Organic Chemistry 202-SNU-RE • CEGEP, Montreal
Benzene is the molecule that refuses to behave as its bonds suggest: three double bonds drawn, and yet no addition, no decolourization of dibromine, and a stability of kJ/mol above the model.
This sheet leaves the lecture material in the course notes and gathers what decides things on an exam: the four conditions to check one by one, the two-step mechanism that restores aromaticity, and the table of directing effects with its exception.
The thread of the chapter
Aromaticity is not an impression given by a drawing but a list of FOUR conditions, and the reactivity that follows from it is a substitution, never an addition.
The four conditions are checked in order, and the first one that fails closes the discussion. Planarity is the one people forget, and it is precisely the one the questions test.
Every reaction has to respect this stabilization: addition would destroy it, substitution preserves it. The whole chapter follows from this single energy comparison.
The intermediate, often called the sigma complex, is NOT aromatic: its positive charge is delocalized over only three carbons, and this delocalization is what explains the directing effects.
Activating or deactivating decides the RATE; ortho-para or meta decides the POSITION. These are two separate questions, and the halogens prove that they do not always go together.
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
“Benzene decolourizes dibromine, like any compound with double bonds.”
What to write
“Benzene does not react with alone: a catalyst is needed, and the reaction is then a SUBSTITUTION, with no decolourization by addition.”
Why: An addition would destroy aromaticity, and so would cost the kJ/mol of resonance energy. Substitution preserves it: it is the only path open.
What not to write
“The ring has three double bonds and an oxygen with two lone pairs, so electrons.”
What to write
“Only the DELOCALIZED electrons in the ring count: a lone pair whose orbital is not parallel to the system does not count.”
Why: An atom can commit only one lone pair to the cloud, the one whose orbital is perpendicular to the plane of the ring. The other stays in the plane and takes part in nothing.
What not to write
“Cyclooctatetraene has eight electrons, so it is antiaromatic.”
What to write
“It is not planar: it adopts a tub shape and escapes the problem, it is simply NON-aromatic.”
Why: Hückel's rule applies only to planar, fully conjugated systems. A molecule that can pucker always does so, precisely to avoid antiaromaticity.
What not to write
“This compound is not aromatic, so it is antiaromatic.”
What to write
“Antiaromatic means DESTABILIZED relative to the open analogue; non-aromatic only means the question does not arise.”
Why: An alkane is not aromatic and is perfectly fine. Antiaromaticity is a real energy penalty, reserved for planar, cyclic, conjugated systems with electrons.
What not to write
“Chlorine is deactivating, so it directs meta.”
What to write
“The halogens are the exception: deactivating by the inductive effect, but ORTHO-PARA directors by the resonance effect.”
Why: Two effects are at odds: the inductive effect slows the reaction, the resonance effect decides the position. It is the only family where the two answers do not go together, and the exam asks about it every session.
What not to write
“Nitrobenzene is acylated with to obtain the meta ketone.”
What to write
“A ring bearing a strongly withdrawing group does not undergo Friedel-Crafts reactions: the reaction simply does not take place.”
Why: The Friedel-Crafts electrophile is weak and demands a ring at least as rich as benzene. A nitro group depletes the ring far below that threshold.
What not to write
“To attach a propyl group, carry out a Friedel-Crafts alkylation with -chloropropane.”
What to write
“The primary carbocation rearranges to isopropyl: acylate first, then reduce the ketone, which gives the straight chain.”
Why: Alkylation suffers from two flaws: rearrangement of the carbocation and polyalkylation, the product being richer, hence more reactive, than the starting material. Acylation has neither.
What not to write
“Brominate then nitrate, or the reverse, it comes to the same thing.”
What to write
“The first group attached controls the position of the second: brominating first gives the para product, nitrating first gives the meta.”
Why: Each substituent imposes its orientation on the next. An aromatic synthesis is therefore planned backward, starting from the desired isomer.
What not to write
“The intermediate carbocation is stabilized by the aromaticity of the ring.”
What to write
“The intermediate has LOST its aromaticity: its charge is delocalized over only three carbons, and that is why the step is slow.”
Why: If the intermediate stayed aromatic, there would be no barrier and benzene would react like an alkene. The temporary loss of aromaticity IS the costly step.
Ask the four questions in this order and stop at the first negative answer.
If the system is not cyclic → non-aromatic, the discussion is closed
Example: an open polyene, even if perfectly conjugated
If an atom of the ring is → conjugation is interrupted, so non-aromatic
Example: a single in the ring is enough
If the ring cannot stay planar → non-aromatic, whatever the electron count
Example: cyclooctatetraene in its tub shape
If cyclic, planar, conjugated, with electrons → AROMATIC
Example: , , or electrons
If cyclic, planar, conjugated, with electrons → ANTIAROMATIC, hence destabilized
Example: cyclobutadiene, with four electrons
A lone pair counts only if it occupies an orbital perpendicular to the plane of the ring. An atom never commits more than one, however many lone pairs it carries.
