Organic Chemistry

A molecular formula gives atom counts but does not uniquely identify an organic compound. A structural formula must make the connectivity unambiguous. For example, CH₃CH(OH)COOH shows a three-carbon chain, an OH group on its middle carbon and a carboxylic acid group at its end. Its molecular formula is C₃H₆O₃, and its empirical formula is CH₂O.

For naming, choose a parent chain containing the principal functional group and number it to give that group the required low locant, then locate double bonds and substituents. Examples include butan-2-ol, 2-bromopropane and propan-2-one. Do not choose a visually longest line if it excludes the functional group that defines the name.

R is a general alkyl or organic group and Ar denotes an aromatic group. They are placeholders, not atoms. For simple acyclic saturated monofunctional examples, alcohols have C_nH_2n+2O, aldehydes/ketones C_nH_2nO and monocarboxylic acids/esters C_nH_2nO₂. A ring, extra multiple bond or additional group changes these relations, so use the structural group when the scope is wider.

Hybridisation is a bonding model: one s and three p orbitals combine into four sp³ orbitals, one s and two p into three sp², or one s and one p into two sp orbitals. The remaining p orbitals are available for π overlap. Count regions of σ bonding around an atom; a double bond still points in one direction in the molecular framework.

Constitutional (structural) isomers have the same molecular formula but different connectivity. They may differ in carbon skeleton, functional-group position or functional-group class. Butane and 2-methylpropane differ in skeleton; propan-1-ol and propan-2-ol differ in OH position; ethanol and methoxymethane have different functional groups despite both being C₂H₆O.

Cis-trans isomerism in an alkene arises because rotating about C=C would disrupt sideways p-orbital overlap and break the π bond. Each double-bond carbon must have two different substituents. If either carbon has two identical groups, there is no such pair. Where the cis/trans names are unambiguous, cis places the matching groups on the same side and trans on opposite sides. E/Z nomenclature is not required.

A chiral centre is commonly a tetrahedral carbon bonded to four different atoms or groups. In butan-2-ol, the OH-bearing carbon is attached to OH, H, CH₃ and CH₂CH₃, giving two non-superimposable mirror-image arrangements called enantiomers. Compare complete groups, not just the first attached atom: methyl and ethyl both begin with carbon but are different groups.

With several chiral centres, inspect the whole molecule. An internal plane of symmetry can make a structure achiral despite containing stereogenic centres. Conversely, a general molecule can be chiral for reasons other than a tetrahedral carbon centre; for the structures supplied, use the actual geometry and symmetry rather than treating a centre count as an infallible rule.

An optically active sample rotates plane-polarised light and contains chiral molecules with a non-cancelling composition. A racemic mixture contains equal amounts of two enantiomers, so their rotations cancel. Optical inactivity alone therefore does not prove that all molecules in the sample are achiral. The sign of rotation must be measured; it is not read directly from a wedge drawing.

Biological receptors and enzymes are chiral. Two stereoisomers may bind differently and therefore differ in activity, side effects or metabolism. This explains why a molecular formula and functional-group list alone cannot establish a drug's behaviour; three-dimensional arrangement matters. No claim that one particular handedness is always beneficial follows from this principle.

A full-headed curly arrow shows the movement of an electron pair; its tail begins at a bond or lone pair and its head points to the atom or bond receiving the pair. A single-barbed fishhook shows movement of one electron. Arrows do not show atoms moving. A positive charge is not an electron source.

For an alcohol or halogenoalkane, primary/secondary/tertiary describes how many carbon groups are attached to the carbon carrying OH or halogen. For amines, it counts carbon groups attached to nitrogen: RNH₂, R₂NH and R₃N. R₄N⁺ is a quaternary ammonium ion. A quaternary carbon is bonded to four other carbons; it is not a neutral nitrogen with four bonds and an extra lone pair.

For otherwise comparable halogenoalkanes, hydrolysis generally becomes easier from RCl to RBr to RI because C-Cl is stronger than C-Br, which is stronger than C-I. The trend is not explained by the C-X polarity alone. Carbon skeleton, solvent and mechanism must also be comparable before interpreting a rate comparison.

