Organic Chemistry 202-SNU-RE • CEGEP, Montreal
Resonance is the chapter that separates those who learn rules from those who understand reactivity. “ goes to the carbon with the most hydrogens” is enough as long as the question is friendly, and it collapses as soon as a rearrangement occurs.
This sheet leaves the course in the textbook and gathers what decides in an exam in Organic Chemistry 202-SNU-RE: the rules for moving electrons, the stability scale, and the tree that settles between substitution and elimination.
The thread of the chapter
A single scale governs the whole chapter, that of carbocation stability: Markovnikov, SN1, E1 and rearrangements are only consequences of it.
The more numerous and EQUIVALENT the resonance structures, the more stabilized the hybrid. This is what explains the acidity of carboxylic acids and the stability of allylic and benzylic cations.
This scale is the one thing in the chapter to know by heart. It decides Markovnikov, the choice between SN1 and SN2, and whether a rearrangement occurs.
The rate law and the optical rotation are the two experimental proofs expected: order and inversion for the SN2, order and partial racemization for the SN1.
With a very bulky base, the major alkene becomes the LEAST substituted: this is the Hofmann exception, and the question always signals it through the name of the base.
These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.
What not to write
“I move the to the other side to get the second resonance structure.”
What to write
“Only the electrons move: the nuclei stay exactly where they are, otherwise it is another compound.”
Why: Two resonance structures describe THE SAME molecule seen differently. Moving an atom makes an isomer, which is a chemical reaction, not a way of writing.
What not to write
“The molecule oscillates between the two structures, half the time in each.”
What to write
“The real molecule is the HYBRID, unique and permanent: the resonance structures do not exist separately.”
Why: An equilibrium is written with two opposite half-arrows and involves two real species. The double-headed arrow of resonance links two DRAWINGS of a single species.
What not to write
“The goes to the carbon with the most hydrogens, so the bromine necessarily goes to the other, and that is that.”
What to write
“The proton attaches so as to generate the MOST STABLE carbocation; that is the rule, and it is what lets you predict a rearrangement.”
Why: The rhyme gives the right result in simple cases and the wrong one as soon as the carbocation formed can rearrange. It describes a consequence, not the cause.
What not to write
“The nucleophile attacks the tertiary carbon from behind and inverts the configuration.”
What to write
“A tertiary carbon is too crowded: the backside attack is impossible, and the reaction goes by an SN1 or an elimination.”
Why: The SN2 requires free access on the side opposite the leaving group. Three substituents block this passage, and it is precisely the case where the carbocation is most stable.
What not to write
“The substrate is primary, but with a good solvent we still get a carbocation.”
What to write
“A primary carbocation is too unstable to form: on a primary, it is an SN2.”
Why: The stability scale fixes what is possible: what cannot form does not form, whatever the solvent. Only allylic and benzylic cations escape this rule, thanks to resonance.
What not to write
“The product of the SN2 is a racemic mixture.”
What to write
“The SN2 gives a COMPLETE INVERSION of configuration: a single enantiomer, the one of opposite configuration.”
Why: The backside attack imposes a single exit geometry. It is the SN1, going through a planar carbocation that can be attacked from both sides, that gives a mixture.
What not to write
“We remove the from the most accessible carbon, so the terminal alkene is major.”
What to write
“Zaitsev's rule: the major alkene is the most SUBSTITUTED, hence the most stable, unless the base is bulky.”
Why: The most substituted alkene is stabilized by hyperconjugation, like the carbocation. The exception appears when the base is too big to reach the internal hydrogen.
What not to write
“There is an on the neighbouring carbon, so the E2 can happen.”
What to write
“An ANTI-PERIPLANAR to the leaving group is needed: on a ring, this requires both to be axial.”
Why: The E2 is concerted: the orbitals must overlap in a single plane. The conformation that allows this is not always the most stable one, and that is the whole point of the question.
What not to write
“The secondary carbocation is formed, the bromine attaches to it, reaction over.”
What to write
“A hydride from the neighbouring carbon can migrate to give a tertiary carbocation, more stable: it is the one that gives the major product.”
Why: A rearrangement occurs as soon as it gains a step on the stability scale, and it is faster than the attack of the nucleophile.
Look at the substrate first, the reagent next, the conditions last.
If methyl or primary substrate, good nucleophile → SN2, with inversion of configuration
Example: with
If primary substrate with a strong BULKY base → E2, and the alkene obtained is the least substituted
Example: potassium tert-butoxide
If tertiary substrate, weak nucleophile, protic solvent → SN1 and E1 in competition, with partial racemization
Example: in water
If secondary or tertiary substrate, strong unhindered base → E2, Zaitsev alkene major
Example: in ethanol, hot
If the temperature is high → elimination wins over substitution, whatever the rest
Example: heating is always a clue in the question
A tertiary carbon never does an SN2, a primary carbon never an SN1: these two prohibitions eliminate half the possibilities before you even read the reagent.
Compare the two drawings criterion by criterion, in this order.
