Industrial Sources of Carboxylic Acids

Carboxylic acids are produced on a large scale for use in the manufacture of soaps, textiles, plastics, drugs, and food preservatives. The important industrial sources are given below.

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.

$\text{R–CH}_2\text{OH} \xrightarrow{[\text{O}]} \text{R–CHO} \xrightarrow{[\text{O}]} \text{R–COOH}$

Example: Ethanol → acetaldehyde → acetic acid. Industrially, acetaldehyde is oxidized by air in the presence of manganese acetate catalyst to give acetic acid.

2. Oxidation of hydrocarbons Higher alkanes (from petroleum) on catalytic air oxidation give fatty acids. This is the source of the long-chain acids used for soap making.

$\text{R–CH}_2\text{–CH}_2\text{–R'} \xrightarrow{\text{air, catalyst}} 2\,\text{R–COOH}$

3. Carbonylation of methanol (Monsanto process) Acetic acid is manufactured industrially by reaction of methanol with carbon monoxide using a rhodium–iodine catalyst.

$\text{CH}_3\text{OH} + \text{CO} \xrightarrow{\text{Rh/I}_2,\ 150\text{–}200^\circ\text{C}} \text{CH}_3\text{COOH}$

4. Hydrolysis of fats and oils Fats and oils (triglycerides) on hydrolysis with water or dilute acid give glycerol and a mixture of long-chain carboxylic acids (stearic, palmitic, oleic acids). This is the oldest industrial route to higher fatty acids.

5. Hydrolysis of nitriles and esters Nitriles on hydrolysis with dilute acid give carboxylic acids; esters on acid hydrolysis also yield acids. These are used when a specific acid is required.

6. Fermentation Vinegar (dilute acetic acid) is obtained by bacterial fermentation of ethanol or sugar solutions.

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

7. From carbon monoxide and alkali Sodium formate is made by passing CO into hot concentrated NaOH under pressure; acidification gives formic acid.

$\text{CO} + \text{NaOH} \rightarrow \text{HCOONa} \xrightarrow{\text{H}^+} \text{HCOOH}$

Conclusion: Industrially, acetic acid is obtained mainly by carbonylation of methanol and by oxidation of acetaldehyde, while higher fatty acids come from fats, oils, and petroleum hydrocarbons.

a. All electron withdrawing groups increase acidity.

Electron withdrawing groups (–NO₂, –CN, –CHO, –COOH, –X, etc.) pull electron density away from the O–H bond of the carboxylic acid through the inductive effect (–I effect) and/or resonance effect (–M effect). This:

• increases the polarity of the O–H bond, and
• stabilises the carboxylate ion (RCOO⁻) formed after losing a proton, by dispersing the negative charge over a larger area.

A more stable conjugate base means the acid ionises more readily, so the acid strength (Ka) increases and pKa decreases. Hence, electron withdrawing groups increase acidity. The effect is stronger when the group is closer to –COOH and when its –I effect is greater (e.g., Cl₃C–COOH > Cl₂CH–COOH > ClCH₂–COOH > CH₃COOH).

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 polar –COOH group dominates. The –COOH group forms strong hydrogen bonds with water molecules, making these acids freely soluble in water.
• As the chain length increases (higher members such as palmitic acid, stearic acid), the non-polar hydrocarbon part becomes large and dominates. The hydrophobic part cannot form hydrogen bonds with water and instead interferes with the water structure. The attractive force between the large hydrocarbon chains (van der Waals forces) also exceeds the acid–water interaction.

Therefore, the hydrophilic character of –COOH is outweighed by the hydrophobic character of the long alkyl chain, and higher members become practically insoluble in water (they dissolve only in organic solvents like benzene, ether, alcohol).

Basicity of amines is the tendency of the lone pair on nitrogen to accept a proton. In water:

$\mathrm{RNH_2 + H_2O \rightleftharpoons RNH_3^+ + OH^-}$

It is measured by $K_b$ or $pK_b$; the smaller the $pK_b$ (the larger the $pK_a$ of the conjugate acid), the stronger the base. It depends on the electron density on nitrogen and on the stability of the conjugate ammonium ion.

Aromatic amines are weaker bases than aliphatic ones (aniline $pK_b\approx 9.4$; ethylamine $pK_b\approx 3.3$) because the lone pair on N is delocalised into the benzene ring by resonance, so it is less available for protonation. Substituents on the ring change this.

1. Electron-donating groups (EDG: $-\mathrm{CH_3}$, $-\mathrm{OCH_3}$, $-\mathrm{NH_2}$) push electron density towards N (by $+I$ or $+R$ effect) and increase basicity. Examples: $p$-toluidine ($pK_b\approx 8.9$) and $p$-anisidine ($pK_b\approx 8.7$) are stronger bases than aniline.

2. Electron-withdrawing groups (EWG: $-\mathrm{NO_2}$, $-\mathrm{CN}$, $-\mathrm{CHO}$, $-\mathrm{COOH}$, halogens) pull electron density away from N (by $-I$ and/or $-R$ effect) and decrease basicity. Examples: $m$-nitroaniline ($pK_b\approx 11.5$) and $p$-nitroaniline ($pK_b\approx 13.0$) are much weaker bases than aniline.

3. Position of the substituent. A $-R$ group such as $-\mathrm{NO_2}$ withdraws most strongly from the para and ortho positions, so the order of basicity is meta $>$ para $>$ ortho: $m$-nitroaniline ($pK_b\approx 11.5$) $>$ $p$-nitroaniline ($\approx 13.0$) $>$ $o$-nitroaniline ($\approx 14.3$). For donor groups such as $-\mathrm{CH_3}$ the para isomer is the strongest base.

4. Ortho effect. Almost any ortho substituent, whether donating or withdrawing, makes the aniline a weaker base than aniline itself. The substituent hinders solvation and stabilisation of the $-\mathrm{NH_3^+}$ group (steric hindrance to protonation) and exerts a close-range inductive effect. Example: $o$-toluidine ($pK_b\approx 9.6$) is a weaker base than aniline ($9.4$) and than $p$-toluidine; $o$-chloroaniline ($pK_a\approx 2.6$) is far weaker than aniline ($pK_a\approx 4.6$).

Order of basicity: $p$-anisidine $>$ $p$-toluidine $>$ aniline $>$ $p$-chloroaniline $>$ $m$-nitroaniline $>$ $p$-nitroaniline.

An azo dye is prepared from aniline in two steps: (1) diazotisation of aniline to benzenediazonium chloride at $0\text{--}5\,^\circ\mathrm{C}$, and (2) azo coupling of the diazonium salt with an activated aromatic compound such as phenol or aniline.

Step 1: Diazotisation

Aniline is dissolved in excess dilute HCl and cooled to $0\text{--}5\,^\circ\mathrm{C}$ in an ice-salt bath. Cold aqueous sodium nitrite is added slowly with stirring. Above $5\,^\circ\mathrm{C}$ the diazonium salt decomposes to phenol and $\mathrm{N_2}$.

$\mathrm{NaNO_2 + HCl \rightarrow HNO_2 + NaCl}$

$\mathrm{C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0-5\,^\circ C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O}$

Mechanism of diazotisation

1. Formation of the nitrosonium ion (electrophile): $\mathrm{HNO_2 + H^+ \rightleftharpoons H_2O^+{-}N{=}O \rightarrow H_2O + NO^+}$
2. The lone pair of aniline nitrogen attacks $\mathrm{NO^+}$, giving the N-nitrosoammonium ion: $\mathrm{C_6H_5NH_2 + NO^+ \rightarrow C_6H_5\overset{+}{N}H_2{-}N{=}O}$
3. Loss of a proton gives the N-nitrosamine: $\mathrm{C_6H_5\overset{+}{N}H_2{-}N{=}O \rightarrow C_6H_5NH{-}N{=}O + H^+}$
4. Tautomerism to the diazohydroxide: $\mathrm{C_6H_5NH{-}N{=}O \rightleftharpoons C_6H_5N{=}N{-}OH}$
5. Protonation and loss of water gives the diazonium ion: $\mathrm{C_6H_5N{=}N{-}OH + H^+ \rightarrow C_6H_5N{=}N{-}OH_2^+ \rightarrow C_6H_5{-}N{\equiv}N^+ + H_2O}$

Step 2: Azo coupling

The diazonium ion is a weak electrophile and attacks only strongly activated rings, at the para position (ortho if para is blocked).

