CBSE Class 12 Chemistry Part 2 Exercise Solutions

(a) Fe (b) Cu (c) Zn (d) Cr

(a) Tetraamminedichloridocobalt(III) chloride (b) Dichloridotetraamminecobalt(III) chloride (c) Tetraamminedichloridocobalt(II) chloride (d) Tetraamminedichlorocobalt(III) chloride

(a) 1° alkyl halide and polar aprotic solvent (b) 3° alkyl halide and polar protic solvent (c) 1° alkyl halide and polar protic solvent (d) 3° alkyl halide and polar aprotic solvent

(a) 1°, 2°, 3° amines (b) 1°, 2°, 3° alcohols (c) aldehydes and ketones (d) aliphatic and aromatic halides

(a) Acetaldehyde (b) Acetone (c) Formaldehyde (d) Propanal

(a) NH₃ > CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N (b) (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃ (c) (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃ (d) CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N > NH₃

(a) Change in optical rotation due to change in temperature (b) Change in specific rotation due to interconversion of α and β anomers in solution (c) Rotation of plane-polarised light by a chiral compound (d) Racemisation of an optically active compound

(a) Vitamin A (b) Vitamin B₁ (c) Vitamin C (d) Vitamin D

(a) 0 (b) 2 (c) 4 (d) 6

(a) Salicylic acid (b) Phenyl benzoate (c) Salicylaldehyde (d) o-Cresol

Transition elements exhibit variable oxidation states because the energy difference between (n-1)d and ns orbitals is very small. Electrons from both the ns and (n-1)d orbitals can participate in bond formation. As the number of available d-electrons varies, different oxidation states become accessible. For example, Mn shows oxidation states from +2 to +7 (3d⁵ 4s² → using 0 to 5 d-electrons plus 2 s-electrons).

IUPAC name: Potassium hexacyanidoferrate(III) Let oxidation state of Fe = x. x + 6(−1) + 3(+1) = 0 → x − 6 + 3 = 0 → x = +3 Iron is in the +3 oxidation state. CN⁻ is a strong field ligand, so the complex is low spin (d⁵ → t₂g⁵ eg⁰) and diamagnetic.

(i) Resonance stabilisation: In chlorobenzene, the lone pair on Cl participates in resonance with the benzene ring, giving the C–Cl bond partial double bond character. This makes it shorter and stronger than in haloalkanes. (ii) sp² hybridisation of carbon: In haloarenes, the carbon attached to halogen is sp² hybridised (more s-character, 33%), making the C–X bond shorter and stronger than the sp³ carbon in haloalkanes (25% s-character).

| Feature | SN1 | SN2 | |---|---|---| | Steps | Two-step (carbocation intermediate) | One-step (concerted) | | Rate law | Rate = k[alkyl halide] (first order) | Rate = k[alkyl halide][nucleophile] (second order) | | Stereochemistry | Racemisation (planar carbocation) | Inversion of configuration (Walden inversion) | | Substrate preference | 3° > 2° > 1° | 1° > 2° > 3° | | Solvent | Polar protic | Polar aprotic |

Lucas test uses a mixture of concentrated HCl and anhydrous ZnCl₂ (Lucas reagent). It is based on the rate of formation of alkyl chlorides, which appear as insoluble turbidity: - 3° alcohols: Immediate turbidity at room temperature (stable carbocation forms instantly) - 2° alcohols: Turbidity within 5 minutes (moderately stable carbocation) - 1° alcohols: No turbidity at room temperature (carbocation too unstable; requires heating)

The Cannizzaro reaction is a disproportionation reaction in which a non-enolisable aldehyde (no α-H) reacts with concentrated alkali to give an alcohol and a carboxylic acid salt simultaneously. 2HCHO + Conc. NaOH → CH₃OH + HCOONa (formaldehyde) (methanol) (sodium formate) One molecule is oxidised to carboxylic acid salt while another is reduced to alcohol. This reaction is not given by aldehydes containing α-hydrogen (which undergo aldol condensation instead).

