IGNOU BCHCT-131 Solved Question Paper PDF Download

IGNOU BCHCT-131 Previous Year Solved Question Paper in Hindi

IGNOU BCHCT-131 Previous Year Solved Question Paper in English

Part—A

Q1. (a) State the limitations of Bohr’s theory. (b) What are matter waves?

Ans. (a) The limitations of Bohr’s theory are as follows:

• Inapplicability to Multi-electron Atoms: Bohr’s model could only successfully explain the spectrum of single-electron species (e.g., H, He⁺, Li²⁺). It failed to account for the spectra of multi-electron atoms.
• Fine Structure of Spectral Lines: When viewed through a high-resolution spectroscope, spectral lines are found to be split into several closely spaced lines, known as the fine structure. Bohr’s theory could not explain this phenomenon.
• Zeeman and Stark Effects: Bohr’s theory was unable to explain the splitting of spectral lines in the presence of a magnetic field (Zeeman effect) or an electric field (Stark effect).
• Violation of Heisenberg’s Uncertainty Principle: According to Bohr, an electron revolves in a circular orbit with a fixed radius and a definite velocity. This contradicts Heisenberg’s Uncertainty Principle, which states that it is impossible to determine simultaneously and precisely both the position and momentum of a particle.
• Explanation of Chemical Bonding: The theory failed to explain how atoms combine via chemical bonds to form molecules.
• Wave-Particle Duality: Bohr’s model treated the electron only as a particle and did not take into account its wave nature.

(b)

Matter Waves:

In 1924, Louis de Broglie proposed that just as light possesses both wave and particle properties (dual nature), moving particles of matter, such as electrons, should also exhibit wave properties. These waves associated with a moving particle are called

matter waves

or

de Broglie waves

.

According to de Broglie, the wavelength (λ) of the wave associated with a particle of mass ‘m’ moving with a velocity ‘v’ is given by:

λ = h / mv = h / p

where ‘h’ is Planck’s constant and ‘p’ is the momentum of the particle. This hypothesis was later experimentally verified by Davisson and Germer, confirming the dual nature of matter.

Q2. (a) Calculate the de-Broglie’s wavelength associated with a bullet having mass 2.2×10⁻³ kg moving with a velocity of 30 ms⁻¹. (b) (i) What is Laplacian operator? (ii) What is Hamiltonian operator? Also give their significance.

Ans. (a) The de-Broglie wavelength (λ) can be calculated using the formula: λ = h / mv Where:

• h = Planck’s constant = 6.626 x 10⁻³⁴ J s
• m = mass of the bullet = 2.2 x 10⁻³ kg
• v = velocity of the bullet = 30 ms⁻¹

Substituting the values into the formula:

λ = (6.626 x 10⁻³⁴ J s) / (2.2 x 10⁻³ kg * 30 ms⁻¹)

λ = (6.626 x 10⁻³⁴) / (66 x 10⁻³) m

λ = (6.626 / 66) x 10⁻³¹ m

λ = 0.10039 x 10⁻³¹ m

λ = 1.004 x 10⁻³² m

The wavelength associated with the bullet is extremely small, which is why the wave properties of macroscopic objects are not observable.

(b) (i) Laplacian Operator (∇²): The Laplacian operator, also called ‘del squared’, is a second-order differential operator. In three-dimensional Cartesian coordinates, it is defined as: ∇² = ∂²/∂x² + ∂²/∂y² + ∂²/∂z² Its significance lies in quantum mechanics where it is a part of the kinetic energy operator of a particle. In the Schrödinger equation, it acts on the wave function (ψ).

(ii) Hamiltonian Operator (Ĥ): The Hamiltonian operator corresponds to the total energy (kinetic energy + potential energy) of a quantum mechanical system. It is represented as: Ĥ = T̂ + V̂ Where T̂ is the kinetic energy operator and V̂ is the potential energy operator. T̂ = – (ħ² / 2m) ∇² (where ħ = h/2π) Therefore, Ĥ = [- (ħ² / 2m) ∇²] + V̂(x, y, z) Significance: The Hamiltonian operator is the central component of the time-independent Schrödinger equation (Ĥψ = Eψ). When the Hamiltonian operator acts on the wave function (ψ) of a system, the result is the same wave function multiplied by the eigenvalue of the total energy (E) of the system. Thus, it allows us to determine the allowed energy levels of a quantum system.

Q3. (a) Explain why O₂ is paramagnetic and O₂²⁻ is diamagnetic. (b) Write Lewis structures of NH₄⁺ and CO₃²⁻.

Ans. (a) The paramagnetic behaviour of O₂ and diamagnetic behaviour of O₂²⁻ can be explained on the basis of Molecular Orbital Theory (MOT) . O₂ molecule: An oxygen atom (Z=8) has the electronic configuration 1s²2s²2p⁴. The O₂ molecule has a total of 16 electrons. The molecular orbital configuration is: σ1s², σ 1s², σ2s², σ 2s², σ2p₂, π2pₓ², π2pᵧ², π 2pₓ¹, π 2pᵧ¹ In this configuration, there are two electrons that are unpaired in the two degenerate π* (antibonding) molecular orbitals. According to Hund’s rule, they occupy separate orbitals. The presence of these unpaired electrons makes the O₂ molecule paramagnetic , meaning it is attracted by an external magnetic field.

