What are transition elements? Point out the differences between 3d and those of 4d and 5d elements.

Transition elements

Transition elements are those elements whose atoms or simple ions have partially filled d-orbitals in their ground state or in any of their common oxidation states. In practice, this means the elements of the d-block — the ten groups lying between the s-block and the p-block in the periodic table (Groups 3 to 12).

According to the modern definition, a transition element is one which has an incompletely filled d-subshell in its atomic state or in one of its stable oxidation states. Zinc, cadmium and mercury are excluded, because in all their compounds the d-subshell is completely filled (d¹⁰).

General characteristics: they are all metals, hard and lustrous, with high melting points, high densities, variable oxidation states, a strong tendency to form complex ions, coloured compounds, catalytic activity, and paramagnetism due to unpaired d-electrons.

Differences between 3d and 4d/5d elements

The most important single reason for the differences is the lanthanide contraction, which makes the 4d and 5d elements nearly the same size and hence very similar in properties, while the 3d elements differ markedly from them.

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OR

How is H₂O₂ manufactured?

Hydrogen peroxide is manufactured mainly by three methods.

1. From barium peroxide (laboratory method)

Cold dilute sulphuric acid is added to hydrated barium peroxide:

$\mathrm{BaO_{2}.8H_{2}O + H_{2}SO_{4} \rightarrow BaSO_{4} \downarrow + H_{2}O_{2} + 8H_{2}O}$

The insoluble barium sulphate is filtered off and the filtrate is concentrated under reduced pressure to obtain H₂O₂.

2. From sodium peroxide

Sodium peroxide is treated with cold dilute sulphuric acid:

$\mathrm{Na_{2}O_{2} + H_{2}SO_{4} \rightarrow Na_{2}SO_{4} + H_{2}O_{2}}$

3. By the auto-oxidation of 2-ethylanthraquinone (industrial method)

2-Ethylanthraquinone is reduced with hydrogen in the presence of a palladium catalyst to 2-ethylanthrahydroquinone. This is then oxidised by blowing air through it, regenerating the quinone and producing hydrogen peroxide:

$\mathrm{2-Ethylanthraquinone + H_{2} \rightarrow 2-Ethylanthrahydroquinone}$

$\mathrm{2-Ethylanthrahydroquinone + O_{2} \rightarrow 2-Ethylanthraquinone + H_{2}O_{2}}$

The H₂O₂ is extracted with water and the quinone is recycled. This is the modern industrial method.

H₂O₂ as an oxidising agent

In H₂O₂ the oxidation number of oxygen is −1, which lies between 0 and −2. Hence it can be reduced to −2 (acting as an oxidising agent) or oxidised to 0 (acting as a reducing agent).

As oxidising agent — it oxidises lead sulphide to lead sulphate:

$\mathrm{PbS + 4H_{2}O_{2} \rightarrow PbSO_{4} + 4H_{2}O}$

It also oxidises acidified potassium iodide to iodine:

$\mathrm{2KI + H_{2}O_{2} + H_{2}SO_{4} \rightarrow K_{2}SO_{4} + 2H_{2}O + I_{2}}$

H₂O₂ as a reducing agent

As reducing agent — it reduces acidified potassium permanganate, decolourising it:

$\mathrm{2KMnO_{4} + 3H_{2}SO_{4} + 5H_{2}O_{2} \rightarrow K_{2}SO_{4} + 2MnSO_{4} + 8H_{2}O + 5O_{2}}$

It also reduces chlorine to hydrogen chloride:

$\mathrm{H_{2}O_{2} + Cl_{2} \rightarrow 2HCl + O_{2}}$

Structure

Hydrogen peroxide is a non-planar, open-book shaped molecule. The two oxygen atoms are joined by a single O–O bond, and each oxygen is bonded to one hydrogen atom. The O–O–H bond angle is about 95° (94.8°) in the gas phase and about 101° in the solid state. The two O–H bonds lie in different planes, giving the molecule a skew (book) structure. The dihedral angle between the two O–O–H planes is about 111° in the gas phase.

Uses

1. As a bleaching agent for silk, wool, cotton, hair and paper.
2. As a disinfectant and antiseptic for wounds (dilute solution).
3. As an oxidising agent in the laboratory and in industry.
4. As a rocket propellant (concentrated H₂O₂).
5. In the manufacture of sodium perborate and other peroxides.

Mond's process is a method used for the purification (refining) of nickel. It is based on the formation and subsequent decomposition of a volatile compound, nickel tetracarbonyl, Ni(CO)₄.

Procedure:

1. Impure nickel is heated in a stream of carbon monoxide at about 330–350 K (60–80 °C).
2. The nickel present in the impure metal combines with carbon monoxide to form volatile nickel tetracarbonyl vapour:

$\mathrm{Ni + 4CO \rightarrow [\sim 330\text{ K}] Ni(CO)_{4} \uparrow}$

3. The volatile vapour is passed into a chamber maintained at a higher temperature of about 450–470 K (180–200 °C).
4. At this higher temperature, nickel tetracarbonyl decomposes to give pure nickel and carbon monoxide:

$\mathrm{Ni(CO)_{4} \rightarrow [\sim 450\text{ K}] Ni + 4CO \uparrow}$

The carbon monoxide released is recycled. The pure nickel so obtained is called Mond's nickel and is about 99.95% pure.

Principle involved: The principle is that impure nickel reacts with carbon monoxide to form a volatile compound (Ni(CO)₄) at a lower temperature, which on heating to a higher temperature decomposes to give pure metal. Since the impurities present in the impure nickel do not form such volatile carbonyls, they are left behind, and the metal is obtained in a very pure state. Thus the process depends on the thermal instability of nickel tetracarbonyl and the difference in the temperatures of formation and decomposition.

Factors influencing the choice of an extraction process (metallurgy)

The method used to extract a metal from its ore is chosen by balancing the following factors.

1. Nature and grade of the ore: whether the metal occurs as oxide, sulphide, carbonate, halide or native, and how rich the ore is. Sulphides are usually concentrated by froth flotation and roasted; oxides can be reduced directly; carbonates are calcined first.
2. Reactivity of the metal: the position of the metal in the electrochemical series and the free energy of formation of its compound (Ellingham diagram). Highly reactive metals (Na, K, Ca, Al) need electrolytic reduction; moderately reactive metals (Zn, Fe, Sn, Pb) can be reduced with carbon or CO; unreactive metals (Au, Ag, Cu) can be obtained by simple heating or by chemical displacement (e.g. cyanide leaching of gold).
3. Nature of the impurities (gangue): a basic or acidic flux is added accordingly to form slag, and the method of concentration (gravity separation, magnetic separation, flotation, leaching) depends on the gangue.
4. Energy requirement and cost: electrolysis consumes much electrical energy (the Hall–Héroult process for aluminium), so it is used only where cheap power is available; a blast furnace needs coke and a supply of fuel.
5. Purity required: metals for electrical or alloy use need further refining (electrolytic refining of copper, zone refining, the van Arkel method), which adds to cost.
6. Availability of reducing agent, flux and water, and the scale of production.
7. Environmental considerations: roasting of sulphide ores gives $SO_2$, so gas must be captured (often to make sulphuric acid), and cyanide or slag wastes must be treated.

Conclusion: the most economical process that gives the metal at the required purity, with acceptable environmental impact, is chosen.

Important biological applications of Ca (calcium):

1. Bone and teeth formation — Calcium combines with phosphate as hydroxyapatite, $\mathrm{Ca_{10}(PO_4)_6(OH)_2}$, giving bones and teeth their hardness and strength; about 99% of body calcium is stored in bone.

2. Blood clotting — $\mathrm{Ca^{2+}}$ acts as an essential cofactor (Factor IV) in the coagulation cascade, needed for the conversion of prothrombin to thrombin and for fibrin formation.

3. Muscle contraction — Release of $\mathrm{Ca^{2+}}$ from the sarcoplasmic reticulum triggers contraction by binding to troponin, and its re-uptake causes relaxation.

4. Nerve impulse transmission — $\mathrm{Ca^{2+}}$ influx at the axon terminal causes release of neurotransmitters (e.g., acetylcholine) across the synapse.

5. Enzyme activation and cell signalling — It activates enzymes such as ATPase, lipase and blood-clotting factors, and serves as a second messenger (via calmodulin) in hormone and cell-signalling pathways.

6. Membrane permeability and heart function — $\mathrm{Ca^{2+}}$ regulates permeability of cell membranes and is required for normal heartbeat (cardiac muscle contraction) and for milk secretion in mammary glands.

2e⁻-3c bonds (two-electron, three-centre bonds)

A 2-electron-3-centre bond is a delocalised bond in which 2 electrons are shared over 3 atoms instead of the usual 2 atoms. The three atoms lie in a line (or nearly so), and the electron pair occupies a bonding molecular orbital spread over all three atoms. Such bonds form when there are fewer electrons than needed for normal 2-centre bonds — i.e. in electron-deficient molecules.

