i. Select the correct choice from the given options. Spectroscopy is a technique in which matter interacts with:
Answer: a) Light
Explanation: Spectroscopy is the study of the interaction between matter and electromagnetic radiation (light). This interaction includes absorption, emission, or scattering of light.
ii. Which technique is used to detect conjugation in an unsaturated hydrocarbon.
Answer: b) UV-Visible spectroscopy
Explanation: UV-Visible spectroscopy is used to detect conjugation in unsaturated hydrocarbons. Extended conjugation lowers the energy gap between HOMO and LUMO, shifting absorption to longer wavelengths (visible region).
iii. Coloured compounds absorb in the wavelength between:
Answer: d) 400 to 800 nm
Explanation: Coloured compounds absorb light in the visible region of the electromagnetic spectrum, which ranges from approximately 400 nm (violet) to 800 nm (red).
iv. IR spectroscopy is used to detect:
Answer: a) Functional group
Explanation: Infrared (IR) spectroscopy is primarily used to identify functional groups in a molecule based on their characteristic
vibrational frequencies (e.g., C=O, O-H, N-H).
v. Which radiations are used in NMR spectroscopy?
Answer: d) Radio waves
Explanation: Nuclear Magnetic Resonance (NMR) spectroscopy uses radiofrequency (RF) radiation to excite the nuclear spins of atoms (typically ¹H or ¹³C) in a strong magnetic field.
vi. Which bond shows greater chemical shift in ¹H NMR?
Answer: a) O-H
Explanation: The O-H proton is highly deshielded because oxygen is highly electronegative, withdrawing electron density. This results in a larger chemical shift (downfield, typically 2-5 ppm, variable) compared to N-H (1-3 ppm), C-H (0.5-2 ppm), and Si-H (3-4 ppm).
i. Write basic principle of IR spectroscopy.
Answer: The basic principle of IR spectroscopy is that molecules absorb infrared radiation at specific frequencies that correspond to the vibrational energies of their chemical bonds. When the frequency of the incident IR radiation matches the natural vibrational frequency of a bond, the bond absorbs the energy and its vibrational amplitude increases. For a bond to absorb IR radiation, there must be a change in the dipole moment during the vibration. The resulting IR spectrum provides a "fingerprint" of the molecule's functional groups.
ii. How can you identify functional group of an organic compound by using IR spectroscopy?
Answer: Functional groups can be identified by IR spectroscopy because each functional group absorbs IR radiation at a characteristic frequency range. By examining the absorption peaks in the IR spectrum, one can identify the presence of specific functional groups:
O-H (alcohol): Broad peak at 3200-3600 cm⁻¹. O-H (carboxylic acid): Very broad peak at 2500-3300 cm⁻¹.
C=O (aldehyde, ketone, ester): Strong peak at 1700-1750 cm⁻¹. N-H (amine, amide): Peak at 3300-3500 cm⁻¹. C[source symbol unavailable]N (nitrile): Peak at 2200-2250 cm⁻¹.
C=C (alkene): Peak at 1620-1680 cm⁻¹.
iii. With the help of UV- Visible spectrum of an unseen compound, how can you predict it's colour?
Answer: The UV-Visible spectrum shows the wavelengths of light absorbed by the compound. The colour of the compound is the complementary colour of the absorbed light. For example:
If a compound absorbs in the violet region (~400-435 nm), it will appear yellow-green. If it absorbs in the green region (~500-560 nm), it will appear red. If it absorbs in the red region (~650-750 nm), it will appear green. By identifying the wavelength of maximum absorption (λmax), one can predict the observed colour using a colour wheel.
iv. Draw a rough UV-Visible spectrum of a compound which is yellow in colour.
Answer: (Diagram Description) A compound that is yellow absorbs in the violet/blue region of the spectrum (approximately 400-435 nm). The UV-Visible spectrum would show a peak (λmax) in this region. The absorbance would be high in the violet/blue region and
low in the yellow region (since yellow light is transmitted). The y-axis is Absorbance, and the x-axis is Wavelength (nm).
v. Explain principles of AES and AAS.
Answer:
AES (Atomic Emission Spectroscopy): The principle is that when atoms are excited by heat or electrical energy, they emit light at characteristic wavelengths as they return to the ground state. The intensity of the emitted light is proportional to the concentration of the element. AAS (Atomic Absorption Spectroscopy): The principle is that when light of a specific wavelength is passed through an atomic vapor, the atoms absorb light at their characteristic wavelengths, transitioning from the ground state to an excited state. The amount of light absorbed is proportional to the concentration of the element in the sample.
vi. How is atomic absorption spectroscopy different from atomic emission spectroscopy?
