Does an absorption near 3300 cm⁻¹ in an infrared spectrum always mean the molecule contains an alcohol O–H group? No. The broad region around 3300–3500 cm⁻¹ can contain several fundamentally different stretching vibrations. A hydrogen-bonded alcohol may have a broad O–H feature in this area. A primary or secondary amine may have one or two relatively sharper N–H stretching features. A terminal alkyne, R–C≡C–H, can show a comparatively sharp, often strong ≡C–H stretch near 3300 cm⁻¹. Band position is useful, but shape, number of bands and complementary functional-group evidence decide which assignment is plausible.
For a JC2 student seeking an H3 Chemistry tutor in Hougang, the immediate benefit is avoiding a costly misconception: one wavelength number is not one unique functional group. An H3 question may give an unfamiliar spectrum or a candidate formula whose composition rules out one of the obvious interpretations. The student should be able to explain why the O–H, N–H and terminal ≡C–H vibrations may occur in similar high-wavenumber regions and then use the rest of the spectrum to decide between them.
The 2027 Singapore–Cambridge H3 Chemistry 9813 syllabus requires students to identify and predict characteristic infrared absorptions, infer structures from IR spectra and understand which vibrational modes are IR-active. This article is a focused comparison of three functional-group clues using those prescribed skills. It is not an instruction to memorise every exact peak position or to identify compounds from one spectral line without corroboration.
- Alcohol O–H: often broad and strong around 3300–3600 cm⁻¹ when hydrogen-bonded; a relatively free OH can be sharper near 3600 cm⁻¹ or above.
- Primary amine –NH₂: often two relatively sharp N–H stretching bands in the 3300–3500 cm⁻¹ region under suitable conditions.
- Secondary amine –NH–: often one N–H stretching band in the same broad region; tertiary amines have no N–H stretch because they have no N–H bond.
- Terminal alkyne ≡C–H: typically a sharp band near 3300 cm⁻¹, often supported by a C≡C stretch in the roughly 2100–2260 cm⁻¹ region.
- Carboxylic acid caution: its O–H band is often extremely broad over roughly 2500–3300 cm⁻¹, rather than a typical simple alcohol band.
- Proton NMR cross-check: OH and NH protons may exchange and have variable shifts, while a terminal alkynyl C–H can resonate around δ 2–3 ppm and does not simply disappear by neutral D₂O exchange in the same way.
- Interpretation rule: band shape suggests a class; molecular formula and independent bands test the hypothesis.
Official SEAB 2027 H3 Chemistry 9813 syllabus · OpenStax: IR spectra of common functional groups · OpenStax: primary, secondary and tertiary amine N–H bands · Hougang JC Chemistry hub.
Navigate: Why bands overlap · Alcohol OH · Primary and secondary amines · Terminal alkyne C–H · Acid OH caution · One practical comparison · Worked unknown spectra · Use proton NMR carefully · A diagnostic plan · Misconceptions · Questions with answers · Teaching and careers · FAQs.
Step 1: Why Different Bonds Can Have Bands Near the Same Wavenumber
A vibrational absorption occurs when the molecule can take up a photon of suitable energy to access an allowed vibrational state. In the simple harmonic-oscillator model, stretching frequencies depend on an effective bond force constant and the masses involved. Bonds between light atoms can vibrate at relatively high frequencies, which is one reason O–H, N–H and terminal C–H stretching may all appear in the higher-wavenumber part of an organic IR spectrum.
But the exact values depend on much more than the labels O, N and C. The carbon in a terminal alkyne is sp-hybridised and connected to the hydrogen by a bond with different character from ordinary alkane sp³ C–H. Alcohol O–H can be strongly affected by hydrogen bonding. Amine N–H frequencies and intensities are sensitive to the local nitrogen substitution and hydrogen-bonding environment.
A band at 3300 cm⁻¹ is therefore evidence for an energetic type of stretch, not a machine-generated statement that the sample is an alcohol. The student must identify which candidate bond could have that vibration and ask whether the rest of the measured spectrum supports the candidate.
