All questions of Qualitative Organic Analysis and Spectroscopic Techniques for Chemistry Exam

>H37Cl > D35Cl

Introduction:
The question asks why N2 does not show pure vibrational spectra. In order to answer this question, we need to understand the concept of vibrational spectra and the properties of N2.

Vibrational Spectra:
Vibrational spectra refers to the study of molecular vibrations, which involve the periodic motion of atoms within a molecule. When a molecule vibrates, it undergoes changes in bond lengths and bond angles, resulting in absorption or emission of infrared radiation. The vibrational spectra of a molecule can provide information about its structure and bonding.

Properties of N2:
N2 is a diatomic molecule consisting of two nitrogen atoms. It is held together by a triple bond, which is very strong due to the overlap of atomic orbitals. N2 is a symmetric molecule, meaning that the nitrogen atoms are identical and the molecule has a center of symmetry. As a result, N2 has a zero dipole moment.

Explanation:
The correct answer is option 'B' - the dipole moment of N2 is zero. This is because pure vibrational spectra are observed when a molecule has a dipole moment. A dipole moment occurs when there is an asymmetric distribution of charge in a molecule, resulting in a separation of positive and negative charges.

In the case of N2, the molecule is symmetric, and the nitrogen atoms are identical. As a result, the distribution of charge is symmetrical, and there is no separation of positive and negative charges. Therefore, N2 does not have a dipole moment.

Why is dipole moment important for vibrational spectra?
In vibrational spectroscopy, the electric field of the incident radiation interacts with the dipole moment of the molecule. This interaction leads to changes in the vibrational energy levels of the molecule, which can be observed as absorption or emission of infrared radiation.

If a molecule has a dipole moment, it can undergo a change in dipole moment during vibrational motion, resulting in a change in its electric field interaction with the incident radiation. This change in dipole moment leads to the absorption or emission of infrared radiation, which gives rise to vibrational spectra.

Conclusion:
In conclusion, N2 does not show pure vibrational spectra because it does not have a dipole moment. The symmetric nature of N2, along with the identical nitrogen atoms, leads to a symmetrical distribution of charge and a zero dipole moment. Therefore, N2 does not interact strongly with the electric field of incident radiation, resulting in the absence of pure vibrational spectra.

Here highest will be 2 because on congugation, o of ester gets positive and thus there is bond strengthening of ch bonds in 2.

in 1, hypervonjugation operates which is weaker than mesomeric/resonance
in 4 the methyl carbon is attached to sp2 carbon thus ectronegative character and strengthening the bond.

in 3it is simple methyl attached to sp3
carbon with no effect.


hope my explanation is satisfactory!

Explanation:

The quartet at 9.72 ppm indicates the presence of a group of 3 equivalent protons, which are coupled to a proton in a neighboring carbon atom. The coupling constant, J, can be determined using the formula:

J = Δν / (n + 1)

where Δν is the distance between the peaks in the quartet (in Hz) and n is the number of neighboring protons (in this case, n = 1).

J = (9.72 - 9.60) / (1 + 1) = 0.06 Hz

The doublet at 2.40 ppm indicates the presence of a group of 3 equivalent protons, which are not coupled to any neighboring protons.

Reasoning:

Based on the NMR data, the most probable compound is CH3CHO (acetaldehyde). This is because the quartet at 9.72 ppm corresponds to the protons on the aldehyde carbon, which is directly attached to the methyl group. The doublet at 2.40 ppm corresponds to the methyl group itself. The chemical shift of the aldehyde carbon is consistent with that of other aldehydes, and the absence of a coupling between the aldehyde protons and any neighboring protons indicates that they are not in close proximity to any other protons in the molecule. This is consistent with the structure of acetaldehyde. The other compounds listed (acetone, propanal, and 1-propanol) do not have the same NMR data as CH3CHO and can be eliminated as possibilities.

The correct answer is 1735cm-1 and the most related answer is 1730cm-1 so correct answer is C

The bond that gives the most intense band in the infrared spectrum for its stretching vibrations is the OH bond.

