Organic Chemistry - Some Basic Principles and Techniques

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XI, Chapter 8.

How can one element make millions of compounds?

Everything that burns, rots, ferments or is alive is built mostly from carbon, hydrogen, oxygen and nitrogen. Yet there are more known carbon compounds than compounds of all other elements put together. The reason is that carbon bonds strongly to other carbon atoms and forms chains, branched chains and rings. This is catenation, and no other element does it as well, because carbon is small, has four valence electrons and its C–C bond is strong. For a long time chemists believed that the compounds of living things needed a mysterious "vital force". In 1828 Wöhler heated ammonium cyanate and got urea, a compound found in urine:

NH₄CNO → NH₂CONH₂ (on heating)

Kolbe's synthesis of acetic acid from inorganic substances in 1845 finished the idea. This chapter gives you the toolkit for the rest of organic chemistry: how carbon bonds, how to draw and name molecules, why isomers exist, how electrons move, and how to purify and analyse a compound.

Beyond the textbook. For the exam: Wöhler's 1828 urea synthesis disproved the vital force theory. In reality: one experiment did not end the idea overnight. Vitalism faded over the following decades as more organic compounds were made in the laboratory from simple inorganic materials. Why it matters: the definition of organic chemistry changed from "chemistry of living things" to "chemistry of carbon compounds", which is how you should define it in the exam.

Carbon's bonding: hybridisation and bond lengths

Carbon is tetravalent. The way it mixes its orbitals decides the shape of the molecule.

More s character holds the electrons closer to the nucleus. So going from sp³ to sp the bond gets shorter and stronger, and the carbon becomes more electronegative. A single bond is one σ bond, a double bond is one σ plus one π, and a triple bond is one σ plus two π. The π bond is weaker and more reactive.

Drawing and classifying molecules

Organic chemists use three kinds of structural formula, all of which appear in questions.

For three dimensional shape, a solid wedge shows a bond pointing towards you, a dashed wedge shows one pointing away, and a plain line lies in the plane of the paper. In methane, two bonds are drawn in the plane, one as a wedge and one dashed, which captures the tetrahedron on paper.

Compounds are classified first by the shape of the carbon skeleton.

Aromatic rings are unusually stable because they have $(4n+2)$ delocalised π electrons (Hückel's rule). Furan, thiophene and pyridine each have six.

A functional group is the atom or group that decides the chemistry. Compounds with the same functional group that differ by one –CH₂– form a homologous series, and their physical properties change steadily along the series (the boiling points of methanol, ethanol and propanol rise step by step).

IUPAC naming: a method that always works

The IUPAC name is built in a fixed order: prefixes (substituents) + root (longest chain) + primary suffix (saturation) + secondary suffix (functional group).

The primary suffix is -ane for single bonds, -ene for a double bond and -yne for a triple bond. The functional group usually becomes the secondary suffix, as in ethanol, propanone, ethanoic acid. Some groups can only be prefixes.

The rows run in order of priority. When a molecule has more than one group, the highest in the table becomes the suffix and takes the lowest possible number, while the others become prefixes. Halo, alkoxy and nitro are always prefixes.

The five steps. First choose the longest continuous chain that contains the principal functional group and any multiple bond. Second, number from the end that gives the lowest locant to the principal group, then to multiple bonds, then to substituents. Third, name the substituents. Fourth, list them in alphabetical order (ignoring di, tri), with a locant before each. Fifth, apply the first point of difference rule: when two sets of locants are possible, compare them number by number from the left and take the set that has the lower number at the first difference.

Worked example: naming practice. Name each compound.

The common wrong step: numbering so that the first substituent gets the lowest number without checking the whole set, or listing substituents in the order you notice them rather than alphabetically.

Cyclic compounds take the prefix cyclo- (cyclohexene). On a disubstituted benzene the positions 1,2 / 1,3 / 1,4 are also called ortho, meta and para. IUPAC keeps some common names, including toluene, aniline, phenol and anisole.

Isomerism: same formula, different molecule

Isomers share a molecular formula but differ in structure. Butane (C₄H₁₀) has 2 isomers, pentane (C₅H₁₂) has 3 and hexane (C₆H₁₄) has 5. The three pentanes are pentane, 2-methylbutane and 2,2-dimethylpropane.

The first five are structural isomerism, and the last two are stereoisomerism. Remember that tautomers are different compounds in equilibrium, whereas resonance structures are different drawings of one molecule. Atoms move in tautomerism and only electrons move in resonance.

Breaking bonds: fission and reaction intermediates

When a covalent bond breaks, the shared pair can split in two ways.

The reactive species formed are called intermediates. Do not mix up their shapes.

Alkyl groups push electrons towards carbon. That helps a carbocation, which is short of electrons, and hurts a carbanion, which already has too many. So the two stability orders are opposite.

Pause and predict: which is more stable, CH₃⁻ or (CH₃)₃C⁻?

