Organic Chemistry - Some Basic Principles and Techniques | ISC Class 11 Chemistry Notes

This note covers carbon bonding, classification, nomenclature, isomerism, electronic effects, reaction mechanisms, purification, and qualitative and quantitative analysis.

Why does carbon form so many organic compounds?

Organic chemistry studies carbon compounds, including compounds with hydrogen, oxygen, nitrogen, sulphur, phosphorus and halogens. Their chemical symbols are C, H, O, N, S and P; the halogens fluorine, chlorine, bromine and iodine are represented by F, Cl, Br and I.

Catenation is the ability of carbon atoms to bond covalently to one another. A covalent bond shares an electron pair. Carbon's tetravalency, a combining capacity of four, permits chains, branches and rings; multiple bonding and isomerism add variety. This diversity and characteristic chemistry justify separate study, with applications in fuels, dyes and medicines.

The vital force theory attributed the formation of organic substances to a special force in living organisms. Wöhler's preparation of urea from ammonium cyanate in 1828 rejected that idea: NH₄CNO → NH₂CONH₂. The arrow means “forms”.

How do bonding and structural representations help?

An orbital describes an electron's spatial wave behaviour. Hybridisation combines orbitals: sp³ uses one s and three p orbitals, sp² one s and two p orbitals, and sp one s and one p orbital.

Methane, CH₄, has tetrahedral sp³ carbon; ethene, CH₂=CH₂, has planar sp² carbons; ethyne, HC≡CH, has linear sp carbons. A double bond contains one sigma bond, formed by head-on overlap, and one pi bond, formed by sideways overlap. Their symbols are σ and π.

A triple bond contains one σ and two π bonds. Double-bond rotation is restricted by sideways overlap. In general, π bonds provide the most reactive centres in molecules containing multiple bonds.

Condensed formulas group atoms without showing every bond. Bond-line formulas imply carbon and its attached hydrogens at corners and ends; other atoms are explicit.

A carbon has four bonds to hydrogen. Normal lines show bonds in the paper, a solid wedge shows a bond towards the observer, and a dashed wedge shows a bond away.

See Fig. 8.1 in your NCERT textbook

How are organic compounds classified into families?

Open-chain or acyclic compounds contain straight or branched carbon chains. Closed-chain or cyclic compounds contain rings. A ring composed entirely of carbon is homocyclic; a ring containing an atom other than carbon is heterocyclic. That different ring atom is called a heteroatom.

Aromatic compounds include benzene and related ring compounds with delocalised electrons, meaning electrons spread over several bonded atoms. A cyclic compound is not automatically aromatic: compare cyclohexane with benzene.

A functional group is an atom or group attached to a carbon framework that gives characteristic chemical properties. Examples include the hydroxyl group, -OH; aldehyde group, -CHO; and carboxyl group, -COOH. Compounds with the same functional group undergo similar reactions.

What makes a homologous series?

A homologous series shares a functional group and general formula. Successive members differ by CH₂, with similar chemical properties and gradually changing physical properties.

Alkanes are saturated hydrocarbons, compounds containing carbon and hydrogen with carbon-carbon single bonds. Their general formula is CₙH₂ₙ₊₂, where n is the number of carbon atoms. Methane and ethane are successive members. Unsaturated hydrocarbons contain a carbon-carbon double or triple bond.

How are systematic names built from organic structures?

IUPAC means International Union of Pure and Applied Chemistry. Its nomenclature links a systematic name to a structure. The parent is the main chain or ring; a substituent replaces hydrogen on it; a locant is a number showing a position.

An alkyl group results from removing hydrogen from an alkane: methyl is -CH₃ and ethyl is -CH₂CH₃. The symbol R, used in general organic formulas, represents an alkyl group unless another meaning is stated.

1. For a branched alkane, identify the longest continuous carbon chain. If equal-length chains are possible, select the one with more side chains.
2. Number the parent so that substituents receive the lowest possible locants, comparing the numbering sequences at their first difference.
3. Name different substituents alphabetically. Indicate repeated groups by di-, tri- or tetra-, meaning two, three or four; ignore these multiplying prefixes in alphabetisation.
4. Separate numbers by commas and numbers from words by hyphens. Join the substituent names to the parent name without spaces.

CH₃CH(CH₃)CH₂CH(CH₃)CH₃ is 2,4-dimethylpentane: five carbons form the parent and methyl groups occupy positions 2 and 4. A saturated single ring takes cyclo-, as in cyclopropane.

How do functional groups change the name?

A carbonyl group contains carbon doubly bonded to oxygen. A ketone has two carbon groups attached to its carbonyl carbon. Number a parent containing the principal functional group to give it the lowest appropriate locant; name subordinate groups as prefixes.

