¹³C NMR: the carbon that holds the oxygen
- The carbinol carbon (–CH–OH) resonates at 50–90 ppm: primary alcohols 59–64, secondary 65–71, tertiary 71–78 ppm.
- Phenol's ipso carbon (the ring carbon bearing –OH) appears at ~150–160 ppm, well downfield within the aromatic region (110–160 ppm).
- Signal counting reveals symmetry: 1-propanol gives 3 carbon signals; 2-propanol gives 2 (the methyls are equivalent).
Mass spectrometry: fragmentation fingerprints
- Alcohols: the molecular ion (M⁺) is often weak or absent because alcohols fragment easily. Two classic fragmentations:
- α-cleavage (cleavage next to the C–O bond): primary alcohols give CH₂=OH⁺ at m/z 31; secondary give CH₃CH=OH⁺ at m/z 45; tertiary give (CH₃)₂C=OH⁺ at m/z 59.
- Loss of water: M⁺ − 18 (M–18 peak) from dehydration.
- Phenols: the aromatic ring stabilizes the cation, so phenols give a strong M⁺. Phenol itself: M⁺ at m/z 94, with loss of CO giving m/z 66.
Putting it together
Combine the methods: IR says "O–H present, C–O at 1050–1200" → alcohol; ¹H NMR counts and places the protons; ¹³C counts unique carbons and checks symmetry; MS gives molar mass and fragmentation class (1°/2°/3°). Isomers of C₄H₁₀O — 1-butanol, 2-butanol, 2-methyl-2-propanol, diethyl ether, methyl propyl ether — are distinguishable by exactly this combination (ethers show no O–H stretch and no exchangeable proton H that swaps with solvent D₂O Full entry →).
How It Works / Step-by-Step Process
Identifying an unknown alcohol or phenol (workflow):
- Get the molecular formula (high-resolution MS or combustion analysis). Compute degrees of unsaturation Rings + π bonds from the formula: (2C + 2 + N - H - X)/2 Full entry →.
- Record the IR spectrum: broad O–H near 3400 cm⁻¹? If yes, alcohol/phenol. Find the C–O band and read off 1°/2°/3° (1050–1085 / 1085–1125 / 1125–1200 cm⁻¹). Aromatic C–H (~3030) + C–O at ~1240 → phenol.
- Run ¹H NMR: integrate the signals (total H must match the formula), note the –CH(OH)– region (3.2–4.0 ppm), and do a D₂O shake to locate the O–H.
- Run ¹³C: count signals (symmetry check) and confirm the carbinol carbon (50–90 ppm) or phenol ipso (~155 ppm).
- Check the MS: M⁺ (molar mass), m/z 31/45/59 (α-cleavage class), M–18 (alcohol), strong M⁺ (phenol).
- Cross-check: does the boiling point match the proposed isomer?
Common Confusions
| Do not confuse | With | Difference |
|---|---|---|
| Broad O–H stretch (3200–3600) | Sharp N–H stretch | O–H is broad from H-bonding; N–H is sharper and –NH₂ shows two bands; C–H appears ~2900–3000, below the O–H region |
| O–H chemical shift (0.5–7 ppm) | A fixed value | It varies with concentration, temperature, and solvent; the D₂O shake, not the shift, identifies it |
| m/z 31 (CH₂=OH⁺) | Proof of any primary alcohol alone | Other compounds (e.g., methyl ethers give CH₃O⁺ = 31) can fragment near there — combine with IR (O–H present?) and M⁺ |
| Phenol ipso carbon (~155 ppm) | Ordinary alcohol carbinol (50–90 ppm) | The aromatic ipso carbon is far downfield; seeing 150–160 ppm plus aromatic H at 6.5–8 ppm says "phenol" |
| One technique alone | Enough to identify a structure | Always combine IR + ¹H/¹³C NMR + MS — single spectra can be ambiguous (e.g., ether vs. alcohol by IR only) |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Think of spectroscopy as giving every molecule an ID card. IR checks whether the molecule has an "O–H" badge — alcohols and phenols show a big, wide smudge at 3200–3600 cm⁻¹ because their O–H groups hold hands (hydrogen-bond) with each other. ¹H NMR counts the hydrogen atoms and tells you what neighborhood each lives in; D₂O is like a magic eraser that wipes out the O–H signal. Mass spec weighs the molecule and its broken pieces. Together they name the molecule without ever opening it up.
Worked example
Example 1: Degrees of unsaturation for C₄H₁₀O
Problem: A compound has formula C₄H₁₀O. How many rings/π bonds does it have?
Formula first: DoU = 2C + 2 + N - H - X2
DoU = 2(4) + 2 - 102 = 02 = 0
Zero rings or π bonds: the compound is saturated. With one oxygen, it must be either an alcohol (1-butanol, 2-butanol, 2-methyl-1-propanol, 2-methyl-2-propanol) or an ether (diethyl ether, methyl propyl ether, isopropyl methyl ether). IR decides: O–H band → alcohol; no O–H band → ether.
Example 2: Telling 1-propanol from 2-propanol by NMR
Problem: Two isomers of C₃H₈O give these ¹³C signal counts: compound A shows 3 signals, compound B shows 2. Which is 1-propanol and which is 2-propanol?
Analysis: 1-propanol (CH₃CH₂CH₂OH) has three unique carbons (CH₃, CH₂, CH₂OH) → 3 signals. 2-propanol ((CH₃)₂CHOH) has two unique carbons (two equivalent CH₃ groups + one CH) → 2 signals.
