CUET PG Chemistry Mock Test - 2 - CUET PG MCQ

CONCEPT:
Reactivity of Aniline in Electrophilic Aromatic Substitution Reactions

• Nitrogen Group Effects: The NH₂ group in aniline is an electron-donating group due to its strong +M (mesomeric) effect, which increases electron density on the ortho and para positions of the benzene ring.
• This makes the NH₂ group ortho and para directing and a powerful activating group, thus facilitating substitution at these positions.

EXPLANATION:

• Statement I:
  • The NH₂ group in aniline is indeed ortho and para directing and a powerful activating group because it donates electrons to the benzene ring through resonance, increasing reactivity at these positions.
  • This statement is correct.

• Statement II:
  • Aniline does not undergo Friedel-Craft’s reactions (alkylation and acylation) because the lone pair of electrons on the nitrogen can coordinate with the AlCl₃ catalyst, forming a complex. This interaction deactivates the benzene ring toward further substitution.
  • This statement is also correct.

Therefore, the most appropriate answer is:

• 1) Both Statement I and Statement II are correct

Concept:
Electrophiles

• Electrophiles are species that are electron-deficient and can accept a pair of electrons from nucleophiles during a chemical reaction.
• They can be positively charged ions (e.g., NO⁺) or neutral species with an electron-deficient atom (e.g., BF₃ or CO₂).
• Non-electrophilic species are those that are not electron-deficient and cannot act as electron acceptors.

Explanation:

• Option 1: NO⁺
  • NO⁺ is a positively charged species and is highly electron-deficient.
  • It acts as a strong electrophile.
• Option 2: BF₃
  • BF₃ is a neutral molecule with an electron-deficient boron atom.
  • It readily accepts electrons, making it an electrophile.
• Option 3: CO₂
  • CO₂ is a neutral molecule with a partial positive charge on the carbon atom.
  • It acts as an electrophile in reactions such as nucleophilic addition.
• Option 4: NH₄⁺
  • NH₄⁺ is a positively charged ion but is not electron-deficient.
  • It cannot accept electrons because all of its orbitals are fully occupied, making it non-electrophilic.

Correct Answer: 4. NH₄⁺

CONCEPT:
Decarboxylation of β-Keto Acids

• β-Keto acids are compounds with a keto group (C=O) at the β position relative to the carboxylic acid group (COOH).
• Upon heating, β-keto acids undergo decarboxylation, which is the loss of a carbon dioxide (CO₂) molecule.
• This reaction is facilitated by the presence of the β-keto group, making it easier to eliminate CO₂ and form a stable enolate intermediate, which tautomerizes to give the final product.

CALCULATION:

• In the given compound, CH₃CH₂C(=O)CH₂C(=O)OH, we observe that there is a β-keto group relative to the carboxylic acid (COOH) group.
• On heating, this β-keto acid undergoes decarboxylation, leading to the loss of a CO₂ molecule.
• The remaining structure after decarboxylation is a ketone: CH₃CH₂C(=O)CH₃.
• Reaction: 

CONCLUSION:
The correct option is: Option 2.

Concept:
Electrophilic Substitution of Pyrrole:

• The electrophilic substitution reaction of Pyrrole occurs at C-2 and C-5 positions.
• The mechanism will be as follows:

Explanation:
Mechanism

• In the first step of the reaction, KOH acts as a base and abstracts the most acidic N-H proton from pyrrole.
• In the next step, it undergoes an electrophilic substitution reaction at the C-2/5 position and gives potassium salt of the carboxylic acid.
• In the last step, it undergoes an acid hydrolysis reaction to give the final product.

​Conclusion:

• Hence, the major product of the following reaction is

Concept:

Infrared (IR) Absorption and Molecular Vibrations

• In IR spectroscopy, molecular vibrations are detected based on changes in the dipole moment of a molecule during vibration.
• Vibrational modes can be classified as:
  • Symmetric Stretch: Both bonds stretch or compress simultaneously without changing the dipole moment.
  • Antisymmetric Stretch: One bond stretches while the other compresses, leading to a change in the dipole moment.
• Key Principle: Vibrational modes that do not result in a change in the dipole moment are IR inactive.

