JAM Organic Basics — Hybridisation and Resonance
Almost every reasoning question in IIT-JAM organic chemistry reduces to two skills: reading the hybridisation off a structure, and ranking resonance contributors correctly. Acidity orders, basicity orders, bond lengths, planarity, aromaticity and the site of nucleophilic attack all fall out of those two. This guide treats them as the working tools they are, not as definitions to memorise.
Assigning hybridisation — the steric number
SN 4 → sp³, ~109.5° · SN 3 → sp², 120° · SN 2 → sp, 180°
Count σ bonds only. A double bond is one σ plus one π; a triple bond is one σ plus two π. The π bonds live in unhybridised p orbitals and never enter the count.
Assign the hybridisation of every heavy atom in CH₃–CH=CH–C≡N.
C1 (methyl): 4 σ, 0 lone pairs → SN 4 → sp³
C2 and C3 (alkene): 3 σ, 0 lone pairs → SN 3 → sp²
C4 (nitrile carbon): 2 σ (to C3 and to N), 0 lone pairs → SN 2 → sp
N: 1 σ + 1 lone pair → SN 2 → sp, and the lone pair sits in an sp
orbital pointing along the molecular axis.
Two cases regularly assigned wrongly. The amide nitrogen has three σ bonds and a lone pair, so naive counting gives SN 4 and sp³ — but it is experimentally planar and sp², because the lone pair is delocalised into the carbonyl π system. Similarly the oxygen of phenol or of an ester behaves as sp² because its lone pair conjugates with the ring or the C=O. Whenever a lone pair is in conjugation, it occupies a p orbital and the atom flattens.
What s character actually does
An sp orbital is 50% s, sp² is 33% s and sp³ is 25% s. Because s orbitals sit closer to the nucleus, higher s character means the electrons are held more tightly. Three consequences are examined constantly:
| Property | sp | sp² | sp³ |
|---|---|---|---|
| s character | 50% | 33% | 25% |
| C–H bond length | ~1.06 Å (shortest) | ~1.09 Å | ~1.09–1.10 Å |
| Effective electronegativity of C | highest | intermediate | lowest |
| Approx. pKa of C–H | ~25 (HC≡CH) | ~44 (CH₂=CH₂) | ~50 (CH₃CH₃) |
| Basicity of a lone pair | least basic | intermediate | most basic |
Why is ethyne so much more acidic than ethene or ethane?
Removing H⁺ leaves the electron pair in the carbon hybrid orbital. In HC≡C⁻ that orbital is sp with 50% s character, so the pair is held close to the nucleus and the anion is stabilised. In CH₃CH₂⁻ the sp³ orbital holds it further out. The pKa gap of roughly 25 units corresponds to an enormous difference in anion stability — enough that a terminal alkyne can be deprotonated by NaNH₂ while an alkane cannot.
The same logic explains basicity in reverse: pyridine, whose nitrogen lone pair is in an sp² orbital in the ring plane and not part of the aromatic sextet, is a normal base; pyrrole, whose lone pair is in the p orbital contributing to the aromatic six π electrons, is barely basic at all — protonating it would destroy the aromaticity.
Resonance — the rules that decide the ranking
Resonance contributors are not real, interconverting species. They are limiting structures whose weighted combination describes one real, delocalised molecule. Two absolute requirements: only electrons move, never atoms, and every contributor must have the same number of unpaired electrons. A structure that moves an atom is a tautomer, not a resonance form — a distinction JAM tests directly.
Rank contributors by these criteria, in order:
- More covalent bonds is better.
- Complete octets on second-row atoms beat incomplete ones.
- Fewer formal charges is better.
- If charges are unavoidable, put negative charge on the more electronegative atom and positive charge on the less electronegative one.
- Separating opposite charges over a distance costs energy.
Criterion 2 outranks criterion 4 in a common trap: for a carbonyl, the structure with a positive carbon and a negative oxygen is a genuine minor contributor precisely because both atoms keep complete octets — and it is exactly that contributor which explains why carbonyl carbon is electrophilic.
