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Organic Chemistry I · Spectroscopy

IR spectroscopy practice problems: read a spectrum in four looks

An IR spectrum looks like a mess of dips, but on an exam you only need a handful of them. IR tells you which functional groups are in a molecule, not the whole structure. Learn where four or five key peaks sit, check them in the same order every time, and most IR questions take under a minute.

Jump to: the four looks · key peak table · wavenumber lookup · worked examples · 11 practice problems · FAQ

The map: four zones, left to right

The x-axis runs backwards, from 4000 cm⁻¹ on the left to about 500 cm⁻¹ on the right. Peaks point down: a deep dip means the molecule absorbs strongly there. Stronger bonds and lighter atoms absorb further left, so the spectrum sorts itself into zones.

  1. 4000–2500Bonds to HO–H, N–H, C–H
  2. 2500–2000Triple bondsC≡C, C≡N
  3. 2000–1500Double bondsC=O, C=C
  4. below 1500FingerprintC–O, bends

Most exam answers come from the three zones on the left. The fingerprint region (below 1500 cm⁻¹) is unique to each compound but hard to read by eye.

The four looks (do them in this order)

  1. 1. Is there a strong, sharp dip near 1700?That's a C=O. If yes, you have an aldehyde, ketone, acid, ester, amide or similar, and steps 2 and 3 tell you which.
  2. 2. What's happening from 3600 down to 2500?A broad, rounded dip around 3300–3600 is an alcohol O–H. A huge, very broad dip from 3300 down to 2500 that swallows the C–H peaks is a carboxylic acid O–H. One or two narrower, weaker dips at 3300–3500 are N–H.
  3. 3. Which side of 3000 are the C–H peaks on?Below 3000 is ordinary sp³ C–H, which nearly everything has. A small peak just above 3000 means C–H on a double bond or a benzene ring. A sharp, strong peak near 3300 is the C–H of a terminal alkyne. Two small peaks near 2720 and 2820 are the aldehyde C–H.
  4. 4. Anything between 2100 and 2260?Almost nothing absorbs there, so any peak is a triple bond: C≡N (stronger, near 2250) or C≡C (weaker, 2100–2260).

Then use the evidence both ways. A missing peak is as useful as a present one: no dip near 1700 rules out every carbonyl compound, and no broad dip above 3000 rules out alcohols and acids.

Key IR peaks to know

Values are from OpenStax Organic Chemistry. Real peaks wander by 10–20 cm⁻¹, so treat these as targets, not exact numbers.

Key IR absorptions by bond
Bond (group)Where (cm⁻¹)What it looks like
O–H (alcohol, phenol)3300–3600Strong and broad because of hydrogen bonding
O–H (carboxylic acid)2500–3300Very broad, often runs into the C–H peaks
N–H (amine)3300–3500Sharper and weaker than O–H. Two peaks for –NH₂, one for N–H, none for a tertiary amine
≡C–H (terminal alkyne)about 3300Strong and sharp
=C–H (alkene) and aromatic C–H3020–3100 (aromatic about 3030)Weak to medium, just above 3000
C–H (sp³, alkane)2850–2960Strong, just below 3000. Almost every organic molecule has it
C–H (aldehyde)2700–2760 and 2800–2860Two weaker peaks. The one near 2720 stands alone and is the giveaway
C≡N (nitrile)about 2250Medium to strong, sharp
C≡C (alkyne)2100–2260Medium to weak; can be very weak or missing in internal alkynes
C=O (carbonyl)1670–1780Strong and sharp, usually the biggest peak on the page
C=C (alkene)about 1640–1680Medium to weak; weak for symmetric alkenes
C=C (aromatic ring)1450–1600Two or three medium peaks
C–O (alcohol, ether, ester)1000–1300 (alcohols near 1050)Strong, but in the crowded fingerprint region

Telling carbonyls apart

The C=O position shifts with what's attached. Conjugation with a C=C or a benzene ring lowers it by about 25–30 cm⁻¹, and ring strain raises it.

