nucleophilllic aromatic substitution

A nucleophilic aromatic substitution is a substitution reaction in organic chemistry in which the nucleophile displaces a goodleaving group, such as a halide, on an aromatic ring. There are 6 nucleophilic substitution mechanisms encountered with aromatic systems:

• the S_NAr (addition-elimination) mechanism

• Summary 
   
  • The generally accepted mechanism for nucleophilic aromatic substitution in nitro-substituted aryl halides is shown by example below:
  
  
  
  • Attack of the strong nucleophile on the halogen substituted aromatic carbon forming an anionic intermediate.
  • Loss of the leaving group, the halide ion restores the aromaticity.
  • Kinetics of the reaction are observed to be second order.
  • The addition step is the rate determining step (loss of aromaticity).
  • Nucleophilic substitution, and therefore reaction rate, is facilitated by the presence of a strong electron withdrawing group (esp. NO₂) ortho or para to the site of substitution, which stabilize the cyclohexadienyl anion through resonance.
  
  
  
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• the aromatic SN1 mechanism encountered with diazonium salts

The wide utility of aryl diazonium ions as synthetic intermediates results from

the excellence of N2 as a leaving group. There are several general mechanisms by

which substitution can occur.One involves unimolecular thermal decomposition of

the diazonium ion, followed by capture of the resulting aryl cation by a nucleophile.

The phenyl cation is very unstable and therefore  Either the solvent or an anion can act as the nucleophile. to react with it and completes nucleophillic aromatic substitution.

• the benzyne mechanism

Elimination-Addition Mechanism: Benzyne

Summary:

• This pathway is followed when the nucleophile is an exceptionally strong base (e.g. amide ion, NH₂^-) and the absence of the strong electron withdrawing groups:

• Nucleophilic substitution can lead to substitution on either
• the same carbon that bore the leaving group (see addition mechanism above)
• or on an adjacent carbon (see addition mechanism below)

 
• This is most readily apparent when the benzyne is substituted:

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Resonance: Most Important Resonance Contributor

Discussion:  Many molecules or ions that participate in an organic reaction have resonance. 

  When deciding which resonance contributor to use, it makes sense to use the one that makes the greatest contribution to the resonance hybrid.

  These rules are based on the idea that if individual resonance contributors did indeed exist, the most thermodynamically stable structures would make more significant contributions to the resonance hybrid. 
Factors that enhance thermodynamic stability are maximization of covalent bonding and minimization of charge.

 Resonance increases stability by increasing the bonding between adjacent atoms and by distributing charge over a greater number of atoms.

Drawing Resonance Forms

There are several things that should be checked before and after drawing the resonance forms. First know where the nonbonding electrons are, keep track of formal charges on atoms, and do not break sigma bonds. Finally, after drawing the resonance form make sure all the atoms have eight electrons in the outer shell. Checking these will make drawing resonance forms easier.

When drawing a resonance structure there are three rules that need to be followed for the structures to be correct:

1. Only electrons move and the nuclei of the atoms never move.
2. Only electrons that can move are pi electrons, single unpaired electrons, and lone pair electrons. 
3. The total number of electrons in the molecule do not change and neither do the number of paired and unpaired electrons. 

Approaches for moving electrons are move pi electrons toward a positive charge or toward an another pi bond. Move a single nonbonding electron towards a pi bond. Move lone pair electrons toward a pi bond and when electrons can be moved in more than one direction, move them to the more electronegative atom. 

Helpful hints

1. Two resonance structures differ in the position of multiple bonds and non bonding electron. The placement of atoms and single bonds always stays the same. 
2. They must make sense and agree to the rules. Hydrogens must have two electrons and elements in the second row cannot have more than 8 electrons. If so, the resonance structure is not valid. Always look at the placement of arrows to make sure they agree. 
3. Electrons move toward a sp2 hybridized atom. The sp2 hybridized atom is either a double-bonded carbon, or a carbon with a positive charge, or it is an unpaired electron. Electrons do not move toward a sp3 hybridized carbon because there is no room for the electrons. 

