Organic Chemistry 2 Quiz: 13c Nmr Key Signal Patterns
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13c Nmr Key Signal PatternsQuestion 1 of 14

The ¹³C NMR spectrum of propyne (CH₃–C≡C–H) shows three signals. The signal for the terminal sp-hybridized carbon (≡C-H) is found significantly upfield (at a lower δ value) of the internal sp-hybridized carbon (–C≡C–). Which factor best explains this observation?

The terminal alkyne proton is highly acidic, which strongly deshields the attached carbon.
The magnetic anisotropy of the triple bond shields the terminal carbon more effectively than the internal carbon.
Hyperconjugation from the methyl group donates electron density to the internal sp-carbon, causing it to be more shielded.
The methyl group exerts a deshielding inductive effect on the adjacent internal sp-carbon, shifting it downfield.
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: 13c Nmr Key Signal Patterns

Practice 13c Nmr Key Signal Patterns 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 13c Nmr Key Signal Patterns, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

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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.

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Question 1

The ¹³C NMR spectrum of propyne (CH₃–C≡C–H) shows three signals. The signal for the terminal sp-hybridized carbon (≡C-H) is found significantly upfield (at a lower δ value) of the internal sp-hybridized carbon (–C≡C–). Which factor best explains this observation?

  1. The terminal alkyne proton is highly acidic, which strongly deshields the attached carbon.
  2. The magnetic anisotropy of the triple bond shields the terminal carbon more effectively than the internal carbon.
  3. Hyperconjugation from the methyl group donates electron density to the internal sp-carbon, causing it to be more shielded.
  4. The methyl group exerts a deshielding inductive effect on the adjacent internal sp-carbon, shifting it downfield. (correct answer)
Explanation: When analyzing ¹³C NMR chemical shifts, you need to consider how different electronic effects influence the magnetic environment around carbon atoms. In propyne, the internal and terminal sp-carbons experience different electronic environments that dramatically affect their chemical shifts. The correct answer is D because the methyl group exerts a deshielding inductive effect on the adjacent internal sp-carbon. Alkyl groups are electron-donating through inductive effects, but this donation actually removes electron density from the carbon's immediate vicinity by pushing electrons toward the more electronegative sp-carbon. This decreased electron density around the internal carbon reduces its shielding, causing it to appear downfield (higher δ value) compared to the terminal carbon. A is incorrect because while the terminal proton is acidic, this actually makes the carbon more electron-deficient and should cause deshielding, not the observed upfield shift. B misapplies magnetic anisotropy concepts. The triple bond's anisotropic effects don't differentially shield the terminal vs. internal carbons in the way described—both carbons are part of the same triple bond system. C confuses the direction of hyperconjugation. If hyperconjugation were significantly donating electron density to the internal carbon, it would shield that carbon and move it upfield, opposite to what's observed. Study tip: Remember that inductive effects in ¹³C NMR often work counterintuitively—electron-donating groups frequently cause deshielding of the carbon they're attached to because they pull electron density away from the carbon's immediate environment. Focus on the local electronic environment around each individual carbon.

Question 2

How many distinct signals are expected in the broadband decoupled ¹³C NMR spectrum of (R)-3-methylcyclohexanone?

  1. 3
  2. 4
  3. 6
  4. 7 (correct answer)
Explanation: The presence of a single chiral center at C3 removes all planes of symmetry and rotational axes that would exist in a simpler molecule like cyclohexanone. As a result, every carbon atom in (R)-3-methylcyclohexanone is in a unique chemical environment. The seven carbons—the carbonyl (C1), the five ring methylene/methine carbons (C2, C3, C4, C5, C6), and the methyl group carbon—are all chemically non-equivalent and will each produce a distinct signal.

Question 3

The ¹³C NMR spectrum of anisole (methoxybenzene) shows signals in the aromatic region at approximately δ 160, 130, 121, and 114 ppm. Which signal is the correct assignment for the para-carbon (C4)?

  1. δ 160 ppm
  2. δ 130 ppm
  3. δ 121 ppm (correct answer)
  4. δ 114 ppm
Explanation: The methoxy group (-OCH₃) is an electron-donating group. The ipso-carbon (C1, attached to the oxygen) is the most deshielded, appearing furthest downfield at δ 160 ppm. The meta-carbons (C3/C5) are least affected by resonance and appear near the value for benzene, at δ 130 ppm. The electron-donating group shields the ortho (C2/C6) and para (C4) positions via resonance, shifting them upfield. The ortho carbons are typically the most shielded, appearing at δ 114 ppm. The remaining signal, δ 121 ppm, corresponds to the para-carbon, which is also shielded but less so than the ortho carbons.

