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Q No 1: Define the followings. (9)

1. Mass Analyzer: A mass analyzer is a component of a mass spectrometer that separates ions based on their mass-to-charge ratio ($m/z$).
2. Base peak: The base peak is the most intense (tallest) peak in a mass spectrum, assigned a relative abundance of 100%. It represents the most stable fragment ion.
3. TOF (Time-of-Flight): Time-of-Flight (TOF) is a type of mass analyzer that measures the time it takes for ions to travel a fixed distance. Lighter ions travel faster than heavier ions.
4. Local Diamagnetic current: In NMR spectroscopy, a local diamagnetic current refers to the circulation of electrons within a molecule induced by an external magnetic field, creating a secondary magnetic field that opposes the external field, leading to shielding.
5. Vicinal Coupling: Vicinal coupling is spin-spin coupling between protons on adjacent carbon atoms (separated by three bonds, $\text{H-C-C-H}$), denoted as $^3J$.
6. Relative Configuration: Relative configuration describes the configuration of one chiral center in a molecule with respect to another chiral center within the same molecule. It indicates whether two chiral centers have the same or opposite configurations (e.g., erythro/threo or cis/trans in cyclic systems).
7. ESI (Electrospray Ionization): Electrospray Ionization (ESI) is a soft ionization technique used in mass spectrometry that produces ions directly from a liquid solution, often used for large, polar, and thermally labile molecules.
8. Nitrogen Rule: The Nitrogen Rule states that a molecule with an odd number of nitrogen atoms will have an odd nominal mass, while a molecule with an even number of nitrogen atoms (or no nitrogen) will have an even nominal mass.
9. Allene: An allene is a compound in which one carbon atom is double-bonded to two other carbon atoms ($\text{C=C=C}$). The central carbon is $sp$ hybridized, and the terminal carbons are $sp^2$ hybridized.

Q No 2 A-Label the following isomers as R or S (4)

1. First molecule:

   • Priorities: 1. $\text{NH}_2$, 2. $\text{COOH}$, 3. $\text{CH}_2CH_3$, 4. $\text{H}$
   • The $\text{H}$ is on a horizontal line (coming out). The path 1-2-3 is counter-clockwise (S). Since $\text{H}$ is on a horizontal line, we reverse the assignment.
   • Answer: $\boxed{R}$

2. Second molecule:

   • Priorities: 1. $\text{OH}$, 2. $\text{COOH}$, 3. $\text{CH}_3$, 4. $\text{H}$
   • The $\text{H}$ is on a dash (going away). The path 1-2-3 is clockwise (R). Since $\text{H}$ is on a dash, we keep the assignment.
   • Answer: $\boxed{R}$

3. Third molecule:

   • C2 (carbon with Cl):
     • Priorities: 1. $\text{Cl}$, 2. $\text{CH}(NH_2)\text{COOH}$, 3. $\text{CH}_3$, 4. $\text{H}$
     • The $\text{H}$ is on a horizontal line (coming out). The path 1-2-3 is counter-clockwise (S). Reverse due to $\text{H}$'s position.
     • Answer: $\boxed{2R}$
   • C3 (carbon with NH2):
     • Priorities: 1. $\text{NH}_2$, 2. $\text{COOH}$, 3. $\text{CH}(Cl)CH_3$, 4. $\text{H}$
     • The $\text{H}$ is on a horizontal line (coming out). The path 1-2-3 is counter-clockwise (S). Reverse due to $\text{H}$'s position.
     • Answer: $\boxed{3R}$
   • Overall: $\boxed{2R, 3R}$

4. Fourth molecule (spiro compound):

   • This is a spiro[2.4]heptane derivative with two methyl groups on the cyclopentane ring. The stereochemistry is not explicitly drawn with wedges/dashes, so we assume the methyl groups are on wedges and hydrogens on dashes for a specific isomer. Let's label the spiro carbon as C1, and the carbons with methyl groups as C2 and C3 of the cyclopentane ring.
   • C2 (left methyl):
     • Priorities: 1. $\text{CH}_3$, 2. $\text{C3}$ (carbon with other methyl), 3. $\text{C1}$ (spiro carbon), 4. $\text{H}$
     • Assuming $\text{CH}_3$ is on a wedge and $\text{H}$ on a dash. The path 1-2-3 is clockwise (R).
     • Answer: $\boxed{R}$
   • C3 (right methyl):
     • Priorities: 1. $\text{CH}_3$, 2. $\text{C2}$ (carbon with other methyl), 3. $\text{C4}$ (carbon adjacent to spiro), 4. $\text{H}$
     • Assuming $\text{CH}_3$ is on a wedge and $\text{H}$ on a dash. The path 1-2-3 is clockwise (R).
     • Answer: $\boxed{R}$
   • Overall: $\boxed{R, R}$ (This assignment is based on the assumption that both methyl groups are on wedges and hydrogens on dashes, as stereochemistry was not explicitly indicated in the drawing).