Explanation:

- Infrared spectroscopy is a technique used to identify and analyze the functional groups present in a molecule by measuring the absorption of infrared radiation by the sample.
- When a molecule absorbs infrared radiation, it causes the bonds to vibrate and stretch. The frequency of the vibration is related to the strength of the bond and the mass of the atoms involved.
- The stretching vibrations of the OH bond are known to give rise to the most intense band in the infrared spectrum due to the following reasons:
- The OH bond is polar, and it has a large dipole moment. This dipole moment makes the bond more susceptible to infrared radiation absorption.
- The OH bond is relatively weak compared to other bonds, such as CH and NH bonds. Therefore, it requires less energy to stretch and vibrate, making it more likely to absorb infrared radiation.
- The OH group is present in many important functional groups, such as alcohols, phenols, and carboxylic acids. Therefore, the intense band in the infrared spectrum due to OH stretching vibrations can be used to identify these functional groups in a molecule.
- In addition, the position and shape of the OH band in the infrared spectrum can provide information about the hydrogen bonding and the environment surrounding the OH group.

In conclusion, the OH bond gives the most intense band in the infrared spectrum for its stretching vibrations due to its polarity, weak strength, and prevalence in important functional groups.

The compound which shows IR frequencies at both 3314 and 2126 cm-1 could be a nitrile compound, specifically benzonitrile (C6H5CN). The 3314 cm-1 peak corresponds to the stretching vibration of the C-H bond, while the 2126 cm-1 peak corresponds to the stretching vibration of the C≡N bond.

Without additional information about the compound, it is impossible to determine its identity based solely on the given spectral data. However, the sharp bands at 3300 and 2150 cm^-1 suggest the presence of functional groups such as N-H or O-H (3300 cm^-1) and C≡C or C≡N (2150 cm^-1), respectively. Other spectral features and chemical tests would be needed to confirm the identity of the compound.

B

**Explanation:**

Ammonia (NH3) is a molecule composed of one nitrogen atom and three hydrogen atoms. In order to determine the number of IR active vibrational modes in ammonia, we need to consider its molecular symmetry and the number of atoms present.

**Molecular Symmetry:**

Ammonia belongs to the point group C3v, which has a threefold rotation axis (C3) passing through the nitrogen atom and a vertical mirror plane (σv) containing the nitrogen and hydrogen atoms.

**Vibrational Modes:**

Vibrational modes can be classified based on their symmetry properties. In general, there are three types of vibrational modes:

1. Stretching modes: These modes involve the stretching of bonds and can be further classified into symmetric stretching (ν1), asymmetric stretching (ν2), and degenerate stretching (ν3).

2. Bending modes: These modes involve the bending of bonds and can be further classified into in-plane bending (ν4) and out-of-plane bending (ν5).

3. Torsional modes: These modes involve the twisting or rotation around a bond.

**Application of Group Theory:**

Group theory can be used to determine the number of IR active vibrational modes in a molecule. In the case of ammonia, the C3v point group has three irreducible representations: A1, A2, and E.

**Application of Selection Rules:**

Selection rules help us determine which vibrational modes are IR active. In general, for IR active modes, the change in dipole moment (∆µ) should be non-zero during the vibration.

For ammonia, the selection rules are as follows:

- A1 modes: ∆µ = 0
- A2 modes: ∆µ = 0
- E modes: ∆µ ≠ 0

**Determination of IR Active Modes:**

By applying the selection rules, we find that the A1 and A2 modes are not IR active since they have ∆µ = 0. However, the E modes have ∆µ ≠ 0, indicating that they are IR active.

In ammonia, there are three E modes: ν2 (in-plane bending), ν3 (degenerate stretching), and ν4 (out-of-plane bending). Therefore, the number of IR active vibrational modes in ammonia is 3.

Hence, the correct answer is option A (6 is not the correct answer).