Hinsberg test uses benzenesulfonyl chloride (C₆H₅SO₂Cl) as the reagent: - 1° amine: RNH₂ + C₆H₅SO₂Cl → C₆H₅SO₂NHR (sulfonamide with one H on N). This H makes the sulfonamide soluble in NaOH (forms salt). Solution remains clear. - 2° amine: R₂NH + C₆H₅SO₂Cl → C₆H₅SO₂NR₂ (sulfonamide with no H on N). Insoluble in NaOH because no H to form salt. Precipitate forms. - 3° amine: R₃N does NOT react with benzenesulfonyl chloride. No precipitate and the amine remains insoluble (oily layer).

| Feature | DNA | RNA | |---|---|---| | Sugar | Deoxyribose (no —OH at C2) | Ribose (—OH at C2) | | Base difference | Thymine (T) pairs with A | Uracil (U) replaces T | | Structure | Double-stranded helix | Usually single-stranded | | Stability | More stable (no 2′-OH) | Less stable (2′-OH makes it susceptible to hydrolysis) |

SN1 Mechanism (Substitution Nucleophilic Unimolecular): Step 1 — Formation of carbocation (slow, rate-determining): R₃C–X → R₃C⁺ + X⁻ The C–X bond breaks heterolytically. This is the slow step, so the rate depends only on [substrate]. Step 2 — Nucleophilic attack (fast): R₃C⁺ + Nu⁻ → R₃C–Nu The nucleophile can attack the planar carbocation from either side. Stereochemical consequence: Since the planar carbocation allows attack from both faces, the product is a racemic mixture (50% retention + 50% inversion). However, in practice, slight inversion preference is observed due to ion-pair effects. Favouring conditions: 3° substrate (stable carbocation), polar protic solvent (stabilises ions), weak nucleophile. SN2 Mechanism (Substitution Nucleophilic Bimolecular): Single-step concerted reaction: Nu⁻ + R–CH₂–X → [Nu···CH₂R···X]⁻‡ → Nu–CH₂–R + X⁻ The nucleophile attacks from the back side (180° to the leaving group). Bond formation and bond breaking occur simultaneously through a pentacoordinate transition state. Stereochemical consequence: Complete inversion of configuration (Walden inversion). If the starting material is (R)-2-bromobutane, the product is (S)-2-butanol with nucleophile substitution. This is a hallmark of SN2. Favouring conditions: 1° substrate (minimal steric hindrance), polar aprotic solvent (enhances nucleophilicity), strong nucleophile. Key differences summary: - SN1: Two steps, carbocation intermediate, racemisation, first-order kinetics, favoured by 3° halides - SN2: One step, no intermediate, inversion, second-order kinetics, favoured by 1° halides - SN1 rearrangements are possible (hydride or alkyl shift); SN2 has no rearrangements

Crystal Field Theory (CFT) Basics: CFT treats ligands as point charges. The approach of ligands towards the central metal ion causes repulsion with the d-electrons, splitting the degenerate d-orbitals into two sets. Octahedral Splitting: In an octahedral field (six ligands along +x, -x, +y, -y, +z, -z): - t₂g orbitals (lower energy): dxy, dyz, dxz — directed between the axes, experience less repulsion - eg orbitals (higher energy): dx²-y², dz² — directed along the axes, experience more repulsion - Crystal field splitting energy: Δ₀ (or 10 Dq) Energy relations: - Energy of t₂g = −0.4 Δ₀ (stabilised) - Energy of eg = +0.6 Δ₀ (destabilised) - Barycentre rule: 4(−0.4) + 2(+0.6) = 0 (average energy preserved) Spectrochemical Series (increasing Δ₀): I⁻ < Br⁻ < SCN⁻ < Cl⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < NH₃ < en < NO₂⁻ < CN⁻ < CO Weak field ligands (left side): Small Δ₀, favour high-spin complexes Strong field ligands (right side): Large Δ₀, favour low-spin complexes High-Spin vs Low-Spin: - If Δ₀ < P (pairing energy): electrons occupy eg before pairing in t₂g → high-spin (more unpaired electrons, paramagnetic) - If Δ₀ > P (pairing energy): electrons pair in t₂g before occupying eg → low-spin (fewer unpaired electrons, may be diamagnetic) - Only relevant for d⁴–d⁷ configurations; d⁰–d³ and d⁸–d¹⁰ have only one possible arrangement CFSE Calculations: [Fe(H₂O)₆]²⁺ (Fe²⁺ = d⁶, H₂O = weak field ligand): Configuration: t₂g⁴ eg² (high spin) CFSE = (−0.4 × 4 + 0.6 × 2)Δ₀ = (−1.6 + 1.2)Δ₀ = −0.4 Δ₀ Unpaired electrons: 4, paramagnetic [Fe(CN)₆]⁴⁻ (Fe²⁺ = d⁶, CN⁻ = strong field ligand): Configuration: t₂g⁶ eg⁰ (low spin) CFSE = (−0.4 × 6 + 0.6 × 0)Δ₀ = −2.4 Δ₀ + P (pairing energy for 2 pairs) Net CFSE = −2.4 Δ₀ (pairing energy cost is more than offset by large Δ₀) Unpaired electrons: 0, diamagnetic

