Organic Chemistry – Some Basic Principles and Techniques

This chapter builds the grammar of organic chemistry in four movements. First, shape and bonding: tetravalence, orbital hybridisation, and the sigma-pi description of single, double and triple bonds, followed by the three ways to draw a structure on paper and in 3D. Second, classification and naming: acyclic versus cyclic, aliphatic versus aromatic, functional groups, homologous series, and the IUPAC rules for alkanes, functionalised chains and substituted benzenes — plus the four kinds of structural isomerism that make careful naming necessary. Third, the electronic machinery of reactions: how bonds break (homolysis, heterolysis), the resulting intermediates (free radicals, carbocations, carbanions), nucleophile and electrophile, and the four electron-displacement effects that decide acid strength, intermediate stability and reaction outcome. Fourth, the laboratory toolkit: purification techniques and the qualitative and quantitative methods used to detect and weigh the elements in an unknown compound. Two derivations walk through bond fission and inductive/resonance reasoning; three worked numericals reproduce NCERT's own Liebig, Dumas and inductive-effect problems.

Tetravalence of carbon and orbital hybridisation

Carbon ($1s^2\,2s^2\,2p^2$) has only two unpaired electrons in the ground state but forms four bonds in almost every stable compound. Promoting one $2s$ electron into the empty $2p_z$ orbital gives four unpaired electrons, and mixing (hybridising) the $2s$ orbital with a chosen number of $2p$ orbitals then produces a new set of equivalent hybrid orbitals pointing where the bonds actually go. In methane, all four orbitals mix to give four $sp^3$ hybrids at $109.5^\circ$ (tetrahedral), each overlapping end-on with an H $1s$ orbital. In ethene, each carbon mixes the $2s$ orbital with only two $2p$ orbitals, giving three coplanar $sp^2$ hybrids at $120^\circ$ plus one leftover, unhybridised $p$ orbital perpendicular to that plane; two $sp^2$ orbitals bond to hydrogen, the third forms the C–C sigma bond, and the two unhybridised $p$ orbitals overlap sideways to give the pi bond — one sigma plus one pi makes the C=C double bond. In ethyne, only one $2p$ orbital mixes with the $2s$ orbital, giving two collinear $sp$ hybrids at $180^\circ$; the two remaining, mutually perpendicular $p$ orbitals form two pi bonds, so the triple bond is one sigma plus two pi bonds. Rule of thumb: $s$-character rises $sp^3(25\%) \to sp^2(33\%) \to sp(50\%)$, and bonds get shorter and stronger in that order.

Sigma and pi bonds

A sigma ($\sigma$) bond forms by head-on (axial) orbital overlap along the internuclear axis; the overlap is extensive and symmetric, and the bond is strong enough to allow free rotation of the two halves of the molecule about it, which is why alkanes are conformationally flexible. A pi ($\pi$) bond forms by sideways (lateral) overlap of two parallel, unhybridised $p$ orbitals, giving two lobes of electron density above and below the sigma-bond axis. The overlap is smaller, the bond is weaker, its electrons are more loosely held and more easily attacked by electrophiles, and rotation about it is restricted, because twisting would break the sideways overlap. This is why alkenes show geometrical (cis-trans) isomerism while alkanes do not: every multiple bond is one sigma bond plus one or more pi bonds, never two identical bonds.

Representing organic molecules: structural formulas and 3D shape

Chemists use three levels of detail for a flat structure. A complete (Lewis) structural formula shows every atom and bond explicitly — accurate but cluttered beyond propane. A condensed formula groups atoms onto one carbon without drawing every bond, e.g. $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$ for butane. A bond-line (skeletal) formula draws only the carbon skeleton as a zig-zag: each vertex or free line-end is one carbon, hydrogens are silently added to bring every carbon to four bonds, and heteroatoms or functional groups are written explicitly. Skeletal formulas are the fastest to draw and read, and are the default in practice. Three-dimensional shape is shown with wedge-and-dash notation: a plain line lies in the plane of the paper, a bold wedge (thick end away from the stem) is a bond coming toward the viewer, and a hashed wedge goes back behind the page — together with the implicit fourth bond, this lets a flat page carry the tetrahedral geometry of an $sp^3$ carbon.