Typical C=O positions
CarbonylC=O (cm⁻¹)Partner peak that confirms it
Acid chlorideabout 1810None needed: the C=O is unusually high
Anhydrideabout 1820 and 1760Two C=O peaks
Cyclopentanoneabout 1750Ring strain pushes it up
Ester1735 (conjugated about 1715–1720)Two strong C–O peaks at 1000–1300
Aldehyde1730 (conjugated about 1705)C–H near 2720 and 2820
Ketone1715 (conjugated about 1685–1690)No O–H, no aldehyde C–H, no strong C–O
Carboxylic acidabout 1710Very broad O–H, 2500–3300
Amideabout 1650–1690N–H peaks at 3100–3500 unless it's N,N-disubstituted

Wavenumber lookup

Type a peak position from your problem set to see what usually absorbs there. It uses the same values as the tables above.

Worked examples

Example 1: Three isomers of C₃H₆O

An unknown with formula C₃H₆O could be propanal, acetone (propan-2-one), or prop-2-en-1-ol (allyl alcohol). Its spectrum:

Sketch of the example 1 spectrum: strong C–H peaks just below 3000 and one very strong peak at 1715; nothing above 3000.4000350030002500200015001000500Wavenumber (cm⁻¹)
  1. Look 1: a very strong peak at 1715 cm⁻¹. That's a C=O, so allyl alcohol (no C=O) is out.
  2. Look 2: nothing broad above 3000. No O–H, which also agrees with allyl alcohol being out.
  3. Look 3: C–H only below 3000, and nothing near 2720. No aldehyde C–H, so propanal is out. Propanal would also put its C=O a little higher, near 1730.

Answer: acetone. A C=O at 1715 with no aldehyde C–H and no O–H is the classic ketone pattern.

Example 2: Hexanoic acid, hexan-1-ol, or hexanal?

All three have a six-carbon chain, so all three show strong sp³ C–H at 2850–2960. The functional group peaks do the sorting:

What each candidate should show
CompoundRegion 3600–2500Near 1700
Hexanoic acidVery broad O–H from 3300 down to 2500Strong C=O near 1710
Hexan-1-olBroad O–H near 3300–3400, C–H peaks stay separateNothing; strong C–O near 1050 instead
HexanalNo O–H; small peaks near 2720 and 2820Strong C=O near 1730

The rule of thumb: O–H and C=O together, with the O–H smeared all the way to 2500, means carboxylic acid. O–H alone means alcohol. C=O plus a lone peak near 2720 means aldehyde.

Example 3: Did the reaction work?

You run hydroboration–oxidation on hex-1-ene to make hexan-1-ol. IR is the quickest check, because the starting material and the product have different functional groups.

  1. Gone if it worked: the =C–H peak just above 3000 (about 3080) and the C=C stretch near 1640.
  2. New if it worked: a broad O–H near 3300–3400 and a strong C–O near 1050.
  3. Unchanged: the sp³ C–H peaks below 3000. They can't tell you anything here.

Answer: look for the broad O–H appearing and the peaks just above 3000 and near 1640 disappearing. If you still see 1640, you have leftover alkene.

Example 4: Hex-1-yne vs hex-2-yne

Both are C₆H₁₀ alkynes. Hex-1-yne has a hydrogen on the triple-bond carbon; hex-2-yne doesn't.

Sketch of the hex-1-yne spectrum: a strong sharp peak near 3300, C–H peaks below 3000, and a weak peak near 2120.4000350030002500200015001000500Wavenumber (cm⁻¹)
  1. The sharp, strong peak near 3300 cm⁻¹ is a ≡C–H. Only a terminal alkyne has one.
  2. The weak peak near 2120 is the C≡C stretch. Don't rely on it alone: in internal alkynes like hex-2-yne it is weak and can be hard to see.

Answer: this is hex-1-yne. Hex-2-yne would show no peak near 3300 and at most a faint one around 2200.

Want more spectra like these? OrgoDrill's spectroscopy chapter has IR and NMR practice, and review brings back what you missed until it's automatic.

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IR practice problems

Each sketch is a simplified drawing of the main peaks, not a measured spectrum, and the table under it lists the peaks that matter. Pick an answer and press Check answer. Every wrong choice tells you what it would have looked like.

