Substituted benzenes Flashcards
Fast rate of electrophilic substitution?
The slower step is the formation of an intermediate carbocation by reaction of the benzene ring with an electrophile the lower the energy of the transition state leading to the carbocation the lower the activation energy and the faster the rate of electrophilic substitution
How to predict the effect of an existing substituent on the rate of electrophilic substitution?
Need to consider whether R stabilises or destabilises the transition state leading to the carbocation
How can you use Hammonds postulate?
For each reaction the transition state leading to the carbocation is closer in energy to the carbocation than to the reactants using Hammonds postulate you can assume that the structure of the transition state resembles the structure of the carbocation in each reaction so the stabilities of the intermediate carbocations can be used to explain the different rates of electrophilic substitution reacts of substituted benzenes, the more stable the intermediate carbocation the faster the rate of electrophilic substitution
Activating groups?
Substituents that stabilise the carbocation make the ring more reactive toe electrophilic substitution than benzene and are called activating groups
Deactivating groups?
Substituents that destabilise the carbocation make the ring less reactive to electrophilic substitution than benzene and are called deactivating groups
What are activating groups?
Groups that donate electrons into the benzene ring by positive inductive effects or positive mesomeric effects, electron donating groups increase the reactivity of the ring to reaction with electrophile because the carbocation is stabilised by electron donating groups.
Common activating groups?
- NH2, -NHR, -NR2, -OH, -OR, -O-, -NHCOR,
- OCOR, -Ph, -CH=CH2, -R
How to activating groups stabilise the ring?
+M effects are stronger than +I effects so electron donating groups with + M effect stabilise the carbocation more effectively than electron donating groups with +I effect. =NH2 group has + M effects and is a strong electron snort whereas the -Me groups has a +I effect and is a weaker electron donor, this explains why aniline reacts faster with electrophile than toluene, electron donating groups with strong +M effects stabilise carbocations more effectively than electron donating groups with weaker +M effects
What are deactivating groups?
Withdraw electrons from the benzene ring by negative inductive effects -I or negative mesomeric effects -M. Electron withdrawing groups decrease the reactivity of the ring to reaction with electrophiles because the carbocation is destabilised by electron withdrawing groups
Common deactivating groups?
- Cl, -Br, -I, -CHO, -COR, -CO2H, -CO2R,
- SO3H, -NO2, -NR3+
How to deactivating groups destabilise the ring?
Electron withdrawning groups with -M effects destabilise carbocations more effectively than electron withdrawing groups with -I effects because -M effects are stronger than -I effects. NO2 group has a -M effect and is a strong electron acceptor whereas the Cl group has a -I effect and is a weaker electron acceptor. This explains why chlorobenzene reacts faster with electrophiles than nitrobenzene. The strongest electron withdrawing groups include SO3H and NO2 groups. These groups contain three or four highly electronegative atoms that exert a strong -I effect and have at least one polar double bond that accepts a pair of electrons and so exerts a strong -M effect that is strong -I and -M effects reinforce one another
Ortho?
2 position
Meta?
3 position
Para?
4 position
Electron donating activating positons?
Electron donating groups direct electrophile to the 2 and 4 positions
Electron withdrawing deactivating positons?
Electron withdrawing groups direct electrophiles to the 3 positions
Exception of halogens?
When the groups is Cl, Br or I electrophiles are directed to the 2 and 4 positions even though these groups are deactivating
Why are activating groups 2,4 directing activators?
all activating substituents with positive inductive +I and positive mesomeric +M effects direct the incoming electrophile to the 2 and 4 positions of the ring this is because the intermediate carbocation is stabilised by the +I or +M effect of the activating substituent, whereas introducing the electrophile at the 3 positions produces an intermediate carbocation that cannot be effectively stabilised by the +I and +M effect of the activating substituent, because carbocations formed from attack at the 2 and 4 positions are the more stable electrophilic substitutions at these positions will have lower Gibbs energy of activation
2,4 directing deactivators?
Cl, Br and I substituents are unique in that although deactivating they direct the incoming electrophile to the 2 and 4 positions of the ring, the halogens are deactivating because they are so highly electronegative that the -I effect o these atoms is stronger than the +M effect. However the weak +M effect of the halogens foes explain why these substituents direct electrophiles to the 2 and 4 positions
3 directing deactivators?
All deactivating substituents with negative inductive -I and negative mesomeric -M effects direct the incoming electrophile to the 3 positions of the ring
Steric effects?
If the substituent is large it may direct the electrophile to a different position
Reactivity of substituted benzenes?
Depends on the existing substituents on the beaned ring, benzene ring containing activating substituents are more reactive to electrophiles than benzene rings containing deactivating substituents. This explains why activated benzene rings rect with weak electrophiles whereas deactivated benzene rings only react with strong electrophiles
Brominate nitrobenzene?
Strong electrophile is needed because the NO2 group in nitrobenzene is so strongly deactivating
Brominate phenol?
Requires only Br2 a weak electrophile because the OH group in phenol is strongly activating. Even 2 and 4-bromophenol react with a further two equivalent of Br2 to form 2,4,6-tribromophenol this is impressive because a Br substituent is a deactivator so the greater the number of Br atom on the ring the less reactive it is to Br2
Why is matching the electrophile to the substituted benzene important in treating nitrobenzene or phenylamine with RCl and AlCl3 or RCOCl and AlCl3?
Neither aniline nor nitrobenzene undergo Fridel Crafts reactions.