Look at the group ALREADY on the ring, never the one that is arriving.
If donor group: , , , alkyl → activating, ortho and para director
Example: phenol reacts much faster than benzene
If withdrawing group: , , , → deactivating, meta director
Example: nitrobenzene reacts slowly, and meta
If halogen → deactivating BUT ortho and para director
Example: it is the only exception, and it comes up often
If two substituents already present → the more ACTIVATING of the two controls the orientation
Example: an wins over a halogen
Activating or deactivating answers “how fast”; ortho-para or meta answers “where”. Answering the two questions separately avoids the halogen trap.
A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.
When to use it: The question asks whether a compound is aromatic, or asks you to rank several rings.
Concluding sentence
“The ring is planar, fully conjugated and has electrons, that is with : it is aromatic.”
The trap: Concluding from the electron count alone without having checked planarity: this is exactly the case of cyclooctatetraene.
Marking: 0.5 mark per condition checked, 1 mark for the justified electron count, 1 mark for the named conclusion.
When to use it: The question asks for a specific isomer, ortho, meta or para, starting from benzene.
Concluding sentence
“To get the meta isomer, nitrate first: the nitro group will direct the bromination meta.”
The trap: Planning in chronological order rather than from the product: you end up with the wrong isomer one time out of two.
Marking: 1 mark for the analysis of the product, 1 mark for the justified order, 1 mark for the feasibility of each step.
Five minutes of checking recover more marks than one more problem started in a hurry.
Does the electron count give an integer n?
Subtract from the count found and divide by : the result must be a nonnegative integer.
electrons give : this is not .
Does the reaction preserve aromaticity?
Look at the announced product: the ring must still be planar, conjugated and have electrons.
An addition product on benzene has lost its aromaticity: it does not form.
Does the orientation match the group already present?
Check that the position of the new group is the one imposed by the initial substituent, halogens included.
A meta product obtained from toluene is impossible: the methyl group directs ortho and para.
Is each step feasible?
Reread the synthesis and check that no Friedel-Crafts reaction is attempted on a ring bearing a strongly withdrawing group.
Acylating nitrobenzene gives nothing, whatever the amount of catalyst.
Benzene, dibromine with , and a nitrating mixture are available.
In what order must you proceed to obtain the META isomer, and what would the reverse order give?
Step 1
Analysis of the product: the two groups are at positions and , so meta to each other.
Why
You always start from the desired product. The relative position alone designates the group that must be attached first.
Step 2
The group is a withdrawing group: only it directs meta. It is therefore the one to attach first.
Why
Bromine, an ortho-para director, would send the second group elsewhere. The reasoning is done on the substituent ALREADY present, never on the one arriving.
Step 3
First step: nitration of benzene with and , giving nitrobenzene.
Why
The electrophile is the nitronium ion, formed by the sulfuric acid. The starting ring is bare benzene, so perfectly reactive.
Step 4
Second step: bromination with and , directed meta by the nitro group.
Why
The ring is deactivated, so the reaction is slow and needs a catalyst, but it remains possible: this is not a Friedel-Crafts reaction.
Step 5
Reverse order: bromination would give bromobenzene, then nitration would take place ortho and para, mainly para.
Why
Bromine is deactivating but an ortho-para director: this is the exception of the course, and it alone explains why the order is not a matter of indifference.
Step 6
Check: neither step is a Friedel-Crafts reaction, so the deactivation blocks nothing.
Why
Nitration and catalyzed halogenation still work on a depleted ring, unlike alkylation and acylation.
The conclusion, written out
“Nitrate first, then brominate: the nitro group directs the bromination meta and gives -bromo--nitrobenzene. The reverse order would lead mainly to the para isomer.”
The classic mistake on this problem: Brominating first on the assumption that bromine, being deactivating, directs meta: this is precisely the exception of the course.
Four, and all together: the system must be cyclic, planar, fully conjugated, and have a number of delocalized electrons of the form four n plus two, so two, six, ten or fourteen. If planarity is missing, the compound is simply non-aromatic; if the count is four n, it is antiaromatic.
Because an addition would destroy its aromaticity, which is worth about one hundred and fifty kilojoules per mole of stabilization. The ring prefers substitution, which replaces a hydrogen and keeps the delocalized system. A catalyst is needed, incidentally, for this substitution to occur.
Because two opposing effects coexist. By the inductive effect, the very electronegative halogen depletes the ring and therefore slows the reaction. By the resonance effect, its lone pair delocalizes toward the ortho and para positions, which become the richest. Rate and position are thus decided by two different effects.
Because alkylation suffers from two flaws. The carbocation formed can rearrange, so that a straight chain becomes branched. And the product, richer than the starting material, reacts a second time, giving polyalkylation. Acylation avoids both, and the ketone obtained is then reduced cleanly.
Start from the desired product and look at the relative position of the two groups. The one that, alone on the ring, directs toward that position must be attached first. For a meta product, you therefore install the withdrawing group first; for an ortho or para product, the donor or the halogen first.
© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-snu-aromatic-compounds. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).
