Structure Determination of Organic Compounds MCQ Quiz in मल्याळम - Objective Question with Answer for Structure Determination of Organic Compounds - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

The answer is A.

Concept:-

• Quantum Numbers and Atomic Orbitals: Describing the size, shape, orientation, and spin of the electron's orbital.
• Total Spin Quantum Number (S): Sum of the spin quantum numbers of all electrons.
• Total Orbital Quantum Number (L): Sum of the orbital quantum numbers of all electrons.
• Total Angular Momentum (J): Sum of the total spin and orbital quantum numbers.
• j-j coupling and Russel-Saunders coupling : Describing the coupling of orbital and spin angular momenta of different electrons.

Explantion:-

So ground trem symbol is ⁴I_9/2

This leads to correct option is A

Simmilary the highest Term symbol is ⁴I_15/2

Conclusion:-

So, the lowest and highest states of Nd3+, is 4I9/2 and 4I15/2

The answer is 28 and 30.

Concept:-

The gyromagnetic ratio (also sometimes known as the magnetogyric ratio in other disciplines) of a particle or system is the ratio of its magnetic moment to its angular momentum, and it is often denoted by the symbol γ, gamma. Its SI unit is the radian per second per tesla (rad⋅s−1⋅T−1)

Explanation:-

A molecule shows two absorptions at 896 and 960 MHz in its 13C NMR 

Applied Magnetic Field = 3T

Chemical shift = shift / operation frequency

operation frequency = magnetogyric ratio x Bo
 

The chemical shift (δ) can be calculated based on the shift in frequency from an applied magnetic field, relative to an operating frequency.

In the case of NMR spectroscopy, the operating frequency (ν_observed) can be determined by using the magnetogyric ratio (γ) and the applied magnetic field (Bo).

The equation for the operating frequency is:

ν_obs= γ x Bo/2π 

Substituting the given magnetogyric ratio for 13C (γ = 6.72 x 10⁷ rad T^-1 s^-1) and the applied magnetic field (Bo = 3T), we find the operating frequency of the instrument:

ν_obs = 6.72 x 10⁷ rad T^-1 s^-1 x 3 T
= 201.6 / 2π 
= 32.10 MHz

The chemical shift (δ) is given by:

Chemical shift = shift / operation frequency

chemical shift (δ) = 896 / 32.1 = 28
chemical shift (δ) = 960 / 32.1 = 30

Conclusion:-

So, chemical shift (δ) for two absorptions at 896 and 960 MHz is 28 and 30

The answe is 900.

Concept:-

• Magnetic Resonance: The fundamental principle of nuclear magnetic resonance (NMR) and magnetic resonance imaging (MRI) is based on the Larmor frequency. In an applied magnetic field, the resonant frequency of a nucleus or an electron is known as the Larmor frequency.
• Gyromagnetic Ratio: The gyromagnetic ratio (also known as the gyromagnetic factor) is the ratio of the magnetic moment to the angular momentum of an atomic particle, such as an electron or proton. It is represented by the symbol γ.
• Larmor Precession: In quantum mechanics, the Larmor precession of a particle is the precession of the magnetic moment around the direction of the magnetic field that is being applied.
• Magnetic Fields and Their Interactions: The effect of a magnetic field on a charged particle, such as a proton, is integral to the concept of Larmor frequency.

Explnation:-

ω=γB

ω is frequency (MHz)
γ is the gyromagnetic ratio is 26.752 × 107 rad T-1 s.
B is magnetic field 

Also we know that the gyromagnetic ratio for Hydrogen is 42.57 MHz T^-1 = 26.752 × 107 rad T-1 s.

putting in above equation

ω=42.57 MHz T-1 x 21.1 T
ω= 898.22 MHz 
which is nearly equal to 900MHz

Conclusion:-

So, magnetic field is, approximately, 900MHz

Concept:-

• ¹³C NMR Spectroscopy: This is a type of nuclear magnetic resonance (NMR) spectroscopic technique that makes use of the carbon-13 isotope. This technique is often used to study organic and inorganic molecules to uncover information about their structure and dynamics.
• Chemically Equivalent Carbons: In regard to NMR spectroscopy, carbons are considered chemically equivalent if they have the same magnetic environment—typically, if they are bonded to the same types and numbers of atoms. Chemically equivalent carbons will produce identical peaks in an NMR spectrum. Therefore, the number of distinct peaks represents the number of unique carbon environments in the molecule.​

Explanation:-

Since The above compound have the plane of symmetry hence in proton decoupled ₁₃C-NMR spectrum recorded at 25C in CDCl₃ will be 8