For example, HOCH₂(CH₂)₃CH₂COCH₃ is 7-hydroxyheptan-2-one: the ketone has priority over the alcohol.

For disubstituted benzene, ortho, meta and para denote positions 1,2; 1,3; and 1,4 respectively, abbreviated o-, m- and p-. Thus 1,3-dibromobenzene is m-dibromobenzene. Use numerical locants for three or more substituents.

How do the different forms of structural isomerism differ?

A molecular formula lists the types and numbers of atoms. Isomers share this formula but differ in properties; structural isomers have different atom connections.

Tautomers are structural isomers in dynamic equilibrium, meaning that interconversion occurs in both directions. The symbol ⇌ represents this reversible process. In keto-enol tautomerism, a carbonyl form interconverts with an enol, a form containing hydroxyl attached to a doubly bonded carbon.

For ordinary keto-enol tautomerism, a hydrogen on the carbon next to the carbonyl carbon can shift to oxygen as the double bond changes position. Such a neighbouring carbon is called an α-carbon, pronounced alpha-carbon, relative to that carbonyl group.

Tautomers differ from resonance contributors. Tautomerism moves a hydrogen nucleus as well as electrons; resonance contributors retain the same nuclear positions and differ in electron arrangement. Resonance contributors are alternative representations of one species, not separate substances in equilibrium.

When do compounds show geometrical isomerism?

Stereoisomers have the same connections between atoms but different arrangements in space. Geometrical and optical isomerism are two types. Geometrical isomerism arises when restricted rotation prevents the interchange of different spatial arrangements.

For an alkene to show geometrical isomerism, each doubly bonded carbon must carry two different groups. Identical groups on either carbon prevent it.

In cis-but-2-ene, methyl groups lie on the same side of the double bond; in trans-but-2-ene they lie on opposite sides. Both have connectivity CH₃CH=CHCH₃ but different physical properties.

What do syn and anti mean in oximes?

An oxime contains the group C=N-OH. In an aldoxime, one substituent on that carbon is hydrogen. Ethanal oxime, CH₃CH=N-OH, provides an example of geometrical isomerism about the carbon-nitrogen double bond.

Using the aldoxime convention, syn means that the carbon-bound hydrogen and the hydroxyl group on nitrogen lie on the same side of C=N; anti means they lie on opposite sides.

How are optical activity and specific rotation explained?

Plane-polarised light has its electric vibrations confined to one plane. A Nicol prism can produce it from ordinary light. An optically active substance rotates its plane of polarisation. A polarimeter measures the angle of this rotation.

A substance rotating the plane clockwise is dextrorotatory, designated d or (+); anticlockwise rotation is laevorotatory, designated l or (−). Rotation signs must be measured, not inferred from configuration.

Chirality means non-superimposability on a mirror image. Non-superimposable mirror-image stereoisomers are enantiomers. A tetrahedral carbon attached to four different groups is an asymmetric carbon or stereocentre. A molecule with one such centre is chiral.

Lactic acid, CH₃CH(OH)COOH, has four different groups around its middle carbon: hydrogen, hydroxyl, methyl and carboxyl. It has a pair of enantiomers. Their rotations have equal magnitudes and opposite signs under identical measurement conditions.

How is rotation measured?

1. Set the polarimeter reference using the solvent and a specified light wavelength.
2. Fill a tube of known length with solution of known concentration, avoiding bubbles.
3. Rotate the analyser, the component used to examine the emerging polarised light, to restore the chosen reference condition.
4. Record the angular change and its sign at a stated temperature, then calculate the specific rotation.

Let α be observed rotation in degrees, l the solution path length in decimetres, and c the concentration in grams per millilitre. One decimetre is 10 centimetres. At specified temperature T and wavelength λ, specific rotation is written [α]ᵀλ.

[α]ᵀλ = α/(lc)

For these units, specific rotation is expressed in degrees millilitre per gram per decimetre. State temperature and wavelength with the measurement.

Why can a substance with stereocentres be inactive?

A racemic mixture, denoted dl or (±), contains equal amounts of two enantiomers. External compensation cancels their opposite rotations between molecules. Racemisation produces this mixture.

Tartaric acid, HOOCCH(OH)CH(OH)COOH, has two stereocentres. It has an enantiomeric pair and a meso form, which is achiral because of internal symmetry. Its optical inactivity is called internal compensation. A meso compound is one substance; a racemate is a mixture.

What happens when covalent bonds break?

A reaction mechanism describes bond breaking, electron movement and bond formation step by step. A reagent reacts with the organic substrate. In a multistep mechanism, an intermediate forms in one step and is consumed in a later step.