Answer:
| Feature | Atomic Absorption Spectroscopy (AAS) | Atomic Emission Spectroscopy (AES) |
| Principle | Measures absorption of light by ground-state atoms. | Measures emission of light by excited-state atoms. |
| Excitation Source | External light source (hollow cathode lamp). | Heat or electrical energy (flame, plasma). |
| Process | Atoms absorb light. | Atoms emit light. |
| Measurement | Decrease in light intensity. | Increase in light intensity. |
| Example | Determining lead in water. | Determining sodium in a flame test. |
vii. What are NMR active nuclei? How can you tell about a nucleus, whether it is NMR active or not?
Answer: NMR active nuclei are those that possess a nuclear spin (I [source symbol unavailable] 0). A nucleus is NMR active if it has an odd number of protons
or an odd number of neutrons, resulting in a non-zero nuclear spin quantum number (I = 12, 1, 32, etc.). Examples include ¹H, ¹³C,
¹⁵N, ¹⁹F, and ³¹P. Nuclei with even numbers of both protons and neutrons (e.g., ¹²C, ¹⁶O) have zero spin (I = 0) and are NMR inactive.
viii. What is the effect of external magnetic field on the spinning nuclei?
Answer: In the absence of an external magnetic field, nuclear spins are randomly oriented. When an external magnetic field (B₀) is applied, the spinning nuclei align themselves either parallel (lower energy, α state) or antiparallel (higher energy, β state) to the applied field. The energy difference between these two states (ΔE) is proportional to the strength of the magnetic field. This splitting of energy levels is the basis of NMR spectroscopy.
ix. Explain chemical shift.
Answer: Chemical shift is the position of an NMR signal on the x-axis of an NMR spectrum, expressed in parts per million (ppm)
relative to a reference standard (usually tetramethylsilane, TMS). It is defined as: δ = νsample - νreferenceνspectrometer × 10⁶. It reflects the electronic environment of the nucleus. Nuclei that are deshielded by electronegative atoms appear at higher chemical
shifts (downfield), while shielded nuclei appear at lower chemical shifts (upfield).
x. Why do the peaks in NMR spectrum split into doublet, triplets or quartets etc?
Answer: The splitting of NMR peaks (spin-spin coupling or J-coupling) is caused by the interaction of the magnetic moment of a proton with the magnetic moments of neighboring non-equivalent protons. The n+1 rule predicts the multiplicity:
A proton with n equivalent neighboring protons will split into n+1 peaks.
Doublet: 1 neighboring proton (n=1).
Triplet: 2 neighboring protons (n=2).
Quartet: 3 neighboring protons (n=3). The spacing between the split peaks is called the coupling constant (J), which is a measure of the interaction strength.
xi. What is deuterium exchange method and what is it's use?
Answer: The deuterium exchange method involves adding a few drops of D₂O (heavy water) to a sample and recording the NMR spectrum. Protons attached to heteroatoms (e.g., -OH, -NH, -COOH) are exchangeable and are replaced by deuterium. Since deuterium is NMR inactive, the signal for that proton disappears. This method is used to identify exchangeable protons (like OH and NH) in a molecule, distinguishing them from non-exchangeable C-H protons.
xii. Why does acetone give a single peak in proton-NMR and two peaks in ¹³C-NMR? Explain with reason.
Answer:
¹H-NMR: Acetone (CH₃COCH₃) has six chemically equivalent hydrogen atoms (the two methyl groups are identical due to symmetry). Therefore, it gives a single peak.
¹³C-NMR: Acetone has two types of carbon atoms: the methyl carbons (CH₃) and the carbonyl carbon (C=O). Although the two methyl carbons are equivalent, they are different from the carbonyl carbon. Therefore, there are two distinct signals in the ¹³C-NMR spectrum.
xiii. What is TMS? Tell its significance?
Answer: TMS stands for Tetramethylsilane (Si(CH₃)₄). It is the standard reference compound used in NMR spectroscopy. Its significance:
It is assigned a chemical shift of 0 ppm. It is inert, volatile, and soluble in most organic solvents. It gives a single, sharp peak (all 12 hydrogens are equivalent). Its signal appears upfield (at 0 ppm) from most organic protons, so it does not interfere with sample signals.
xiv. Why deuterated solvents are used in proton NMR spectroscopy?
Answer: Deuterated solvents (e.g., CDCl₃, D₂O, DMSO-d₆) are used in ¹H-NMR spectroscopy because they do not contain protons (they contain deuterium, which is NMR inactive for ¹H). If ordinary solvents were used, their protons would give a huge signal that would swamp the signals of the sample. Deuterated solvents provide a "transparent" medium for observing the sample's protons.
i. Write the details of IR spectroscopy including its principle, spectrum and uses.