Classification, functional groups and homologous series

Organic compounds are first split by the shape of the carbon framework. Acyclic (open-chain) or aliphatic compounds have a straight or branched chain with no ring; cyclic compounds split further into alicyclic (carbon rings behaving much like open chains, e.g. cyclohexane), benzenoid aromatic (built from a benzene ring), non-benzenoid aromatic, and heterocyclic compounds, whose ring contains an atom other than carbon (nitrogen in pyridine, oxygen in furan). Cutting across this is the functional group — the atom or group chiefly responsible for a molecule's characteristic chemistry, e.g. $-\text{OH}$ (alcohol), $-\text{CHO}$ (aldehyde), $>\text{C}=\text{O}$ (ketone), $-\text{COOH}$ (acid), $-\text{NH}_2$ (amine), $-X$ (haloalkane). Compounds sharing one functional group, one general formula, and differing only by a repeating $-\text{CH}_2-$ unit form a homologous series — alkanes ($\text{C}_n\text{H}_{2n+2}$), alcohols ($\text{C}_n\text{H}_{2n+1}\text{OH}$), and so on. Because each step adds a fixed $14\ \text{u}$, physical properties change gradually along the series while chemical properties, set by the shared functional group, stay essentially the same.

IUPAC nomenclature

An unambiguous IUPAC name has four pieces: a word root for the carbon count (meth-, eth-, prop-, but- for one to four carbons, pent- through dec- for five to ten), a primary suffix for saturation ($-\text{ane}$, $-\text{ene}$, $-\text{yne}$), a secondary suffix for the principal group if present ($-\text{ol}$ alcohol, $-\text{al}$ aldehyde, $-\text{one}$ ketone, $-\text{oic acid}$ acid), and prefixes for every other substituent, cited alphabetically with a locant. Straight-chain alkanes just take the word root plus $-\text{ane}$ ($\text{CH}_4$ methane up to $\text{C}_{10}\text{H}_{22}$ decane, Table 8.2). Removing one H gives an alkyl group, named by changing $-\text{ane}$ to $-\text{yl}$ (methyl, ethyl, and the branched isopropyl, sec-butyl, isobutyl, tert-butyl, neopentyl). A branched-chain alkane follows four rules in order: longest chain is the parent; number it so branch carbons get the lowest locants, breaking ties alphabetically; cite repeated substituents with di-, tri-, tetra- while alphabetising by the substituent name — iso- and neo- count for alphabetical order, sec- and tert- do not; write all prefixes alphabetically before the parent name. When a functional group is present, the longest chain containing it becomes the parent, numbered for its lowest locant. NCERT's decreasing priority is: carboxylic acid $>$ sulphonic acid $>$ ester $>$ acid halide $>$ amide $>$ nitrile $>$ aldehyde $>$ ketone $>$ alcohol $>$ amine $>$ ether; halogens and $-\text{NO}_2$ are always prefixes, never a suffix. Substituted benzenes work the same way: one substituent prefixes "benzene" (chlorobenzene), though several keep retained names (toluene, phenol, aniline, benzaldehyde, benzoic acid); two substituents are ortho (1,2-), meta (1,3-) or para (1,4-), and three or more take the lowest overall locant set.