Key features:

• 3 atomic orbitals combine to give 3 molecular orbitals: one bonding, one non-bonding, and one antibonding.
• The 2 available electrons fill only the bonding MO, so the bond order over the three centres is ½ per pair of atoms (total bond order = 1 spread over 3 centres).
• The atoms involved are usually B, Al, Be or H (bridge atoms), and the geometry is planar or bridged.

Example: Diborane, B₂H₆

Diborane has 12 valence electrons, but 8 are needed for the four terminal B–H bonds and 4 for the two B–H–B bridges. The two bridging hydrogen atoms each form a 2e⁻-3c bond:

$\text{B} \cdots \text{H} \cdots \text{B}$

Each B–H–B bridge uses 2 electrons shared by 2 boron atoms and 1 hydrogen atom, giving diborane its characteristic banana-shaped bridge bonds and its D₂ₕ structure. This is why B₂H₆ exists as a dimer rather than as monomeric BH₃ (which would be electron-deficient with only 6 electrons on boron).

Other examples: Al₂(CH₃)₆ (trimethylaluminium dimer) and Be(CH₃)₂ in the gas phase.

Fullerenes are a third allotropic form of carbon (after diamond and graphite) in which carbon atoms are arranged in closed cage-like molecules. The most common is buckminsterfullerene, C₆₀, a soccer-ball-shaped molecule of 60 carbon atoms forming 20 hexagons and 12 pentagons, with each carbon bonded to three others (sp² hybridised). Related members include C₇₀, C₇₆, C₈₄ and higher fullerenes, collectively called "buckyballs."

Discovery:

• In 1985, Harold Kroto (UK), Robert Curl and Richard Smalley (USA) vaporised graphite with a focused laser beam in a stream of helium gas and analysed the resulting carbon clusters by mass spectrometry.
• They observed a remarkably stable cluster with mass 720 amu, corresponding to C₆₀.
• The molecule was named buckminsterfullerene after the geodesic dome designed by architect R. Buckminster Fuller, whose structure it resembles.
• For this work the three scientists were awarded the Nobel Prize in Chemistry in 1996.

Example of $d\pi - p\pi$ bonding

A $d\pi - p\pi$ bond is formed by sideways overlap of a filled $p$ orbital (lone pair) on a small atom with a vacant $d$ orbital on a second-period-or-lower atom such as Si, P, S or Cl. The electron density flows from the $p$ orbital into the $d$ orbital, which adds partial double-bond character to a bond that is otherwise a single $\sigma$ bond.

Example: trisilylamine, $\mathrm{N(SiH_3)_3}$

• The nitrogen atom carries a lone pair in a $2p$ orbital, and the silicon atoms have vacant $3d$ orbitals of suitable energy and symmetry.
• Sideways overlap of the N $2p$ lone pair with the Si $3d$ orbitals gives $d\pi - p\pi$ bonding.
• As a result the nitrogen is planar (Si–N–Si $\approx 120^\circ$, nearly $sp^2$) and the molecule is a very weak base and poor donor, since the lone pair is delocalised. By contrast, trimethylamine $\mathrm{N(CH_3)_3}$ is pyramidal and a good base because carbon has no low-lying $d$ orbitals.

Other examples: the P–O bond in $\mathrm{R_3P{=}O}$, and the S–O and Cl–O bonds in $\mathrm{SO_4^{2-}}$, $\mathrm{PO_4^{3-}}$ and $\mathrm{ClO_4^-}$, where oxygen $p$ lone pairs overlap with vacant $d$ orbitals of P, S and Cl; these bonds are shorter than a pure single bond.

E.A.N. Rule (Effective Atomic Number Rule)

Statement: The E.A.N. rule states that a central metal atom in a complex tends to be most stable when the total number of electrons surrounding it — its own electrons plus the electrons donated by the ligands — equals the effective atomic number of the next noble gas.

$\text{E.A.N.} = Z - \text{(oxidation state)} + 2 \times \text{(number of ligand donor atoms)}$

where $Z$ is the atomic number of the metal. Each sigma-donor ligand contributes 2 electrons (one lone pair) to the metal.

Example: $\left[\text{Co(NH}_3\text{)}_6\right]^{3+}$

• Metal: Co, $Z = 27$
• Oxidation state of Co = $+3$
• Number of ligand donor atoms = 6 (six NH₃ molecules, each donating one lone pair)

$\text{E.A.N.} = 27 - 3 + 2 \times 6 = 24 + 12 = 36$

The effective atomic number 36 is the atomic number of the noble gas krypton (Kr), so the complex is stable and obeys the E.A.N. rule.

Example: $\left[\text{Ni(CO)}_4\right]$

• Metal: Ni, $Z = 28$
• Oxidation state of Ni = $0$
• Number of ligand donor atoms = 4 (four CO molecules)

$\text{E.A.N.} = 28 - 0 + 2 \times 4 = 28 + 8 = 36$

Again 36, the atomic number of Kr — hence $\left[\text{Ni(CO)}_4\right]$ is stable and obeys the rule.

Liquid ammonia as a solvent — two examples

Liquid ammonia (bp 239 K) is a widely used non-aqueous solvent. Like water it is self-ionising:

$2\text{NH}_3 \rightleftharpoons \text{NH}_4^+ + \text{NH}_2^-$

so any substance that gives $\text{NH}_4^+$ behaves as an acid and one that gives $\text{NH}_2^-$ behaves as a base in this solvent.

Example 1 — Dissolution of an alkali metal (solvated electron formation) Sodium dissolves in liquid ammonia without evolution of hydrogen, giving a deep blue, electrically conducting solution:

$\text{Na} + (x+y)\text{NH}_3 \rightarrow \text{Na}^+(\text{NH}_3)_x + e^-(\text{NH}_3)_y$

The blue colour and conductivity are due to the ammoniated (solvated) electron. On evaporation of ammonia, the metal is recovered unchanged — this shows that liquid ammonia, unlike water, does not oxidise the metal.

Example 2 — Neutralisation (acid–base) reaction in liquid ammonia Ammonium salts act as acids and metal amides as bases, and they neutralise each other just as an acid and a base do in water:

$\text{NH}_4\text{Cl} + \text{NaNH}_2 \rightarrow 2\text{NH}_3 + \text{NaCl}$

Here $\text{NH}_4^+$ is the acid and $\text{NH}_2^-$ the base; the reaction is the ammono-analogue of $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$.

Similarly, precipitation reactions occur in liquid ammonia, e.g. $\text{AgNO}_3$ and $\text{BaCl}_2$ give insoluble $\text{AgCl}$, which confirms the solvent behaviour of liquid ammonia.

In KMnO₄ the Mn is in d⁰ state yet it is highly coloured — explanation:

In KMnO₄, manganese is in the +7 oxidation state. The electronic configuration of Mn⁷⁺ is [Ar]3d⁰4s⁰, i.e. it has no d electrons (d⁰ state). Since there are no d electrons, d–d transitions are not possible. Yet KMnO₄ is intensely purple. This is because its colour arises from charge transfer (CT) transitions, not d–d transitions.

Reason:

1. Charge transfer from O to Mn: In the permanganate ion (MnO₄⁻), Mn⁷⁺ is a strong electron acceptor (high oxidation state, high charge density) and O²⁻ is an electron donor. An electron from a filled p-orbital of an oxygen atom is transferred to an empty d-orbital (or the empty molecular orbital) of Mn⁷⁺.

2. Ligand-to-metal charge transfer (LMCT): This is called a ligand-to-metal charge transfer transition. The energy required for this transfer is moderate and falls in the visible region of the spectrum.

3. Absorption and colour: The MnO₄⁻ ion strongly absorbs green–yellow light (around 500–550 nm) due to this O→Mn charge transfer. The transmitted light is rich in the complementary colour, so the compound appears deep purple/violet.

4. High intensity: Charge transfer transitions are fully allowed (Laporte and spin allowed), so the molar absorptivity (ε) is very large (≈ 2000–3000 L mol⁻¹ cm⁻¹). Hence even a small amount of KMnO₄ gives an intense colour.

Conclusion: The deep purple colour of KMnO₄ is due to ligand-to-metal charge transfer (O→Mn) and not due to d–d transitions, so the d⁰ configuration of Mn⁷⁺ does not prevent it from being highly coloured.

Mechanisms of the given reactions

(a) Cannizzaro reaction

$2C_6H_5CHO + NaOH \rightarrow C_6H_5COO^-Na^+ + C_6H_5CH_2OH$

Benzaldehyde has no $\alpha$-hydrogen, so in concentrated alkali it undergoes disproportionation: one molecule is oxidised to the acid (as its salt) and another is reduced to the alcohol.

Step 1: Nucleophilic addition. $OH^-$ attacks the carbonyl carbon to give a tetrahedral alkoxide anion.

