1h NMR and 13c NMR Spectroscopy MCQ Quiz - Objective Question with Answer for 1h NMR and 13c NMR Spectroscopy - Download Free PDF

CONCEPT:

NMR Chemical Shifts for Vinyl Protons in Ethyl Vinyl Ether

• In an ethyl vinyl ether, the vinyl group protons (Ha, Hb, Hc) exhibit distinct chemical shifts due to different environments and coupling constants.
• Typical patterns:
  • Ha: Attached directly to the oxygen-bonded vinyl carbon → more deshielded → appears downfield (highest δ)
  • Hb: Trans to Ha → slightly upfield from Ha but shows large coupling (J ≈ 13–17 Hz)
  • Hc: Cis or geminal proton → further upfield (lowest δ), smaller coupling

• Ha: most downfield, large J (13 Hz) and smaller coupling (7 Hz) → i
• Hb: coupled to Ha (13 Hz) and Hc (2 Hz) → iii
• Hc: lowest δ, small couplings (7 and 2 Hz) → ii

So the correct match is A (Ha) → I, B (Hb) → iii, C (Hc) → ii

CONCEPT:

1H NMR Splitting Patterns – AB Quartet vs. Quartet

• In proton NMR, the splitting pattern of methylene (–CH₂–) protons depends on the chemical environment and whether the two protons are magnetically equivalent or not.
• Quartet: Arises when methylene protons are equivalent and coupled to a –CH₃ (3 equivalent protons), following the n+1 rule.
• AB Quartet: Occurs when two geminal (same carbon) protons are in different chemical environments (i.e., not magnetically equivalent), leading to complex splitting due to different coupling constants.

EXPLANATION:

• Compound X: ClCH₂–C(Cl)(Br)CH₃
  • The CH₂ group is next to a highly asymmetric center (with Cl, Br, and CH₃).
  • This causes the two protons on CH₂ to be in nonequivalent environments.
  • Hence, the splitting pattern is a typical AB quartet.
    There will be a doublet for the methylene protons (CH₂), a doublet of doublets for the methine proton (CH), and a doublet for the methyl protons (CH₃). 
• 
• Compound Y: CH₃CH₂–CCl₂CH₃
  • The CH₂ group is between a CH₃ and a symmetrical CCl₂ group.
  • The environment is more symmetric, and the two CH₂ protons are equivalent.
  • This leads to a normal quartet splitting due to coupling with the adjacent CH₃.
    The 1H NMR spectrum of CH3CH2CCl2CH3 will show three signals due to the different chemical environments of the protons. The CH3 groups (both CH3CH2 and CH3C) will appear as distinct signals, and the CH2 group will also be Quartet 
• 

Therefore, the correct answer is X = AB quartet; Y = quartet.

Concept:

NMR Splitting Patterns

• The splitting pattern in NMR spectroscopy is caused by spin-spin coupling between adjacent nuclei. The number of splits (multiplet) is determined by the number of adjacent protons (n) according to the n+1 rule.

Explanation:

• Understanding the Structure
  • 1, 2-Dibromoethane (BrCH2CH2Br) is a symmetrical molecule. This means that the two CH2 groups are chemically equivalent. In NMR spectroscopy, chemically equivalent protons do not split each other.   
• Splitting Pattern
  • Since all four protons are chemically equivalent, they will appear as a single peak (a singlet) in the NMR spectrum. There will be no splitting observed.
• Reason
  • Splitting occurs due to the interaction of non-equivalent protons on adjacent carbon atoms. In 1, 2-dibromoethane, all the protons are in the same chemical environment, so there are no neighboring protons to cause splitting.   

Why Splitting Occurs

• NMR splitting happens because of the interaction between non-equivalent protons on neighboring carbons.
• If protons are chemically equivalent, they resonate at the same frequency. Since they have the same resonance, they cannot split each other. It's like two identical tuning forks – they'll vibrate together, but one won't cause the other to split its sound.
• Think of it this way:
• Imagine you have two identical twins. They look the same, act the same, and have the same experiences. If you tried to distinguish them based on their appearance or behavior, it would be impossible. Similarly, the protons in 1,2-dibromoethane are like those identical twins – NMR can't tell them apart.

Therefore, the correct answer is: No splitting.

Concept:

NMR Signals and Chemical Environments

• Nuclear Magnetic Resonance (NMR) spectroscopy is a powerful technique for determining the structure of organic compounds. The number of NMR signals corresponds to the number of distinct chemical environments in the molecule.
• Each signal in the proton NMR (1H NMR) spectrum corresponds to a set of protons (hydrogen atoms) that are in equivalent chemical environments. Equivalent protons give rise to the same signal.
• To determine the number of signals:
  • Look for non-equivalent protons—those that are in different environments due to factors like neighboring electronegative atoms or functional groups.
  • Protons attached to the same carbon or in the same environment will give a single signal.