(a) With phenol (mildly alkaline medium, pH 9-10): phenol is converted into the more reactive phenoxide ion.

$\mathrm{C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{NaOH} p\text{-}HO{-}C_6H_4{-}N{=}N{-}C_6H_5 + HCl}$

The product is p-hydroxyazobenzene (orange dye).

(b) With aniline (weakly acidic medium, pH 4-5):

$\mathrm{C_6H_5N_2^+Cl^- + C_6H_5NH_2 \rightarrow p\text{-}H_2N{-}C_6H_4{-}N{=}N{-}C_6H_5 + HCl}$

The product is p-aminoazobenzene (yellow dye).

Mechanism of coupling (electrophilic aromatic substitution):

1. The para carbon of the phenoxide ion (or aniline) attacks the terminal nitrogen of $\mathrm{Ar{-}N{\equiv}N^+}$, forming a resonance-stabilised sigma complex (arenium ion).
2. Loss of $\mathrm{H^+}$ from the para carbon restores aromaticity, giving the azo compound $\mathrm{Ar{-}N{=}N{-}Ar'}$.

The extended conjugation through the $-\mathrm{N{=}N-}$ chromophore, with $-\mathrm{OH}$ or $-\mathrm{NH_2}$ as auxochrome, makes these compounds coloured.

Medium: a strongly acidic medium converts phenol or aniline into non-reactive forms (aniline to anilinium), and a strongly alkaline medium converts the diazonium ion into diazotate. Hence the pH must be controlled.

(a) Picric acid is a stronger acid than phenol.

Acid strength depends on the stability of the conjugate base. Picric acid is 2,4,6-trinitrophenol. The three nitro groups at the ortho and para positions are strongly electron-withdrawing by both $-I$ and $-M$ (resonance) effects. They:

1. polarise the O-H bond, so the proton is lost more easily, and
2. delocalise the negative charge of the picrate ion from oxygen into the ring and onto the oxygen atoms of the nitro groups, so the picrate ion is greatly stabilised.

In phenol there is no such group, so the phenoxide ion is much less stabilised.

$\text{Picric acid }(pK_a\approx0.4)\ \gg\ \text{phenol }(pK_a\approx10)$

(b) The C-OH bond in phenol is shorter than the C-OH bond in alcohols.

1. Resonance: in phenol, a lone pair on oxygen is conjugated with the $\pi$-electrons of the benzene ring: $\mathrm{Ar{-}\ddot{O}H \longleftrightarrow\ \text{structures with } C{=}\overset{+}{O}H \text{ and negative charge on ring carbons}}$ This gives the C-O bond partial double-bond character, which shortens it.
2. Hybridisation: in phenol, oxygen is attached to an $sp^2$ carbon (about 33% s-character), whereas in alcohols it is attached to an $sp^3$ carbon (25% s-character). The more s-character, the more tightly held and shorter the bond.

Hence the C-O bond length is about $1.36\ \text{\AA}$ in phenol, against about $1.43\ \text{\AA}$ in methanol.

GROUP "C" (PHYSICAL)

Comprehensive Question

(a) Transport number (2 marks)

The transport number (Hittorf number or transference number) of an ion is the fraction of the total current carried by that ion during electrolysis.

$t_+ = \frac{\text{current carried by cation}}{\text{total current}} = \frac{u_+}{u_+ + u_-}, \qquad t_- = \frac{u_-}{u_+ + u_-}$

where $u_+$ and $u_-$ are the ionic mobilities. Clearly $t_+ + t_- = 1$.

(b) Determination of transport number by the moving boundary method (4 marks)

Principle. When two electrolytes having a common ion are placed in contact in a vertical tube and a current is passed, the boundary between them moves. The distance moved by the boundary in a given time measures the transport number of the common ion.

Apparatus and setup.

• Leading solution — the electrolyte whose transport number is required, e.g. HCl, placed in the lower part of the tube.
• Indicator solution — an electrolyte sharing the common ion (Cl⁻) but having a slower cation, e.g. CdCl₂, placed above the HCl.
• Cathode at the top (in CdCl₂), anode at the bottom (in HCl). The sharp boundary is observed with a travelling microscope/telescope (it is visible due to the difference in refractive index).

Working. On passing current, H⁺ migrates upward and Cl⁻ downward. H⁺ ions enter the CdCl₂ region and the boundary moves upward. Since Cd²⁺ is slower than H⁺, it cannot overtake the H⁺ ions, so the boundary stays sharp and moves with the speed of the H⁺ ions.

Derivation. Let the boundary move a distance $x$ cm in time $t$ s when a current $I$ A is passed. Let $A$ be the area of cross-section (cm²) and $c$ the concentration of HCl in mol cm⁻³.

• Volume swept by the boundary $= xA$ cm³
• Moles of H⁺ that have crossed the boundary $= xAc$
• Charge carried by these H⁺ ions $= xAcF$ coulomb

But the total charge passed is $Q = It$, and the charge carried by H⁺ is $t_+It$. Hence

$t_+ \, It = xAcF \quad\Rightarrow\quad \boxed{t_+ = \frac{xAcF}{It}}$

If $c$ is taken in mol L⁻¹, then $c_{\text{mol cm}^{-3}} = c/1000$, so

$t_+ = \frac{xAcF}{1000\,It}$

Similarly, using an indicator with a common cation but a slower anion, the anionic boundary moves downward and

$t_- = \frac{x'AcF}{1000\,It}$

where $x'$ is its displacement. Since $t_+ + t_- = 1$, either value gives the other.

Conditions for a sharp boundary.

1. The two electrolytes must have a common ion.
2. The indicator ion must have lower mobility than the leading ion.
3. The concentration of the indicator solution must be suitably adjusted (Kohlrausch's adjustment).
4. Current and temperature must be kept constant.

Advantages. Accurate results; applicable to any concentration; unaffected by electrode reactions.

(c) Numerical (4 marks)

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

To find: $\Lambda^\circ(\text{CH}_3\text{COOH})$

Formula (Kohlrausch's law of independent migration of ions):

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

Substitution:

$\Lambda^\circ(\text{CH}_3\text{COOH}) = 426 + 91 - 126 = 391\ \text{S cm}^2\text{ mol}^{-1}$

Result: The molar conductance of acetic acid at infinite dilution is 391 S cm² mol⁻¹.

(Note: the third value in the question is printed as CH₃COOH; it must be read as CH₃COONa — sodium acetate — for the calculation to be meaningful.)

6.1 Hardy–Schulz rule

The coagulating power of an electrolyte depends mainly on the valency of the ion opposite in charge to the colloidal particles. Higher the valency of the oppositely charged ion, greater is its coagulating power. This is the Hardy–Schulz rule (also called Schulze–Hardy rule).

For As₂S₃ sol, the colloidal particles are negatively charged because they adsorb sulphide ions, $\mathrm{S^{2-}}$, from the medium. Therefore, coagulation is caused by cations.

• $\mathrm{KCl}$ supplies $\mathrm{K^+}$ ions, valency $+1$.
• $\mathrm{MgCl_2}$ supplies $\mathrm{Mg^{2+}}$ ions, valency $+2$.

According to Hardy–Schulz rule, coagulating power increases sharply with valency. The coagulating value is roughly inversely proportional to $z^6$, where $z$ is the valency of the oppositely charged ion. Hence $\mathrm{Mg^{2+}}$ is much more effective than $\mathrm{K^+}$. Hence $\mathrm{MgCl_2}$ is a better coagulant than $\mathrm{KCl}$ for negatively charged $\mathrm{As_2S_3}$ sol.

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6.2 Origin of charge on colloidal solution and zeta-potential

Colloidal particles acquire charge mainly by:

1. Preferential adsorption of ions from the dispersion medium.
   Example: $\mathrm{As_2S_3}$ sol adsorbs $\mathrm{S^{2-}}$ ions and becomes negatively charged.