No reaction is observed with nitrobenzene because the ring is too unreactive to react with the coordination complex or cabochon no substituted benzene with a deactivating group undergoes Friedel Crafts alkylation of acylation reactions
AlCl3 reacts with the NH2 group of phenylamine in preference to RCl or RCOCl, the NH2 group donates a lone pair of electrons to AlCl3 and this produces a coordination complies which has appositely charged nitrogen atom, the psotiviely charged nitrogen atom exerts a strong -I effect the ring becomes deactivated and does not react in Friedel crafts alkylation or acylation reactions
How is it easy to separate the desired structural isomer?
In synthesis where only one structural isomer is required the desired structural isomer is usually easily separated from other structural isomers because the stratal isomers normally have different physical properties, often reactions conditions are varied in an attempt to maximise the yield o the desired structural isomer - this is especially important in multistep synthesis
Polysubstitution?
Other potential byproducts in electrophilic substitutions are formed by polysubstitution of the benzene ring, the likelihood of polysubstitution depends on the nativity of the first formed products and reactants to electrophiles.
Introducing substituents not the ring?
Oxidation and reduction of substituents on benzene rings is very useful in synthesis, in the Clemmensen reduction zinc amalgam and HCl are used to reduce a ketone group to a CH2 group, similarly a nitro substituent NO2 is reduced to an amine substituent NH2 using tin and hCl. Both reductions convert a deactivating substitute into an activating substituent
Oxidation of a CH3 substitute to a carboxylic acid using KMnO4, oxidation coverts an activating substituent to a deactivating substituent
Aryl diazonium?
Prepared by reacting aromatic amines with NO+ ion between 0 and 5 degrees, the NO+ ion is generated from HNO2 because nitrous acid is unstable it is formed in situ by reacting NaNO2 with HCl at 0 degrees.
NO+?
Strong electrophile and it reacts with a nucleophilic amine at 0 to 5 degrees C to form a compound with a new N-N bond. After deprotonation a nitrosamine is formed and this reacts by the series of protonations and deprotonations to form the aryl diazonium ion and water, a process called diazotization. The diazonium ion is formed with an accompanying anion and if HCl is used in the synthesis an arenediazonium chloride salt is formed
Aryl diazonium salts?
Unstable and on warming to room temperature or above they lose N2, this forms a very unstable aryl cation that reacts with nucleophiles in an SN1 reaction, reaction with water introduces an OH substituent on the ring and this is an important way of making phenols, they cannot be prepared directly by electrophilic substitution reactions because HO+ cannot be generated
Sandmeyer reactions?
Reactions using copper(I) salts
Aryl diazonium salt reacts with CuCN to give a CN substituent
Aryl diazonium salt reacts with CuBr to give a Br substituent
Aryl diazonium salt reacts with CuCl to give a Cl substituent
Likely these reactions involve aryl radical intermediate rather than aryl cations
Aryl diazonium salt reacts with KI?
To give a I substituent
Aryl diazonium salt reacts with H3PO2?
To give a H substituent
Aryl diazonium salt reacts with H2O and heat?
To give a OH substituent
Aryl diazonium salt reacts with HBF4 and heat?
To give a F substituent
Aryl diazonium ions react with phenol or anilines?
Because aryl diazonium ions are positively charged they act as electrophiles and react with strongly activated substituted benzenes to form highly coloured ago compounds in coupling reactions. The lone pair of electrons on oxygen in phenol attacks through the benzene ring, if the OH group attacked the diazonium ion directly the product would be a less stable product with a weak O-N bond. Also notice that phenol attacks the terminal nitrogen atom in the diazonium ion, attack at the more hindered positively charged nitrogen atom would require nitrogen to be able to form five bonds
Birch reduction?
Na or Li in ammonia metal becomes oxidised and electrons are released into solution, solvated electrons as into low energy anti bonding orbitals of the aromatic to produce a dianion which repel one another so sit on opposite carbons on either side of the ring, dianion is then promptly protonated by an alcohol also present in solution to afford the unconjugated diene, when an electron withering group is present on the aromatic the ipso and para position are reduced, when an electron donating group is present the orthodox and meta carbon are reduced on opposite sides of the ring
Nucleophilic aromatic substitution conditions?
Harsh, basic conditions aromatics which possess a good leaving group such as a halogen undergo nucleophilic substitution. strong bases such as sodium amide or potassium test but oxide are required and in this instance the base also acts as a nucleophile.
Process of SnAr?
Two step elimination addition mechanism, strong base deprotonates the proton ortho to the leaving group to give an aryl anion, syn addition elimination then occurs to give an aromatic intermediate with a triple bond - benzyne. Addition of the nucleophile to the electrophilic triple bond and subsequent deprotonation gives the substituted product
Benzyne?
not true alkyne, carbon atoms within the bond are not sp hybridised and the bond is not linear, the carbons atoms are still sp2 hybridised and a pi bond is still incorporated in the aromatic ring. the other pi bond is perpendicular to the pi cloud of the aromatic system and is formed by overlap of the two adjacent sp2 orbitals. The overlap is poor so this bond is very weak and is readily stacked by nucleophiles to give a substituted product
Benzyne + R2NH?
NR2 substituent
Benzyne + ROH?
OR substituent
Benzyne + RCOOH?
COOR substituent
Benzyne + RMgBr?
R substituent
Benzyne + -TMSF-OTF?
2 trimethylsilyl phenyl triflate substituent
Benzyne + CO2 and N2 nd heat?
COO and N triple bond N substituent
