Organic Chemistry 2 Quiz: Ir Spectroscopy Functional Group Identification
18 questions · exam conditions
0:00
Ir Spectroscopy Functional Group IdentificationQuestion 1 of 18

Compound X, an acyclic ketone, displays a C=O stretch at 1715 cm⁻¹. Compound Y, an isomeric acyclic ketone, displays a C=O stretch at 1685 cm⁻¹. What is the most plausible structural difference explaining this observation?

The carbonyl in compound Y is conjugated with a π system, while the carbonyl in X is not.
The carbonyl in compound X is conjugated with a π system, while the carbonyl in Y is not.
Compound X is a methyl ketone, whereas compound Y is an ethyl ketone.
The carbonyl in compound Y is more sterically hindered than the carbonyl in X.
← Back to quizzes

Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Ir Spectroscopy Functional Group Identification

Practice Ir Spectroscopy Functional Group Identification in Organic Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Ir Spectroscopy Functional Group Identification, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Compound X, an acyclic ketone, displays a C=O stretch at 1715 cm⁻¹. Compound Y, an isomeric acyclic ketone, displays a C=O stretch at 1685 cm⁻¹. What is the most plausible structural difference explaining this observation?

  1. The carbonyl in compound Y is conjugated with a π system, while the carbonyl in X is not. (correct answer)
  2. The carbonyl in compound X is conjugated with a π system, while the carbonyl in Y is not.
  3. Compound X is a methyl ketone, whereas compound Y is an ethyl ketone.
  4. The carbonyl in compound Y is more sterically hindered than the carbonyl in X.
Explanation: Conjugation of a carbonyl group with a π system (like a C=C bond or an aromatic ring) delocalizes electron density, gives the C=O bond more single-bond character, and weakens it. This results in a lower vibrational frequency. A typical non-conjugated ketone absorbs around 1715 cm⁻¹, while a conjugated ketone absorbs ~20-30 cm⁻¹ lower, around 1685 cm⁻¹. Therefore, Y must be the conjugated ketone.

Question 2

Ring strain significantly affects the stretching frequency of a carbonyl group in cyclic ketones. Which of the following ketones is expected to exhibit the C=O stretching absorption at the highest wavenumber (cm⁻¹)?

  1. Cyclohexanone
  2. Cyclopentanone
  3. Cyclobutanone (correct answer)
  4. Cycloheptanone
Explanation: Increased ring strain forces more s-character into the exocyclic C=O bond, strengthening it and increasing its vibrational frequency. Cyclobutanone has significant angle strain, leading to a high C=O frequency (~1780 cm⁻¹). Cyclopentanone also has some strain (~1750 cm⁻¹). Cyclohexanone is nearly strain-free and serves as a baseline (~1715 cm⁻¹). Cycloheptanone has slightly lower frequency than cyclohexanone (~1705 cm⁻¹). Thus, cyclobutanone has the highest frequency.

Question 3

An unknown compound with the molecular formula C₄H₁₁N is analyzed by IR spectroscopy. The spectrum shows two distinct, medium-intensity peaks in the region of 3300-3400 cm⁻¹. Which of the following best describes the compound?

  1. It is a primary amine. (correct answer)
  2. It is a secondary amine.
  3. It is a tertiary amine.
  4. It is a quaternary ammonium salt.
Explanation: The N-H stretching region is diagnostic for amine classes. Primary amines (R-NH₂) have two N-H bonds and exhibit two peaks in this region, corresponding to symmetric and asymmetric stretching modes. Secondary amines (R₂NH) have one N-H bond and show only one peak. Tertiary amines (R₃N) and quaternary ammonium salts (R₄N⁺) have no N-H bonds and therefore show no peaks in this region.

Question 4

An unknown compound has the molecular formula C₄H₈O₂. Its IR spectrum shows a strong, sharp absorption at 1740 cm⁻¹ and a strong C-O absorption at 1240 cm⁻¹. There is no broad absorption above 3000 cm⁻¹. Which of the following is the most likely structure of the compound?