The shape of an absorption is especially useful. Hydrogen bonding can produce a broad distribution of O–H stretching energies, while a terminal alkyne ≡C–H often produces a comparatively narrow feature. An N–H stretch may be sharper and less intense than the common broad alcohol band, although real spectra can be complicated by hydrogen bonding and overlap.
Step 2: Alcohol O–H Stretching—Broad When Hydrogen Bonded
An alcohol has the group R–O–H. The oxygen–hydrogen bond is polar, and its stretching motion changes the molecular dipole moment, allowing a characteristic IR absorption. In common condensed-phase samples, OH groups hydrogen-bond to neighbouring molecules, and the band is often strong and broad in a region around 3300–3600 cm⁻¹.
The broad feature is not merely an instrument fault. Hydrogen bonding changes the local O–H vibrational energy, and the sample contains a distribution of environments and interactions. The recorded spectrum therefore combines a range of frequencies rather than one identical isolated O–H oscillation.
In relatively free or non-hydrogen-bonded OH environments, an O–H stretch may be much sharper and at a somewhat higher wavenumber, commonly around 3600–3650 cm⁻¹ for ordinary alcohol teaching examples. This is a useful caution: not every OH absorption must look broad. Band shape should be interpreted in the context of phase, concentration and hydrogen bonding.
Example: Cyclohexanol
Cyclohexanol is a six-membered saturated ring containing an OH substituent. An ordinary liquid-phase IR spectrum can show a broad O–H feature in the 3300–3500-cm⁻¹ region and a C–O stretching feature in parts of the fingerprint region, often near 1050–1150 cm⁻¹ depending on structure. It does not inherently contain a C≡C triple bond or nitrogen, so the molecular formula and other bands help distinguish it from a terminal alkyne or amine.
A learner should not report ‘the broad 3400 peak proves cyclohexanol’. It supports an alcohol or other OH-containing hypothesis, but many other alcohols and phenols also contain OH. Additional carbon skeleton evidence is needed for a unique structure.
Hydrogen bonding and change of conditions
A sample’s OH absorption may shift or change shape with solvent, temperature and concentration because the balance of intermolecular and intramolecular hydrogen bonding can change. Thus a narrower OH band in a different sample preparation does not automatically imply the OH bond has chemically disappeared. The scientist must control conditions before treating a shape change as a structural transformation.
For an integrated comparison of alcohol environments and chemical-shift effects, read hydrogen bonding and variable OH proton NMR. The present guide addresses the infrared ambiguity around 3300 cm⁻¹ rather than only one OH signal.
Step 3: Primary and Secondary Amines—Count N–H Bonds
A primary amine has the structure R–NH₂, with two hydrogen atoms attached to nitrogen. Those two N–H bonds can participate in symmetric and asymmetric stretching normal modes, giving two characteristic N–H absorption features under suitable conditions. Common educational reference values are around 3350 and 3450 cm⁻¹, while the broader amine N–H region lies approximately 3300–3500 cm⁻¹.
A secondary amine has the structure R₂N–H, with only one N–H bond. In the simple reference pattern it produces one N–H stretching feature, often around 3350 cm⁻¹. A tertiary amine, R₃N, has no N–H bonds and therefore cannot produce an intrinsic N–H stretching vibration.
Do not interpret this as a statement that tertiary amines are spectroscopically silent. They contain C–N bonds and other molecular vibrations and may have characteristic features elsewhere in the IR spectrum. The correct narrow conclusion is that they lack N–H stretching from an N–H bond.
The two-band primary amine pattern is particularly useful in a clean reference spectrum. However, real N–H features can overlap, broaden or change with hydrogen bonding and instrument resolution. A primary amine should not be categorically rejected because a low-quality spectrum resolves only one apparent high-wavenumber feature.
