Organic Chemistry Complete Study Notes Grade 11 – 12

Introduction to Organic Chemistry

Organic chemistry is defined as the scientific study of compounds that contain the element carbon. Carbon is particularly unique in the periodic table because of its ability to form long chains, rings, and branched structures, which has led to the existence of millions of different identified compounds. Historically, organic chemistry was a term that referred only to those compounds produced by living organisms. In the modern scientific context, it includes all carbon-containing compounds with a few specific exceptions that are classified as inorganic. These exceptions include carbon dioxide ($CO_2$), carbon monoxide ($CO$), carbonates, and carbides.

Carbon holds a special place in chemistry due to several specific characteristics. First, carbon possesses 4 valence electrons, allowing it to form 4 covalent bonds with other atoms. Second, carbon can bond to other carbon atoms to form extensive chains, a process known as catenation. Third, the bonds between carbon atoms can vary in order, including single bonds ($C-C$), double bonds ($C=C$), and triple bonds ($C\equiv C$). Finally, carbon is capable of forming strong, stable covalent bonds with a variety of other elements, specifically hydrogen, oxygen, nitrogen, sulfur, and the halogens.

Organic compounds are derived from several primary sources. The main industrial source is petroleum, also known as crude oil. Other significant sources include natural gas, coal, and various living organisms such as plants and animals. To represent these molecules, chemists use various formula types. The molecular formula shows the actual number of each atom in a molecule, such as $C_2H_6$ for ethane. The structural formula provides a complete visual representation of every bond between every atom, drawing out all $C-H$ and $C-C$ bonds. The semi-structural formula groups hydrogen atoms with their respective carbons, such as $CH_3CH_3$. The condensed structural formula is a more compressed version, often identical to the molecular formula in simple cases ($C_2H_6$). Finally, the skeletal or line-bond formula uses zigzag lines where each vertex or end represents a carbon atom, and hydrogen atoms are implied rather than drawn.

Functional Groups and Homologous Series

A functional group is defined as a specific atom or a group of atoms within a molecule that is responsible for the characteristic chemical reactions of that specific molecule. A homologous series refers to a group of organic compounds that share the same functional group and general formula, show a gradual change in physical properties, possess similar chemical properties, and differ from each other by a $-CH_2-$ unit.

Alkanes are characterized by single $C-C$ bonds only, follow the general formula $C_nH_{2n+2}$, and use the suffix "-ane". Alkenes contain at least one $C=C$ double bond, follow the general formula $C_nH_{2n}$, and use the suffix "-ene". Alkynes contain at least one $C\equiv C$ triple bond, follow the general formula $C_nH_{2n-2}$, and use the suffix "-yne". Haloalkanes contain a carbon-halogen bond ($C-X$, where $X = F, Cl, Br, I$) and are named using prefixes like fluoro-, chloro-, bromo-, or iodo-.

Alcohols contain the hydroxyl ($-OH$) functional group, follow the general formula $C_nH_{2n+1}OH$, and use the suffix "-ol". Aldehydes feature a carbonyl group at the end of the chain ($-CHO$), follow the general formula $C_nH_{2n}O$, and use the suffix "-al". Ketones feature a carbonyl group located within the carbon chain ($-CO-$), follow the general formula $C_nH_{2n}O$, and use the suffix "-one". Carboxylic acids contain the carboxyl functional group ($-COOH$), follow the general formula $C_nH_{2n}O_2$, and use the suffix "-oic acid". Esters contain the $-COO-$ functional group and are named with the suffix "-yl …-oate". Ethers feature an oxygen atom bonded between two carbons ($-O-$) and use the "-oxy-" suffix. Amines contain the $-NH_2$ group and use "-amine" or "-amino-", while amides contain the $-CONH_2$ group and use the suffix "-amide". It is a critical exam requirement to be able to identify these functional groups from structural formulas.

Detailed Study of Homologous Series

Alkanes are saturated hydrocarbons, meaning they contain only single bonds. The series begins with methane ($CH_4$), followed by ethane ($C_2H_6$), propane ($C_3H_8$), butane ($C_4H_{10}$), pentane ($C_5H_{12}$), hexane ($C_6H_{14}$), heptane ($C_7H_{16}$), octane ($C_8H_{18}$), nonane ($C_9H_{20}$), and decane ($C_{10}H_{22}$). Alkenes are unsaturated hydrocarbons with at least one double bond. The simplest alkene is ethene ($C_2H_4$). For chains with 4 or more carbons, the position of the double bond must be specified, such as in but-1-ene or but-2-ene. Alkynes are also unsaturated, featuring at least one triple bond. Ethyne ($C_2H_2$) is the simplest member, and triple bonds are noted for being highly reactive.