Isomerism

Compounds sharing one molecular formula but differing in structure are isomers. NCERT divides structural (constitutional) isomerism into four kinds. Chain isomerism differs in skeleton branching, e.g. n-butane and isobutane ($\text{C}_4\text{H}_{10}$). Position isomerism keeps the same skeleton and functional group but moves the group to a different carbon, e.g. propan-1-ol and propan-2-ol. Functional group isomerism puts entirely different groups on the same formula, e.g. ethanol and dimethyl ether ($\text{C}_2\text{H}_6\text{O}$), or an aldehyde and ketone sharing $\text{C}_3\text{H}_6\text{O}$. Metamerism arises when a functional group sits between two alkyl (or aryl) groups of different size, e.g. diethyl ether and methyl propyl ether ($\text{C}_4\text{H}_{10}\text{O}$). Separately, stereoisomerism covers identical connectivity with different spatial arrangement — geometrical (cis-trans) isomerism at a restricted-rotation double bond, and optical isomerism at a chiral carbon — introduced here and developed later.

Fission of covalent bonds, electrophiles and nucleophiles

A reaction mechanism is the step-by-step account of how a reaction happens: which bond breaks first, how electrons move, how fast each step goes. Every mechanism starts with fission of a covalent bond, and there are exactly two ways to break one. In homolytic fission, the bonded pair splits symmetrically, drawn with single-barbed (fish-hook) arrows, typically needing heat, light or a non-polar medium; it produces neutral, unpaired-electron species called free radicals ($\text{Cl}-\text{Cl} \rightarrow \text{Cl}^{\bullet} + \text{Cl}^{\bullet}$). In heterolytic fission, the more electronegative atom keeps both electrons, drawn with one full, double-barbed arrow, typically in a polar solvent, giving oppositely charged fragments. A carbon left electron-deficient, positively charged, $sp^2$ and planar with an empty $p$ orbital is a carbocation; a carbon that keeps the pair, becoming negatively charged, $sp^3$ and pyramidal with a lone pair, is a carbanion. The organic molecule supplying the carbon skeleton is the substrate; the attacking species is the reagent. An electron-rich species that donates a pair to an electron-poor centre is a nucleophile (Lewis base: $\text{OH}^-$, $\text{CN}^-$, $\text{NH}_3$); an electron-deficient species that accepts a pair is an electrophile (Lewis acid: $\text{H}^+$, carbocations, $\text{BF}_3$).

Electron displacement effects: inductive, resonance, electromeric, hyperconjugation

Four related but distinct effects describe how electron density moves, and telling them apart is one of the most tested skills here. The inductive effect is a permanent sigma-bond polarisation from an electronegativity difference, transmitted along a chain and fading rapidly beyond the third carbon; electronegative groups (halogens, $-\text{NO}_2$, $-\text{CN}$, $-\text{COOH}$) show $-I$, alkyl groups show $+I$, increasing $\text{CH}_3 < \text{CH}_3\text{CH}_2 < (\text{CH}_3)_2\text{CH} < (\text{CH}_3)_3\text{C}$. The resonance (mesomeric) effect operates through a conjugated system and is also permanent: $+R$ groups donate a lone pair into the pi system ($-\text{OH}$, $-\text{OR}$, $-\text{NH}_2$, halogens on a ring), $-R$ groups withdraw density from it ($-\text{NO}_2$, $-\text{CHO}$, $-\text{COOH}$, $-\text{CN}$). The electromeric effect is temporary, appearing only during attack by a reagent, and is the complete transfer of a shared pi-electron pair to one atom of a multiple bond; it vanishes once the reagent is removed, unlike the permanent inductive and resonance effects. Hyperconjugation ("no-bond resonance") delocalises sigma C–H electrons adjacent to a carbocation, radical, or multiple bond into an empty $p$ orbital or pi system; it needs at least one alpha hydrogen, and more of them mean more stabilisation — exactly why carbocation and radical stability runs tertiary $>$ secondary $>$ primary $>$ methyl.