In homolytic fission, each atom receives one bonding electron, forming free radicals: neutral species with an unpaired electron. A dot denotes that electron in CH₃•, the methyl radical. A half-headed curved arrow shows single-electron movement.

In heterolytic fission, one fragment receives both electrons: CH₃Br → CH₃⁺ + Br⁻ forms methyl carbocation and bromide. Superscripts + and − denote ionic charges. A full-headed curved arrow shows electron-pair movement.

Primary, secondary and tertiary mean that one, two or three carbon groups are directly attached to the reactive carbon. The sign > means “more stable than” in this table. The trends concern simple alkyl species.

How do nucleophiles differ from electrophiles?

A nucleophile donates an electron pair. Examples include hydroxide, OH⁻; cyanide, CN⁻; and neutral ammonia, NH₃, with its lone pair of non-bonding electrons.

An electrophile accepts an electron pair. Carbocations are charged examples. Boron trifluoride, BF₃, where B denotes boron, is a neutral example with an electron-deficient boron centre. Charge alone does not distinguish these roles.

How do electronic effects influence organic reactions?

The inductive effect transmits polarisation, unequal charge distribution, through σ bonds because atoms attract electrons differently. Electronegativity means an atom's attraction for bonding electrons. In chloroethane, CH₃CH₂Cl, chlorine draws electron density towards itself.

The symbols δ+ and δ− mean partial positive and partial negative charge. The inductive effect decreases rapidly as intervening bonds increase and becomes vanishingly small after three bonds. Electron withdrawal relative to hydrogen is the −I effect; donation is the +I effect.

Halogens and the nitro group, -NO₂, withdraw electron density inductively. Alkyl groups are usually considered electron-donating groups. Their donation helps stabilise positive charge on a neighbouring carbocation.

How do resonance and electromeric effects differ?

Resonance represents electron delocalisation using contributing structures with the same nuclear positions and the same number of unpaired electrons. The actual molecule is a resonance hybrid with energy lower than any individual contributor. Benzene is an example.

A conjugated system allows delocalisation through adjacent orbitals, commonly with alternating single and double bonds. Resonance or mesomeric effects involve interacting π bonds or a π bond and an adjacent lone pair.

Donation towards the conjugated system is the +R or +M effect, as with -OH and -NH₂. Withdrawal is the −R or −M effect, as with -NO₂ and -CHO. Here R and M label resonance and mesomeric effects, rather than an alkyl group.

The electromeric effect, labelled E, is temporary complete transfer of a π-electron pair under the influence of an attacking reagent. It disappears when that reagent is removed. In +E, the pair moves to the atom receiving the reagent; in −E, it moves away.

Hydrogen-ion addition to an alkene illustrates +E. Attack by cyanide on a carbonyl carbon illustrates −E as the carbonyl π pair moves to oxygen. When inductive and electromeric effects operate oppositely, the electromeric effect predominates.

What is hyperconjugation?

Hyperconjugation permanently delocalises adjacent carbon-hydrogen σ electrons into an unsaturated system or suitable p orbital. This stabilising interaction is also called no-bond resonance.

In the ethyl carbocation, CH₃CH₂⁺, adjacent carbon-hydrogen bonds can interact with the empty p orbital. Methyl carbocation, CH₃⁺, lacks this interaction. In general, more alkyl groups attached to the positively charged carbon give greater hyperconjugative stabilisation.

How do substitution, addition and elimination mechanisms work?

Substitution replaces an atom or group; addition joins atoms or groups across a multiple bond; elimination removes atoms or groups to create a multiple bond. For example, ethene adds bromine: CH₂=CH₂ + Br₂ → BrCH₂CH₂Br, producing 1,2-dibromoethane.

How does a free-radical chain operate?

Methane chlorination is supposed to proceed through a free-radical chain mechanism. Initiation produces radicals, propagation regenerates them while products form, and termination consumes radicals. Light or heat initiates homolysis of chlorine.

• Initiation: Cl₂ → 2Cl•.
• Propagation: Cl• + CH₄ → HCl + CH₃•; then CH₃• + Cl₂ → CH₃Cl + Cl•. HCl is hydrogen chloride; CH₃Cl is chloromethane.
• Termination can occur through Cl• + Cl• → Cl₂, CH₃• + CH₃• → CH₃CH₃, or CH₃• + Cl• → CH₃Cl.

How do SN1 and SN2 compare?

SN denotes nucleophilic substitution. The numerals distinguish unimolecular and bimolecular mechanisms. A leaving group departs with the bonding electron pair. Here k is the rate constant, square brackets denote molar concentration (moles of solute per litre of solution), and X in RX denotes halogen.

