How can you distinguish butan-1-ol from diethyl ether when both have molecular formula C₄H₁₀O? Use two different kinds of evidence. Butan-1-ol, CH₃CH₂CH₂CH₂OH, is an alcohol and can show a broad O–H stretching feature in its infrared spectrum, commonly in the 3200–3600 cm⁻¹ region when hydrogen bonding is present. Diethyl ether, CH₃CH₂OCH₂CH₃, has an oxygen atom but no O–H bond, so its pure-molecule spectrum should not contain an alcohol O–H stretch. Then look at proton NMR: butan-1-ol has five principal hydrogen environments in a conventional simple solution model; diethyl ether has just two, because its two ethyl groups are symmetry-related.
The ether’s two sets give a particularly useful 6H triplet from equivalent terminal methyl groups and a 4H quartet from equivalent oxygen-bound CH₂ groups, under appropriate first-order coupling conditions. In butan-1-ol, by contrast, the carbon chain has a terminal 3H methyl, three distinguishable 2H methylene groups and a variable 1H OH feature. A student who can reconstruct that difference has learned how connectivity and molecular symmetry shape the spectrum.
For families seeking an H3 Chemistry tutor in Hougang, the lesson is about evidence, not memorising these particular two names. The 2027 Singapore–Cambridge H3 Chemistry 9813 syllabus includes interpretation of infrared functional-group bands, ¹H NMR chemical shifts, equivalent protons, integration, first-order coupling and combined spectroscopic deductions. This C₄H₁₀O pair is a focused teaching application; it is not a claim that SEAB requires students to memorise named spectra for butan-1-ol and ethoxyethane.
- Same molecular formula: C₄H₁₀O, nominal mass 74, degree of unsaturation zero.
- Butan-1-ol IR: broad alcohol O–H absorption when hydrogen-bonded, plus C–O-related fingerprint bands.
- Diethyl ether IR: ether C–O–C stretching features but no intrinsic hydroxyl absorption.
- Butan-1-ol proton environments: 3H methyl, three distinct 2H methylenes, and 1H OH = five environments.
- Diethyl ether proton environments: 6H methyl triplet and 4H O–CH₂ quartet = two environments in a suitable simple spectrum.
- Most useful inference: alcohol versus ether from O–H evidence, then structural symmetry and proton multiplicity for a cross-check.
- Critical limit: exact OH shifts, hydrogen bonding, impurities and overlapping multiplets can complicate observed spectra.
Official SEAB 2027 H3 Chemistry 9813 syllabus · How to count all seven C₄H₁₀O alcohol and ether isomers · Hougang JC Chemistry learning hub.
Guide: Formula and DBE · Draw the two structures · Infrared comparison · Butan-1-ol proton NMR · Diethyl ether proton NMR · Why triplet and quartet? · Two worked unknowns · The evidence ladder · Common errors · Original practice answers · Teaching and careers · FAQs.
Step 1: What the Molecular Formula and Mass Can Actually Prove
Both molecules contain four carbon atoms, ten hydrogens and one oxygen atom. Using simple nominal masses, C₄H₁₀O has mass 4(12)+10(1)+16 = 74. If an intact singly charged molecular ion is observed under a suitable ionisation method, a peak at nominal m/z 74 can be consistent with either compound. The mere presence of that peak cannot determine whether oxygen bonds to hydrogen or lies between two carbon atoms.
The standard double-bond equivalent for an ordinary neutral C/H/O formula is DBE = (2C+2−H)/2. Here DBE = (8+2−10)/2 = 0. That supports a saturated acyclic framework under conventional valence assumptions. It rules out a requirement for a C=C or C=O double bond, but says nothing about the position of the oxygen or whether it carries hydrogen.
An M+1 satellite arising predominantly from natural carbon-13 might have a simplified intensity ratio of roughly 4×1.1% ≈ 4.4% relative to M, provided the molecular ion is correctly assigned. This again suggests four carbon atoms but cannot distinguish constitutional isomers sharing the same formula. A measurement may be accurate but insufficient for a particular structural question.
This first stage models scientific restraint. We know enough to reject many wrong molecular formulas, but not enough to choose between the proposed alcohol and ether. The next observation has to be sensitive to functional-group connectivity.
Step 2: Draw Both Structures in Full
Butan-1-ol: oxygen at the terminal carbon
Write CH₃–CH₂–CH₂–CH₂–OH. The oxygen bonds to the last carbon and to hydrogen, forming the alcohol O–H group. There is one terminal methyl carbon and three chemically distinct methylene carbons in the straight chain. The OH hydrogen is an additional, potentially exchangeable environment.