Purification techniques

Every crude organic product needs purifying, and the method chosen always exploits a genuine physical difference between the compound and its impurities. Sublimation purifies a volatile solid (naphthalene, camphor, benzoic acid) by heating it so it passes directly to vapour and re-solidifies on a cold surface, skipping the liquid state. Crystallisation exploits a solubility difference: the impure solid is dissolved in the minimum hot solvent, filtered hot, and cooled slowly so pure crystals separate while impurities stay in the mother liquor. Distillation, simply, separates a volatile liquid from a non-volatile impurity or from a liquid boiling more than about $25^\circ\text{C}$ away, using the vapour-pressure difference. Fractional distillation separates miscible liquids with close boiling points using a fractionating column of repeated vaporisation-condensation cycles (theoretical plates) — the method behind refining crude petroleum. Vacuum distillation suits liquids that decompose at their normal boiling point, since lowering pressure lowers the boiling temperature (concentrating sugarcane juice, purifying glycerol). Steam distillation separates a steam-volatile, water-immiscible compound (aniline, essential oils) by co-distilling with steam, then splitting the immiscible layers with a separating funnel. Differential extraction separates a compound from water by shaking with an immiscible organic solvent it favourably partitions into; several small extractions recover more than one large one. Chromatography separates and tests purity using differential movement between a stationary and a mobile phase. Adsorption chromatography relies on components adsorbing to different extents on silica gel or alumina: column chromatography washes the mixture through a packed column with solvents of increasing polarity, emerging as separated bands; thin layer chromatography (TLC) coats the adsorbent on a plate, the sample spotted near the bottom, and stood in a closed jar so the solvent front climbs by capillary action, carrying each component a different distance, captured by the retardation factor $$R_f = \frac{\text{distance moved by the component}}{\text{distance moved by the solvent front}},$$ always between $0$ and $1$. Partition chromatography (paper chromatography) relies instead on continuous partitioning between a stationary liquid phase on an inert support and a moving liquid phase — the same physics as extraction, run continuously.

Qualitative and quantitative analysis of organic compounds

Detecting carbon and hydrogen is done by heating the compound with dry copper(II) oxide: carbon is oxidised to $\text{CO}_2$ (turns lime water milky), hydrogen to $\text{H}_2\text{O}$ (turns anhydrous copper sulphate blue). Nitrogen, sulphur, halogens and phosphorus cannot be detected directly, so the compound is first converted to water-soluble ionic form by Lassaigne's test (sodium fusion): sodium metal is heated strongly with the compound in a fusion tube, then the red-hot tube is plunged into water and filtered to give the sodium fusion extract. Nitrogen (as cyanide) is tested by boiling the extract with ferrous sulphate, acidifying, and looking for a Prussian-blue colouration; if sulphur is also present, thiocyanate forms instead, giving a blood-red colour with ferric chloride. Sulphur alone gives a violet colour with sodium nitroprusside, or a black lead sulphide precipitate with lead acetate. Halogens are confirmed by acidifying with nitric acid (boiling first with concentrated acid if N or S is present) and adding silver nitrate: curdy white (soluble in ammonia) is chlorine, pale yellow is bromine, yellow (insoluble) is iodine. Phosphorus gives a yellow colouration with ammonium molybdate after oxidative fusion. Quantitative analysis weighs each element as a percentage of the compound's mass. In Liebig's method for C and H, a known mass is burnt completely over hot copper oxide; the water and carbon dioxide produced are absorbed and weighed separately, giving $$\%C = \frac{12 \times m_{\text{CO}_2} \times 100}{44 \times m}, \qquad \%H = \frac{2 \times m_{\text{H}_2\text{O}} \times 100}{18 \times m}.$$ Nitrogen uses either of two methods. In the Dumas method, the compound is heated with excess CuO in a $\text{CO}_2$ atmosphere, converting nitrogen to free $\text{N}_2$ gas, collected over KOH and its volume corrected to STP: $$\%N = \frac{28 \times V \times 100}{22400 \times m}.$$ In the Kjeldahl method, the compound is digested with concentrated $\text{H}_2\text{SO}_4$ to ammonium sulphate, then boiled with excess NaOH to liberate ammonia, which is absorbed in a known volume of standard acid: $$\%N = \frac{1.4 \times M \times V}{m},$$ with $V$ the volume of acid neutralised. This method fails for ring nitrogen (pyridine) or nitro/azo nitrogen, which is not converted to ammonium sulphate. Halogens, sulphur and phosphorus all use the Carius method — heating with fuming nitric acid in a sealed tube — giving, respectively, $$\%X = \frac{\text{at. mass}_X \times m_{\text{AgX}} \times 100}{\text{mol. mass}_{\text{AgX}} \times m}, \quad \%S = \frac{32 \times m_{\text{BaSO}_4} \times 100}{233 \times m}, \quad \%P = \frac{62 \times m_{\text{Mg}_2\text{P}_2\text{O}_7} \times 100}{222 \times m}.$$ Oxygen is rarely measured directly; it is obtained by subtracting the sum of every other element's percentage from $100$.