  1. Butanoic acid
  2. Ethyl acetate (correct answer)
  3. 3-Hydroxy-2-butanone
  4. 1,4-Dioxane
Explanation: The strong peak at 1740 cm⁻¹ is characteristic of a saturated ester's C=O stretch. The absence of a broad O-H absorption rules out butanoic acid (A) and 3-hydroxy-2-butanone (C). 1,4-Dioxane (D) is an ether and would not have a C=O absorption at all. Ethyl acetate is an ester with the correct formula and spectral features.

Question 5

A sample of acetone is prepared using exclusively the oxygen-18 isotope instead of the naturally abundant oxygen-16. How will the C=O stretching frequency of this ¹⁸O-labeled acetone compare to that of normal ¹⁶O-acetone?

  1. The frequency will be higher.
  2. The frequency will be lower. (correct answer)
  3. The frequency will be unchanged because the bond order is the same.
  4. The peak will be absent because ¹⁸O is not IR-active.
Explanation: The vibrational frequency of a bond is described by Hooke's Law, where frequency is inversely proportional to the square root of the reduced mass of the two atoms. By replacing the lighter ¹⁶O with the heavier ¹⁸O isotope, the reduced mass of the C=O bond increases. An increase in mass leads to a decrease in vibrational frequency. Therefore, the peak will shift to a lower wavenumber.

Question 6

A student attempts to prepare ethyl benzoate via Fischer esterification of benzoic acid with ethanol. The IR spectrum of the product shows the expected ester C=O peak at ~1720 cm⁻¹ and C-O peak at ~1250 cm⁻¹, but also an unusually broad absorption spanning 2500-3300 cm⁻¹. What is the most likely reason for this unexpected peak?

  1. The product is contaminated with a significant amount of unreacted benzoic acid. (correct answer)
  2. The product is contaminated with unreacted ethanol.
  3. The product is wet with water from the aqueous workup.
  4. The ester has rearranged to form a product with an alcohol and a ketone.
Explanation: Fischer esterification is an equilibrium reaction. The presence of a very broad absorption from 2500-3300 cm⁻¹ is the characteristic signature of a carboxylic acid O-H stretch. This indicates that the reaction did not go to completion and a significant amount of the starting material, benzoic acid, remains in the product mixture. While ethanol (B) and water (C) have O-H groups, their IR peaks are typically less broad and centered at a higher frequency than a carboxylic acid's O-H.

Question 7

An IR spectrum shows a very strong, sharp peak at 2250 cm⁻¹ and medium peaks at 1605 cm⁻¹ and 1490 cm⁻¹. No peaks corresponding to N-H or O-H bonds are observed. If the compound is an aromatic derivative with the formula C₇H₅N, what is its structure?

  1. Phenylacetylene
  2. Aniline
  3. Benzonitrile (correct answer)
  4. Benzaldehyde
Explanation: The strong, sharp peak at 2250 cm⁻¹ is highly characteristic of a nitrile (C≡N) group. The peaks at 1605 cm⁻¹ and 1490 cm⁻¹ are typical for C=C stretching in an aromatic ring. The formula C₇H₅N is consistent with a nitrile group attached to a phenyl ring. Phenylacetylene (A) has the wrong formula (C₈H₆). Aniline (B) has the wrong formula (C₆H₇N) and would show N-H stretches. Benzaldehyde (D) has the wrong formula (C₇H₆O) and would show a C=O stretch.

Question 8

A compound has the molecular formula C₆H₁₀O. Its IR spectrum shows a strong, broad peak at 3350 cm⁻¹, a strong, sharp peak at 3300 cm⁻¹, and a medium peak at 2120 cm⁻¹. There are no significant absorptions between 1600-1800 cm⁻¹. Which structure is consistent with this data?