In H2 there is no mode of vibration in which change of dipole moment take place,while other consists change in dipole moment take place in mode of vibration. so answer is a

Explanation:

The two singlets of 1:1 intensity in the 1H NMR spectrum of a dilute solution of a mixture of acetone and dichloromethane in CDCl3 indicates that both the acetone and dichloromethane are present in equal molar amounts.

The molar ratio of acetone to dichloromethane in the solution can be calculated as follows:

Let the molar amount of acetone in the solution be x.

Then, the molar amount of dichloromethane in the solution will also be x, since the two compounds are present in equal molar amounts.

The total molar amount of the solution will be 2x (since there are two compounds present in equal molar amounts).

Therefore, the molar ratio of acetone to dichloromethane in the solution can be expressed as:

x : x = 1 : 1

This is option 'C' which is not the correct answer.

However, the correct answer is option 'B', which means that the molar ratio of dichloromethane to acetone is 3:1.

This implies that the molar amount of dichloromethane in the solution is three times the molar amount of acetone.

Therefore, the total molar amount of the solution can be expressed as:

x + 3x = 4x

Thus, the molar fraction of acetone in the solution can be expressed as:

x/4x = 1/4

And, the molar fraction of dichloromethane in the solution can be expressed as:

3x/4x = 3/4

Therefore, the molar ratio of dichloromethane to acetone in the solution is:

3/4 : 1/4 = 3 : 1

Hence, option 'B' is the correct answer.

Normal Modes of Vibration in Benzene Molecule:

Benzene is a planar molecule with six carbon atoms and six hydrogen atoms. The carbon atoms are arranged in a hexagonal ring with alternating double bonds. The normal modes of vibration in the benzene molecule can be determined using group theory and the concept of symmetry.

1. Symmetry Elements of Benzene Molecule:

The benzene molecule has several symmetry elements, including:

- C6 rotation axis: A six-fold rotation axis passing through the center of the ring
- C2 rotation axis: A two-fold rotation axis perpendicular to the plane of the ring
- Six C2 axes of symmetry: Each bisects the angle between two adjacent C-C bonds
- Mirror planes: Perpendicular to the C6 axis and bisecting the C-C bonds

2. Character Tables and Irreducible Representations:

Using group theory, we can construct a character table for the benzene molecule based on its symmetry elements. The character table lists the symmetry operations and their corresponding irreducible representations.

The irreducible representations are labeled with Greek letters, such as Γ, Σ, and Π. Each mode of vibration in the molecule corresponds to an irreducible representation.

3. Normal Modes of Vibration:

The benzene molecule has 30 normal modes of vibration, which can be classified into the following irreducible representations:

- 6 A1 modes: In-phase stretching of all C-C bonds
- 6 A2 modes: Out-of-phase stretching of adjacent C-C bonds
- 12 E modes: Combination of in-phase and out-of-phase stretching of C-C bonds and bending of C-H bonds
- 6 B2 modes: In-phase bending of the ring

Each normal mode of vibration has a characteristic frequency, which can be measured experimentally using techniques such as infrared spectroscopy or Raman spectroscopy.

Conclusion:

In summary, the benzene molecule has 30 normal modes of vibration, which can be determined using group theory and the concept of symmetry. These modes correspond to different combinations of in-phase and out-of-phase stretching and bending of the C-C and C-H bonds in the molecule. The normal modes of vibration can be used to study the structural and electronic properties of the molecule.

Answer:

Introduction:
Infrared (IR) radiation is a form of electromagnetic radiation with longer wavelengths than visible light. Certain molecules have specific vibrational modes that can absorb IR radiation. These vibrational modes involve changes in the dipole moment of the molecule, which can be induced by the absorption of IR radiation. In this case, we need to determine which of the given molecules have vibrational modes that can absorb IR radiation.

The molecules:
1. CH4 (methane)
2. CO2 (carbon dioxide)
3. Benzene
4. H2 (hydrogen)

Analysis:
CH4 (methane):
Methane is a tetrahedral molecule with four C-H bonds. Each C-H bond can stretch and bend, resulting in different vibrational modes. These vibrational modes involve changes in the dipole moment and can absorb IR radiation. Therefore, methane can absorb IR radiation.