O₂²⁻ ion (Peroxide ion): The O₂²⁻ ion is formed by adding two extra electrons to the O₂ molecule, making a total of 18 electrons. These two additional electrons go into the π* antibonding orbitals. The molecular orbital configuration is: σ1s², σ 1s², σ2s², σ 2s², σ2p₂, π2pₓ², π2pᵧ², π 2pₓ², π 2pᵧ² In the O₂²⁻ ion, all molecular orbitals are completely filled, and there are no unpaired electrons . All electrons are paired. Therefore, the O₂²⁻ ion is diamagnetic , meaning it is weakly repelled by an external magnetic field.

(b) Lewis Structures: (i) Ammonium ion (NH₄⁺):

1. Total valence electrons: N (5) + 4xH (4×1) – 1 (for the positive charge) = 8 electrons.
2. The central atom is Nitrogen (N), surrounded by four Hydrogen (H) atoms.
3. 4 x 2 = 8 electrons are used to form a single bond with each H.
4. All 8 electrons are used. Nitrogen has a full octet, and each hydrogen has a full duet.
5. The overall structure has a formal charge of +1.

Structure: [ H-N(H)-H ]⁺ (with an H above the N)

[Image of NH₄⁺ Lewis structure showing N bonded to 4 H atoms, enclosed in brackets with a +1 charge]

(ii) Carbonate ion (CO₃²⁻):

1. Total valence electrons: C (4) + 3xO (3×6) + 2 (for the negative charge) = 4 + 18 + 2 = 24 electrons.
2. The central atom is Carbon (C), surrounded by three Oxygen (O) atoms.
3. Connect C to each O with a single bond (3 x 2 = 6 electrons).
4. Place the remaining 24 – 6 = 18 electrons as lone pairs on the outer atoms (O). Each O gets 6 electrons (3 lone pairs).
5. Now, the central carbon atom has only 6 electrons (incomplete octet). To complete carbon’s octet, move a lone pair from one of the oxygens to form a C-O double bond.
6. This leads to resonance. The carbonate ion is a hybrid of three resonance structures.

Structure: [ O=C(O⁻)-O⁻ ] ↔ [ O⁻-C(O⁻)=O ] ↔ [ O⁻-C(=O)-O⁻ ] (All structures in brackets with a 2- charge)

[Image of the three resonance structures of CO₃²⁻, showing the double bond moving between the three C-O bonds, and the overall 2- charge]

Q4. (a) Define electron affinity. Write units used to measure it. (b) Arrange the following in the increasing order of ionisation energy: Be, B, N, O. Also give reasons for your answer.

Ans. (a) Electron Affinity: Electron affinity is defined as the amount of energy released when an electron is added to the outermost shell of a neutral, gaseous atom. This process forms a negative ion (anion). It can be represented as: X(g) + e⁻ → X⁻(g) + Energy (ΔH = -ve) A high electron affinity (more negative value) indicates that the atom has a strong tendency to accept an electron. Elements like halogens have high electron affinities because gaining one electron gives them a stable, filled valence shell. Units: Electron affinity is typically measured in the following units:

• Kilojoules per mole (kJ/mol)
• Electron volts per atom (eV/atom)

(b) Increasing Order of Ionisation Energy: The correct increasing order of ionisation energy for the given elements (Be, B, N, O) is: B < Be < O < N

Reasons: Ionisation energy generally increases from left to right across a period because the nuclear charge increases and atomic size decreases, making it more difficult to remove the outermost electron. However, there are two exceptions in this series:

1. Why is the ionisation energy of Be greater than B?
   • Beryllium (Be) has the electronic configuration 1s² 2s² . Its 2s orbital is completely filled, which gives it an extra stable configuration.
   • Boron (B) has the electronic configuration 1s² 2s² 2p¹ . Its outermost electron is in the 2p orbital, which is of higher energy and is shielded by the 2s orbital.
   • Therefore, more energy is required to remove an electron from the stable, fully-filled 2s orbital of Be than to remove the single 2p electron of B.
2. Why is the ionisation energy of N greater than O?
   • Nitrogen (N) has the electronic configuration 1s² 2s² 2p³ . Its 2p orbital is exactly half-filled, which is a particularly stable configuration.
   • Oxygen (O) has the electronic configuration 1s² 2s² 2p⁴ . The 2p orbital contains one paired electron. Removing this paired electron results in a stable, half-filled configuration (2p³). The inter-electronic repulsion between the paired electrons also makes this electron easier to remove.
   • Therefore, more energy is required to remove an electron from the stable, half-filled 2p³ configuration of Nitrogen than to remove one from the 2p⁴ configuration of Oxygen.