Example: Propan-1-amine
Propan-1-amine, CH₃CH₂CH₂NH₂, has a primary amino group. It can show a pair of N–H stretching features around the ordinary 3300–3500-cm⁻¹ region, with shape and intensity dependent on conditions. A molecular formula containing nitrogen is consistent with the functional group; an established formula containing only C, H and O would not be consistent with a normal amine.
Amine N–H hydrogens can also be exchangeable in proton NMR. Their positions may be broad or variable, and counting them as a neat n+1 splitting partner in every spectrum is unsafe. IR and NMR are complementary, but each needs its own interpretation rules.
Step 4: Terminal Alkyne ≡C–H—Sharp Near 3300 cm⁻¹
A terminal alkyne has the fragment R–C≡C–H, where the final carbon of the triple bond carries one hydrogen. That hydrogen is attached to an sp-hybridised carbon, producing a C–H stretching vibration at a substantially higher wavenumber than an ordinary alkyl sp³ C–H stretch.
A characteristic terminal ≡C–H band appears near 3300 cm⁻¹, often relatively sharp and conspicuous in a suitable IR spectrum. It can superficially resemble a sharper N–H absorption or a relatively free OH stretch if only a small portion of the graph is shown. The structure must be checked using the rest of the spectrum.
The carbon–carbon triple bond can also give a C≡C stretch in the roughly 2100–2260-cm⁻¹ region, although intensity depends on substitution and molecular symmetry. Terminal alkynes commonly show more noticeable C≡C-related features than some symmetric internal alkynes, whose corresponding stretch can be weak or absent under ordinary electric-dipole IR conditions.
Example: Propyne
Propyne, CH₃–C≡CH, contains one terminal alkynyl hydrogen. A suitable IR spectrum can show the sharp ≡C–H stretch near 3300 cm⁻¹ together with an absorption attributable to its C≡C bond in the triple-bond region. Unlike an alcohol or primary amine, it has neither oxygen nor nitrogen in its formula C₃H₄.
A proton NMR spectrum can provide an extra check: the terminal acetylene hydrogen often appears around δ 2–3 ppm in a broadly characteristic alkyne environment, rather than the variable exchangeable behaviour of many OH or NH protons. The exact position and splitting require reference to the real molecular structure.
This is also a good reminder that a peak at 3300 cm⁻¹ does not always correspond to a heteroatom–hydrogen bond. A carbon-bound hydrogen can vibrate in that region when the carbon has the appropriate orbital hybridisation and bond character.
Step 5: The Carboxylic Acid Exception—Very Broad OH Extending Below 3300
Carboxylic acids contain –C(=O)–OH and often form strong hydrogen-bonded associations in condensed-phase samples. Their O–H absorption can be unusually broad, extending across much of 2500–3300 cm⁻¹, sometimes overlapping typical C–H stretches. A strong carbonyl absorption accompanies this functional group.
Therefore, if the 3300-cm⁻¹ region is part of a very broad envelope extending well below 3000 cm⁻¹ and the spectrum also contains an appropriate C=O band, a carboxylic acid is worth considering. That is different from identifying every broad band near 3400 as a simple alcohol.
An acid proton can also appear far downfield, often around δ 10–12 ppm in proton NMR under suitable conditions, but exchange and hydrogen bonding make the position variable. The combined IR and NMR evidence is more persuasive than a single band.
For a complete acid-versus-ester case, see butanoic acid versus ethyl ethanoate IR and ¹H NMR. The present guide focuses on differentiating high-wavenumber X–H stretching features.
Step 6: A Practical Comparison Table in Words
- Hydrogen-bonded alcohol: broad and often strong O–H near 3300–3600 cm⁻¹; C–O fingerprint support; formula contains O.
- Free alcohol O–H: may be relatively sharp at higher OH wavenumbers; do not assume broad shape is compulsory.
- Primary amine RNH₂: often two relatively sharp N–H absorptions around 3350 and 3450 cm⁻¹; formula contains N.