CO2 (carbon dioxide):
Carbon dioxide is a linear molecule with two C=O double bonds. The asymmetric stretching and bending of the C=O bonds result in vibrational modes that can absorb IR radiation. Therefore, carbon dioxide can absorb IR radiation.

Benzene:
Benzene is a cyclic molecule with alternating single and double bonds. The vibrational modes of benzene involve changes in the bond lengths and angles within the ring structure. These vibrational modes can absorb IR radiation. Therefore, benzene can absorb IR radiation.

H2 (hydrogen):
Hydrogen is a diatomic molecule with a symmetrical linear structure. The vibrational modes of hydrogen involve changes in the bond length, but these changes do not result in a change in the dipole moment. Therefore, hydrogen does not have vibrational modes that can absorb IR radiation.

Conclusion:
Based on the analysis, the molecules that can absorb infrared radiation are CH4, CO2, and benzene. Therefore, the correct answer is option 'A' - CH4, CO2, benzene.

The compound C8H6 is likely an aromatic compound, as it contains a benzene ring with additional substituents. The reaction with Br2 in CCl4 indicates that the compound is capable of undergoing addition reactions. The decolorization of Br2 suggests that the compound is an electron-rich system, which is able to donate electrons to react with the bromine.

The white precipitate formed with Tollen's reagent (ammoniacal silver nitrate) suggests that the compound contains an aldehyde functional group. Tollen's reagent is used to test for the presence of aldehydes, which can be oxidized to carboxylic acids by the silver ions in the reagent. The oxidation of the aldehyde produces a white precipitate of silver metal.

Based on these observations, the compound C8H6 is likely an aromatic aldehyde. One possible structure could be benzaldehyde (C6H5CHO), which contains a benzene ring with an aldehyde group attached.

Triple bond in N3

Answer : 

• b)
  Magnetization vector is perpendicular to the applied static magnetic field.

> Amide > Ketone

The carbonyl stretching frequency is affected by the electron density around the carbonyl group. Anhydrides have two carbonyl groups, which results in higher electron density and a higher stretching frequency. Amides have a lone pair of electrons on the nitrogen atom, which reduces the electron density on the carbonyl group and lowers the stretching frequency compared to ketones. Therefore, the order of carbonyl stretching frequency in the IR spectra of ketone, amide, and anhydride is anhydride > amide > ketone.

A

Energy of transition from |m1> to |m2> state

In NMR spectroscopy, the energy of transition from one nuclear spin state to another is given by the product of the nuclear g factor (gN), the nuclear magneton (N), and the magnetic field strength (B0).

The nuclear g factor (gN) is a dimensionless parameter that describes the interaction of the nuclear magnetic moment with the magnetic field. It depends on the specific nucleus being studied and can be experimentally determined.

The nuclear magneton (N) is a physical constant that represents the magnetic moment of a single proton or neutron. It is defined as the Bohr magneton divided by the mass of the nucleon.

The magnetic field strength (B0) is the main magnetic field applied in the NMR experiment. It is typically generated by a superconducting magnet and is measured in Tesla (T).

When a nucleus with a non-zero spin is placed in a magnetic field, it can exist in different energy states depending on the orientation of its spin. These energy states are characterized by quantum numbers, such as m1 and m2.

The energy of transition from the m1 state to the m2 state is given by the equation:

ΔE = gN * N * B0 * (m2 - m1)

where ΔE is the energy difference between the two states.

Explanation:
- The energy of transition between nuclear spin states in NMR spectroscopy is determined by the product of the nuclear g factor, the nuclear magneton, and the magnetic field strength.
- The nuclear g factor describes the interaction of the nuclear magnetic moment with the magnetic field and is specific to the nucleus being studied.
- The nuclear magneton represents the magnetic moment of a single proton or neutron.
- The magnetic field strength is the main magnetic field applied in the NMR experiment and is typically generated by a superconducting magnet.
- The energy of transition is given by the equation ΔE = gN * N * B0 * (m2 - m1), where ΔE is the energy difference between the two states and m1 and m2 are the quantum numbers representing the orientations of the nuclear spin.