Q5. (a) State Aufbau’s principle with example. (b) Write the electronic configuration of Copper (At. no. of Cu = 29). (c) Write the hybridisation of each carbon atom in the following: H₃C — CH = CH — CH₃

Ans. (a) Aufbau’s Principle: The Aufbau principle (from German ‘Aufbau’ meaning ‘building up’) states that, in its ground state, the electrons of an atom first occupy the orbitals with the lowest available energy, and then progressively enter orbitals of higher energy. The order in which orbitals are filled is in the increasing order of their energy, which is generally determined by the (n + l) rule :

1. Orbitals with a lower value of (n + l) are filled first. (n = principal quantum number, l = azimuthal quantum number).
2. If two orbitals have the same value of (n + l), the orbital with the lower value of n is filled first.

The order of filling is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, …

Example: Potassium (K, Z = 19)

Potassium has 19 electrons. According to the Aufbau principle, the electrons will fill as follows:

• 1s² (2 electrons)
• 2s² (2 electrons)
• 2p⁶ (6 electrons)
• 3s² (2 electrons)
• 3p⁶ (6 electrons)

So far, 18 electrons are filled. The 19th electron will go into either 3d or 4s.

• For 4s: n=4, l=0, so n+l = 4
• For 3d: n=3, l=2, so n+l = 5

Since the (n+l) value is lower for 4s, the 19th electron will enter the 4s orbital.

Thus, the electronic configuration of Potassium is:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

.

(b) Electronic Configuration of Copper (Cu) (Z = 29): Copper is a notable exception to the Aufbau principle. The expected configuration based on the Aufbau principle would be: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁹. However, due to the extra stability of completely filled (d¹⁰) and half-filled (d⁵) subshells, one electron moves from the 4s orbital to the 3d orbital to make the 3d subshell completely filled. This leads to a more stable configuration. Therefore, the actual electronic configuration of Copper is: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ or [Ar] 3d¹⁰ 4s¹

(c) Hybridisation in H₃C — CH = CH — CH₃ (But-2-ene): Let’s number the carbon atoms from C1 to C4 in the given molecule: H₃ C¹ — C² H = C³ H — C⁴ H₃

• C¹ and C⁴: Both of these carbon atoms are attached to four other atoms via single bonds (three C-H and one C-C). They are sp³ hybridised . (Geometry: tetrahedral).
• C² and C³: Both of these carbon atoms are part of a double bond (C=C) and are also attached to two other atoms via single bonds (one C-C and one C-H). They are sp² hybridised . (Geometry: trigonal planar).

Q6. (a) Write the possible values of all the quantum numbers for all the electrons present in a 2p orbital. (b) Using VSEPR theory, predict the shape of SF₄.

Ans. (a) For a 2p orbital , the quantum numbers are as follows:

• The principal quantum number (n) = 2 (because it’s the second shell).
• The azimuthal quantum number (l) = 1 (because it’s a ‘p’ subshell).

The p subshell contains three orbitals, corresponding to the different values of the magnetic quantum number (mₗ).

• Possible values for the magnetic quantum number (mₗ) are from -l to +l, i.e., -1, 0, +1 .

Each orbital (each mₗ value) can hold two electrons, which have opposite spin quantum numbers (mₛ).

• Possible values for the spin quantum number (mₛ) are +1/2 and -1/2 .

Since the 2p subshell has 3 orbitals, it can accommodate a total of 6 electrons. The possible sets of all four quantum numbers for all 6 electrons in a 2p subshell are:

(b) Shape of SF₄ using VSEPR Theory: The shape of SF₄ can be predicted using the Valence Shell Electron Pair Repulsion (VSEPR) theory.

1. Central Atom: Sulfur (S)
2. Number of valence electrons of the central atom: Sulfur is in Group 16, so it has 6 valence electrons.
3. Electrons used in bonding: Sulfur forms four single bonds with four Fluorine (F) atoms. Each bond uses 1 electron. So, 4 electrons are used in bonding.
4. Number of lone pair electrons: 6 (total) – 4 (bonding) = 2 electrons. These 2 electrons form one lone pair .
5. Total Electron Pairs: 4 bonding pairs + 1 lone pair = 5 electron pairs.
6. Electron Geometry: For 5 electron pairs, the electron geometry that minimizes repulsion is trigonal bipyramidal .
7. Molecular Shape: The lone pair occupies more space and causes more repulsion than bonding pairs. To minimize repulsion, the lone pair occupies an equatorial position in the trigonal bipyramidal geometry. The remaining four bonding pairs occupy the two axial and two equatorial positions. This results in a distorted shape called a see-saw .

Thus, according to VSEPR theory, the shape of SF₄ is a

see-saw

.

Q7. (a) Write the resonance structures of NO₃⁻ ion. (b) Draw the molecular orbitals formed by the combination of s-s atomic orbitals. Also name these orbitals.