$C_6H_5CHO + OH^- \rightleftharpoons C_6H_5-CH(OH)-O^-$

Step 2: Hydride transfer (rate-determining). This anion transfers a hydride ion from the C–H bond to the carbonyl carbon of a second benzaldehyde molecule. The donor becomes benzoic acid and the acceptor becomes benzyl alkoxide:

$C_6H_5-CH(OH)-O^- + C_6H_5CHO \rightarrow C_6H_5COOH + C_6H_5CH_2O^-$

Step 3: Proton transfer. The strongly basic alkoxide takes a proton from benzoic acid, giving the alcohol and the benzoate ion, which is the sodium salt in the medium:

$C_6H_5COOH + C_6H_5CH_2O^- \rightarrow C_6H_5COO^- + C_6H_5CH_2OH$

(b) Friedel–Crafts acylation

$Ar{-}H + RCOCl \xrightarrow{AlCl_3} Ar{-}COR + HCl$

Step 1: Formation of the electrophile. $AlCl_3$ abstracts $Cl^-$ from the acid chloride to give the resonance-stabilised acylium ion:

$RCOCl + AlCl_3 \rightarrow [R-C\equiv O]^+ + AlCl_4^-$

Step 2: Electrophilic attack. The $\pi$ electrons of the aromatic ring attack the acylium ion, forming a resonance-stabilised arenium ion ($\sigma$-complex).

Step 3: Loss of proton. $AlCl_4^-$ removes the proton from the $sp^3$ carbon, restoring aromaticity:

$\text{arenium ion} + AlCl_4^- \rightarrow ArCOR + HCl + AlCl_3$

The ketone remains complexed with $AlCl_3$ and is freed on hydrolysis, so slightly more than one equivalent of $AlCl_3$ is needed.

(c) The structure for part (c) is not given in the question text (it was an image), so its mechanism cannot be written here.

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OR: Baeyer's strain theory, boat and chair forms of cyclohexane

Baeyer's strain theory (1885). Baeyer assumed that the carbon atoms in a ring lie in one plane and that the normal tetrahedral angle is $109^\circ28'$. Any deviation of the ring bond angle from this causes angle strain, which makes the ring less stable. The deviation per bond is

$d=\tfrac{1}{2}\left(109^\circ28' - \text{ring angle}\right)$

Failure and modification. The theory explains the reactivity of small rings but fails for larger ones, because rings of six or more carbons are not planar. Sachse (1890) and Mohr (1918) proposed that cyclohexane can be puckered, with all angles close to $109^\circ28'$, so it is strain-free and stable. Further points:

• Cyclopropane is held together by bent ("banana") bonds, with overlap off the internuclear axis, which also explains its reactivity.
• Besides angle strain there are torsional strain (eclipsing, Pitzer) and steric strain (van der Waals repulsion between close groups).
• Cyclopentane is slightly puckered (envelope form) to relieve torsional strain.

Chair and boat conformations of cyclohexane

• Chair: C–C–C angles are about $111^\circ$, all C–H bonds on adjacent carbons are staggered, and no hydrogens are close together. It is free of angle, torsional and steric strain.
• Boat: angles are near tetrahedral, but the C–H bonds on C-2/C-3 and C-5/C-6 are eclipsed (torsional strain), and the "flagpole" hydrogens on C-1 and C-4 point towards each other and repel (steric strain).

Stability: the chair is more stable, by about $29\ \text{kJ mol}^{-1}$, because the boat suffers from both eclipsing and flagpole repulsion. At room temperature cyclohexane exists almost entirely in the chair form.

Cycloparaffins (cycloalkanes) are saturated hydrocarbons in which carbon atoms are joined by single bonds in a closed ring. Their general formula is $C_nH_{2n}$ ($n \ge 3$). They are alicyclic compounds that resemble alkanes in chemical behaviour. Examples are cyclopropane, cyclobutane, cyclopentane and cyclohexane.

Preparation of cyclohexane from diethyl pimelate by Dieckmann condensation

Diethyl pimelate (diethyl heptanedioate) is $\mathrm{EtOOC-(CH_2)_5-COOEt}$, a 1,7-diester. Its intramolecular Dieckmann condensation closes a six-membered ring (the ring contains C-2 to C-6 together with the carbonyl carbon of the attacked ester).

Step 1: Dieckmann condensation. Sodium ethoxide removes an $\alpha$-hydrogen from one end of the chain. The carbanion attacks the ester carbonyl at the other end, and loss of $\mathrm{EtO^-}$ gives the cyclic $\beta$-keto ester, ethyl 2-oxocyclohexanecarboxylate (2-carbethoxycyclohexanone).

$\mathrm{EtOOC(CH_2)_5COOEt \xrightarrow{NaOEt} 2\text{-carbethoxycyclohexanone} + EtOH}$

Step 2: Hydrolysis and decarboxylation. Hydrolysis with dilute acid or alkali, followed by heating, gives the $\beta$-keto acid, which loses $CO_2$ to give cyclohexanone.

$\text{2-carbethoxycyclohexanone} \xrightarrow{\text{H}_3\text{O}^+,\ \Delta} \text{cyclohexanone} + CO_2 + EtOH$

Step 3: Reduction. Cyclohexanone is reduced to cyclohexane by Clemmensen reduction ($\mathrm{Zn{-}Hg/conc.\ HCl}$) or Wolff–Kishner reduction ($\mathrm{NH_2NH_2/KOH}$, heat).

$\text{cyclohexanone} \xrightarrow{\text{Zn-Hg/HCl}} \text{cyclohexane}$

Overall: diethyl pimelate $\rightarrow$ 2-carbethoxycyclohexanone $\rightarrow$ cyclohexanone $\rightarrow$ cyclohexane.

(a) All the carbon-to-carbon bond lengths in benzene are intermediate between those of a double and a single bond.

• X-ray and spectroscopic studies show that benzene is a regular planar hexagon in which all six C–C bond lengths are equal, measuring about 139 pm.
• This value lies between the typical C–C single bond length (154 pm) and the C=C double bond length (134 pm).
• The reason is resonance (delocalisation): benzene is a hybrid of two equivalent Kekulé structures (and other contributing structures). The six π-electrons are delocalised over all six carbon atoms, so no bond is purely single or purely double.
• The bond order of each C–C bond is therefore about 1.5, giving every bond the same intermediate length. This equalisation of bond lengths is strong evidence for the delocalised structure of benzene.

(b) Nitration of toluene is easier than that of benzene.

• The methyl group (–CH₃) attached to the ring is an electron-releasing (activating) group. It releases electron density into the ring by the +I (inductive) effect and, more importantly, by hyperconjugation.
• This increased electron density at the ring, especially at the ortho and para positions, makes the ring more nucleophilic and hence more reactive towards the electrophile NO₂⁺ (nitronium ion) generated in the nitrating mixture (conc. HNO₃ + conc. H₂SO₄).
• Consequently, toluene reacts faster than benzene and requires milder conditions (lower temperature and dilute acid) for nitration.
• The methyl group is ortho- and para-directing, so nitration of toluene gives mainly o-nitrotoluene and p-nitrotoluene (with a small amount of m-nitrotoluene), whereas benzene gives only nitrobenzene.

Aldol Condensation

Aldol condensation is the base-catalysed reaction in which two molecules of an aldehyde or ketone containing at least one $\alpha$-hydrogen condense in the presence of dilute alkali (NaOH/KOH) to form a $\beta$-hydroxy aldehyde or (\beta)-hydroxy ketone (an aldol), which on heating loses a molecule of water to give an (\alpha,\beta)-unsaturated carbonyl compound.

General reaction

$2\,\mathrm{CH_3CHO} \xrightarrow[\text{dil.}]{\text{NaOH}} \mathrm{CH_3CH(OH)CH_2CHO} \xrightarrow{\Delta} \mathrm{CH_3CH=CHCHO} + \mathrm{H_2O}$

(Acetaldehyde → 3-hydroxybutanal → crotonaldehyde)

Mechanism (base-catalysed, taking acetaldehyde)

Step 1 — Formation of the enolate ion (removal of $\alpha$-H):

$\mathrm{CH_3CHO} + \mathrm{OH^-} \rightleftharpoons \mathrm{CH_2^-CHO} \text{ (enolate)} + \mathrm{H_2O}$

The base abstracts the $\alpha$-hydrogen, giving a resonance-stabilised carbanion (enolate):

$\mathrm{CH_2^-CHO} \leftrightarrow \mathrm{CH_2=CH\text{-}O^-}$

Step 2 — Nucleophilic attack of the enolate on the carbonyl carbon of the second aldehyde molecule:

$\mathrm{CH_2^-CHO} + \mathrm{CH_3CHO} \rightarrow \mathrm{CH_3CH(O^-)CH_2CHO}$

Step 3 — Protonation of the alkoxide ion by water:

$\mathrm{CH_3CH(O^-)CH_2CHO} + \mathrm{H_2O} \rightarrow \mathrm{CH_3CH(OH)CH_2CHO} + \mathrm{OH^-}$

This gives the aldol (3-hydroxybutanal).

Step 4 — Dehydration (on heating):

$\mathrm{CH_3CH(OH)CH_2CHO} \xrightarrow{\Delta} \mathrm{CH_3CH=CHCHO} + \mathrm{H_2O}$

The $\alpha$-hydrogen is removed by base and the $\beta$-OH leaves, forming the conjugated (\alpha,\beta)-unsaturated aldehyde (crotonaldehyde). The conjugation makes the product stable.