Alcohols are categorized based on the position of the hydroxyl group. A primary alcohol has the $-OH$ on a carbon bonded to only one other carbon (e.g., ethanol). A secondary alcohol has the $-OH$ on a carbon bonded to two other carbons (e.g., propan-2-ol). A tertiary alcohol has the $-OH$ on a carbon bonded to three other carbons. Examples include methanol ($CH_3OH$), ethanol ($C_2H_5OH$), and butan-1-ol ($C_4H_9OH$). Carboxylic acids, such as methanoic acid ($HCOOH$) and ethanoic acid ($CH_3COOH$), act as weak acids by donating $H^+$ ions in solution. Esters are the product of the reaction between a carboxylic acid and an alcohol. They are named by taking the alkyl group from the alcohol and changing the acid's name to end in "-oate". For instance, the reaction of ethanol and ethanoic acid produces ethyl ethanoate and water.

IUPAC Nomenclature

The International Union of Pure and Applied Chemistry (IUPAC) provides systematic rules for naming organic compounds. To name a straight-chain compound, first count the longest continuous carbon chain to determine the parent name. The prefixes are: 1 carbon = meth, 2 = eth, 3 = prop, 4 = but, 5 = pent, 6 = hex, 7 = hept, 8 = oct, 9 = non, and 10 = dec. Second, identify the functional group to determine the suffix. Third, number the chain from the end closest to the functional group or the first branch. Fourth, name and number all substituents or branches. Substituent names include methyl ($-CH_3$), ethyl ($-C_2H_5$), propyl ($-C_3H_7$), chloro ($-Cl$), bromo ($-Br$), fluoro ($-F$), iodo ($-I$), hydroxy ($-OH$), and amino ($-NH_2$). Finally, assemble the name using numbers followed by substituents in alphabetical order, then the parent chain and the suffix.

Common naming rules specify that if a substituent appears multiple times, prefixes like di-, tri-, or tetra- are used (e.g., 2,3-dimethylbutane). Substituents are listed alphabetically, ignoring the di/tri prefixes. Commas must be used between numbers, and hyphens must be used between numbers and letters. For alcohols, the carbon bearing the $-OH$ group gets the lowest possible number. For alkenes, numbering depends on the first carbon of the double bond, and for ketones, the carbonyl carbon must receive the lowest possible number. A common mistake is counting from the wrong end; one must always prioritize the functional group first, then the branches if the functional group is equidistant.

Physical Properties of Organic Compounds

The physical properties of these compounds are determined by intermolecular forces (IMF), not by the covalent bonds within the molecules. London (dispersion) forces occur in all molecules and increase with molar mass and surface area due to temporary induced dipoles. Dipole-dipole forces occur between polar molecules, such as haloalkanes, ketones, and aldehydes. Hydrogen bonding is the strongest IMF and occurs when hydrogen is bonded to oxygen, nitrogen, or fluorine, and interacts with a lone pair on another such atom. Hydrogen bonding is present in alcohols, carboxylic acids, and water.

Boiling point trends are influenced by three factors. First, longer chains (higher molar mass) lead to more electrons and stronger London forces, resulting in higher boiling points. Second, branching makes a molecule more compact and spherical, reducing the surface area and weakening London forces, which lowers the boiling point (e.g., pentane has a higher boiling point than 2,2-dimethylpropane). Third, the functional group dictates the type of IMF. Carboxylic acids have the highest boiling points because they can form hydrogen bonds and dimers. Alcohols follow with high boiling points due to hydrogen bonding. Aldehydes and ketones have moderate boiling points due to dipole-dipole forces, and alkanes have the lowest boiling points because they only possess London forces.

Solubility follows the "like dissolves like" principle. Polar solvents dissolve polar solutes. Short-chain alcohols and carboxylic acids are soluble in water because they can form hydrogen bonds with water molecules. As the carbon chain length increases, the non-polar hydrocarbon portion becomes dominant, making the compound less soluble in water. Alkanes and alkenes are non-polar and thus insoluble in water. Vapour pressure is inversely related to IMF; molecules with higher IMF have lower vapour pressure because it is harder for them to escape the liquid phase.