The structure is neither butan-2-ol nor an ether. Moving OH from C1 to C2 makes a different alcohol constitutional isomer. Replacing an alcohol C–O–H group with a C–O–C connection makes an ether. The molecular formula may remain C₄H₁₀O, but the local bonding changes.
Diethyl ether: oxygen between two equivalent ethyl groups
Write CH₃–CH₂–O–CH₂–CH₃, whose systematic name is ethoxyethane. The central oxygen bonds to two carbon atoms; it does not carry a hydroxyl hydrogen. One ethyl group is the mirror-related counterpart of the other in the molecular connectivity, so the terminal methyl groups are chemically equivalent and the two O–CH₂ groups are chemically equivalent in an ordinary freely averaging, achiral solution.
That equivalence is why diethyl ether is a particularly good H3 NMR example. There are ten hydrogens, but only two proton chemical environments under the uncomplicated model. The number of signals is not the number of atoms, carbon groups or printed methyl fragments: it is the number of chemically distinct hydrogen environments.
Step 3: IR Evidence—An Alcohol OH Bond Versus an Ether C–O–C Link
An OH bond undergoes stretching vibrations that can produce an infrared absorption. In common hydrogen-bonded samples of alcohols, the O–H stretching feature is broad, often across roughly 3200–3600 cm⁻¹. The width reflects a spread of hydrogen-bonding environments and associated vibrational energies. More weakly associated or gas-phase OH absorptions can be relatively sharp and at different positions, so the ‘broad’ observation should be tied to conditions.
Butan-1-ol contains an O–H bond and is therefore capable of showing that feature. Diethyl ether has no O–H bond in its pure molecular structure and cannot produce its own intrinsic alcohol O–H stretch. If an ether sample displays a broad OH-like feature, contamination by water or alcohol, sample handling or another component should be considered.
Both structures have C–O bonds. Ether C–O–C and alcohol C–O stretching contribute to the fingerprint region, often with absorptions between roughly 1000 and 1300 cm⁻¹ depending on structure. A band in that region alone is not a unique ether identifier. It must be interpreted with the absence or presence of OH and the molecular formula.
Why a broad OH feature is evidence, not an absolute verdict
A very dilute gas-phase measurement of an alcohol may not have the same broad hydrogen-bonded O–H absorption as a condensed-phase sample. Conversely, a wet ether sample may show an OH feature from water. Thus the correct scientific explanation is: a clean, suitably measured pure butan-1-ol sample has an O–H vibration and the pure ether structure does not. The actual spectrum still deserves inspection rather than an unconditional slogan.
One useful procedural habit is to compare the unknown spectrum with its solvent or sample background. Spectroscopy is a measurement of a real sample, not a theoretical list of all bonds on a page. A spectral inference is strongest when the sample conditions are compatible with the proposed functional group.
Step 4: Assign Every Proton Environment of Butan-1-ol
Number the butan-1-ol chain from the OH end if that helps: HO–CH₂(C1)–CH₂(C2)–CH₂(C3)–CH₃(C4). The molecule has three distinct CH₂ groups, one CH₃ group and OH. The expected integrations are 2:2:2:3:1 in that order, summing to ten hydrogens. Chemical shifts are approximate and should be explained by the local bonding.
O–CH₂ at C1: around δ 3.6 ppm, two hydrogens
C1 is directly attached to electronegative oxygen. Its two hydrogens are comparatively deshielded relative to ordinary saturated alkyl CH₂ hydrogens, so an illustrative resonance is often around δ 3.5–3.7 ppm. C2 bears two neighbouring protons, and the uncomplicated first-order carbon–carbon coupling may make C1 appear as a 2H triplet. Coupling to the exchangeable OH proton may be weak, broadened or absent in routine measurements.
CH₂ at C2: around δ 1.5–1.7 ppm, two hydrogens
C2 is between O–CH₂ and another CH₂. It couples to two distinguishable neighbouring two-hydrogen groups. If the relevant coupling constants happen to be similar, its signal can approximate a five-line quintet-like multiplet. But the two neighbour sets are not chemically identical, so a more complicated multiplet may be observed. Its integration remains 2H.
CH₂ at C3: around δ 1.2–1.5 ppm, two hydrogens
C3 lies between a CH₂ group and terminal CH₃. It can couple to two protons on one side and three on the other, giving a complex multiplet, sometimes approximately sextet-like when the coupling constants are close enough. Do not mechanically write a perfect sextet without checking the stated first-order assumptions and measured resolution.