Rate(SN1) = k[RX]

Rate(SN2) = k[RX][nucleophile]

SN1 generally occurs in polar protic solvents, such as water or alcohol, which can hydrogen-bond through oxygen-hydrogen groups. Tert-butyl bromide first ionises slowly: (CH₃)₃CBr → (CH₃)₃C⁺ + Br⁻. Hydroxide then attacks rapidly, giving (CH₃)₃COH, tert-butyl alcohol.

Slow ionisation controls rate; carbocation stability gives tertiary > secondary > primary SN1 reactivity among simple alkyl halides. A planar carbocation can be attacked from either face, explaining racemisation when an appropriate chiral substrate reacts.

SN2 occurs in one step: OH⁻ + CH₃Cl → CH₃OH + Cl⁻. Backside attack forms the carbon-oxygen bond while the carbon-chlorine bond breaks. The transition state is a fleeting, high-energy arrangement, not an isolable intermediate.

Attack inverts the arrangement at the reacting carbon, called inversion of configuration. Bulky groups hinder the nucleophile: methyl reacts fastest, followed by primary and secondary substrates, with tertiary least reactive. Strong nucleophiles and polar aprotic solvents, lacking hydrogen-bond-donating groups, commonly favour SN2.

How do E1 and E2 compare?

In E1, unimolecular elimination, slow leaving-group departure forms a carbocation. A base, a proton acceptor, then removes a β-hydrogen from a neighbouring carbon and the carbon-carbon double bond forms. Here α-carbon bears the leaving group and β-carbon is adjacent to it.

Rate(E1) = k[RX]

Heated tert-butyl bromide in aqueous ethanol can eliminate through E1: (CH₃)₃CBr → (CH₃)₃C⁺ + Br⁻; loss of a β-proton then gives CH₂=C(CH₃)₂, 2-methylpropene. Heat, a stable carbocation and an ionising protic solvent favour this route; SN1 can compete.

In E2, bimolecular elimination, base removes β-hydrogen as the leaving group departs in one step. The carbon-hydrogen electrons form the π bond. An anti-periplanar arrangement, with the departing bonds coplanar and oppositely directed, favours elimination.

Rate(E2) = k[RX][base]

Heating bromoethane with alcoholic potassium hydroxide, KOH, gives CH₃CH₂Br + KOH → CH₂=CH₂ + KBr + H₂O. K denotes potassium; KBr is potassium bromide. Strong base and heat favour E2. Substitution and elimination compete according to substrate, reagent and conditions.

How is a suitable purification method selected?

Purification separates a desired compound from impurities by exploiting physical differences. Most pure compounds have sharp melting and boiling points. Choose according to volatility, solubility and stability on heating.

How does crystallisation work?

1. Select a solvent in which the compound is sparingly soluble at room temperature but appreciably soluble when hot.
2. Dissolve the impure compound and concentrate the solution until nearly saturated.
3. Cool the solution so that crystals separate.
4. Filter off the crystals. The mother liquor, the remaining solution, retains impurities and a small quantity of compound.

Activated charcoal can adsorb coloured impurities. Adsorption is accumulation at a surface. Repeated crystallisation may be needed when impurities have comparable solubilities. A fractionating column instead provides surfaces for repeated heat exchange between rising vapour and descending liquid.

A heated round-bottomed flask connects to a thermometer and a sloping condenser. The condenser has a lower water inlet and an upper outlet to the sink; condensed liquid collects in a conical flask.

See Fig. 8.5 in your NCERT textbook

Differential extraction transfers a dissolved substance into another solvent in which it is more soluble. The two solvents must be immiscible, meaning they form separate layers. Separate these layers in a separating funnel, then remove the extracting solvent to recover the compound.

How does chromatography separate a mixture?

Chromatography separates components through different interactions with a stationary phase and a moving phase. The stationary phase remains in place; the mobile phase is a moving solvent, solvent mixture or gas.

In adsorption chromatography, components bind to a surface to different degrees. Silica gel and alumina are common adsorbents. Column chromatography packs the adsorbent into a tube; the eluant, the liquid mobile phase, passes through it. More strongly adsorbed substances remain nearer the top.

Thin-layer chromatography, abbreviated TLC, uses adsorbent coated on a glass plate. Spot the mixture near the base. In a solvent-containing jar, the rising solvent carries components different distances.

The retardation factor, Rf, divides substance travel distance x by solvent-front distance y, measured from the baseline in identical units. The ratio has no unit.

Rf = x/y

The developing plate stands in a jar with the sample dot above the solvent. The developed chromatogram labels the baseline, spot and solvent front; x reaches the spot and y reaches the solvent front.