Answer:
Principle: IR spectroscopy is based on the absorption of infrared radiation by molecules, causing transitions between vibrational energy levels. For a vibration to be IR active, there must be a change in the dipole moment of the molecule.
Spectrum: An IR spectrum is a plot of transmittance (%T) versus wavenumber (cm⁻¹). The x-axis typically ranges from 4000 cm⁻¹ to 400 cm⁻¹. Absorption peaks appear as downward spikes (valleys) in the transmittance plot. The position, intensity, and shape of these peaks provide information about the functional groups and bonding in the molecule.
Uses:
Identification of Functional Groups: Each functional group has a characteristic absorption frequency (e.g., C=O at ~1700 cm⁻¹, O-H at ~3300 cm⁻¹). Structure Elucidation: The "fingerprint region" (below 1500 cm⁻¹) is unique to each molecule and can be used to confirm identity. Detection of Impurities: Extra peaks indicate impurities. Studying Reaction Progress: Disappearance of reactant peaks and appearance of product peaks. Quantitative Analysis: Beer-Lambert law can be applied to determine concentrations.
In summary, IR spectroscopy is a powerful tool for identifying functional groups and determining molecular structure.
ii. Explain the uses of UV-visible spectroscopy.
Answer: UV-Visible spectroscopy has numerous uses in chemistry and related fields:
Identification of Conjugation: It detects conjugated systems (alternating single and double bonds) in molecules. Extended conjugation shifts absorption to longer wavelengths (visible region).
Quantitative Analysis: Using the Beer-Lambert Law (A = εcl), the concentration of a substance in solution can be determined by measuring its absorbance at a specific wavelength.
Determination of λmax: The wavelength of maximum absorption gives information about the electronic structure of the molecule. Detection of Impurities: The presence of impurities can be detected by comparing the spectrum with that of a pure sample. Studying Reaction Kinetics: The change in absorbance over time can be used to monitor the rate of a reaction. Determination of pKa: The UV-Visible spectrum changes with pH, allowing determination of acid dissociation constants. Analysis of Pharmaceuticals: Used for quality control and drug analysis.
In summary, UV-Visible spectroscopy is widely used for both qualitative and quantitative analysis in chemistry, biochemistry, and industry.
iii. Nuclear magnetic resonance is a powerful tool for the elucidation of structures of organic compounds. Defend the statement.
Answer: Nuclear Magnetic Resonance (NMR) spectroscopy is indeed a powerful tool for structure elucidation because it provides detailed information about the carbon-hydrogen framework of organic molecules.
Number of Signals: The number of signals in a ¹H-NMR spectrum indicates the number of chemically non-equivalent proton environments. In ¹³C-NMR, it indicates the number of non-equivalent carbon atoms. Chemical Shift: The position of each signal (chemical shift) provides information about the electronic environment of the proton or carbon. It indicates the presence of electronegative atoms, functional groups, and hybridization. Integration: The area under each ¹H-NMR signal (integration) is proportional to the number of protons in that environment, giving the ratio of different types of protons. Splitting Pattern: The splitting of signals (multiplicity) due to spin-spin coupling reveals the number of neighboring non- equivalent protons, providing information about the connectivity of the molecule. Coupling Constants: The magnitude of the coupling constant (J) provides information about the spatial relationship between coupled protons (e.g., cis/trans in alkenes). 2D NMR: Advanced techniques (COSY, HSQC, HMBC) provide correlations between coupled nuclei, allowing for the
complete assembly of complex molecular structures.
In summary, NMR provides a wealth of information (number, type, ratio, and connectivity of atoms) that is essential for determining the structure of organic compounds.
iv. How is PMR different from CMR?
Answer:
| Feature | PMR (Proton Magnetic Resonance) / ¹H-NMR | CMR (Carbon Magnetic Resonance) / ¹³C-NMR |
| Nucleus | Proton (¹H) | Carbon-13 (¹³C) |
| Natural Abundance | ~99.98% | ~1.1% |
| Sensitivity | High | Low (due to low abundance) |
| Chemical Shift Range | ~0-12 ppm | ~0-200 ppm |
| Integration | Area is proportional to number of protons. | Area is not always proportional to number of carbons (unless decoupled). |
| Splitting | Splitting occurs due to neighboring protons | Proton-decoupled ¹³C-NMR gives singlets for each carbon. |
| (n+1 rule). | ||
| Information | Number and type of protons. | Number and type of carbons. |
| Reference | TMS (0 ppm) | TMS (0 ppm) |
In summary, PMR provides information about hydrogen atoms, while CMR provides information about the carbon skeleton of the molecule.