- Secondary amine R₂NH: generally one N–H absorption in the 3300–3500-cm⁻¹ region; formula contains N.
- Tertiary amine R₃N: no intrinsic N–H stretch; may contain other C–N or molecular features.
- Terminal alkyne R–C≡CH: relatively sharp ≡C–H near 3300 cm⁻¹, often with a C≡C-related absorption around 2100–2260 cm⁻¹.
- Internal alkyne R–C≡C–R′: no ≡C–H stretch, even though C≡C exists; its C≡C feature may be weak.
- Carboxylic acid: very broad OH region extending towards 2500 cm⁻¹ plus an appropriate C=O band.
The student should treat the table as the first screening step, not the final verdict. If a strong sharp band at 3300 appears, the formula contains only carbon and hydrogen and a second feature occurs in the alkyne region, a terminal alkyne becomes plausible. If the formula contains nitrogen and the 3400 region has two N–H-type bands, a primary amine becomes more plausible. If an OH-containing formula gives a broad absorption whose width changes with hydrogen bonding, an alcohol becomes plausible.
No single one of these observations proves every atom’s location. IR functional groups are valuable because they exclude inconsistent candidate structures and suggest the next measurement, not because a spectrometer automatically prints the exact compound name.
Four Worked Unknown-Spectrum Cases
Unknown A: The broad 3400-cm⁻¹ band
A fictional pure sample contains C, H and O, but no nitrogen. Its IR spectrum shows a broad, strong absorption centred around 3400 cm⁻¹ and a characteristic C–O feature. There is no meaningful C≡C absorption, and a suitable proton NMR test identifies an exchangeable hydrogen.
Reasoned conclusion: An alcohol is supported among the supplied simple candidates. The broad O–H stretch and oxygen-containing formula fit. A terminal alkyne would need a suitable carbon triple bond and would not itself explain the oxygen-related evidence. The complete alcohol structure still needs independent atom-connectivity information.
Unknown B: Two narrower bands around 3350 and 3450
Another invented pure sample contains nitrogen and no oxygen. The spectrum shows two relatively sharp moderate N–H features in the 3300–3500-cm⁻¹ range and additional evidence consistent with a carbon–nitrogen bond.
Reasoned conclusion: A primary amine is supported in the standard interpretation, because a –NH₂ group has symmetric and asymmetric N–H stretching modes. A single exact amine identity is not established by those two bands alone.
Unknown C: A sharp 3300-cm⁻¹ peak plus a triple-bond feature
A third hypothetical sample contains only carbon and hydrogen and gives a relatively sharp peak at 3300 cm⁻¹ together with an absorption around 2100–2150 cm⁻¹. Its formula is consistent with a terminal alkyne, and its proton NMR contains a one-hydrogen resonance in the ordinary acetylene region.
Reasoned conclusion: A terminal alkyne is well supported. The ≡C–H and C≡C features are mutually consistent, and the formula excludes an ordinary OH or NH bond. The exact alkyl chain or substituents require more information.
Unknown D: A broad 2500–3300-cm⁻¹ envelope with C=O
A fourth fictional sample shows a strong C=O absorption and an unusually broad OH-like region extending well below 3000 cm⁻¹. Its formula contains two oxygens and is consistent with a –COOH group.
Reasoned conclusion: A carboxylic acid is strongly plausible. The broad band should not be casually assigned to a terminal alkyne merely because part of it lies near 3300 cm⁻¹.
These are original educational examples, not transcriptions of actual past-year examination questions. They illustrate how several independent observations support or weaken candidate functional-group assignments. In a real experiment the sample history and spectrum quality must also be examined.
Step 7: Use Proton NMR Without Confusing Exchangeable and Carbon-Bound Hydrogens
OH and NH hydrogens can exchange with suitable proton sources or deuterated reagents. Their proton NMR resonances may broaden and shift with solvent, concentration, hydrogen bonding and temperature. Under appropriate conditions, a D₂O exchange experiment can cause an OH or NH signal to weaken or disappear when the proton is replaced by deuterium.
