Organic Chemistry

A molecular formula tells you how many of each atom there are. It does not tell you which organic compound you have. A structural formula must show clearly which atoms are joined. For example, CH₃CH(OH)COOH shows a chain of three carbons, an OH group on the middle carbon and a carboxylic acid group at the end. Its molecular formula is C₃H₆O₃, and its empirical formula is CH₂O.

To name a compound, choose the parent chain that contains the main functional group. Number the chain so that this group gets the lowest number it can. Then give the positions of double bonds and other groups. Examples are butan-2-ol, 2-bromopropane and propan-2-one. Do not pick the longest line on the page if it leaves out the functional group that gives the name.

R stands for any alkyl or organic group, and Ar stands for an aromatic group. They are labels, not atoms. For simple open-chain saturated compounds with one functional group, alcohols are C_nH_2n+2O, aldehydes and ketones are C_nH_2nO, and carboxylic acids and esters are C_nH_2nO₂. A ring, an extra double bond or another group changes these formulae, so use the structure when the molecule is more complex.

Hybridisation is a model of bonding. One s and three p orbitals combine to make four sp³ orbitals. One s and two p orbitals make three sp² orbitals. One s and one p orbital make two sp orbitals. Any p orbitals left over can overlap sideways to form π bonds. To decide the shape, count the regions of σ bonding around an atom. A double bond still counts as one direction.

Constitutional (structural) isomers have the same molecular formula but their atoms are joined differently. They can differ in the carbon chain, the position of a functional group, or the type of functional group. Butane and 2-methylpropane have different chains. Propan-1-ol and propan-2-ol have the OH in different places. Ethanol and methoxymethane are both C₂H₆O but have different functional groups.

Cis-trans isomerism happens in alkenes because the C=C bond cannot rotate. Rotating it would break the sideways overlap of p orbitals, which is the π bond. Each carbon of the double bond must carry two different groups. If either carbon has two identical groups, there are no cis and trans forms. In cis, the matching groups are on the same side. In trans, they are on opposite sides. You do not need E/Z names.

A chiral centre is usually a carbon bonded to four different atoms or groups. In butan-2-ol, the carbon with the OH is joined to OH, H, CH₃ and CH₂CH₃. So there are two arrangements that are mirror images and cannot be placed on top of each other. These are enantiomers. Compare whole groups, not just the first atom. Methyl and ethyl both start with carbon, but they are different groups.

When a molecule has several chiral centres, look at the whole molecule. A plane of symmetry inside it can make it achiral, even though it has chiral centres. A molecule can also be chiral for other reasons. So use the real shape and symmetry of the structure you are given. Do not just count centres.

An optically active sample rotates plane-polarised light. It contains chiral molecules whose rotations do not cancel. A racemic mixture has equal amounts of two enantiomers, so their rotations cancel. So a sample that does not rotate light may still contain chiral molecules. The direction of rotation must be measured. You cannot read it from a wedge drawing.

Receptors and enzymes in the body are chiral. Two stereoisomers can bind to them differently, so they can differ in activity, side effects or how the body breaks them down. That is why the molecular formula and functional groups alone cannot tell you how a drug behaves. The three-dimensional shape matters too. This does not mean one particular form is always the better one.

A curly arrow with a full head shows an electron pair moving. Its tail starts at a bond or a lone pair. Its head points to the atom or bond that receives the pair. A half-headed fishhook arrow shows one electron moving. Arrows show electrons moving, not atoms. A positive charge is never the start of an arrow, because it is not a source of electrons.

For an alcohol or halogenoalkane, primary, secondary and tertiary tell you how many carbon groups are joined to the carbon that carries the OH or halogen. For amines, count the carbon groups joined to the nitrogen: RNH₂, R₂NH and R₃N. R₄N⁺ is a quaternary ammonium ion. A quaternary carbon is a carbon bonded to four other carbons.

For similar halogenoalkanes, hydrolysis gets easier from RCl to RBr to RI. This is because the C-Cl bond is the strongest and the C-I bond is the weakest. Bond polarity alone does not explain the trend. Before you compare rates, make sure the carbon chain, solvent and mechanism are the same.