In chlorobenzene, a chlorine lone pair overlaps with the ring π system. The C-Cl bond has partial double-bond character and is harder to break than in a comparable halogenoalkane. The rigid ring framework hinders the usual backside-attack geometry at the carbon bearing chlorine, while direct ionisation would require a very unstable phenyl cation. Chlorobenzene therefore resists ordinary nucleophilic substitution under the hydrolysis conditions used for halogenoalkanes.

Benzene still has electron density that can attack an electrophile, but disrupting its delocalised system costs stabilisation. A stronger electrophile or catalyst is commonly needed than for alkene addition. Subsequent loss of H⁺ restores aromaticity, explaining substitution rather than permanent addition across the ring.

Ethane reacts with chlorine under ultraviolet light at room temperature to form chloroethane and HCl. The UV light causes homolytic fission of Cl-Cl, not heterolysis to ions. One chlorine radical can sustain many propagation cycles before being removed in a termination event.

Adding the propagation steps cancels the ethyl and chlorine radicals, giving CH₃CH₃ + Cl₂ → CH₃CH₂Cl + HCl. Further substitution can replace more hydrogens, so a mixture forms; using excess ethane reduces the chance that the already-substituted product is attacked. Bromination under UV follows the same stage pattern with bromine radicals.

The ethene π cloud polarises an approaching Br₂ molecule. Electron density is donated towards its δ⁺ bromine as the Br-Br bond breaks heterolytically. The chemically accurate intermediate is a bridged bromonium ion: one Br is bonded to both original alkene carbons and carries positive formal charge. The other bromine leaves as Br^-.

Addition of HX follows a different intermediate: the π bond accepts H⁺ from H-X as the H-X bond pair moves to X, forming a carbocation; X^- then donates a pair to the positive carbon. For propene + HBr, protonation that gives the secondary carbocation is favoured over the primary alternative, so 2-bromopropane is the major product.

Markovnikov's rule predicts the usual major product of HX addition to an unsymmetrical alkene under the stated polar mechanism. It does not mean that a minor product is impossible, nor should a free carbocation automatically be substituted for the bromonium intermediate in Br₂ addition. Radical peroxide pathways are outside the named mechanism here.

For monobromination, use Br₂ with anhydrous AlBr₃. The Lewis acid accepts electron density and strongly polarises Br₂. A useful formal representation is Br₂ + AlBr₃ ⇌ Br⁺ + AlBr₄^-; this is electron bookkeeping for the activated electrophile rather than a claim that a bottle contains free isolated Br⁺.

Overall, C₆H₆ + Br₂ → C₆H₅Br + HBr. Substitution replaces H while restoring the delocalised system. Permanent addition would lose that aromatic stabilisation, explaining benzene's preference for substitution. Do not draw a full circle inside the sigma complex or leave the catalyst consumed in the final overall equation.

In S_N2, a nucleophile donates a pair to the carbon bearing the leaving group while the C-X pair moves to X. These changes occur together through one transition state, with no carbocation intermediate. The nucleophile approaches opposite the leaving group, so bulky groups around the reacting carbon strongly hinder the pathway.

For an elementary S_N2 step, rate = k[halogenoalkane][nucleophile]. Methyl and primary substrates are generally more accessible than secondary ones; tertiary substrates are strongly hindered. These are pathway preferences, not a claim that solvent, nucleophile and competing elimination never matter.

Inversion of configuration means the three remaining groups turn through the transition-state geometry as the nucleophile replaces the leaving group from the opposite side. An initially single enantiomer gives the inverted arrangement for this pathway, rather than a racemic mixture. Merely rotating the product drawing on the page is not inversion.

The nucleophile then donates a lone pair to the positive carbon. With OH^-, this gives the alcohol directly; with H₂O, it first gives a protonated alcohol that must lose H⁺. The simple rate law is rate = k[halogenoalkane], because the slow ionisation precedes nucleophile attack.

For simple alkyl carbocations, tertiary is generally more stable than secondary, which is more stable than primary. Alkyl groups donate electron density towards the electron-deficient centre. A substrate that would form an unstable primary carbocation is therefore unlikely to hydrolyse by a simple S_N1 route.