Terminal CH₃ at C4: around δ 0.9 ppm, three hydrogens
The terminal methyl group is adjacent to C3 CH₂ with two hydrogens. It usually yields an approximate 3H triplet under the familiar n+1 rule. Its position is relatively upfield because it is remote from the directly oxygen-bound carbon. This triplet by itself does not prove the alcohol; many ethyl and propyl chains have a similar terminal methyl pattern.
The OH proton: variable chemical shift, one hydrogen
The hydroxyl hydrogen is exchangeable, and its observed chemical shift may vary widely with hydrogen bonding, solvent, concentration and temperature. It commonly gives a broad or singlet-like one-hydrogen resonance in introductory spectra, but may not integrate or resolve as neatly as a carbon-bound hydrogen. A D₂O exchange experiment can help test whether a feature corresponds to an exchangeable OH proton.
The complete prediction is therefore five chemically distinct proton environments in the ordinary model: four carbon-bound groups and one OH group. Real signals from C2 and C3 may overlap; a weak OH signal can be difficult to assign. The structural environment count remains a useful model even when an imperfect spectrum does not show five perfectly isolated peaks.
Step 5: Predict the Two Proton Environments of Diethyl Ether
Ethoxyethane, CH₃CH₂OCH₂CH₃, consists of two equivalent ethyl fragments linked by oxygen. Because the groups are related by symmetry and rapid molecular motions in ordinary solution, the two CH₃ groups contribute one NMR environment and the two CH₂ groups adjacent to oxygen contribute another. The total is two proton environments, not four.
Two equivalent CH₃ groups: one 6H triplet
Each terminal methyl is adjacent to a CH₂ group with two neighbouring hydrogens. In the simple first-order picture, n+1 = 2+1 gives a triplet. Both terminal methyl groups are chemically equivalent, so their contributions coincide, integrating to 6H altogether. A representative chemical shift is around δ 1.1–1.3 ppm.
Two equivalent O–CH₂ groups: one 4H quartet
Each oxygen-bound CH₂ group is adjacent to a CH₃ group containing three equivalent neighbouring hydrogens. Under the first-order model, n+1 = 3+1 gives a quartet. The two CH₂ groups are chemically equivalent, so together they yield one 4H quartet, commonly around δ 3.3–3.6 ppm.
This pattern is particularly diagnostic when the formula and sample context support a simple symmetrical ether: 6H triplet + 4H quartet. The triplet and quartet arise from an ethyl coupling system. The appropriate coupling intervals expressed in hertz should agree within experimental precision if the first-order interpretation is suitable.
There is no intrinsic OH signal. Diethyl ether therefore accounts for all ten hydrogens using only the two carbon-bound environments, and its IR spectrum does not require an alcohol O–H stretch.
Step 6: Why Integration and Multiplicity Are Different
A six-hydrogen signal is not automatically a sextet. The integration describes how many chemically equivalent hydrogens contribute to one environment; the multiplicity reflects interactions with suitable neighbouring nuclear spins. In diethyl ether, the six methyl hydrogens collectively give a triplet because each equivalent CH₃ group is coupled mainly to a neighbouring two-hydrogen CH₂ group.
Similarly, a 4H quartet contains four line components in the ideal first-order treatment, but the four lines do not each represent a distinct proton. One chemical environment consists of four hydrogens spread across the two symmetry-equivalent O–CH₂ groups. The quartet shape comes from coupling to three neighbouring methyl protons.
A good answer will therefore report signal chemical shift, integration, multiplicity and structural assignment as separate pieces of information. For example: ‘δ approximately 3.45 (q, 4H), corresponding to the two equivalent O–CH₂ groups coupled to terminal CH₃.’ That gives a reader a chemically meaningful explanation.
Two Worked Structure-Elucidation Unknowns
Unknown A: Broad OH plus five environments
A fictional pure sample has formula C₄H₁₀O and nominal molecular mass 74. Its IR spectrum includes a broad O–H stretching feature, and its ¹H NMR shows a terminal 3H triplet, several distinct 2H alkyl signals, an oxygen-bound 2H resonance, and an exchangeable OH signal. Which candidate is supported?
Answer: Butan-1-ol. Its structural formula accounts for the O–H bond, all three different CH₂ groups, the terminal CH₃ and the total ten protons. The ether has no O–H bond and predicts a simpler two-environment NMR pattern.