Conclusion:-

The number of peaks will be 8

The correct answer is option 1

Concept :-

• Molecular Formula: This gives the overall summary of the numbers of each kind of atom in a molecule. In the case given, C10H14 means that there are 10 carbon atoms and 14 hydrogen atoms in the molecule.
• Degree of Unsaturation or Double Bond Equivalents: It provides the total number of pi bonds and rings present in the compound. Each degree of unsaturation could be a pi bond (as in a double or triple bond), or a ring. You can calculate this degree by using the formula (2C+2+N-X-H)/2, where C represents number of carbons, N represents number of nitrogens, X represents number of halogens, and H represents number of hydrogens in the formula of the compound.
• Nuclear Magnetic Resonance (NMR) Spectroscopy: This is a powerful tool used by chemists to determine the structure of organic compounds. It works on the principle that certain atomic nuclei have a property termed 'spin' when placed in an external magnetic field, and the nucleus absorbs energy in the radio frequency of the electromagnetic spectrum.
• ¹H NMR Spectroscopy: Proton (Hydrogen-1) NMR spectroscopy specifically looks at the hydrogen atoms in a molecule. It provides information about the number of hydrogens, their types, and their environment within the molecules.
• ¹³C NMR Spectroscopy: Carbon-13 NMR spectroscopy focuses on the carbon atoms in a molecule. It provides similar information as ^1H NMR but about the carbon atoms — their quantity and their chemical environment.
• Chemical Shift in NMR: In an NMR spectrum, the position of a signal is called its chemical shift, and it gives information about the type of chemical environment of the protons or carbons. For example, protons attached to carbons in an aromatic ring will give a signal at a different position than those attached to a carbon in an alkane.
• Signal Splitting in NMR: The splitting of signals in an NMR spectrum provides information about the number of directly neighboring protons (hydrogen atoms). A singlet means no neighboring protons (for H NMR) or no directly linked protons (for C NMR).
• Symmetry in Molecules: Molecules can have symmetry, and in the case of this molecule, biphenyl, due to its symmetry, has fewer unique atom environments, resulting in fewer signals in the NMR spectrum.

Explanation:-

DBE = (number of carbons + 1) - (number of monovalent atoms)/2+ (number of trivalent atoms)/2

DBE= 4 (contains one benzene ring)

 C10H14 exhibited two singlets in the 1H NMR spectrum means only two chemically equivalent hydrogen which only possible in option 1 i.e. Ha , Hb

 C10H14 exhibited three signals in the 13C NMR spectrum means three chemically equivalent Carbon which only possible in option 1 i.e. C1,C2,C3

Conclusion:-

So, An organic compound having the molecular formula C10H14 exhibited two singlets in the 1H NMR spectrum, and three signals in the 13C NMR spectrum. The compound is in option 1

The correct answer is a triplet with an intensity ratio of 1:1:1

Concept:

A deuterated compound will show a strong peak in deuterium NMR but not proton NMR.

Coupling between proton and deuterium is most commonly seen in the solvent signals of the 1H spectrum and sometimes in the residual water signal that has become partially deuterated.

Explanation:

Here I for D is 1

so according to the formula

2nI+1

= 2×1×1+1 = 3( triplet).

For I = 1, and n =1 ratio of intensities will be 1:1:1

The 13C signals can be shown as

Concept:

• Mass spectrometry is an analytical tool useful for measuring the mass-to-charge ratio (m/z) of one or more molecules present in a sample.

• A mass spectrum is a histogram plot of intensity vs. mass-to-charge ratio (m/z) in a chemical sample, usually acquired using an instrument called a mass spectrometer.

• In a typical MS procedure, a sample, which may be solid, liquid, or gaseous, is ionized, for example by bombarding it with a beam of electrons. This may cause some of the sample's molecules to break up into positively charged fragments or simply become positively charged without fragmenting.

• This process is known as electron ionization (EI, formerly known as electron impact ionization and electron bombardment ionization).

• These ions (fragments) are then separated according to their mass-to-charge ratio, for example by accelerating them and subjecting them to an electric or magnetic field: ions of the same mass-to-charge ratio will undergo the same amount of deflection.​

​Explanation:-

• The mass value of any ion is its true mass, i.e., the sum of the masses of each atom (most common isotope) in that single ion (accurate), and not its molecular weight calculated from chemical atomic weights (integral atomic mass, weighted averages of weights of all of the isotopes).
• The precise mass of some elements are given below on the C-12 scale

Isotope peaks:

• Many elements have more than one natural Isotope. Therefore isotope peaks are often seen in MS.
• Relative to [M]+ ion = 100%, calculation of isotope peaks.
• The abundance of [M+1]+ ion

= (number of C x 1.1) +(number of H x 0.016) + (number of N x 0.37) + (number of O x 0.04) + (number of S x 0.8)

• The intensity ratio of the peaks of  [M]+, [M+2]+ and [M+4]+ will be in the ratio 1: 2: 1 as shown below:

• The compound has two -Br atom, so the compound is

Concept:

• Mass spectrometry is an analytical tool useful for measuring the mass-to-charge ratio (m/z) of one or more molecules present in a sample.