Question 4

An unknown compound with the molecular formula C₈H₁₀O has a broadband decoupled ¹³C NMR spectrum showing 6 signals. Its DEPT-135 spectrum shows 3 positive signals and 1 negative signal. How many quaternary carbons does the compound contain?

  1. 1
  2. 2 (correct answer)
  3. 3
  4. 4
Explanation: The total number of unique carbon environments is given by the broadband ¹³C NMR spectrum, which is 6. The DEPT-135 spectrum only shows signals for protonated carbons: CH and CH₃ groups appear as positive signals, and CH₂ groups appear as negative signals. Quaternary carbons (C) are absent. In this case, the DEPT-135 spectrum shows a total of 3 (positive) + 1 (negative) = 4 signals from protonated carbons. The number of quaternary carbons is the difference between the total signals and the signals observed in the DEPT spectrum: 6 (total signals) - 4 (DEPT signals) = 2 quaternary carbons.

Question 5

Cyclohexanone is treated with LiAlH₄ followed by an H₃O⁺ workup to form cyclohexanol. Which of the following changes would be expected when comparing the ¹³C NMR spectrum of the product to that of the starting material?

  1. The signal around δ 210 ppm will disappear, and a new signal will appear around δ 70 ppm. (correct answer)
  2. The total number of signals will decrease from 6 to 4 due to increased symmetry in the product.
  3. A new signal will appear around δ 175 ppm corresponding to the new C-O single bond.
  4. The signal around δ 210 ppm will shift upfield to approximately δ 120 ppm.
Explanation: The reaction is the reduction of a ketone to a secondary alcohol. Cyclohexanone, the starting material, has a characteristic ketone carbonyl signal around δ 210 ppm. The product, cyclohexanol, lacks a carbonyl group but has a new sp³ carbon bonded to an oxygen (a C-OH group). This type of carbon typically appears in the δ 60-80 ppm range. Therefore, the most significant change is the disappearance of the ketone signal and the appearance of a new signal in the C-O region. Both cyclohexanone and cyclohexanol have a plane of symmetry and exhibit 4 signals, so the number of signals does not change.

Question 6

A compound C₄H₆O shows a strong IR peak at 1690 cm⁻¹ and a medium peak at 1620 cm⁻¹. Its ¹³C NMR spectrum contains four signals at δ 198, 137, 129, and 26. A DEPT-135 experiment shows a positive peak at 137, a negative peak at 129, and a positive peak at 26. What is the structure?

  1. 3-Buten-2-one (methyl vinyl ketone) (correct answer)
  2. 2-Butenal (crotonaldehyde)
  3. Cyclobutanone
  4. 2-Butyn-1-ol
Explanation: The data suggest an α,β-unsaturated ketone. DU=2. IR at 1690 cm⁻¹ (conjugated C=O) and 1620 cm⁻¹ (C=C) confirms this. The ¹³C shifts also support this: δ 198 (ketone C=O), 137/129 (alkene C=C), and 26 (alkyl C). The DEPT-135 data is decisive. The signal at δ 198 is absent (quaternary C=O). The negative signal at δ 129 corresponds to a CH₂ group. The positive signals at 137 and 26 correspond to CH or CH₃ groups.
  • 3-Buten-2-one (CH₃-CO-CH=CH₂) has a quaternary C=O, a CH₃ group, a CH group, and a CH₂ group. This perfectly matches the DEPT data (1 absent, 1 negative, 2 positive).
  • 2-Butenal (CH₃CH=CHCHO) is an aldehyde; its carbonyl carbon is a CH and would be positive in DEPT-135. It also lacks a CH₂ group.

Question 7

A student obtains a broadband decoupled ¹³C NMR spectrum for a pure compound and observes two signals of very different intensities. Why would it be unreliable to conclude that the more intense peak corresponds to more carbons than the less intense peak?

  1. The signal intensity is determined by the number of attached halogens, not the number of carbons.
  2. The Nuclear Overhauser Effect (NOE) enhances signals for protonated carbons to varying degrees, making peak intensities non-proportional to the number of carbons. (correct answer)
  3. Quaternary carbons exhibit much more intense signals than protonated carbons due to their longer relaxation times.
  4. Integration is only valid in DEPT spectra where the pulse sequence corrects for non-uniform signal enhancement.
Explanation: Standard broadband decoupled ¹³C NMR spectra are not quantitative. The primary reason is the Nuclear Overhauser Effect (NOE), where the decoupling of protons irradiates them and transfers energy to nearby carbons, enhancing their signals. The magnitude of this enhancement depends on the number of attached protons, making the intensities non-uniform across different carbon environments. Additionally, different carbons have different spin-lattice relaxation times (T₁), which also affects signal intensity. Quaternary carbons, in particular, often have long T₁ times and show weaker, not stronger, signals.

