Benzene and aromatics

Question 1

Structure of benzene?

Answer 1

six sp2 hybrisidsed carbon atoms link together to form a planar ring, all six C bond angles equal to 120 degrees, all same length bond. Each sp2 carbon has an unhybridised p orbital at right angles to plane of the ring containing a single electron. P orbitals are parallel and close enough to overlap side on, pi electrons. Overallping produces two continuous rings of pi electrons above and below the plane of the benzene ring, electrons are delocalised.

Question 2

Why does benzene not have alternating double and single bonds?

Answer 2

In the Kekule structure the three C=C bond would be shorter than the thing C-C bonds but all sic bonds in benzene are the same length
In the Kekule structure the three C=C bond would be expected to react with electrophiles in addition reactions however benzene undergoes electrophilic substitution reactions
Beneze is more stable than the Kekule structure suggests, because p electrons are delocalised around the ring rather than being localised in three pi bonds

Question 3

How can stability of benzene be measured?

Answer 3

By measuring enthalpy changes that occur on hydrogenation of cyclohexene, cylochexa-1,3-diene and benzene

Question 4

Hydrogenation of cyclohexane?

Answer 4

Hydrogenation of cyclohexene is exothermic because two C-H bonds are formed in cyclohexane and these bonds are stronger than the C=C and H-H bonds that are broken, more energy realised when bonds are formed than when broken.

Question 5

Hydrogenation of cylochexa-1,3-diene?

Answer 5

Hydrogenation of cylochexa-1,3-diene is about double this

Question 6

Hydrogenation of Kekule structure?

Answer 6

Applying this to the Kekule structure you would expect the hydrogenation to be -360 KJmol-1 but it is actually a lot less than this, benzene is actually more stable than the Kekule structure, the difference between the two enthalpy changes is 152KJmol-1, this is called the resonance energy or the delocalisation energy of benzene

Question 7

Stability of benzene?

Answer 7

Stability is due to the delocalisation of the electrons in the ring and this is an example of resonance, the two Kekule structures are the resonance forms of benzene

Question 8

When is a molecules aromatic?

Answer 8

Must be:
Cyclic
Planar (allow overlap of p orbitals)
Contained an uninterrupted ring of pi electrons
Number of pi electrons must be equal to 4n+2 (Huckels rule)

Question 9

Annulenes?

Answer 9

Conjugated cyclic hydrocarbons, benzene [6]annulene and [14]annulene are both aromatic they are both planar and they have 6 and 14 pi electrons. An expectational case in [10] annulene which should be aromatic since it has 10 pi electrons but is unstable because of a combination of steric and angle strain

Question 10

Heteroatoms?

Answer 10

Heteroatoms can be part of the ring in some compounds a lone pair of electrons on the heteroatom is part of the ring of pi electrons, eg in pyrrole lone pairs electron on nitrogen is in a p orbital that is part of the pi ring system, contrast to pyridine where the one pair on the nitrogen is not part of the pi ring system

Question 11

Charges?

Answer 11

Aromatic compounds can be neutral or charged, eg in C5H5- two electrons on the negatively charged carbon are in a p orbital that is part of the pi ring system

Question 12

Antiaromatic compounds?

Answer 12

Compounds containing 4n pi electrons are antiarotmatic,

Question 13

Non aromatic compounds?

Answer 13

Cyclic molecules that are nonplanar and have an interrupted or uninterrupted ring of pi electrons are nonaromatic compounds, similar stability to related compounds that are not cyclic

Question 14

MO theory for benzene?

Answer 14

Six p orbitals combine to form six MOs, the three lower energy MOs which have lower energies than the six p AOs are the bonding MOs, notice that two of the bonding MOs have the same energy. The three higher energy MOs are anti bonding orbitals

Question 15

Electrons in MOs?