Board level - percentage of carbon and hydrogen by Liebig's combustion method

On complete combustion, $0.246\ \text{g}$ of an organic compound gave $0.198\ \text{g}$ of $\text{CO}_2$ and $0.1014\ \text{g}$ of $\text{H}_2\text{O}$. Determine the percentage of carbon and hydrogen in the compound.

Step 1. Recall the Liebig formulas: $\%C = \dfrac{12 \times m_{\text{CO}_2} \times 100}{44 \times m}$ and $\%H = \dfrac{2 \times m_{\text{H}_2\text{O}} \times 100}{18 \times m}$, where $12/44$ and $2/18$ convert the mass of product to the mass of the element it came from.

Step 2. Substitute for carbon: $\%C = \dfrac{12 \times 0.198 \times 100}{44 \times 0.246} = \dfrac{237.6}{10.824} = 21.95\%$.

Step 3. Substitute for hydrogen: $\%H = \dfrac{2 \times 0.1014 \times 100}{18 \times 0.246} = \dfrac{20.28}{4.428} = 4.58\%$. Answer: the compound is $21.95\%$ carbon and $4.58\%$ hydrogen by mass.

Competitive level (JEE / NEET) - percentage of nitrogen by the Dumas method

In a Dumas nitrogen estimation, $0.3\ \text{g}$ of an organic compound gave $50\ \text{mL}$ of moist nitrogen collected at $27^\circ\text{C}$ ($300\ \text{K}$) and $700\ \text{mm}$ pressure. Calculate the percentage of nitrogen, given that the volume of nitrogen at STP works out to $41.9\ \text{mL}$.

Step 1. Correct the collected volume to STP using $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$, with $P_1 = 700\ \text{mm}$, $V_1 = 50\ \text{mL}$, $T_1 = 300\ \text{K}$, $P_2 = 760\ \text{mm}$, $T_2 = 273\ \text{K}$: $V_2 = \dfrac{700 \times 50 \times 273}{300 \times 760} = 41.9\ \text{mL}$.

Step 2. Convert this STP volume to a mass of $\text{N}_2$ using molar volume $22{,}400\ \text{mL}$ and molar mass $28\ \text{g}$: mass $= \dfrac{28 \times 41.9}{22400} = 0.05236\ \text{g}$.

Step 3. Express as a percentage of the compound taken: $\%N = \dfrac{28 \times 41.9 \times 100}{22400 \times 0.3} = 17.46\%$. Answer: the compound contains $17.46\%$ nitrogen by mass. Timing note: about 90 seconds once the STP-corrected volume is known. The examiner-favourite trap is skipping the $P_1V_1/T_1 = P_2V_2/T_2$ correction and plugging the raw collected volume straight into the percentage formula — check whether the volume quoted is at STP or as collected first.

Competitive level (JEE / NEET) - ranking acid strength and carbocation stability by inductive and hyperconjugative reasoning

(a) Arrange $\text{CH}_3\text{COOH}$, $\text{FCH}_2\text{COOH}$ and $\text{ClCH}_2\text{COOH}$ in increasing order of acid strength. (b) Arrange the carbocations $\text{CH}_3^+$, $\text{CH}_3\text{CH}_2^+$, $(\text{CH}_3)_2\text{CH}^+$ and $(\text{CH}_3)_3\text{C}^+$ in increasing order of stability.