Explanation:

• Option 1: Symmetric CO₂ Stretch
  • This vibrational mode involves the symmetric stretching of C=O bonds in CO₂, which does not change the dipole moment.
  • Hence, it is IR inactive and shows no absorption bands.
• Option 2: Antisymmetric CO₂ Stretch
  • This mode causes a change in the dipole moment as one bond stretches and the other compresses.
  • It is IR active and shows absorption bands.
• Option 3: Symmetric O=C=S Stretch
  • This vibration involves stretching of O=C=S bonds and changes the dipole moment.
  • It is IR active and shows absorption bands.
• Option 4: CH₃-C≡CH Stretch
  • The stretching of this bond alters the dipole moment and is IR active.

Correct Answer: 1. Symmetric CO₂ Stretch

Concept: -

• Isomerism is the name given to a phenomenon in which more than one compounds have the same chemical formula but a different chemical structure. Compounds having the same molecular formula but different arrangements of atoms are called isomers.
• In isomerism, the chemical formula of compounds is the same but their arrangement is different.
• As the name suggests functional isomers are isomers that have the same chemical formula but the functional group present in the compound is different. 
• Structural isomers are divided into chain, positional, functional, metamerism, tautomerism, and ring chain isomers.
• Structural isomerism is commonly called constitutional isomerism.
  • In these isomers, functional groups and atoms in the molecules are connected in different ways.

• Stereoisomers: 
  • The compounds having the same molecular as well as the same structural formulae but differing in the relative arrangement of the atoms or groups in space are called stereoisomers.
  • This phenomenon is termed as stereoisomerism. 
  • Optically Active Stereoisomers: The stereoisomers that can rotate plane-polarized light are called as optically active whereas an optically inactive compound is not capable of optical rotation.
• Diastereomers:
  • Stereoisomers that are not mirror images of each other.
  • Diastereomers have the same configuration at least at one chiral centre and the opposite configuration at the remaining chiral centres.
• Enantiomers:
  • Stereoisomers are mirror images of each other.
  • Enantiomers have opposite configurations at all the chiral centres.

Explanation: 

• As we can see from the structure the linkage of atoms is the same in both thus, it rules out the possibility of constitution isomerism.
• Let's draw a mirror between two structures given to check whether they are mirror images of each other or not -
  
• As we can see both are mirror images of each other.
• Now, let's check whether they are superimposable or not -
  
• Thus, both structures as identical.

Conclusion: -
Both structures are superimposable mirror images of each other
Thus, they are identical.
Hence, the correct option is (1).
Alternate Method

• Check the symmetry elements of the structure.
• We can see that structure has a two-fold improper axis of symmetry.
• Thus, is optically inactive.
• We know the rotation of 180º in the plane of paper of a structure does not bring any change in configuration.
• Hence, both are homomers or identical.

CONCEPT:
Uranocene (U(C₈H₈)₂)

• Uranocene is an organometallic compound of uranium.
• It consists of uranium metal sandwiched between two cyclooctatetraenide ligands (COT^2-).
• The bonding in uranocene typically involves uranium in its +4 oxidation state.
• Each cyclooctatetraenide ligand contributes 10 π-electrons in total (8 from the ring and 2 for the negative charges).
• Uranocene is an example of a "sandwich" compound as both COT rings are positioned parallel to each other above and below the U center.

EXPLANATION:

• (A) Oxidation state of uranium is ‘+4’: This is correct. In uranocene, uranium is in the +4 oxidation state (U(IV)).
• (B) It has cyclooctatetraenide ligands: This is also correct. Uranocene contains two cyclooctatetraenide (COT^2-) ligands.
• (C) It is a bent sandwich compound: This is incorrect. Uranocene is a sandwich compound, but it is not bent; it has a linear or nearly planar sandwich structure with symmetric ligands.
• (D) It has a ‘-2’ charge: This is incorrect. Uranocene is a neutral compound, with no overall charge.

Therefore, the correct statements about uranocene are  (a) A and B.

Concept:

• Cyclohexane has a cyclic structure.
• It consists of six sp3 hybridized C atoms that constitute the ring, each of the C atoms are linked to two other C atoms and two H atoms by sigma bonds.
• Naturally, the C-C-C, H-C-C, and H-C-H bond angles are roughly 109.280.
• Thus the molecule is nonplanar, puckered so that all the C atoms do not belong to one and the same plane.
• ​Boat conformation is less stable than the chair and the chair conformation of cyclohexane is the most stable conformation and derivatives of cyclohexane almost always exist in the chair conformation.
• The "A values" in the context of cyclohexane derivatives refer to the effect of different substituents on the equilibrium between the chair conformations of the cyclohexane ring. These values are used to describe how bulky or electron-withdrawing a substituent is and how it influences the stability of the chair conformations.