See Fig. 8.12 in your NCERT textbook

A chromatogram is the developed pattern. Coloured spots are visible directly; some colourless substances fluoresce under ultraviolet light or become visible with iodine or spray reagents.

In partition chromatography, components distribute differently between two phases. Paper chromatography uses water trapped in the paper as the stationary phase and a moving solvent as the mobile phase. Different partitioning separates the components into spots.

How are elements detected in an organic compound?

Qualitative analysis identifies which elements are present. Heating a compound with copper(II) oxide, CuO, converts carbon to carbon dioxide, CO₂, and hydrogen to water, H₂O. Cu denotes copper; the Roman numeral II identifies its +2 oxidation state.

Carbon dioxide makes limewater, aqueous calcium hydroxide, Ca(OH)₂, turbid by forming calcium carbonate, CaCO₃. Water turns white anhydrous copper sulphate, CuSO₄, blue by hydration. Anhydrous means without water of crystallisation; Ca denotes calcium.

C + 2CuO → 2Cu + CO₂; H₂ + CuO → Cu + H₂O; CO₂ + Ca(OH)₂ → CaCO₃ + H₂O.

Why is sodium fusion needed?

Lassaigne's test converts covalently bound nitrogen, sulphur and halogens into water-soluble ionic sodium compounds. Na denotes sodium. Fuse the compound with sodium, then extract the fused mass with distilled water to obtain the sodium fusion extract.

Na + C + N → NaCN; 2Na + S → Na₂S; Na + X → NaX. Here X is chlorine, bromine or iodine; the products are sodium cyanide, sodium sulphide and sodium halide respectively.

A precipitate is an insoluble solid formed from solution. Fe, Pb and Ag denote iron, lead and silver. Relevant ionic equations are 6CN⁻ + Fe²⁺ → [Fe(CN)₆]⁴⁻ and 3[Fe(CN)₆]⁴⁻ + 4Fe³⁺ → Fe₄[Fe(CN)₆]₃, the Prussian blue compound.

The sulphur and halogen equations include S²⁻ + Pb²⁺ → PbS and X⁻ + Ag⁺ → AgX. Sodium nitroprusside gives S²⁻ + [Fe(CN)₅NO]²⁻ → [Fe(CN)₅NOS]⁴⁻, the violet species. Brackets here enclose groups acting together as complex ions.

When nitrogen and sulphur occur together, sodium thiocyanate, NaSCN, can form: Na + C + N + S → NaSCN. Its thiocyanate ion gives a blood-red complex: Fe³⁺ + SCN⁻ → [Fe(SCN)]²⁺. Excess sodium converts it further: NaSCN + 2Na → NaCN + Na₂S.

Note: If nitrogen or sulphur is present, boil the extract with concentrated nitric acid before adding silver nitrate. This removes cyanide or sulphide that would otherwise interfere with the halogen test.

How are carbon, hydrogen and nitrogen estimated quantitatively?

Quantitative analysis measures how much of each element is present. Mass percentage means the mass of an element per hundred parts by mass of compound. In Liebig's combustion method, a weighed sample burns in excess oxygen with copper(II) oxide.

Water is trapped in weighed anhydrous calcium chloride, CaCl₂; carbon dioxide is then absorbed in weighed potassium hydroxide solution. Increases in absorber masses give the masses of water and carbon dioxide. Their order prevents water from adding to the apparent carbon dioxide mass.

Derivation: Carbon and hydrogen percentages

Let m be sample mass, mC the collected carbon dioxide mass and mW the collected water mass, all in grams, symbol g. Using relative atomic masses C = 12, H = 1 and O = 16:

1. Carbon dioxide has relative molecular mass 44, so carbon mass is (12/44)mC.
2. Water has relative molecular mass 18, so hydrogen mass is (2/18)mW.
3. Divide each elemental mass by m and multiply by 100 to obtain its mass percentage.

Carbon percentage = 12mC × 100/(44m); Hydrogen percentage = 2mW × 100/(18m).

Worked example 1. Complete combustion of 0.246 g compound gives 0.198 g CO₂ and 0.1014 g H₂O. Using C = 12, H = 1 and O = 16, calculate carbon and hydrogen percentages.

Formula: Carbon percentage = 12mC × 100/(44m); Hydrogen percentage = 2mW × 100/(18m).

Substitute: carbon = 12 × 0.198 × 100/(44 × 0.246); hydrogen = 2 × 0.1014 × 100/(18 × 0.246).

Answer: Carbon mass = 0.054 g; hydrogen mass = 0.01127 g approximately. Using unrounded values gives 21.95% carbon and 4.58% hydrogen.