After drawing resonance structures check the net charge of all the structures. For example, if a structure has a net charge of +1 then all other structures must also have a net charge of +1. If not, the structure is not correct. Always check the net charge after each structure. These important details can ensure success in drawing any Resonance structure.

Rules for drawing contributing resonance structures.  Some rules must be considered when drawing resonance structures.

Rule 1: All resonance structures must have the same number of valence electrons.  Electrons are not created or destroyed, nor are they lost or gained to other molecules or ions during resonance.





The rule is violated because structure E has 12 valence electrons (four bonding pairs and two lone pairs), whereas structure F has 14 valence electrons (five bonding pairs and two lone pairs).  Therefore these cannot be resonance structures of the same ion.  (Structure F also violates rule 2.)

Rule 2:  The octet rule must be obeyed.  Hydrogen may never have more than two valence electrons.  Lithium through fluorine may never have more than eight valence electrons. (A carbon with five attachments, often called a "pentavalent carbon" has ten valence electrons.  This is forbidden because carbon does not have the space in its orbitals to accommodate ten electrons.  You should take care to avoid this common mistake made by inexperienced organic chemistry students.)  Certain elements commonly encountered in organic chemistry may have ten valance electrons.  These elements are in periods three and higher in the periodic table, and include chlorine, bromine, iodine, phosphorus and silicon.  Of all the atoms commonly encountered in organic chemistry, only sulfur can routinely expand its octet to include twelve valence electrons. These atoms expand their octets so as to improve the importance of the resonance structure.

Resonance structure G is acceptable.  Structure H is not acceptable because the carbon has ten valance electrons.


Structures I and J are both acceptable resonance contributors for bisulfate ion, the conjugate base of sulfuric acid.  The sulfur atom of structure J has 12 valence electrons, an expanded octet.  Sulfur is a third row element, so an expanded octet is allowed.  Structure J is the more important resonance structure because it maximizes the number of covalent bonds and minimizes the number atoms with a nonzero formal charge.

Rule 3: Nuclei do not change positions in space between resonance structures.  Resonance structures differ only in the arrangement of valence electrons.


Structures K and L are both acceptable Lewis structures, but they are not related by resonance because the circled hydrogen atom has changed position in space.

Preference 1: The most important contributor has the maximum number of atoms with full octets.
This preference gets priority over the other three rules for determining the most important resonance contributor.
The carbon of structure A has an open octet.  All the atoms of structure B have full octets.  Therefore contributor B is more important than contributor A, despite the fact that the positive charge is on the more electronegative oxygen atom instead of the less electronegative carbon atom.
Preference 2: If a resonance contributor must have formal charge, the most importnat contributor has these charge(s) on the atoms most willing to accommodate them.  Negative charges are best accommodated on more electronegative atoms, whereas positive charges are best accommodated on the least electronegative atoms.
All atoms of resonance contributors C and D have a complete octet, so we turn to other preferences to determine the most important resonance contributor.  A negative charge is best accommodated by a more electronegative atom.  Because oxygen is more electronegative than carbon, contributor D is more important than contributor C.  (If the ion shown above was a cation, then the resonance contributor with the positive charge on carbon would be more important than the contributor with the positive charge on oxygen.)

Preference 3: The most significant contributor has the maximum number of covalent bonds.  Contributor B (above) is more important than contributor A because B has the carbon-oxygen p bond absent in A.

Preference 4: The most significant contributor will have the least number of formal charges.
Resonance contributor F is more significant than contributor E because F has no atoms with formal charges, whereas E has two atoms with formal charges.  (Contributor F is also favored by Preference 3 as well.)

Preference 5: The most significant contributor has the least number of unpaired electrons.

For example, contributors G and H each have one unpaired electron, and thus are preferred over contributor I which has three unpaired electrons.  Resonance contributors that include avoidable unpaired electrons are rarely of any consequence and thus should not be considered.  There is one common exception: molecular oxygen.  Due to molecular orbital considerations, molecular oxygen is best described as having two unpaired electrons and an oxygen-oxygen single bond (contributor J) and not as lacking unpaired electrons with an oxygen-oxygen double bond bond (contributor K).