For an initially chiral substrate, the carbocation is approximately trigonal planar. Attack can occur from either face, producing both enantiomers and racemisation in the ideal model. Real ion pairs can partly shield a face, so exact equality is an idealisation; the key contrast is loss of stereochemical specificity versus S_N2 inversion. The illustrated tert-butyl example itself is achiral because its three methyl groups are identical.

Use HCN with a small amount of KCN as catalyst. HCN alone supplies only a low concentration of CN^-; the cyanide ion is the nucleophile. Its carbon end donates a lone pair to the δ⁺ carbonyl carbon, while the C=O π pair moves onto oxygen. Protonation of the resulting alkoxide by HCN produces the hydroxynitrile and regenerates CN^-.

An aldehyde RCHO gives RCH(OH)CN; a ketone R₂CO gives R₂C(OH)CN. Attack on either face of a planar carbonyl can form a racemic pair if a new chiral centre is created in an otherwise achiral system. Ethanal creates such a centre; propanone does not, because the product has two identical methyl groups.

Complete combustion of ethane is 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. With inadequate oxygen, carbon monoxide and/or carbon can form. Halogenation is a different reaction: Cl₂ or Br₂ under UV at room temperature replaces C-H by C-X through the radical chain described earlier.

The specified Br₂/CCl₄ mechanism gives a vicinal dibromide; do not silently treat water as an inert solvent in all halogen additions. For a bromine-water test, the rapid disappearance of colour is the useful observation, but it is not unique to alkenes because activated aromatic compounds can also consume bromine.

A cyclic alkene can open into one molecule containing two oxygenated ends rather than two separate molecules. Count the carbon atoms in every product, including CO₂, before proposing the original structure. Cold alkaline and hot acidified manganate(VII) are not interchangeable conditions: one gives a diol, the other cleaves the double bond.

Concentrated sulfuric acid protonates nitric acid, enabling formation of NO₂⁺: HNO₃ + H₂SO₄ ⇌ NO₂⁺ + HSO₄^- + H₂O. Nitration then follows electrophilic substitution and the acid catalyst is regenerated. The syllabus explicitly specifies 30 °C for methylbenzene and 50 °C for benzene; these temperatures are part of these named conditions.

Activation describes how fast the ring reacts relative to benzene; direction describes where substitution occurs. They are not the same property. Steric hindrance can reduce substitution next to a bulky group, so do not assume that every allowed position is formed in equal amounts.

Ordinary vigorous side-chain oxidation requires a hydrogen on the carbon directly attached to the ring. Methylbenzene satisfies this condition: C₆H₅CH₃ + 3[O] → C₆H₅COOH + H₂O. In alkaline oxidation the carboxylate forms first, so acidification is needed to isolate the carboxylic acid.

Gases such as CO₂, CH₄ and N₂O absorb outgoing infrared radiation. Increasing their atmospheric abundance strengthens the greenhouse effect and changes the climate energy balance. Distinguish this mechanism from ozone depletion and from local toxicity: a pollutant can contribute to more than one problem, but the chemical causes are not interchangeable.

A catalyst needs suitable temperature and contact with active sites. Removing CO, NO_x and hydrocarbons improves exhaust composition, while fuel use and carbon content still determine the associated carbon dioxide production. Assess a proposed fuel using energy output, incomplete-combustion products and wider emissions rather than one chemical label alone.

A beta hydrogen is on a carbon adjacent to the carbon bearing the leaving group. Elimination may give more than one alkene if different adjacent positions are available. Draw the carbon skeleton and mark eligible beta carbons before predicting products; do not simply erase Br and one arbitrarily placed H.

The primary amine product still has a nucleophilic lone pair and can react further with halogenoalkane to give secondary and tertiary amines, then a quaternary ammonium salt. Excess ammonia favours encounters with NH₃ and improves primary-amine formation. It does not make further substitution logically impossible.