Answer 15

Each MO can accommodate two electrons if the spins of the electrons are opposed so adding the six electrons to the MOs starting with the lower energy molecules orbital results in all three bonding MOs being filled. The three higher antibonding MOs are empty. Because all three bonding MOs are filled benzene is said to have a closed bonding shell and it is this that makes benzene so stable

Question 16

Frost circle?

Answer 16

Simple way to find reactive differences in energy between the MOs

Question 17

How does benzene react?

Answer 17

Electron rich benzene ring reacts with strong electrophiles in substitution rather than addition, in an electrophilic substitution reaction an atom or group of atoms replaces a hydrogen atom on the benzene ring so that the product retains the stab aromatic ring.

Question 18

Why does benzene not react in addition reactions?

Answer 18

The aromatic ring would be destroyed and the products would be much less stable than benzene

Question 19

What happens in substitution reactions?

Answer 19

In step 1 which is the slower step of the reaction two electrons from the aromatic ring of benzene are attracted towards the electrophile, the aromaticity is broken as a new C-E bond forms and this produces a none aromatic carbocation, although the carbocation is not aromatic it is stabilised by resonance, the three resonance forms contribute to the resonance hybrid and show the delocalisation of the electrons in the carbocation, in step 2 the carbocation loses a proton usually by reacting with a base and the two electrons in the C-H bond move into the ring to restore aromaticity. The non aromatic carbocation always loses the H+ from the carbon atom that forms a bonds to the electrophile, overall the electrophilic substitution of benzene produces a substituted benzene

Question 20

Why is the first step of the electrophilic substitution slow?

Answer 20

Because the Gibbs energy of activation for formation of the intermediate carbocation is much greater than the Gibbs energy of activation for deprotonation of the carbocation

Question 21

Why does the intermediate carbocation have a much higher Gibbs energy than the reactants or the products?

Answer 21

Because the carbocation is non aromatic whereas benzene and the substituted benzene are aromatic

Question 22

Gibbs energy change overall for reaction?

Answer 22

Negative as energy of products is less than that of the reactants so the products will be favoured over reactants at equilibrium

Question 23

Step 1 of halogenation of benzene?

Answer 23

A brome atom in Br2 donates a pair of electrons to the Lewis acid to form a coordinated complex. The halogen atom that donated the electrons becomes positively charged and the Br-Br bond becomes polarised the pair of electrons in the Br-Br bond is attracted to the positive charge and this produces a partial positive charge on the single bonded bromine atom. By forming a complex with the Lewis acid, the Br2 is converted into a stronger electrophile, without the lewis acid Br2 does not react with benzene

Question 24

Step 2 of halogenation of benzene?

Answer 24

Benzene acts as a nucleophile and attacks the partially positive bromine atom in the coordination complex, this produces a non aromatic carbocation together with [FeBr4]-

Question 25

Step 3 of halogenation of benzene?

Answer 25

[FeBr4]- removes proton from the nonaromatic carbocation the aromatic ring is reformed to give bromobenzene together with HBr and FeBr3 since the FeBr3 is regenerated it acts as a catalysts for the bromination of benzene

Question 26

What reactants are used for bromination of benzene?

Answer 26

FeBr3

Question 27

What reactants are used for chlorination of benzene?

Answer 27

Cl2/FeCl3 or Cl3/AlCl3

Question 28

Why problems with fluorination of benzene?

Answer 28

Fluorine reacts extremely rapidly with benzene and it is difficult to control the number of fluorine atoms that’s are introduced on the benzene ring, a number of fluorination products are formed so the reaction is not used in synthesis

Question 29

Why problems with iodination of benzene?

Answer 29

Does not react with benzene, requires an oxidising agent

Question 30

Reactants needed for nitration of benzene?

Answer 30

Concentrated nitric acid HNO3 and concentrated sulphuric acid H2SO4

Question 31

Step 1 of nitration of benzene?

Answer 31

Sulphuric acids donates a proton to nitric acid in an acid base reaction

Question 32

Step 2 of nitration of benzene?