Resonance: Most Important Resonance Structure
examples
When determining the most important resonance structure, we consider full octets above other preferences.  In neither of these resonance structures do all atoms have full octets.  When looking to rank structures, we must look for differences.  In each structure, the positive charge resides on the exactly the same type of atom, each has the same number of covalent bonds, and each has the same number of formal charges.  Thus, there is no preference between these two structures.  Resonance structures which are deemed to be of equal importance (equal energy) are termed degenerate.  Structures A and Bcontribute equally to the resonance hybrid structure.

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Considering full octets first, we see that all atoms of structures C and D have full octets.  The have no formal charge, and the same number of covalent bonds.  These structures are of equal importance.

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Each of these structures has one carbon atom with an open octet (the carbon with the positive formal charge).  The charge always resides on a carbon, and there is an equal number of formal charges and covalent bonds for each structure.  At first glance, we would conclude that all five structures are of equal importance.  (We will learn in section 6.3 of the text that a carbon bearing a positive charge, called a carbocation, is most stable when the carbon bearing the charge has the greatest number of other carbons attached to it.  We will also learn in sections 9.1 and 9.2 of the special stability associated with having three alternating carbon-carbon double bonds in a six-membered ring, called aromaticity.  In this case, aromaticity is the most important factor, so degenerate structures E and F are the most important.)

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The carbon of resonance structure J has an open octet.  All atoms of structure K have complete octets.  Although structure K has formal charges, and the positive charge is on the more electronegative atom (oxygen vs. carbon), the full octet preference dominates.  Structure K is therefore the most important resonance structure for carbon monoxide.

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The carbon bearing the positive charge of structure M has an open octet.  All of the atoms of structures L and N have full octets, so structure M is less important.  Structure M also has more formal charges than either of the other two structures.  Structures L and N have the same number of formal charges and covalent bonds, so the difference is the position of the formal charge.  A negative charge is best accommodated by a more electronegative atom, so structure N is more important than structure L.

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The carbon bearing the positive charge of structure P has an open octet.  All of the atoms of structures O and Q have full octets, so structure P is less important.  Structures O and Q have the same number of covalent bonds.  Structure Q has two formal charges whereas structure O has no formal charges.  Thus structure O is more important than structures P or Q.

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All atoms in all of these resonance structures have a complete octet.  (Recall that while phosphorus can expand its octet to ten or twelve valence electrons, it still needs only eight electrons to fill its octet.)  The negative charges always reside on oxygen and the positive charges always reside on phosphorus, so we cannot rank importance of these structures using Preference 2.  Structure R has four covalent bonds, whereas structures S - V each have five covalent bonds (four single and one double), so structure R is less important than structures S - V.  Structures S - V all have three oxygens with formal charges.  Thus we conclude that structures S - V are degenerate, and are equally important.

mesomeric effect

Resonance & Mesomeric Effect:






There are many organic molecules which can not be represented by a single lewis structure. In turn, they are assigned more than one structure called canonical forms or contributing of resonating structures. The phenomenon exhibited by such compounds is called resonance. For example, 1, 3 – butadiene has following resonance structure.


The two essential conditions are


There must be conjugation in the molecule. Conjugation is defined as the presence of alternate double and single bonds in the compound like 


The part of the molecules having conjugation must be essentially planar or nearly planar. The first condition of conjugation is not only confined to the one mentioned above but some other systems are also categorized under conjugation. These are




Resonance (mesomeric) effect is of two types.

(i) If the atom or group of atoms is giving electrons through resonance, it is called +R or +M effect. For example,



















Other groups that shows +M effect are -NHR, -NR2, -OH, -OR, -NHCOR, -Cl, -Br,-I etc.










(ii) If the atom or group of atoms is withdrawing electrons through resonance, it is called ¾R or ¾M effect. For example,














(-M effect of -NO2 group)The -NO2 group in nitrobenzene has -M effect.

