Put ammonia and chloromethane in the same flask. You can predict what happens without knowing the name of the reaction. Ammonia’s nitrogen has a lone pair that it is not using. Chloromethane’s carbon is δ+, because chlorine pulls the C–Cl bonding pair toward itself. So the lone pair goes to the carbon and forms a new N–C bond, and chloride leaves with the old C–Cl pair.
That prediction came from a single question: where is the electron density, and where is it missing? Nucleophiles and Electrophiles looked at each half of a reaction on its own. This page puts the two halves together, so that you can scan any molecule and find both kinds of site.
One lens for everything
An electron-rich atom has electron density it can share. It shows at least one of three signs: a lone pair, a negative charge or a π bond. An electron-rich atom tends to act as a nucleophile, a species that donates an electron pair to form a new bond.
An electron-poor atom is short of electron density. It shows at least one of three other signs: a δ+ caused by a more electronegative neighbor, a positive charge, or an empty orbital. An electron-poor atom tends to act as an electrophile, a species that accepts an electron pair to form a new bond.
In a polar step, one in which electrons move in pairs, a rich site always meets a poor one. The curved arrow starts at the rich site and points at the poor one.
None of this is a new principle. It is the electronegativity and bond polarity of Foundations, used to predict what a molecule will do instead of only describing how it is built.
Electron-poor is not enough
Compare chloroethane, CH₃CH₂Cl, with ethanol, CH₃CH₂OH. In both, the CH₂ carbon is bonded to a more electronegative atom, so in both it is δ+. Yet nucleophiles replace the chlorine of chloroethane, while the OH of ethanol stays put.
The difference lies in the bonds the carbon already has. A carbon that already has four bonds can take a new pair only if it gives up one of its bonds at the same time. In chloroethane, the C–Cl bond breaks, and chlorine leaves as chloride ion, Cl⁻, taking the bonding pair with it. A group that departs this way is a leaving group, as Electrophiles described. In ethanol, the group would have to leave as hydroxide ion, HO⁻, and hydroxide does not leave on its own.
For now, a short list is enough. Chlorine, bromine and iodine on a carbon with four single bonds leave as halide ions. Hydroxide (HO⁻), alkoxides (RO⁻, the group an ether would have to lose) and the amide ion (H₂N⁻) do not leave on their own. Leaving groups, the next page, explains why.
A carbon with a π bond has a second way to take a pair. In acetaldehyde, CH₃CHO, the C=O carbon is δ+. When a nucleophile bonds to it, the C=O π bond breaks and its pair moves onto oxygen. The carbon still ends up with four bonds, and nothing has to leave.
So an electron-poor carbon is a reaction site only if it passes a second test. It must hold a leaving group, or a π bond whose pair can move onto a more electronegative atom.
Scanning a whole molecule
Take 1-chloro-2-ethoxyethane, CH₃–CH₂–O–CH₂–CH₂–Cl. Walk along the chain and ask the two questions at every atom. Is it electron-poor? And does it have a way to take a new pair?
The CH₃ on the left. It is bonded only to carbon and hydrogen, so nothing makes it δ+. It is not a site.
The oxygen. It has two lone pairs and is δ−. It is electron-rich, so it could act as a nucleophile itself, but it is not a target.
The two CH₂ groups bonded to oxygen. Oxygen pulls on both, so both are δ+. But the group either one would have to lose is an alkoxide, and alkoxides do not leave. Neither is a site.
The CH₂ bonded to chlorine. Chlorine makes it δ+, and chloride can leave. This carbon passes both tests, so this is where a nucleophile attacks.
Notice what the answer does not rest on. You never had to decide whether the carbon next to chlorine is more δ+ than the carbons next to oxygen. Oxygen is the more electronegative atom (3.44 against 3.16 for chlorine). But electronegativity alone cannot settle the comparison, because bond length matters too, as Bond polarity showed. The second test decides the site. That is why “which atom is attacked?” and “which atom is most electron-poor?” are different questions, and exams like to blur them.
Now scan 5-aminopentan-2-one, H₂N–CH₂–CH₂–CH₂–CO–CH₃. This molecule has a rich site and a poor site of its own. Its carbons are numbered from the CH₃ next to the C=O, so that CH₃ is C1, the C=O carbon is C2, and the nitrogen sits on C5.
The nitrogen. It has a lone pair, and no π bond sits next to it to spread that pair out. It is electron-rich, and it is the molecule’s nucleophile.
C5, the carbon bonded to nitrogen. Nitrogen pulls on the C–N bonding pair, so C5 is slightly δ+. But it could react only by losing H₂N⁻, one of the worst leaving groups there is. A δ+ carbon with no way to take a pair is not a reaction site.
C4, C3 and C1. Each of these carbons is at least two bonds from the nitrogen and from the oxygen, so it is at most slightly δ+, and none of them carries anything that can leave. None is a site. (The hydrogens on C3 and C1, the carbons next to the C=O, are slightly acidic, but that story belongs to Enolate Chemistry.)
C2, the carbonyl carbon. The C=O makes it strongly δ+, and it needs no leaving group, because its π pair can move onto oxygen. This is the molecule’s electrophile.