Check your answer: CH₃⁻. The three methyl groups in (CH₃)₃C⁻ push even more electron density onto a carbon that is already negative, which destabilises it.

Reagents. A nucleophile is electron rich and seeks a positive centre: OH⁻, CN⁻, RO⁻, H₂O, NH₃, R₂NH. An electrophile is electron poor and seeks an electron rich centre: carbocations, AlCl₃, BF₃, NO₂⁺, SO₃. Nucleophiles attack the carbon of a carbocation, which is why it is attacked from either face of its flat shape.

A two step substitution. The same ideas explain a reaction you will meet in Class XII.

Four general types of organic reaction cover almost everything: substitution (CH₃Br + OH⁻ → CH₃OH + Br⁻), addition (CH₂=CH₂ + HBr → CH₃CH₂Br), elimination (CH₃CH₂Br loses HBr to give CH₂=CH₂) and rearrangement, in which a group shifts inside the molecule. Their detailed mechanisms come in Class XII.

How electrons are pushed around: electronic effects

Bond polarity and electron shifts explain why molecules behave as they do. Four effects matter, and they differ in permanence and in how the electrons travel.

Inductive effect. A more electronegative atom pulls the σ electrons along the chain. The pull is strongest at the nearest bond and fades quickly, and is negligible beyond about the third carbon. Groups that withdraw electrons show a −I effect (–NO₂, –CN, –COOH, the halogens –F, –Cl, –Br, –I, –OH, –NH₂). Alkyl groups push electrons, a +I effect, in the order (CH₃)₃C– > (CH₃)₂CH– > CH₃CH₂– > CH₃–. The effect explains why acid strength rises with the number of chlorines: Cl₃CCOOH is stronger than Cl₂CHCOOH, which is stronger than ClCH₂COOH, which is stronger than CH₃COOH. Chlorine pulls electrons away from the O–H bond and stabilises the carboxylate ion.

Resonance effect. When a lone pair or π bond can spread over several atoms, the molecule is described by several canonical structures that differ only in the position of electrons. The real molecule is a resonance hybrid, which is more stable than any one of them. The rules are strict: the atoms stay put, only π electrons and lone pairs move, and the number of paired electrons stays the same. Acrolein shows it:

CH₂=CH–CH=O ↔ ⁺CH₂–CH=CH–O⁻

On a benzene ring, +R groups direct incoming electrophiles to the ortho and para positions and −R groups to the meta position (the Hydrocarbons chapter uses this). Groups such as –OH and –NH₂ are −I but +R: the resonance effect usually wins on the ring. Halogens are −I and +R, but here the −I effect is stronger, so they slightly deactivate the ring while still directing ortho and para.

Beyond the textbook. For the exam: benzene is represented by two resonance structures with alternate double and single bonds. In reality: neither structure exists on its own. Benzene is a single hybrid in which all six carbon–carbon bonds have the same length, about 139 pm, between a typical single bond and a double bond. Why it matters: if benzene really flipped between two Kekulé forms, the molecule would show bonds of two lengths. Resonance is an improvement in how we draw it, not a reaction that occurs.

Electromeric effect. A temporary shift of the π electrons of a multiple bond only when an attacking reagent approaches, and it disappears when the reagent goes. If the π pair moves towards the atom that the reagent attaches to, it is +E (an alkene with H⁺). If it moves away from that atom, it is −E (CN⁻ adding to a carbonyl group, where the π electrons go to oxygen).

Hyperconjugation. Electrons in a σ C–H bond on the carbon next to a carbocation, a double bond or a radical spread slightly into the empty or π orbital. It is also called no bond resonance. It needs at least one α hydrogen, which means one on the carbon directly attached to the electron deficient centre. The more α hydrogens, the more stabilisation.

So tertiary carbocations are the most stable because they gain from both three +I alkyl groups and many hyperconjugation structures. The same logic says that an alkene with more alkyl groups on the double bond is more stable.

Purifying an organic compound

A freshly made compound is always mixed with by-products. The method of purification is chosen according to which property differs.

Why steam distillation works. Water and the organic liquid do not mix, so each exerts its own vapour pressure and the total pressure is the sum of the two. The mixture boils when this sum reaches atmospheric pressure. That happens below 373 K, so the compound distils well below its own boiling point and does not decompose. It is not that the compound alone boils below 100 °C.

Chromatography. A mixture moves along a stationary phase carried by a mobile phase. In adsorption chromatography (column, thin layer) the stationary phase is a solid such as alumina or silica gel, and compounds are held back according to how strongly they stick. In partition chromatography (paper) the stationary phase is the water trapped in the cellulose. The result on a TLC plate is quantified by the retardation factor:

$$R_f = \dfrac{\text{distance moved by the substance}}{\text{distance moved by the solvent front}}$$

Both distances are measured from the starting line. $R_f$ is always between 0 and 1 and depends on the solvent and adsorbent.