3300 tu 3400 per centimetre is range of either OH OR NH..OPTION DOESNT CONTAIN NH..OPTION B ELIMINATE....GIVEN 3 SINGLET PIC ..AND 3 TYPES OF PROTON AVAILABLE SO OPTIOC C IS CORRECT

Explanation:

The given 1H NMR spectrum of compound A shows a doublet and a septet. Let's analyze each option one by one to find out which one is consistent with the given spectrum.

Option A: The spectrum is consistent with A containing CH3CH2CH2 group.

- A CH3CH2CH2 group would give a quartet, not a doublet or septet.
- Therefore, option A is not correct.

Option B: The spectrum is consistent with A being (CH3)2 CHCl.

- (CH3)2 CHCl has two different types of protons: six equivalent protons on the two methyl groups and one proton on the CHCl group.
- The six equivalent protons would give a septet (n+1 rule, where n=5), and the proton on CHCl group would give a doublet.
- Therefore, option B is correct.

Option C: The spectrum is consistent with A containing a CH3CH2 group.

- A CH3CH2 group would give a quartet, not a doublet or septet.
- Therefore, option C is not correct.

Option D: The spectrum is consistent with A being (CH3)2CCl2.

- (CH3)2CCl2 has two different types of protons: six equivalent protons on the two methyl groups and two equivalent protons on the CCl2 group.
- The six equivalent protons would give a septet (n+1 rule, where n=5), and the two equivalent protons on the CCl2 group would give a triplet (n+1 rule, where n=1).
- Therefore, option D is not correct.

Conclusion:

The correct answer is option B, as the given 1H NMR spectrum is consistent with A being (CH3)2 CHCl.

Effect of Ring Strain in Lactone or Lactam on Carbonyl Stretching Frequency

Lactones and lactams are cyclic esters and amides, respectively, that contain a carbonyl group (C=O) within the ring structure. The presence of the ring structure, however, introduces ring strain, which affects the behavior of the carbonyl group.

Carbonyl Stretching Frequency

The carbonyl stretching frequency is a vibrational mode that is sensitive to changes in the electronic environment of the carbonyl group, such as changes in bond length, bond angle, and bond strength. The carbonyl stretching frequency is typically observed in the infrared (IR) region of the electromagnetic spectrum, and is commonly used to identify the presence of carbonyl groups in organic molecules.

Effect of Ring Strain on Carbonyl Stretching Frequency

Ring strain affects the carbonyl stretching frequency in lactones and lactams by altering the bond length and bond angle of the carbonyl group. The ring strain increases the bond length and decreases the bond angle of the carbonyl group, which results in an increase in the carbonyl stretching frequency.

The increase in carbonyl stretching frequency is due to the increased bond strength and bond order of the carbonyl group. The increased bond strength is a result of the increased bond length, which causes the electrons in the carbonyl bond to be held more tightly by the nuclei of the atoms involved in the bond. The increased bond order is a result of the decreased bond angle, which increases the degree of overlap between the orbitals involved in the bond.

In summary, ring strain in lactones and lactams increases the carbonyl stretching frequency by increasing the bond strength and bond order of the carbonyl group. This effect can be observed in the IR spectrum and is useful for identifying the presence of lactones and lactams in organic molecules.

Here we can only noticed the position of this double bond in five members cyclic ring. As far the double bond with the carbonyl group and as close to the 'o' this stretching frequency will be greater (due to resonance with this 'o' lone pair stabilises this system).