  1. Hex-5-yn-1-ol (correct answer)
  2. Cyclohex-2-en-1-ol
  3. Hex-1-en-3-ol
  4. 2-Methylcyclopentanone
Explanation: The spectral data indicate several key features: a strong, broad peak at 3350 cm⁻¹ is characteristic of an alcohol O-H stretch. A strong, sharp peak at 3300 cm⁻¹ is the ≡C-H stretch of a terminal alkyne. A medium peak at 2120 cm⁻¹ is the C≡C stretch. The absence of peaks from 1600-1800 cm⁻¹ rules out carbonyls and alkenes. Hex-5-yn-1-ol is the only structure that contains both an alcohol and a terminal alkyne functional group and fits the molecular formula.

Question 9

A chemist needs to distinguish between 2-pentanone and pentanal. Both compounds are expected to show a strong C=O absorption near 1720 cm⁻¹. What additional spectral information would unambiguously identify the sample as pentanal?

  1. A broad absorption centered near 3400 cm⁻¹.
  2. A pair of weak to medium absorptions near 2720 cm⁻¹ and 2820 cm⁻¹. (correct answer)
  3. A strong absorption between 1000-1200 cm⁻¹.
  4. The absence of any absorptions above 3000 cm⁻¹.
Explanation: Aldehydes have a unique feature that distinguishes them from ketones: the C-H bond of the aldehyde group gives rise to two characteristic stretching absorptions, often called 'aldehyde Fermi doublets', near 2720 cm⁻¹ and 2820 cm⁻¹. Ketones lack this hydrogen and will not show these peaks. An O-H peak (A) indicates an alcohol. A C-O stretch (C) is not unique. Both molecules have sp³ C-H bonds that absorb just below 3000 cm⁻¹, so D is incorrect.

Question 10

An IR spectrum of an unknown compound (C₃H₅ClO₂) shows an extremely broad absorption from 2500-3300 cm⁻¹ that partially obscures the C-H stretches, and a strong, sharp absorption at 1710 cm⁻¹. Which structure is most consistent with this data?

  1. 3-Chloro-1,2-propanediol
  2. Methyl 2-chloroacetate
  3. 3-Chloropropanoic acid (correct answer)
  4. 1-Chloro-2-propanone
Explanation: The combination of an extremely broad O-H stretch (2500-3300 cm⁻¹) and a C=O stretch (~1710 cm⁻¹) is the definitive signature of a carboxylic acid, resulting from the hydrogen-bonded dimer. 3-Chloropropanoic acid fits this description and the molecular formula. Methyl 2-chloroacetate (B) is an ester and lacks the O-H group. 3-Chloro-1,2-propanediol (A) has O-H groups but no C=O group. 1-Chloro-2-propanone (D) is a ketone and lacks the O-H group.

Question 11

A compound with the formula C₄H₆O₃ is known to be an acyclic acid derivative. Its IR spectrum is notable for two very strong absorptions in the carbonyl region, one near 1820 cm⁻¹ and another near 1760 cm⁻¹. Which functional group is present?

  1. An α-keto ester
  2. An acid anhydride (correct answer)
  3. A β-dicarbonyl
  4. A carboxylic acid dimer
Explanation: The presence of two distinct, strong carbonyl absorptions at high frequencies (~1820 and ~1760 cm⁻¹) is the classic signature of an acid anhydride. These two peaks arise from symmetric and asymmetric stretching modes of the two coupled C=O bonds. Other dicarbonyl compounds, like α-keto esters, would have two C=O peaks but at different, lower frequencies (e.g., ~1745 and ~1720 cm⁻¹). A carboxylic acid dimer shows only one C=O peak, although it is broad.

Question 12

Comparing the C=O stretching frequencies of acetophenone (1685 cm⁻¹) and acetone (1715 cm⁻¹), one can infer a difference in the electrophilicity of the carbonyl carbon. Which statement is the most accurate conclusion?