Ans. (a) Resonance Structures of the Nitrate Ion (NO₃⁻): The nitrate ion consists of a central nitrogen atom bonded to three oxygen atoms, with an overall charge of -1. Its Lewis structure involves one N=O double bond and two N-O single bonds. The double bond can be formed with any of the three oxygen atoms, leading to three equivalent resonance structures.

1. Total valence electrons: N(5) + 3xO(6) + 1(charge) = 5 + 18 + 1 = 24 electrons.
2. The actual structure is a resonance hybrid of these three resonance forms.
3. In the resonance hybrid, all three N-O bonds are identical in length and strength, intermediate between a single and a double bond. The negative charge of -1 is delocalized equally over the three oxygen atoms.

The three resonance structures are as follows:

[Image showing the three resonance structures of the nitrate ion, NO₃⁻.

Structure 1: N double-bonded to the top O, single-bonded to the left and right O. The single-bonded Os each have a -1 formal charge.

Structure 2: N double-bonded to the left O, single-bonded to the top and right O. The single-bonded Os each have a -1 formal charge.

Structure 3: N double-bonded to the right O, single-bonded to the top and left O. The single-bonded Os each have a -1 formal charge.

Double-headed arrows (↔) are drawn between the structures.]

(b) Molecular Orbitals from s-s Atomic Orbital Combination: When two s-atomic orbitals (AOs) approach each other, they combine linearly to form two molecular orbitals (MOs). This combination can happen in two ways:

1. Bonding Molecular Orbital: This is formed when the two s-orbitals undergo additive overlap or constructive interference. The resulting MO has increased electron density between the two nuclei. This increased density holds the nuclei together, stabilizing the system. This molecular orbital is lower in energy than the original atomic orbitals. It is called a sigma (σₛ) molecular orbital because it is symmetrical about the internuclear axis. ψ(σₛ) = ψ(s₁) + ψ(s₂) [Image showing two s-orbitals approaching, overlapping in-phase (same shading) to form a larger, lower-energy sigma (σₛ) bonding molecular orbital with high electron density between the nuclei.]
2. Antibonding Molecular Orbital: This is formed when the two s-orbitals undergo subtractive overlap or destructive interference. The resulting MO has a nodal plane between the two nuclei, where the electron density is zero. The lack of electron density between the nuclei leads to nuclear repulsion, destabilizing the system. This molecular orbital is higher in energy than the original atomic orbitals. It is called a sigma-star (σ*ₛ) molecular orbital. ψ(σ*ₛ) = ψ(s₁) – ψ(s₂) [Image showing two s-orbitals approaching, overlapping out-of-phase (different shading) to form a higher-energy sigma-star (σ*ₛ) antibonding molecular orbital with a nodal plane between the nuclei.]

Names:

• The lower energy orbital: Sigma (σₛ) bonding molecular orbital .
• The higher energy orbital: Sigma-star (σ*ₛ) antibonding molecular orbital .

Part—B

Q8. (a) Define Lewis acids. Which of the following are Lewis acids? CH₃⁺, CH₃OCH₃, CH₃CH=CH₂, BF₃ (b) (i) Arrange the following groups in an ascending order of priority in assigning Z/E configuration: –CH(CH₃)₂, –C(CH₃)₃, –CH₂–CH₂CH₃ (ii) Draw the structure of isomer of CH₃FC=CFCH₃ which has a higher dipole moment.

Ans. (a) Lewis Acids: According to G.N. Lewis, a Lewis acid is any species (molecule or ion) that can accept a pair of electrons to form a covalent bond. Lewis acids are electron-deficient. They are typically one of the following:

• Molecules with an incomplete octet (e.g., BF₃, AlCl₃).
• Cations (e.g., H⁺, CH₃⁺).
• Molecules where the central atom can expand its octet using vacant d-orbitals (e.g., SiF₄, SnCl₄).

Lewis acids from the given compounds:

• CH₃⁺ (Methyl carbocation): This is a cation with only 6 electrons on the carbon. It has a vacant p-orbital and can readily accept a pair of electrons. Therefore, it is a Lewis acid .
• CH₃OCH₃ (Dimethyl ether): The oxygen atom has two lone pairs of electrons, which it can donate. Therefore, it is a Lewis base, not a Lewis acid.
• CH₃CH=CH₂ (Propene): The pi (π) bond has electron density which it can donate to an electrophile. It typically acts as a Lewis base.
• BF₃ (Boron trifluoride): The central boron atom has only 6 valence electrons around it, giving it an incomplete octet. It can complete its octet by accepting an electron pair. Therefore, it is a strong Lewis acid .

Thus,

CH₃⁺

and

BF₃

are Lewis acids.

(b) (i) Ascending Order of Priority: Priority for Z/E configuration is determined by the Cahn-Ingold-Prelog (CIP) rules. Priority is based on the atomic number of the atom directly attached. If they are the same, we proceed along the chain until the first point of difference.