Q No 2 B-Draw all possible Optical and Geometrical Stereoisomerism of the given molecule. (3) Molecule: 1-bromo-4-(prop-1-en-1-yl)cyclohexane. This molecule has one chiral center at C1 of the cyclohexane ring (due to the bromine and the prop-1-en-1-yl group) and one double bond with E/Z isomerism in the prop-1-en-1-yl group. Therefore, there are $2^2 = 4$ possible stereoisomers.

1. (1R, E)-1-bromo-4-(prop-1-en-1-yl)cyclohexane

   \begin{center} \includegraphics[width=0.4\textwidth]{1R_E_isomer_en.png} \end{center}

   (Here, Br is shown on a wedge at C1, H on a dash, and the double bond has E configuration)

2. (1S, E)-1-bromo-4-(prop-1-en-1-yl)cyclohexane

   \begin{center} \includegraphics[width=0.4\textwidth]{1S_E_isomer_en.png} \end{center}

   (Here, Br is shown on a dash at C1, H on a wedge, and the double bond has E configuration)

3. (1R, Z)-1-bromo-4-(prop-1-en-1-yl)cyclohexane

   \begin{center} \includegraphics[width=0.4\textwidth]{1R_Z_isomer_en.png} \end{center}

   (Here, Br is shown on a wedge at C1, H on a dash, and the double bond has Z configuration)

4. (1S, Z)-1-bromo-4-(prop-1-en-1-yl)cyclohexane

   \begin{center} \includegraphics[width=0.4\textwidth]{1S_Z_isomer_en.png} \end{center}

   (Here, Br is shown on a dash at C1, H on a wedge, and the double bond has Z configuration)

Q No 2 C-Draw structure for (1R,3R)-1,3-dibromo-1,3-dimethylcyclohexane. (1) For (1R,3R)-1,3-dibromo-1,3-dimethylcyclohexane, both C1 and C3 are chiral centers with R configuration. At C1: Priorities are Br (1), C2 (2), $\text{CH}_3$ (3), H (4). For R, if H is on a dash, the path 1-2-3 is clockwise. So, Br is on a wedge, $\text{CH}_3$ on a dash. At C3: Priorities are Br (1), C4 (2), $\text{CH}_3$ (3), H (4). For R, if H is on a dash, the path 1-2-3 is clockwise. So, Br is on a wedge, $\text{CH}_3$ on a dash.

Q No 2 D-Draw favoured and unfavoured chair form of cis-4-tert-butylcyclohexanol. (1) In cis-4-tert-butylcyclohexanol, the tert-butyl group and the hydroxyl group are on the same side of the cyclohexane ring. The bulky tert-butyl group strongly prefers the equatorial position.

• Favoured chair form: The tert-butyl group is in the equatorial position, and since the hydroxyl group is cis to it, it will also be in an equatorial position.

  \begin{center} \includegraphics[width=0.4\textwidth]{favoured_cis_4_tert_butylcyclohexanol_en.png} \end{center}

• Unfavoured chair form: This form results from a ring flip of the favoured form. The tert-butyl group would be in the axial position, and the hydroxyl group would also be in the axial position. This form is highly disfavoured due to significant 1,3-diaxial interactions involving the large tert-butyl group.

  \begin{center} \includegraphics[width=0.4\textwidth]{unfavoured_cis_4_tert_butylcyclohexanol_en.png} \end{center}

Q No 3: Attempt the questions. (2x 7=14)

I. How MALDI is different from FAB ionization method?

MALDI (Matrix-Assisted Laser Desorption/Ionization) and FAB (Fast Atom Bombardment) are both soft ionization techniques used in mass spectrometry, but they differ significantly in their mechanisms and applications.

• MALDI involves mixing the sample with a matrix that absorbs laser energy. The laser pulse causes the matrix to vaporize, carrying the analyte molecules into the gas phase as ions. It is highly effective for large, non-volatile, and thermally labile molecules like proteins, peptides, and polymers, producing minimal fragmentation. It typically works with samples in a solid or dried-spot state.
• FAB involves dissolving the sample in a liquid matrix (e.g., glycerol) and then bombarding it with a beam of high-energy atoms (e.g., argon or xenon). This bombardment transfers energy to the sample, causing it to desorb and ionize. FAB is also used for larger molecules but generally produces more fragmentation than MALDI and is less sensitive for very large biomolecules. It operates with samples in a liquid state.

II. What is the importance of HRMS?

HRMS (High-Resolution Mass Spectrometry) is crucial due to its ability to measure the mass of ions with extremely high accuracy, typically to four or more decimal places. Its importance lies in several key areas:

• Elemental Composition Determination: The precise mass measurement allows for the unambiguous determination of a molecule's exact elemental composition (e.g., distinguishing between $\text{C}_{10}H_{14}\text{O}$ and $\text{C}_{11}H_{18}$), which is not possible with low-resolution mass spectrometry.
• Isomer Identification: HRMS can differentiate between isomers that have the same nominal mass but slightly different exact masses due to variations in isotopic composition or bond energies.
• Structure Elucidation of Unknown Compounds: It is an indispensable tool for elucidating the structures of unknown compounds in fields such as organic chemistry, natural product chemistry, and metabolomics, by providing highly accurate molecular formulas.

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