2. Dissociation/ionisation of surface groups on colloidal particles.
   Example: protein sols contain $-\mathrm{COO^-}$ and $-\mathrm{NH_3^+}$ groups; their net charge depends on pH.

3. Frictional electrification in some cases, though it is less important.

Zeta-potential is the potential difference between the slipping plane (shear plane) and the bulk dispersion medium. It is also called electrokinetic potential. It determines the stability of a colloid. A high magnitude of zeta-potential means strong electrostatic repulsion between particles, so the colloid is stable. When zeta-potential becomes zero, coagulation occurs; this point is called the isoelectric point.

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6.3 Fluorescence and phosphorescence

Both are types of photoluminescence.

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6.4 Thermochemical vs photochemical reaction

Quantum yield of a photochemical reaction is the number of molecules reacted or product formed per photon absorbed.

$$\phi = \frac{\text{number of molecules reacted}}{\text{number of quanta absorbed}}$$

It may be less than, equal to, or greater than 1. For chain reactions, $\phi > 1$.

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6.5 Enzyme catalysis

Enzyme catalysis is catalysis in which enzymes act as biological catalysts and increase the rate of biochemical reactions by lowering activation energy.

Characteristics:

1. High specificity: an enzyme usually catalyses only one reaction or a particular type of substrate.
2. Very high efficiency: small amounts catalyse large amounts of substrate.
3. Mild conditions: work at ordinary temperature and pressure.
4. pH sensitive: each enzyme has an optimum pH.
5. Temperature sensitive: activity increases up to optimum temperature, then decreases; high temperature denatures enzymes.
6. Cofactors/coenzymes: many enzymes require non-protein cofactors or coenzymes.
7. Not consumed in the reaction; they remain unchanged at the end.

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6.6 Conductometric titration

Conductometric titration is a titration in which the electrical conductance of the solution is measured as titrant is added. The end point is obtained from the break or intersection in the conductance vs volume curve.

(a) HCl vs NaOH
Strong acid–strong base titration. Initially conductance is high due to fast $\mathrm{H^+}$ and $\mathrm{Cl^-}$ ions. On adding $\mathrm{NaOH}$, $\mathrm{H^+}$ is replaced by slower $\mathrm{Na^+}$:

$$\mathrm{H^+ + Cl^- + Na^+ + OH^- \to Na^+ + Cl^- + H_2O}$$

So conductance decreases linearly up to equivalence point. After equivalence, excess $\mathrm{OH^-}$ and $\mathrm{Na^+}$ increase conductance. The curve is V-shaped with a minimum at the equivalence point.

(b) CH₃COOH vs NaOH
Weak acid–strong base titration. Initially conductance is low because $\mathrm{CH_3COOH}$ is weakly ionised. On adding $\mathrm{NaOH}$, highly ionised $\mathrm{CH_3COONa}$ is formed, so conductance increases. After equivalence, excess $\mathrm{OH^-}$ causes a steeper increase. The curve has two straight-line portions; their intersection gives the end point.

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6.7 Cell potential

Cell:

$$\mathrm{Zn(s)\,|\,Zn^{2+}(0.01\,M)\,||\,Fe^{2+}(0.01\,M),Fe^{3+}(0.1\,M)\,|\,Pt}$$

Left half-cell is anode:

$$\mathrm{Zn \to Zn^{2+} + 2e^-}$$

Right half-cell is cathode:

$$\mathrm{Fe^{3+} + e^- \to Fe^{2+}}$$

Overall reaction:

$$\mathrm{Zn + 2Fe^{3+} \to Zn^{2+} + 2Fe^{2+}}$$

Here $n=2$.

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

Nernst equation:

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log Q$$

where

$$Q = \frac{[\mathrm{Zn^{2+}}][\mathrm{Fe^{2+}}]^2}{[\mathrm{Fe^{3+}}]^2} = \frac{(0.01)(0.01)^2}{(0.1)^2} = 1.0 \times 10^{-4}$$ $$E_{\text{cell}} = 1.53 - \frac{0.0591}{2}\log(10^{-4})$$ $$= 1.53 - 0.02955(-4) = 1.53 + 0.1182 = 1.6482\,\mathrm{V}$$ $$\boxed{E_{\text{cell}} \approx 1.648\,\mathrm{V}}$$

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6.8 Entropy in terms of randomness and unavailable energy

Entropy ($S$) is a state function that measures the degree of randomness or disorder of a system. The more ways the particles can be arranged (the larger the thermodynamic probability $W$), the greater the entropy, as in Boltzmann's relation $S = k\ln W$. Thus $S$(solid) $<$ $S$(liquid) $<$ $S$(gas), and entropy rises when a gas expands or a solid dissolves.

In terms of unavailable energy, for a reversible change $dS = \dfrac{\delta q_{rev}}{T}$. Of the heat absorbed by a system, only part can be converted into work; the rest, $T\Delta S$, is energy that is unavailable for work. The greater the entropy, the larger the share of the system's energy that is unavailable. For any spontaneous change in an isolated system, $\Delta S > 0$.

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6.9 Adiabatic reversible expansion

Given: $n = 2$ mol, $T_1 = 273$ K, $P_1 = 20$ atm, $P_2 = 2$ atm, $C_p = \tfrac{5}{2}R$.

$C_v = C_p - R = \tfrac{3}{2}R = 12.47\ \mathrm{J\,K^{-1}\,mol^{-1}}$, so $\gamma = \dfrac{C_p}{C_v} = \dfrac{5}{3}$.

For a reversible adiabatic change of an ideal gas:

$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(\gamma-1)/\gamma} = \left(\frac{2}{20}\right)^{0.4}$

$T_2 = 273 \times (0.1)^{0.4} = 273 \times 0.398 \approx 109\ \mathrm{K}\ (\approx -164\,^\circ\mathrm{C})$

Work done by the gas (since $q = 0$, $W = -\Delta U$):

$W = nC_v(T_1 - T_2) = 2 \times 12.47 \times (273 - 108.7) \approx 4.1\times10^{3}\ \mathrm{J}$

Final temperature $\approx 109$ K; work done by the gas $\approx 4.1$ kJ.

The Haloform reaction (often written as "haloform reaction") is the reaction in which a methyl ketone (or acetaldehyde, or any compound that can be oxidised to a methyl ketone) is treated with a halogen in the presence of a base to give a haloform (CHX₃) and the carboxylate salt of the remaining part of the molecule.

General reaction:

$\mathrm{R-CO-CH_{3} + 3X_{2} + 4NaOH \rightarrow R-COONa + CHX_{3} + 3NaX + 3H_{2}O}$

where X = Cl, Br, or I. The haloform formed is chloroform (CHCl₃), bromoform (CHBr₃) or iodoform (CHI₃). The iodoform test (yellow precipitate of CHI₃) is used to detect methyl ketones and alcohols that give methyl ketones on oxidation.

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Mechanism

The reaction proceeds in three stages.

Step 1 — Base-catalysed halogenation of the methyl group (three times)

The base removes an α-hydrogen from the methyl group to give an enolate, which attacks the halogen:

$\mathrm{R-CO-CH_{3} + OH- \rightleftharpoons R-CO-CH_{2}- + H_{2}O}$

$\mathrm{R-CO-CH_{2}- + X_{2} \rightarrow R-CO-CH_{2}X + X-}$

This repeats until all three α-hydrogens of the methyl group are replaced:

$\mathrm{R-CO-CHX_{2} + OH- \rightleftharpoons R-CO-CX_{2}- + H_{2}O}$

$\mathrm{R-CO-CX_{2}- + X_{2} \rightarrow R-CO-CX_{3} + X-}$

The three electron-withdrawing halogen atoms make the remaining α-hydrogen strongly acidic, so the final enolate is formed very readily.

