Unit I Inorganic Chemistry · 12 Marks

General Principles of Metallurgy

Extraction, purification, and thermodynamic principles of metal extraction using Ellingham diagrams, electrode potentials, and key industrial processes.

1. Ellingham Diagram

An Ellingham diagram is a graphical representation of the standard Gibbs free energy change (ΔG°) versus temperature (T) for the oxidation reactions of metals. It helps predict the thermodynamic feasibility of reducing a metal oxide using a particular reducing agent.

Fig 1. Ellingham Diagram — ΔG° vs Temperature for oxide formation. CO line has negative slope (key feature).

Key Features of Ellingham Diagram

Feature	Reason	Implication
Positive slope (most metal oxides)	O₂ consumed → ΔS < 0, so −TΔS term becomes more positive with T	ΔG° increases with T → less stable at high T
Negative slope of CO line	2C + O₂ → 2CO produces 2 moles gas from 1 → ΔS > 0	CO becomes better reducing agent at high T
Nearly zero slope of CO₂	C + O₂ → CO₂: equal moles of gas, ΔS ≈ 0	CO₂ effectiveness barely changes with T
Crossover point	Where ΔG° of two reactions are equal	Above crossover T, lower substance reduces upper one
Al₂O₃ line very low	Al–O bond very strong; Al is highly reactive	Al can reduce most metal oxides (thermite)

📌

Exam Key Point: The CO line has a NEGATIVE slope because converting 1 mol O₂ → 2 mol CO increases entropy (ΔS > 0). This is the most frequently asked concept from Unit I.

2. Purification of Metals

After extraction, crude metals contain impurities and must be refined. Different methods are used based on the physical/chemical properties of the metal and its impurities.

Method	Metal	Principle	Key Equation
Mond Process	Nickel (Ni)	Formation & decomposition of volatile Ni(CO)₄	Ni + 4CO ⇌ Ni(CO)₄
Van Arkel–de Boer	Ti, Zr, Hf, B	Volatile iodide formed & thermally decomposed	Ti + 2I₂ ⇌ TiI₄
Kroll Process	Titanium (Ti)	TiCl₄ reduced by Mg metal	TiCl₄ + 2Mg → Ti + 2MgCl₂
Zone Refining	Si, Ge, Ga, In	Impurities more soluble in molten zone	Impurities swept to one end
Electrolytic Refining	Cu, Al, Zn, Ag, Au	Electrodeposition of pure metal at cathode	Crude = anode; pure deposits at cathode
Liquation	Sn, Bi, Pb	Low-MP metal flows away from impurities on inclined hearth	Physical separation

Mond Process — Detailed Mechanism

1

Carbonyl Formation (50°C / 323 K): Impure nickel is heated with CO gas at low temperature. Nickel uniquely reacts to form volatile nickel tetracarbonyl.

Ni(impure) + 4CO ──(323 K)──→ Ni(CO)₄ [volatile, bp = 43°C]

2

Transfer: The volatile Ni(CO)₄ gas is carried away, leaving solid impurities (Cu, Co, Fe) behind — they don't form stable volatile carbonyls at this temperature.

3

Decomposition (230°C / 503 K): Ni(CO)₄ is heated to higher temperature → pure Ni deposited; CO recycled.

Ni(CO)₄ ──(503 K)──→ Ni(pure) + 4CO↑ [CO recycled]

💡 Memory Trick

"Mond = Ni se CO baat karta hai dono temperatures pe" — Mond Process uses CO and Ni, one temperature to form Ni(CO)₄, another to decompose it. Think of it as Ni "going on a CO trip and coming back pure."

Van Arkel–de Boer Process (Ti purification)

1

Crude titanium is heated with iodine at moderate temperature to form volatile TiI₄

Ti(crude) + 2I₂ ──(~500°C)──→ TiI₄ [volatile red solid/gas]

2

TiI₄ vapour passed over a hot tungsten filament (≈1400°C) → decomposes → pure Ti deposits on filament; I₂ recycled

TiI₄ ──(1400°C, W filament)──→ Ti(pure) + 2I₂ [I₂ recycled]

⚠️

PYQ Direct Question: "What is Kroll process and for which metal?" — Answer: TiCl₄ + 2Mg → Ti + 2MgCl₂. Used for extraction of Titanium. Mg is the reducing agent.

Unit II Inorganic Chemistry · 12 Marks

s- and p-Block Elements

Periodicity, effective nuclear charge, ionization energy, anomalous behaviour, and special features of s- and p-block elements.

1. Effective Nuclear Charge (Z_eff)

The effective nuclear charge is the net positive charge experienced by a valence electron after accounting for the shielding effect of inner electrons.

Z_eff = Z − σ where Z = atomic number, σ = shielding constant (Slater's rules)

Slater's Rules for Shielding (σ)

Electron group shielding the valence e⁻	Shielding contributed (σ)
Electrons in the same group (same n, same l)	0.35 each (0.30 for 1s)
Electrons in (n−1) shell	0.85 each
Electrons in (n−2) or inner shells	1.00 each

✅

PYQ Answer: Increasing order of Z_eff — Na < Mg < Al < Si (Z_eff increases across a period as Z increases but shielding changes less)

2. Ionization Energy Trends

Ionization energy (IE) is the minimum energy required to remove the outermost electron from a gaseous atom in its ground state.