1. –CH₂–CH₂CH₃ (n-propyl): The first carbon is attached to H, H, C.
2. –CH(CH₃)₂ (isopropyl): The first carbon is attached to H, C, C.
3. –C(CH₃)₃ (t-butyl): The first carbon is attached to C, C, C.

Comparing them:

• The t-butyl group has its first carbon attached to three other carbons, giving it the highest priority.
• The isopropyl group has its first carbon attached to two carbons and one hydrogen, giving it intermediate priority.
• The n-propyl group has its first carbon attached to only one carbon and two hydrogens, giving it the lowest priority.

The ascending order of priority is:

–CH₂–CH₂CH₃ < –CH(CH₃)₂ < –C(CH₃)₃

(ii) Isomer with Higher Dipole Moment: The molecule CH₃FC=CFCH₃ (1,2-difluoro-but-2-ene) exhibits geometric isomerism (Z/E).

• E-isomer (trans): In this, the two polar C-F bonds are on opposite sides of the double bond. The individual bond dipoles cancel each other out almost completely, resulting in a very small or zero net dipole moment.
• Z-isomer (cis): In this, the two polar C-F bonds are on the same side of the double bond. The individual bond dipoles add up, resulting in a significant, non-zero net dipole moment .

Therefore, the

Z-isomer

has the higher dipole moment.

Structure of the Z-isomer:

[Image of the Z-isomer of 1,2-difluoro-but-2-ene. It shows a C=C double bond. The two F atoms are on the same side of the double bond’s axis, and the two CH₃ groups are also on the same side.]

Q9. (a) Explain with suitable structures as to why is allyl carbocation less stable than benzyl cation. (b) What do you understand by Anti-Markownikoff’s addition? Give one example of such a reaction.

Ans. (a) The stability of a carbocation depends on the extent of delocalization of the positive charge. The greater the delocalization, the more the charge is spread out, and the more stable the cation. Both allyl and benzyl carbocations are stabilized by resonance, but the difference in their stability arises from the extent of delocalization.

Allyl Carbocation (CH₂=CH-CH₂⁺): In the allyl carbocation, the positive charge is delocalized over two carbon atoms, giving rise to two resonance structures: CH₂=CH—CH₂⁺ ↔ ⁺CH₂—CH=CH₂ [Image showing the two resonance structures of the allyl carbocation.] The positive charge is shared between only two adjacent carbon atoms.

Benzyl Carbocation (C₆H₅CH₂⁺): In the benzyl carbocation, the positive charge is delocalized into the benzene ring. The positive charge can be spread over the ortho and para positions of the ring, resulting in a total of five resonance structures (one structure with no ring delocalization and four with delocalization into the ring). [Image showing the five resonance structures of the benzyl carbocation, with the positive charge delocalized from the benzylic carbon to the ortho, para, and then the other ortho position of the benzene ring.] Starting from the CH₂⁺ group, the charge moves to the ortho, then para, then the other ortho carbon.

Conclusion: Since the benzyl carbocation has five resonance structures while the allyl carbocation has only two , the delocalization of positive charge is much greater in the benzyl carbocation. This greater dispersal of charge makes the benzyl carbocation significantly more stable than the allyl carbocation.

(b) Anti-Markovnikov Addition: Anti-Markovnikov addition, also known as the ‘peroxide effect’ or ‘Kharasch effect’, is an addition reaction of hydrogen bromide (HBr) to an unsymmetrical alkene that takes place in the presence of a peroxide (e.g., R-O-O-R) . It is the opposite of Markovnikov’s rule. The anti-Markovnikov rule states that the negative part of HBr (Br) adds to the carbon atom of the double bond that has the greater number of hydrogen atoms , and the positive part (H) adds to the carbon with fewer hydrogen atoms. This reaction proceeds via a free-radical chain mechanism , not an electrophilic addition mechanism. This effect is observed only with HBr, not with HCl or HI.

Example: Addition of HBr to Propene In the presence of peroxide, the addition of HBr to propene (CH₃-CH=CH₂) gives 1-bromopropane. CH₃—CH=CH₂ + HBr –(Peroxide)–> CH₃—CH₂—CH₂Br (Propene) (1-Bromopropane – Major product) In this reaction, the bromine (Br) attaches to the terminal carbon (CH₂), which has more hydrogens, and the hydrogen (H) attaches to the central carbon (CH).

Q10. (a) Complete the following: CH₃COO⁻Na⁺ –(Electrolysis)–> A –(Dimerisation)–> B (Alkene) + C (Gas) (b) Write the products of the following reactions: (i) CH₃CH₂C(=O)Cl + H₂/Pd-BaSO₄ → ? (ii) CH₃CH₂C≡CH + NaNH₂ → ? –(CH₃CH₂I)–> ?