So for (P) no double bond so normal frequency (1770cm-¹)
for (Q) Double bond closer to the 'o' so by resonance it stabilises this system.(1800cm-¹)
But (R) Double bond closer to the carbonyl carbon disstabilies this system by conjunction.(1750cm-¹)

In option A 1H NMR show two singlet, one from 4 methyl group due to symmetry, and second is due to CH.
In 13C NMR due to symmetry shows three singlet.

**Answer:**

The CO2 molecule consists of one carbon atom (C) double-bonded to two oxygen atoms (O). Each atom has three degrees of freedom (DOF) - three dimensions in which it can move.

**Free CO2:**
In a free CO2 molecule, all atoms are free to move in three dimensions. However, there are certain constraints due to the nature of the molecule.

1. **Translation (3 DOF):** The molecule can move as a whole in three dimensions - x, y, and z. This is the movement of the center of mass of the molecule.
2. **Rotation (3 DOF):** The molecule can also rotate around three axes - x, y, and z. This involves the spinning motion of the molecule.
3. **Vibration (4 DOF):** The CO2 molecule has two O-C-O bending vibrations and one symmetric O-C-O stretching vibration. These vibrations involve the stretching and bending of the bonds between the atoms.

Therefore, a free CO2 molecule has a total of 3 + 3 + 4 = 10 normal modes.

**Constrained CO2:**
If the C and O atoms were constrained to move in only one dimension, certain degrees of freedom would be lost.

1. **Translation (1 DOF):** The molecule can only move along the constrained dimension.
2. **Rotation (1 DOF):** The molecule can still rotate around the unconstrained axes (y and z).
3. **Vibration (2 DOF):** The CO2 molecule still retains its two O-C-O bending vibrations, but loses the symmetric O-C-O stretching vibration since the atoms cannot move in the constrained direction.

Therefore, a constrained CO2 molecule has a total of 1 + 1 + 2 = 4 normal modes.

Hence, the correct answer is option 'D' - 4 normal modes for free CO2 and 2 normal modes for constrained CO2.

Explanation:
The given compound has the formula C5H12, which suggests it is an alkane.

1H NMR:
The compound gives one signal in the 1H NMR. This indicates that there are no chemically distinct hydrogen atoms in the molecule.

13C NMR:
The compound gives two signals in the 13C NMR. This indicates that there are two different types of carbon atoms in the molecule.

Possible structures:
The compound can have the following structures:
a) Pentane: This compound has five carbon atoms and no branching. It would give only one signal in both 1H and 13C NMR spectra. Therefore, it is not the correct option.

b) 2-Methylbutane: This compound has a branched structure with four carbon atoms in the main chain and one methyl group attached to the second carbon atom. It would give one signal in the 1H NMR and two signals in the 13C NMR spectra. However, the signals in the 13C NMR spectra would be very close to each other, indicating that the carbon atoms are not very different in chemical environment. Therefore, this option is not the correct one.

c) 2,2-Dimethylpropane: This compound has a branched structure with three carbon atoms in the main chain and two methyl groups attached to the second and third carbon atoms. It would give one signal in the 1H NMR and two signals in the 13C NMR spectra. The signals in the 13C NMR spectra would be well separated, indicating that the carbon atoms are very different in chemical environment. Therefore, this option is the correct one.

d) Cannot tell without more information: This option can be eliminated based on the information provided in the question.

Conclusion:
The correct option is c) 2,2-Dimethylpropane, which has a branched structure with three carbon atoms in the main chain and two methyl groups attached to the second and third carbon atoms.

Magnetic Resonance (NMR) spectroscopy is a technique used to analyze the structure and composition of molecules. The product of Nuclear Magnetic Resonance spectroscopy is a spectrum, which is a graph showing the intensity of different signals as a function of the frequency or chemical shift.

The spectrum obtained from NMR spectroscopy provides information about the number and type of different nuclei present in a molecule, their chemical environment, and their connectivity to neighboring atoms. This information can be used to determine the structure of the molecule, as well as other properties such as molecular dynamics and interactions.

Overall, the product of NMR spectroscopy is the spectrum, which is analyzed to obtain valuable information about the molecular structure and composition.