The carbonyl oxygen. It has two lone pairs and is δ−, so it is electron-rich too. But a small, very electronegative atom holds its pairs tightly, so it is a much weaker nucleophile than the nitrogen. Sorting atoms into rich and poor is only the first cut. Among the rich atoms, the ones that hold their pairs loosely react first.
The oxygen is also a very weak base. An acid would protonate the nitrogen long before it, and only a strong acid protonates a carbonyl oxygen at all. The few carbonyl groups that do get protonated react much faster, which is why acids speed up so many carbonyl reactions.
The verdict. The nitrogen is the nucleophile, C2 is the electrophile, and nothing in the molecule can leave. Because both sites are in the same molecule, the chain can curl around until the nitrogen reaches C2. The new N–C2 bond then closes a ring of five atoms: N, C5, C4, C3 and C2. Five-membered rings close easily. With one CH₂ fewer, the chain would have to close a four-membered ring, which is strained and forms far less easily.
You are not expected to know the final product. You are expected to say which two atoms bond.
Induction: a pull through σ bonds
One part of a molecule can change the electron density at another in two ways, induction and resonance. The next two sections take them one at a time.
Look at 1-chloropropane, Cl–CH₂–CH₂–CH₃. Chlorine pulls the C–Cl bonding pair toward itself, so C1 is δ+. C1 in turn pulls a little on its bond to C2, so C2 is slightly δ+ as well. By C3 almost nothing is left.
This pull, passed along the σ bonds, is induction, the same effect that Factors affecting acidity used to rank acids. Electronegativity drives it. It needs no curved arrows and no new structure, because it only tilts the pairs in the bonds that are already there. And it weakens quickly with distance, fading out within two or three bonds.
Acid strengths put numbers on the fade. Chloroacetic acid, ClCH₂COOH, is about 80 times more acidic than acetic acid (pKa 2.86 against 4.76). Move the chlorine three carbons away from the acid group, as in 4-chlorobutanoic acid, and the pKa is 4.52, close to butanoic acid’s 4.82.
Resonance can reach further than you would expect
Now take propenal, H₂C=CH–CHO, an aldehyde whose C=O is joined directly to a C=C. The carbon bonded to the C=O carbon is the alpha (α) carbon. The next one along, at the far end of the C=C, is the beta (β) carbon. A compound built this way is an α,β-unsaturated aldehyde.
Push the arrows. The C=O π pair moves onto oxygen, and the C=C π pair moves over to form a new π bond between the α carbon and the C=O carbon. The result is a resonance structure with a − on oxygen and a + on the β carbon.
The β carbon is three bonds from the oxygen and has no electronegative neighbor of its own, so induction cannot reach it. Resonance can. That structure contributes to the hybrid, and it makes a testable prediction: nucleophiles attack the β carbon. This is conjugate addition, also called 1,4-addition. When the nucleophile is an enolate (a carbon nucleophile made next to a C=O), the reaction is the Michael reaction of Enolate Chemistry, one of the most important ways to make carbon–carbon bonds.
In general, resonance moves electron pairs through a conjugated system, an unbroken chain of overlapping p orbitals, as Resonance showed. You draw it with curved arrows, and each push gives a new resonance structure. It barely weakens along the conjugated path, but it stops abruptly at the first carbon with four single bonds (an sp³ carbon), because that carbon has no p orbital to pass the pair along.
When induction and resonance disagree
Benzene is the six-carbon ring drawn with three alternating double bonds. Aniline is benzene with an NH₂ group in place of one hydrogen. Nitrogen is more electronegative than carbon, so by induction it pulls a little density out of the ring through the C–N σ bond.
But nitrogen’s lone pair sits right next to the ring’s π bonds, so it can also push density into the ring by resonance. Push the arrows and you get a resonance structure with a + on nitrogen and a − on the ring carbon next to the one bonded to nitrogen. That neighbor is an ortho carbon.
Keep pushing the ring’s π pairs and the − moves on to the carbon opposite the NH₂ group, the para carbon. Pushing the other way around the ring puts it on the second ortho carbon. The two carbons in between, the meta carbons, never carry it.
The resonance push is much larger than the inductive pull. The ring ends up richer in electrons than benzene, and aniline reacts with electrophiles far faster than benzene does. This one comparison explains which ring positions react first, the directing effects of Aromatic Chemistry.
In most molecules resonance wins like this. One rough way to see why is that a resonance structure moves a whole electron pair onto another atom, while induction only tilts a bonding pair a little. Treat that as a rule of thumb, not a law, because the real molecule is only partly that resonance structure. Halogens are the best-known exception: on a benzene ring, chlorine’s inductive pull outweighs its resonance push, as Aromatic Chemistry will show. The table sums up the two effects.
| Induction | Resonance | |
|---|---|---|
| Through | σ bonds | overlapping p orbitals |
| Driven by | electronegativity | conjugation |
| Arrows? | no | yes |
| Range | 2–3 bonds, fading | whole conjugated system |
| Strength | weaker | usually stronger |
What carries forward
This lens is the working method for the rest of the course. It tells you which carbon is attacked in Substitution & Elimination, why the carbonyl carbon is the target in Carbonyl Chemistry, which position on a ring reacts in Aromatic Chemistry, and which proton is removed in Enolate Chemistry. When a later page says a site is “activated” or “deactivated,” it is talking about electron density, and the explanation will always turn out to be induction, resonance, or both.