Chemical Reactions of Organic Compounds

Organic reactions are categorized into several types. Addition reactions involve two molecules joining to form one product as a double or triple bond opens, turning an unsaturated molecule into a saturated one. Elimination reactions involve one molecule losing atoms to form a double or triple bond, moving from saturated to unsaturated. Substitution reactions involve one atom or group replacing another. Combustion is the burning of a compound in oxygen to produce $CO_2$ and $H_2O$. Esterification is the reaction between an alcohol and a carboxylic acid to produce an ester and water. Hydrolysis is the breaking of a bond using water, the reverse of esterification. Oxidation involves the addition of oxygen or removal of hydrogen. Polymerisation involves many small monomers joining to form a large polymer.

In addition reactions for alkenes, several subtypes exist. Hydrogenation adds $H_2$ to an alkene to form an alkane using a $(Ni, Pt, \text{ or } Pd)$ catalyst under high temperature and pressure. Halogenation adds $X_2$ (like $Cl_2$ or $Br_2$) to form a dihaloalkane; this is used as a test for unsaturation where orange bromine water decolourises. Hydrohalogenation adds $HX$ to form a haloalkane. Hydration adds $H_2O$ using a phosphoric acid ($H_3PO_4$) catalyst at high temperature to form an alcohol. Markovnikov's Rule states that when $HX$ adds to an unsymmetrical alkene, the $H$ attaches to the carbon that already has more hydrogen atoms, while the $X$ attaches to the carbon with fewer hydrogen atoms. For example, the reaction of propene and $HBr$ primarily produces 2-bromopropane.

Elimination reactions include dehydrohalogenation, where a haloalkane reacts with hot, concentrated $KOH$ in an ethanol solvent to form an alkene, $KX$, and water. Dehydration involves an alcohol reacting with concentrated $H_2SO_4$ or $H_3PO_4$ at high temperatures (approximately $170\,^∘C$ for ethanol) to form an alkene and water. Notably, $KOH$ in water promotes substitution, while $KOH$ in ethanol promotes elimination. Substitution reactions include the conversion of a haloalkane to an alcohol using dilute $KOH$ in water at room temperature. Free radical substitution occurs when an alkane reacts with a halogen in the presence of UV light to form a haloalkane and a hydrogen halide.

Combustion can be complete or incomplete. Complete combustion in excess oxygen produces $CO_2$ and $H_2O$. For example: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$. Incomplete combustion in limited oxygen produces carbon monoxide ($CO$) or soot ($C$) and water. Carbon monoxide is dangerous because it is colorless, odorless, and binds to hemoglobin more strongly than oxygen. Alcohol oxidation uses acidified potassium dichromate ($K_2Cr_2O_7/H_2SO_4$) or potassium permanganate ($KMnO_4/H_2SO_4$). Primary alcohols are oxidized to aldehydes and then to carboxylic acids. Secondary alcohols are oxidized to ketones, while tertiary alcohols are resistant to oxidation. The color change for potassium dichromate is orange to green.

Isomers and Polymers

Isomers are compounds with the same molecular formula but different structural arrangements. Structural isomers include chain isomers (different chain lengths), positional isomers (different functional group positions), and functional group isomers (different functional groups entirely, such as ethanol and methoxymethane). Stereoisomers, specifically cis-trans or geometric isomers, occur around a double bond. Cis isomers have the same groups on the same side of the double bond, while trans isomers have them on opposite sides. This is only possible if both carbons of the $C=C$ bond have two different groups attached.

Polymers are large molecules made of repeating units called monomers. Addition polymerisation involves alkene monomers joining as the double bond opens, producing polymers like polyethene (PE) for plastic bags, polypropene (PP) for containers, polyvinyl chloride (PVC) for pipes, and polystyrene (PS) for packaging. Condensation polymerisation joins monomers with the loss of a small molecule, usually water. Polyesters are formed from a diol and a dicarboxylic acid, while polyamides like Nylon-6,6 are formed from a diamine and a dicarboxylic acid. Natural polymers include starch (glucose monomer), cellulose (glucose monomer), proteins (amino acid monomers), DNA (nucleotide monomers), and natural rubber (isoprene monomer).

Biological Macromolecules

Carbohydrates consist of carbon, hydrogen, and oxygen ($C_n(H_2O)_n$). They are classified as monosaccharides (glucose), disaccharides (sucrose), or polysaccharides (starch, cellulose). Proteins are polymers of amino acids linked by peptide bonds ($-CO-NH-$). They have four levels of structure: primary (sequence), secondary ($\alpha$-helix/$\beta$-sheet), tertiary (3D folding via H-bonds, ionic bonds, disulfide bridges), and quaternary (multiple chains). Denaturation occurs when heat or pH changes disrupt the tertiary/quaternary structure, causing a loss of function while the primary sequence remains unchanged.