Why only the $\alpha$-hydrogen is involved

1. Acidity: The $\alpha$-hydrogen lies on the carbon directly attached to the carbonyl group. The carbonyl carbon is electron-withdrawing (both inductive and resonance effects), so the $\alpha$-C–H bond is polarised and the $\alpha$-hydrogen is acidic. It can therefore be removed easily by a base.

2. Stability of the enolate: Loss of the $\alpha$-hydrogen gives a carbanion whose negative charge is delocalised onto the carbonyl oxygen by resonance. This resonance stabilisation makes the enolate stable enough to form. Hydrogens on other carbons (β, γ, etc.) are far from the carbonyl group, are not acidic, and give no such stabilised anion.

3. Reactivity of the enolate: The resonance-stabilised enolate is a good nucleophile and attacks the carbonyl carbon of the second molecule. Only the $\alpha$-carbon carries this nucleophilic character, so condensation occurs only at the $\alpha$-position.

Hence, aldol condensation requires at least one $\alpha$-hydrogen, and only the $\alpha$-hydrogen is abstracted and involved in the reaction.

(a) Acid catalysis of nucleophilic addition reactions of aldehydes and ketones

A nucleophile (Nu⁻ or NuH) attacks the electrophilic carbonyl carbon. The carbonyl carbon is only weakly electrophilic because the C=O oxygen pulls electron density toward itself but the carbon still bears only a partial positive charge. Protonation of the carbonyl oxygen by an acid converts the C=O into a C=OH⁺ (oxocarbenium) group:

$\mathrm{R_{2}C=O + H+ \rightleftharpoons R_{2}C=OH+}$

This protonated carbonyl has a full positive charge on carbon, so it is far more electrophilic and the nucleophile attacks much more readily. The acid also converts a weak nucleophile into a stronger one in some cases (e.g. an alcohol ROH into ROH₂⁺, which then loses H₂O to give the product). After attack, loss of a proton gives the neutral addition product and regenerates the acid catalyst. Thus the acid is not consumed — it lowers the activation energy of the rate-determining nucleophilic attack step, so the reaction is acid-catalyzed. (Strong acid must be avoided for nucleophiles that are themselves acid-sensitive.)

(b) Wolff–Kishner reduction preferred over Clemmensen reduction

Both methods convert C=O to CH₂, but the Wolff–Kishner reduction is preferred for the following reasons:

1. Basic medium: Wolff–Kishner (hydrazine + KOH/NaOH, ethylene glycol, ~200 °C) works under strongly basic conditions. Clemmensen uses Zn(Hg)/conc. HCl, i.e. strongly acidic conditions.
2. Acid-sensitive groups survive: Substrates containing groups that are destroyed or altered by acid — alcohols, esters, acetals, acid-labile protecting groups, and especially groups that would undergo rearrangement or dehydration — cannot be reduced by Clemmensen. Wolff–Kishner leaves them intact.
3. No rearrangement: In acid, ketones that can form stable carbocations (e.g. those with a tertiary or benzylic centre) may rearrange during Clemmensen reduction. The basic Wolff–Kishner conditions avoid carbocation intermediates, so the carbon skeleton is retained.
4. Cleaner and cheaper reagents: Hydrazine and alkali are easier to handle and give fewer side products than amalgamated zinc and concentrated hydrochloric acid.
5. Better yields for hindered and aromatic ketones: For many aryl and sterically hindered ketones, Wolff–Kishner gives higher yields.

Hence, unless the substrate is base-sensitive (in which case Clemmensen is used), Wolff–Kishner is the method of choice.

Industrial Sources of Carboxylic Acids

Carboxylic acids are produced on a large scale industrially by several important methods:

1. Oxidation of alcohols and aldehydes Primary alcohols and aldehydes are oxidized to carboxylic acids using oxidizing agents such as alkaline KMnO₄ or acidified K₂Cr₂O₇. Industrially, ethanol is oxidized to acetic acid.

$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{[O]}} \text{CH}_3\text{COOH}$

2. Oxidation of hydrocarbons Higher alkanes obtained from petroleum are oxidized in the presence of catalysts (e.g., cobalt or manganese salts) to give a mixture of carboxylic acids. This is a major industrial route to acetic acid from butane/butene.

$\text{C}_4\text{H}_{10} + \tfrac{5}{2}\text{O}_2 \xrightarrow{\text{Co/Mn salts}} 2\text{CH}_3\text{COOH} + \text{H}_2\text{O}$

3. Hydrolysis of nitriles and esters Nitriles (alkyl cyanides) on hydrolysis with dilute acid or alkali give carboxylic acids. Esters also undergo acid or alkaline hydrolysis (saponification) to yield acids — the latter is the basis of soap manufacture from fats and oils.

$\text{RCN} + 2\text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{RCOOH} + \text{NH}_3$

4. Fermentation Vinegar (dilute acetic acid) is produced industrially by the bacterial fermentation of ethanol or sugar solutions in the presence of Acetobacter.

$\text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{Acetobacter}} \text{CH}_3\text{COOH}$

5. From fats and oils Hydrolysis of triglycerides (fats/oils) with steam under pressure gives long-chain carboxylic acids (fatty acids) such as stearic and palmitic acids, along with glycerol. This is a large-scale industrial source of higher fatty acids.

$\text{Fat} + \text{H}_2\text{O} \rightarrow \text{Glycerol} + \text{Fatty acids}$

Conclusion: Among these, oxidation of hydrocarbons and fermentation are the most important industrial routes for acetic acid, while hydrolysis of fats is the chief source of higher fatty acids used in soap and candle industries.

a. All electron withdrawing groups increase acidity.

Acidity of a carboxylic acid depends on the stability of the carboxylate ion (RCOO⁻) formed after losing H⁺. An electron withdrawing group (–NO₂, –Cl, –CN, –CHO, –COOH, etc.) pulls electron density away from the O–H bond through the inductive effect (–I effect) and/or resonance (–M effect). This:

• weakens the O–H bond, making the proton easier to release, and
• disperses the negative charge on the carboxylate ion, thereby stabilising it by delocalisation.

The more stable the conjugate base, the stronger the acid. Hence electron withdrawing groups increase acidity. The effect rises with the –I power of the group and falls with distance from –COOH, e.g. acidity order:

$\text{Cl}_3\text{CCOOH} > \text{Cl}_2\text{CHCOOH} > \text{ClCH}_2\text{COOH} > \text{CH}_3\text{COOH}$

b. Lower members of carboxylic acids are soluble in water but higher members not.

Solubility in water depends on the balance between the polar –COOH group (which forms hydrogen bonds with water) and the non-polar hydrocarbon chain (which is hydrophobic).

• In lower members (formic acid, acetic acid, propionic acid, etc.) the hydrocarbon part is small, so the –COOH group can form strong intermolecular hydrogen bonds with water molecules. The hydrophilic part dominates and the acid dissolves readily — formic, acetic and propionic acids are miscible with water in all proportions.
• As the chain length increases (butyric acid onwards), the non-polar alkyl part becomes larger. The hydrophobic hydrocarbon portion cannot form hydrogen bonds with water and instead breaks the water–water hydrogen bonding. This hydrophobic effect outweighs the solvation of the –COOH group, so solubility decreases sharply.
• From about C₆ (caproic acid) onwards the acids are practically insoluble in water, though they dissolve in organic solvents like benzene, ether and alcohol.

Thus solubility decreases with increasing molecular mass (chain length) of the carboxylic acid.

Basicity of amines

Amines are basic because the nitrogen atom has a lone pair of electrons that can accept a proton (Lewis/Brønsted base). The basicity of an amine is its tendency to accept a proton:

$RNH_2 + H_2O \rightleftharpoons RNH_3^+ + OH^-, \qquad K_b=\frac{[RNH_3^+][OH^-]}{[RNH_2]}$

A larger $K_b$ (smaller $pK_b = -\log K_b$) means a stronger base. Basicity depends on how available the nitrogen lone pair is, and on how well the resulting ammonium ion is stabilised.

Basicity of aromatic amines

In aniline the lone pair on nitrogen is delocalised into the benzene ring by resonance, so it is less available for protonation. Aniline ($pK_b \approx 9.4$) is therefore a much weaker base than ammonia ($pK_b \approx 4.75$) and cyclohexylamine ($pK_b \approx 3.3$).

Effect of substituents

1. Electron-donating groups (EDG) such as $-CH_3$, $-OCH_3$, $-NH_2$ at the para or meta position increase the electron density on nitrogen and stabilise the anilinium ion, so they increase basicity. For example, $p$-toluidine ($pK_b \approx 8.9$) is more basic than aniline.
2. Electron-withdrawing groups (EWG) such as $-NO_2$, $-CN$, $-Cl$, $-COOH$ pull electron density away from nitrogen (by $-I$ and $-R$ effects), making the lone pair less available, so they decrease basicity. The order is: $p$-nitroaniline ($pK_b \approx 13$) $<$ $m$-nitroaniline $<$ aniline. The $-R$ effect acts only at ortho and para positions, so a para nitro group reduces basicity more than a meta nitro group.
3. Position of the substituent: because of the resonance effect, the para isomer of an EDG like $-OCH_3$ is more basic than the meta isomer, whereas for $-NO_2$ the para isomer is the weakest base.
4. Ortho effect: nearly any ortho substituent (even a donor such as $-CH_3$) makes the aniline a weaker base than aniline itself. Steric hindrance to solvation of the $-NH_3^+$ ion, and steric/electronic interference with the protonation, destabilise the conjugate acid; e.g. $o$-toluidine ($pK_b \approx 9.6$) is weaker than aniline.

