Organic Chemistry 2 Quiz: 1h Nmr Chemical Shift Splitting Integration
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1h Nmr Chemical Shift Splitting IntegrationQuestion 1 of 15

A compound with the molecular formula C₄H₇ClO has the following ¹H NMR data: δ 3.70 (s, 3H), δ 3.55 (t, 2H), δ 2.80 (t, 2H). Which structure is most consistent with this data?

Methyl 3-chloropropanoate
Ethyl 2-chloroacetate
4-Chloro-2-butanone
1-Chloro-4-methoxy-2-butanone
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: 1h Nmr Chemical Shift Splitting Integration

Practice 1h Nmr Chemical Shift Splitting Integration in Organic Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on 1h Nmr Chemical Shift Splitting Integration, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

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Question 1

A compound with the molecular formula C₄H₇ClO has the following ¹H NMR data: δ 3.70 (s, 3H), δ 3.55 (t, 2H), δ 2.80 (t, 2H). Which structure is most consistent with this data?

  1. Methyl 3-chloropropanoate (correct answer)
  2. Ethyl 2-chloroacetate
  3. 4-Chloro-2-butanone
  4. 1-Chloro-4-methoxy-2-butanone
Explanation: Let's analyze the data. C₄H₇ClO has a degree of unsaturation of 1. δ 3.70 (s, 3H) strongly suggests a methoxy group (-OCH₃). δ 3.55 (t, 2H) is a CH₂ group next to another CH₂, likely attached to the electronegative Cl. δ 2.80 (t, 2H) is another CH₂ group next to a CH₂, likely alpha to a carbonyl group (which accounts for the DU). Combining these pieces gives Cl-CH₂-CH₂-C(=O)-OCH₃, which is methyl 3-chloropropanoate. Distractor C, 4-chloro-2-butanone, would have two triplets and a singlet, but the singlet for a methyl ketone (CH₃-C=O) should be around δ 2.1, not δ 3.70.

Question 2

An unknown compound C₈H₁₀O gives the following ¹H NMR spectrum: δ 7.2 (m, 5H), δ 4.5 (s, 2H), δ 1.9 (s, 3H). Shaking with D₂O causes no change to the spectrum. Which structure is consistent with this data?

  1. 2-Phenylethanol
  2. 1-Phenoxy-2-propanone (correct answer)
  3. 2-Phenyl-2-propanol
  4. Methoxybenzyl ether
Explanation: The signal at δ 7.2 (m, 5H) indicates a monosubstituted benzene ring (C₆H₅-). The formula C₈H₁₀O leaves C₂H₅O for the substituent. The absence of change with D₂O indicates there is no -OH or -NH group. The signal at δ 4.5 (s, 2H) is a methylene group with no adjacent protons, highly deshielded, suggesting it's next to both the phenyl ring and an oxygen (e.g., -O-CH₂-Ph). The signal at δ 1.9 (s, 3H) is a methyl group with no adjacent protons, likely a methyl ketone (CH₃-C=O). Combining these gives C₆H₅-O-CH₂-C(=O)-CH₃, which is 1-phenoxy-2-propanone. This structure matches all data points. 2-Phenylethanol and 2-Phenyl-2-propanol both have -OH groups and would be affected by D₂O.

Question 3

A researcher synthesizes a compound believed to be either 1,2-dimethoxybenzene or 1,3-dimethoxybenzene. The ¹H NMR spectrum shows a sharp singlet at δ 3.8 (6H) and aromatic signals between δ 6.8-7.3 (4H). What feature in the aromatic region would best distinguish between these two isomers?

  1. 1,2-dimethoxybenzene would show two aromatic signals while 1,3-dimethoxybenzene would show three. (correct answer)
  2. 1,3-dimethoxybenzene would show two aromatic signals while 1,2-dimethoxybenzene would show three.
  3. Both would show identical aromatic splitting patterns due to similar substitution.
  4. The chemical shifts would be identical, making them indistinguishable by ¹H NMR.
Explanation: The key distinction lies in the aromatic region's pattern, which reflects the symmetry of the substitution. In 1,2-dimethoxybenzene, a plane of symmetry makes H3 equivalent to H6, and H4 equivalent to H5, resulting in two signals in the aromatic region. In 1,3-dimethoxybenzene, the plane of symmetry makes H4 equivalent to H6, but H2 and H5 are unique, resulting in three distinct signals in the aromatic region. Both isomers would show a 6H singlet for the methoxy groups due to their equivalence.