Calculate the degree of unsaturation in:
a. Benzene (C₆H₆)
Answer: Degree of Unsaturation = 4 (3 double bonds + 1 ring).
b. Ethyne (C₂H₂)
Answer: Degree of Unsaturation = 2 (1 triple bond).
Q1. A compound absorbs the light with a wavelength of 500nm-560nm. What colour do you expect for this compound?
Answer: The compound absorbs green light (500-560 nm) and will appear red/magenta (the complementary color).
Q2. What wavelength do you suggest about the absorption of light by the components of air?
Answer: The components of air (N₂, O₂, Ar) do not absorb visible light. They absorb in the ultraviolet region (below 200 nm).
Q1. How many peaks do you expect in the proton NMR of benzene?
Answer: 1 peak (all six hydrogens are equivalent).
Q2. Why the peak of two protons attached with alpha carbon of ethyl alcohol have greater chemical shift than the ethylene group of ethyl chloride?
Answer: The -OH group in ethanol is more electronegative than the -Cl group in ethyl chloride, causing greater deshielding of the adjacent CH₂ protons and a higher chemical shift.
Q3. Given is a ¹H-NMR spectrum of an organic compound "x" with a molecular formula C₄H₈O. Deduce its structural formula by using its ¹H-NMR spectrum.
Answer: The spectrum (quartet at ~4.0 ppm, triplet at ~1.0 ppm) indicates an ethyl group attached to an electronegative atom. The
structure is ethyl ethanoate (ethyl acetate), CH₃COOCH₂CH₃.
Q1. How many peaks do you expect in the NMR spectrum of acetone? If it is a single peak describe the reason.
Answer: 1 peak. All six hydrogen atoms in acetone (CH₃COCH₃) are chemically equivalent.
Q2. What would be the difference between the proton NMR spectrum of ethyl alcohol and ethyl chloride? Why is the chemical shift of OH proton in alcohol greater?
Answer: Ethanol shows three signals (CH₃, CH₂, OH); ethyl chloride shows two signals (CH₃, CH₂). The OH proton in ethanol is deshielded by the electronegative oxygen and hydrogen bonding, causing a higher chemical shift.
Q3. By using the information given in the C-13 NMR of the compound "y" answer the given questions. (Spectrum shows peaks at 8.9, 27.6, and 181.5 ppm)
a. How many carbon atoms are present in compound "y"?
Answer: 3 (three distinct signals).
b. What is the chemical shift of CH₃ carbon out of these three values?
Answer: 8.9 ppm.
c. One C is dirty attached to two oxygen atoms, can you tell it's chemical shift?
Answer: 181.5 ppm (carbonyl carbon).
d. Deduce the possible structure of compound "y" if its molecular mass is 74 amu.
Answer: Propanoic acid (CH₃CH₂COOH).
Q1. How many types of C atoms are present in ethoxy ethane (diethyl ether)? Draw an approximate ¹³C-NMR spectrum of this molecule taking help from the given table of chemical shifts.
Answer: 2 types of carbon atoms (CH₃ and CH₂). The spectrum shows two peaks: one at ~15 ppm (CH₃) and one at ~65 ppm (CH₂).
Q2. Why are the chemical shifts in ¹³C-NMR spectrum much greater as compared to the ¹H-NMR?
Answer: Carbon atoms experience a wider range of electronic environments due to different hybridizations, functional groups, and electronegative substituents, resulting in a larger chemical shift range (0-200 ppm vs 0-12 ppm for ¹H).
Draw approximate proton NMR spectrum and ¹³C-NMR spectrum of 3-pentanone.
Answer: Structure: 3-Pentanone is CH₃CH₂COCH₂CH₃ (diethyl ketone).
¹H-NMR Spectrum (Approximate):
Triplet at ~1.0 ppm: For the two terminal CH₃ groups (6H, split by adjacent CH₂, n+1 = 3).
Quartet at ~2.4 ppm: For the two CH₂ groups (4H, split by adjacent CH₃, n+1 = 4). Integration Ratio: 6:4 (or 3:2).
¹³C-NMR Spectrum (Approximate):
Peak at ~8 ppm: For the two equivalent CH₃ carbons. Peak at ~35 ppm: For the two equivalent CH₂ carbons.
Peak at ~210 ppm: For the carbonyl carbon (C=O).
Number of Signals: 3 signals (CH₃, CH₂, C=O).