Question 8

Two isomers, A and B, both with the molecular formula C₈H₁₀, are analyzed by ¹³C NMR. Isomer A shows 3 signals, while isomer B shows 4 signals. Which of the following pairs correctly identifies A and B?

  1. A = p-xylene, B = o-xylene (correct answer)
  2. A = o-xylene, B = p-xylene
  3. A = m-xylene, B = ethylbenzene
  4. A = ethylbenzene, B = p-xylene
Explanation: This question requires analyzing the symmetry of xylene isomers.
  • p-xylene (1,4-dimethylbenzene) has two perpendicular planes of symmetry, making it highly symmetric. It has only 3 unique carbon environments: the two equivalent methyl carbons, the two equivalent aromatic carbons bonded to the methyl groups, and the four equivalent aromatic CH carbons. Thus, A is p-xylene.
  • o-xylene (1,2-dimethylbenzene) has one plane of symmetry. It has 4 unique carbon environments: the two equivalent methyl carbons, the two equivalent aromatic carbons bonded to the methyl groups, and two different pairs of equivalent aromatic CH carbons. Thus, B is o-xylene.
  • m-xylene has 5 signals, and ethylbenzene has 6 signals.

Question 9

A compound with the formula C₄H₈O₂ shows a strong, sharp IR absorption at 1740 cm⁻¹ and the absence of any broad absorption from 2500-3300 cm⁻¹. Its ¹³C NMR spectrum includes a prominent signal at δ 172 ppm. Which structure is most consistent with this data?

  1. Ethyl acetate (correct answer)
  2. Butanoic acid
  3. 3-Hydroxy-2-butanone
  4. 1,4-Dioxane
Explanation: The ¹³C NMR signal at δ 172 ppm is characteristic of a carboxylic acid derivative like an ester or acid. The IR absorption at 1740 cm⁻¹ is also in the carbonyl region. The key piece of information is the absence of a broad absorption from 2500-3300 cm⁻¹, which rules out the presence of a carboxylic acid's O-H group. Therefore, the compound is an ester. Ethyl acetate fits the formula C₄H₈O₂ and is an ester. Butanoic acid is a carboxylic acid. 3-Hydroxy-2-butanone is a ketone (C=O shift ~200 ppm). 1,4-Dioxane is an ether and has no carbonyl group.

Question 10

A compound with formula C₉H₁₂ displays 6 signals in its ¹³C NMR spectrum. The DEPT-135 spectrum shows 5 positive signals and 1 absent signal. Which of the following is the correct structure?

  1. n-Propylbenzene
  2. Isopropylbenzene (correct answer)
  3. 1,3,5-Trimethylbenzene
  4. 1,2,4-Trimethylbenzene
Explanation: Let's analyze the options:
  • n-Propylbenzene has 7 unique carbons (ipso, ortho, meta, para, and three in the propyl chain), so it would show 7 signals.
  • Isopropylbenzene (cumene) has a plane of symmetry through the isopropyl group and the benzene ring. It has 6 unique carbons: ipso-C, ortho-C (x2), meta-C (x2), para-C, benzylic CH, and methyl C (x2). This matches the 6 signals. Its DEPT-135 spectrum would show the quaternary ipso-C as absent, and all other carbons (4 aromatic CH, 1 benzylic CH, 1 methyl type) as positive, giving 5 positive signals. This perfectly matches the data.
  • 1,3,5-Trimethylbenzene is highly symmetric and shows only 3 signals.
  • 1,2,4-Trimethylbenzene has no symmetry and would show 9 signals.

Question 11

A ¹³C NMR spectrum of a sample dissolved in CDCl₃ is recorded. A small, distinct signal appears as a triplet centered at δ 77 ppm. What is the origin of this signal?

  1. An impurity of undeuterated chloroform (CHCl₃) in the solvent.
  2. The signal for the carbon atom in the deuterated chloroform (CDCl₃) solvent. (correct answer)
  3. A reference signal from tetramethylsilane (TMS) added to the sample.
  4. An artifact from the spectrometer known as a spinning sideband.
Explanation: The signal at δ 77 ppm in a spectrum run in CDCl₃ is the signal from the solvent itself. The carbon atom (¹³C, spin I=1/2) is bonded to a deuterium atom (D, spin I=1). The coupling between them follows the 2nI+1 rule, where n=1 (number of deuterium atoms) and I=1 (spin of deuterium). This results in a multiplicity of 2(1)(1) + 1 = 3, a triplet. Residual CHCl₃ would appear as a doublet due to coupling with a single proton (I=1/2). TMS appears at δ 0 ppm.