Question 4

Why are the two vinylic protons in cis-1,2-dichloroethene chemically equivalent, resulting in a single ¹H NMR signal, while the two vinylic protons in trans-1,2-dichloroethene are also chemically equivalent and give a single signal?

  1. Free rotation around the C=C bond averages their environments.
  2. Both molecules possess a C₂ axis of symmetry that relates the two protons. (correct answer)
  3. The inductive effects of the two chlorine atoms cancel each other out perfectly.
  4. They are only equivalent if the magnetic field strength is low.
Explanation: Chemical equivalence is determined by symmetry operations. Protons that can be interchanged by a rotational axis (Cₙ) or a plane of symmetry (σ) are chemically equivalent. In cis-1,2-dichloroethene, there is a C₂ axis perpendicular to the C=C bond axis that passes through its midpoint, interchanging the two protons. In trans-1,2-dichloroethene, there is also a C₂ axis (passing through the C=C bond midpoint and perpendicular to the plane of the molecule) that interchanges the protons. Therefore, in both cases, the protons are equivalent due to a C₂ symmetry element. Free rotation around a C=C bond (A) does not occur. Inductive effects (C) influence chemical shift but not equivalence. Equivalence is independent of field strength (D).

Question 5

A sample of an unknown alcohol with formula C₄H₁₀O is analyzed by ¹H NMR. A notable feature is a doublet with an integration of 6H at approximately δ 1.2. After shaking the sample with D₂O, this doublet remains unchanged, but a broad signal elsewhere in the spectrum disappears. Which structure is most consistent with these observations?

  1. 1-Butanol
  2. 2-Butanol
  3. 2-Methyl-1-propanol (correct answer)
  4. 2-Methyl-2-propanol (tert-butanol)
Explanation: A 6H doublet indicates two equivalent methyl groups (2 x CH₃) that are adjacent to a single proton (-CH). This structural feature is an isopropyl group, -CH(CH₃)₂. Shaking with D₂O removes the signal for the exchangeable O-H proton but does not affect C-H signals. 2-Methyl-1-propanol, (CH₃)₂CHCH₂OH, contains this isopropyl group; its two methyl groups would appear as a 6H doublet. 1-Butanol would show a 3H triplet. 2-Butanol has two methyl groups, but one is a triplet and one is a doublet, not a combined 6H doublet. 2-Methyl-2-propanol has three equivalent methyl groups which would appear as a 9H singlet, and it has no C-H adjacent to the OH group.

Question 6

In the ¹H NMR spectrum of 1,1,2-trichloroethane (Cl₂CH-CH₂Cl), which statement accurately describes the expected signals?

  1. A singlet at high field and a singlet at low field, with a 2:1 integration ratio.
  2. A doublet with an integration of 2H and a triplet with an integration of 1H. (correct answer)
  3. A triplet with an integration of 2H and a doublet with an integration of 1H.
  4. A quartet with an integration of 2H and a singlet with an integration of 1H.
Explanation: The molecule has two distinct proton environments: the single proton on C1 (Cl₂CH-) and the two protons on C2 (-CH₂Cl). The C1 proton is adjacent to the two C2 protons, so its signal will be split into a triplet (n+1 = 2+1 = 3). The two equivalent C2 protons are adjacent to the single C1 proton, so their signal will be split into a doublet (n+1 = 1+1 = 2). Therefore, the spectrum will show a doublet (2H integration) and a triplet (1H integration). The triplet for the C1 proton will be further downfield due to being on a carbon with two chlorine atoms.

Question 7

The ¹H NMR spectrum of (R)-3-methylpentan-2-one is analyzed. What is the expected multiplicity of the signal for the protons of the C4 methylene group (-CH₂-)?

  1. A quartet, due to coupling with the adjacent C5 methyl group.
  2. A doublet, due to coupling with the single proton at the C3 chiral center.
  3. A triplet, incorrectly assuming coupling to two equivalent adjacent protons.
  4. A complex multiplet, because the C4 protons are diastereotopic. (correct answer)
Explanation: The C4 methylene group is adjacent to a stereocenter at C3. This makes the two protons on C4 chemically non-equivalent (diastereotopic). Let's call them Hₐ and Hₑ. They will have slightly different chemical shifts. Each proton couples to the single proton on C3 and to the three protons on C5, as well as to each other (geminal coupling). For Hₐ, it is split by Hₑ, the C3 proton, and the C5 protons, resulting in a complex pattern. The same applies to Hₑ. The combined result for the C4 position is a complex multiplet, often unresolvable into simple patterns.