Step 2 — Nucleophilic attack of OH⁻ on the carbonyl carbon

$\mathrm{R-CO-CX_{3} + OH- \rightarrow R-C(OH)(CX_{3})-O-}$

(tetrahedral intermediate)

Step 3 — Cleavage of the C–C bond and formation of haloform

The CX₃ group is a good leaving group (stabilised by the three halogens). The C–C bond breaks, expelling the CX₃ carbanion and giving the carboxylate:

$\mathrm{R-C(OH)(CX_{3})-O- \rightarrow R-COO- + CX_{3}-}$

The carbanion CX₃⁻ abstracts a proton from water (or from RCOOH) to give the haloform:

$\mathrm{CX_{3}- + H_{2}O \rightarrow CHX_{3} + OH-}$

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Utility in organic chemistry

1. Detection and identification of methyl ketones and acetaldehyde — the iodoform test gives a yellow precipitate of CHI₃ with a characteristic melting point (119 °C), confirming the presence of a –COCH₃ group.

2. Detection of alcohols — ethanol, isopropanol and other alcohols containing the CH₃–CH(OH)– group are first oxidised to the corresponding methyl ketone and then give a positive iodoform test. Thus it distinguishes ethanol from methanol.

3. Preparation of chloroform and bromoform — haloforms are obtained from acetone or ethanol on a laboratory/industrial scale.

4. Synthetic use in degradation — the reaction converts a methyl ketone into a carboxylic acid with one carbon less, so it is used to shorten a carbon chain and to locate the position of a keto group in a molecule.

5. Distinguishing methyl ketones from other ketones — only compounds with a $\mathrm{CH_3CO-}$ group respond. Acetone and acetophenone ($\mathrm{C_6H_5COCH_3}$) give a yellow precipitate of $\mathrm{CHI_3}$, whereas benzophenone ($\mathrm{C_6H_5COC_6H_5}$) and diethyl ketone (3-pentanone) do not.

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 is not nucleophilic enough to attack the carbonyl carbon directly. An acid catalyst overcomes this in two ways:

1. Activation of the carbonyl group: The acid protonates the carbonyl oxygen, giving a resonance-stabilised oxonium ion (C=O⁺H). The positive charge on oxygen strongly withdraws electron density from the carbonyl carbon, making it far more electrophilic (more susceptible to nucleophilic attack).

2. Generation of a stronger nucleophile (in some cases): For neutral nucleophiles like alcohols, protonation of the nucleophile is avoided, but the acid medium also helps convert poor nucleophiles into better ones (e.g. an alcohol is used in excess with dry HCl to give the protonated species that attacks readily).

Thus, by increasing the electrophilicity of the carbonyl carbon (and, where required, by activating the nucleophile), the acid catalyst lowers the activation energy of the rate-determining nucleophilic attack and hence catalyses the addition reaction. (Note: strong base, on the other hand, catalyses the reaction by converting the nucleophile into a more powerful anionic nucleophile, e.g. CN⁻ from HCN.)

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

Fehling's solution (copper(II) tartrate in alkaline medium) is a mild oxidising agent that oxidises only aliphatic aldehydes; it does not oxidise aromatic aldehydes.

• Propionaldehyde (CH₃CH₂CHO) is an aliphatic aldehyde. It has no group that interferes with oxidation, so it readily reduces Cu²⁺ to red-brown Cu₂O:

$\mathrm{CH_{3}CH_{2}CHO + 2Cu^2+ + 5OH- \rightarrow CH_{3}CH_{2}COO- + Cu_{2}O \downarrow + 3H_{2}O}$

• Benzaldehyde (C₆H₅CHO) is an aromatic aldehyde. The –CHO group is directly attached to the benzene ring, and the ring exerts a strong −I and +M (resonance) electron-releasing effect that increases the electron density on the carbonyl carbon. This makes the C–H bond of the –CHO group stronger and the carbonyl carbon less electrophilic, so benzaldehyde is not easily oxidised by the mild oxidising agent Fehling's solution. (It also has no α-hydrogen, and the aromatic ring stabilises the aldehyde, further resisting oxidation.)

Hence, under the mild oxidising conditions of Fehling's solution, only propionaldehyde (aliphatic) is oxidised, while benzaldehyde (aromatic) is not.

Preparation of aromatic carboxylic acids by oxidation of alkylbenzene

Alkylbenzenes having at least one benzylic hydrogen are oxidised to aromatic carboxylic acids. The entire side chain is converted to a –COOH group, regardless of its length.

Reagents: alkaline KMnO₄ (KMnO₄ + NaOH) with heating, or hot dilute HNO₃, or chromic acid (Na₂Cr₂O₇/H₂SO₄).

General reaction:

$\text{Ar–R} \xrightarrow[\text{heat}]{\text{KMnO}_4/\text{OH}^-} \text{Ar–COOH}$

Examples:

$\text{C}_6\text{H}_5\text{CH}_3 \xrightarrow[\text{heat}]{\text{KMnO}_4/\text{OH}^-} \text{C}_6\text{H}_5\text{COOH}$

$\text{C}_6\text{H}_5\text{CH}_2\text{CH}_3 \xrightarrow[\text{heat}]{\text{KMnO}_4/\text{OH}^-} \text{C}_6\text{H}_5\text{COOH}$

$\text{C}_6\text{H}_5\text{CH}(\text{CH}_3)_2 \xrightarrow[\text{heat}]{\text{KMnO}_4/\text{OH}^-} \text{C}_6\text{H}_5\text{COOH}$

Mechanism (in brief): The benzylic C–H bond is attacked first, giving a benzylic radical/carbanion which is oxidised to the benzyl alcohol, then to benzaldehyde, and finally to the carboxylic acid. In alkaline medium the acid is obtained as its salt (benzoate), so the reaction mixture is acidified (HCl/H₂SO₄) at the end to liberate the free acid.

Conditions and limitations:

• Only alkylbenzenes with benzylic hydrogen react. tert-Butylbenzene, C₆H₅C(CH₃)₃, has no benzylic H and is not oxidised.
• Fused-ring compounds behave similarly: naphthalene → phthalic acid; phenanthrene → phenanthrenequinone.
• Strong oxidants may also attack the ring if the ring is activated (e.g. phenols), so milder reagents are chosen in such cases.

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Comparison of acidities: aromatic vs aliphatic carboxylic acids

Reason: In benzoic acid the –COOH group is attached to an sp² carbon, which is more electronegative than the sp³ carbon of acetic acid. This exerts a –I (electron-withdrawing) effect that stabilises the carboxylate anion and increases acidity. The carboxylate charge itself is not delocalised into the ring, so the inductive effect is the main reason. Hence benzoic acid (pKa 4.20) is stronger than acetic acid (pKa 4.76). Remember: lower pKa means stronger acid. Among substituted benzoic acids, –NO₂ and –Cl increase acidity (pKa of p-nitrobenzoic acid ≈ 3.44), while –CH₃ and –OCH₃ decrease it (pKa of p-methoxybenzoic acid ≈ 4.47).

Summary: Aromatic carboxylic acids are prepared by oxidising alkylbenzenes with KMnO₄/OH⁻ or hot HNO₃; they are stronger acids than aliphatic carboxylic acids because the sp² carbon and the ring withdraw electron density, stabilising the carboxylate ion, and this effect is modified by substituents on the ring.

a. Carboxylic acids are usually stronger acids than phenols.

Acid strength depends on the stability of the conjugate base: the more stable the anion left after losing $\mathrm{H^+}$, the stronger the acid.

• In a carboxylic acid, the conjugate base $\mathrm{RCOO^-}$ is stabilised by resonance in which the negative charge is shared equally by two equivalent oxygen atoms:

$\mathrm{R-C(=O)-O^- \;\longleftrightarrow\; R-C(-O^-)=O}$

The two structures have equal energy and the charge sits on electronegative oxygen in both, so the carboxylate ion is strongly stabilised.

• In phenol, the phenoxide ion $\mathrm{C_6H_5O^-}$ is also resonance-stabilised, but the negative charge is spread onto the ortho and para carbon atoms of the ring. These structures are not equivalent, put the charge on less electronegative carbon, and disrupt the aromatic sextet, so the stabilisation is only partial.

Because the carboxylate ion is stabilised much better than the phenoxide ion, carboxylic acids lose $\mathrm{H^+}$ more easily: $\mathrm{p}K_a \approx 4\text{–}5$ for carboxylic acids against about $10$ for phenols.