General order: $p$-anisidine $>$ $p$-toluidine $>$ aniline $>$ $p$-chloroaniline $>$ $p$-nitroaniline (decreasing basicity).

Preparation of an azo dye from aniline (diazo coupling reaction)

Aniline is first converted to benzene diazonium chloride by diazotization, and this diazonium salt is then coupled with a phenol or an aromatic amine to give the coloured azo dye.

Step 1 — Diazotization

Aniline is dissolved in dilute HCl and cooled to 0–5 °C in an ice bath. A cold solution of sodium nitrite is added slowly with stirring. Nitrous acid is generated in situ, which reacts with aniline to give benzene diazonium chloride.

$\mathrm{C_{6}H_{5}NH_{2} + NaNO_{2} + 2HCl \rightarrow [0-5^\circ C] C_{6}H_{5}N_{2}^+Cl^- + NaCl + 2H_{2}O}$

The temperature must be kept below 5 °C, otherwise the diazonium salt decomposes to phenol.

Step 2 — Coupling (azo dye formation)

The cold diazonium salt solution is added to an alkaline solution of phenol (or to a solution of aniline in acetic acid). Coupling occurs at the para position of the phenol ring, giving p-hydroxyazobenzene — an orange dye.

$\mathrm{C_{6}H_{5}N_{2}^+Cl^- + C_{6}H_{5}OH \rightarrow [NaOH] p-HO-C_{6}H_{4}-N=N-C_{6}H_{5} + HCl}$

If aniline is used as the coupling component in weakly acidic medium, p-aminoazobenzene (yellow dye) is obtained:

$\mathrm{C_{6}H_{5}N_{2}^+Cl^- + C_{6}H_{5}NH_{2} \rightarrow [CH_{3}COOH] p-H_{2}N-C_{6}H_{4}-N=N-C_{6}H_{5} + HCl}$

Mechanism of coupling

1. Formation of the electrophile: The diazonium ion $\mathrm{C_{6}H_{5}N_{2}^+}$ (or its resonance form $\mathrm{C_{6}H_{5}-N^{+}\equiv N}$) acts as the electrophile. The terminal nitrogen carries a positive charge and is electron-deficient.

2. Activation of the ring: In alkaline medium, phenol forms the phenoxide ion $\mathrm{C_{6}H_{5}O^-}$. The lone pair on oxygen donates electrons into the ring, making it strongly nucleophilic. The para position is the most electron-rich (ortho/para directing, para favoured for steric reasons).

3. Electrophilic substitution: The diazonium ion attacks the para carbon of the phenoxide ring. A Wheland intermediate (arenium ion) is formed, in which the positive charge is delocalised over the ring and stabilised by the oxygen lone pair.

4. Loss of proton: The intermediate loses a proton to the base, restoring aromaticity and giving the azo compound $\mathrm{p-HO-C_{6}H_{4}-N=N-C_{6}H_{5}}$.

The —N=N— group is the chromophore responsible for the intense colour. The dye is isolated by filtration, washed and recrystallised.

a. Picric acid is a stronger acid than phenol.

Picric acid (2,4,6-trinitrophenol) is much stronger than phenol because of the three nitro (–NO₂) groups present at the ortho and para positions.

• The nitro group is a powerful electron-withdrawing group. By its –I (inductive) effect and –M (resonance) effect, it pulls electron density away from the benzene ring and from the O–H bond.
• This withdrawal of electrons stabilises the phenoxide ion (the conjugate base) formed after losing the proton, since the negative charge on the oxygen is delocalised into the ring and further onto the nitro groups.
• The greater the stability of the conjugate base, the stronger the acid. Hence picric acid (pKa ≈ 0.4) is far stronger than phenol (pKa ≈ 10).

In phenol there is no such electron-withdrawing group, so its phenoxide ion is far less stabilised and phenol is only weakly acidic.

b. The C–OH bond in phenol is shorter than the C–OH bond in alcohol.

• In phenol, the oxygen atom is attached directly to an sp²-hybridised carbon of the benzene ring. The C(sp²)–O bond has more s-character, so the bond length is shorter (about 1.36 Å).
• The lone pair of electrons on the oxygen is conjugated with the π-electrons of the ring (resonance), which gives the C–O bond partial double-bond character and shortens it further.
• In alcohols, the oxygen is attached to an sp³-hybridised carbon (bond length ≈ 1.43 Å). There is no resonance with the alkyl group, so the bond remains a pure single bond and is longer.

Hence the C–OH bond in phenol is shorter than in alcohols.

Part A: Determination of Transport Number by the Moving Boundary Method

Principle: When a steady current is passed through an electrolyte solution, the ions migrate towards the oppositely charged electrodes. If a sharp boundary is set up between the solution under study (the leading solution) and another solution (the indicator solution) having an ion in common, the boundary moves with the speed of the faster ion. By measuring the displacement of this boundary, the transport number can be calculated.

Apparatus: A vertical glass tube of uniform cross-section. The anode is at the top and the cathode at the bottom. The indicator solution (e.g. CdCl₂) is placed at the bottom and the leading solution (e.g. HCl) above it, so that a sharp boundary is formed between them.

Working: On passing a steady current, H⁺ ions move towards the cathode (downward) and Cl⁻ ions move towards the anode (upward). Since H⁺ is faster than Cd²⁺, the boundary moves downward at the speed of H⁺. The displacement $x$ of the boundary in time $t$ is measured with a cathetometer.

Derivation: Let

• $c$ = concentration of HCl (mol cm⁻³)
• $A$ = area of cross-section (cm²)
• $x$ = displacement of boundary (cm)
• $I$ = current (A), $t$ = time (s)
• $F$ = 96500 C mol⁻¹

Volume swept by the boundary $= A\,x$.
Moles of HCl in this volume $= c\,A\,x$.
Charge passed $= I\,t$.
Charge carried by H⁺ $= t_+ I\,t$, so moles of H⁺ that have crossed the boundary $= \dfrac{t_+ I\,t}{F}$.

Equating the two:

$\frac{t_+ I\,t}{F} = c\,A\,x$

$\boxed{t_+ = \frac{c\,A\,x\,F}{I\,t}}$

Similarly, using a suitable indicator for the anion, $t_- = \dfrac{c\,A\,x'\,F}{I\,t}$, and $t_+ + t_- = 1$.

Conditions for a sharp boundary:

1. The indicator solution must have an ion in common with the leading solution.
2. The mobility of the leading ion must be greater than that of the corresponding indicator ion.
3. The density of the indicator solution must be less than that of the leading solution.
4. The current must be steady and the temperature constant.

Advantages: Direct and accurate method.
Limitations: Not applicable to weak electrolytes; a suitable indicator is required; the boundary may become diffuse.

---

Part B: Numerical

The data are the limiting molar conductances of HCl, NaCl and sodium acetate, $\mathrm{CH_3COONa}$ (the value 91 belongs to $\mathrm{CH_3COONa}$; the acetic acid value is what is to be found).

Given:

$\Lambda^\circ_{\text{HCl}} = 426,\quad \Lambda^\circ_{\text{NaCl}} = 126,\quad \Lambda^\circ_{\text{CH}_3\text{COONa}} = 91\ \ \text{S cm}^2\text{mol}^{-1}$

By Kohlrausch's law of independent migration of ions, the limiting molar conductance of a weak electrolyte is the sum of the limiting ionic conductances:

$\Lambda^\circ_{\text{CH}_3\text{COOH}} = \lambda^\circ_{\text{CH}_3\text{COO}^-} + \lambda^\circ_{\text{H}^+}$

$\Lambda^\circ_{\text{CH}_3\text{COOH}} = \Lambda^\circ_{\text{HCl}} + \Lambda^\circ_{\text{CH}_3\text{COONa}} - \Lambda^\circ_{\text{NaCl}}$

$= 426 + 91 - 126 = 391\ \text{S cm}^2\text{mol}^{-1}$

Hence, the molar conductance of acetic acid at infinite dilution is $391\ \text{S cm}^2\,\text{mol}^{-1}$.

Hardy–Schulz rule

The Hardy–Schulz rule states that the coagulating power of an electrolyte is due mainly to the ion carrying a charge opposite to that of the colloidal particles, and that this coagulating power increases very sharply with the valency of that oppositely charged ion. The greater the valency of the active (counter) ion, the smaller the amount of electrolyte needed to bring about coagulation. For a negatively charged sol, the coagulating power follows the order:

$\text{Al}^{3+} > \text{Ba}^{2+} > \text{Na}^{+} \quad (\text{i.e. trivalent} > \text{divalent} > \text{monovalent})$

Why $\text{MgCl}_2$ is a better coagulant than KCl for $\text{As}_2\text{S}_3$ sol

$\text{As}_2\text{S}_3$ sol is a negatively charged sol (it adsorbs $\text{S}^{2-}$ ions from $\text{H}_2\text{S}$), so the coagulating ion must be a cation.