Question 8

A compound with formula C₅H₁₀O₂ gives a ¹H NMR spectrum with signals having relative integration values of 35.4, 17.7, and 5.9. Which of the following is the most likely structure?

  1. Isopropyl acetate (correct answer)
  2. tert-Butyl formate
  3. Ethyl propanoate
  4. Pentanoic acid
Explanation: First, determine the simplest whole-number ratio of the protons by dividing each integration value by the smallest value (5.9). Ratio ≈ 35.4/5.9 : 17.7/5.9 : 5.9/5.9 ≈ 6 : 3 : 1. The total number of protons in the formula is 10. The sum of the ratio units (6+3+1) is 10, so the proton counts for the signals are 6H, 3H, and 1H. Isopropyl acetate (CH₃COOCH(CH₃)₂) has a 3H singlet for the acetyl methyl, a 6H doublet for the two equivalent isopropyl methyls, and a 1H septet for the isopropyl CH. This matches the 6:3:1 integration pattern. The other isomers do not: tert-butyl formate (9H, 1H), ethyl propanoate (3H, 2H, 2H, 3H), pentanoic acid (3H, 2H, 2H, 2H, 1H).

Question 9

In the ¹H NMR spectrum of 1,3,5-trimethylbenzene (mesitylene), only two sharp singlets are observed. Which principle best explains this simple spectrum?

  1. The anisotropic effect of the ring current causes all proton signals to become degenerate.
  2. Rapid interconversion between resonance structures averages all proton environments.
  3. High molecular symmetry makes all methyl protons and all aromatic protons chemically equivalent within their respective groups.
  4. High molecular symmetry makes all nine methyl protons chemically equivalent, and all three aromatic protons chemically equivalent. (correct answer)
Explanation: Mesitylene has a C₃ rotational axis and three planes of symmetry. Due to this high symmetry, all three methyl groups are in identical environments, making their 9 protons chemically equivalent (appearing as one singlet). Likewise, all three aromatic protons are chemically equivalent (appearing as another singlet). Since neither group has adjacent non-equivalent protons, both signals appear as singlets. Option C is less precise, as it doesn't specify that the methyl and aromatic protons form two separate equivalent sets.

Question 10

The protons on the central carbon of propane (CH₃-CH₂-CH₃) appear as a septet in the ¹H NMR spectrum. What is the correct explanation for this splitting pattern?

  1. The two protons on the central carbon split each other's signals.
  2. The central CH₂ protons are coupled to the six equivalent protons of the two adjacent methyl groups. (correct answer)
  3. The signal is split by the seven total protons on the adjacent carbons.
  4. The central CH₂ protons are coupled to the six protons of one methyl group and the single proton of an impurity.
Explanation: The multiplicity of a signal is determined by the number of adjacent, non-equivalent protons (n) according to the n+1 rule. The two protons of the central CH₂ group are equivalent to each other. They are adjacent to two methyl groups. Due to symmetry, the two methyl groups are equivalent, meaning all six of their protons are chemically equivalent neighbors to the central CH₂. Therefore, n = 6. The signal for the central CH₂ protons is split into n+1 = 6+1 = 7 peaks, which is a septet.

Question 11

A signal in a ¹H NMR spectrum is observed as a triplet with a coupling constant J = 7 Hz. Which of the following structural fragments is most likely responsible for this signal?

  1. An ethyl group (-CH₂CH₃), observing the -CH₃. (correct answer)
  2. An isopropyl group (-CH(CH₃)₂), observing the -CH.
  3. A -CH₂- group situated between a -CH- and a -CH₃ group.
  4. A vinyl group (-CH=CH₂), observing the CH proton.
Explanation: A triplet indicates that the proton(s) giving the signal are adjacent to a group with two protons (n=2, so n+1=3). A coupling constant of 7 Hz is typical for vicinal coupling in a freely rotating alkyl chain. In an ethyl group (-CH₂CH₃), the three protons of the methyl group (-CH₃) are adjacent to the two protons of the methylene group (-CH₂). Therefore, the methyl signal appears as a triplet. An isopropyl group's CH would be a septet. A -CH₂- between -CH- and -CH₃ would be a complex multiplet. A vinyl CH would be a doublet of doublets.