Worked example 2. A 0.20 g compound contains 69% carbon and 4.8% hydrogen. Find the combustion-product masses, using relative atomic masses C = 12, H = 1 and O = 16.

Formula: mC = m × (carbon percentage/100) × 44/12; mW = m × (hydrogen percentage/100) × 18/2.

Substitute: mC = 0.20 × 0.69 × 44/12; mW = 0.20 × 0.048 × 18/2.

Answer: 0.506 g CO₂ and 0.0864 g H₂O.

How does Kjeldahl's method work?

1. Digest a known sample mass with concentrated sulphuric acid, H₂SO₄, converting its nitrogen into ammonium sulphate, (NH₄)₂SO₄.
2. Add excess sodium hydroxide, NaOH, and heat to liberate ammonia: (NH₄)₂SO₄ + 2NaOH → Na₂SO₄ + 2NH₃ + 2H₂O.
3. Absorb ammonia in excess standard sulphuric acid: 2NH₃ + H₂SO₄ → (NH₄)₂SO₄. A standard solution has known concentration.
4. Titrate the remaining acid with standard alkali. Titration measures the reacting volume; subtraction gives the acid consumed by ammonia.

Molarity, the molar concentration defined earlier, is written mol L⁻¹ or M. The mole, symbol mol, measures chemical amount. One litre, L, contains 1000 millilitres, mL. Nitrogen has molar mass 14 g mol⁻¹.

Derivation: Nitrogen from back-titration

Let a be initial moles of sulphuric acid and b the moles of sodium hydroxide used to titrate residual acid. Here m remains sample mass in grams.

1. One mole of sulphuric acid reacts with two moles of sodium hydroxide, so residual acid is b/2 moles.
2. Acid consumed by ammonia is a − b/2 moles; therefore ammonia contains 2a − b moles of nitrogen.
3. Multiply by 14 g mol⁻¹ and divide by sample mass to obtain the nitrogen fraction.

Nitrogen percentage = 14(2a − b) × 100/m

Worked example 3. Ammonia from 0.5 g compound neutralises 10 mL of 1 M H₂SO₄. Use nitrogen molar mass 14 g mol⁻¹ and the ratio two moles NH₃ per mole H₂SO₄.

Answer: Acid consumed = 1 × 10/1000 = 0.010 mol; nitrogen = 0.020 × 14 = 0.280 g. Nitrogen percentage = 0.280 × 100/0.5 = 56.0%.

Worked example 4. Ammonia from 0.50 g compound enters 50 mL of 0.5 M H₂SO₄. Residual acid needs 60 mL of 0.5 M NaOH. Use nitrogen molar mass 14 g mol⁻¹ and the reactions above.

Answer: a = 0.5 × 50/1000 = 0.025 mol; b = 0.5 × 60/1000 = 0.030 mol. Nitrogen mass = 14 × (0.050 − 0.030) = 0.280 g; nitrogen percentage = 56%.

Kjeldahl's limitation: nitrogen in nitro groups, azo groups containing -N=N-, and rings such as pyridine is not converted to ammonium sulphate under these conditions. The method is therefore not applicable to those compounds.

How does Carius analysis estimate halogens, sulphur and phosphorus?

In the Carius method for halogens, heat a known mass of organic compound with fuming nitric acid and silver nitrate in a hard-glass Carius tube. Carbon and hydrogen are oxidised; the halogen forms a silver halide, which is filtered, washed, dried and weighed.

Let m be sample mass, mX silver-halide mass in the same unit, A(X) the halogen's relative atomic mass and Mr(AgX) the silver salt's relative formula mass. Each AgX unit contains one halogen atom.

Halogen percentage = A(X)mX × 100/[Mr(AgX)m]

Worked example 5. A 0.15 g compound gives 0.12 g silver bromide in Carius analysis. Use relative atomic masses Ag = 108 and Br = 80.

Answer: Relative formula mass of AgBr = 188; bromine mass = 80 × 0.12/188 = 0.05106 g approximately. Using the unrounded mass, bromine percentage = 80 × 0.12 × 100/(188 × 0.15) = 34.04%.

How are sulphur and phosphorus converted to weighable products?

Oxidise sulphur with fuming nitric acid or sodium peroxide in a Carius tube to sulphuric acid. Excess barium chloride, BaCl₂, precipitates barium sulphate, BaSO₄. Ba denotes barium. Filter, wash, dry and weigh this solid.

Let mS be BaSO₄ mass and m the sample mass, both in grams. Relative formula mass of BaSO₄ is 233, containing sulphur mass 32 on the same scale.