Answer 32

The protonated OH group of nitric acid provides a good leaving group, loss of H2O produces a +NO2 ion which is a strong electrophile

Question 33

Step 3 of nitration of benzene?

Answer 33

Benzene donates a pair of electrons to the +NO2 ion to form a nonaromatic carbocation intermediate two electrons move form the benzene ring to form a new C-N bond and at the same time a pair of pi electrons in one of the N=O bonds in the nitronium ion moves onto one of the O atoms

Question 34

Step 4 of nitration of benzene?

Answer 34

Water deprotonates the carbocation to form nitrobenzene together with H3O+

Question 35

Reactants needed for suffocation of benzene?

Answer 35

Fuming sulphuric acid (oleum), mixture of H2SO4 and SO3

Question 36

Step 1 of sulfonation of benzene?

Answer 36

Sulphuric acid rects with sulphur trioxide in an acid base reaction the protonated sulfur trioxide molecules is a strong electrophile because the electrons in the S=+OH bond are attracted to the positive charge and this produces a partial positive charge on the sulfur atom

Question 37

Step 2 of sulfonation of benzene?

Answer 37

The HSO3+ ion react with benzene to form a nonaromatic carbocation two electrons move from the benzene ring to form a new C-S bond and at the same time a pair of electrons in the S=+OH bond moves to the positively charged oxygen

Question 38

Step 3 of sulfonation of benzene?

Answer 38

The -OSO3H ion deprotonates the carbocation to form benzenesulfonic acid and regenerate sulphuric acid

Question 39

Reverse of sulfonation?

Answer 39

All steps are reversible in the sulfonation of benzene unlike other electrophilic aromatic substitution reactions, when benzenesulfonic acid is heated with dilute sulphuric acid the reverse reaction occurs and benzenesulfonic acid is converted into benzene in a desulfonation reaction

Question 40

Mechanism of desulfonisation of benzene?

Answer 40

Starts by protonation of the aromatic ring of benzenesulfonic acid to give a nonaromatic carbocation in step 1, in step 2 the carbocation loses the HSO3+ ion to form benzene. The HSO3+ ion could then react with benzene to reform benzenesulfonic acid but in dilute sulphuric acid it is more likely to react with water in step 3, to give sulphuric acid in step 4 this is because dilute sulphuric acid contains a relatively high concentration of water and is a stronger nucleophile than benzene

Question 41

Alkylation of benzene?

Answer 41

Reactions of benzene with chloro or bromo alkanes and a Lewis acid catalysts such as AlCl3, FeCl3 or FeBr3 produces alkylbenzenes in Friedel Crafts alkylation reactions.

Question 42

Step 1 of alkylation of benzene with a primary chloroalkane?

Answer 42

In step 1 the primary chloroalkane reacts with AlCl3 to form a coordination complex, by forming a coordination complex with the Lewis acid the primary chloroalkane is converted into a stronger electrophile because the electrons in the C-Cl bond of the coordination complex are strongly attracted to the positively charged chlorine

Question 43

Step 2 of alkylation of benzene with a primary chloroalkane?

Answer 43

Benzene attacks the partially positive carbon atom in the complex to produce a nonaromatic carbocation together with [AlCl4]-

Question 44

Step 3 of alkylation of benzene with a primary chloroalkane?

Answer 44

[AlCl4]- removes a proton from the nonaromatic carbocation to give an alkylbeonznene together with HCl and AlCl3 since AlCl3 is regenerated it acts as a catalyst for alkylation of benzene

Question 45

Step 1 of alkylation of benzene with a secondary or tertiary chloroalkane?

Answer 45

The coordination complex formed by region of a secondary or tertiary chloroalkane with AlCl3 in step one usually break down to form a carbocation and [AlCl4]- in step 2. The secondary or tertiary carbocation is a strong electrophile and in step 3 two electrons move from the benzene ring to the carbocation, the mechanism is then the same as for the reaction with a primary chloroalkane

Question 46

Why are tertiary or secondary carbocations more stable?