Compare hydrolysis rates only with controlled temperature, solvent, concentrations and comparable carbon skeletons. For the same skeleton, RI usually hydrolyses faster than RBr, then RCl, because bond strength increases in the reverse order. A direct test of the original organic liquid is not a valid assumption that all covalently bound halogen is already present as X^-.

Strong C-F bonds contribute to the relative chemical inertness of many fluoroalkanes and fluorohalogenoalkanes. That stability has supported uses such as refrigerants and specialised fluids; suitability also depends on physical properties. Inertness near ground level can also allow a substance to persist long enough to reach other parts of the atmosphere.

The syllabus requires the environmental distinctions, not the detailed CFC/HCFC ozone-depletion radical mechanism. Avoid treating lower-atmosphere inertness, ozone impact and greenhouse effect as one single property.

Acidified dichromate(VI) changes from orange to green as it is reduced; acidified manganate(VII) loses its purple colour as Mn²⁺ forms. The colour change indicates oxidation has occurred but does not alone distinguish a primary from a secondary alcohol. Identify the product using a carbonyl test and, where needed, an aldehyde-specific test.

Distillation allows a volatile aldehyde to leave the oxidising mixture, limiting further oxidation. Reflux condenses vapour back into the flask, allowing prolonged heating without losing volatile reactants. Neither term itself identifies a reagent: write both the oxidant and the apparatus condition.

Aldehydes reduce to primary alcohols and ketones to secondary alcohols using LiAlH₄ in dry ether followed by aqueous work-up, or H₂/Ni under suitable hydrogenation conditions. For example, CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃. The carbon skeleton is retained.

Warm an appropriate alcohol with alkaline aqueous iodine. A yellow precipitate of tri-iodomethane, CHI₃, indicates an oxidisable CH₃CH(OH)R unit: oxidation first gives CH₃COR, which undergoes the iodoform reaction. Ethanol is included with R = H; propan-2-ol is positive, whereas propan-1-ol and a typical tertiary alcohol are negative.

In aqueous medium the acidity order is phenol > water > ethanol. Phenoxide is stabilised by delocalisation of negative charge through the aromatic system; ethoxide has no corresponding delocalisation and its electron-donating ethyl group destabilises the negative charge relative to hydroxide. Compare the conjugate bases, not just the O-H bond polarity.

Phenol reacts with NaOH but does not normally release CO₂ from aqueous hydrogencarbonate, whereas a carboxylic acid does. Ethanol does not react appreciably with aqueous NaOH to form ethoxide, although dry ethanol reacts with sodium metal. Metal reaction and aqueous base reaction are different tests.

Aldehyde oxidation is RCHO + [O] → RCOOH in acidic conditions; in alkaline test reagents, the carboxylate is the organic oxidation product. Ketones do not have the aldehydic H and resist these mild oxidations without carbon-carbon cleavage. Tollens and Fehling observations are therefore complementary evidence, not universal tests for every possible reducing organic compound.

The carbonyl iodoform transformation can be represented as RCOCH₃ + 3I₂ + 4OH^- → RCOO^- + CHI₃ + 3I^- + 3H₂O. The methyl carbon becomes CHI₃; the other carbonyl fragment becomes a carboxylate. Use that carbon accounting when deducing an unknown.

Carboxylic acids donate H⁺ from O-H. In the carboxylate, negative charge is delocalised over two oxygen atoms, making the C-O bonds equivalent in the resonance hybrid. This stabilises the conjugate base more effectively than in an alcohol, and generally more than in phenol, so carboxylic acids are appreciably stronger acids.

Chlorine substituents withdraw electron density through σ bonds and stabilise the carboxylate anion. Thus chloroethanoic acid is stronger than ethanoic acid; adding more chlorine atoms strengthens the electron-withdrawing effect. The inductive effect weakens with distance, so a chlorine closer to COOH generally increases acidity more than one farther away on an otherwise comparable chain.

Carboxylic acids usually release CO₂ with hydrogencarbonate or carbonate, unlike phenol under the ordinary test conditions. The acid-base products of benzoic acid follow the same pattern as ethanoic acid. In esterification, the acid-derived part keeps its carbonyl and the alcohol contributes the group attached through oxygen.