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b. Most carboxylic acids exist as cyclic dimers.

The $\mathrm{-COOH}$ group has a polar $\mathrm{O-H}$ bond (hydrogen carries a partial positive charge) and a polar $\mathrm{C=O}$ group (oxygen carries a partial negative charge). Two molecules therefore hold each other by two intermolecular hydrogen bonds, the $\mathrm{O-H}$ of each molecule bonding to the $\mathrm{C=O}$ oxygen of the other. This closes a planar eight-membered ring, the cyclic dimer:

         O···H–O
       //        \
   R–C            C–R
       \        //
        O–H···O

Two hydrogen bonds per pair make the association strong and energetically favourable. As a result, in the vapour state and in non-polar solvents such as benzene or $\mathrm{CCl_4}$, carboxylic acids exist mostly as dimers, and their molecular mass found by freezing-point depression or vapour density is nearly double the formula mass. In water the acid forms hydrogen bonds with the solvent instead, so the dimers break up.

Part 1: 1,3,5-Tribromobenzene from p-nitroaniline

All three bromines are meta to one another, but bromine is ortho/para-directing, so direct bromination of benzene cannot give this compound. The strongly activating $\mathrm{-NH_2}$ group is used to place the bromines and is then removed.

1. Bromination. p-Nitroaniline $\xrightarrow{\ \mathrm{Br_2/CH_3COOH}\ }$ 2,6-dibromo-4-nitroaniline. The $\mathrm{-NH_2}$ directs both bromines to its ortho positions (the para position is blocked by $\mathrm{-NO_2}$).
2. Deamination. Diazotise with $\mathrm{NaNO_2/HCl}$ at 0–5 °C, then warm with hypophosphorous acid ($\mathrm{H_3PO_2}$), which replaces $\mathrm{N_2^+}$ by H: this gives 1,3-dibromo-5-nitrobenzene (3,5-dibromonitrobenzene).
3. Reduction. $\mathrm{Fe/HCl}$ (or $\mathrm{Sn/HCl}$) reduces $\mathrm{-NO_2}$ to $\mathrm{-NH_2}$: 3,5-dibromoaniline.
4. Sandmeyer reaction. Diazotise with $\mathrm{NaNO_2/HCl}$ (0–5 °C), then treat with $\mathrm{CuBr/HBr}$: 1,3,5-tribromobenzene.

$\text{p-nitroaniline}\xrightarrow{\mathrm{Br_2}}\text{2,6-dibromo-4-nitroaniline}\xrightarrow{\mathrm{1)\,HNO_2;\ 2)\,H_3PO_2}}\text{3,5-dibromonitrobenzene}\xrightarrow{\mathrm{Fe/HCl}}\text{3,5-dibromoaniline}\xrightarrow{\mathrm{1)\,HNO_2;\ 2)\,CuBr}}\text{1,3,5-tribromobenzene}$

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Part 2: p-Chloroaniline from chlorobenzene

1. Nitration. Chlorobenzene $\xrightarrow{\ \mathrm{conc.\ HNO_3/H_2SO_4},\ 50\text{–}60\,^\circ\mathrm{C}\ }$ a mixture of p-nitrochlorobenzene (major) and o-nitrochlorobenzene (minor), because $\mathrm{-Cl}$ is ortho/para-directing. The para isomer is separated by crystallisation, since it has the higher melting point and lower solubility.
2. Reduction. p-Nitrochlorobenzene $\xrightarrow{\ \mathrm{Sn/HCl}\ \text{(or Fe/HCl)}\ }$ p-chloroaniline.

a. Aniline is a much weaker base than cyclohexylamine.

Basicity depends on the availability of the lone pair of electrons on the nitrogen atom for protonation. In cyclohexylamine, the nitrogen is bonded to an sp³-hybridised carbon of the cyclohexane ring. The lone pair on nitrogen is localised and freely available, so cyclohexylamine readily accepts a proton and behaves as a strong base (pK_b ≈ 3.3).

In aniline, the nitrogen lone pair is delocalised into the benzene ring by resonance. The lone pair is donated into the π-system of the ring, giving resonance structures in which the nitrogen acquires a positive charge and negative charge develops at the ortho and para positions:

        NH2            NH2(+)          NH2(+)      NH2(+)
         |               |               |            |
    C6H5 ring  <-->  ring with (-) at ortho/para positions

Because the lone pair is partly engaged in resonance with the ring, it is not fully available for protonation. This makes aniline a much weaker base than cyclohexylamine (pK_b ≈ 9.4). The sp² character of the C–N bond in aniline (greater s-character) also holds the lone pair more tightly, further reducing basicity. Hence aniline is a much weaker base than cyclohexylamine.

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b. It is not desirable to nitrate aniline directly to prepare nitroaniline.

Direct nitration of aniline with a mixture of concentrated HNO₃ and concentrated H₂SO₄ is not used because:

1. Formation of anilinium ion: In strongly acidic medium, aniline is protonated to form the anilinium ion (–NH₃⁺). The –NH₃⁺ group is strongly electron-withdrawing and meta-directing. As a result, a large amount of m-nitroaniline is formed along with o- and p-nitroaniline. Thus the product is a mixture, and the yield of the desired o-/p-nitroaniline is poor.

2. Oxidation of aniline: Concentrated HNO₃ is a strong oxidising agent. It oxidises aniline to dark-coloured, tarry oxidation products (e.g., quinones and polymeric material), which contaminate the product and reduce the yield.

Therefore, direct nitration is not suitable. Instead, aniline is first protected by acetylation (converting it to acetanilide, which is less reactive towards oxidation and not protonated in acid), then nitrated, and finally the acetyl group is removed by hydrolysis to obtain pure o- and p-nitroaniline.

a. o-Nitrophenol is steam volatile but m- and p-nitrophenols are not.

In o-nitrophenol, the –OH group and the –NO₂ group are on adjacent carbon atoms, so the –OH group forms an intramolecular hydrogen bond with the oxygen of the –NO₂ group (chelation). Because the hydrogen is already bonded internally, the molecule cannot form intermolecular hydrogen bonds with neighbouring molecules. As a result, the molecules are held together only by weak van der Waals forces, the intermolecular attraction is low, and the compound has a relatively low boiling point and appreciable vapour pressure — hence it is steam volatile.

In m- and p-nitrophenols, the –OH and –NO₂ groups are too far apart to form intramolecular hydrogen bonds. Instead, the –OH group forms intermolecular hydrogen bonds with the –OH groups of neighbouring molecules, producing a strongly associated, polymeric network. This raises the boiling point and lowers the vapour pressure sharply, so these isomers are not steam volatile.

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b. p-Nitrophenol is a stronger acid than phenol.

Acidity depends on the stability of the conjugate base (the phenoxide ion) — the more stable the anion, the stronger the acid.

In p-nitrophenol, the –NO₂ group at the para position is a strong electron-withdrawing group (it has a high −I effect and, more importantly, a strong −M or resonance effect). It withdraws electron density from the ring and, in particular, delocalises the negative charge of the phenoxide ion onto its own oxygen atoms through resonance:

$\text{O}_2\text{N–C}_6\text{H}_4\text{–O}^- \;\longleftrightarrow\; {}^-\text{O–C}_6\text{H}_4\text{–N}^+(\text{=O})\text{O}^-$

This spreads the negative charge over several atoms and greatly stabilises the conjugate base. The O–H bond therefore ionises more readily, giving p-nitrophenol a much lower pKₐ (≈ 7.15) than phenol (pKₐ ≈ 10).

In phenol there is no such electron-withdrawing substituent; the phenoxide ion is stabilised only by delocalisation into the ring, so phenol is a weaker acid. Hence p-nitrophenol is a stronger acid than phenol.

First Alternative

1. Reference electrode

A reference electrode is an electrode whose potential is known, constant and reproducible under fixed conditions. It is used as a standard against which the potential of another electrode is measured. Common examples are the standard hydrogen electrode (SHE), calomel electrode and Ag/AgCl electrode.