Answer: A = 1-propanol, B = 2-propanol. ¹H NMR confirms: 1-propanol shows a 3H triplet (CH₃), a 2H sextet (CH₂), a 2H triplet (CH₂OH, ~3.6 ppm); 2-propanol shows a 6H doublet (two CH₃) and a 1H septet (~4.0 ppm). Each O–H is a broad exchangeable signal that disappears on D₂O shake.
Example 3: Reading a mass spectrum
Problem: An unknown alcohol has M⁺ = 60 and a base peak at m/z 31. Is it 1-propanol or 2-propanol?
Analysis: The α-cleavage ion CH₂=OH⁺ has m/z 31 and is the diagnostic fragment of a primary alcohol. 2-Propanol would give CH₃CH=OH⁺ at m/z 45 instead. M⁺ = 60 matches C₃H₈O (propanol isomers).
Answer: 1-propanol. The M–18 peak (m/z 42, loss of water) is also expected for alcohols, but m/z 31 is the clincher for primary.
Example 4: Proving an O–H signal with D₂O
Problem: ¹H NMR of phenol (C₆H₅OH) in CDCl₃ shows 5 aromatic protons (6.8–7.3 ppm) and a broad 1H signal at 5.2 ppm. How do you prove the 5.2 ppm signal is the O–H?
Answer: Add D₂O and re-run. The O–H proton exchanges with deuterium (C₆H₅OH + D₂O → C₆H₅OD + HOD), so the 5.2 ppm signal vanishes while the aromatic signals stay. No other proton in the molecule is exchangeable, so this is conclusive. (The O–H shift also moves with concentration — another hint that it's labile.)
Key takeaways
- IR: broad O–H 3200–3600 cm⁻¹ = alcohol/phenol (sharp ~3600 only if non-H-bonded).
- C–O stretch tells substitution: 1° ≈ 1050–1085, 2° ≈ 1085–1125, 3° ≈ 1125–1200 cm⁻¹; phenols ~1230–1260 cm⁻¹.
- ¹H NMR: O–H is broad, exchangeable (disappears with D₂O), and variable (0.5–5 ppm alcohols; 4–7 ppm phenols); –CH(OH)– protons at 3.2–4.0 ppm; aromatic H at 6.5–8 ppm.
- ¹³C NMR: carbinol carbon 50–90 ppm; phenol ipso carbon ~150–160 ppm.
- MS: primary alcohol α-cleavage ion m/z 31 (CH₂=OH⁺); secondary m/z 45; tertiary m/z 59; M–18 (loss of H₂O); phenols give a strong M⁺.
- D₂O shake is the fastest way to prove a signal is O–H (or N–H).
- Isomeric ethers show no O–H IR band and no D₂O-exchangeable proton — the fastest way to tell an alcohol from an ether.
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
Why is the O–H IR stretch of an alcohol broad while that of a dilute (non-H-bonded) sample is sharp?
Show answer
Hydrogen bonding: each O–H is H-bonded to different extents, so the O–H bonds absorb over a range of energies, smearing the band into a broad envelope. In dilute solution (or gas phase) the molecules are mostly non-H-bonded and all absorb at nearly the same energy — a sharp peak near 3600 cm⁻¹.
A compound C₄H₁₀O shows no O–H band in IR and no D₂O-exchangeable NMR proton. What class of compound is it?
Show answer
An ether. No O–H means it cannot be an alcohol; the formula (DoU = 0) permits only alcohols and ethers, and IR/NMR rule out the alcohol.
How many ¹³C signals does 2-propanol show, and why?
Show answer
Two: the two methyl groups are equivalent by symmetry, so (CH₃)₂CHOH has only two unique carbon environments (the CH₃ carbons and the CH carbon).
Which fragment ion identifies a primary alcohol in the mass spectrum, and what is its m/z?
Show answer
CH₂=OH⁺ from α-cleavage of a primary alcohol — m/z 31. Secondary gives m/z 45 (CH₃CH=OH⁺); tertiary gives m/z 59 ((CH₃)₂C=OH⁺).
A phenol's ¹³C NMR shows a signal at ~155 ppm. What carbon is this, and why is it so downfield?
Show answer
The ipso carbon — the aromatic ring carbon bearing the –OH. It's downfield because the oxygen withdraws electron density (inductively) and resonance places partial positive character on it, deshielding it relative to other aromatic carbons.
Describe the D₂O-shake experiment and what it proves.
Show answer
Add D₂O and re-run the ¹H NMR: labile protons (O–H, N–H) exchange with deuterium and their signals disappear, while C–H signals remain. It proves a given signal is an exchangeable proton, i.e., an O–H or N–H.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- IR spectroscopy
- Measures bond vibrations by infrared absorption
- wavenumber (cm⁻¹)
- Number of wave cycles per centimeter
- hydrogen bonding
- Attraction between an O–H and a lone pair
- ¹H NMR
- Counts and characterizes protons by magnetic environment
- chemical shift (ppm)
- Position of an NMR signal relative to TMS
- D₂O shake
- Adding deuterium oxide to exchange labile H
- exchangeable proton
- H that swaps with solvent D₂O
- ¹³C NMR
- Counts unique carbons and reports their environments
- mass spectrometry
- Ionizes and weighs molecules/fragments
- α-cleavage
- Bond break next to the C–O bond in MS
- degrees of unsaturation
- Rings + π bonds from the formula: (2C + 2 + N - H - X)/2
Sources & references
This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.
Educational content only. It is not medical, legal or professional advice. Found an error? Tell us.