Explanation:

​

• • In total, we can expect 4 distinct chemical environments, leading to 4 NMR signals.

Hence, the correct answer is 4 signals.

Concept:

NMR Signal Splitting Pattern

• In proton NMR spectroscopy, splitting patterns arise due to spin-spin coupling between nonequivalent hydrogen atoms on adjacent carbon atoms.
• The splitting pattern follows the n+1 rule, where n is the number of equivalent protons on the adjacent carbon atom.
• Common splitting patterns include singlet (no adjacent protons), doublet (one adjacent proton), and triplet (two adjacent protons).

Explanation:

• In the given compound:
  • The central CH proton is split into a doublet due to coupling with one of the CH₂ protons.
  • The protons on the CH₂ groups are split into a triplet due to coupling with the single central CH proton.
• Hence, the splitting pattern in increasing order is a doublet followed by a triplet.

Therefore, the correct answer is "Doublet followed by triplet."

Concept:

• The number of lines expected in 1H NMR spectrum is:
  • Protons of the same group do not interact among themselves to cause observable splitting, so they give one signal for example, the hydrogen atoms of the CH₃ group do not interact with each other.
  • The multiplicity of the peak of a group of equivalent protons is determined by the neighbouring protons. 
  • In general, if 'n' equivalents of protons interact or couple with the protons on an adjacent carbon atom, the resonance peak is split into 'n+1' peaks or signals.
  • The intensities are symmetric about the mid-point of the group and the intensities of the n+1 peaks are given by the coefficients of the binomial expansion of the order 'n', (1 + x)ⁿ.
  • These coefficients can be arranged according to Pascal's triangle as shown below:

• It can be noted that the spin spin interaction is independent of the applied magnetic field strength but the chemical shift depends on the field strength.

Explanation:

Chemical equivalent Nuclei:

• If the nucleus has exactly the same chemical shift, they are called chemical shift equivalent nuclei and designated by the same letter.
• Chemical shift equivalent nuclei can be interchanged by one of the symmetry operations of a molecule. For example, all the protons in benzene are chemically equivalent in nature.
• If nuclei are coupled in exactly the same way to every other nucleus in the molecule, they are called magnetically equivalent nuclei.
• The number of different protons in the given molecule is:

Hence, the number of signals is Three.

Explanation:-

• Enantiotropic protons are protons that are replaced by another group of protons.
• They have the same mirror image and chemical shift in the NMR spectrum.
• Therefore, Here Each Proton will give one signal, and five protons will give five signals. 
• The number of 1H NMR signals observed for the following compound is 5.

• Protons (a) and (b) are enantiotopic since replacing each gives a pair of enantiomers.

• 5 and 6 marked are enantiotropic therefor they will show 2 different peak. 

Concept:

• The number of lines expected in 1H NMR spectrum is:
  • Protons of the same group do not interact among themselves to cause observable splitting, so they give one signal, for example, the hydrogen atoms of the CH3 group do not interact with each other.
  • The multiplicity of the peak of a group of equivalent protons is determined by the neighboring protons. 
  • In general, if 'n' equivalents of protons interact or couple with the protons on an adjacent carbon atom, the resonance peak is split into 'n+1' peaks or signals.
  • The intensities are symmetric about the mid-point of the group and the intensities of the n+1 peaks are given by the coefficients of the binomial expansion of the order 'n', (1 + x)n.
  • These coefficients can be arranged according to Pascal's triangle as shown below:

• It can be noted that the spin-spin interaction is independent of the applied magnetic field strength but the chemical shift depends on the field strength.

Explanation: -

• We discussed if 'n' equivalents of protons interact or couple with the protons on an adjacent carbon atom, the resonance peak is split into 'n+1' peaks or signals. But this generalization is only valid when only protium is present as a neighbor.
• In general for all the other neighboring groups than protium, the formula used to find spin multiplicity is 2nI +1, where 'I' is the quantum spin number.
• the intensity of the splitter signal but depends only upon the n+1 rule.

Calculation:

Given,

Spin Quantum number of deuterium ²H = 1

So,  2nI +1 =  2(1)(1) +1 = 3

Thus, a triple will be produced,

The intensity depends only on n+1 is 2 hence ratios will be 1:1:1.

Conclusion:-

Hence, the correct answer is Triplet of 1 : 1 : 1 ratio i.e, option 3.

Explanation: -

The Signals in the 1H NMR Spectrum are δ 2.4 (s, 3H), 3.9 (s, 3H), 7.25 (d, J = 7 Hz, 2H), 7.95 (d, J = 7 Hz, 2H) ppm.