In chlorobenzene, a lone pair on chlorine overlaps with the π electrons of the ring. This gives the C-Cl bond some double-bond character, so it is harder to break than in a halogenoalkane. The flat ring also blocks a nucleophile from attacking the back of that carbon. Losing chloride first would make a very unstable phenyl cation. So chlorobenzene does not undergo nucleophilic substitution under the conditions that hydrolyse halogenoalkanes.

Benzene has electrons that can attack an electrophile. But breaking up its delocalised ring costs a lot of stabilisation energy. So benzene needs a stronger electrophile, often made with a catalyst, than an alkene does. Afterwards, the intermediate loses H⁺ and the delocalised ring is restored. That is why benzene undergoes substitution, not addition.

Ethane reacts with chlorine in ultraviolet light at room temperature to form chloroethane and HCl. The UV light splits Cl-Cl evenly, one electron to each atom. This is homolytic fission, and it makes radicals, not ions. One chlorine radical can keep many propagation steps going before a termination step removes it.

Adding the propagation steps cancels the ethyl and chlorine radicals. The overall equation is CH₃CH₃ + Cl₂ → CH₃CH₂Cl + HCl. More hydrogens can also be replaced, so a mixture of products forms. Using excess ethane makes it less likely that a radical attacks the product. Bromination in UV light follows the same three stages with bromine radicals.

An alkene reacts with electrophiles because of its π bond. The π electrons sit above and below the C=C bond, further from the carbon nuclei than the σ electrons. They are held less tightly, so the C=C bond is an electron-rich site that is easy to reach. An electrophile is an electron-pair acceptor. It is attracted to this electron-rich region, and the π bond donates a pair of electrons to it.

Step 1 (slow). Br₂ is non-polar on its own. As it comes close to ethene, the electron-rich π bond repels the electrons in the Br-Br bond. This gives Br₂ an induced dipole: the nearer Br becomes δ+ and the further Br becomes δ-. The π electrons move to form a new C-Br bond with the δ+ bromine. At the same time the Br-Br bond breaks by heterolytic fission: both bonding electrons go to the δ- bromine, which leaves as Br^-. The other carbon has lost its share of the π pair, so it now has only three bonds and a positive charge. This ion is a carbocation. The step is slow because it breaks the π bond and the Br-Br bond and creates two ions.

Step 2 (fast). The bromide ion donates a lone pair to the positive carbon. This forms the second C-Br bond. Oppositely charged ions attract strongly and no bond has to break, so this step is fast. The product is BrCH₂CH₂Br, 1,2-dibromoethane.

Addition of HX follows the same two-step pattern. The π bond accepts H⁺ from the δ+ H of H-X while the H-X bond pair moves onto X. This forms a carbocation and X^-. Then X^- donates a lone pair to the positive carbon. For propene + HBr, the step that gives the secondary carbocation is favoured over the one that gives the primary carbocation, so 2-bromopropane is the major product.

Markovnikov's rule predicts the major product. It does not mean the minor product is impossible: a little 1-bromopropane still forms through the primary carbocation. The same carbocation argument decides orientation when the electrophile is δ+ Br from Br₂, as the worked example below shows. Radical addition in the presence of peroxides is outside this mechanism.

To add one bromine to benzene, use Br₂ with anhydrous AlBr₃. AlBr₃ is a Lewis acid. It accepts electron density from Br₂ and makes it strongly polar. You can write this as Br₂ + AlBr₃ ⇌ Br⁺ + AlBr₄^-. This equation shows where the electrons go. It does not mean free Br⁺ ions sit in the bottle.

Overall, C₆H₆ + Br₂ → C₆H₅Br + HBr. Substitution replaces an H and keeps the delocalised ring. Addition would destroy the ring and its extra stability, which is why benzene prefers substitution. Do not draw a full circle inside the intermediate. The catalyst is made again, so it must not appear as used up in the overall equation.