Fig 2. IE₁ values for Period 2. Orange bars = anomalous dips (Be>B, N>O)

IE Anomalies — Most Important for Exams

Be > B (Anomaly)

Be: 2s² — completely filled, extra stable
B: 2p¹ — p-orbital electron, easier to remove
IE: Be (899) > B (800) kJ/mol

N > O (Anomaly)

N: 2p³ — half-filled, extra exchange energy
O: 2p⁴ — one paired e⁻, inter-electronic repulsion ↑
IE: N (1402) > O (1314) kJ/mol

Factors Affecting IE

Nuclear charge (Z) ↑ → IE ↑
Atomic radius ↑ → IE ↓
Shielding ↑ → IE ↓
Penetration: s>p>d>f
Half-filled/full subshell → IE anomalously high

3. Special Concepts

Concept	Definition	Example	Exam Relevance
Inert Pair Effect	ns² electrons in heavier p-block reluctant to participate in bonding due to poor shielding by d/f electrons	Tl(I) stable; Pb(II) stable; Bi(III) stable	Why Pb²⁺ more stable than Pb⁴⁺
Diagonal Relationship	1st period element resembles 2nd period element diagonally placed (similar charge/radius ratio)	Li↔Mg, Be↔Al, B↔Si	Be and Al: amphoteric, form covalent halides
Anomalous Behaviour	1st member of group differs due to: small size, high charge density, no d-orbitals available	Li, Be, B, C, N, O, F vs. their heavier congeners	Li floats on oil (density), F most electronegative
Allotropy	Existence of same element in two or more physical forms	C: diamond/graphite/C₆₀; S: rhombic/monoclinic; P: white/red/black	Compare structure and properties

Unit II Hydrides — High Exam Weightage

Hydrides — Classification & Properties

Type	Nature of Bonding	Examples	Properties
Ionic (Saline)	H⁻ ion present; ionic lattice	NaH, KH, CaH₂, MgH₂	High MP/BP, conduct electricity in melt, react vigorously with water releasing H₂: MgH₂ + 2H₂O → Mg(OH)₂ + 2H₂
Covalent (Molecular)	Covalent E–H bonds; molecular	HF, HCl, H₂O, NH₃, CH₄, GeH₄	Volatile, low MP/BP, non-conductors; sub-types: electron precise (CH₄), electron rich (H₂O, NH₃), electron deficient (B₂H₆)
Metallic (Interstitial)	H atoms occupy lattice gaps in metal	TiH₂, PdHₓ (x≤1), VH₂	Non-stoichiometric, metallic lustre, good conductors; Pd absorbs H₂ reversibly (H₂ storage)

✅

PYQ MCQ Answer: MgH₂ = Ionic | GeH₄ = Electron precise (covalent) | B₂H₆ = Electron deficient | HF = Electron rich (extra lone pairs)

Diborane (B₂H₆) — Multicenter Bonding

Diborane is the simplest boron hydride. It cannot be described by simple 2-centre 2-electron (2c-2e) bonds because there are only 12 electrons available for 8 bonds (which would need 16 electrons). It uses 3-centre 2-electron (3c-2e) banana bonds.

Fig 3. Structure of Diborane — 4 terminal B–H (2c-2e) bonds and 2 bridging B–H–B (3c-2e) banana bonds

Feature	Detail
Total electrons available	12 (2×3 from B + 6×1 from H)
Terminal B–H bonds	4 bonds × 2e = 8 electrons used
Bridge B–H–B bonds	2 bonds × 2e = 4 electrons (each bond shared over 3 atoms)
Hybridization of B	sp³ (approximately)
Terminal H–B–H angle	121.5°
Bridge B–H–B angle	83.5°
Classification	Electron deficient (fewer e⁻ than needed for 2c-2e bonds)

Oxoacids of Sulphur

Acid	Formula	Structure Features	O–O linkage?	S oxidation state
Sulphurous acid	H₂SO₃	S with 2 OH, 1 =O	No	+4
Sulphuric acid	H₂SO₄	S with 2 OH, 2 =O	No	+6
Peroxodisulphuric acid	H₂S₂O₈	HO₃S–O–O–SO₃H; peroxy bridge	YES ✓	+7
Pyrosulphuric acid (Oleum)	H₂S₂O₇	HO₃S–O–SO₃H; S–O–S bridge	No (S–O–S)	+6
Thiosulphuric acid	H₂S₂O₃	Like sulfate with one S replacing =O	No	+2, 0

✅

Direct PYQ Answer: H₂S₂O₈ (peroxodisulphuric acid / Marshall's acid) has the –O–O– (peroxy) linkage.

Unit III Organic Chemistry · 12 Marks

Carboxylic Acids & Their Derivatives

Reactions, mechanisms, and preparation of carboxylic acids, acid chlorides, esters, anhydrides, and amides — with focus on name reactions.

1. Hell–Volhard–Zelinsky (HVZ) Reaction

The HVZ reaction introduces a bromine atom at the α-carbon of a carboxylic acid using Br₂ and PCl₃ (or P + Br₂) as reagents. Only acids with α-hydrogen atoms undergo this reaction.

Reaction: CH₃COOH + Br₂ ──(PCl₃ or P, Δ)──→ BrCH₂COOH + HBr (Acetic acid) (Bromoacetic acid) Mechanism (4 steps): Step 1: PCl₃ converts acid → acid chloride (in situ) CH₃COOH + PCl₃ → CH₃COCl + H₃PO₃ Step 2: Acid chloride forms enol tautomer CH₃COCl ⇌ CH₂=C(OH)Cl (enol form) Step 3: Br₂ attacks the nucleophilic α-carbon of enol CH₂=C(OH)Cl + Br₂ → BrCH₂–C(OH)Cl–Br → BrCH₂COCl + HBr Step 4: Hydrolysis of α-bromo acid chloride BrCH₂COCl + H₂O → BrCH₂COOH + HCl

⚠️

Important: Formic acid (HCOOH) has no α-hydrogen → does NOT undergo HVZ reaction. This is a common MCQ trap!

