Introduction To Organic Chemistry

## Explain carbon skeleton diversity - Carbon commonly forms four covalent bonds. Strong carbon–carbon bonds allow connected chains, branches and rings; this capacity is catenation. - The skeletal comparison shows three different carbon frameworks with the same total carbon count. Line ends and junctions represent carbon atoms unless another element is written. - An ordinary displayed or skeletal formula is a connectivity model. A zigzag drawing does not mean a saturated chain is geometrically straight or all its atoms lie in one plane. ## Recognise a homologous series - Members of a specified homologous series share a characteristic functional group or bonding pattern and a common general formula. Neighbouring members differ by CH₂. - They have similar characteristic chemistry rather than identical behaviour under every condition. Molecular size also changes physical properties. - The two alcohol examples have one hydroxyl group each. A formula alone may include other families, so inspect connectivity before assigning a series. ## Interpret a physical-property pattern - The supplied values show boiling point rising from methane to hexane at comparable pressure. Larger electron clouds generally produce stronger London interactions. - Boiling separates molecules; it does not normally break the covalent carbon skeleton. Members still share the alkane bonding pattern. - Do not claim that every physical property changes monotonically. The melting temperatures in the same record show packing effects and do not rise at every step. ## Separate hydrocarbons from other carbon compounds - A hydrocarbon contains only carbon and hydrogen. Methane and ethene meet that definition; ethanol contains oxygen as well. - Organic chemistry largely concerns carbon compounds, but common carbon oxides and carbonates are conventionally treated as inorganic. Carbon content alone is not a complete classification test. - Inspect the elements and bonding in the supplied formulas. A hydrocarbon need not be an alkane:it can contain multiple bonds or rings. ## Identify local functional-group patterns - A functional group is an atom or group of atoms associated with characteristic reactions. Its local bonding matters, not just the list of elements. - Alkenes contain C=C; alcohols have an OH bonded to a suitable saturated carbon; halogenoalkanes have a carbon–halogen bond in the relevant framework. - The displayed ethanol formula shows that carbon is bonded to oxygen, which is bonded to hydrogen. Writing C–H–O would assign hydrogen an impossible two-bond role in this ordinary neutral model. ## Distinguish carbonyl-containing families - Aldehydes include a carbonyl carbon bonded to H; ketones have carbon groups on both sides of the carbonyl carbon. - Carboxylic acids contain C(=O)OH. Esters contain C(=O)O joined onward to another carbon group. A carbonyl alone does not distinguish all these families. - Count the carbonyl carbon when counting the carbon framework. Inspect which atoms are directly connected to it before choosing the family or suffix. ## Count nitrogen-containing functional groups - A primary amine can be represented by RNH₂ with R an appropriate carbon group. Other amines have more carbon groups bonded to nitrogen. - A nitrile contains a carbon triple-bonded to nitrogen. That nitrile carbon belongs to the total carbon count and usually to the parent name. - R is a context-defined placeholder, not an element symbol or always any arbitrary atom. For these examples it denotes an appropriate hydrocarbon group. ## Relate molecular and empirical formulas - A molecular formula records the actual count of each element in one molecule. It does not generally specify which atoms are bonded together. - An empirical formula gives the simplest whole-number element ratio. Reduce all subscripts by their common divisor; do not remove an element. - Ethanoic acid has molecular formula C₂H₄O₂ and empirical formula CH₂O. A different molecular formula can have the same empirical ratio without being an isomer. ## State the limits of a general formula - Acyclic saturated alkanes obey CₙH₂ₙ₊₂. Acyclic hydrocarbons with one C=C and no other unsaturation obey CₙH₂ₙ. - A saturated monocyclic hydrocarbon also has CₙH₂ₙ. The same formula pattern can describe different bonding arrangements. - Always state the family constraints before using a formula to infer hydrogen count. Rings or additional multiple bonds change the relation. ## Read a displayed formula completely - A displayed formula shows every atom symbol and every covalent bond, including the C–H bonds. Double bonds require two bond lines. - The ethanol illustration has two carbons, one oxygen and six hydrogens. Its single O–H bond is part of the displayed connectivity. - Displayed formulas are not complete three-dimensional geometry measurements. A flat arrangement of C–H bonds is not a claim that a tetrahedral carbon is square planar. ## Expand brackets without losing atoms - A condensed structural formula groups atoms while retaining enough sequence and brackets to show connectivity. CH₃CH(CH₃)CH₂CH₃ has a methyl