• In $\text{MgCl}_2$ the active cation is $\text{Mg}^{2+}$, which is divalent.
• In KCl the active cation is $\text{K}^{+}$, which is monovalent.

According to the Hardy–Schulz rule, coagulating power rises sharply with the valency of the oppositely charged ion. Hence $\text{Mg}^{2+}$ (divalent) neutralises the negative charge of the $\text{As}_2\text{S}_3$ particles far more effectively than $\text{K}^{+}$ (monovalent). Therefore $\text{MgCl}_2$ coagulates the sol at a much lower concentration and is a better coagulant than KCl.

Origin of charge on colloidal particles

Colloidal particles carry an electric charge because of the following reasons:

1. Preferential adsorption of ions from the medium: Colloidal particles have a large surface area and adsorb ions of a particular sign from the dispersion medium. For example, when AgNO₃ is added to KI, the precipitated AgI adsorbs I⁻ ions (common to the precipitate) and becomes negatively charged. If KI is added to AgNO₃, AgI adsorbs Ag⁺ ions and becomes positively charged.

2. Ionisation of surface groups: The molecules on the surface of the colloidal particle may ionise, leaving charged groups on the particle. For example, soap molecules ionise to give RCOO⁻ ions, so soap sols are negatively charged. Proteins contain –COOH and –NH₂ groups; at a pH above the isoelectric point they give up H⁺ and become negatively charged, while below it they become positively charged.

3. Friction between the dispersed phase and the dispersion medium: When the dispersed phase and the medium are rubbed against each other during the preparation of the sol, electrons may be transferred from one phase to the other, leaving one of them charged.

4. Presence of a charged double layer: The charge on the particle is balanced by an equal and opposite charge in the medium, so the system as a whole remains electrically neutral.

Zeta-potential

The colloidal particle, together with the firmly adsorbed layer of ions on its surface, forms the fixed layer (Stern layer). Around this fixed layer there is a loosely held diffused layer of oppositely charged ions extending into the medium. The boundary between the fixed layer and the diffused layer is called the shear plane (or slipping plane).

The zeta-potential (electrokinetic potential) is the potential difference between the fixed layer of the colloidal particle and the bulk of the dispersion medium at the shear plane. It is denoted by ζ and is measured in volts (usually in millivolts).

Significance: Zeta-potential measures the effective charge on the colloidal particle. A high value of zeta-potential (positive or negative) means strong repulsion between particles, so the sol is stable. When the zeta-potential is reduced to zero (the isoelectric point) by adding an electrolyte, the repulsion disappears, the particles come together and the sol coagulates. Thus zeta-potential decides the stability of a colloidal solution.

Fluorescence and Phosphorescence

Both are forms of photoluminescence — light emitted by a substance after it absorbs energy (usually UV or visible light). The difference lies in the nature of the excited state and the duration of emission.

Fluorescence

• The molecule absorbs a photon and is excited to a singlet state (all electron spins paired).
• It returns to the ground state by emitting a photon.
• The return is spin-allowed, so it is very fast.
• Lifetime: about 10⁻⁹ to 10⁻⁷ s (nanoseconds).
• Emission stops immediately when the exciting source is removed.
• Example: quinine in tonic water, fluorescein, chlorophyll.

Phosphorescence

• The molecule is excited to a singlet state, then undergoes intersystem crossing to a triplet state (two electron spins parallel).
• Return to the ground state from the triplet state is spin-forbidden, so it is slow.
• Lifetime: about 10⁻³ s to several seconds (or longer).
• Emission continues for some time after the source is removed (afterglow).
• Example: glow-in-the-dark paints (zinc sulfide doped with copper), watch dials.

Key difference in one line: In fluorescence the emission is from a singlet excited state and is short-lived; in phosphorescence it is from a triplet excited state and is long-lived.

Difference Between Thermochemical and Photochemical Reactions

Quantum Yield (Φ)

Definition: Quantum yield of a photochemical reaction is the ratio of the number of molecules of reactant consumed (or product formed) to the number of photons (quanta) of light absorbed by the system in the same time interval.

$\Phi = \frac{\text{Number of molecules reacted (or product formed)}}{\text{Number of photons absorbed}}$

Since one mole contains $N_A$ molecules and one mole of photons (one Einstein) also contains $N_A$ photons, it can also be written as:

$\Phi = \frac{\text{Number of moles of substance reacted}}{\text{Number of Einsteins of light absorbed}}$

Cases:

• Φ = 1 — one photon causes reaction of one molecule (e.g. primary step of $H_2 + Cl_2$).
• Φ > 1 — chain reaction; one photon triggers many molecules (e.g. Φ for $H_2 + Cl_2$ is about $10^6$).
• Φ < 1 — some absorbed photons are wasted by de-excitation or recombination (e.g. Φ for $HBr$ formation is about 0.2).

Enzyme catalysis is the process by which enzymes — biological macromolecules, almost always proteins — speed up the rate of a biochemical reaction by lowering its activation energy, without themselves being consumed or permanently altered in the reaction. The enzyme binds the reactant (substrate) at its active site to form an enzyme–substrate complex, which then breaks down to give the product and regenerate the free enzyme:

$E + S \rightleftharpoons ES \rightarrow E + P$

Characteristics of enzyme catalysis:

1. High efficiency — a single enzyme molecule can convert thousands to millions of substrate molecules per second; even a very small amount of enzyme catalyses a large amount of reaction.

2. High specificity — an enzyme usually acts on only one substrate or a group of closely related substrates (e.g., urease acts only on urea). This is due to the precise shape and chemical nature of the active site.

3. Mild conditions — enzymes work under mild conditions of temperature (about 25–40 °C), pH (a narrow optimum range) and atmospheric pressure, unlike industrial catalysts that need high temperature and pressure.

4. Sensitivity to temperature and pH — activity increases with temperature up to an optimum, then falls sharply because the protein is denatured; similarly, each enzyme has an optimum pH (e.g., pepsin ≈ 2, trypsin ≈ 8).

5. Reversibility — enzymes catalyse both the forward and backward reactions and are not consumed; they are recovered unchanged at the end of the reaction.

6. Regulation — enzyme activity can be controlled by activators, inhibitors, cofactors and coenzymes, and by feedback inhibition.

7. Colloidal nature — enzymes are colloidal in nature and cannot pass through semipermeable membranes.

Conductometric titration is a volumetric method in which the end point is detected by measuring the change in electrical conductivity of the solution as the titrant is added. The conductivity depends on the number and mobility of ions present; as the reaction proceeds, ions are replaced by other ions of different mobility, so the conductance of the solution changes. A plot of conductance against volume of titrant gives two straight lines whose intersection gives the end point. It is especially useful for weak acids, coloured or turbid solutions where indicators fail.

a) HCl vs NaOH

• Initially the solution contains highly mobile H⁺ and Cl⁻ ions, so conductance is high.
• On adding NaOH, H⁺ is replaced by the less mobile Na⁺ (H⁺ + OH⁻ → H₂O), so conductance decreases linearly.
• At the equivalence point, only NaCl remains, giving minimum conductance.
• Beyond the end point, excess OH⁻ (highly mobile) and Na⁺ are added, so conductance increases linearly.
• The curve is V-shaped with a sharp minimum at the end point.

b) CH₃COOH vs NaOH

• Initially the weak acid is only slightly ionised, so conductance is low.
• On adding NaOH, the highly conducting salt CH₃COONa is formed (CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O), so conductance increases (the common-ion suppression of acetic acid is outweighed by salt formation).
• The increase continues up to the equivalence point, where all acid is converted to salt.
• Beyond the end point, excess OH⁻ is added, so conductance increases more steeply.
• The curve is two straight lines of different slopes intersecting at the end point; the end point is obtained by extrapolating the two lines.

Cell reaction

At anode (oxidation): $Zn \rightarrow Zn^{2+} + 2e^-$

At cathode (reduction): $Fe^{3+} + e^- \rightarrow Fe^{2+}$

Overall balanced cell reaction: $Zn + 2Fe^{3+} \rightarrow Zn^{2+} + 2Fe^{2+}, \quad n=2$

Standard cell potential $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ $= 0.77 - (-0.76) = 1.53\ \text{V}$

Reaction quotient $Q = \frac{[Zn^{2+}][Fe^{2+}]^2}{[Fe^{3+}]^2}$ $= \frac{(0.01)(0.01)^2}{(0.1)^2}$ $= \frac{0.01 \times 10^{-4}}{0.01} = 10^{-4}$

Nernst equation $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log Q$ $= 1.53 - \frac{0.0591}{2}\log(10^{-4})$ $= 1.53 - 0.02955(-4)$ $= 1.53 + 0.1182$ $= 1.6482\ \text{V}$

$\boxed{E_{\text{cell}} \approx 1.648\ \text{V}}$

Entropy is a thermodynamic state function that measures the degree of disorder or randomness of a system and, equivalently, the amount of energy that is unavailable to do useful work.