For a reversible esterification, using one reactant in excess or removing a product can improve equilibrium yield. Concentrated sulfuric acid catalyses the reaction; changing catalyst amount alone does not change the equilibrium constant at fixed temperature. Reflux supports heating, while subsequent separation, washing, drying and distillation may be selected according to the product's solubility and volatility.

Phenyl benzoate is C₆H₅COOC₆H₅, formed from benzoyl chloride and phenol/phenoxide. The phenyl group after the ester oxygen comes from phenol; the benzoyl portion comes from the acid chloride. Phenol does not esterify readily by the ordinary carboxylic-acid/alcohol method, so the acyl-chloride route is a useful distinction.

In an acyl chloride, nucleophilic attack occurs at the strongly δ⁺ carbonyl carbon; addition can be followed by loss of chloride. Hydrolysis is consequently much easier than in an alkyl chloride, which requires substitution at a saturated carbon and usually heating with an aqueous nucleophile. An aryl chloride such as chlorobenzene is still more resistant under those ordinary conditions because delocalisation strengthens C-Cl and the ring hinders the usual substitution pathways.

To identify an ester from hydrolysis, the acid-side carbonyl stays with the carboxylic fragment and the group after oxygen becomes the alcohol or phenol. Phenyl benzoate gives benzoic acid and phenol in acidic hydrolysis. In excess strong alkali, its products are benzoate and phenoxide because the phenol product is also deprotonated; acidification recovers the neutral compounds.

For nitrobenzene reduction, the formal redox change is C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O. In the acidic reaction medium the amine is protonated, so the later NaOH step is chemically necessary when the target is the free amine. Reduction changes NO₂ into NH₂; it does not put nitrogen into the aromatic ring.

An amine lone pair accepts H⁺ to form a salt: C₂H₅NH₂ + HCl → C₂H₅NH₃⁺Cl^-. Phenylamine similarly gives phenylammonium salts. Adding aqueous base reverses protonation. This change between neutral amine and charged salt can help separate an amine from neutral organic impurities.

Amines are Lewis bases because nitrogen can donate its lone pair to an electron-pair acceptor such as H⁺. In the gaseous phase, increasing alkyl substitution generally strengthens the basicity of comparable primary, secondary and tertiary alkylamines: tertiary > secondary > primary > ammonia. Alkyl electron donation increases nitrogen electron density and stabilises the protonated ion without competing hydration effects.

In aqueous medium, the specified order is ethylamine > ammonia > phenylamine. The ethyl group donates electron density, increasing lone-pair availability relative to ammonia. In phenylamine, the nitrogen lone pair is delocalised into the aromatic ring and is less available to accept a proton. Solvation of the base and its conjugate acid also contributes to an aqueous basicity, so do not transfer a gas-phase primary/secondary/tertiary order unchanged into water.

The same lone-pair donation that weakens phenylamine's basicity activates its aromatic ring towards electrophilic substitution. Aqueous bromine gives a white precipitate of 2,4,6-tribromophenylamine and loses its colour, without requiring an AlBr₃ catalyst. The NH₂ group directs substitution to its two ortho positions and para position.

A primary amine RNH₂ condenses with an acyl chloride R'COCl to form R'CONHR. The nitrogen remains attached to its original R group, and the acyl group comes from the chloride. For example, ethanoyl chloride + methylamine gives N-methylethanamide, CH₃CONHCH₃. A second equivalent of amine, or another base, can remove the HCl formed.

The amide nitrogen lone pair delocalises towards C=O. It is therefore much less available to accept H⁺ than an amine lone pair, and an ordinary amide is essentially neutral in water. The C-N bond has partial double-bond character. This does not mean an amide can never be protonated under strongly acidic conditions; it explains its lack of ordinary aqueous basic behaviour.

For an N-substituted amide, hydrolysis releases the corresponding amine or its ammonium ion instead of necessarily releasing NH₃/NH₄⁺. Reduction retains the N-substituent. For example, CH₃CONHCH₃ reduces to CH₃CH₂NHCH₃, a secondary amine, rather than losing its N-methyl group.