A terminal alkyne hydrogen is attached to carbon. It can be more acidic than a typical saturated hydrocarbon hydrogen in a suitable strong-base reaction, but it does not ordinarily disappear from a routine neutral D₂O shake simply by the same facile exchange mechanism used to identify labile O–H or N–H protons. This is a useful distinction, with the caveat that strongly basic or catalysed conditions can permit different isotope-exchange reactions.
A terminal alkynyl hydrogen often appears around δ 2–3 ppm; an ordinary alcohol OH hydrogen can occupy a broad range; an amine NH signal can be broad or variable. Thus a lone NMR chemical shift is not universally decisive. D₂O exchange, structural formula, integration and supporting IR bands combine to make the identification more credible.
The conventional first-order n+1 rule should also be applied cautiously to exchangeable OH/NH hydrogens, because exchange can suppress otherwise possible coupling. A broad NH or OH resonance is not evidence of a separate carbon skeleton without further context.
The separate D₂O exchange and OH proton NMR guide explains exchange in detail. Here it serves only as one additional test among several IR hypotheses.
An Evidence-Based Decision Method for a 3300-cm⁻¹ Band
- 1. Confirm the axis. Is the graph plotted in absorbance or percent transmittance? In a transmittance graph, downward features indicate absorption.
- 2. Inspect the band shape. Broad and strong may support hydrogen-bonded OH; one or two relatively sharp bands may support NH or terminal ≡C–H.
- 3. Count the features. Two well-resolved N–H stretches can support a primary amine; one may support a secondary amine or other candidates.
- 4. Search for other bands. C–O, C–N, C≡C and C=O features constrain the interpretation.
- 5. Check the formula. A molecule without oxygen cannot have an alcohol OH; one without nitrogen cannot have an amine N–H.
- 6. Check the molecular connectivity. An internal alkyne has no terminal ≡C–H, even if it has a C≡C bond.
- 7. Use NMR as a cross-check. Integration, exchangeability and carbon-bound proton environments provide different information.
- 8. State the result at the right confidence. An observation may support a functional group without proving one unique molecular structure.
This procedure is especially valuable when the spectrum is unfamiliar. The correct approach is to restrict the hypothesis space and ask for the next diagnostic observation. Do not choose one familiar label immediately and then reinterpret all other peaks to fit it.
Ten Common IR Misinterpretations at 3300 cm⁻¹
- ‘Every 3300 peak is O–H’. N–H and terminal ≡C–H can also absorb nearby.
- ‘Every O–H peak is broad’. A relatively unassociated alcohol OH can be much sharper.
- ‘Two peaks around 3400 must mean two different molecules’. A primary amine can have two N–H stretching normal modes in one molecule.
- ‘One N–H band proves a tertiary amine’. A tertiary amine has no N–H bond; a secondary amine commonly has one.
- ‘All alkynes have a 3300 C–H band’. Internal alkynes lack terminal ≡C–H bonds.
- ‘A weak C≡C peak disproves an alkyne’. Band intensity depends on structure and symmetry; some internal alkyne stretches are weak.
- ‘A broad acid OH feature is a typical simple alcohol’. The 2500–3300-cm⁻¹ acid envelope and carbonyl evidence should be considered together.
- ‘Any hydrogen near 3300 exchanges with D₂O’. Bond identity and reaction conditions govern exchange behaviour.
- ‘The tallest band gives the exact chemical name’. An IR band often supports one functional group, not unique connectivity.
- ‘A high-frequency absorption means high concentration’. Wavenumber is a frequency/energy-related quantity; intensity and concentration are separate.
Sixteen Original H3 Chemistry Practice Questions with Answers
Question 1: Alcohol
A broad strong IR feature around 3400 cm⁻¹ occurs in an oxygen-containing compound. What group is plausible?
Answer: An OH-containing group such as an alcohol, subject to formula and other supporting data.