Ans. (a) This reaction is the Kolbe’s electrolytic reaction . When an aqueous solution of sodium acetate (CH₃COO⁻Na⁺) is electrolyzed:

At the Anode (Oxidation): The acetate ion (CH₃COO⁻) loses an electron to form an acetate free radical, which immediately decomposes to give a methyl free radical and carbon dioxide gas. CH₃COO⁻ → CH₃COO• + e⁻ CH₃COO• → CH₃• (A) + CO₂ (C) So, A = CH₃• (Methyl radical) and C = CO₂ (Carbon dioxide) .

Dimerisation: Two methyl radicals (A) formed at the anode combine to form ethane (an alkane). (Note: the question says B is an Alkane, not Alkene) . CH₃• + •CH₃ → CH₃—CH₃ (B) So, B = CH₃—CH₃ (Ethane) .

Summary of the reaction: A = CH₃• (Methyl radical) B = CH₃—CH₃ (Ethane) C = CO₂ (Carbon dioxide)

(b) Products of the reactions:

(i) CH₃CH₂C(=O)Cl + H₂/Pd-BaSO₄ → ? This is the Rosenmund reduction . It is a selective catalytic hydrogenation of an acyl chloride to an aldehyde. The catalyst is palladium (Pd) supported on barium sulfate (BaSO₄) and partially deactivated (poisoned) with something like sulfur or quinoline. The poison prevents the aldehyde from being further reduced to an alcohol. The reduction of propanoyl chloride will yield propanal. CH₃CH₂C(=O)Cl + H₂ –(Pd/BaSO₄)–> CH₃CH₂CHO + HCl Product: Propanal

(ii) CH₃CH₂C≡CH + NaNH₂ → ? –(CH₃CH₂I)–> ? This reaction occurs in two steps: Step 1: Deprotonation of the terminal alkyne But-1-yne (CH₃CH₂C≡CH) has an acidic terminal hydrogen. Sodium amide (NaNH₂) is a strong base that removes this proton to form an alkynide ion (a nucleophile). CH₃CH₂C≡CH + NaNH₂ → CH₃CH₂C≡C⁻Na⁺ + NH₃ The first product is sodium but-1-ynide.

Step 2: Nucleophilic substitution (Alkylation) The resulting alkynide ion then undergoes an Sₙ2 reaction with an alkyl halide, such as iodoethane (CH₃CH₂I). The alkynide acts as the nucleophile and displaces the iodide ion, forming a longer internal alkyne. CH₃CH₂C≡C⁻Na⁺ + CH₃CH₂I → CH₃CH₂C≡CCH₂CH₃ + NaI Final product: Hex-3-yne

Q11. (a) Give one example each of the following with its structure: (i) A bicyclic aromatic compound (ii) A heterocyclic aromatic compound (iii) An aromatic compound containing 14π electrons (b) What is Sabatier-Sabatier reaction? Give one example.

Ans. (a) Examples and Structures: (i) A bicyclic aromatic compound: This is an aromatic compound containing two fused rings. The most common example is naphthalene . Example: Naphthalene (C₁₀H₈) Structure: [Image of the structure of Naphthalene, showing two fused benzene rings.] It obeys Hückel’s rule (4n+2)π, with 10π electrons (n=2).

(ii) A heterocyclic aromatic compound: This is a cyclic compound that is aromatic and contains at least one other element (a heteroatom), such as nitrogen, oxygen, or sulfur, in the ring besides carbon. Example: Pyridine (C₅H₅N) Structure: [Image of the structure of Pyridine, showing a six-membered ring with five carbons, one nitrogen, and alternating double bonds. The nitrogen has a lone pair.] Pyridine has 6π electrons and is aromatic. The lone pair on nitrogen is not part of the π system.

(iii) An aromatic compound containing 14π electrons: This obeys Hückel’s rule (4n+2)π, where 4n+2 = 14, so n=3. This means it is an aromatic compound. Example: Anthracene (C₁₄H₁₀) Structure: [Image of the structure of Anthracene, showing three benzene rings fused in a linear fashion.] Anthracene consists of three benzene rings fused linearly and has a total of 14π electrons.

(b) Sabatier-Sabatier Reaction: The Sabatier-Sabatier reaction is the hydrogenation of unsaturated hydrocarbons (alkenes or alkynes) to saturated hydrocarbons (alkanes) . The reaction uses hydrogen gas (H₂) in the presence of a finely divided Nickel (Ni) catalyst at high temperatures (around 200-300 °C) and pressure. Other catalysts like Palladium (Pd) or Platinum (Pt) can also be used. It is a highly exothermic addition reaction. General Reaction: R-CH=CH-R’ + H₂ –(Ni, Δ)–> R-CH₂-CH₂-R’ (Alkene to Alkane) R-C≡C-R’ + 2H₂ –(Ni, Δ)–> R-CH₂-CH₂-R’ (Alkyne to Alkane)

Example: Hydrogenation of Ethene When ethene gas is mixed with hydrogen and passed over a heated nickel catalyst, it is reduced to form ethane. CH₂=CH₂ + H₂ –(Ni, 250°C)–> CH₃—CH₃ (Ethene) (Ethane)