Pause and predict: a spot travelled 3.0 cm while the solvent front travelled 12.0 cm. What is $R_f$, and can it ever be 1.4?

Check your answer: $R_f = 3.0/12.0 = 0.25$. It can never exceed 1, because a substance cannot travel further than the solvent that carries it.

Qualitative analysis: which elements are present?

Carbon and hydrogen. Heat the compound with dry copper(II) oxide. Carbon becomes CO₂, which turns lime water milky, and hydrogen becomes H₂O, which turns anhydrous CuSO₄ blue.

CₓH_y + (2x + y/2) CuO → x CO₂ + (y/2) H₂O + (2x + y/2) Cu

Nitrogen, sulphur, halogens: Lassaigne's test. These elements are covalently bound and cannot be detected directly. Fusing the compound with sodium metal turns them into ionic salts that dissolve in water: Na + C + N → NaCN, 2Na + S → Na₂S and Na + X → NaX. The water extract of the fusion mass (the sodium fusion extract) is then tested.

The three silver halides are separated by their solubility in ammonia: AgCl dissolves in NH₄OH, AgBr is sparingly soluble and AgI is insoluble. Acidifying with HNO₃ and boiling before adding AgNO₃ decomposes any NaCN and Na₂S, which would otherwise give false precipitates. When nitrogen and sulphur are both present, the SCN⁻ ion forms and gives blood red with Fe³⁺, and no Prussian blue appears. Oxygen has no direct qualitative test. It is found by difference.

Beyond the textbook. For the exam: Kjeldahl's method estimates the nitrogen in an organic compound. In reality: food laboratories use the same method to estimate protein in milk, grain and feed. They find the nitrogen and multiply by a conversion factor (6.25 is the usual one, because proteins are about 16 % nitrogen). Why it matters: it also explains a known weakness: the method counts all nitrogen, not just protein nitrogen.

Quantitative analysis: how much of each element?

Each method turns the element into something you can weigh or measure, and the percentage follows from the mass relationships.

The formulas are below.

$$\%\ \text{C} = \dfrac{12}{44}\times\dfrac{\text{mass of CO}_2}{\text{mass of compound}}\times 100, \qquad \%\ \text{H} = \dfrac{2}{18}\times\dfrac{\text{mass of H}_2\text{O}}{\text{mass of compound}}\times 100$$

$$\text{Dumas: } \%\ \text{N} = \dfrac{28 \times V}{22400 \times m}\times 100 \quad (V \text{ in mL of N}_2 \text{ at STP})$$

$$\text{Kjeldahl: } \%\ \text{N} = \dfrac{1.4 \times M \times 2 \times V}{m} \quad (M \text{ and } V \text{ for the H}_2\text{SO}_4 \text{ neutralised by NH}_3)$$

$$\text{Carius: } \%\ \text{X} = \dfrac{\text{atomic mass of X}}{\text{molar mass of AgX}}\times\dfrac{m_{\text{AgX}}}{m}\times 100, \qquad \%\ \text{S} = \dfrac{32}{233}\times\dfrac{m_{\text{BaSO}_4}}{m}\times 100$$

$$\%\ \text{P} = \dfrac{62}{222}\times\dfrac{m_{\text{Mg}_2\text{P}_2\text{O}_7}}{m}\times 100$$

In all of them $m$ is the mass of the compound in grams. In Kjeldahl's method the volume $V$ is in mL, and the factor 2 appears because each H₂SO₄ neutralises two NH₃. The method fails for nitrogen in nitro (–NO₂), azo (–N=N–) and ring nitrogen compounds such as pyridine, because the nitrogen is not converted completely to ammonium sulphate.

Worked example: Liebig's method. 0.30 g of an organic compound gave 0.44 g of CO₂ and 0.18 g of H₂O. Find the percentage composition and the empirical formula.

Worked example: Kjeldahl's method. A 0.50 g sample was digested and the ammonia evolved neutralised 20 mL of 0.10 M H₂SO₄. Find the percentage of nitrogen.

The common wrong step: leaving out the factor 2. The same answer follows from the formula: $1.4 \times 0.10 \times 2 \times 20 / 0.50 = 11.2$.

Worked example: Carius method. 0.15 g of an organic compound gave 0.12 g of AgBr. Find % Br (Ag = 108, Br = 80).

Tests and recognition practice

Use this to link an observation to its conclusion.

Exam traps

Key reactions and ideas

60-second revision

Where this chapter connects

• Before this: Chemical Bonding and Molecular Structure gives you hybridisation and resonance, and Redox Reactions gives the idea of electron movement that organic mechanisms build on.
• Next: Hydrocarbons puts naming, isomerism and electronic effects to work on alkanes, alkenes, alkynes and benzene.
• Forward link: Haloalkanes and Haloarenes uses carbocations and the substitution mechanism, Alcohols, Phenols and Ethers and Amines rely on the inductive and resonance effects for acidity and basicity.