Keq = [equatorial conformer] / [axial conformer]
Now, A = -ΔG° = RT lnKeq
The more than the value of Keq, the more the value of A.

• Thus, more amount of equatorial isomer will lead to a higher value of A.
• The larger the substituent on a cyclohexane ring, the more the equatorial substituted conformer will be favored.

Explanation:-

• The correct order of magnitude for the given substituents (tBu, Br, Me, F) in cyclohexane derivatives is as follows, from most destabilizing (largest A value) to least destabilizing (smallest A value):

^tBu > Br > Me > F
Here's a brief explanation of each substituent's effect on the cyclohexane ring conformation:

• ^tBu (tert-butyl): The tert-butyl group is highly bulky and leads to significant steric hindrance, which destabilizes the chair conformation of cyclohexane. It has the largest A value.

• CH(Me)₂: It is less bulky substituent than tBu, but bulkier than Me and F. Thus, the A value for -CH(Me)2 is second highest among the four substituents.

• Me (methyl): The methyl group is a relatively small substituent and has a moderate destabilizing effect on the chair conformation compared to tBu and Br.

• F (fluorine): Fluorine is a small substituent and has the least destabilizing effect among the given substituents. Its A value is the smallest in this series.

Conclusion:-

• Hence, the correct order of magnitude of ‘A values’ for the given substituents in cyclohexane derivatives is

tBu > CH(Me)2 > Me > F

Concept:

• Boron is an element of period III with atomic number 5.
• The electronic configuration of boron is 1s22s22p1.
• Elemental Boron exists in a solid-state at room temperature either as black solid monoclinic crystal or yellow or brown amorphous powders.
• Elemental Borron was first synthesized by the electrical decomposition of borates.
• Two crystalline modification of boron exists as α rhombohedral and β-rhombohedral.
• The chemical properties of Boron are dominated by its small size and high electron affinity.
• Boron has a high affinity for oxygen thus forming borates and oxo-compounds.
• Crystalline boron on the other hand is very inert in nature.
• Boron can form three to four covalent bonds and has the special ability to form 3-centre 2 electron bonds.
• 3 centre two-electron bonds consist of 3 atoms sharing two electrons forming two bonds, observed in B2H6.

Explanation:
Aminoboranes:

• Aminoboranes contain B-N bonds in which the boron is trigonally hybridized.
• The lone pair on nitrogen may be donated to a vacant p orbital on boron, giving rise to B-N bond order greater than one.
• Aminoboranes are prototypes of alkenes.
• Aminoboranes can be prepared from amine boranes.
• The reaction of diborane with ammonia gives borazines. the reaction is:

3B2H6 + 2NH3 →​ B2H6.2NH3 + 12H2
Additional Information:

• When reacted with excess NH3, the reaction is:

B2H6 + Excess NH3 →​ (BN)x + H2

• The reaction takes place at high temperatures with the evolution of hydrogen.

Hence, we can prepare (BN)x from B2H6 by the reaction B2H6 + Excess NH3 →​ (BN)x

Concept:
Identifying the Structure Using NMR Data

• The number of signals in NMR indicates the number of distinct hydrogen environments in the molecule.
• The chemical shifts (δ values) provide information about the electronic environment of hydrogens:
  • δ ~ 7.1 ppm: Hydrogens on an aromatic ring.
  • δ ~ 2.2 ppm: Hydrogens attached to a carbon adjacent to an aromatic ring or part of a methyl group near an electron-withdrawing group.
  • δ ~ 1.5 ppm: Hydrogens of a methylene (-CH₂) group in an aliphatic chain.
  • δ ~ 0.9 ppm: Hydrogens of a methyl (-CH₃) group in an aliphatic chain.
• The integration values of the peaks correspond to the relative number of hydrogens producing each signal.