Preparation of Phenol: 1. Cumene process (industrial method): Propylene + benzene → cumene (isopropylbenzene) in presence of H₃PO₄ Cumene + O₂ → cumene hydroperoxide (auto-oxidation) Cumene hydroperoxide → Phenol + Acetone (acid-catalysed rearrangement) This is the most economical method and co-produces acetone. 2. From chlorobenzene (Dow's process): Chlorobenzene + NaOH (623 K, 300 atm) → Sodium phenoxide → Phenol (with HCl) NaOH attacks via nucleophilic substitution; high temperature and pressure needed due to resonance stabilisation of C–Cl bond. 3. From diazonium salt: Benzenediazonium chloride + H₂O (warm) → Phenol + N₂ + HCl This is a useful laboratory method. Acidic nature of phenol: Phenol is more acidic than alcohol because the phenoxide ion is stabilised by resonance (delocalisation of negative charge into the ring). However, it is less acidic than carboxylic acids. pKa: Phenol (~10) > Ethanol (~16) > Water (15.7) > Acetic acid (4.76) Electron-withdrawing groups (NO₂) increase acidity of phenol; electron-donating groups (CH₃) decrease it. Kolbe's reaction (Kolbe-Schmitt reaction): Phenol + NaOH → Sodium phenoxide Sodium phenoxide + CO₂ (100–125°C, 100 atm) → Sodium salicylate Sodium salicylate + H⁺ → Salicylic acid (ortho-hydroxybenzoic acid) The electrophilic substitution occurs at the ortho position because NaO⁻ is a strong activating group. Reimer-Tiemann reaction: Phenol + CHCl₃ + NaOH (heat) → Salicylaldehyde (ortho-hydroxybenzaldehyde) Mechanism: NaOH abstracts H from CHCl₃, generating :CCl₂ (dichlorocarbene) which acts as the electrophile. The carbenoid carbon bonds to the ortho position of the phenoxide ring. Hydrolysis gives the aldehyde. With excess CHCl₃: 2-hydroxy-4-chlorobenzaldehyde (both ortho positions substituted). Bromination of phenol: - With bromine water at room temperature: Phenol → 2,4,6-tribromophenol (white precipitate) + 3HBr The —OH group is strongly activating; bromine water provides Br₂ without a catalyst, and all three activated positions (2, 4, 6) are substituted. - With bromine in CS₂ or CHCl₃ at low temperature: Monobromination occurs, giving a mixture of ortho and para bromophenol.

Cannizzaro Reaction: It is a disproportionation (self-oxidation-reduction) reaction of non-enolisable aldehydes (those without α-hydrogen) in the presence of concentrated alkali. General reaction: 2RCHO + Conc. NaOH → RCH₂OH + RCOONa Mechanism (with formaldehyde): Step 1 — Nucleophilic addition of OH⁻: HCHO + OH⁻ → HCH(OH)O⁻ (hydroxymethoxide anion) Step 2 — Hydride transfer: The hydroxymethoxide anion (which has a negatively charged O and an H on the same carbon) acts as a hydride donor. A second molecule of formaldehyde acts as the hydride acceptor: HCH(OH)O⁻ + HCHO → HCOO⁻ + CH₃OH (hydride donor) (hydride acceptor) (formate) (methanol) The C–H bond in HCH(OH)O⁻ is weakened by the adjacent negative charge, making it a good hydride donor. The hydride (H⁻) transfers from the donor to the carbonyl carbon of the acceptor. Cross-Cannizzaro Reaction: When two different non-enolisable aldehydes are mixed with concentrated NaOH, the Cannizzaro reaction occurs between them. Typically, formaldehyde (the most electrophilic) acts as the hydride acceptor (gets reduced to methanol), while the other aldehyde acts as the hydride donor (gets oxidised to carboxylic acid salt). HCHO + C₆H₅CHO + Conc. NaOH → CH₃OH + C₆H₅COONa (formaldehyde is reduced) (benzaldehyde is oxidised) This is industrially useful because it ensures one aldehyde is selectively oxidised and the other reduced, rather than getting a statistical mixture of four products. Why acetaldehyde (CH₃CHO) does NOT give Cannizzaro reaction: Acetaldehyde has α-hydrogen atoms (the 3 H's on the —CH₃ group). In the presence of dilute NaOH, these α-H atoms are abstracted, leading to the aldol condensation pathway instead. Only aldehydes without α-H (formaldehyde, trimethylacetaldehyde, benzaldehyde) undergo Cannizzaro reaction. Acetaldehyde undergoes aldol condensation: 2CH₃CHO + Dil. NaOH → CH₃CH(OH)CH₂CHO (3-hydroxybutanal, aldol product)