Three worked rankings.
Carbonate, CO₃²⁻. Three equivalent contributors, each with one C=O and two C–O⁻. Because they are equivalent, they contribute equally, and all three C–O bonds are identical in length — intermediate between a single and a double bond. Equivalence is the strongest possible form of resonance stabilisation.
An enolate, CH₂=CH–O⁻ ↔ ⁻CH₂–CH=O. Both have the same number of bonds and complete octets, so ranking falls to criterion 4: the contributor with the negative charge on oxygen is the major one. This is why enolates alkylate on carbon but protonate readily on oxygen, and why the carbonyl α-C–H (pKa ≈ 20 for a ketone) is far more acidic than an ordinary C–H.
An amide, R–CO–NR′₂. The contributor with C=N⁺ and O⁻ is significant enough to give the C–N bond substantial double-bond character. The consequences are all observable: the nitrogen is planar, rotation about C–N is restricted with a barrier of roughly 75–90 kJ mol⁻¹ (which is why the two N-methyls of DMF give separate NMR signals at room temperature), and an amide protonates on oxygen, not nitrogen, because the nitrogen lone pair is already committed.
Resonance needs geometry — the coplanarity requirement
Delocalisation requires the p orbitals to overlap, which requires them to be roughly parallel. Anything that twists the system out of plane switches the resonance off. This is why 2,6-disubstituted nitrobenzenes behave as though the nitro group were far weaker than it is — the two ortho groups force the NO₂ out of the ring plane. Any time a question offers a "sterically hindered" substituent and asks for an anomalous acidity or reactivity, check planarity first.
Acidity orders you should be able to justify
| Compound | Approx. pKa | Reason |
|---|---|---|
| CH₃COOH | 4.76 | Conjugate base has two equivalent resonance forms |
| p-nitrophenol | 7.15 | Charge delocalised onto the nitro oxygens as well as the ring |
| Phenol | 9.95 | Phenoxide charge delocalised into the ring |
| Ethanol | ~16 | Ethoxide charge localised on one oxygen |
| Ethyne | ~25 | sp carbanion, high s character |
| Ethane | ~50 | sp³ carbanion, no stabilisation |
The acetic acid / ethanol pair is the cleanest demonstration of resonance in the whole syllabus: both lose a proton from oxygen, but the carboxylate spreads the charge over two equivalent oxygens and the alkoxide cannot.
Aromaticity in one line
All four conditions, not three. Benzene has six π electrons (n = 1) and a delocalisation energy of roughly 150 kJ mol⁻¹. Cyclobutadiene has 4n and is antiaromatic; cyclooctatetraene has 8 π electrons but escapes antiaromaticity by adopting a non-planar tub shape, so it is simply non-aromatic. Count only the electrons in the conjugated ring system — the exocyclic lone pairs of pyridine's nitrogen do not enter, while pyrrole's do.
Common mistakes
- Counting π bonds towards the steric number. Only σ bonds and lone pairs count.
- Calling the amide nitrogen sp³. Conjugation makes it planar sp².
- Drawing a "resonance structure" that moves a hydrogen. That is tautomerism — a different compound in equilibrium, not a contributor.
- Treating all contributors as equally weighted. Only equivalent ones are; otherwise apply the ranking criteria.
- Confusing hyperconjugation with resonance. Hyperconjugation involves σ bonds donating into an empty or π orbital, and it is much weaker.
- Including pyridine's nitrogen lone pair in the aromatic sextet. It is in the ring plane, perpendicular to the π system, and is exactly why pyridine is basic.
- Ignoring geometry. A twisted system does not delocalise, no matter how good the drawn structure looks on paper.
Electronegativity decides half of these rankings. Criterion 4 for resonance, the direction of every inductive effect and the acidity order all rest on comparing electronegativities — worth checking against real values rather than remembering an order.
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