2. Reformatsky Reaction

An α-bromo ester reacts with an aldehyde or ketone in the presence of zinc metal in dry ether to give a β-hydroxy ester (after hydrolysis).

Reaction: R–CHO + BrCH₂COOC₂H₅ ──(Zn, dry ether)──→ β-hydroxy ester Step 1: Zn inserts into C–Br bond (oxidative addition) Zn + BrCH₂COOC₂H₅ → BrZnCH₂COOC₂H₅ [Reformatsky reagent = zinc enolate] Step 2: Nucleophilic addition to carbonyl BrZnCH₂COOC₂H₅ + RCHO → R–CH(OZnBr)–CH₂COOC₂H₅ Step 3: Aqueous acid workup (hydrolysis) ──(H₃O⁺)──→ R–CH(OH)–CH₂COOC₂H₅ (β-hydroxy ester)

3. Perkin Condensation

An aromatic aldehyde undergoes condensation with an acid anhydride in the presence of the potassium/sodium salt of the corresponding acid (as base) to form an α,β-unsaturated aromatic acid.

Reaction (classic example): C₆H₅CHO + (CH₃CO)₂O ──(CH₃COONa, Δ)──→ C₆H₅–CH=CH–COOH + CH₃COOH (Benzaldehyde) (Acetic anhydride) (Cinnamic acid) Mechanism: Step 1: Base (CH₃COO⁻) deprotonates acetic anhydride at α-position → forms carbanion (enolate of acetic anhydride) Step 2: Nucleophilic addition of enolate to benzaldehyde (C=O) Step 3: Aldol-type condensation Step 4: Elimination (loss of –OH) → α,β-unsaturated intermediate Step 5: Hydrolysis of anhydride portion → cinnamic acid

4. Carboxylic Acid Derivatives — Preparation & Reactivity

Fig 4. Decreasing reactivity order of acyl derivatives. Cl⁻ is best leaving group; NH₂⁻ is worst.

Derivative	Preparation	Leaving Group	Reactivity Rank
Acid Chloride (RCOCl)	RCOOH + SOCl₂ → RCOCl + SO₂ + HCl (best method, gaseous byproducts)	Cl⁻ (weakest base = best leaving group)	1st (most reactive)
Acid Anhydride (RCO)₂O	RCOOH + P₂O₅ or 2RCOOK → heat	RCOO⁻	2nd
Ester (RCOOR')	Fischer esterification: RCOOH + R'OH ⇌ RCOOR' + H₂O [H⁺ cat.]	R'O⁻	3rd
Amide (RCONH₂)	RCOCl + 2NH₃ → RCONH₂ + NH₄Cl	NH₂⁻ (strongest base = worst leaving group)	4th (least reactive)

Ester Hydrolysis — Acidic vs Alkaline

Acidic Hydrolysis (Reversible)

Catalyst: H⁺ (H₂SO₄ or HCl)
RCOOR' + H₂O ⇌ RCOOH + R'OH
Mechanism: Protonation of C=O → nucleophilic attack of H₂O → tetrahedral intermediate → elimination of R'OH
Reversible — equilibrium exists

Alkaline Hydrolysis (Saponification — Irreversible)

Catalyst: NaOH (strong base)
RCOOR' + NaOH → RCOONa + R'OH
Mechanism: OH⁻ attacks C=O directly → tetrahedral int. → R'O⁻ leaves → RCOOH → deprotonated to RCOO⁻
Irreversible (product RCOO⁻ is stable, not electrophilic)

Unit IV Highest Weightage — Every PYQ Paper

Amines — Aliphatic & Aromatic

Classification, preparation methods, reactions, and identification tests for aliphatic and aromatic amines — the highest-scoring unit in PYQs.

1. Classification & Basicity

Class	General Formula	Example	pKb	Basicity (aqueous)
Primary (1°)	RNH₂	CH₃NH₂ (methylamine)	3.36	High
Secondary (2°)	R₂NH	(CH₃)₂NH (dimethylamine)	3.23	Highest (in aq.)
Tertiary (3°)	R₃N	(CH₃)₃N (trimethylamine)	4.19	Lower than 2° (solvation effect)
Aromatic	ArNH₂	C₆H₅NH₂ (aniline)	9.13	Lowest (delocalization)

📌

Basicity order (aqueous): (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃ > C₆H₅NH₂
Aromatic amine is weakest because the lone pair on N is delocalized into the benzene ring by resonance → less available for donation to H⁺.

2. Gabriel's Phthalimide Synthesis

This is the most important method for preparing pure primary amines without contamination from secondary or tertiary amines.

Purpose: Synthesis of pure primary aliphatic amines (R–NH₂) Step 1: Phthalimide is treated with KOH → Potassium phthalimide (K-salt) Phthalimide (NH, acidic) + KOH → Potassium phthalimide + H₂O [The NH proton of phthalimide is acidic due to adjacent C=O groups] Step 2: N-alkylation with alkyl halide (SN2 reaction) K-phthalimide + R–X ──(SN2)──→ N-alkylphthalimide + KX Step 3: Hydrolysis (Ing-Manske modification: use hydrazine) N-alkylphthalimide + H₂NNH₂ → R–NH₂ + phthalhydrazide OR (acidic/basic hydrolysis): N-alkylphthalimide + H₂O/H⁺ → R–NH₂ + phthalic acid

⚠️

Limitation: Cannot prepare aromatic amines (ArNH₂) because ArX doesn't undergo SN2 at sp² C. Cannot prepare secondary or tertiary amines.
PYQ Answer: n-Butylamine can be made by Gabriel synthesis. Triethylamine (3°), neo-pentylamine (hindered SN2), aniline (aromatic) — CANNOT.