Fats and oils, or lipids, are triglycerides formed from glycerol and three fatty acids. Saturated fats have no $C=C$ bonds and are typically solid animal fats. Unsaturated fats have $C=C$ bonds and are liquid plant oils. Saponification is the alkaline hydrolysis of a fat with $NaOH$ to produce glycerol and soap. Soap molecules form micelles in water to dissolve grease. Nucleic acids, DNA and RNA, are polymers of nucleotides containing a sugar, a phosphate, and a nitrogenous base. DNA uses deoxyribose and bases A, T, G, C (forming a double helix with $A-T$ and $G-C$ pairings), while RNA uses ribose and bases A, U, G, C.

Questions & Discussion

Q1: Give the IUPAC name of $CH_3CH(Cl)CH_2CH_3$ Answer: Longest chain = 4 carbons, which is butane. Chloro group is on carbon 2 to give the lowest locant. Name: 2-chlorobutane.

Q2: Draw the structural formula of 2-methylpropan-1-ol Answer: This is a 3-carbon chain (propan) with a methyl branch at C2 and an $-OH$ on C1. Structure: $(CH_3)_2CHCH_2OH$ or fully drawn as $CH_3-CH(CH_3)-CH_2-OH$.

Q3: Name and classify the isomers of $C_4H_8$ Answer: (1) But-1-ene: $CH_2=CHCH_2CH_3$; (2) But-2-ene: $CH_3CH=CHCH_3$; (3) 2-methylpropene: $CH_2=C(CH_3)_2$; (4) Cyclobutane. (1) and (2) are positional isomers. (1)/(2) and (3) are chain isomers.

Q4: What is the product when but-1-ene reacts with $HBr$? Name the product. Answer: Apply Markovnikov's Rule: $H$ adds to C1 and $Br$ adds to C2. Product: $CH_3CH_2CHBrCH_3$ which is 2-bromobutane.

Q5: Write the equation for the complete combustion of propane. Answer: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

Q6: What conditions are needed to convert butan-2-ol to but-2-ene? Answer: Concentrated $H_2SO_4$ or $H_3PO_4$ at high temperature (approximately $170\,^∘C$). This is a dehydration (elimination) reaction. Equation: $CH_3CH(OH)CH_2CH_3 \rightarrow CH_3CH=CHCH_3 + H_2O$.

Q7: Describe the test for an alkene. What do you observe? Answer: Add bromine water (orange/brown) to the compound. If a $C=C$ double bond is present, the bromine water is decolourised (turns colourless) due to the addition of $Br_2$ across the double bond.

Q8: Explain why ethanol has a higher boiling point than propane, even though propane has a higher molar mass. Answer: Ethanol ($CH_3CH_2OH$) has hydrogen bonding between molecules because of the $-OH$ group. Propane ($C_3H_8$) has only weak London dispersion forces as it lacks a polar group. The hydrogen bonds in ethanol are much stronger than the London forces in propane, resulting in a higher boiling point for ethanol.

Q9: Arrange in order of increasing boiling point: pentane, pentan-1-ol, pentanoic acid. Answer: Order is pentane < pentan-1-ol < pentanoic acid. Pentane has only London forces. Pentan-1-ol has hydrogen bonding. Pentanoic acid has hydrogen bonding and forms dimers ($2 -COOH$ groups bond together), requiring more energy to separate molecules.

Q10: Distinguish between addition and condensation polymerisation. Answer: Addition polymerisation involves monomers with a double bond ($C=C$) adding together without releasing by-products, requiring only one type of monomer. Example: polyethene. Condensation polymerisation involves two different monomers with two functional groups reacting, releasing a small molecule like $H_2O$ for each bond and producing ester or amide linkages. Example: PET.

Q11: Identify the monomer and type of polymerisation for PVC. Answer: Monomer: chloroethene (vinyl chloride) $CH_2=CHCl$. Type: Addition polymerisation. Repeat unit: $-[CH_2-CHCl]_n-$.

Q12: What is the difference between a saturated and unsaturated fat? Answer: Saturated fats have only single $C-C$ bonds in fatty acid chains, are solid at room temperature, and are mainly animal-based. Unsaturated fats have at least one $C=C$ double bond, are liquid at room temperature (oils), and are mainly plant or fish-based.

Q13: What happens during saponification? Write a general equation. Answer: Saponification is the alkaline hydrolysis of a fat/oil to produce glycerol and soap. Equation: Triglyceride + $3NaOH \rightarrow \text{glycerol} + 3 \text{ sodium fatty acid salts (soap)}$. Soaps work because the non-polar tail dissolves in grease and the ionic head is attracted to water, forming micelles.