Aminoethanoic acid contains an amino group and a carboxylic acid group. Proton transfer can give the zwitterion H₃N⁺CH₂COO^-, with a positive ammonium group and a negative carboxylate group. These groups explain its amphoteric acid-base behaviour and the ionic character of the solid.

Amino acids with additional ionisable side-chain groups can have further acid-base steps. For an unfamiliar structure, identify every acid and base site and use the stated pH or pK values. Do not assume that all amino acids have exactly the same dominant form at one numerical pH.

Polymers are macromolecules built from monomers. The syllabus uses an average relative molecular mass of at least 1000 or at least 100 repeat units as its recognition convention. A polymer sample usually contains chains of different lengths, so an average relative molecular mass is meaningful.

A chain needs monomers with enough reactive sites to continue linking. A diol plus a dicarboxylic acid can extend at both ends; a monofunctional alcohol plus a monocarboxylic acid mainly gives a small ester. In an exact finite-chain condensation equation, the number of small molecules lost depends on the number of links and end groups, so do not infer it blindly from the repeat-unit bracket.

An α-amino acid has its amino group on the carbon next to COOH: H₂NCH(R)COOH. Condensation links these monomers through -C(=O)-NH- peptide bonds, giving a backbone of repeating -NH-CH(R)-CO- units. The R groups vary along a protein and do not replace the peptide backbone.

Heat proteins with aqueous acid or aqueous alkali to hydrolyse peptide links. Acidic hydrolysis produces amino-acid forms with protonated amino groups; alkaline hydrolysis produces carboxylate forms. For a simple amino-acid residue without extra ionisable side chains, these are H₃N⁺CH(R)COOH in sufficiently acidic solution and H₂NCH(R)COO^- in sufficiently alkaline solution.

Poly(alkenes) have strong C-C and C-H bonds and lack readily hydrolysable functional groups in their backbones. Their relative chemical inertness makes biological breakdown difficult. Cutting a plastic into smaller fragments does not by itself convert it into small harmless molecules; fragmentation and biodegradation are different processes.

Polyesters and polyamides contain ester or amide links that can be hydrolysed, so they are generally more susceptible to biodegradation by hydrolytic pathways than a comparable poly(alkene) backbone. Actual degradation still depends on temperature, water access, crystallinity, chain structure and suitable biological activity. The presence of an ester or amide does not guarantee rapid decay in every natural environment.

Materials and their feedstocks are finite resources. Reuse and effective recycling can reduce demand for new material, but mixed polymers, additives and contamination can make recovery difficult. A material derived from biological feedstock is not automatically biodegradable, and a biodegradable material is not automatically suitable for every recycling stream. State the tradeoff supported by the data rather than treating one label as a complete environmental verdict.

For preparation questions, name each reagent, essential medium and condition, plus the main product. For work-up, ask what remains dissolved in each phase: an acid/base wash can change ionisation and solubility; extraction separates phases; drying removes water; distillation separates sufficiently different volatilities. Choose these steps from the stated mixture instead of memorising one universal purification sequence.

Check every reagent with its essential condition: aqueous versus ethanolic NaOH; UV versus Lewis-acid halogenation; cold alkaline versus hot acidified KMnO₄; distillation versus reflux; acidic versus alkaline hydrolysis. Nitration specifically uses 50 °C for benzene and 30 °C for methylbenzene.

For structure deductions, combine observations: 2,4-DNPH detects aldehyde/ketone carbonyls; Tollens and Fehling help distinguish aldehydes; iodoform identifies the relevant methyl-carbonyl or oxidisable methyl-carbinol unit. Bromine-water decolourisation is not unique to an alkene. Track carbon count through CN substitution and every subsequent hydrolysis or reduction.

For mechanisms, arrows begin at electrons, intermediates must have valid bonds and charges, and catalysts must be regenerated. For stereochemistry, distinguish connectivity, restricted C=C rotation, tetrahedral chirality, S_N2 inversion and S_N1 loss of configuration. For polymers, draw bonds through repeat-unit brackets and identify the bond that hydrolysis actually cleaves.