Problem 1 · C₄H₈OWhich C₄H₈O isomer matches this spectrum? Sketch for problem 1: C–H peaks just below 3000 and a very strong peak at 1715. Nothing above 3000.4000350030002500200015001000500Wavenumber (cm⁻¹)
Problem 1 peaks
cm⁻¹Look
2880–2975strong
1715very strong, sharp
above 3000nothing

Show solution

Butan-2-one. Look 1 finds a C=O at 1715, which removes tetrahydrofuran and but-3-en-1-ol. Look 3 finds C–H only below 3000 and nothing near 2720, which removes butanal. 1715 is right on the value for a saturated ketone.

Problem 2 · C₄H₈OSame formula, different spectrum. Which isomer is it? Sketch for problem 2: broad strong peak centered near 3340, a small peak near 3080, C–H peaks below 3000, a medium peak near 1645 and a strong peak near 1050.4000350030002500200015001000500Wavenumber (cm⁻¹)
Problem 2 peaks
cm⁻¹Look
3340strong, broad
3080weak
2880–2935medium
1645medium
1050strong

Show solution

But-3-en-1-ol. Look 1: no C=O, so every aldehyde and ketone is out. Look 2: broad O–H at 3340 with a strong C–O at 1050, so it's an alcohol. Look 3: a small C–H just above 3000 (3080) and a C=C at 1645 mean a double bond. One C=C plus the alcohol accounts for the one degree of unsaturation in C₄H₈O.

Problem 3 · C₄H₈O₂Which compound fits? Sketch for problem 3: a very broad strong absorption from about 3300 down to 2500 with C–H peaks on top, and a strong peak near 1710.4000350030002500200015001000500Wavenumber (cm⁻¹)
Problem 3 peaks
cm⁻¹Look
2500–3300strong, very broad, C–H peaks sit on top
1710very strong
1290medium

Show solution

Butanoic acid. The two esters and the acid all share the formula C₄H₈O₂, so look 2 decides it. Carboxylic acids form hydrogen-bonded pairs, which smear the O–H into a very broad band from 3300 to 2500 that swallows the C–H peaks. The C=O of the paired acid sits near 1710, below the 1735 of an ester.

Problem 4 · C₆H₁₀Which C₆H₁₀ compound gives this spectrum? Sketch for problem 4: a strong sharp peak at 3310, C–H peaks below 3000, and a weak peak at 2120.4000350030002500200015001000500Wavenumber (cm⁻¹)
Problem 4 peaks
cm⁻¹Look
3310strong, sharp
2870–2960strong
2120weak

Show solution

Hex-1-yne. Look 4 finds a peak at 2120, in the triple-bond zone, so it's an alkyne. Look 3 finds a sharp, strong ≡C–H at 3310, which only a terminal alkyne has. Don't confuse it with an O–H: an O–H near 3300 is broad and rounded, while ≡C–H is narrow.

Problem 5 · Aromatic unknownAn unknown has a benzene ring and one other functional group. Which compound is it? Sketch for problem 5: weak peaks near 3060, two small peaks at 2820 and 2730, a very strong peak at 1705, and medium peaks near 1600, 1585 and 1455.4000350030002500200015001000500Wavenumber (cm⁻¹)
Problem 5 peaks
cm⁻¹Look
3065, 3035weak
2820, 2730weak, two peaks
1705very strong
1600, 1585, 1455medium

Show solution

Benzaldehyde. Look 1: strong C=O at 1705. Look 2: no O–H, so not the acid or the alcohol. Look 3: aromatic C–H just above 3000, no sp³ C–H below 3000, and the two aldehyde C–H peaks at 2820 and 2730. The ring peaks at 1600, 1585 and 1455 confirm the benzene ring. A C=O at 1705 rather than 1730 is what conjugation with the ring does to an aldehyde.

Problem 6 · Nitrogen compoundWhich compound fits? Sketch for problem 6: two medium, fairly narrow peaks at 3370 and 3290, C–H peaks below 3000, and a medium peak near 1600. Nothing near 1700.4000350030002500200015001000500Wavenumber (cm⁻¹)
Problem 6 peaks
cm⁻¹Look
3370 and 3290medium, two peaks
2875–2960strong
1600medium
1650–1780nothing

Show solution

Propan-1-amine. The two peaks at 3370 and 3290 are narrower and weaker than an O–H band would be, which points to N–H. Two N–H peaks means an –NH₂ group (primary amine or primary amide). There is no C=O, so it isn't the amide. The peak near 1600 is the N–H bending of the NH₂ group, not a carbonyl.