Question 12

Which of the following isomeric pentenes (C₅H₁₀) would exhibit the smallest number of signals in its broadband decoupled ¹³C NMR spectrum?

  1. 1-Pentene
  2. (E)-2-Pentene
  3. 2-Methyl-2-butene (correct answer)
  4. 2-Methyl-1-butene
Explanation: The number of signals corresponds to the number of chemically unique carbons.
  • 1-Pentene (CH₂=CHCH₂CH₂CH₃) has 5 unique carbons.
  • (E)-2-Pentene (CH₃CH=CHCH₂CH₃) has 5 unique carbons.
  • 2-Methyl-2-butene ((CH₃)₂C=CHCH₃) has symmetry. The two methyl groups on C2 are equivalent. The unique carbons are: the two equivalent methyls on C2, C2 itself, the CH at C3, and the methyl on C3. This results in 4 signals.
  • 2-Methyl-1-butene (CH₂=C(CH₃)CH₂CH₃) has 5 unique carbons. Therefore, 2-methyl-2-butene has the smallest number of signals (4) among the choices.

Question 13

Which of the following sets of approximate ¹³C NMR chemical shifts and signal counts best represents the spectrum of 4-heptanone?

  1. 7 signals: ~210, ~45, ~40, ~30, ~25, ~20, ~14 ppm
  2. 3 signals: ~210, ~45, ~14 ppm
  3. 4 signals: ~170, ~60, ~30, ~15 ppm
  4. 4 signals: ~210, ~45, ~18, ~14 ppm (correct answer)
Explanation: When analyzing ¹³C NMR spectra, you need to consider both the number of unique carbon environments (signals) and their expected chemical shifts based on electronic environment and functional groups. 4-Heptanone has the structure CH₃-CH₂-CH₂-CO-CH₂-CH₂-CH₃. To determine the number of signals, look for symmetry. This molecule has a plane of symmetry through the carbonyl carbon, making carbons 1 and 7 equivalent, carbons 2 and 6 equivalent, and carbons 3 and 5 equivalent. This gives you exactly 4 unique carbon environments: the carbonyl carbon (C4), the α-carbons (C3/C5), the β-carbons (C2/C6), and the terminal methyls (C1/C7). For chemical shifts, the carbonyl carbon appears around 200-220 ppm (typical for ketones), the α-carbons (next to C=O) appear around 40-50 ppm due to the electron-withdrawing effect of the carbonyl, the β-carbons appear around 15-25 ppm, and terminal methyls typically appear around 10-15 ppm. Answer D correctly shows 4 signals at ~210, ~45, ~18, and ~14 ppm, matching this analysis perfectly. Answer A shows 7 signals, ignoring the molecular symmetry entirely. Answer B only shows 3 signals, incorrectly grouping some of the distinct carbon environments. Answer C shows the wrong chemical shifts, particularly ~170 ppm (typical of esters or amides) and ~60 ppm (typical of carbons attached to oxygen), suggesting confusion with other functional groups. Remember: always check for molecular symmetry to determine the correct number of ¹³C signals, and match chemical shifts to the specific electronic environment of each carbon type.

Question 14

Which of the following describes the expected DEPT-135 NMR spectrum for isobutyl acetate, (CH₃)₂CHCH₂OCOCH₃?

  1. 2 positive signals, 1 negative signal
  2. 3 positive signals, 1 negative signal (correct answer)
  3. 2 positive signals, 2 negative signals
  4. 4 positive signals, 0 negative signals
Explanation: First, identify the different types of carbons in isobutyl acetate. The structure is CH₃-C(=O)-O-CH₂-CH(CH₃)₂.
  • Quaternary carbons (absent in DEPT): The carbonyl carbon C=O.
  • CH₃ carbons (positive): There are two types: the acetyl methyl group and the two equivalent methyl groups of the isobutyl part. This gives 2 positive signals from methyls.
  • CH₂ carbons (negative): The O-CH₂ group. This gives 1 negative signal.
  • CH carbons (positive): The CH group in the isobutyl part. This gives 1 positive signal. In total, there are 2 (from CH₃) + 1 (from CH) = 3 unique positive signals and 1 unique negative signal.