Question 12

Compound X has the molecular formula C₉H₁₀O. Its ¹H NMR spectrum shows signals at δ 7.5-7.9 (m, 5H), δ 3.0 (q, 2H), and δ 1.2 (t, 3H). What is the structure of Compound X?

  1. Phenylacetone
  2. Propiophenone (correct answer)
  3. Ethyl benzoate
  4. 1-Phenyl-1-propanol
Explanation: The formula C₉H₁₀O has a degree of unsaturation of 5. The signal at δ 7.5-7.9 (m, 5H) indicates a monosubstituted benzene ring (C₆H₅-), which accounts for 4 degrees of unsaturation. The remaining degree must be a C=O or C=C bond. The remaining atoms are C₃H₅O. The signals at δ 3.0 (q, 2H) and δ 1.2 (t, 3H) are characteristic of an ethyl group (-CH₂CH₃). The quartet at δ 3.0 is significantly downfield, suggesting it is adjacent to a carbonyl group. Assembling the pieces (C₆H₅-, C=O, and -CH₂CH₃) gives propiophenone (C₆H₅COCH₂CH₃). This structure is fully consistent with all the data.

Question 13

The ¹H NMR spectrum of 2-furaldehyde (furan-2-carbaldehyde) shows three signals for the furan ring protons, each appearing as a doublet of doublets. What is the reason for this complex splitting?

  1. Each proton is coupled to the single aldehyde proton.
  2. The molecule exists as two rapidly interconverting resonance structures.
  3. Each furan proton is coupled to two other non-equivalent furan protons. (correct answer)
  4. The furan ring undergoes rapid ring-puckering, splitting the signals.
Explanation: The furan ring in 2-furaldehyde has three protons at positions 3, 4, and 5. Let's call them H-3, H-4, and H-5. Each proton has two neighbors on the ring to which it can couple. For example, H-4 is adjacent to both H-3 and H-5. Since the electronic environment of H-3 and H-5 are different, they are non-equivalent neighbors to H-4. Therefore, H-4 is split into a doublet by H-3, and each of those peaks is split into another doublet by H-5, resulting in a doublet of doublets. A similar logic applies to H-3 and H-5, which couple to their respective neighbors. Coupling to the aldehyde proton over four bonds (A) is usually very small or non-existent.

Question 14

Which of the following isomeric ketones with formula C₆H₁₂O is expected to have the fewest signals in its ¹H NMR spectrum?

  1. 2-Hexanone
  2. 3-Hexanone
  3. 3,3-Dimethyl-2-butanone (correct answer)
  4. 4-Methyl-2-pentanone
Explanation: The number of signals in a ¹H NMR spectrum corresponds to the number of non-equivalent proton environments. We need to find the isomer with the highest degree of symmetry. A) 2-Hexanone (CH₃COCH₂CH₂CH₂CH₃) has 5 signals. B) 3-Hexanone (CH₃CH₂COCH₂CH₂CH₃) has 4 signals due to symmetry around the carbonyl. C) 3,3-Dimethyl-2-butanone (CH₃COC(CH₃)₃) has a tert-butyl group, where all 9 protons are equivalent, and a methyl group. It has only 2 signals (a 9H singlet and a 3H singlet). D) 4-Methyl-2-pentanone (CH₃COCH₂CH(CH₃)₂) has 4 signals. Therefore, 3,3-dimethyl-2-butanone has the fewest signals.

Question 15

The ¹H NMR spectrum of pure, anhydrous ethanol (CH₃CH₂OH) is recorded at low temperature. What is the expected splitting pattern for the signal corresponding to the methylene (-CH₂-) protons?

  1. A quartet, from coupling only to the methyl protons.
  2. A triplet, from coupling only to the hydroxyl proton.
  3. A doublet of quartets. (correct answer)
  4. A quintet (or pentet).
Explanation: At low temperature and in an anhydrous solvent, the rate of proton exchange for the hydroxyl group is slow enough to observe spin-spin coupling to it. The methylene (-CH₂-) protons are coupled to two non-equivalent sets of neighbors: the three protons of the methyl group (-CH₃) and the single proton of the hydroxyl group (-OH). Coupling to the three methyl protons (n=3) splits the signal into a quartet. Each peak of this quartet is then further split by the single hydroxyl proton (n=1) into a doublet. The resulting pattern is a doublet of quartets.