Problem 7 · Following a reactionYou oxidize cyclohexanol to cyclohexanone. Which change in the IR spectrum shows the reaction went to completion?

Show solution

The O–H disappears and a C=O near 1715 appears. Cyclohexanol is a secondary alcohol, so oxidation gives a ketone. Completion means two things at once: the broad O–H band is gone (no starting material left) and a strong C=O is there. Six-membered ring ketones absorb at 1715, the same as open-chain ketones; only smaller rings shift higher.

Problem 8 · Which carbonyl?A ketone shows its C=O at 1750 cm⁻¹, with only sp³ C–H peaks below 3000. Which ketone is it most likely to be?

Show solution

Cyclopentanone. Start from 1715 for a plain ketone. Conjugation moves the C=O down (acetophenone, about 1690) and ring strain moves it up: cyclopentanone about 1750, cyclobutanone about 1785. 1750 with no aromatic C–H fits the five-membered ring.

Problem 9 · C₄H₈O₂Which compound gives this spectrum? Sketch for problem 9: C–H peaks just below 3000, a very strong peak at 1740, and two strong peaks at 1240 and 1045. Nothing above 3000.4000350030002500200015001000500Wavenumber (cm⁻¹)
Problem 9 peaks
cm⁻¹Look
2940–2985medium
1740very strong
1240 and 1045strong
above 3000nothing

Show solution

Ethyl acetate. Look 1: C=O at 1740. Look 2: no O–H, which removes the acid and the hydroxy ketone. An ester's C=O is a bit higher than a ketone's, and esters add two strong C–O stretches in the 1000–1300 range. Here they're at 1240 and 1045.

Problem 10 · The triple-bond zoneAn unknown shows sp³ C–H peaks below 3000, a sharp medium-to-strong peak at 2250, and nothing else above 1500. Which is it?

Show solution

Butanenitrile. Look 4 finds a peak in the triple-bond zone. Both C≡N and C≡C absorb there, so use the details: a nitrile's C≡N is intense and near 2250, while an alkyne's C≡C is weak. A terminal alkyne would also give a sharp ≡C–H at 3300, which is missing.

Problem 11 · Know the limitsWhich pair would be hardest to tell apart using only the main IR peaks?

Show solution

Hexan-1-ol and hexan-2-ol. IR identifies functional groups. Two molecules with the same functional groups, like two alcohol isomers, give nearly the same main peaks. The fingerprint region below 1500 does differ, but you'd need reference spectra to compare. ¹H NMR, which shows how the carbons and hydrogens are connected, separates them easily.

Frequently asked questions

What IR peaks do I need to memorize for organic chemistry?

Start with seven: C=O near 1700 (strong and sharp), alcohol O–H at 3300–3600 (broad), carboxylic acid O–H at 2500–3300 (very broad), N–H at 3300–3500, C≡C and C≡N at 2100–2260, C–H just above 3000 for alkenes and aromatics vs just below 3000 for sp³ carbons, and the aldehyde C–H pair near 2720 and 2820. Those cover most exam questions.

Why does the 3000 cm⁻¹ line matter?

C–H bonds on sp² carbons (alkenes and benzene rings) absorb just above 3000 cm⁻¹, and C–H bonds on sp³ carbons absorb just below it. A terminal alkyne's sp C–H is higher still, near 3300. More s character makes a stronger C–H bond, and stronger bonds absorb at higher wavenumbers.

Why is the O–H peak broad?

Alcohol molecules hydrogen-bond to each other, so in any sample there's a range of O–H bond strengths, and each absorbs at a slightly different wavenumber. Together they make one broad band. Carboxylic acids hydrogen-bond in pairs even more strongly, which spreads their O–H from 3300 all the way down to 2500.

Do I need to read the fingerprint region?

Usually not on an exam. Below 1500 cm⁻¹ there are many overlapping bends and single-bond stretches, and the pattern is unique to each compound. It is useful for matching an unknown to a known reference spectrum. The one peak worth noticing there is the strong C–O stretch of alcohols, ethers and esters between 1000 and 1300.

Sources

The spectrum sketches on this page are simplified drawings made for practice, not measured spectra.

IR is a few patterns repeated many times. Practice them until they're automatic, along with the rest of Orgo I and II.

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