branch. - The main four-carbon path and the extra methyl carbon make five carbons in total. Do not confuse the parent-chain length with total molecular carbon count. - Expand the bracket at its attachment carbon before deriving the molecular formula or choosing an isomer comparison. ## Recover atoms from a skeletal formula - In a skeletal formula, unlabelled line ends and corners normally represent carbon. Hydrogens attached to carbon are omitted and recovered from carbon valence four. - Heteroatoms and relevant heteroatom-bound hydrogens remain explicit in the simple examples here. A labelled O endpoint is oxygen rather than an extra carbon. - The ethanol skeletal drawing retains the OH group. A single unlabelled line ending at O cannot be counted as carbon at that labelled endpoint. ## Choose a parent chain before naming branches - The parent must include the relevant principal group and required multiple bond in the naming cases considered. Then use the appropriate longest allowed chain. - For a simple branched alkane, identify the longest continuous carbon path. It may bend across the page rather than follow a visually horizontal line. - The stem counts the chosen parent carbons, not every carbon in branches:meth, eth, prop, but, pent and hex correspond to one through six. ## Compare locant sets at the first difference - Choose numbering according to the appropriate naming priorities. For equivalent parent chains of simple substituted alkanes, compare ordered locant sets at the first point of difference. - Do not choose numbering merely because the sum of locants is smaller. A first lower locant can decide the comparison before later numbers are considered. - Repeated identical branches use di, tri or tetra. Commas separate locants; hyphens separate locants from words. ## Order different substituent names alphabetically - After locating substituents, list different simple substituent names alphabetically. Multipliers such as di and tri do not decide this ordering. - The supplied compound has an ethyl group and two methyl groups on a six-carbon parent. Ethyl is listed before methyl despite methyl’s multiplicity. - Punctuation communicates the numbered structure. Keep commas between repeated locants and hyphens between numbers and words. ## Give the principal group its numbering priority - An instruction to choose the smallest numbers is incomplete without stating what has priority. In the simple alcohol case, the hydroxyl suffix locant precedes a methyl substituent locant. - For CH₃CH(OH)CH(CH₃)CH₃, the preferred alcohol numbering gives OH at 2 and methyl at 3. Reversing would give OH at 3. - A carbon–carbon double-bond locant identifies its lower-numbered carbon. Use names such as but-2-ene and butan-2-ol with the correct stem and suffix placement. ## Include terminal functional-group carbons - For a simple aldehyde, the carbonyl carbon carrying H belongs to the parent and is carbon 1. A simple monocarboxylic acid similarly counts its carboxyl carbon as 1. - CH₃CH₂CHO is propanal, not ethanal:the CHO carbon is the third carbon in the molecular count. CH₃CH₂COOH is propanoic acid. - An internal ketone needs a locant where necessary. Naming identifies connectivity rather than merely counting visible CH₃ or CH₂ groups. ## Count both sides of an ester and the nitrile carbon - An ester name starts with the alkyl group attached to oxygen, then identifies the acid-derived alkanoate portion. Count the carbonyl carbon in that second portion. - Ethyl ethanoate has two carbons on the alcohol-derived side and two on the acid-derived side. The molecular formula is C₄H₈O₂. - In CH₃CH₂CN, the nitrile carbon is counted, so the name is propanenitrile. Nitrogen is not a carbon site and does not add another carbon to the parent. ## Require molecular identity before calling compounds isomers - Structural isomers have the same molecular formula but different atom connectivity. Matching empirical formulas alone is insufficient. - Molecular formulas can conceal chain, position or functional-group differences. Reversing a drawing without changing connectivity does not make a new structural isomer. - Always establish actual atom counts first, then compare bonded neighbours. Spatial rearrangement with unchanged connectivity belongs to stereoisomerism. ## Compare carbon skeletons rather than drawing shapes - The three skeletal panels represent pentane,2-methylbutane and 2,2-dimethylpropane from top to bottom. Each contains five carbon sites and gives C₅H₁₂. - They are chain isomers because the carbon connectivity differs. A branch can change the longest path, but longest-chain length alone is not a complete definition of chain isomerism. - The cross in the bottom panel represents one four-connected carbon with four terminal carbons. It is a connectivity notation, not square-planar carbon geometry. ## Recognise position isomerism on a common framework - Position isomers retain the same functional-group type and carbon framework while its position changes. Butan-1-ol and butan-2-ol both have C₄H₁₀O. - Number from the appropriate end before comparing locants. Butan-3-ol is the same connectivity