If one has more complete octets → it is major, this criterion takes priority over all the others
Example: a six-electron carbon weighs heavily against a structure
If both have their octets, but one carries fewer charges → the one with the fewest separated charges is major
Example: a neutral structure beats a two-charge structure
If the charges are equal in number → the negative charge must be on the MOST electronegative atom, the positive on the least
Example: the negative on oxygen rather than on carbon
If the two structures can be superimposed by symmetry → they are EQUIVALENT and the stabilization is maximal
Example: the two structures of the carboxylate
Resonance stabilization explains almost all the acidities of the chapter: the more delocalized the conjugate base, the stronger the corresponding acid.
A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.
When to use it: The question asks for the mechanism, or to “justify the major product”.
Concluding sentence
“The substrate is tertiary and the nucleophile weak: this is an SN1, whose slow step is the formation of the carbocation, which explains the racemization observed.”
The trap: Starting a curved arrow from a positive charge or from an atom: an arrow starts from a lone pair or a bond, never from anything else.
Marking: 1 mark for the mechanism named and justified, 1 mark per step with its arrows, 1 mark for the intermediate named, 1 mark for the stereochemical conclusion.
When to use it: The question asks to compare two acidities, or to explain why a carboxylic acid is more acidic than an alcohol.
Concluding sentence
“The conjugate base of the carboxylic acid has two equivalent resonance structures, which stabilizes it strongly: the carboxylic acid is therefore much more acidic than the alcohol.”
The trap: Comparing the acids themselves instead of their conjugate bases: it is the stability of the ANION that decides.
Marking: 1 mark for the conjugate bases written, 1 mark for the resonance structures, 1 mark for the conclusion in the right direction.
Five minutes of checking recover more marks than one more problem started in a hurry.
The total charge is conserved
Compare the overall charge of each resonance structure: it must be identical, and identical to that of the starting species.
A neutral structure proposed for an anion: an electron has been lost on the way.
No atom has moved
Mentally superimpose the skeletons of the two resonance structures: the positions of the nuclei must coincide exactly.
A moved hydrogen turns a resonance structure into an isomer.
The octet is not exceeded
Count the electrons around each carbon, nitrogen and oxygen: never more than eight.
A carbon with five bonds is impossible, the structure is to be rejected.
Is the carbocation the most stable one accessible?
Before letting a carbocation react, check whether a neighbouring hydride or methyl would make it tertiary.
A secondary cation next to a tertiary carbon will rearrange, and the product changes.
Hydrogen bromide is added to 3-methylbut-1-ene, of formula .
Give the major product and justify each step of the mechanism.
Step 1
Protonation: attaches to carbon , terminal and with the most hydrogens.
Why
This choice generates the most stable carbocation accessible at this step, a secondary cation on carbon . This is Markovnikov's rule in its useful version.
Step 2
Neighbouring carbon is tertiary: a hydride migrates from it to carbon , by a 1,2-hydride shift.
Why
The rearrangement gains a full step on the stability scale, from secondary to tertiary. It is faster than the attack of the bromide, so it takes place.
Step 3
The tertiary carbocation thus formed is attacked by .
Why
The anion attacks the charged carbon, now carbon . Since the carbocation is planar, the two faces are equivalent, but this carbon is not asymmetric here.
Step 4
Major product: -bromo--methylbutane.
Why
It is the product coming from the most stable cation. The product without rearrangement, -bromo--methylbutane, remains minor.
Step 5
Formula check: gives , whose degree of unsaturation is .
Why
The addition consumes the double bond: the product is saturated, which the zero degree of unsaturation confirms. A product that was still unsaturated would signal an elimination.
The conclusion, written out
“The major product is -bromo--methylbutane: protonation gives a secondary cation, which rearranges to a tertiary cation by hydride shift before the attack of the bromide.”
The classic mistake on this problem: Announcing -bromo--methylbutane by applying Markovnikov without looking for the rearrangement: it is the minor product.
Different drawings of one and the same molecule, not species that would alternate. The real molecule is the hybrid of these structures, unique, permanent, and more stable than any one of them. This is why the two carbon-oxygen bonds of a carboxylate ion have exactly the same length, intermediate between single and double.
The proton attaches to the carbon that leads to the most stable carbocation. The common formulation, according to which the hydrogen goes to the carbon with the most hydrogens, is only a consequence of it in simple cases. It becomes false as soon as a rearrangement of the carbocation is possible.
Look at the substrate first. A methyl or primary carbon can only do a one-step substitution, with inversion of configuration. A tertiary carbon can only take the path through a carbocation, with racemization. On a secondary carbon, it is the nucleophile and the solvent that decide.
Because the nucleophile attacks from exactly the side opposite the leaving group, the only accessible side. The other three bonds then flip over like an umbrella in the wind. The product therefore has the configuration opposite to that of the reactant, and a single enantiomer is obtained.
As soon as moving a hydrogen or a methyl from a neighbouring carbon moves it up a step on the stability scale, for example from secondary to tertiary. This shift is faster than the attack of the nucleophile, so the major product comes from the rearranged cation.
© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-snu-resonance-mechanisms. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).