• In the given signals,  7.25 (d, J = 7 Hz, 2H) matches with the values of 1H NMR Spectrum of the benzene ring.
• 7.95 (d, J = 7 Hz, 2H) is also an aromatic pick that is de-shielded hence at a higher frequency hence some electron withdrawing group is attached to the benzene ring.
• We know that signal for the alcoholic group is in range of 3.2 to 2.8, this indicates the presence of the methoxy group in the compound

Both options 1 and 2 satisfy the above points.

 or

But, in option one a methoxy group is attached to the benzene ring which will de-shield the benzene ring to a higher δ value which is not happing as the signal is obtained at the normal range.

Conclusion: -

The NMR of the compound is of: -

Hence, the correct option is (3).

Concept:

• Infrared (IR) absorption bands are specific regions in the infrared spectrum where molecules absorb electromagnetic radiation at characteristic frequencies. These absorption bands arise due to the vibrational motion of atoms within a molecule. As molecules absorb infrared radiation, they undergo vibrational transitions between different energy levels, leading to the absorption of specific frequencies of light. Different types of chemical bonds and functional groups have characteristic vibrational frequencies. By analyzing the positions and intensities of IR absorption bands, we can identify the presence of certain functional groups in a molecule.
• ¹H NMR is a fundamental technique in organic chemistry, used for compound identification, structural elucidation, and quantitative analysis. It's particularly valuable in determining molecular structure and monitoring chemical reactions.
• In mass spectrometry, peak intensity refers to the height or size of a peak on a mass spectrum. Mass spectrometry is an analytical technique used to determine the mass-to-charge ratio (m/z) of ions in a sample. The resulting mass spectrum provides information about the composition and structure of molecules in the sample.

Explanation:

1.    IR absorption band of Ester is ~ 1740 cm^-1.

2. No H-atom is present in C-2 carbon atom, therefore neighboring carbon atoms will show singlet in ¹H NMR.

Conclusion:

Hence, the correct match for the following molecules is A-iii, B-ii, C-i.

Concept:

The local magnetic field experienced by a proton in NMR depends on its chemical shift and the external magnetic field applied. The relationship between the local magnetic field ( B_local ) and the chemical shift ( \(\delta\) ) is:

Formula: \(B_{\text{local}} = \delta \times B_0\) 

• Chemical Shift: The chemical shift is measured in parts per million (ppm), and it reflects the difference in the local magnetic field experienced by different protons in the molecule.

• External Magnetic Field: The external magnetic field ( B₀ ) is given in Tesla (T). In this case, B₀ = 2T.

• Local Magnetic Field: The local magnetic field experienced by each proton can be found by multiplying the chemical shift by the external magnetic field.

• Absolute Difference: The absolute difference in local magnetic fields between two protons can be calculated by subtracting the local magnetic fields calculated for each proton.

Explanation: 

• Step 1: Local magnetic field for the CH₃ proton with a chemical shift of 1.15 ppm.

  • ​\(B_{\text{CH3}} = \delta_1 \times B_0 = 1.15 \times 10^{-6} \times 2 = 2.3 \times 10^{-6} \, \text{T} \)

• Step 2: Local magnetic field for the CH₂ proton with a chemical shift of 3.35 ppm.

  • \(B_{\text{CH2}} = \delta_2 \times B_0 = 3.35 \times 10^{-6} \times 2 = 6.7 \times 10^{-6} \, \text{T}\)

• Step 3: Absolute difference in the local magnetic fields between the two protons.

  • \(\Delta B = B_{\text{CH2}} - B_{\text{CH3}} = 6.7 \times 10^{-6} - 2.3 \times 10^{-6} = 4.4 \times 10^{-6} \, \text{T}\)

Conclusion:

The absolute difference in the local magnetic fields for the two protons is 4.4 × 10^-6 T.

Concept:

→ Pyridine is a six-membered aromatic heterocycle containing five carbon atoms and one nitrogen atom. The carbon atoms in pyridine are labeled as C1 to C6, with C1 being the carbon atom attached to the nitrogen atom.

→ The chemical shift values are expressed in parts per million (ppm) and are measured relative to a reference compound, usually tetramethylsilane (TMS), which is assigned a chemical shift value of 0 ppm. The chemical shift is calculated using the following formula:

Chemical shift (δ) =\(\frac{(frequency of resonance for the nucleus - frequency of resonance for TMS)}{frequency of resonance for TMS\times10^{6}}\)

→ The frequency of resonance for the nucleus is determined by the magnetic field strength and the gyromagnetic ratio of the nucleus, which is a fundamental property of the nucleus. The magnetic field strength is usually expressed in units of tesla (T) or gauss (G), and the frequency of resonance is expressed in units of hertz (Hz).

Explanation:

→ The correct match of ¹³C NMR chemical shift values (δ ppm) for pyridine is C2: 150; C3: 124; C4: 136. This assignment is consistent with the expected chemical shifts for the carbon atoms in pyridine based on their local electronic environment.