as butan-2-ol viewed from the other end. - A shifted double bond or halogen can also give positional isomers where symmetry does not make the drawings identical. ## Change the functional group while retaining the formula - Butan-1-ol and ethoxyethane both have C₄H₁₀O, but one has an alcohol group and the other an ether linkage. They are functional-group isomers. - An aldehyde and a ketone can likewise share a formula. Different local bonding often produces different characteristic reactions. - Functional-group isomerism is not a claim that every property differs in every comparison. Use the actual structures and conditions to predict chemistry. ## Enumerate structural isomers without symmetry duplicates - For dibromopropane, keep the three-carbon skeleton and place two Br substitutions systematically. Reversing the chain removes duplicate locant patterns. - The distinct structural patterns are 1,1;1,2;1,3; and 2,2. Patterns 2,3 and 3,3 duplicate 1,2 and 1,1 by reversal. - This is a structural count. The possible optical stereoisomers of a structure are a separate question and must not be silently added to it. ## Distinguish conformations from new connectivity - Rotation around a carbon–carbon single bond can change the appearance of a chain while leaving bonded neighbours unchanged. - Two bent drawings of propane are not automatically different structural isomers. Trace the atom connections and identify symmetry before counting. - Stereochemical classification must be used carefully:ordinary bond rotations can describe conformations, but they do not turn butane into 2-methylpropane. ## Define stereoisomerism using connectivity and space - Stereoisomers have the same molecular formula and atom connectivity but differ in spatial arrangement. E/Z alkene isomers are one example. - The two but-2-ene structures preserve CH₃–CH=CH–CH₃ connectivity. They differ in how substituents are arranged across the double bond. - The distinction from structural isomerism is the unchanged bonded-neighbour pattern. Optical isomerism is developed in its separate later lesson. ## Check both ends before assigning E or Z - The carbon–carbon double bond restricts rotation because twisting disrupts the sideways overlap responsible for the π bond. - For E/Z isomerism at the ordinary alkene in this course, each double-bond carbon must carry two different substituents. Restricted rotation alone is insufficient. - The two substituents at one end may match groups at the other end. Four mutually different groups are not required:but-2-ene already satisfies the condition. ## Rank the first directly bonded atoms - CIP priority begins with the atomic number of the atom directly attached to the double-bond carbon. Compare the two substituents at each carbon separately. - For Br versus H, the directly attached bromine outranks hydrogen. For OH versus CH₃, the oxygen outranks carbon. - Do not choose priority from group size, visual position or total group mass. The first point of difference in the prescribed sequence decides. ## Resolve equal first atoms at the next difference - If both directly attached atoms are carbon, compare their next attached atoms in descending atomic-number order, excluding the alkene carbon being traced back to. - CH₂Br supplies Br, H, H whereas CH₂CH₃ supplies C, H, H; the bromine entry wins at the first difference. CH₂OH similarly outranks CH₃ because O outranks H at that next layer. - Compare corresponding entries until a difference appears. Counting the total hydrogens or the longest remote chain is not the priority rule. ## Assign the spatial descriptor after ranking - Rank the groups independently at each alkene carbon, then compare the two higher-priority groups. Same side gives Z; opposite sides gives E. - In the illustrated but-2-ene pair, the methyl groups have higher priority than H. The left structure is Z and the right is E. - For this simple pair, cis and trans also track the methyl arrangement. Do not assume cis always translates to Z in every substituted system; E/Z explicitly uses CIP priority. ## Track electron pairs in a reaction mechanism - A mechanism describes bond changes step by step. A full-headed curly arrow tracks an electron pair from a lone pair or covalent bond to its destination. - For bond formation, the arrow starts at the electron source rather than a positive charge. For heterolytic bond cleavage, the arrow starts on the bond and points toward the atom retaining the pair. - The supplied input describes a lone pair on ammonia approaching a proton. Forming NH₄⁺ uses that pair without changing the total atom count or charge. ## Use dots and balanced steps for free radicals - A radical has an unpaired electron, shown by a dot. Homolytic splitting of Cl₂ gives two Cl• species, one electron from the original bond to each atom. - Propagation can conserve a radical overall while producing a different radical intermediate. For example, Cl•+CH₄→HCl+CH₃• conserves all atoms. - AQA does not require curly arrows for the radical mechanisms here. Do not replace radical dots with ionic charge signs or confuse homolysis with heterolytic Cl⁺/Cl⁻ formation.