In S_N2, a nucleophile gives a pair of electrons to the carbon that carries the leaving group. At the same time, the C-X bond pair moves onto X. Both changes happen together, in one step with one transition state. No carbocation forms. The nucleophile comes in from the side opposite the leaving group, so bulky groups around that carbon get in the way and slow the reaction a lot.

For an S_N2 step, rate = k[halogenoalkane][nucleophile]. Methyl and primary halogenoalkanes are easier to reach than secondary ones. Tertiary ones are strongly blocked. These are general preferences. The solvent, the nucleophile and competing elimination still matter.

Inversion of configuration means the other three groups flip over, like an umbrella in the wind, as the nucleophile replaces the leaving group from the opposite side. So a single enantiomer gives the inverted form, not a racemic mixture. Turning the drawing of the product round on the page is not inversion.

The nucleophile then gives a lone pair to the positive carbon. With OH^-, this gives the alcohol directly. With H₂O, it first gives a protonated alcohol, which then loses H⁺. The rate equation is rate = k[halogenoalkane], because the slow step, ionisation, happens before the nucleophile attacks.

For simple carbocations, tertiary is more stable than secondary, which is more stable than primary. Alkyl groups push electron density towards the positive carbon and spread out its charge. So a halogenoalkane that would form an unstable primary carbocation is unlikely to react by S_N1.

The carbocation is flat (trigonal planar). If the starting compound is chiral, the nucleophile can attack from either face. In the simple model this gives equal amounts of both enantiomers, which is racemisation. In reality the leaving ion can partly block one face, so the mixture may not be exactly equal. The key contrast is with S_N2, which gives inversion. The tert-butyl example shown is itself achiral, because its three methyl groups are identical.

Use HCN with a small amount of KCN as a catalyst. HCN alone gives very little CN^-, and the cyanide ion is the nucleophile. The carbon of CN^- gives a lone pair to the δ⁺ carbon of the carbonyl group. At the same time, the C=O π pair moves onto the oxygen. HCN then gives H⁺ to the negative oxygen. This forms the hydroxynitrile and makes CN^- again.

An aldehyde RCHO gives RCH(OH)CN. A ketone R₂CO gives R₂C(OH)CN. The carbonyl group is flat, so CN^- can attack from either face. If this creates a new chiral centre, a racemic mixture forms. Ethanal creates a chiral centre. Propanone does not, because its product has two identical methyl groups.

The complete combustion of ethane is 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. With too little oxygen, carbon monoxide or carbon can form. Halogenation is a different reaction. Cl₂ or Br₂ in UV light at room temperature replaces a C-H bond with C-X, by the radical chain described earlier.

In CCl₄, the only nucleophile is Br^-, so the product is the dibromide. In water, water molecules attack the carbocation, so the main product is a bromoalcohol. Always check which nucleophiles are present before you write the product. In the bromine-water test, the colour disappears quickly. This is a useful sign, but not only alkenes do it: some activated aromatic compounds also use up bromine.

A cyclic alkene opens into one molecule with two oxygen-containing ends, not two separate molecules. Count the carbon atoms in every product, including CO₂, before you suggest the starting structure. Cold alkaline and hot acidified manganate(VII) give different results. Cold alkaline gives a diol. Hot acidified breaks the double bond.

Concentrated sulfuric acid protonates nitric acid, which forms NO₂⁺: HNO₃ + H₂SO₄ ⇌ NO₂⁺ + HSO₄^- + H₂O. NO₂⁺ is the electrophile. Nitration then follows electrophilic substitution, and the acid catalyst is made again. The syllabus states 30 °C for methylbenzene and 50 °C for benzene, so learn these temperatures as part of the conditions.

Activation tells you how fast the ring reacts compared with benzene. Direction tells you where the new group goes. These are two different properties. A bulky group can block the positions next to it, so do not assume every allowed position forms the same amount of product.

Strong oxidation of a side chain needs a hydrogen on the carbon joined directly to the ring. Methylbenzene has one: C₆H₅CH₃ + 3[O] → C₆H₅COOH + H₂O. In alkaline conditions the carboxylate ion forms first, so you must add acid to get the carboxylic acid.