Step 1. For (a), both fluorine and chlorine are $-I$ groups that stabilise the carboxylate conjugate base, so both haloacetic acids beat plain acetic acid, whose only neighbours are weakly $+I$ hydrogens.

Step 2. Fluorine, more electronegative than chlorine, exerts the stronger $-I$ pull at the same position, so it stabilises the conjugate base more: acid strength is $\text{CH}_3\text{COOH} < \text{ClCH}_2\text{COOH} < \text{FCH}_2\text{COOH}$.

Step 3. For (b), carbocation stability rises with alpha hydrogens available for hyperconjugation and $+I$ donation: methyl has none, ethyl has three, isopropyl six, tert-butyl nine. Answer: acid strength $\text{CH}_3\text{COOH} < \text{ClCH}_2\text{COOH} < \text{FCH}_2\text{COOH}$; carbocation stability $\text{CH}_3^+ < \text{CH}_3\text{CH}_2^+ < (\text{CH}_3)_2\text{CH}^+ < (\text{CH}_3)_3\text{C}^+$. Timing note: under a minute, since both rankings reduce to counting the same thing — electron-releasing alpha hydrogens, or the strength of a nearby electron-withdrawing puller. The trap is ranking halogens by atomic size instead of electronegativity, which is what actually governs inductive strength for F vs Cl.

Students lose marks on this chapter in a handful of specific, repeatable ways. Confusing a sigma bond with a pi bond is the most common: every double bond is one sigma plus one pi, every triple bond is one sigma plus two pi — never two identical bonds. Writing a bond-line formula with the wrong hydrogen count is a frequent slip; count every line at a vertex, subtract from four, and that remainder is the hidden hydrogen count. In naming branched alkanes, the recurring errors are numbering the parent chain to favour the substituents' lowest locants (not just the longest chain) and alphabetising sec- and tert- as if they counted, when only iso- and neo- do. Students routinely confuse the four kinds of structural isomerism, especially position isomerism versus metamerism (metamerism specifically needs a functional group flanked by two different alkyl or aryl groups, most often in ethers, esters and secondary amines — not just "the group is somewhere else"). The inductive, resonance and electromeric effects are treated as interchangeable, but only the electromeric effect is temporary, present only during attack by a reagent; inductive and resonance effects are permanent ground-state features. Finally, a common analytical error is applying the Kjeldahl formula to a compound with ring or nitro nitrogen, where the method is known to fail, or forgetting to correct a Dumas nitrogen volume to STP before using the percentage formula.

• Predict the hybridisation ($sp^3$, $sp^2$, $sp$) and bond angle at any carbon, and describe the sigma-pi composition of single, double and triple bonds.
• Convert between complete, condensed and bond-line structural formulas, and draw 3D shape using wedge-and-dash notation.
• Classify a compound as acyclic/aliphatic, alicyclic, benzenoid aromatic, non-benzenoid aromatic or heterocyclic, and name its functional group and homologous series.
• Apply IUPAC rules to name a branched alkane or functionalised chain: parent chain, lowest locants, alphabetical prefixes.
• Distinguish chain, position, functional group and metameric isomerism from stereoisomerism.
• Draw the curved-arrow mechanism for homolytic and heterolytic fission, and name the resulting free radical, carbocation or carbanion.
• Identify a nucleophile versus an electrophile, and label a substituent's effect as inductive, resonance, electromeric or hyperconjugative.
• Rank acids, carbocations or radicals by stability using inductive and hyperconjugative reasoning.
• Match a purification technique to the physical property it exploits, and compute an $R_f$ value from a chromatogram.
• Carry out the Liebig, Dumas, Kjeldahl and Carius percentage-composition calculations, including STP corrections.

Ask the tutor: Want a quick drill on naming five branched-chain IUPAC structures or on identifying inductive versus resonance effects in a set of substituted acids?