2. Measurement of standard electrode potential of zinc using SHE

The cell is set up as:

$$\mathrm{Zn(s)\;|\;Zn^{2+}(1\,M)\;||\;H^+(1\,M)\;|\;H_2(g,1\,atm)\;|\;Pt}$$

Zinc has a negative electrode potential, so it acts as the anode and SHE acts as the cathode.

Anode half-reaction:

$$\mathrm{Zn \rightarrow Zn^{2+} + 2e^-}$$

Cathode half-reaction:

$$\mathrm{2H^+ + 2e^- \rightarrow H_2}$$

Overall cell reaction:

$$\mathrm{Zn + 2H^+ \rightarrow Zn^{2+} + H_2}$$

The measured cell potential is:

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$

Since (E^\circ_{\text{SHE}} = 0),

$$E^\circ_{\text{cell}} = 0 - E^\circ_{\mathrm{Zn^{2+}/Zn}}$$

Experimentally,

$$E^\circ_{\text{cell}} = +0.76\ \mathrm{V}$$

Therefore,

$$E^\circ_{\mathrm{Zn^{2+}/Zn}} = -0.76\ \mathrm{V}$$

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3. For the cell

$$\mathrm{Zn(s)\;|\;Zn^{2+}(0.01)\;||\;Fe^{2+}(0.01),\;Fe^{3+}(0.1)\;|\;Pt}$$

Cell reaction

At anode:

$$\mathrm{Zn \rightarrow Zn^{2+} + 2e^-}$$

At cathode:

$$\mathrm{Fe^{3+} + e^- \rightarrow Fe^{2+}}$$

Multiplying the cathode reaction by 2 to balance electrons:

$$\mathrm{2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+}}$$

Overall cell reaction:

$$\boxed{\mathrm{Zn + 2Fe^{3+} \rightarrow Zn^{2+} + 2Fe^{2+}}}$$

Here, (n = 2).

Standard cell potential

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

Cell potential by Nernst equation

For the reaction,

$$Q = \frac{[\mathrm{Zn^{2+}}][\mathrm{Fe^{2+}}]^2}{[\mathrm{Fe^{3+}}]^2}$$

Substituting the given concentrations:

$$Q = \frac{(0.01)(0.01)^2}{(0.1)^2}$$ $$Q = \frac{(0.01)(0.0001)}{0.01} = 1 \times 10^{-4}$$

Using the Nernst equation at (25^\circ\mathrm{C}),

$$E = E^\circ - \frac{0.0591}{n}\log Q$$ $$E = 1.53 - \frac{0.0591}{2}\log(1 \times 10^{-4})$$ $$E = 1.53 - \frac{0.0591}{2}(-4)$$ $$E = 1.53 + 0.1182 = 1.6482\ \mathrm{V}$$ $$\boxed{E \approx 1.648\ \mathrm{V}}$$

Work done by the cell

$$\Delta G = -nFE$$ $$\Delta G = -2 \times 96500 \times 1.6482$$ $$\Delta G = -318102.6\ \mathrm{J} = -318.1\ \mathrm{kJ}$$

Maximum work done by the cell is:

$$W_{\max} = -\Delta G = nFE$$ $$W_{\max} = 2 \times 96500 \times 1.6482 = 318102.6\ \mathrm{J}$$ $$\boxed{W_{\max} \approx 318.1\ \mathrm{kJ}}$$

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OR Alternative

1. Adiabatic expansion of an ideal gas

An adiabatic expansion is a process in which the gas expands without exchange of heat with the surroundings, i.e.

$$q = 0$$

For an ideal gas, the internal energy change is:

$$dU = nC_V dT$$

From the first law,

$$dU = q + w$$

For reversible expansion,

$$w = -P\,dV$$

So,

$$nC_V dT = -P\,dV$$

Using (PV = nRT),

$$P = \frac{nRT}{V}$$

Thus,

$$nC_V dT = -\frac{nRT}{V}dV$$ $$C_V \frac{dT}{T} = -R \frac{dV}{V}$$

Integrating between states 1 and 2,

$$C_V \ln\frac{T_2}{T_1} = -R \ln\frac{V_2}{V_1}$$ $$\ln\frac{T_2}{T_1} = -\frac{R}{C_V}\ln\frac{V_2}{V_1}$$ $$T V^{R/C_V} = \text{constant}$$

Since,

$$C_P - C_V = R$$ $$\frac{R}{C_V} = \frac{C_P - C_V}{C_V} = \gamma - 1$$

Therefore,

$$\boxed{T V^{\gamma - 1} = \text{constant}}$$

Using (T \propto PV),

$$(PV)V^{\gamma - 1} = \text{constant}$$ $$\boxed{P V^\gamma = \text{constant}}$$

This is the required pressure-volume relation for adiabatic expansion of an ideal gas.

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2. Numerical

Given:

$$n = 2\ \mathrm{mol}$$ $$T_1 = 25^\circ\mathrm{C} = 298\ \mathrm{K}$$ $$P_1 = 10\ \mathrm{atm}$$ $$P_2 = 2\ \mathrm{atm}$$ $$C_P = \frac{7R}{2}$$

For an ideal gas,

$$C_V = C_P - R = \frac{7R}{2} - R = \frac{5R}{2}$$ $$\gamma = \frac{C_P}{C_V} = \frac{7R/2}{5R/2} = \frac{7}{5} = 1.4$$

For adiabatic reversible expansion,

$$T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{\frac{\gamma - 1}{\gamma}}$$ $$T_2 = 298 \left(\frac{2}{10}\right)^{\frac{0.4}{1.4}}$$ $$T_2 = 298 (0.2)^{2/7}$$ $$T_2 = 188.15\ \mathrm{K}$$ $$\boxed{T_2 \approx 188\ \mathrm{K}}$$

Maximum work done by the gas in adiabatic reversible expansion is:

$$W_{\max} = nC_V(T_1 - T_2)$$ $$W_{\max} = 2 \times \frac{5R}{2} \times (298 - 188.15)$$ $$W_{\max} = 5R \times 109.85$$ $$W_{\max} = 5 \times 8.314 \times 109.85$$ $$W_{\max} = 4566\ \mathrm{J}$$ $$\boxed{W_{\max} \approx 4.57\ \mathrm{kJ}}$$

So, the final temperature is approximately 188 K and the maximum work done by the gas is approximately 4.57 kJ.

TU BSc 2nd Year — Short Answer Questions

6.1 Colloidal Solution; Optical and Kinetic Properties

Colloidal solution: A heterogeneous, two-phase system in which particles of one substance (dispersed phase), of size 1 nm – 1000 nm ($10^{-9}$ to $10^{-6}$ m), are distributed throughout another substance (dispersion medium). It is intermediate between a true solution and a suspension. Examples: As₂S₃ sol, milk, smoke, blood.

Optical properties

1. Tyndall effect (Faraday–Tyndall effect): When a beam of light is passed through a colloidal solution kept in a dark room, the path of the beam becomes visible because the colloidal particles scatter light. The particle size is comparable to the wavelength of light. True solutions do not show this; suspensions scatter only weakly. The scattered light is polarised.
2. Ultramicroscope: Using the Tyndall cone, individual colloidal particles appear as bright spots against a dark background, so their number and size can be counted.
3. Colour: Colloidal solutions are often coloured due to scattering and absorption — gold sol is red, As₂S₃ sol is yellow.

Kinetic properties

1. Brownian movement: Random zig-zag motion of colloidal particles caused by unequal bombardment by molecules of the dispersion medium. It is greater for smaller particles and at higher temperature, and it keeps the sol stable against settling.
2. Diffusion: Colloidal particles diffuse from higher to lower concentration, but very slowly because of their large size (Fick's law).
3. Osmotic pressure: Colloidal solutions show osmotic pressure, but it is very low compared with true solutions of the same concentration, since the number of particles is small. It is used to find molecular mass by van't Hoff's equation.
4. Sedimentation: Particles settle very slowly under gravity; an ultracentrifuge is used to sediment them and determine particle size and mass.