• A mass spectrum is a histogram plot of intensity vs. mass-to-charge ratio (m/z) in a chemical sample, usually acquired using an instrument called a mass spectrometer.

• In a typical MS procedure, a sample, which may be solid, liquid, or gaseous, is ionized, for example by bombarding it with a beam of electrons. This may cause some of the sample's molecules to break up into positively charged fragments or simply become positively charged without fragmenting.

• This process is known as electron ionization (EI, formerly known as electron impact ionization and electron bombardment ionization).

• These ions (fragments) are then separated according to their mass-to-charge ratio, for example by accelerating them and subjecting them to an electric or magnetic field: ions of the same mass-to-charge ratio will undergo the same amount of deflection.​

​Explana​tion:-

• The mass value of any ion is its true mass, i.e., the sum of the masses of each atom (most common isotope) in that single ion (accurate), and not its molecular weight calculated from chemical atomic weights (integral atomic mass, weighted averages of weights of all of the isotopes).
• The precise mass of some elements are given below on the C-12 scale

Isotope peaks:

• Many elements have more than one natural Isotope. Therefore isotope peaks are often seen in MS.
• Relative to [M]+ion = 100%, calculation of isotope peaks.
• The abundance of [M+1]⁺ ion

= (number of C x 1.1) +(number of H x 0.016) + (number of N x 0.37) + (number of O x 0.04) + (number of S x 0.8)

• The abundance of [M+2]+ ion

= (number of C x 1.1)2/200 + (number of O x 0.2) + (number of S x 4.4)

• For Halogen isotopes:-
  • Chlorine has two isotopes Cl-35 and Cl-37 (3:1 ratio)
  • Bromine has two isotopes Br-79 and Br-81 (1:1 ratio)
  • Both fluorine and iodine are mono-isotopic F-19 and I-127
  • Compounds containing Cland Br are easily identified in mass spectrometry because of the prominent isotope Peaks separated by two mass units [M] and [M+2].
• The intensity ratio can be calculated as coefficients of the polynomial (a+b)ⁿ where a and b are relative abundances of the two

isotopes and n is the number of halogen atoms.

• If two halogens are present then

(a+b)ⁿ (c+d)^m

• For 2Br and 1Cl

= (a+b)²(c+d)  [n = 2 and m = 1]

= (a²+2ab+b²)(c+d)

• Thus for Br2Cl, the intensity ratio is

​3:7:5:1

Conclusion:-

• Hence, option 1 is the correct answer.

Concept:

The stretching frequency of a chemical bond depends on the masses of the atoms involved and the strength of the bond. In the case of O-H and O-D bonds, the deuterium atom is heavier than the hydrogen atom, which means the O-D bond is stronger than the O-H bond.

Explanation:

→ When two atoms are bound together by a chemical bond, they vibrate around their equilibrium position due to the influence of the neighboring atoms.

→ The stretching frequency of a bond is the frequency at which it vibrates when subjected to an external force or energy, such as infrared radiation.

→ In the case of O-H and O-D bonds, the stretching frequency is influenced by the mass of the atom attached to the oxygen atom. The mass of deuterium (D) is greater than the mass of hydrogen (H), so the O-D bond is stronger than the O-H bond.

→The frequency at which a bond absorbs in an IR spectrum depends on the mass of the atoms as well as the strength of the bond. Deuterium (D) is an isotope of hydrogen that is approximately twice as heavy.

• The stretch frequency of O-D versus O-H decreases approximately as the square root of the mass increase, since the frequency (ν) of vibration can be approximated by the equation:
• ν = 1/(2π) * √(k/μ)
•  
• In this equation, k is the force constant (analogous to bond strength) and μ is the reduced mass of the system. Deuterium is about twice as massive as hydrogen, so we expect the O-D stretch frequency to decrease by about the square root of 2 which is approximately 1.414.
• When we divide the O-H stretch frequency by this number (∼3600 cm-1 / 1.414), we get approximately 2600 cm-1.

→ A stronger bond requires more energy to stretch, which corresponds to a lower stretching frequency. Therefore, the O-D stretch frequency is lower than the O-H stretch frequency.

→ Experimental studies have shown that the O-D stretching frequency is typically around 2500-2700 cm^-1, which is lower than the O-H stretching frequency of around 3600 cm^-1.

Conclusion:
So, The closest option to the O-D stretching frequency would be option 2) 2600 cm^-1.