Sulphur percentage = 32mS × 100/(233m)

Worked example 6. A 0.157 g compound gives 0.4813 g BaSO₄. Use relative atomic masses Ba = 137, S = 32 and O = 16, giving BaSO₄ formula mass 233.

Answer: Sulphur mass = 32 × 0.4813/233 = 0.06610 g approximately. Using the unrounded mass, sulphur percentage = 32 × 0.4813 × 100/(233 × 0.157) = 42.10%.

For phosphorus estimation, fuming nitric acid oxidises phosphorus to phosphoric acid. Adding ammonia and ammonium molybdate precipitates ammonium phosphomolybdate, (NH₄)₃PO₄·12MoO₃, where Mo denotes molybdenum. Its relative formula mass is 1877 and it contains one phosphorus atom of relative mass 31.

Alternatively, magnesia mixture, a magnesium salt reagent, precipitates magnesium ammonium phosphate, MgNH₄PO₄. Ignition, strong heating, converts it to magnesium pyrophosphate, Mg₂P₂O₇. Mg denotes magnesium. This solid has relative formula mass 222 and contains two phosphorus atoms with combined relative mass 62.

Let mP be the mass of the stated phosphorus-containing precipitate, in the same unit as sample mass m.

Phosphorus percentage = 31mP × 100/(1877m) for ammonium phosphomolybdate.

Phosphorus percentage = 62mP × 100/(222m) for magnesium pyrophosphate.

Glossary

• Catenation — The ability of atoms of an element to bond to one another, forming chains or rings.
• Functional group — An atom or group responsible for characteristic chemical properties within an organic compound.
• Homologous series — A family with a common general formula whose successive members differ by a CH₂ unit.
• Structural isomers — Compounds with the same molecular formula but different connections between their constituent atoms.
• Stereoisomers — Compounds with identical atom connectivity but different arrangements of atoms or groups in space.
• Enantiomers — Stereoisomers that are mirror images and cannot be superimposed upon one another.
• Racemic mixture — Equal quantities of two enantiomers whose opposite optical rotations cancel each other.
• Meso compound — An achiral compound containing stereocentres but showing internal symmetry and no optical rotation.
• Nucleophile — A species that donates an electron pair to form a new covalent bond.
• Electrophile — A species that accepts an electron pair during formation of a covalent bond.
• Resonance hybrid — The actual delocalised structure represented collectively by contributing structures with unchanged nuclear positions.
• Chromatography — Separation based on different interactions of mixture components with stationary and mobile phases.

Common errors and misconceptions

• Misconception: Every carbon ring is aromatic. Correct: Cyclohexane is alicyclic; benzene is aromatic.
• Misconception: Resonance contributors rapidly interconvert. Correct: They represent one hybrid; tautomers are distinct interconverting structures.
• Misconception: Any alkene has cis-trans isomers. Correct: Each doubly bonded carbon must carry two different groups.
• Misconception: A meso compound is a racemate. Correct: A meso compound is one achiral substance; a racemate contains two enantiomers.
• Misconception: All electrophiles are positively charged. Correct: Neutral boron trifluoride accepts electron pairs.
• Misconception: SN2 forms a carbocation first. Correct: Bond formation and cleavage occur together through a transition state.
• Misconception: Kjeldahl's method estimates every form of organic nitrogen. Correct: Nitro, azo and ring nitrogen are exceptions.

Exam-style questions with model answers

Q1. Name CH₃CH(CH₃)CH₂CH(CH₃)CH₃ and explain the numbering. [2 marks]

1. The name is 2,4-dimethylpentane: the longest continuous chain contains five carbon atoms and bears two methyl substituents.
2. Numbering places the methyl groups at carbon atoms 2 and 4; the prefix di- indicates that two identical substituents are present.

Q2. Explain chirality, enantiomerism, external compensation and internal compensation, using lactic acid and tartaric acid where appropriate. [4 marks]

1. Chirality is non-superimposability on a mirror image. Lactic acid, CH₃CH(OH)COOH, is chiral because its middle carbon bears four different groups.
2. Enantiomers are non-superimposable mirror-image stereoisomers, such as the two optical forms of lactic acid.
3. External compensation occurs in an equal mixture of two enantiomers, whose opposite rotations cancel to give a racemic mixture.
4. Internal compensation describes the optical inactivity of the meso form of tartaric acid: internal symmetry makes it achiral despite its stereocentres.

Q3. Distinguish the inductive effect, resonance effect and electromeric effect by electron involvement and persistence. [3 marks]

1. The inductive effect is permanent polarisation transmitted through σ bonds. It decreases rapidly with increasing numbers of intervening bonds.
2. The resonance effect involves delocalisation through interacting π bonds or a π bond and an adjacent lone pair; it persists without an attacking reagent.
3. The electromeric effect is temporary complete transfer of a π-electron pair under an attacking reagent's influence. It disappears when the reagent is removed.