Explanation:

• The molecular formula C₉H₁₂ suggests the compound contains an aromatic ring (C₆H₅) and a propyl group (C₃H₇).
• The NMR signals can be assigned as follows:
  • δ ~ 7.1 ppm: Aromatic protons (5 hydrogens).
  • δ ~ 2.2 ppm: Benzyl methylene group (-CH₂-Ar, 2 hydrogens).
  • δ ~ 1.5 ppm: Methylene group (-CH₂, 2 hydrogens) in the propyl chain.
  • δ ~ 0.9 ppm: Terminal methyl group (-CH₃, 3 hydrogens).
• The structure that matches this description is n-propylbenzene.
• 

Correct Answer: Option 2.

Concept:
S_N1 and S_N2 Reactions

• S_N1 reaction: A unimolecular nucleophilic substitution reaction involving the formation of a carbocation intermediate. The rate depends only on the concentration of the substrate.
• S_N2 reaction: A bimolecular nucleophilic substitution reaction occurring in a single step. The rate depends on both the substrate and nucleophile concentrations.

Explanation:

• Statement (A): Correct. S_N1 is unimolecular and follows first-order kinetics, while S_N2 is bimolecular and follows second-order kinetics.
• Statement (B): Incorrect. S_N1 involves carbocation formation, whereas S_N2 does not.
• Statement (C): Correct. S_N2 is stereospecific, producing products with inversion of configuration due to the backside attack mechanism.
• Statement (D): Correct. S_N1 is favored by weak bases or poor nucleophiles due to its reliance on carbocation stability.

Correct Answer: (A), (C), and (D)

Concept and Explanation:
For the complex GdCl₃.6H₂O:
It exists as [GdCl₂(H₂O)₆]Cl. Thus the coordination number of Gd is 8.
Conclusion:
The coordination number of Gd in GdCl₃.6H₂O is 8.

CONCEPT:
Aldol Condensation in Aromatic Compounds with α, β-Unsaturated Carbonyl Products

• Aldol condensation occurs between two carbonyl compounds (aldehydes or ketones) in the presence of a strong base, such as NaOH, resulting in the formation of a β-hydroxy aldehyde or ketone (aldol).
• In an aldol condensation, one carbonyl compound (acting as the nucleophile) is deprotonated to form an enolate ion, which then attacks the carbonyl carbon of another molecule, forming a new C-C bond.
• After the initial aldol formation, the β-hydroxy product typically undergoes dehydration (loss of H₂O) to form an α, β-unsaturated carbonyl compound, a common feature in conjugated systems.

REACTION MECHANISM:

• Step 1: Enolate Formation – The base (NaOH) abstracts a proton from the α-carbon of one of the aldehyde groups, creating an enolate ion.
• Step 2: Nucleophilic Attack – The enolate ion then attacks the carbonyl carbon of the second aldehyde, forming a β-hydroxy aldehyde intermediate.
• Step 3: Dehydration – The intermediate β-hydroxy aldehyde loses a molecule of water, resulting in the formation of an α, β-unsaturated carbonyl compound.

CONCLUSION:
The correct option is: Option 2

Concept:
Classification of Cyclic Ethers

• Cyclic ethers are compounds containing oxygen within a ring structure.
• They are classified based on the number of atoms in the ring:
  • Oxirane: A three-membered cyclic ether (e.g., ethylene oxide).
  • Oxetane: A four-membered cyclic ether.
  • Oxolane: A five-membered cyclic ether (e.g., tetrahydrofuran).
  • Oxepane: A seven-membered cyclic ether.

Explanation:​
Matching:

• A → I
• B → III
• C → IV
• D → II

Therefore, the correct option is 4.

CONCEPT:
Number of ¹H Signals in NMR Spectroscopy

• The number of ¹H signals in proton nuclear magnetic resonance (¹H NMR) spectroscopy reflects the number of distinct hydrogen environments in a molecule.
• Each unique hydrogen (proton) environment, where the protons are chemically different (due to different bonding or electronic effects), gives rise to a separate signal in the ¹H NMR spectrum.
• The number of signals corresponds to the number of different proton environments, not the number of protons in the molecule.
• Hydrogens that are in the same chemical environment (i.e., they experience the same electronic environment) will produce the same signal.
• Factors that influence the number of signals include symmetry, equivalent protons, and any functional groups that affect the proton’s environment.

EXPLANATION:

Therefore, this molecule has 9 numbers of ¹H signals.