Other groups showing -M effect are -CN, -CHO, -COR, -CO2H, -CO2R, -CONH2, -SO3H, -COCl etc.

electrophillic aromatic substitution

The reactions of aromatic compounds are closely related to the stabilization effects given from the conjugated pi system. 


The non-equivalence of the carbon atoms in the benzene ring leads to the unique properties and different substitution effects. 

Each of these groups represent a slightly different chemical environment.

Substitution to an aromatic benzene ring


Benzene almost always reacts via a substitution mechanism. This is due to the energy favourability in maintaining the aromatic ring after the reaction. This means that there is usually a loss of a hydrogen atoms from the ring re-forming the aromatic ring.

             Electrophilic substitution

Electrophilic substitution is one of the most common ways that benzene undergoes aromatic substitution. This is when a positively charged particle is reacted with the electron dense benzene causing a positively charged benzene to form. These reactions then involve the loss of the hydrogen on the benzene to maintain aromaticity.

This means that a positive charge is introduced to the benzene enviornement which then undergoes a reaction. The position that this electrophile attacks is based on where the position of any substituents are in the reaction. The chemical nature of these substituents effect the electron density on the carbon positions of the benzene making some more electron rich than others. Electrophiles will react where there can be best stabalisation of the charge produced.

The diagram below doesn't need to be memorised although having a good idea of the boundaries is very useful.

A better understanding of why each of these substituents have the following properties by looking at the different resonance states for each of the substituents.

The stability of each can be shown after electrophilic substitution by the following diagrams:

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Reactions of benzene and its derivatives

There are a number of reaction of benzene and its derivatives that need to be taken into account. The final two mechanisms show how to achieve a reasonable level of selectivity through the reactions.

Friedel-Crafts addition

Friedel-Crafts addition is an effective way to add alkyl chains onto a benzene group. There is the problem of carbocation rearrangement meaning the addition of secondary or tertiary alkyl groups becomes impossible. 

Friedel-Crafts acylation

This reaction has the advantage of not allowing carbocation rearrangement to occur. This means that the use of this reaction is usually favoured.

These are two very effective methods although past simple alkyl groups they do not allow the addition of anything more complex. This is seen very strongly in the attempted production of the following compound. This is because the acyl chloride derivative of formaldehyde does not exist.  

This means a different reaction mechanism needs to be used. A good example of this being the use of the Vilsmever – Haack reaction. This is where an amide group can be reacted with a halogenated phosphoxide group. This forms a strong electrophile which can then react with the benzene ring. This forms an imine which can then be hydrolysed forming an aldehyde group.

The Mannick reaction also utilises this imine intermediate. 

This shows how there is great number of uses for a nitrogen ion. These have the possibility to react easily as either an electrophile or a nucleophile. This places great importance on the formation of the electrophilic amine group.

The Sand Meyer reactions are also important this involves the addition of a nitrogen molecule onto a benzene ring and then the addition of a number of different reactants to form a number of products. The cation for this reaction can be produced by the following method.

These diazonium ions can be easily removed. This produces the reactions known as the SandMeyer reactions.

Protecting groups

It is important in a lot of these reactions that the electrophile attacks the correct location on the benzene ring. Ortho protection can be carried out by reacting an acyl chloride with the amine group. This stops ortho reactions by kinetic effects.

This shows how a protecting group can be used to allow the formation of a para positioned electrophile on the benzene ring.

It is also worth remembering that for meta substitution there needs to be a powerful electron withdrawing group. This means that the initial nitro substituted benzene can be present for the nitro addition and then the addition and then reduction forming the amine.

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Heteroaromatic reactions

Pyridine

Pyridine is very similar to benzene and is likewise an aromatic compound. The nitrogen lone pair is not involved in the aromatic π system. 

So this leaves the need to understand whether pryridine acts as an ortho/para director or a meta director. This is found by looking at the resonance structures and remembering that the nitrogen is unstable when positively charged.  

There are other heteroaromatic systems such as pyrrole, furan and thiophene. These tend to react at the alpha carbon.

It is important to note that pyrrole in an acidic mixture reacts to form a polymeric tar.