1. Entropy as randomness (disorder): Every system tends to move spontaneously from an ordered state to a more disordered state. Entropy quantifies this disorder — the greater the randomness of molecular arrangement and motion, the greater the entropy. For example:

• A solid has molecules fixed in an ordered lattice → low entropy
• A liquid has loosely arranged, mobile molecules → higher entropy
• A gas has freely moving, randomly distributed molecules → highest entropy

Thus melting, vaporisation, and mixing of gases all increase entropy. This is why the entropy of a substance follows the order: solid < liquid < gas.

2. Entropy as unavailable energy: When energy is converted (e.g., heat flows from a hot body to a cold body), not all of it can be converted into work. The portion of energy that becomes degraded and can no longer perform useful work is measured by entropy. In the relation

$dS = \frac{dQ_{\text{rev}}}{T}$

the entropy change equals the heat transferred divided by the absolute temperature. A rise in entropy means more energy has been dissipated into random molecular motion and is therefore unavailable for work. This is why the efficiency of any real heat engine is always less than 100 % — some energy is inevitably lost as unavailable energy, and the total entropy of the universe always increases (second law of thermodynamics).

Conclusion: Entropy is thus a single concept with two faces — it is a measure of molecular randomness and, at the same time, a measure of the energy that has become unavailable for doing useful work.

Given:
$n = 2$ mol, $T_1 = 0^\circ\text{C} = 273\ \text{K}$,
$P_1 = 20$ atm, $P_2 = 2$ atm,
$C_p = \dfrac{5R}{2}$

For an ideal gas,
$C_v = C_p - R = \frac{5R}{2} - R = \frac{3R}{2}$

So,
$\gamma = \frac{C_p}{C_v} = \frac{5R/2}{3R/2} = \frac{5}{3}$

For reversible adiabatic expansion,
$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma - 1}{\gamma}}$

Substituting values:
$\frac{T_2}{273} = \left(\frac{2}{20}\right)^{\frac{2/3}{5/3}} = (0.1)^{2/5}$

$T_2 = 273 \times (0.1)^{0.4} = 273 \times 0.3981 \approx 108.7\ \text{K}$

$\boxed{T_2 \approx 108.7\ \text{K} \approx -164.3^\circ\text{C}}$

Work done by the gas:
For adiabatic process, $Q = 0$, so
$W = -\Delta U = nC_v(T_1 - T_2)$

$W = 2 \times \frac{3R}{2} \times (273 - 108.7)$

$W = 3R \times 164.3 = 492.9R$

Using $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$,
$W = 492.9 \times 8.314 \approx 4098\ \text{J}$

$\boxed{W \approx 4.10\ \text{kJ}}$

So, the final temperature is about 108.7 K and the work done by the gas during expansion is about 4.10 kJ.

The "Nairom reaction" appears to be a mis-spelling of the Nierenstein reaction, so the answer below is written for that reaction.

Nierenstein reaction. The reaction of an acid chloride with diazomethane to give an α-chloromethyl ketone (chloromethyl ketone) through a diazoketone intermediate. The reaction is named after M. Nierenstein.

$\mathrm{R{-}COCl+CH_2N_2\longrightarrow R{-}CO{-}CHN_2+HCl}$

$\mathrm{R{-}CO{-}CHN_2+HCl\longrightarrow R{-}CO{-}CH_2Cl+N_2}$

Overall (with one equivalent of $\mathrm{CH_2N_2}$, so that the HCl liberated is not removed):

$\mathrm{R{-}COCl+CH_2N_2\longrightarrow R{-}CO{-}CH_2Cl+N_2}$

Mechanism

1. Nucleophilic acyl substitution. The nucleophilic carbon of diazomethane ($\mathrm{{}^-CH_2{-}N_2^+}$) attacks the carbonyl carbon of the acid chloride to form a tetrahedral intermediate.
2. Loss of chloride. The carbonyl is regenerated with expulsion of $\mathrm{Cl^-}$, giving $\mathrm{R{-}CO{-}CH_2{-}N_2^+}$.
3. Deprotonation. $\mathrm{Cl^-}$ (or a second molecule of $\mathrm{CH_2N_2}$) removes a proton to give the diazoketone $\mathrm{R{-}CO{-}CHN_2}$ and HCl.
4. Protonation. The HCl formed protonates the diazoketone at carbon, giving $\mathrm{R{-}CO{-}CH_2{-}N_2^+}$.
5. SN2 displacement. $\mathrm{Cl^-}$ attacks the $\mathrm{CH_2}$ carbon, the excellent leaving group $\mathrm{N_2}$ departs, and the product $\mathrm{R{-}CO{-}CH_2Cl}$ is formed.

If two equivalents of diazomethane are used, the second equivalent consumes the HCl (forming $\mathrm{CH_3Cl}$), and the diazoketone is isolated instead of the chloromethyl ketone.

Utility in organic chemistry

• It gives α-halo (chloromethyl) ketones, which are useful building blocks, for example in the synthesis of thiazoles and other heterocycles and of α-hydroxy and α-amino ketones.
• The diazoketone intermediate is the starting point of the Arndt-Eistert synthesis: Wolff rearrangement (Ag$_2$O, $\mathrm{H_2O}$) converts $\mathrm{R{-}COCl}$ into $\mathrm{R{-}CH_2{-}COOH}$, a one-carbon homologation of a carboxylic acid.
• It extends the carbon chain by one carbon in a mild, general way.
• Diazoketones are also used to prepare cyclopropanes and C-H insertion products through carbene chemistry.

(a) Nucleophilic addition reactions of aldehydes and ketones are catalyzed by acids.

The carbonyl carbon of an aldehyde or ketone is electron-deficient (δ+) and is attacked by a nucleophile. However, a weak nucleophile such as water, an alcohol or a thiol cannot attack the carbonyl carbon efficiently on its own. An acid catalyst activates the carbonyl group by protonating the carbonyl oxygen:

$\mathrm{R_{2}C=O + H+ \rightleftharpoons R_{2}C=OH+}$

Protonation of the oxygen pulls electron density away from the carbonyl carbon through the inductive effect, making the carbon more electrophilic (more δ+) and therefore far more susceptible to nucleophilic attack. The acid also converts a weak nucleophile into a stronger one in some cases (e.g. an alcohol is not a good nucleophile, but the neutral alcohol attacks the activated carbonyl and the product is deprotonated to give the acetal). Thus the acid lowers the activation energy of the rate-determining nucleophilic attack step, so the reaction is catalyzed by acids. (Note: strong acid is avoided in some cases because it can protonate the nucleophile itself, but in general acid catalysis operates by carbonyl activation.)

(b) Propionaldehyde reduces Fehling's solution but benzaldehyde does not.

Fehling's solution (alkaline cupric tartrate containing $\mathrm{Cu^2+}$) is reduced to red cuprous oxide ($\mathrm{Cu_{2}O}$) only by aliphatic aldehydes; the aldehyde is oxidized to the corresponding carboxylic acid (carboxylate).

• Propionaldehyde $\mathrm{CH_{3}CH_{2}CHO}$ is an aliphatic aldehyde. It undergoes the oxidation readily, reducing $\mathrm{Cu^2+}$ to $\mathrm{Cu+}$, so a red precipitate of $\mathrm{Cu_{2}O}$ appears.

• Benzaldehyde $\mathrm{C_{6}H_{5}CHO}$ is an aromatic aldehyde. The $\mathrm{-CHO}$ group is attached directly to the benzene ring. The ring withdraws electron density by resonance and inductive effects, so the carbonyl carbon is less positive and the aldehyde is much less easily oxidized. Consequently it fails to reduce Fehling's solution and no red precipitate is formed. (Benzaldehyde does, however, reduce Tollens' reagent, and it undergoes Cannizzaro's reaction with concentrated alkali.)

Preparation of Aromatic Carboxylic Acids by Oxidation of Alkylbenzenes

Alkylbenzenes (toluene, ethylbenzene, isopropylbenzene, etc.) on vigorous oxidation give aromatic carboxylic acids. The whole side chain is oxidized down to a –COOH group attached directly to the ring.