  1. Acetophenone is more electrophilic because its lower frequency indicates a weaker, more reactive bond.
  2. Acetophenone is more electrophilic due to the electron-withdrawing nature of the attached phenyl group.
  3. Both are equally electrophilic; the frequency difference is due only to the difference in molecular weight.
  4. Acetone is more electrophilic because its higher frequency indicates less electron delocalization and a more partial positive carbon. (correct answer)
Explanation: When you encounter IR spectroscopy questions involving C=O stretching frequencies, focus on the relationship between bond strength, electron delocalization, and electrophilicity. Lower frequencies indicate weaker bonds due to electron delocalization, while higher frequencies suggest stronger, more polarized bonds. Acetone's higher frequency (1715 cm⁻¹) indicates a stronger, more polarized C=O bond with less electron delocalization. This creates a more electropositive carbonyl carbon, making it more electrophilic and reactive toward nucleophiles. Acetophenone's lower frequency (1685 cm⁻¹) results from resonance between the carbonyl and the benzene ring, which delocalizes electron density and reduces the partial positive charge on the carbonyl carbon. Option A incorrectly assumes that weaker bonds mean higher electrophilicity—it's actually the opposite. The weaker bond in acetophenone results from electron delocalization that reduces electrophilicity. Option B has the relationship backwards; while the phenyl group does affect the carbonyl, it donates electron density through resonance, making the carbon less electrophilic, not more. Option C dismisses the frequency difference as merely due to molecular weight, ignoring the crucial electronic effects that IR spectroscopy reveals about bond character. Option D correctly identifies that acetone's higher frequency reflects less electron delocalization and greater carbonyl polarization, resulting in higher electrophilicity. Study tip: Remember that in carbonyl chemistry, resonance stabilization (lower IR frequency) typically decreases electrophilicity, while isolated carbonyls with higher frequencies are generally more reactive toward nucleophiles.

Question 13

Benzene is treated with acetyl chloride (CH₃COCl) and AlCl₃. The resulting product is then treated with NaBH₄ in ethanol, followed by a dilute acid workup. The final purified product would exhibit which of the following sets of key IR absorptions?

  1. A strong peak at ~1690 cm⁻¹ and aromatic C-H peaks > 3000 cm⁻¹.
  2. A strong peak at ~1740 cm⁻¹ and sp³ C-H peaks < 3000 cm⁻¹.
  3. A broad peak at ~3000 cm⁻¹ and a strong peak at ~1710 cm⁻¹.
  4. A broad peak at ~3400 cm⁻¹ and aromatic C-H peaks > 3000 cm⁻¹. (correct answer)
Explanation: This question tests your ability to predict IR spectroscopy results for a two-step reaction sequence involving Friedel-Crafts acylation followed by reduction. When benzene reacts with acetyl chloride and AlCl₃, you get a Friedel-Crafts acylation that introduces an acetyl group (CH₃CO-) onto the benzene ring, forming acetophenone. This ketone has a characteristic C=O stretch around 1690 cm⁻¹. However, the reaction doesn't stop there—the product is then treated with NaBH₄ in ethanol followed by acid workup. NaBH₄ is a reducing agent that specifically reduces aldehydes and ketones to alcohols. So acetophenone gets reduced to 1-phenylethanol (C₆H₅CH(OH)CH₃). This alcohol will show a broad O-H stretch around 3400 cm⁻¹ (due to hydrogen bonding) and aromatic C-H peaks above 3000 cm⁻¹ from the benzene ring. Choice A describes the intermediate acetophenone before reduction—it shows the ketone peak at 1690 cm⁻¹ but doesn't account for the reduction step. Choice B suggests an ester (1740 cm⁻¹), which isn't formed in this reaction sequence. Choice C incorrectly shows both alcohol and ketone peaks, suggesting incomplete reduction, but the broad peak at 3000 cm⁻¹ (rather than 3400 cm⁻¹) mischaracterizes the O-H stretch. The correct answer is D, showing the alcohol product with its characteristic broad O-H stretch and aromatic C-H peaks. Study tip: Always track each reaction step completely—don't stop at the first transformation when multiple steps are given.