3. Hofmann Bromamide (Degradation) Reaction

Converts a primary amide (RCONH₂) to a primary amine (RNH₂) with one less carbon. This is a degradation reaction involving a rearrangement.

1

N-bromination: Br₂ in NaOH brominates the amide nitrogen

RCONH₂ + Br₂ + NaOH → RCONHBr + NaBr + H₂O [N-bromoamide]

2

Deprotonation: NaOH removes the remaining NH proton

RCONHBr + NaOH → RCONBr⁻ + NaBr + H₂O [anion formed]

3

Rearrangement (KEY STEP): Br⁻ leaves as nitrene forms; R migrates from C to N (1,2-shift)

RCONBr⁻ → [R–N=C=O] + Br⁻ [isocyanate formed by 1,2-migration of R]

4

Hydrolysis: Isocyanate hydrolyzed → carbamic acid → decarboxylation → primary amine

R–N=C=O + H₂O → R–NH–COOH → R–NH₂ + CO₂↑

🎯 Net result: RCONH₂ (n carbons) → RNH₂ (n−1 carbons) — one carbon is lost as CO₂. This is the hallmark of Hofmann Bromamide reaction.

4. Amine Identification Tests

Test	Reagent	Primary Amine (1°)	Secondary Amine (2°)	Tertiary Amine (3°)
Hinsberg Test	C₆H₅SO₂Cl + KOH (aq.)	Sulfonamide formed → soluble in KOH (−NH remains acidic)	Sulfonamide formed → insoluble in KOH (no −NH)	No reaction (no N–H to react)
Carbylamine Test	CHCl₃ + KOH (alcoholic)	Foul-smelling isocyanide (R–NC) formed ✓	No reaction	No reaction
Nitrous Acid Test	NaNO₂ + HCl (0–5°C)	Unstable diazonium salt → N₂ gas evolved + yellow oily alcohol	N-nitroso compound (yellow oily)	Nitrosamine (yellow oil, on N-nitroso)

Carbylamine Test — Mechanism

R–NH₂ + CHCl₃ + 3KOH → R–N≡C: (isocyanide) + 3KCl + 3H₂O Mechanism: Step 1: CHCl₃ + KOH → :CCl₂ (dichlorocarbene, highly reactive) Step 2: :CCl₂ + R–NH₂ → R–NH–CCl₂ (addition to carbene) Step 3: Base removes 2 HCl sequentially: R–NH–CCl₂ → R–N=CCl₂ → R–N≡C: (isocyanide, foul smell) Positive test = foul smell = confirms primary amine

5. Hofmann vs Saytzeff Elimination

Property	Hofmann Elimination	Saytzeff (Zaitsev) Elimination
Substrate	Quaternary ammonium hydroxide [R₄N⁺OH⁻]	Alkyl halide (RX)
Base	Ag₂O/H₂O or moist Ag₂O (bulky base)	KOH/alcohol or NaOEt (small base)
Major product	Less substituted alkene (anti-Markovnikov)	More substituted alkene (Markovnikov)
Reason	Bulky leaving group (NR₃) makes bulky base remove least hindered H	Small base removes β-H from most substituted carbon
Stability	Less stable alkene (thermodynamic paradox)	More stable alkene formed
Mechanism	E2 (anti-periplanar)	E2 (anti-periplanar)

6. Schotten–Baumann Reaction

Acylation of an amine using acid chloride in the presence of aqueous NaOH (Schotten-Baumann conditions). NaOH neutralizes HCl formed, preventing protonation of amine.

C₆H₅–NH₂ + CH₃COCl ──(NaOH, H₂O, 0°C)──→ C₆H₅–NHCOCH₃ + NaCl (Aniline) (Acetyl chloride) (Acetanilide) Why NaOH: HCl produced would protonate amine → R–NH₃⁺ → amine no longer nucleophilic NaOH scavenges HCl: HCl + NaOH → NaCl + H₂O → amine remains free for reaction

Unit IV Diazonium Salts

Diazonium Salts

1. Preparation (Diazotization)

Ar–NH₂ + NaNO₂ + HCl ──(0–5°C)──→ Ar–N₂⁺Cl⁻ + NaCl + H₂O (Primary arylamine) (Diazonium chloride) ⚠️ Must maintain 0–5°C (ice bath) — at room temperature diazonium salt decomposes to phenol! Mechanism: Step 1: NaNO₂ + HCl → HNO₂ + NaCl (nitrous acid formed in situ) Step 2: HNO₂ + H⁺ → H₂O + +NO (nitrosonium ion) Step 3: ArNH₂ + +NO → Ar–NH–N=O (N-nitrosoamine) Step 4: Protonation + loss of H₂O → Ar–N₂⁺ (diazonium ion)

2. Reactions of Diazonium Salts

Reaction	Reagent/Condition	Product	Notes
Sandmeyer (→ Cl)	CuCl + HCl, Δ	ArCl	Cu(I) salt catalyst
Sandmeyer (→ Br)	CuBr + HBr, Δ	ArBr	Cu(I) salt catalyst
Sandmeyer (→ CN)	CuCN + KCN	ArCN	Extension of chain by 1C
Gattermann (→ Cl/Br)	Cu powder + HCl or HBr	ArCl / ArBr	Cu metal (not salt) — less pure product
Balz-Schiemann (→ F)	HBF₄ → dry heat	ArF	Fluorobenzene — only good route to ArF
→ Phenol	H₂O, warm (H⁺ cat.)	ArOH + N₂	Hydrolysis
→ Arene (deamination)	H₃PO₂ (hypophosphorous acid)	Ar–H + N₂	Reductive removal of –N₂⁺
Coupling → Azo dyes	β-naphthol (NaOH, 0°C) or N,N-dimethylaniline	Ar–N=N–Ar' (azo compound)	Electrophilic aromatic substitution; forms colored dyes

🔍

Sandmeyer vs Gattermann:
Sandmeyer = Cu(I) salts (CuCl, CuBr, CuCN) — better yield
Gattermann = Cu metal/powder (Cu + HCl or HBr) — cheaper but lower yield
Both give haloarenes from diazonium salts.