CONCEPT:
Magnetic Moment and Unpaired Electrons

• The magnetic moment of a complex is a measure of its paramagnetism, which arises due to the presence of unpaired electrons. It is measured in Bohr Magneton (BM) units and is given by:
• Magnetic Moment (µ) = √n(n+2) BM
• Here, 'n' represents the number of unpaired electrons in the metal ion. The more unpaired electrons present, the higher the magnetic moment.
• For instance, if n = 2 (two unpaired electrons), the magnetic moment is:
  μ = √2(2 + 2) = √8 ≈ 2.83 BM

• This means a magnetic moment of 2.83 BM corresponds to a complex ion with exactly two unpaired electrons.

EXPLANATION

•  [V(H2O)6]3+
  • Vanadium in the +3 oxidation state has the electron configuration [Ar] 3d².
  • This means there are two unpaired electrons in the d-orbital.
  • Using the formula for magnetic moment, we calculate:

  • μ = √2(2 + 2) = √8 ≈ 2.83 BM
    Therefore, the magnetic moment of 2.83 BM matches with this complex, making it the correct option.

  • [Cr(H2O)6]³⁺:
    Chromium in the +3 oxidation state has the electron configuration [Ar] 3d³, meaning it has three unpaired electrons.
    The magnetic moment for 3 unpaired electrons would be:
    μ = √3(3 + 2) = √15 ≈ 3.87 BM
    This is significantly higher than 2.83 BM, making it an incorrect choice.

  • [Cu(CN)₄]²⁻:
    Copper in the +2 oxidation state has the electron configuration [Ar] 3d⁹. However, cyanide (CN⁻) is a strong field ligand, leading to pairing of electrons and forming a low-spin, diamagnetic complex.
    This means there are no unpaired electrons, and the magnetic moment is 0 BM, making this option incorrect.

  • [MnCl₄]²⁻:
    Manganese in the +2 oxidation state has the electron configuration [Ar] 3d⁵, meaning it has 5 unpaired electrons.
    The magnetic moment for 5 unpaired electrons is:
    μ = √5(5 + 2) = √35 ≈ 5.92 BM
    This is much higher than 2.83 BM, making it an incorrect option as well.

ADDITIONAL INFO:

• Magnetic moments help in understanding the electronic structure and the number of unpaired electrons in a metal complex.
• Paramagnetic complexes have unpaired electrons and exhibit a measurable magnetic moment, whereas diamagnetic complexes have no unpaired electrons and have zero magnetic moment.
• The higher the number of unpaired electrons, the stronger the paramagnetic behavior and the larger the magnetic moment.

CONCLUSION:
The correct answer is: Option 1 - [V(H₂O)₆]³⁺

Concept:
Conformations of Cyclohexane

• Cyclohexane exists in various conformations, including chair, boat, twist boat, and half-chair forms.
• The stability of these conformations depends on:
  • Steric Strain: Repulsion between atoms or groups due to close proximity.
  • Angle Strain: Deviation from the ideal bond angle (109.5° for sp³ hybridized carbon).
  • Torsional Strain: Resistance due to eclipsing of bonds.
• Order of Stability:
  • Chair: Most stable due to minimal steric and torsional strain.
  • Twist Boat: Less stable than chair but more stable than boat due to reduced torsional strain.
  • Boat: Less stable due to steric strain from flagpole interactions and torsional strain.
  • Half Chair: Least stable due to maximum torsional and angle strain.

Explanation:

• (A) Chair: Most stable conformation due to minimal strain.
• (B) Boat: Less stable than chair due to steric and torsional strain.
• (C) Twist Boat: Intermediate stability as it reduces strain compared to the boat conformation.
• (D) Half Chair: Least stable due to high torsional and angle strain.

Correct Order of Stability: (A) Chair > (C) Twist Boat > (B) Boat > (D) Half Chair

Concept:
Acidic Strength of Phenolic Compounds

• The acidic strength of phenols is influenced by the substituents attached to the benzene ring.
• Electron-withdrawing groups (EWGs), such as -NO₂ or -SO₃H, increase acidic strength by stabilizing the phenoxide ion formed after the loss of H⁺.
• Electron-donating groups (EDGs), such as -OCH₃ or -CH₃, reduce acidic strength by destabilizing the phenoxide ion.