Question 14

A reaction is expected to produce either heptan-2-one (a ketone) or heptanal (an aldehyde). The IR spectrum of the product shows a strong peak at 1715 cm⁻¹ but is inconclusive. A chemist then checks the 2700-2900 cm⁻¹ region. What observation in this region would confirm the product is heptan-2-one?

  1. The presence of two weak peaks at ~2720 cm⁻¹ and ~2820 cm⁻¹.
  2. The complete absence of any peaks in this region.
  3. The presence of typical sp³ C-H stretches but the absence of peaks at ~2720 cm⁻¹ and ~2820 cm⁻¹. (correct answer)
  4. The presence of a single, strong, broad peak centered at 2800 cm⁻¹.
Explanation: The key to distinguishing an aldehyde from a ketone is the aldehydic C-H stretch. Aldehydes show two characteristic weak peaks near 2720 and 2820 cm⁻¹. Ketones lack this C-H bond and do not show these peaks. Both molecules have many sp³ C-H bonds, which absorb in the 2850-3000 cm⁻¹ range. Therefore, to confirm the product is the ketone, one must observe the normal sp³ C-H stretches but see a clear absence of the two specific aldehyde C-H peaks.

Question 15

The IR spectrum of ethanol (CH₃CH₂OH) is recorded under two conditions: first as a pure liquid, and second as a very dilute solution in hexane. What is the expected difference in the O-H stretching region of the two spectra?

  1. The pure liquid shows a sharp peak at ~3600 cm⁻¹, while the dilute solution shows a broad peak at ~3350 cm⁻¹.
  2. The pure liquid will show no O-H peak, but the dilute solution will show a sharp peak at ~3600 cm⁻¹.
  3. Both spectra will show an identical broad peak at ~3350 cm⁻¹, as the functional group is the same.
  4. The pure liquid shows a broad peak at ~3350 cm⁻¹, while the dilute solution shows a sharp peak at ~3600 cm⁻¹. (correct answer)
Explanation: When analyzing IR spectra of alcohols, you need to consider how intermolecular hydrogen bonding affects the O-H stretching frequency. The key insight is that hydrogen bonding weakens the O-H bond, causing it to stretch at lower frequencies. In pure liquid ethanol, alcohol molecules are densely packed and form extensive intermolecular hydrogen bonds between the O-H groups. This hydrogen bonding significantly weakens the O-H bonds, causing them to absorb at lower frequencies around 3350 cm⁻¹. The peak appears broad because the hydrogen bonds have varying strengths, creating a range of slightly different O-H stretching frequencies. When ethanol is highly diluted in hexane (a nonpolar solvent), the alcohol molecules are far apart and cannot form intermolecular hydrogen bonds with each other. The isolated O-H groups are stronger and stretch at their natural, higher frequency around 3600 cm⁻¹. This peak is sharp because all the non-hydrogen-bonded O-H groups vibrate at essentially the same frequency. Looking at the choices: A reverses the conditions—it incorrectly assigns the sharp peak to pure liquid and broad peak to dilute solution. B suggests pure liquid shows no O-H peak, which is impossible since ethanol always has an O-H group. C incorrectly claims both spectra would be identical, ignoring the dramatic effect of hydrogen bonding on IR frequencies. The correct answer is D: pure liquid shows a broad peak at ~3350 cm⁻¹ (hydrogen-bonded), while dilute solution shows a sharp peak at ~3600 cm⁻¹ (free O-H). Remember: hydrogen bonding always shifts O-H stretches to lower frequencies and broadens the peaks.

Question 16

The successful conversion of cyclohexanol to cyclohexanone using an oxidizing agent like PCC would be best confirmed by which of the following changes in the IR spectrum?