Electrophilic Substitution on Aniline

Reaction	Conditions	Product	Note
Bromination	Br₂(aq), no catalyst	2,4,6-tribromoaniline ↓ (white ppt)	–NH₂ strongly activates ring; no AlBr₃ needed
Nitration	HNO₃/H₂SO₄ (dilute, cold)	o- and p-nitroaniline (major)	In conc. acid: NH₃⁺ formed → meta director!
Sulphonation	H₂SO₄ (200°C)	Sulphanilic acid (p-NH₂C₆H₄SO₃H)	Used in dye industry; zwitter ion form
Acylation	CH₃COCl + NaOH (Schotten-Baumann)	Acetanilide	Protects –NH₂ during synthesis

Unit V Amino Acids — 12 Marks

Amino Acids

Structure, zwitter ion, isoelectric point, electrophoresis, synthesis (Strecker), and chemical reactions of amino acids.

1. Zwitter Ion & Isoelectric Point

An α-amino acid carries both an acidic –COOH and a basic –NH₂ group. In solution, an internal proton transfer occurs to give a dipolar ion (zwitter ion).

Fig 5. Amino acid forms at different pH values. Electrophoresis migration depends on net charge.

2. Strecker Synthesis

1

Condensation: Aldehyde reacts with ammonia → aldimine (Schiff's base)

RCHO + NH₃ → RCH=NH + H₂O [aldimine]

2

Addition of HCN: HCN adds to the C=N bond (nucleophilic addition)

RCH=NH + HCN → RCH(NH₂)–CN [α-aminonitrile]

3

Hydrolysis: Nitrile hydrolyzed to carboxylic acid → α-amino acid

RCH(NH₂)–CN + 2H₂O/H⁺ → RCH(NH₂)–COOH + NH₃ [α-amino acid]

📝

Example: CH₃CHO (acetaldehyde) → CH₃CH(NH₂)CN → CH₃CH(NH₂)COOH = Alanine (L-alanine)

3. Chemical Reactions of Amino Acids

Reaction	Group Reacting	Reagent	Product
Esterification	–COOH	R'OH + HCl (dry)	H₂N–CHR–COOR' + HCl (as H₃N⁺–CHR–COOR')
Acetylation	–NH₂	(CH₃CO)₂O or CH₃COCl	CH₃CO–NH–CHR–COOH (N-acetyl amino acid)
Ninhydrin test	both –NH₂ + –COOH	Ninhydrin (triketone)	Blue-violet color (Ruhemann's purple) — all α-amino acids
Cu²⁺ complexation	–NH₂ + –COO⁻	CuSO₄ solution	Deep blue chelate complex (bidentate ligand)

Unit V Carbohydrates

Carbohydrates

1. Classification

Class	Units	Examples	Reducing?	Hydrolysis
Monosaccharides	1 sugar unit	Glucose, Fructose, Galactose, Ribose	Yes (all)	Cannot be hydrolyzed further
Disaccharides	2 sugar units	Sucrose, Maltose, Lactose	Maltose ✓, Lactose ✓; Sucrose ✗	Gives 2 monosaccharides
Polysaccharides	Many units	Starch, Cellulose, Glycogen	No	Gives many monosaccharides

✅

PYQ Answer: Galactose is NOT a disaccharide — it is a monosaccharide (aldohexose, epimer of glucose at C4). Sucrose, Maltose, Lactose are all disaccharides.

2. Glucose — Open Chain & Cyclic Structure

Fig 6. D-Glucose — Fischer projection (open chain) and Haworth pyranose ring form

3. Mutarotation

The change in specific optical rotation observed when a pure anomer of glucose is dissolved in water and allowed to reach equilibrium is called mutarotation. This occurs due to interconversion of α and β anomers through the open-chain form.

α-D-Glucose ⇌ Open-chain form ⇌ β-D-Glucose [α]D = +112° (pure) [α]D = +18.7° (pure) At equilibrium (room temperature): 36% α-form + 64% β-form [α]D(equilibrium) = +52.7° Mutarotation is catalyzed by: H⁺, OH⁻, and the enzyme mutarotase β-D-glucose is more stable (–OH at C1 equatorial in chair conformation)

4. Disaccharides Comparison

Property	Sucrose	Maltose	Lactose
Components	Glucose + Fructose	Glucose + Glucose	Galactose + Glucose
Linkage	α(1→2)β glycosidic	α(1→4) glycosidic	β(1→4) glycosidic
Reducing sugar?	No (both anomeric C blocked)	Yes (free C1)	Yes (free C1)
Mutarotation?	No	Yes	Yes
Source	Sugarcane, sugar beet	Malted grain (beer)	Milk sugar
Hydrolysis product	Glucose + Fructose (invert sugar)	2 × Glucose	Galactose + Glucose

⚠️

Why sucrose is non-reducing: In sucrose, C1 of α-D-glucose and C2 of β-D-fructose are both involved in the glycosidic bond → no free anomeric OH group → cannot open to aldehyde form → non-reducing.