Explanation:

• Compound A: Contains three -NO₂ groups (EWGs) which significantly enhance acidic strength due to strong resonance. This is the most acidic among all.
• Compound B: Contains a -OCH₃ group (+M) (EDG), which decreases acidic strength by destabilizing the phenoxide ion. This is the least acidic among all.
• Compound C: Contains a -SO₃H group (EWG), which increases acidic strength due to its strong electron-withdrawing effect.
• Compound D: Contains a -CH₃ group (+I) (EDG), which slightly reduces acidic strength compared to the other compounds. This is more acidic than B.

Order of Acidic Strength:
A > C > D > B
Hence, the decreasing order of acidic strength is A > C > D > B.

The correct answer is Square pyramidal
Concept:
The Valence Shell Electron Pair Repulsion (VSEPR) theory helps predict the geometry of molecules based on the repulsion between electron pairs (both bonding pairs and lone pairs) around a central atom. According to VSEPR theory, electron pairs will arrange themselves as far apart as possible to minimize repulsion, determining the shape of the molecule.
Shapes based on VSEPR Theory:

Explanation:-
The given compound [TeF5]– has five bond-pair electrons and one lone-pair, therefore according to VSEPR theory, the shape will be square pyramidal. 

Conclusion:-
Thus, we can conclude that the correct shape of [TeF5]– ion based on VSEPR theory is Square pyramidal.

Concept:

Electrophilic Substitution on Benzene Derivatives

• Benzene derivatives undergo electrophilic aromatic substitution reactions, where the nature of the substituent influences the reactivity and orientation of the reaction.
• The -CCl₃ group is an electron-withdrawing group due to its inductive and resonance effects, which deactivate the aromatic ring towards electrophilic substitution.
• Electron-withdrawing groups are meta-directing, meaning substitution typically occurs at the meta position relative to the substituent.
• Chlorination in the presence of FeCl₃ generates the electrophile Cl⁺, which reacts with the deactivated benzene ring.

Explanation:

• The aromatic ring has a deactivating -CCl₃ group, making the meta position the preferred site for electrophilic substitution.
• The chlorine electrophile (Cl⁺) substitutes at the meta position relative to the -CCl₃ group.
• Ortho and para substitution are disfavored due to steric hindrance and lower electron density at those positions.

The major product is a meta-chlorinated derivative of benzotrichloride.

Concept:

Molar Conductance at Infinite Dilution (Λ_m⁰)

• The molar conductance at infinite dilution (Λ_m⁰) is the sum of the ionic conductances of the cation and anion at infinite dilution.
• For Ba(OH)₂, the dissociation can be represented as:

  Ba(OH)₂ → Ba²⁺ + 2OH^-

• The molar conductance at infinite dilution is given by:

  Λ_m⁰(Ba(OH)₂) = λ⁰(Ba²⁺) + 2 × λ⁰(OH^-)

Explanation:

Given:

• Λ_m⁰(NaOH) = 248.1 × 10^-4 S m² mol^-1
• Λ_m⁰(NaCl) = 126.5 × 10^-4 S m² mol^-1
• Λ_m⁰(BaCl₂) = 280.0 × 10^-4 S m² mol^-1

Now:

Substituting into the equation:

Λ_m⁰(Ba(OH)₂) = Λm0(BaCl2) - 2Λm0(NaCl) + 2Λm0(NaOH)

= 280.0× 10-4 - 2×126.5× 10-4 + 2× 248.1× 10-4

= 523.2 × 10^-4 S m² mol^-1

Correct Answer: 1. 523.2 × 10^-4 S m² mol^-1

CONCEPT:
Carbonyl Stretching Frequencies in IR Spectroscopy

• The carbonyl (C = O) stretching frequency in infrared (IR) spectroscopy depends on the structure of the compound and the electronic environment around the carbonyl group.
• Key factors influencing C = O stretching frequency:
  • Ring strain: Smaller ring sizes (e.g., five-membered vs. six-membered) increase the stretching frequency due to increased bond strength.
  • Conjugation: Conjugation with a double bond or aromatic system decreases the frequency due to delocalization of electrons.
  • Substituents: Electron-withdrawing groups increase the stretching frequency.