Gases such as CO₂, CH₄ and N₂O absorb infrared radiation given off by the Earth. More of these gases in the air strengthens the greenhouse effect and changes the climate. This is a different cause from ozone depletion and from local poisoning. One pollutant can add to more than one problem, but the chemical causes are different.

A catalytic converter needs to be hot enough, and the gases must reach its active sites. Removing CO, NO_x and unburnt hydrocarbons makes exhaust cleaner. But the amount of fuel burned and its carbon content still decide how much CO₂ is made. To judge a fuel, look at its energy output, its incomplete-combustion products and its other emissions, not one label alone.

A beta hydrogen is on a carbon next to the carbon that carries the leaving group. If there are different neighbouring carbons with hydrogens, elimination can give more than one alkene. Draw the carbon chain and mark each beta carbon before you predict the products. Do not just remove Br and any H you choose.

The primary amine formed still has a lone pair, so it is still a nucleophile. It can react with more halogenoalkane to give secondary and tertiary amines, and then a quaternary ammonium salt. Using excess ammonia means the halogenoalkane is more likely to meet NH₃ than the amine, so more primary amine forms. Further substitution is still possible.

Compare hydrolysis rates only when the temperature, solvent, concentrations and carbon chains are the same. For the same chain, RI usually hydrolyses faster than RBr, which is faster than RCl, because the bond strength rises in the opposite order. Testing the organic liquid directly does not work, because the halogen is bonded to carbon, not present as X^- ions.

Strong C-F bonds help make many fluoroalkanes and fluorohalogenoalkanes unreactive. That stability made them useful as refrigerants and special fluids, together with suitable physical properties. But because they are so unreactive near the ground, they can last long enough to reach the upper atmosphere.

The syllabus asks for these environmental differences, not the detailed radical mechanism of ozone depletion by CFCs and HCFCs. Do not treat low reactivity near the ground, harm to ozone and the greenhouse effect as one single property.

Acidified dichromate(VI) turns from orange to green as it is reduced. Acidified manganate(VII) loses its purple colour as Mn²⁺ forms. Either colour change shows that oxidation happened. It does not tell a primary alcohol from a secondary one. To identify the product, use a carbonyl test and, if needed, a test that only aldehydes pass.

In distillation, the aldehyde boils off as it forms and leaves the oxidising mixture, so it is not oxidised further. In reflux, the vapour condenses and drips back into the flask, so you can heat for a long time without losing volatile substances. Neither word names a reagent, so write both the oxidising agent and the apparatus condition.

Aldehydes are reduced to primary alcohols, and ketones to secondary alcohols. Use LiAlH₄ in dry ether followed by water, or H₂ with a nickel catalyst. For example, CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃. The carbon chain stays the same.

Warm the alcohol with alkaline aqueous iodine. A yellow precipitate of tri-iodomethane, CHI₃, shows that the alcohol has a CH₃CH(OH)R unit. The alcohol is first oxidised to CH₃COR, which then gives the iodoform reaction. Ethanol counts, with R = H. Propan-2-ol gives a positive test. Propan-1-ol and a typical tertiary alcohol do not.

In water, the order of acid strength is phenol > water > ethanol. In the phenoxide ion, the negative charge spreads into the benzene ring, which stabilises it. The ethoxide ion cannot spread its charge. Its ethyl group also pushes electrons onto the oxygen, which makes the ion less stable than hydroxide. Compare how stable the anions are, not just how polar the O-H bond is.

Phenol reacts with NaOH, but it does not release CO₂ from aqueous hydrogencarbonate. A carboxylic acid does both. Ethanol hardly reacts with aqueous NaOH, although dry ethanol does react with sodium metal. Reacting with a metal and reacting with an aqueous base are different tests.

In acidic conditions, an aldehyde is oxidised to a carboxylic acid: RCHO + [O] → RCOOH. In the alkaline test reagents, the product is the carboxylate ion. Ketones have no H on the carbonyl carbon, so these mild oxidising agents do not oxidise them. Tollens and Fehling results support each other. They are not tests for every organic compound that can act as a reducing agent.