Question 2: Primary amine
A pure sample gives two relatively sharp bands around 3350 and 3450 cm⁻¹. What structural group is worth considering?
Answer: A primary amine with two N–H stretching modes.
Question 3: Secondary amine
How many ordinary N–H bonds does a secondary amine R₂NH have?
Answer: One, so it typically has one N–H stretching feature.
Question 4: Tertiary amine
Why does a tertiary amine R₃N not produce its own N–H stretching band?
Answer: It has no N–H bond.
Question 5: Terminal alkyne
A sample shows a relatively sharp band around 3300 cm⁻¹ and a C≡C-region feature near 2100–2260 cm⁻¹. Which group is plausible?
Answer: A terminal alkyne R–C≡C–H.
Question 6: Internal alkyne
Does R–C≡C–R′ contain the ≡C–H group?
Answer: No. It has no terminal alkyne hydrogen and should not have the corresponding intrinsic ≡C–H stretch.
Question 7: An acid OH
A very broad absorption extends from roughly 2500 to 3300 cm⁻¹ and a strong carbonyl band is present. What group should be tested?
Answer: A carboxylic acid under suitable sample conditions.
Question 8: Formula constraint
An established molecular formula contains C and H only. Could an O–H or N–H stretch arise from the pure molecule?
Answer: No, because oxygen and nitrogen are absent. Consider a carbon-bound hydrogen feature such as terminal ≡C–H if the connectivity fits.
Question 9: Hydroxy proton exchange
A broad 3400-cm⁻¹ band is accompanied by a labile ¹H NMR signal that disappears after a suitable D₂O experiment. Does this support an OH/NH hypothesis?
Answer: Yes, though it does not automatically distinguish OH from NH; formula and other evidence remain relevant.
Question 10: Terminal alkyne exchange
Will a terminal ≡C–H proton ordinarily behave exactly like an alcohol OH proton in a neutral D₂O shake?
Answer: No. Carbon-bound alkyne hydrogens do not usually undergo the same facile neutral exchange, though strong-base or catalysed conditions can change behaviour.
Question 11: Band broadening
Why does hydrogen bonding commonly broaden an alcohol O–H absorption?
Answer: Molecules experience a distribution of O–H interaction strengths and thus a spread of stretching energies.
Question 12: NMR chemical shift
What approximate region commonly contains a terminal alkynyl C–H proton in ¹H NMR?
Answer: Roughly δ 2–3 ppm, with variation by molecular environment.
Question 13: Number of amine bands
Could a primary amine’s two N–H bands overlap into one observed envelope in a poor-quality spectrum?
Answer: Yes. Resolution and hydrogen bonding may complicate the neat textbook two-band picture.
Question 14: Another absorption
A 3300-cm⁻¹ peak is observed. Why is checking 2100–2260 cm⁻¹ useful?
Answer: It can reveal a C≡C-related feature that supports a terminal alkyne explanation.
Question 15: IR axis
A transmittance graph has a deep downward feature near 3400 cm⁻¹. Is that an absorption?
Answer: Yes. Lower transmitted fraction at that wavenumber corresponds to greater absorption.
Question 16: Uniqueness
Can one sharp 3300-cm⁻¹ feature identify the exact entire molecular structure?
Answer: No. It provides a functional-group clue whose interpretation needs molecular formula, supporting bands and other spectral evidence.
Question 17: Propanol versus propyne
An unknown has formula C₃H₄ and a sharp 3300-cm⁻¹ band. Why is propan-1-ol not the correct formula-based candidate?
Answer: Propan-1-ol has formula C₃H₈O. It contains oxygen and four more hydrogens, so it contradicts the established formula; propyne C₃H₄ is compatible with terminal alkyne analysis.
Question 18: Primary versus secondary
Which N–H observation is more typical of a primary amine than a secondary amine?
Answer: Two relatively sharp N–H stretching features rather than one, under conditions in which the modes are resolved.