Unknown B: A 6H triplet and 4H quartet
A second C₄H₁₀O unknown has no credible alcohol O–H absorption after suitable blank correction. Its proton NMR contains only two well-resolved signal groups: δ 1.20 triplet integrating to 6H and δ 3.45 quartet integrating to 4H. What structure fits?
Answer: Diethyl ether, or ethoxyethane. The two ethyl groups are chemically equivalent and explain the summed 6H triplet and 4H O–CH₂ quartet. No hydroxyl proton is required.
Both datasets are invented for instruction, not measurements from a particular laboratory sample or past-year paper. The important point is to use the whole assignment rather than one conveniently named peak.
A Reusable Evidence Ladder for Alcohol and Ether Isomers
- 1. Establish the molecular formula. C₄H₁₀O and nominal mass 74 are compatible with both candidates.
- 2. Check unsaturation. DBE zero supports a saturated acyclic structure but cannot locate oxygen.
- 3. Inspect IR for OH. A suitable alcohol OH band supports butan-1-ol; a pure ether lacks that bond.
- 4. Count ¹H environments. A linear alcohol has five distinguishable groups; symmetrical ethoxyethane has two in the simple model.
- 5. Use integration. Butan-1-ol predicts 3:2:2:2:1; ethoxyethane predicts 6:4.
- 6. Check splitting. The ether’s 6H triplet and 4H quartet form a mutually coupled ethyl pattern.
- 7. State limitations. OH exchange, impurities and overlap can change how many distinct features are visibly resolved.
- 8. Explain why the rejected candidate fails. A defensible identification accounts for every meaningful observation and contradiction.
This method transfers to other C₄H₁₀O structural isomers. The H2 Chemistry library includes the other possible butanols, 2-methylpropanols and ethers. The H3 extension is not about memorising all seven names; it is about generating a spectrum prediction from an unseen structure and critically evaluating evidence.
Seven Avoidable H3 NMR and IR Mistakes
- Using m/z 74 as a unique structure label. Both compounds have the same nominal molecular mass.
- Calling every oxygen-containing compound an alcohol. Ether oxygen is bonded to two carbons and has no O–H.
- Counting diethyl ether’s two identical ethyl groups as four different environments. Molecular symmetry can make groups equivalent.
- Calling a 6H triplet a six-line multiplet. Integration and splitting describe different quantities.
- Calling an OH peak’s position a fixed constant. Exchange and hydrogen bonding change its shift and line shape.
- Forcing exact n+1 splitting on every middle CH₂. Coupling to distinct neighbour groups can produce complex multiplets.
- Ignoring water contamination in ether IR. An OH-like band may belong to an impurity rather than the pure ether structure.
Original H3 Chemistry Practice Problems With Answers
Question 1: Formula
Calculate the nominal molecular mass of C₄H₁₀O.
Answer: 74 from 48+10+16.
Question 2: DBE
Calculate DBE for an ordinary neutral formula C₄H₁₀O.
Answer: 0.
Question 3: Functional group
Which candidate contains an O–H bond: butan-1-ol or ethoxyethane?
Answer: Butan-1-ol.
Question 4: IR
A broad O–H stretch appears for a pure sample under hydrogen-bonding conditions. Which candidate is supported?
Answer: Butan-1-ol.
Question 5: Environment count
How many principal proton environments does diethyl ether have in the ordinary solution model?
Answer: Two—combined terminal methyls and combined O–CH₂ groups.
Question 6: Triplet integral
Why does the ether’s methyl triplet integrate to 6H?
Answer: Both terminal methyl groups are chemically equivalent and each contains three hydrogens.
Question 7: Quartet integral
Why does the ether’s oxygen-bound CH₂ quartet integrate to 4H?
Answer: Two chemically equivalent O–CH₂ groups contribute two hydrogens each.
Question 8: Butanol count
How many chemically distinct carbon-bound CH₂ groups are present in butan-1-ol?
Answer: Three.
Question 9: OH exchange
An OH signal disappears after a suitable D₂O exchange test. Does that alone identify butan-1-ol uniquely among all C₄H₁₀O isomers?
Answer: No. Several alcohol isomers contain exchangeable OH protons. Use the carbon-bound NMR pattern to identify the specific alcohol.
Question 10: Distinguish a quartet from four protons
A quartet shows four component lines. Does it always represent four hydrogens?
Answer: No. Multiplicity counts split components; integration is a separate measure of hydrogen number.