The iodoform reaction of a methyl ketone is RCOCH₃ + 3I₂ + 4OH^- → RCOO^- + CHI₃ + 3I^- + 3H₂O. The methyl carbon becomes CHI₃, and the rest of the molecule becomes a carboxylate with one carbon fewer. Use this carbon count when you work out an unknown.

Carboxylic acids release H⁺ from their O-H group. In the carboxylate ion, the negative charge is spread over two oxygen atoms, so the two C-O bonds become the same. This makes the anion more stable than the anion of an alcohol, and usually more stable than phenoxide. So carboxylic acids are clearly stronger acids.

Chlorine atoms pull electron density away through the σ bonds and spread out the negative charge of the carboxylate ion. This makes the anion more stable, so chloroethanoic acid is stronger than ethanoic acid. More chlorine atoms pull more strongly and make the acid stronger still. The effect weakens with distance, so a chlorine closer to the COOH group raises the acid strength more.

Carboxylic acids release CO₂ with hydrogencarbonate or carbonate. Phenol does not, under the usual test conditions. Benzoic acid forms the same kinds of products as ethanoic acid. In esterification, the part from the acid keeps its C=O, and the alcohol provides the group joined through oxygen.

Esterification is reversible. Using one reactant in excess, or removing a product, raises the equilibrium yield. Concentrated sulfuric acid is the catalyst. Using more catalyst does not change the equilibrium constant at a fixed temperature. Heat under reflux. Then choose separation, washing, drying and distillation steps that suit how soluble and volatile the product is.

Phenyl benzoate is C₆H₅COOC₆H₅. It forms from benzoyl chloride and phenol or phenoxide. The phenyl group after the ester oxygen comes from phenol, and the benzoyl part comes from the acyl chloride. Phenol does not form esters easily with a carboxylic acid, so the acyl chloride route is needed.

In an acyl chloride, the carbonyl carbon is strongly δ⁺, so a nucleophile attacks it. The nucleophile adds, then chloride leaves. So an acyl chloride hydrolyses much more easily than an alkyl chloride, which needs substitution at a saturated carbon and usually heating with an aqueous nucleophile. An aryl chloride such as chlorobenzene is even harder to hydrolyse. Delocalisation strengthens its C-Cl bond and the ring blocks the usual substitution routes.

To identify an ester from its hydrolysis, remember that the C=O stays with the acid part, and the group after the oxygen becomes the alcohol or phenol. Phenyl benzoate gives benzoic acid and phenol with acid. In excess strong alkali, it gives benzoate and phenoxide ions, because phenol also loses its proton. Adding acid afterwards gives back the neutral compounds.

The reduction of nitrobenzene is C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O. The reaction mixture is acidic, so the amine forms as its protonated salt. That is why you add NaOH afterwards to get the free amine. Reduction turns the NO₂ group into NH₂. It does not put nitrogen into the ring.

The lone pair on an amine nitrogen accepts H⁺ to form a salt: C₂H₅NH₂ + HCl → C₂H₅NH₃⁺Cl^-. Phenylamine forms phenylammonium salts in the same way. Adding aqueous base reverses this. Switching between the neutral amine and its 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 gas phase, more alkyl groups make a stronger base: tertiary > secondary > primary > ammonia. Alkyl groups push electron density onto nitrogen and stabilise the protonated ion. In the gas phase there is no water to complicate this.

In water, the order you need is ethylamine > ammonia > phenylamine. The ethyl group pushes electron density onto nitrogen, so its lone pair is more available than in ammonia. In phenylamine, the lone pair spreads into the benzene ring, so it is less available to accept a proton. In water, how well the ions are surrounded by water molecules also matters. So do not use the gas-phase order of primary, secondary and tertiary amines for water.