EXPLANATION:

• Compound P:
  • P is a saturated ketone with no conjugation and no significant ring strain.
  • The C = O stretching frequency is around 1750 cm^-1 (X).
• Compound Q:
  • Q is a conjugated ketone with additional stabilization due to delocalization.
  • The C = O stretching frequency is slightly lower, around 1770 cm^-1 (Y).
• Compound R:
  • R has significant ring strain, leading to a higher stretching frequency.
  • The C = O stretching frequency is around 1800 cm^-1 (Z).
• Correct Matching: P → X (1750 cm^-1), Q → Z (1770 cm^-1), R → Y (1800 cm^-1).

The correct match is P-Y, Q-Z, R-X.

Explanation:-
Matching Complexes with Their Colors

• Fe₄[Fe(CN)₆]₃ .xH₂O:
  • Commonly known as Prussian Blue.
  • Color: Deep blue.
• [Fe(CN)₅NOS]^4–:
  • Ferricyanide complex with nitrosyl ligand.
  • Color: Violet.
• [Fe(SCN)]²⁺:
  • Iron complex with thiocyanate ligand.
  • Color: Blood Red.
• (NH₄)₃PO₄.12MoO₃:
  • Ammonium phosphomolybdate.
  • Color: Yellow.
• Fe₄[Fe(CN)₆]₃ .xH₂O → III. Prussian Blue
• [Fe(CN)₅NOS]^4– → I. Violet
• [Fe(SCN)]²⁺ → II. Blood Red
• (NH₄)₃PO₄.12MoO₃ → IV. Yellow

The correct answer is: 1) A - III, B - I, C - II, D - IV

CONCEPT:
Oxidation of Alkenes with Potassium Permanganate (KMnO₄)

• Potassium permanganate (KMnO₄) is a strong oxidizing agent used to oxidize alkenes.
• In acidic medium, KMnO₄ cleaves the double bond and oxidizes the carbon atoms to the highest possible oxidation state.
• For terminal alkenes, this usually results in the formation of carboxylic acids and/or carbon dioxide.

Explanation:-:

• The given compound is 1-methylcyclohexene.
• When 1-methylcyclohexene is treated with KMnO₄ in acidic medium, the double bond is cleaved, and the resulting carbon atoms are oxidized.
• For terminal alkenes like 1-methylcyclohexene, the oxidation results in the formation of a carboxylic acid at the terminal position.
• The methyl group is also oxidized to a carboxylic acid group.

The reaction can be represented as follows:

• Starting compound: 1-methylcyclohexene
• Oxidation with KMnO₄/H⁺ results in cleavage of the double bond.
• Both the methyl group and the carbon of the double bond are oxidized to carboxylic acids.
• The major product 'P' is 1-methylcyclohexane-1,1-dicarboxylic acid.

Conclusion

• Product P is 1-methylcyclohexane-1,1-dicarboxylic acid.

Correct answer: 4)
Concept:

• The Friedel–Crafts acylation is the reaction of an arene with acyl chlorides or anhydrides using a strong Lewis acid catalyst.
• This reaction proceeds via electrophilic aromatic substitution to form monoacetylated products.
• The Friedel–Crafts alkylation is the reaction of an arene with alkyl halides using a strong Lewis acid catalyst.
• This reaction proceeds via electrophilic aromatic substitution to form monoacetylated products such as toluene.

Explanation:

• When phenol is treated with chloromethane in presence of AlCl₃, we get a mixture of o & p-cresol (o-hydroxy toluene and p-hydroxy toluene).
• It is an example of Friedel Craft's alkylation reaction.

Conclusion:
Thus, the product of the given reaction is o-hydroxy toluene and p-hydroxy toluene.

Concept:
Curtius Rearrangement
Curtius Rearrangement is also called Curtius degradation or Curtius reaction. Curtius rearrangement is a thermal decomposition of acyl acid to form isocyanate with a loss of nitrogen as stated by Theodor Curtius in the year 1885. It is also known as Curtius degradation or Curtius reaction. This reaction is identical to the Schmidt reaction.
Isocyanates are subjected to attack by various nucleophiles namely alcohols, water, and amines which in turn output urea derivative or carbamate and essential amines.

Explanation:
Mechanism

Therefore, the correct option is 2.