Gabriel Phthalimide Synthesis: Steps: 1. Phthalimide + KOH → Potassium phthalimide 2. Potassium phthalimide + R–X → N-Alkyl phthalimide + KX 3. N-Alkyl phthalimide + NaOH (hydrolysis) → 1° amine + Phthalic acid salt (or) N-Alkyl phthalimide + H₂NNH₂ (hydrazinolysis) → 1° amine + Phthalhydrazide Scope: Produces pure 1° amines. No 2° or 3° amines are formed as by-products. Limitations: - Only 1° amines can be prepared (phthalimide can be alkylated only once). - Best with 1° alkyl halides; aryl halides do not undergo nucleophilic substitution with phthalimide (due to resonance stabilisation of C–X bond). - 2° and 3° alkyl halides give elimination products. Hoffmann Bromamide Degradation: Steps: R–CONH₂ + Br₂ + 4NaOH → R–NH₂ + Na₂CO₃ + 2NaBr + 2H₂O Mechanism: 1. R–CONH₂ + Br₂ → R–CONHBr (N-bromoamide) 2. R–CONHBr + NaOH → R–CONBr⁻Na⁺ (bromamide anion) 3. Rearrangement: Alkyl group migrates from C=O to N with loss of CO₂ (1,2-shift) 4. R–N=C=O (isocyanate) + H₂O → R–NH₂ + CO₂ Scope: Produces pure 1° amines with one fewer carbon atom than the starting amide. Useful for stepping down the carbon chain. Limitations: - Only 1° amines are obtained. - One carbon is lost (as CO₂), so the product has one less carbon. - Not suitable for preparing methylamine (no amide with zero carbons to degrade from, though formamide gives methylamine). - Aryl amines (anilines) can be prepared from benzamide. Comparison: | Feature | Gabriel Synthesis | Hoffmann Degradation | |---|---|---| | Starting material | Alkyl halide + phthalimide | Amide + Br₂/NaOH | | Product | 1° amine (same carbon chain) | 1° amine (one less carbon) | | Aryl amines | Cannot prepare | Can prepare (from aryl amide) | | By-products | Phthalic acid | Na₂CO₃, NaBr | | Purity | Pure 1° amine (no 2°/3°) | Pure 1° amine |

Open Chain Structure of Glucose (D-Glucose): Glucose has the molecular formula C₆H₁₂O₆. Based on evidence (reduction to sorbitol, oxidation to gluconic acid/saccharic acid, acetylation to penta-acetate), the open chain structure was proposed: CHO–CHOH–CHOH–CHOH–CHOH–CH₂OH (aldehyde group at C1, hydroxyl groups at C2–C5, primary alcohol at C6) D-Glucose has the same configuration as D-glyceraldehyde at the lowest chiral centre (C5). Limitations of the Open Chain Structure: 1. Glucose does NOT restore the full Schiff's test (weak aldehyde reaction) — expected of a free aldehyde group. 2. Glucose does NOT form a bisulfite addition product with NaHSO₃. 3. Glucose penta-acetate does NOT react with NH₂OH — no free aldehyde group available. 4. Two crystalline forms of glucose exist: α-D-glucose (mp 419 K, specific rotation +112°) and β-D-glucose (mp 423 K, specific rotation +19°). The open chain structure cannot explain two forms. 5. Both α and β forms in solution equilibrate to a constant specific rotation of +52.7° (mutarotation). Cyclic Structure (Haworth Projection): Glucose cyclises through nucleophilic attack of the C5 —OH on the C1 carbonyl, forming a six-membered pyranose ring (hemiacetal). The new asymmetric centre at C1 is called the anomeric carbon. - α-D-Glucopyranose: —OH at C1 is BELOW the ring plane (trans to CH₂OH at C5) - β-D-Glucopyranose: —OH at C1 is ABOVE the ring plane (cis to CH₂OH at C5) - In both forms, all large substituents are equatorial in the most stable chair conformation. Mutarotation: When crystalline α-D-glucose (specific rotation +112°) or β-D-glucose (specific rotation +19°) is dissolved in water, the specific rotation gradually changes and stabilises at +52.7°. This phenomenon is called mutarotation. Mechanism: The cyclic forms interconvert through the open chain form. At equilibrium, approximately 36% is α-D-glucose, 64% is β-D-glucose, and less than 0.01% is the open chain form. The β form predominates because all large groups are equatorial, making it more stable. The open chain form exists in trace amounts, which explains why glucose shows typical aldehyde reactions (Tollen's test, Fehling's test) but with less intensity than expected for a free aldehyde.