Q4. Compare SN1 and SN2 mechanisms in terms of steps, rate dependence, substrate preference and stereochemical result. [4 marks]

1. SN1 involves slow carbocation formation followed by nucleophile attack. SN2 forms and breaks bonds simultaneously in one step, without a carbocation intermediate.
2. SN1 rate depends on substrate concentration; SN2 rate depends on both substrate and nucleophile concentrations.
3. Tertiary alkyl halides favour SN1 through carbocation stability. Methyl and primary substrates favour SN2 because they obstruct backside attack less.
4. At an appropriate stereocentre, SN1 gives racemisation through attack on a planar carbocation, whereas SN2 produces inversion of configuration through backside attack.

Q5. Explain sodium fusion and the subsequent tests for nitrogen, sulphur and halogens, including the interference precaution before the halogen test. [5 marks]

1. Sodium fusion converts covalently bound nitrogen, sulphur and halogens into sodium cyanide, sulphide and halide. Extraction with distilled water gives the sodium fusion extract.
2. For nitrogen, boil extract with iron(II) sulphate and acidify with concentrated sulphuric acid. Prussian blue iron(III) hexacyanidoferrate(II) confirms nitrogen.
3. For sulphur, acidify extract with acetic acid and add lead acetate. Black lead sulphide indicates sulphur.
4. For halogens, acidify with nitric acid and add silver nitrate. Chlorine, bromine and iodine give white, yellowish and yellow precipitates respectively.
5. If nitrogen or sulphur is present, first boil extract with concentrated nitric acid to remove cyanide and sulphide, preventing interference with silver nitrate.

Q6. A 0.246 g organic sample burns completely in excess oxygen with copper(II) oxide. The products are 0.1014 g water and 0.198 g carbon dioxide. Explain their absorption and calculate carbon and hydrogen percentages. Use relative atomic masses C = 12, H = 1 and O = 16. [6 marks]

1. Pass the combustion gases through weighed anhydrous calcium chloride first. It absorbs water, so its mass increase measures the water produced.
2. Pass the remaining gases through weighed potassium hydroxide solution. It absorbs carbon dioxide, and its mass increase measures carbon dioxide produced.
3. Carbon dioxide has relative molecular mass 12 + 2 × 16 = 44. The carbon mass is therefore 0.198 × 12/44 = 0.054 g.
4. Carbon percentage is the carbon mass divided by the original sample mass, multiplied by 100: 0.054 × 100/0.246 = 21.95%.
5. Water has relative molecular mass 18. Its hydrogen mass fraction is 2/18, giving hydrogen mass 0.1014 × 2/18 = 0.01127 g approximately.
6. Using the unrounded hydrogen mass, hydrogen percentage is 0.1014 × 2 × 100/(18 × 0.246) = 4.58%.

Key takeaways

• Carbon's tetravalency, catenation, multiple bonding and isomerism account for the extensive variety of organic compounds.
• Systematic naming identifies the parent, principal functional group, substituents and their positions in a structure.
• Structural isomers differ in connectivity; stereoisomers retain connectivity but differ in the arrangement of groups in space.
• Inductive, resonance, electromeric and hyperconjugative effects explain electron distribution and help interpret stability and reactivity.
• SN1 and E1 involve carbocations; SN2 and E2 are one-step mechanisms whose outcomes depend on substrate and conditions.
• Purification separates compounds, qualitative analysis identifies elements, and quantitative analysis calculates elemental mass percentages from measured products.

Test yourself

Why does methoxypropane show metamerism with ethoxyethane?

Both have molecular formula C₄H₁₀O, but distribute their carbon groups differently on either side of the ether oxygen.

What distinguishes a tautomer from a resonance contributor?

Tautomers differ in hydrogen position and bond arrangement; resonance contributors retain nuclear positions and differ in electron distribution.

What does syn mean for ethanal oxime?

The carbon-bound hydrogen and hydroxyl group on nitrogen lie on the same side of the carbon-nitrogen double bond.

Why is tertiary carbocation formation favoured over methyl carbocation formation?

Alkyl groups stabilise positive charge by inductive donation and hyperconjugation; the methyl carbocation lacks adjacent carbon-hydrogen hyperconjugation.

What are the two phases in paper chromatography?

Water trapped in paper is the stationary phase; the solvent moving along the paper is the mobile phase.

Why is the mass of silver bromide not the mass of bromine?

Silver bromide contains silver as well as bromine, so multiply its mass by bromine's mass fraction in the salt.