How I Would Teach the 3300-cm⁻¹ Ambiguity
I begin by writing ‘3300 cm⁻¹’ in the middle of the board and asking learners for more than one possible bond. If they say OH, I ask whether a carbon–hydrogen bond could stretch that high. Then I give an alkyne structure. Next I show an amine structure and ask why a primary amine may show two N–H bands. This deliberately replaces one-to-one memorisation with comparative chemical reasoning.
In the second stage, each student examines an unfamiliar spectrum and draws a plausible molecular fragment, not necessarily a complete molecular name. We compare the shape of the high-frequency band, any C≡C-related feature, a C–O or C–N fingerprint region and the established elemental formula. Each learner must identify what the evidence supports and what remains uncertain.
Next I add a proton NMR result. Does the sample show a labile OH/NH resonance? Is there a carbon-bound terminal alkynyl proton? Could impurities or overlapping peaks explain a difference? Students should not force NMR to answer more than its signal permits.
Finally, we reverse the task. Students invent three small structures—one alcohol, one primary amine and one terminal alkyne—and predict which IR observations would distinguish them. A learner who can construct both a sensible candidate and a useful falsifying test is becoming independent.
For families searching from Hougang, the eduKate JC2 H3 Chemistry small-group tutorial guide explains a teaching approach focused on correcting individual reasoning. Confirm the actual specialist class availability, the student’s simultaneous H2 requirements, the lesson timetable and the venue near Sixth Avenue MRT. The Hougang title is for the readership and does not claim a separate classroom there.
Why High-Wavenumber Band Interpretation Matters in Applied Chemistry
Infrared spectroscopy is used to monitor organic synthesis, check product identity, examine materials and identify functional groups in samples. An OH group, an amine and a terminal alkyne can have very different reactivity even though their X–H stretching bands occur in overlapping regions. An analyst who recognises the limitations of one high-wavenumber band is less likely to misidentify a reaction product.
Real laboratory practice combines IR with NMR, mass spectrometry, chromatography and knowledge of sample preparation when necessary. A reported peak is a measured feature, and each candidate molecular structure must be checked against all available evidence. That habit is precisely what H3 Chemistry’s emphasis on interpreting unfamiliar information is designed to encourage.
IR at 3300 cm⁻¹: Frequently Asked Questions
Does 3300 cm⁻¹ always mean alcohol OH?
No. It can also be consistent with N–H or a terminal alkyne ≡C–H stretch, depending on band shape and supporting evidence.
How can an alcohol O–H band differ from a terminal alkyne C–H band?
Hydrogen-bonded alcohol OH is often broad, while terminal ≡C–H is commonly sharp near 3300 cm⁻¹; an accompanying C≡C feature may support the alkyne.
Why can primary amines show two N–H bands?
Their NH₂ group has symmetric and asymmetric stretching normal modes, which can be separately observed.
Do tertiary amines give N–H stretches?
No. A tertiary amine has no N–H bond to stretch.
Is a sharp 3300 band proof of a terminal alkyne?
No. Some N–H or relatively free OH absorptions can also be sharp. Look for C≡C and other supporting observations.
What is the acid OH clue?
A very broad band extending toward 2500 cm⁻¹ together with a C=O feature can support a carboxylic acid.
Can an alkyne proton be tested by D₂O like an OH?
Not generally under the same ordinary neutral exchange conditions. Terminal alkyne C–H isotope exchange needs a different chemical context.
Is this within H3 Chemistry 9813?
Yes as an application of the 2027 syllabus outcomes for characteristic infrared absorptions and structural interpretation. The three-way comparison is a useful original exercise, not an extra named compulsory chapter.
The real skill is not recognising one wavenumber but deciding which bond, molecular structure and physical conditions could have produced it. A broad OH feature, a pair of N–H modes and a sharp terminal alkyne C–H feature are all clues that may occupy neighbouring parts of the same IR plot. Once the learner knows how to test them, the spectrum becomes a chemical argument rather than a guessing contest.
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