Reagents used:

1. Alkaline KMnO₄ (hot, dilute) followed by acidification
2. Hot chromic acid (Na₂Cr₂O₇ / H₂SO₄)
3. Dilute HNO₃ (for toluene)

General reaction:

$$\text{Ar–R} \xrightarrow[\text{(i) KMnO}_4/\text{OH}^-,\ \Delta]{\text{(ii) H}^+} \text{Ar–COOH}$$

Examples:

$$\text{C}_6\text{H}_5\text{CH}_3 \xrightarrow{\text{KMnO}_4/\text{OH}^-,\ \Delta} \text{C}_6\text{H}_5\text{COO}^- \xrightarrow{\text{H}^+} \text{C}_6\text{H}_5\text{COOH}$$ $$\text{C}_6\text{H}_5\text{CH}_2\text{CH}_3 \xrightarrow{\text{Na}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4,\ \Delta} \text{C}_6\text{H}_5\text{COOH}$$ $$\text{C}_6\text{H}_5\text{CH}(\text{CH}_3)_2 \xrightarrow{\text{KMnO}_4/\text{OH}^-,\ \Delta} \text{C}_6\text{H}_5\text{COOH}$$

Key points:

• Only alkyl groups bearing at least one α-hydrogen are oxidized. Thus tert-butylbenzene, C₆H₅–C(CH₃)₃, is not oxidized (no α-H).
• The aromatic ring itself is unaffected under these conditions.
• Oxidation stops at the carboxylic acid stage; the –COOH carbon comes from the side chain, not the ring.
• If two alkyl groups are present, both are oxidized, giving a dicarboxylic acid, e.g. p-xylene → terephthalic acid.

Comparison of Acidities: Aromatic vs Aliphatic Carboxylic Acids

Aromatic carboxylic acids (e.g. benzoic acid) are stronger acids than aliphatic carboxylic acids (e.g. acetic acid).

$$\text{C}_6\text{H}_5\text{COOH} \quad (pK_a \approx 4.20) \qquad \text{CH}_3\text{COOH} \quad (pK_a \approx 4.76)$$

Reason:

1. Resonance stabilisation of the carboxylate ion: In the benzoate ion, the negative charge is delocalised not only over the two oxygen atoms of –COO⁻ but also into the benzene ring. This extra delocalisation stabilises the anion, so benzoic acid loses H⁺ more readily.

2. –I effect of the phenyl ring: The sp²-hybridised phenyl carbon is more electronegative than the sp³ carbon of the methyl group. It therefore exerts a stronger electron-withdrawing (–I) effect, which further stabilises the conjugate base.

3. In aliphatic acids: The methyl group exerts a +I (electron-releasing) effect, which destabilises the carboxylate ion and makes acetic acid weaker.

Effect of substituents on the ring:

• Electron-withdrawing groups (–NO₂, –CN, –Cl, –CHO) increase acidity. Example: p-nitrobenzoic acid, pK_a ≈ 3.44, is much stronger than benzoic acid.
• Electron-releasing groups (–CH₃, –OCH₃, –NH₂) decrease acidity. Example: p-methoxybenzoic acid, pK_a ≈ 4.47, is weaker than benzoic acid.
• The effect is strongest when the substituent is at the ortho or para position, because these positions allow resonance/inductive effects to reach the –COOH group directly.

Summary order of acidity:

$$\underbrace{p\text{-NO}_2\text{C}_6\text{H}_4\text{COOH}}_{\text{strongest}} > \text{C}_6\text{H}_5\text{COOH} > \text{CH}_3\text{COOH} > \underbrace{p\text{-CH}_3\text{OC}_6\text{H}_4\text{COOH}}_{\text{weakest}}$$

Thus, aromatic carboxylic acids are distinctly stronger acids than their aliphatic counterparts, mainly due to resonance stabilisation of the benzoate ion by the aromatic ring.

a. Carboxylic acids are usually stronger acids than phenols.

A carboxylic acid, R–COOH, ionises to give a carboxylate ion, RCOO⁻, in which the negative charge is delocalised equally over two equivalent oxygen atoms by resonance:

$\mathrm{R-COOH \rightleftharpoons R-COO^- + H+}$

The two resonance structures of RCOO⁻ are identical in energy, so the resonance stabilisation is large. The resulting carboxylate ion is therefore highly stable, which makes the O–H proton easy to release — carboxylic acids have pKa ≈ 4–5.

In phenol, C₆H₅OH, the phenoxide ion C₆H₅O⁻ is stabilised only by delocalisation of the negative charge into the benzene ring, and the contributing structures are not all equivalent (some place the charge on carbon). The stabilisation is thus smaller than in a carboxylate ion. Moreover, in phenol the O–H bond is attached to an sp² carbon of the ring, but the deciding factor is the poorer resonance stabilisation of the anion. Hence phenol is a weaker acid, pKa ≈ 10.

Since a stronger acid is the one whose conjugate base is more stable, carboxylic acids (pKa ≈ 4–5) are much stronger acids than phenols (pKa ≈ 10).

b. Most of the carboxylic acids exist as cyclic dimers.

Carboxylic acids have a strongly polar –COOH group in which the C=O oxygen is a good hydrogen-bond acceptor and the O–H hydrogen is a good hydrogen-bond donor. In the vapour phase and in non-polar solvents, two molecules therefore link through two intermolecular hydrogen bonds to form a cyclic (eight-membered ring) dimer:

$\mathrm{R-COOH \cdots O=C(OH)R \quad (two H-bonds, ring of 8 atoms)}$

The dimer is favoured because:

1. Two hydrogen bonds per dimer are formed, so the stabilisation energy is roughly double that of a single H-bond.
2. The cyclic arrangement is planar and strain-free, and the ring is held rigidly, giving a large negative enthalpy of dimerisation.
3. The bulky R groups point away from each other, so there is little steric hindrance.

This is why the observed molecular mass of acetic acid in benzene or in the vapour state is about 120 (twice 60), corresponding to the formula (CH₃COOH)₂, and why carboxylic acids have abnormally high boiling points.

Synthesis of 1,3,5-Tribromobenzene from p-Nitroaniline

Step 1: Diazotization of p-nitroaniline

Dissolve p-nitroaniline in dilute HCl and cool to 0–5 °C. Add sodium nitrite solution dropwise with stirring. p-Nitroaniline forms the diazonium salt:

$\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0-5\,^\circ\text{C}} \text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}$

Step 2: Replacement of the diazonium group by bromine

Add the cold diazonium salt solution to a warm solution of cuprous bromide in HBr (Sandmeyer reaction), or to a mixture of CuBr and KBr:

$\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{N}_2^+\text{Cl}^- \xrightarrow{\text{CuBr / HBr}} \text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{Br} + \text{N}_2$

This gives p-bromonitrobenzene.

Step 3: Reduction of the nitro group

Reduce p-bromonitrobenzene with Sn/HCl (or Fe/HCl) to get p-bromoaniline:

$\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{Br} \xrightarrow{\text{Sn/HCl}} \text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{Br}$

Step 4: Acetylation to protect the amino group

Treat p-bromoaniline with acetic anhydride to form p-bromoacetanilide, protecting the –NH₂ group so that it directs incoming bromine to the ortho positions:

$\text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{Br} \xrightarrow{(\text{CH}_3\text{CO})_2\text{O}} \text{CH}_3\text{CONH}-\text{C}_6\text{H}_4-\text{Br}$

Step 5: Bromination

Brominate p-bromoacetanilide with Br₂ in acetic acid. The acetamido group is o,p-directing, so bromine enters both positions ortho to it, giving 2,4,6-tribromoacetanilide:

$\text{CH}_3\text{CONH}-\text{C}_6\text{H}_4-\text{Br} \xrightarrow{\text{Br}_2/\text{CH}_3\text{COOH}} \text{CH}_3\text{CONH}-\text{C}_6\text{H}_2\text{Br}_3$

Step 6: Hydrolysis (removal of the acetyl group)

Hydrolyse the tribromoacetanilide with dilute HCl or NaOH to obtain 2,4,6-tribromoaniline:

$\text{CH}_3\text{CONH}-\text{C}_6\text{H}_2\text{Br}_3 \xrightarrow{\text{H}^+/\text{H}_2\text{O}} \text{H}_2\text{N}-\text{C}_6\text{H}_2\text{Br}_3 + \text{CH}_3\text{COOH}$

Step 7: Diazotization and deamination

Diazotize 2,4,6-tribromoaniline with NaNO₂/HCl at 0–5 °C, then treat the diazonium salt with hypophosphorous acid (H₃PO₂) or ethanol to replace the –N₂⁺ group by hydrogen:

$\text{H}_2\text{N}-\text{C}_6\text{H}_2\text{Br}_3 \xrightarrow{\text{NaNO}_2/\text{HCl}} \text{Br}_3\text{C}_6\text{H}_2-\text{N}_2^+\text{Cl}^- \xrightarrow{\text{H}_3\text{PO}_2} \text{C}_6\text{H}_3\text{Br}_3 + \text{N}_2 + \text{H}_3\text{PO}_3$

The product is 1,3,5-tribromobenzene.

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Synthesis of p-Chloroaniline from Chlorobenzene

Step 1: Nitration of chlorobenzene

Treat chlorobenzene with a mixture of concentrated HNO₃ and concentrated H₂SO₄ at about 60 °C. Chlorine is o,p-directing, so the para isomer predominates:

$\text{C}_6\text{H}_5\text{Cl} \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{Cl} + \text{H}_2\text{O}$

The product is p-chloronitrobenzene (along with some o-isomer, which is separated by steam distillation or crystallization).

Step 2: Reduction of the nitro group

Reduce p-chloronitrobenzene with Sn/HCl or Fe/HCl:

$\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{Cl} \xrightarrow{\text{Sn/HCl}} \text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{Cl}$

The product is p-chloroaniline.

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Summary of routes