Q12. (a) What do you understand by the following? (i) Resolution (ii) Laevorotatory compound (iii) Racemic mixture (b) Write the reactions for the following: (i) Products of ozonolysis of butyne-2 (ii) Products of ozonolysis of butene-2

Ans. (a) (i) Resolution: Resolution is the process by which a racemic mixture (a 50:50 mixture of two enantiomers) is separated into its individual, pure enantiomers, the (+)-form and the (-)-form. Since enantiomers have identical physical properties (boiling point, melting point, solubility), they cannot be separated by simple physical methods like fractional distillation or crystallization. Resolution typically involves converting the enantiomers into a pair of diastereomers by reacting them with a pure chiral agent (called a resolving agent). Diastereomers have different physical properties, so they can be separated by fractional crystallization or chromatography. After separation, the resolving agent is removed from the diastereomers to regenerate the original enantiomers.

(ii) Laevorotatory compound: This is an optically active compound that rotates the plane of plane-polarized light to the left (in a counter-clockwise direction) . This property is called laevorotation. It is designated by a small letter ‘l’ or a minus sign (-) , written in parentheses before the name of the compound, e.g., (-)-lactic acid. The angle of rotation is measured with an instrument called a polarimeter. The enantiomer of a laevorotatory compound is always dextrorotatory, rotating the plane of light by an equal amount to the right.

(iii) Racemic mixture: A racemic mixture is an equimolar (50:50) mixture of the two enantiomers of a chiral compound. Since the mixture contains equal amounts of the dextrorotatory (+) enantiomer and the laevorotatory (-) enantiomer, the rotation caused by one enantiomer is exactly cancelled by the rotation caused by the other. As a result, a racemic mixture is optically inactive . It is often designated by the prefix (±)-, e.g., (±)-lactic acid.

(b) (i) Products of ozonolysis of but-2-yne: Ozonolysis of an alkyne (reaction with O₃) followed by aqueous workup (H₂O) cleaves the triple bond and forms two carboxylic acids. But-2-yne (CH₃-C≡C-CH₃) is a symmetrical alkyne. CH₃—C≡C—CH₃ + O₃ → [Ozonide intermediate] –(H₂O)–> 2 CH₃COOH The products are two molecules of acetic acid (ethanoic acid) . Reaction: CH₃—C≡C—CH₃ –(1. O₃)–> –(2. H₂O)–> 2 CH₃COOH

(ii) Products of ozonolysis of but-2-ene: Ozonolysis of an alkene cleaves the double bond. The products depend on the nature of the workup (reductive or oxidative). If not specified in the question, a reductive workup (e.g., with Zn/H₂O) is usually assumed, which gives aldehydes or ketones. But-2-ene (CH₃-CH=CH-CH₃) is a symmetrical alkene. CH₃—CH=CH—CH₃ + O₃ → [Ozonide intermediate] –(Zn/H₂O)–> 2 CH₃CHO With a reductive workup, the products are two molecules of acetaldehyde (ethanal) . Reaction: CH₃—CH=CH—CH₃ –(1. O₃)–> –(2. Zn/H₂O)–> 2 CH₃CHO

Q13. (a) Draw and name the most stable and the least stable conformations of butane. (b) Define and give one example each of the following: (i) An electrophile (ii) A nucleophile (iii) A free radical

Ans. (a) The conformations of butane (CH₃-CH₂-CH₂-CH₃) arise from rotation about the C2-C3 bond. They are represented using Newman projections.

Most Stable Conformation: Anti-periplanar or Anti-staggered This is the lowest-energy, and therefore most stable, conformation of butane. In it, the two large methyl (CH₃) groups are at a dihedral angle of 180° to each other. This arrangement minimizes steric hindrance. Name: Anti-periplanar or Anti conformation Structure (Newman Projection): [Image of the anti-conformation of butane as a Newman projection. The front carbon has a methyl group pointing down, and two hydrogens pointing up-left and up-right. The back carbon has a methyl group pointing up, and two hydrogens pointing down-left and down-right. The two methyl groups are 180° apart.]

Least Stable Conformation: Syn-periplanar or Fully eclipsed This is the highest-energy, and therefore least stable, conformation of butane. In it, the two large methyl (CH₃) groups are eclipsing each other (dihedral angle 0°). This arrangement causes maximum steric hindrance and torsional strain between the two bulky groups. Name: Syn-periplanar or Fully eclipsed conformation Structure (Newman Projection): [Image of the syn-periplanar conformation of butane. The front carbon has a methyl group pointing down. The back carbon’s methyl group is directly behind it, also pointing down. The hydrogens on both carbons are also eclipsing each other.]