Answer 46

he rate of formation of the carbocation from the coordination complex depends on the stability of the carbocation. As tertiary carbocation are more stable than seocdar and particularly primary carbocation tertiary carbocation are formed more raptly from coordination complexes

Question 47

Rearrangements in Fridel Crafts alkylations and polyalkylation?

Answer 47

Generally gives more than one product particularly when using chloro or bromo alkanes and this restricts the use of the reaction is synthesis

Question 48

What is one reason why Friedel Crafts give so many products?

Answer 48

The intermediate coordination complex rearranges. Primary carbocation can undergo a 1,2-hydride shift to from a secondary carbocation, it is this secondary carbocation that reacts with benzene. The driving force for the 1,2-hydride shift is the stability of the secondary carbocation. Because the coordination complex has a partial positive charge on a primary carbon atom the coordination complex resembles a primary carbocation and the coordination complex undergoes a 1,2-hydride shift to produce a more stable secondary carbocation, conversation of the coronation to the secondary carbocation is described as a rearrangement and the substituted benzene formed from the second carbocation is called the rearrangement product

Question 49

What is second reason why Friedel Crafts give so many products?

Answer 49

Polyalylation takes place, difficult to introduce just one alkyl substituent on to a benzene ring because replacing a H atom by an electron donating alkyl substituent makes the ring more nucleophilic and therefore more reactive to further electrophilic substitution There comes a point when the size of the alkyl substituents on the benzene ring hinders the approach of the coordination complex so it is extremely difficult to substitute all six hydrogen atoms on benzene for butyl groups, the maximum number of alkyl groups that are introduced will depend on the size of the groups and the reaction conditions

Question 50

Acylation of benzene?

Answer 50

Reaction of benzene with acyl chlorides and a Lewis acid typically FeCl3 and AlCl3 produces acylbenzenes in Friedel Crafts acylation reactions

Question 51

Step 1 and 2 of the acylation of benzene?

Answer 51

The acyl chloride reacts with AlCl3 to form a coordination complex which breaks down in step 2 to form an acylium ion and [AlCl4]-

Question 52

Step 3 of the acylation of benzene?

Answer 52

The positively charged acylium ion is stabilised by resonance it is a stronger electrophile than the acyl chloride so the asylum ion reacts with benzene in step 3 to produce a nonaromatic carbocation

Question 53

Step 4 of the acylation of benzene?

Answer 53

In step 4 [AlCl4]- removes a proton from the nonaromatic carbocation to give an acylbenzene together with HCl and AlCl3

Question 54

Step 5 of the acylation of benzene?

Answer 54

The regenerated AlCl3 forms a coordination complex with the acylbenzene

Question 55

Step 6 of the acylation of benzene?

Answer 55

When the reaction is completed water is added to the reaction mixture in step 6 AlCl3 is hydrolysed to Al(oh)3 and the acylbenzene is realised from the complex

Question 56

How to covert benzene into benzaldehyde?

Answer 56

Requires reaction with the acylium ion, however it cannot be bored from HCOCl and AlCl2 because HCOCl is not readily viable as it is highly unstable. To overcome this problem a mixture of benzene CO, HCl and AlCl3 is reacted under high pressure in the Gattermann Koch reactions. At high pressure these reactants produce the acylium ion which then reacts with benzene to form benzaldehyde

Question 57

How many products does acylation form?

Answer 57

Only one acylbenzene product, this is because acylium ion do not rearrange, also an acylbenzene is less reactive to electrophilic substitution than benzene because the acyl group is electron withdrawing so the only product from acylation of benzene is the unrearranged monoacylated benzene - therefore it is more useful synthetically than alkylation

Question 58

Clemmensen reduction?

Answer 58

Acylbenzene can be reacted with a reducing agent that converted the C=O bond into a CH2 group, a common reducing agent for this transformation is an acidic solution of zinc dissolved in mercury