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PYQ NED-7677-Z Exact Question Answers

Previous Year Question — Model Answers

University-exam style answers for each question from the NED-7677-Z paper. Write exactly like this in your exam.

Section A — MCQ Answers

Answer: Ag < Cu < Ni < B < K < Li (increasing reducing power)

Reducing power ∝ tendency to lose electrons = decreasing standard electrode potential (E°). Li has E° = −3.04 V (strongest reducing agent in water) due to very high hydration energy of Li⁺ despite high IE. Ag has E° = +0.80 V (weakest).

Answer: Na < Mg < Al < Si

Across a period, Z increases by 1 each step while shielding remains roughly constant (same shell). Z_eff = Z − σ: Na (Z_eff ≈ 2.51), Mg (3.31), Al (4.07), Si (4.29).

A. MgH₂ → (iii) Ionic
B. GeH₄ → (i) Electron precise
C. B₂H₆ → (ii) Electron deficient
D. HF → (iv) Electron rich

Electron precise = exact electrons for 2c-2e bonds (CH₄, GeH₄). Electron deficient = fewer electrons (B₂H₆, uses 3c-2e). Electron rich = extra lone pairs (H₂O, NH₃, HF).

Answer: H₂S₂O₈ — Peroxodisulphuric acid (Marshall's acid)

Structure: HO₃S–O–O–SO₃H. The peroxy (–O–O–) bridge connects two SO₄ units. S is in +7 oxidation state. It is dibasic. H₂S₂O₇ has –S–O–S– bridge (not –O–O–).

Answer: n-Butylamine (CH₃CH₂CH₂CH₂NH₂)

Gabriel synthesis only gives primary aliphatic amines via SN2 alkylation. Triethylamine is tertiary. Neo-pentylamine has extreme steric hindrance at C for SN2. Aniline is aromatic (ArX cannot undergo SN2). So only n-butylamine (1°, aliphatic, unhindered) can be synthesized.

Answer: Galactose — it is a monosaccharide

Galactose (C₆H₁₂O₆) is an aldohexose monosaccharide — epimer of glucose at C4. Maltose, Sucrose, and Lactose are all disaccharides (C₁₂H₂₂O₁₁) made of two monosaccharide units joined by glycosidic bonds.

Section B — Short Answer (5 Marks Each)

Q1. Factors influencing ionization energy of an atom 5 marks

Introduction

Ionization energy (IE) is the minimum energy required to remove the most loosely bound electron from a gaseous atom in its ground state: M(g) → M⁺(g) + e⁻

Factors Affecting IE

Nuclear Charge (Z): Higher nuclear charge pulls electrons more strongly. IE increases across a period (e.g., Na < Mg < Al < Si < P < S < Cl < Ar).
Atomic Radius: Larger radius → valence electrons farther from nucleus → weaker attraction → lower IE. IE decreases down a group.
Shielding Effect: Inner electrons shield valence electrons from nucleus. More shielding → lower effective Z → lower IE. Shielding increases down a group.
Penetration of Orbitals: Penetration order: s > p > d > f. Greater penetration → greater felt nuclear charge → higher IE. Hence IE(2s) > IE(2p) for same period.
Stability of Electronic Configuration: Half-filled (p³, d⁵) and fully filled (p⁶, d¹⁰) subshells have extra stability → anomalously high IE. Example: N (2p³) > O (2p⁴); Be (2s²) > B (2p¹).

Thus, ionization energy is a resultant of all these competing factors. The anomalies (Be>B and N>O) are regularly asked in exams and arise from subshell stability.

Q2. Geometry, basicity, reducing power of NH₃ and N₂H₄ 5 marks

Ammonia (NH₃)

Geometry: Trigonal pyramidal (sp³ hybridization; 3 bond pairs + 1 lone pair; bond angle = 107°)
Basicity: Kb = 1.8 × 10⁻⁵; moderately basic due to lone pair on N
Reducing power: Weak reducing agent; oxidized to N₂ or NO under strong conditions

Hydrazine (N₂H₄)

Geometry: Two sp³ N atoms in gauche conformation (like H₂O₂ analog); N–N single bond
Basicity: Kb1 = 1.0 × 10⁻⁶; weaker base than NH₃ (electron density on N reduced by adjacent N)
Reducing power: Strong reducing agent (used in rocket fuel); N is in −2 state: N₂H₄ → N₂ + 4H⁺ + 4e⁻ (E° = −0.23 V)

N₂H₄ is a better reducing agent than NH₃ but a weaker base. Both show lone pair basicity at N.

Q4. Two redox reactions — which metal dissolves? 5 marks

Given Reactions

Fe²⁺ + 2e⁻ → Fe, E° = −0.44 V | Zn²⁺ + 2e⁻ → Zn, E° = −0.76 V

Analysis

A metal with more negative E° has greater tendency to lose electrons (oxidized) → stronger reducing agent → will dissolve in acid
Zn: E° = −0.76 V (more negative) → Zn dissolves preferentially
Fe: E° = −0.44 V → Fe also dissolves in acid but less readily than Zn
Zn displaces Fe from FeSO₄ solution: Zn + FeSO₄ → ZnSO₄ + Fe

Zn will dissolve more readily. Any metal with negative E° will dissolve in dilute acid. A metal with positive E° (e.g., Cu = +0.34 V) will NOT dissolve in dilute acid.