(b) (i) An electrophile: Definition: An electrophile (‘electron-loving’) is a reagent that is electron-deficient and forms a new covalent bond by accepting a pair of electrons during a chemical reaction. Electrophiles are either positively charged or are neutral molecules with an electron-deficient center (e.g., incomplete octet). They act as Lewis acids. Examples:

• Positively charged: H⁺ (proton), NO₂⁺ (nitronium ion), CH₃⁺ (carbocation)
• Neutral: BF₃ , SO₃ , AlCl₃

(ii) A nucleophile: Definition: A nucleophile (‘nucleus-loving’) is a reagent that is electron-rich and forms a new covalent bond by donating a pair of electrons during a chemical reaction. Nucleophiles are either negatively charged or are neutral molecules containing at least one lone pair of electrons. They act as Lewis bases. Examples:

• Negatively charged: OH⁻ (hydroxide ion), CN⁻ (cyanide ion), Cl⁻ (chloride ion)
• Neutral: H₂O (water), NH₃ (ammonia), R-OH (alcohols)

(iii) A free radical: Definition: A free radical is an atom, molecule, or ion that has one or more unpaired valence electrons . The unpaired electron makes free radicals highly reactive and unstable. They are typically formed by the homolytic cleavage of covalent bonds, where the two electrons of the bond are split equally between the separating species. Free radicals are denoted by a dot (•). Examples:

• Cl• (Chlorine free radical): Cl₂ → 2Cl• (in the presence of UV light)
• CH₃• (Methyl radical)
• Br• (Bromine radical)

Q14. (a) Which of the following will be the strongest acid and why? CCl₃COOH, CH₃CH₂COOH, CH₂ClCOOH (b) Complete the following reaction: Ph₃P + CH₃CH(X)CH₃ → A –(C₄H₉Li)–> B (Ylide) + R₂C=O → C (Alkene)

Ans. (a) Strongest Acid and Reason: Of the given carboxylic acids, CCl₃COOH (Trichloroacetic acid) will be the strongest acid.

Reason: The strength of an acid depends on the stability of its conjugate base (the carboxylate ion, RCOO⁻). The more stable the conjugate base, the stronger the acid. The stability of the carboxylate ion depends on how effectively the negative charge is delocalized or stabilized.

1. Inductive Effect (-I effect): Electron-withdrawing groups, like chlorine, exert a strong negative inductive effect (-I effect). When a -I group is present near the carboxylate group, it pulls electron density towards itself, thereby dispersing the negative charge on the oxygen atoms. This dispersal of charge stabilizes the conjugate base.
2. Comparison of Groups:
   • CCl₃COOH: This has three chlorine atoms. The combined -I effect of the three -Cl atoms is very strong, making the conjugate base (CCl₃COO⁻) highly stable.
   • CH₂ClCOOH: This has only one chlorine atom. Its -I effect is present, but much weaker than in CCl₃COOH. It stabilizes CH₂ClCOO⁻, but less so than CCl₃COO⁻.
   • CH₃CH₂COOH: This has an ethyl group (CH₃CH₂), which is an electron-donating group. It exerts a positive inductive effect (+I effect), which intensifies the negative charge by pushing electron density onto the oxygens. This destabilizes the conjugate base (CH₃CH₂COO⁻), making it the weakest acid.

Order of Acidity:

CCl₃COOH > CH₂ClCOOH > CH₃CH₂COOH

Therefore, CCl₃COOH is the strongest acid.

(b) This is a general scheme for the Wittig Reaction , which converts an aldehyde or ketone into an alkene. Let’s assume X = Br (Bromine) and for the last step, let’s use a generic carbonyl compound R₂C=O as shown.

Step 1: Formation of Phosphonium Salt (A) Triphenylphosphine (Ph₃P) acts as a nucleophile and performs an Sₙ2 reaction on 2-bromopropane (CH₃CH(Br)CH₃) to form a phosphonium salt (A). Ph₃P + CH₃CH(Br)CH₃ → [Ph₃P⁺—CH(CH₃)₂]Br⁻ (A) A = Isopropyltriphenylphosphonium bromide

Step 2: Formation of Ylide (B) A strong base, such as butyllithium (C₄H₉Li), removes a proton from the phosphonium salt. This proton is removed from the carbon adjacent to the phosphorus, creating a phosphorus ylide (B). [Ph₃P⁺—CH(CH₃)₂]Br⁻ + C₄H₉Li → Ph₃P=C(CH₃)₂ (B) + LiBr + C₄H₁₀ B = Isopropylidenetriphenylphosphorane (an ylide)

Step 3: Wittig Reaction and Alkene Formation (C) The ylide (B) reacts with a carbonyl compound, R₂C=O. This proceeds via a cyclic intermediate (an oxaphosphetane) which decomposes to give an alkene (C) and triphenylphosphine oxide. Ph₃P=C(CH₃)₂ (B) + R₂C=O → R₂C=C(CH₃)₂ (C) + Ph₃P=O C = The corresponding alkene (e.g., if R₂C=O was acetone, C would be 2,3-dimethylbut-2-ene).

Completed Reaction: A = [Ph₃P⁺—CH(CH₃)₂]X⁻ B = Ph₃P=C(CH₃)₂ C = R₂C=C(CH₃)₂