CONCEPT:
Microstates and Electron Configurations

• A microstate is a specific arrangement of electrons in the orbitals of an atom or molecule. The number of microstates depends on the electron configuration and the number of available orbitals.
• The number of microstates for a given electron configuration can be calculated using the formula:
  Number of Microstates = n! / [r!(n - r)!], where n is the total number of available orbitals and r is the number of electrons to be placed.
• For a p³ configuration, there are 3 electrons to be placed in 3 p orbitals px, py, and pz.

EXPLANATION:

• For the p³ configuration, we calculate the number of microstates as follows:
  • Total number of available orbitals (n) = 3 (one for each of px, py, pz)
  • Number of electrons to be placed (r) = 3
  • The number of microstates is calculated using the combination formula:
    Number of Microstates = 6! / [3!(6 - 3)!]
    = (6 × 5 × 4 × 3!) / (3! × 3!)
    = 20
• However, this formula is for combinations in general, but for a p³ configuration specifically, the correct approach is to use the formula based on available microstates for the given electron arrangement.
• The correct calculation leads to 20 microstates for a p³ configuration.

Conclusion:
The correct answer is 20.

Concept:-

• If a substituted biphenyl is divided into two halves by passing a horizontal plane along the σ bond joining the two phenyl rings and if any one of the rings shows a symmetry about that plane then that biphenyl derivative is optically inactive.

Explanation:-

• For the given compound, the substituent present at  6,6' and 2,2' are substituted with identical substituents (-Me).
• The biphenyl can be divided into two halves by passing a horizontal plane along the σ bond joining the two phenyl rings.

¹³C NMR → Total 5-signal
Conclusion:-
Hence, the number of signals observed in the proton decoupled 13C NMR spectrum of the following compound is 5.

Concept:

Conductometric Titration

• Conductometric titration measures the conductance of a solution during the titration process.
• During the titration of HCl (strong acid) with NH₄OH (weak base):
  • Initially, H⁺ ions from HCl dominate the solution, leading to high conductance.
  • As NH₄OH is added, H⁺ ions react with OH^- to form water, reducing the conductance.
  • At the equivalence point, NH₄Cl is formed, and conductance becomes relatively constant because NH₄⁺ and Cl^- ions do not contribute significantly to conductance.
  • After the endpoint, adding excess NH₄OH results in negligible change in conductance because NH₄OH is a weak base with low ionization.

Explanation:

• Option 1: Incorrect. The conductance decreases as HCl is neutralized and does not increase till the endpoint.
• Option 2: Incorrect. Conductance decreases during the reaction and remains constant only after the endpoint.
• Option 3: Correct. After the endpoint, conductance remains relatively constant due to weak ionization of NH₄OH.
• Option 4: Incorrect. Conductance does not increase significantly after the endpoint because NH₄OH is a weak base.

Correct Answer: 3. The conductance remains more or less constant after the endpoint.

Concept:
In mass spectrometry, the base peak represents the most intense ion peak, typically corresponding to the most stable fragment of the molecule. The Electron Ionization (EI) process results in the ionization and subsequent fragmentation of molecules. The fragmentation pattern depends on the structure of the molecule and the stability of the resulting ions.
Fragmentation and Base Peak Formation:

• For compound P: The fragmentation occurs primarily due to the loss of the hydroxyl group (OH), leading to the formation of a stable tropylium ion (m/z = 104). This ion is stabilized by resonance, making it the base peak in the mass spectrum of compound P.
• For compound Q: The hydroxyl group and the benzylic position play a key role in the fragmentation. The molecular ion loses a CH₃ group, resulting in a stable fragment with m/z = 107. This is the base peak for compound Q.
• Stability of the Fragments: The base peak typically corresponds to the most stable cation formed during fragmentation. For P, the tropylium ion (m/z = 104) is highly stable due to resonance. For Q, the benzylic cation (m/z = 107) is also quite stable, leading to it being the base peak.

Explanation: 

• Compound P undergoes fragmentation to form a tropylium ion at m/z = 104 as the base peak. The loss of the hydroxyl group leads to this stable ion.
• Compound Q undergoes fragmentation, where the loss of a CH₃ group from the benzylic position leads to a stable ion with m/z = 107 as the base peak.
  
• The base peaks are formed due to the stability of the resulting fragments, making 104 the base peak for P and 107 the base peak for Q.

Conclusion:
The correct base peaks for compound P and compound Q are 104 and 107, respectively.