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6.2 Origin of Charge on Colloidal Solutions; Zeta Potential

Origin of charge

1. Preferential adsorption of ions (Paneth–Fajans rule): The colloidal particle adsorbs that ion from the medium which is common to it. As₂S₃ sol adsorbs S²⁻ (or HS⁻) and becomes negatively charged; Fe(OH)₃ sol adsorbs Fe³⁺ and becomes positively charged.
2. Dissociation of surface molecules: Soap (RCOONa) dissociates to give RCOO⁻, which forms the negatively charged micelle, Na⁺ acting as counter-ion. Proteins behave similarly.
3. Ionisation of surface groups: In SiO₂ sol, surface Si–OH groups ionise to give SiO⁻ and H⁺, leaving a negative charge.
4. Isomorphic replacement: In clay/soil colloids, Si⁴⁺ in the lattice is replaced by Al³⁺ (or Al³⁺ by Mg²⁺) of lower charge, leaving a net negative charge.
5. Capture of electrons from the medium in some cases.

Zeta potential ($\zeta$)

The colloidal particle is surrounded by a double layer: a tightly held fixed (Stern) layer of counter-ions and a diffuse layer. The potential difference between the fixed layer and the bulk of the dispersion medium (i.e., at the slipping/shear plane) is called the electrokinetic or zeta potential.

• It is the effective potential that decides the stability of the sol: the greater the magnitude of $\zeta$, the greater the repulsion between particles and the more stable the sol.
• When $\zeta = 0$ (the isoelectric point), the particles coagulate.
• It is measured by electrophoresis.

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6.3 Homogeneous Catalysis; Intermediate Compound Formation Theory

Homogeneous catalysis: When the catalyst is in the same physical phase (state) as the reactants, the catalysis is called homogeneous. Examples: esterification catalysed by H₂SO₄ (liquid), hydrolysis of ethyl acetate by acid, oxidation of SO₂ by NO gas.

Intermediate compound formation theory

The catalyst first combines with one of the reactants to form an unstable intermediate compound, which then reacts with the other reactant (or decomposes) to give the product and regenerates the catalyst. The intermediate path has a lower activation energy than the uncatalysed path, so the reaction becomes faster.

Example — oxidation of SO₂ in the lead chamber process (catalyst NO):

$\mathrm{2NO + O_{2} \rightarrow 2NO_{2}}\quad\text{(intermediate)}$ $\mathrm{2NO_{2} + 2SO_{2} \rightarrow 2SO_{3} + 2NO}\quad\text{(catalyst regenerated)}$

Overall: $\mathrm{2SO_{2} + O_{2} \rightarrow 2SO_{3}}$

Example — hydrolysis of ethyl acetate (catalyst H⁺):

$\mathrm{CH_{3}COOC_{2}H_{5} + H+ \rightarrow [CH_{3}COOC_{2}H_{5}H]+ \rightarrow CH_{3}COOH + C_{2}H_{5}OH + H+}$

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6.4 Enzyme Catalysis and its Characteristics

Enzyme catalysis: Catalysis in which enzymes — complex nitrogenous organic (protein) substances produced by living cells — act as catalysts. The substance acted upon is the substrate. The enzyme combines with the substrate to form an enzyme–substrate complex, which decomposes to give the product and free enzyme. Example: invertase catalyses the hydrolysis of sucrose; urease catalyses the hydrolysis of urea.

Characteristics

1. High specificity: Each enzyme catalyses only one reaction or a group of closely related reactions (lock-and-key model). Urease acts only on urea.
2. High efficiency: Very small amounts catalyse large amounts; the turnover number is very high (catalase ≈ 4 × 10⁷ molecules per second).
3. Optimum temperature: Maximum activity at about 37 °C; above 50–60 °C the enzyme is denatured and loses activity.
4. Optimum pH: Each enzyme has a definite pH of maximum activity — pepsin pH ≈ 2, trypsin pH ≈ 8.
5. Sensitivity to poisons: Heavy metal ions (Ag⁺, Hg²⁺, Pb²⁺), cyanide, etc. deactivate enzymes.
6. Reversibility: Enzymes catalyse reversible reactions and remain chemically unchanged at the end.
7. Colloidal nature: Enzymes are colloids; activity depends on surface area.
8. They speed up reactions by lowering the activation energy.

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6.5 Quantum Yield; Reasons for High and Low Values

Quantum yield ($\phi$): The number of molecules of reactant consumed (or product formed) per quantum of light absorbed.

$\phi = \frac{\text{number of molecules reacted}}{\text{number of quanta absorbed}}$

Reasons for high quantum yield ($\phi > 1$)

• Chain reactions: One absorbed photon produces a reactive species (atom/radical) that starts a chain of steps, so many molecules react per photon. Example: $\mathrm{H_{2} + Cl_{2} \rightarrow 2HCl}$, $\phi \approx 10^6$.
• Secondary reactions following the primary absorption also consume extra molecules.

Reasons for low quantum yield ($\phi < 1$)

• Deactivation of excited molecules by collision, fluorescence, phosphorescence or internal conversion, so the absorbed energy is wasted as heat/light.
• Recombination of primary products before they can react. Example: $\mathrm{H_{2} + Br_{2} \rightarrow 2HBr}$, $\phi \approx 0.01$.
• The excited molecule may not react at all, or the primary products may undergo a reverse reaction.

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6.6 Kohlrausch's Law; Molar Conductance of Acetic Acid at Infinite Dilution

Kohlrausch's law of independent migration of ions: At infinite dilution each ion migrates independently of the other ions present, so the limiting molar conductance of an electrolyte is the sum of the limiting ionic conductances of its cation and anion:

$\Lambda_m^\infty = \lambda_+^\infty + \lambda_-^\infty$

Calculation. The given values 426, 126 and 91 S cm² mol⁻¹ belong to HCl, NaCl and sodium acetate (CH₃COONa), which are the three strong electrolytes needed to build acetic acid:

$\Lambda^\infty(\mathrm{CH_3COOH}) = \Lambda^\infty(\mathrm{HCl}) + \Lambda^\infty(\mathrm{CH_3COONa}) - \Lambda^\infty(\mathrm{NaCl})$

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

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6.7 Antimony–Antimony Oxide Electrode and pH

Sb–Sb₂O₃ electrode: a metal–metal oxide electrode made of a rod of antimony coated with its sparingly soluble oxide, dipped in the solution whose pH is required: $\mathrm{Sb\,|\,Sb_2O_3(s)\,|\,H^+}$.

Electrode reaction:

$\mathrm{Sb_2O_3 + 6H^+ + 6e^- \rightleftharpoons 2Sb + 3H_2O}$

Electrode potential (25 °C):

$E_{\mathrm{Sb}} = E^\circ - \frac{0.0591}{1}\,\mathrm{pH} = E^\circ - 0.0591\,\mathrm{pH}$

Measurement of pH: The electrode is combined with a reference electrode of known potential, such as the saturated calomel electrode (SCE, $E = 0.2415$ V), to form the cell

$\mathrm{Sb\,|\,Sb_2O_3\,|\,test\ solution\,\|\,KCl(sat)\,|\,Hg_2Cl_2\,|\,Hg}$

The emf $E_{cell} = E_{\mathrm{SCE}} - E_{\mathrm{Sb}}$ is measured with a potentiometer, and

$\mathrm{pH} = \frac{E_{cell} - E_{\mathrm{SCE}} + E^\circ}{0.0591}$

In practice the electrode is first calibrated with a buffer of known pH. It is a robust electrode, useful where a glass electrode is unsuitable (for example in hydrofluoric acid solutions or for rough field work), but it works well only in the pH range of about 2 to 10 and is disturbed by oxidising and reducing agents.

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6.8 Second Law of Thermodynamics

1. Kelvin–Planck statement: It is impossible to construct a machine working in a cycle that absorbs heat from a single reservoir and converts all of it into work without any other effect.
2. Clausius statement: It is impossible for heat to flow spontaneously from a colder body to a hotter body without external work being done. A refrigerator therefore needs work to pump heat from cold to hot.