Q5(a). Prepare acid chloride and amide from carboxylic acid + mechanism 5 marks

Preparation of Acid Chloride

Method (best): RCOOH + SOCl₂ → RCOCl + SO₂↑ + HCl↑

Advantage: Gaseous byproducts (SO₂ and HCl) escape easily → pure acid chloride obtained without further purification.

Preparation of Amide from Acid Chloride

RCOCl + 2NH₃ → RCONH₂ + NH₄Cl

Mechanism (Nucleophilic Acyl Substitution)

NH₃ (nucleophile) attacks electrophilic carbonyl carbon of RCOCl
Tetrahedral intermediate formed: R–C(=O)(NH₃)(Cl)⁻ → R–C(Cl)(OH)(NH₂)
Cl⁻ leaves (best leaving group) → RCONH₂ (amide) + HCl
Second NH₃ neutralizes HCl: HCl + NH₃ → NH₄Cl

Acid chlorides are the most reactive acyl derivatives, making them ideal starting materials for amide synthesis.

Q6(b). Write chemical conversion: straight chain → cyclic monosaccharides; configuration 8 marks

Conversion of Open Chain to Cyclic Form

5-carbon monosaccharides (pentoses): The –OH at C4 attacks the aldehyde C1 → 5-membered furanose ring (e.g., ribofuranose)
6-carbon monosaccharides (hexoses): The –OH at C5 attacks the aldehyde C1 → 6-membered pyranose ring (e.g., glucopyranose)

Determining Configuration of Monosaccharides

D-series: In Fischer projection, –OH at the penultimate carbon (second from bottom = C5 for hexoses, C4 for pentoses) is on the RIGHT
L-series: –OH at penultimate C is on the LEFT
Anomers: When ring closes, a new stereocenter forms at C1 (anomeric carbon)
- α-anomer: –OH at C1 is on the SAME side as the ring oxygen reference (below ring in Haworth for D-glucose)
- β-anomer: –OH at C1 is on the OPPOSITE side (above ring in Haworth for D-glucose)

The β-anomer of D-glucose is more stable because the –OH at C1 is equatorial in the chair conformation, reducing steric strain.

Section C — Long Answers (10 Marks Each)

Q1(a). Ellingham Diagram — complete answer 10 marks

Definition

An Ellingham diagram is a graphical plot of standard Gibbs free energy (ΔG°) versus temperature (T in K) for the oxidation reactions of metals. It is used to predict the thermodynamic feasibility of reduction of metal oxides.

Basis

The reactions plotted are: 2xM + O₂ → 2MₓO (normalized to 1 mole O₂)
ΔG° = ΔH° − TΔS° (at constant pressure)
These are plotted as straight lines with slope = −ΔS°

Salient Features

Positive slope (most metal oxides): The reaction consumes O₂ (gas) → ΔS < 0 → slope (−ΔS°) is positive → ΔG° increases with T
Negative slope of CO line: 2C + O₂ → 2CO produces 2 mol gas from 1 → ΔS > 0 → slope is negative → CO becomes a better reducing agent at higher temperatures
CO₂ line: nearly zero slope — equal moles gas consumed and produced (ΔS ≈ 0)
Crossover point: Below the crossover T, the upper substance reduces the lower one. Above crossover T, positions reverse
Al₂O₃ line is very low: Al–O bond is very strong; Al can reduce most metal oxides (thermite: Al + Fe₂O₃ → Al₂O₃ + Fe)

Applications

Choosing the right reducing agent and temperature for metal extraction
Predicting whether a given metal oxide can be reduced by C or CO
Understanding why some metals (Ti, Al) require special processes (not simple coke reduction)

Limitations

Only gives thermodynamic feasibility — not kinetic information (reaction rate)
Assumes pure substances; alloys/solutions not included
Phase transitions cause kinks in the lines

The Ellingham diagram is the most powerful tool for understanding and predicting metal extraction processes in pyrometallurgy.

Q2(a). Hydrides — classification with suitable examples 10 marks

Definition

Binary compounds of hydrogen with other elements are called hydrides. Based on the nature of bonding, hydrides are classified into three types:

1. Ionic (Saline) Hydrides

Formed by s-block metals (alkali and alkaline earth metals except Be, Mg)
Contain H⁻ (hydride ion) in ionic lattice
Examples: NaH, KH, CaH₂, MgH₂
Properties: High MP, ionic conductors in melt, react vigorously with water: NaH + H₂O → NaOH + H₂↑
Strong reducing agents (H⁻ donates electrons)

2. Covalent (Molecular) Hydrides

Formed by p-block elements; covalent E–H bonds
Sub-types:
- Electron precise: CH₄, GeH₄ (equal e⁻ for 2c-2e bonds)
- Electron rich: H₂O, NH₃, HF (extra lone pairs)
- Electron deficient: B₂H₆ (fewer e⁻; uses 3c-2e bonds)
Properties: Volatile, low MP/BP, non-conductors

3. Metallic (Interstitial) Hydrides

Formed by d- and f-block metals (transition metals)
H atoms occupy interstitial gaps in metal lattice
Examples: TiH₂, PdHₓ (x ≤ 1), VH₂, CeH₂.₇
Properties: Non-stoichiometric, metallic appearance, good conductors, H₂ can be released on heating (H₂ storage)

The type of hydride formed depends on the electronegativity difference between the element and hydrogen. Ionic hydrides form when the difference is large; covalent when small; metallic when d-orbitals participate.

