Electrophilic Aromatic Substitution Flashcards
Which molecule shown is the strongest base
b
the methyl group is electron donating and hence will increase the availability of LP on nitrogen
What makes aromatic systems relatively unreactive?
Because of the high stability of aromatic systems due to resonance stabilisation
(breaking the aromaticity requires a lot of energy
What is the difference between the way an isolated alkene vs benzene reactions with bromine
- The isolated alkene reacts through addition
- Benzene reacts through substitution but needs a iron bromide catalyst
Electron-rich aromatics react with…
Electrophiles
Electron-Poor aromatics react with
Nucleophiles
How does the mechanism of electrophilic addition of Bromine to alkenes work
- Electron density from the double bond attacks the δ+ bromine
- Bromide remaining attack carbocation forming a dibromo compound
For both alkenes and aromatics, a what intermediate is formed
Cationic Intermediate
In electrophilic substitution of Benzene using bromine, how is the electrophile generated
- The LP on one of the bromines attacks the iron of FeBr₃ (making bromine even more δ+)
What is the mechanism for electrophilic substitution of the Bromination of Benzene
- Electron density from C=C attacks the δ+ bromine
- Hydrogen breaks it bond to carbon and electrons are used to reform C=C bond
What is the intermediate formed in electrophilic armoatic substitution called
A Wheland intermediate (stabilised cation as the positive charge can delocalise around the ring)
The formation of the Wheland intermediate is…
…rate determining
(due to the intermediate not being aromatic - hence higher in energy)
In the Nitration of Benzene, what are the two typical conditions needed
Conc HNO₃ and H₂SO₄
How do you generate the electrophile in a Nitration?
- The LP on the hydroxyl of HNO₃, will attack the hydrogen of the hydroxyl on H₂SO₄
- The LP on the oxygen forms a N=O bond and the N-O bond is broken forming water
- result in the formation of a nitronium ion
How does the nitronium ion reaction with benzene in a Nitration reaction
- Electrons from C=C on benzene attack the positive nitrogen on the nitronium
- The C-H bond breaks allowing the C=C bond to be reformed (restoring aromaticity)
What are the typical conditions for the Sulfonation of Benzene
Conc. H₂SO₄ saturated with SO₃
How does the generation of the electrphile occur in sulfonation
- The LP on the Oxygen of the hydroxyl group of H₂SO₄ 1, will attack the hydrogen of the other hydroxyl group on H₂SO₄ 2
- The LP on the other hydroxyl of the H₂SO₄ will form a S=O bond causing the S-O bond to break forming water
How does Sulfonation of Benzene occur
- Electrons from C=C on benzene will attack the S (on sulfonium) resulting in the S=OH to become a single bond
- The C-H bond breaks allowing the C=C double bond to reform (restoring aromaticity)
A Friedel-Crafts Alkylation involves?
The addition of an Alkyl group (alkane missing 1H) to a benzene ring
How is the electrophile generated in Friedel Crafts Alkylation?
- The LP on Cl of the R-Cl, will attack Aluminium forming a Cl-Al bond (making the alkyl group more δ+)
- The R-Cl bond breaks forming a carbocation electrophile
How does the Friedel-Craft Alkylation occur
- Electrons from the C=C on benzenes attacks the carbocation
- The C-H bond breaks, reforming the C=C bond (restoring aromaticity)
What are some disadvantages of the Friedel-Craft Alkylation reaction
- The products of Fridel-Crafts Alkylation reactions are more reactive than the starting materials. This is due to alkyl groups being EDG resulting in over-alkylations occuring
- Friedel-Craft Alkylations are only useful with tertiary alkyl halide electrophiles (rearrangement occuring in primary and secondary)
How is the electrophile formed in a Friedel-Crafts Acylation?
- A LP on Cl of Acyl chloride will attack the Al of AlCl₃, resulting in Al-Cl bonding forming
- Electrons from oxygen of carbonyl form a triple C-O bond causing chlorine to leave
- Forms an acylium ion
How does the benzene and acylium ion reaction in a Friedel-Crafts Acylation?
- Electrons from C=C attack the positive carbon on acylium ion, breaking the C-O triple bond
- The C-H bond will break allowing the C=C to reform (restore aromaticity)
Following the trend, Toluene (methyl substituent to benzene) will be nitrated faster than benzene
What are the two reasons for this?
- weak inductive effects (electron donating) will activate the ring towards electrophilic attack
- AND methyl group donates electrons by hyperconjugation
What is the reason over-alkylation can occur in Friedel-Crafts alkylation
Because the substituent added can hyperconjugate with aromatic system
Why does Phenol react much fast than benzene in a nitration reaction (x1000) rate
Because the LP on OH can delocalise into the aromatic system creating stabilisation through multiple resonance forms (particuarlly in the ortho and para positions) = Mesomeric effects
Bigger effect than inductive effects
Why does Nitrobenzene react much slower than benzene
- The nitro group is Mesomerically electron withdrawing (pull aromatic electron density towards itself resulting in positive charge in ortho/para positions)
- AND inductively electron withdrawing
Why does methoxybezene react so much faster than benzene
- Methoxy with mesomerically and inductively donating
Which product(s) would predominatly form if methoxybenzene was nitrated?
a and c
(negative chare on ortho and para positions in the resonance forms)
Why are electron-donating groups ortho and para directing
Because intermediate is more stable due to have multiple resonance forms
Hence the transition state from ortho and para attack is lower in energy
(be careful however cuz para might be preferred due to sterics)
Why are electron-withdrawing groups meta-directing
Meta-products formed because intermediate is more stable as there is the least electron density withdrawn from the carbocationic intermediate + more intermediates can be drawn
(however it is the lest bad route)
Why are Halogens ortho/para directing
Halogens are inductively withdrawing (hence deactivating as removing e density)
BUT there is an opposite mesomerical electron donating effect resulting in resonance structures with -ve charge in ortho/para positions
How should we make para-nitrochlorobenzene?
By chlorinating nitrobenzene or nitrating chlorobenzene?
Nitrating chlorobenzene (as para-directing)