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6.9 Spontaneity of Combustion of Methane

$\Delta G = \Delta H - T\Delta S$

Given $\Delta H = -890$ kJ, $\Delta S = +242$ J K⁻¹ = 0.242 kJ K⁻¹ and $T = 298$ K:

$\Delta G = -890 - (298 \times 0.242) = -890 - 72.1 = -962.1\ \mathrm{kJ}$

Since $\Delta G$ is negative (and both the enthalpy and entropy terms favour the reaction), the combustion of methane at 298 K is spontaneous.

Why Acidity Must Be Carefully Controlled in Addition Reactions of Carbonyl Compounds with Ammonia Derivatives

Ammonia derivatives (such as hydroxylamine, hydrazine, semicarbazide, and 2,4-dinitrophenylhydrazine) react with aldehydes and ketones to form imines, oximes, hydrazones, and semicarbazones by nucleophilic addition–elimination. The acidity of the medium must be carefully controlled because both too little and too much acid slow down or prevent the reaction.

Role of Acid

The carbonyl carbon is electrophilic but not electrophilic enough for a weak nucleophile like the nitrogen of an ammonia derivative. A small amount of acid protonates the carbonyl oxygen:

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

This increases the positive character of the carbonyl carbon and makes it more susceptible to nucleophilic attack by the ammonia derivative.

Why Excess Acid Is Harmful

If the medium is too acidic, the ammonia derivative itself gets protonated:

$\mathrm{NH_{2}-NH_{2} + H+ \rightarrow NH_{3}+-NH_{2}}$

The protonated nitrogen has no lone pair available for nucleophilic attack, so it loses its nucleophilic character. The reagent is thus converted into an unreactive species and the reaction becomes very slow or stops.

Why Too Little Acid Is Also Harmful

If the medium is not acidic enough, the carbonyl carbon remains only weakly electrophilic. The weak nucleophile cannot attack it effectively, so the rate of addition is very low. Moreover, the final step — dehydration of the intermediate carbinolamine to give the C=N bond — also requires acid catalysis:

$\mathrm{R_{2}C(OH)-NH-NH_{2} \rightarrow [H+] R_{2}C=N-NH_{2} + H_{2}O}$

Without sufficient acid, this dehydration step is slow and the reaction does not go to completion.

Conclusion

The reaction therefore needs an optimum pH (mildly acidic, about pH 4–5). At this acidity:

• the carbonyl carbon is sufficiently activated, and
• the ammonia derivative is still largely unprotonated and nucleophilic.

This is why acidity must be carefully controlled — too much acid destroys the nucleophile, while too little acid fails to activate the carbonyl group and to catalyse dehydration.

Acidity of carboxylic acids vs alcohols

The –OH group of a carboxylic acid releases H⁺ far more readily than the –OH of an alcohol because the resulting carboxylate ion (RCOO⁻) is resonance-stabilised, whereas the alkoxide ion (RO⁻) is not.

1. Resonance stabilisation of the carboxylate ion

When a carboxylic acid ionises:

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

the negative charge on the carboxylate ion is delocalised over two equivalent oxygen atoms by two equivalent resonance structures:

$\mathrm{R-C(=O)-O^- \leftrightarrow R-C(-O^-)=O}$

Both structures are identical in energy, so the resonance hybrid is much more stable than either canonical form. This delocalisation lowers the energy of the anion and hence favours ionisation.

In an alcohol:

$\mathrm{R-OH \rightleftharpoons R-O^- + H^+}$

the alkoxide ion has the negative charge localised on a single oxygen atom; there is no possibility of resonance delocalisation. The anion is therefore high in energy and the equilibrium lies far to the left.

2. Inductive effect of the carbonyl group

In RCOOH the –OH is attached to a carbon that already bears a C=O group. The strongly electron-withdrawing carbonyl carbon pulls electron density away from the O–H bond (inductive effect), polarising it further and weakening the O–H bond. This makes the proton easier to remove. In an alcohol the –OH is attached to an sp³ carbon bearing only alkyl groups, which are electron-releasing; the O–H bond is therefore less polar and the proton is held more firmly.

3. Stability of the anion — the deciding factor

Acid strength depends mainly on the stability of the conjugate base. The greater the stability of the anion, the stronger the acid. Since the carboxylate ion is stabilised by resonance (and partly by the inductive effect of the carbonyl group) while the alkoxide ion is not, carboxylic acids are much stronger acids.

Quantitative comparison

The difference of about 11 pKa units corresponds to carboxylic acids being roughly 10¹¹ times stronger as acids than alcohols.

Conclusion

The –OH of a carboxylic acid loses H⁺ readily because the conjugate base (RCOO⁻) is resonance-stabilised by delocalisation of the negative charge over two oxygen atoms, whereas the –OH of an alcohol gives an alkoxide ion (RO⁻) in which the charge is localised on one oxygen. This difference in conjugate-base stability is the fundamental reason for the marked difference in acidity.

Grignard Synthesis and Nitrile Synthesis of Carboxylic Acids

1. Grignard Synthesis of Carboxylic Acids

Reaction:

$$\mathrm{R-MgX + CO_{2} \rightarrow [dry ether] R-COO^-MgX+ \rightarrow [H_{3}O+] R-COOH}$$

Carbon dioxide is passed through a solution of the Grignard reagent in dry ether; the resulting magnesium salt is hydrolysed with dilute acid to give the carboxylic acid. The product has one carbon more than the alkyl halide used.

Mechanism:

Step 1 — Nucleophilic attack on CO₂

The carbanion-like carbon of the Grignard reagent attacks the electrophilic carbonyl carbon of carbon dioxide:

$$\mathrm{R-MgX + O=C=O \rightarrow R-C(=O)-O-MgX}$$

The π electrons of the C=O bond shift onto oxygen, giving the carboxylate salt:

$$\mathrm{R-C(=O)-O^- \ MgX+}$$

Step 2 — Protonation (hydrolysis)

Addition of dilute HCl or H₂SO₄ protonates the carboxylate:

$$\mathrm{R-COO^-MgX+ + H_{3}O+ \rightarrow R-COOH + MgX(OH)}$$

Example:

$$\mathrm{CH_{3}MgBr + CO_{2} \rightarrow [dry ether] CH_{3}COO^-MgBr+ \rightarrow [H_{3}O+] CH_{3}COOH}$$

(Acetic acid)

Limitations: The method fails for Grignard reagents containing acidic or reactive groups (–OH, –NH₂, –COOH, –C≡CH). Only one of the three R groups is utilised; the rest are wasted.

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2. Nitrile Synthesis of Carboxylic Acids

Reaction:

$$\mathrm{R-X + KCN \rightarrow R-CN + KX}$$ $$\mathrm{R-CN + H_{2}O \rightarrow [H+ or OH^-] R-COOH + NH_{3}}$$

The alkyl halide is first converted to a nitrile (cyanide), which is then hydrolysed to the carboxylic acid. The product again has one carbon more than the alkyl halide.

Mechanism:

Step 1 — Sₙ2 formation of the nitrile

The cyanide ion acts as a nucleophile and displaces the halide from the primary alkyl halide:

$$\mathrm{CN^- + R-X \rightarrow [S_{N}2] R-CN + X^-}$$

Step 2 — Hydrolysis of the nitrile

Acid hydrolysis:

Protonation of the nitrile nitrogen, addition of water, and tautomerisation give the amide, which is further hydrolysed:

$$\mathrm{R-CN \rightarrow [H_{3}O+] R-C(=NH)-OH \rightarrow R-C(=O)-NH_{2} \rightarrow [H_{3}O+] R-COOH + NH_{4}+}$$

Base hydrolysis:

Hydroxide attacks the nitrile carbon, giving an imidic acid salt that rearranges to the amide and finally the carboxylate:

$$\mathrm{R-CN + OH^- \rightarrow R-C(=NH)-O^- \rightarrow R-CONH_{2} \rightarrow [OH^-] R-COO^- \rightarrow [H_{3}O+] R-COOH}$$

Example:

$$\mathrm{CH_{3}CH_{2}Br \rightarrow [KCN] CH_{3}CH_{2}CN \rightarrow [H_{3}O+] CH_{3}CH_{2}COOH}$$

(Propanoic acid)

Note: The nitrile route is preferred over the Grignard route when the molecule contains groups that would destroy the Grignard reagent.