Exam Prep Last-Minute Revision

Quick Revision Sheet

Must-Know Name Reactions

Reaction	Starting Material	Reagent	Product
HVZ	RCOOH (with α-H)	Br₂ + PCl₃	α-Bromo carboxylic acid
Reformatsky	R-CHO + α-bromo ester	Zn, dry ether	β-Hydroxy ester
Perkin	ArCHO + acid anhydride	K-salt of acid, heat	α,β-unsaturated acid (cinnamic acid)
Gabriel	Phthalimide + RX	KOH, SN2	Primary amine only
Hofmann Bromamide	RCONH₂	Br₂ + NaOH	RNH₂ (−1 carbon)
Carbylamine	1° Amine	CHCl₃ + KOH	Isocyanide (foul smell)
Sandmeyer	ArN₂⁺ (diazonium)	CuX + HX (Cu salt)	ArX (haloarene)
Gattermann	ArN₂⁺ (diazonium)	Cu powder + HX	ArX (haloarene)
Strecker	RCHO	NH₃ + HCN, then H₂O	α-Amino acid
Mond Process	Impure Ni	CO (50°C then 230°C)	Pure Ni
Kroll Process	TiCl₄	Mg (inert atm.)	Pure Ti
Van Arkel	Crude Ti/Zr	I₂, hot W filament	Pure Ti/Zr

Common Exam Mistakes to Avoid

❌

Confusing Hofmann Bromamide (amide→amine, −1C) with Hofmann Elimination (E2)
Saying Sucrose is a reducing sugar — it is NOT (both anomeric C are blocked)
Writing Carbylamine test positive for 2° or 3° amines — it is ONLY for 1° amines
Saying Gabriel synthesis gives 2° or aromatic amines — ONLY primary aliphatic
Confusing H₂S₂O₈ (O–O linkage) with H₂S₂O₇ (S–O–S, no O–O)
Saying CO₂ line has negative slope in Ellingham — it is the CO line that is negative
Writing amino acid migrates at pI — at pI it has ZERO net charge and does NOT migrate
Writing B₂H₆ has 4 bridging bonds — it has only 2 bridging B–H–B bonds
Sandmeyer = Cu metal (WRONG); Sandmeyer = Cu salt; Gattermann = Cu metal
Saying 3° amine is most basic in aqueous — 2° amine is most basic in aqueous solution

Expected Questions — Upcoming Exam

Ellingham diagram + CO slope explanation Kroll process + Mond process Diborane structure + 3c-2e bonding Gabriel phthalimide — mechanism Hofmann bromamide — mechanism + rearrangement Diazonium salt reactions table Strecker synthesis with mechanism Mutarotation of glucose Sucrose = non-reducing — explain PCl₃ nucleophilic property + geometry Amino acid zwitter ion + electrophoresis Ionization energy anomalies (Be>B, N>O) Carbylamine test mechanism Schotten-Baumann reaction Hinsberg test — distinguish 1°/2°/3° HVZ reaction mechanism

📝

Exam Strategy: Section C (10 marks each) — always write: Definition → Principle → Mechanism (stepwise) → Equation → Diagram/Table → Conclusion. Section B (5 marks) — Definition + 3–4 points + equation + brief conclusion. This format earns maximum marks.

CHE-DSM-411Inorganic & OrganicChemistry

General Principles of Metallurgy

1. Ellingham Diagram

Key Features of Ellingham Diagram

2. Purification of Metals

Mond Process — Detailed Mechanism

Van Arkel–de Boer Process (Ti purification)

s- and p-Block Elements

1. Effective Nuclear Charge (Zeff)

Slater's Rules for Shielding (σ)

2. Ionization Energy Trends

IE Anomalies — Most Important for Exams

3. Special Concepts

Hydrides — Classification & Properties

Diborane (B₂H₆) — Multicenter Bonding

Oxoacids of Sulphur

Carboxylic Acids & Their Derivatives

1. Hell–Volhard–Zelinsky (HVZ) Reaction

2. Reformatsky Reaction

3. Perkin Condensation

4. Carboxylic Acid Derivatives — Preparation & Reactivity

Ester Hydrolysis — Acidic vs Alkaline

Amines — Aliphatic & Aromatic

1. Classification & Basicity

2. Gabriel's Phthalimide Synthesis

3. Hofmann Bromamide (Degradation) Reaction

4. Amine Identification Tests

Carbylamine Test — Mechanism

5. Hofmann vs Saytzeff Elimination

6. Schotten–Baumann Reaction

Diazonium Salts

1. Preparation (Diazotization)

2. Reactions of Diazonium Salts

Electrophilic Substitution on Aniline

Amino Acids

1. Zwitter Ion & Isoelectric Point

2. Strecker Synthesis

3. Chemical Reactions of Amino Acids

Carbohydrates

1. Classification

2. Glucose — Open Chain & Cyclic Structure

3. Mutarotation

4. Disaccharides Comparison

Interactive MCQ Bank

Previous Year Question — Model Answers

Section A — MCQ Answers

Section B — Short Answer (5 Marks Each)

Introduction

Factors Affecting IE

Ammonia (NH₃)

Hydrazine (N₂H₄)

Given Reactions

Analysis

Preparation of Acid Chloride

Preparation of Amide from Acid Chloride

Mechanism (Nucleophilic Acyl Substitution)

Conversion of Open Chain to Cyclic Form

Determining Configuration of Monosaccharides

Section C — Long Answers (10 Marks Each)

Definition

Basis

Salient Features

Applications

Limitations

Definition

1. Ionic (Saline) Hydrides

2. Covalent (Molecular) Hydrides

3. Metallic (Interstitial) Hydrides

Quick Revision Sheet

Must-Know Name Reactions

Common Exam Mistakes to Avoid

Expected Questions — Upcoming Exam

CHE-DSM-411
Inorganic & Organic
Chemistry

1. Effective Nuclear Charge (Z_eff)