  1. The disappearance of a strong, sharp peak at ~1715 cm⁻¹ and the appearance of a broad peak at ~3300 cm⁻¹.
  2. The disappearance of a broad peak at ~3300 cm⁻¹ and the appearance of a strong, sharp peak at ~1715 cm⁻¹. (correct answer)
  3. The appearance of two weak peaks at ~2720 cm⁻¹ and ~2820 cm⁻¹, in addition to a strong peak at ~1715 cm⁻¹.
  4. The disappearance of a C-O stretch at ~1100 cm⁻¹ and the appearance of a C=C stretch at ~1650 cm⁻¹.
Explanation: The starting material, cyclohexanol (an alcohol), has a characteristic broad O-H stretch around 3300 cm⁻¹. The product, cyclohexanone (a ketone), has a characteristic strong C=O stretch around 1715 cm⁻¹. A successful reaction would show the starting material's peak disappearing and the product's peak appearing. Choice A describes the reverse reaction (reduction). Choice C describes the formation of an aldehyde, which is incorrect for the oxidation of a secondary alcohol. Choice D incorrectly identifies the key functional group changes.

Question 17

Which of the following absorption peaks would be ABSENT from the IR spectrum of N,N-dimethylpropanamide?

  1. A strong C=O stretch between 1630-1680 cm⁻¹.
  2. C-N stretches in the fingerprint region.
  3. sp³ C-H stretches between 2850-3000 cm⁻¹.
  4. An N-H stretch between 3200-3400 cm⁻¹. (correct answer)
Explanation: When analyzing IR spectra of amides, you need to focus on the functional groups present and their characteristic absorptions. The key insight here is understanding the structural difference between primary/secondary amides and tertiary amides. N,N-dimethylpropanamide is a tertiary amide, meaning the nitrogen atom is bonded to three carbon atoms (the carbonyl carbon plus two methyl groups). This structural detail is crucial because it determines which IR absorptions will appear. The correct answer is D because tertiary amides have no N-H bonds to stretch. Since both hydrogen atoms on the nitrogen have been replaced by methyl groups, there's no N-H functionality present in the molecule. Without N-H bonds, you cannot observe N-H stretching vibrations in the 3200-3400 cm⁻¹ region. Option A is incorrect because all amides contain a C=O group, which produces a characteristic strong absorption between 1630-1680 cm⁻¹ (slightly lower than ketones due to resonance). Option B is wrong because the C-N bond is definitely present in this tertiary amide and will show stretching vibrations in the fingerprint region. Option C is incorrect because the molecule contains multiple sp³ C-H bonds (in the propyl chain and N-methyl groups) that will absorb between 2850-3000 cm⁻¹. Study tip: Always determine the degree of substitution on the nitrogen when analyzing amide IR spectra. Primary amides show two N-H stretches, secondary amides show one N-H stretch, and tertiary amides show no N-H stretches at all.

Question 18

A primary amide, a secondary amide, and a tertiary amide of similar molecular weight are analyzed by IR spectroscopy. Which feature allows for the unambiguous differentiation of all three?

  1. The position of the C=O stretch, which is highest for primary and lowest for tertiary amides.
  2. The intensity of the C=O stretch, which is strongest for tertiary and weakest for primary amides.
  3. The number of peaks in the N-H stretching region (3100-3500 cm⁻¹). (correct answer)
  4. The presence or absence of a C-N stretch in the fingerprint region.
Explanation: The number of N-H bonds is unique to each class of amide. A primary amide (R-CONH₂) has two N-H bonds and shows two peaks in the N-H stretching region. A secondary amide (R-CONHR') has one N-H bond and shows one peak. A tertiary amide (R-CONR'₂) has no N-H bonds and shows zero peaks in this region. This provides a clear method to distinguish all three. The C=O position (A) and intensity (B) vary with substitution and H-bonding but are not as definitive. All amides have a C-N bond (D).