→ In an aromatic system like pyridine, the carbon atoms experience a shielding effect due to the circulation of π electrons in the ring. This results in a deshielding effect on the carbon atoms outside the ring, which are connected to hydrogen atoms. This effect is known as the anisotropic effect.

→ The carbon atom at position C2 in pyridine is directly attached to the nitrogen atom, which has a higher electronegativity than carbon. This leads to an upfield shift in the ¹³C NMR spectrum, resulting in a chemical shift value of 150 ppm.

→ The carbon atom at position C3 in pyridine is adjacent to the nitrogen atom and is therefore also affected by the electronegativity of the nitrogen atom. However, this effect is less pronounced than for C2, resulting in a slightly downfield shift in the ¹³C NMR spectrum and a chemical shift value of 124 ppm.

→ The carbon atom at position C4 in pyridine is not directly attached to the nitrogen atom and is therefore less affected by the nitrogen's electronegativity. However, it is still located in the aromatic ring and is affected by the shielding effect of the π electrons. This results in a slightly upfield shift in the ¹³C NMR spectrum and a chemical shift value of 136 ppm.

Conclusion:
Overall, the correct match of ¹³C NMR chemical shift values (δ ppm) for pyridine is C2: 150; C3: 124; C4: 136 based on the local electronic environment of the carbon atoms in the molecule.

Concept:

¹³C NMR:

• The Carbon-13 nuclear magnetic resonance or ¹³C NMR spectroscopy is the application of nuclear magnetic resonance spectroscopy to carbon (C). ¹³C NMR spectroscopy helps to identify the nature or environment of C atoms in organic molecules just as 1H NMR spectroscopy helps to detect the H atoms.
• The nature or environment of C atoms in organic molecules can be identified by chemical shift (δ) values.
• The chemical shift values of some common functional groups are as follows:     

Explanation:-

Camphor

geraniol structure

carvone structure

• From the chemical shift values of some common functional groups, we can conclude that the natural product that gives a signal at δ 218 ppm in its 13C NMR spectrum must contain a carbonyl group.
• Apart from camphor, the other molecules only carvone contain a carbonyl functional group but it was α,β-Unsaturated carbonyl it will around 150cm^-1.
• Thus, the natural product that gives a signal at δ 218 ppm in its 13C NMR spectrum is camphor.

Conclusion:

Hence, the natural product that gives a signal at δ 218 ppm in its 13C NMR spectrum is camphor.

Concept:

• We can examine the NMR properties of Carbon atom in a molecule to determine its structure.
• About 99% of the Carbon atoms are C-12 and thus cannot be seen in an NMR.
• However, 1% of the carbon atoms are C-13 and can be observed in NMR due to an odd number of protons and neutrons.
• The number of peaks observed depends on the types of Carbon atoms present.
• Carbon atoms in a different environment will produce different peaks. Carbon atoms in the same environment will give a single peak.
• The intensity of the peak doesn't depend on the number of carbon atoms.
• Typical values of chemical shift depend on:
  • Hybridization
  • Electronegativity of the attached atom
• When ERGs are attached, they resonate at lower frequencies and when EWGs are attached, they resonate at higher frequencies.

Explanation:

• The given compound goes reaction via the following way to form P.

​

• The molecule P has a plane of symmetry and has 8 carbon atoms, so the number of peaks observed for the molecule is eight.​

Concept:

In ¹H NMR spectroscopy, each proton or group of protons in a molecule gives rise to a signal, which can be split into multiplets depending on the number of adjacent protons (coupling partners). The coupling pattern is described by multiplicity (e.g., singlet, doublet, triplet, etc.) and the coupling constant (J value), which gives insight into the nature of proton-proton interactions.

• Chemical Shift (δ): The chemical shift (δ) indicates the electronic environment around the proton. It is influenced by the electron density around the proton and neighboring atoms or functional groups.

• Coupling Constant (J): The J value gives the strength of the coupling interaction between adjacent protons, measured in Hz. Larger J values typically correspond to protons in an axial-axial relationship, while smaller J values indicate weaker interactions, such as axial-equatorial relationships.

• Multiplicity: The multiplicity depends on the number of protons coupled to the observed proton, with each coupling partner splitting the signal into n+1 peaks. For example, a proton coupled to two non-equivalent protons will result in a doublet of doublets (dd).

Explanation: 

• 

• H_b splits into doublet of doublet due to H_a and H_c

• δ for H_b is 3.15 due to J value, 7.0 Hz from H_a and 2.5 Hz from H_c.

• to identify the J value, H_b also coupled with both H_a and H_c with the same J value.

Conclusion:

Thus, the correct option is Option 4: 3.15 (dd, J = 7.0, 2.5 Hz).