The same lone-pair donation that makes phenylamine a weak base also makes its ring more reactive towards electrophiles. Aqueous bromine gives a white precipitate of 2,4,6-tribromophenylamine and loses its colour. No AlBr₃ catalyst is needed. The NH₂ group sends the bromine to the two positions next to it and the position opposite it.

A primary amine, RNH₂, reacts with an acyl chloride, R'COCl, to form R'CONHR. The nitrogen keeps its own R group, and the acyl group comes from the chloride. For example, ethanoyl chloride and methylamine give N-methylethanamide, CH₃CONHCH₃. A second amine molecule, or another base, removes the HCl formed.

In an amide, the nitrogen lone pair spreads towards the C=O group. So it is much less available to accept H⁺ than the lone pair of an amine, and an amide is almost neutral in water. The C-N bond has some double-bond character. An amide can still be protonated in strong acid; it just does not act as a base in ordinary aqueous conditions.

When an N-substituted amide is hydrolysed, it releases the matching amine or its ammonium ion, not NH₃ or NH₄⁺. Reduction keeps the group on nitrogen too. For example, CH₃CONHCH₃ is reduced to CH₃CH₂NHCH₃, a secondary amine. It does not lose its N-methyl group.

Aminoethanoic acid has an amino group and a carboxylic acid group. A proton can move from one to the other, giving the zwitterion H₃N⁺CH₂COO^-. It has a positive ammonium group and a negative carboxylate group. These groups explain why it reacts with both acids and bases, and why the solid behaves like an ionic substance.

Some amino acids have extra acidic or basic groups in their side chains, which add further steps. For an unfamiliar structure, find every acid and base site and use the pH or pK values given. Do not assume all amino acids have the same main form at one pH value.

Polymers are very large molecules built from monomers. The syllabus counts a molecule as a polymer if its average relative molecular mass is at least 1000, or if it has at least 100 repeat units. A polymer sample contains chains of different lengths, which is why we use an average relative molecular mass.

To make a chain, each monomer needs at least two reactive groups so that linking can continue. A diol and a dicarboxylic acid can both react at each end, so the chain grows. An alcohol and a carboxylic acid with one group each just make a small ester. In an exact equation for a chain of set length, the number of small molecules lost depends on the number of links and end groups, so do not read it straight from the repeat unit.

An α-amino acid has its amino group on the carbon next to the COOH group: H₂NCH(R)COOH. Condensation joins these monomers through -C(=O)-NH- peptide bonds. The backbone repeats -NH-CH(R)-CO- units. The R groups vary along a protein, but the peptide backbone stays the same.

To hydrolyse the peptide links in a protein, heat it with aqueous acid or aqueous alkali. In acid, the amino groups of the amino acids are protonated. In alkali, the carboxylic acid groups become carboxylate ions. For a simple amino acid with no extra acidic or basic side group, you get H₃N⁺CH(R)COOH in acid and H₂NCH(R)COO^- in alkali.

Poly(alkenes) have strong C-C and C-H bonds and no groups in the chain that water can easily split. Because they are so unreactive, living things find them hard to break down. Cutting plastic into smaller pieces does not turn it into small harmless molecules. Breaking into fragments and biodegrading are different processes.

Polyesters and polyamides contain ester or amide links that can be hydrolysed. So they usually biodegrade more easily than a poly(alkene) chain. How fast they actually break down still depends on temperature, water, how crystalline the polymer is, the chain structure and suitable microorganisms. Having ester or amide links does not guarantee fast decay everywhere.

The materials and the raw materials used to make them are limited. Reusing and recycling can reduce the need for new material. But mixed polymers, additives and dirt can make recovery hard. A material made from plants is not automatically biodegradable. A biodegradable material is not suitable for every recycling stream. State the trade-off that the data support, and do not judge a material by one label alone.

In preparation questions, name each reagent, the medium, the conditions and the main product. For separating the product, ask what dissolves in each layer. Washing with acid or base can change whether a substance is ionised, and so how soluble it is. Extraction separates layers. Drying removes water. Distillation separates liquids with different boiling points. Choose the steps that suit the mixture given, rather than learning one fixed 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.