Question 11: Formula check
What integration totals should the two candidate structures sum to?
Answer: Ten hydrogens for either C₄H₁₀O structure.
Question 12: Real spectrum
A diethyl ether sample shows an unexpected broad absorption around 3400 cm⁻¹. Is that definitive evidence the ether molecule contains O–H?
Answer: No. Consider water or other OH-containing impurities, background and sample history.
Question 13: Symmetry
Why can ethoxyethane have fewer NMR environments than butan-1-ol even though the formulas are identical?
Answer: The two ethyl halves of ethoxyethane are symmetry-related, while the alcohol chain has a terminal OH end and several distinguishable carbon environments.
Question 14: Middle CH₂ splitting
Why might butan-1-ol’s central methylene signals appear as multiplets rather than perfect n+1 patterns?
Answer: They couple to chemically different neighbouring proton groups, whose J values need not be equal, and the observed lines may overlap.
Question 15: One measurement
Can molecular mass 74 plus DBE zero by itself prove that an unknown is diethyl ether?
Answer: No. Many saturated C₄H₁₀O alcohols and ethers satisfy both constraints.
What a Good H3 Chemistry Tutorial Should Reveal
I begin by giving the student two unlabelled condensed formulas and asking them to count NMR environments without spectra. The student then assigns integration totals and predicts the influence of nearby oxygen. This makes the molecular structure do the work before a reference table is introduced.
Next we reverse the process. I provide a two-signal spectrum integrating 6:4, with first-order triplet and quartet multiplicities. The learner must draw a suitable structure rather than guess a compound name. Only then is the C₄H₁₀O molecular formula and IR spectrum revealed as an independent cross-check.
In a third round, I give a broad OH signal with imperfectly resolved carbon-bound multiplets. I ask whether the learner can still identify an alcohol and what extra evidence is needed to choose a particular butanol isomer. If the student says ‘OH proves butan-1-ol exactly’, we have found a useful reasoning gap to repair.
A small-group lesson should require each student to produce a complete independent assignment of every hydrogen. Group discussion can help identify ambiguity, but it should not allow one confident learner to do the interpretation for everyone else.
For Hougang families considering specialist support, the eduKate JC2 H3 Chemistry small-group overview describes a tutorial approach near Sixth Avenue MRT. Confirm current availability of an appropriately qualified H3 group, the student’s concurrent H2 Chemistry workload, lesson timetable and actual venue. The search phrase ‘Hougang’ does not imply that the classroom itself is in Hougang.
Why Identifying Alcohols and Ethers Matters Beyond JC
Different functional groups can have very different boiling points, solubilities, hydrogen-bonding behaviour and reaction chemistry even when the elemental formula matches. An alcohol can donate hydrogen bonds through its OH group, whereas an ordinary ether cannot donate an O–H hydrogen bond but can accept certain hydrogen bonds through oxygen. That affects intermolecular interactions as well as spectroscopy.
Organic synthesis laboratories, chemical manufacturing and product-quality analysis use NMR and IR together to determine whether intended functional-group transformations occurred. An H3 student who understands why the spectra differ has begun learning a professional habit: use a physical model to predict measurements, then evaluate the real evidence critically.
Butan-1-ol and Diethyl Ether: FAQs
Do they have the same molecular formula?
Yes. Both have formula C₄H₁₀O and nominal molecular mass 74.
Which one shows an O–H stretch?
Butan-1-ol. Pure diethyl ether has no intrinsic O–H bond.
How many NMR environments does diethyl ether have?
Two in the usual solution model: 6H methyl and 4H oxygen-bound methylene.
Why does diethyl ether give a 6H triplet?
Two equivalent CH₃ groups are each coupled to two CH₂ hydrogens.
Why does it give a 4H quartet?
Two equivalent O–CH₂ groups are each coupled to a neighbouring three-hydrogen methyl group.
How many proton environments does butan-1-ol have?
Five chemically distinct environments in the ordinary model: terminal CH₃, three CH₂ groups and OH.
Is this pair a named compulsory 2027 H3 Chemistry topic?
No. It illustrates the 2027 syllabus’s specified IR, proton NMR and structure-elucidation learning outcomes. Students should prioritise those methods over memorising a named compound pair.
The main achievement is not recognising a 6H triplet. It is understanding why a symmetrical ether can give two hydrogen environments while an alcohol of the same formula gives five—and why an infrared OH band supplies a different, independent kind of evidence. That is how H3 Chemistry turns a spectrum into a reasoned description of a molecule.
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