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CH10135 > Aromatic chemistry > Flashcards

Aromatic chemistry Flashcards

1
Q

Why was the Kekule structure of benzene criticised?

A
  1. It couldn’t explain why 1,2-disubstituted benzenes had no isomers.
  2. It couldn’t explain why the heat of hydrogenation of benzene was lower than expected.
  3. It couldn’t explain why benzene was unreactive under typical addition reaction conditions.
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2
Q

How many pi electrons does benzene have and where are they in the structure?

A

6 pi electrons.

They are free to travel across all 6 carbon atoms = delocalised over the whole conjugated system.

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3
Q

How long are the carbon-carbon bonds in benzene?

A

140 pm - almost halfway between typical single and double carbon-carbon bond lengths, therefore no di-substituted structures can exist.

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4
Q

What is the typical heat of hydrogenation of a carbon-carbon bond?

A

-120 kj/mol

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5
Q

Why does benzene have a lower heat of hydrogenation than expected?

A

The stabilisation effect.

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6
Q

Why does the stabilisation effect prevent typical addition reactions from occurring?

A

For compounds to react with each other, activation energy must be overcome.
Typical alkenes undergo bromination as ΔG is relatively small, but benzene has a prohibitively large ΔG value due to the aromatic stabilisation energy which must be overcome.
When forced, a substituted product is formed (not expected product) to maintain the stabilisation obtained by aromaticity.

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7
Q

What are Huckel’s rules?

A

requirements for aromaticity:
1. molecule must be cyclic.
2. molecule must be planar (all atoms in cycle are sp2 hybridised).
3. molecule must be fully conjugated.
molecule must contain 4n+2 pi-electrons (2,6,10,14,etc)

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8
Q

How many MOs does benzene have?

A

the 6 carbon p-orbitals combine to give 6 pi MOs

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9
Q

What can you use a Frost Circle Mnemonic to tell you?

A

can be used to identify whether a cyclic, planar, fully-conjugated molecule is likely to be aromatic or antiaromatic.

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10
Q

Rules for using a Frost Circle Mnemonic?

A
  1. draw a circle, place a horizontal line through the centre (MOs on the line will be non-bonding, MOs above the line will be anti-bonding, MOs below the line will be bonding).
  2. draw the polygon with the same number of sides as the molecule in question in the circle, apex pointing down.
  3. where the polygon touches the circle, draw horizontal lines to represent the relative energy of a MO.
  4. Fill the MOs obeying Hund’s rule and the Aufbau principle.
  5. If all electrons are paired = aromatic molecule.
    If there are unpaired electrons in the non-bonding or antibonding MOs = antiaromatic molecule.
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11
Q

What is each position on a benzene ring called?

A 
C1 = ipso (para)
C2 = ortho
C3 = meta
C4 = para
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12
Q

How to name polysubstituted aromatics?

A
  1. Find highest priority root name. Carbon bonded next to the highest priority f.g. is the number 1 position.
  2. Number the carbons either clockwise or anticlockwise depending on which way provides the lowest number-labelled substituent.
  3. Substituents are arranged in alphabetical order.
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13
Q

What are the most common heteroatoms found in aromatic heterocycles?

A

N
O
S

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14
Q

What is a polycyclic aromatic hydrocarbon (PAH)?

A

benzene can fuse to itself to give an array of different 2D and 3D structures, known as polycyclic aromatic hydrocarbons.
e.g. fullerenes

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15
Q

What is the simplest polycyclic aromatic hydrocarbon?

A

Naphthalene

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16
Q

What is a pi-sextet?

A

a pi-sextet is classed as 3 double bonds delocalised within a 6-membered ring within a PAH resonance structure.

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17
Q

What is Clar’s theory?

A

the resonance structure with the largest number of disjoint aromatic pi-sextets is the most important for a PAH’s characterisation.

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18
Q

What is the relationship between the number of pi-sextets in PAHs and kinetic stability?

A

PAHs with more pi-sextets in resonance structures are kinetically more stable than those with less.

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19
Q

What is the relationship between the number of pi-sextets in rings and aromatic character?

A

rings which have the most pi-sextets in resonance structures are more aromatic in character than other rings and are therefore more stable.

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20
Q

What does it mean if a compound has a resonance structure that is described as “fully benzenoid”?

A

when a compound has a resonance structure where all alternate rings contain pi-sextets with no isolated or fully conjugated double bonds in other rings.

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21
Q

What is an acene?

A 
a class of polycyclic aromatic hydrocarbon (PAH) that is made up of linearly fused benzene rings.
only one pi-sextet can exist in any resonance structure which gets spread more thinly across more rings going up the series.
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22
Q

In acenes, what is the relationship between the size of the acene and its stability?

A

the larger the acene, the more unstable it is since the stable pi-sextet has to be shared among more rings.

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23
Q

What are phenacenes? Discuss their arrangement.

A

isomers of acenes and are also formed of fused benzene rings but is arranged in a zig-zag pattern rather than a line.
This arrangement of atoms, leads to there being more than one pi-sextet which can now occur - this greatly increases the stability of phenacenes over acenes.

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24
Q

How can we “force” benzene to react with bromine?

A

in a electrophilic aromatic substitution reaction, in the presence of a Lewis acid (e.g. FeBr3).
It gives a substituted product rather than an addition product.

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25
Q

Describe the SEAR mechanism:

A
  1. Aromatic ring breaks aromaticity, and attacks a very reactive electrophile which is often cationic and usually needs to be generated in situ.
  2. the intermediate cation is known as “the Wheland intermediate” and is relatively stable due to resonance.
  3. the proton at the site of electrophilic attack is lost to restore aromaticity.
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26
Q

What can you do to lower the activation energy of the reaction?

A
  1. increase the energy of the HOMO of the nucleophile, i.e. make the nucleophile more nucleophilic.
  2. decrease the energy of the LUMO of the electrophile, i.e. make the electrophile more electrophilic.
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27
Q

How can we achieve bromination?

A

achieved using bromine and catalytic iron (III) bromide (FeBr3).

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28
Q

Describe the mechanism for the bromination of benzene:

A
  1. a bromine atom first donates a lone pair of electrons to the Lewis acidic iron centre to give a zwitterion.
  2. this is now a highly electrophilic intermediate and is reactive enough to react with benzene, which does so at the neutral bromine atom to release FeBr4^-.
  3. The Wheland intermediate then rapidly loses a proton to restore aromaticity and give the desired halogenated product.
  4. By-products, FeBr4^- and H^+, then combine to restore the FeBr3 catalyst and release HBr gas.
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29
Q

How can we achieve chlorination?

A

achieved using chlorine and catalytic aluminium trichloride (AlCl3).

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30
Q

When does iodination occur?

A

only in the presence of very electron-rich aromatics.

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31
Q

Is fluorination accessible by SEAR?

A

no - fluorine is too reactive and difficult to handle.

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32
Q

How can we achieve nitration? What is required?

A

achieved via the SEAR mechanism, it requires the generation of a powerful nitronium ion (NO2^+) electrophile which can be dine using conc nitric acid in the presence of conc sulfuric acid catalyst.

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33
Q

What happens when you reduce a nitro group?

A

Nitro groups can be easily reduced to an amino group to form anilines.

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34
Q

What are the 2 ways you can reduce a nitro group?

A
  1. the Bechamp reduction (tin in dilute HCl)

2. hydrogenation over a palladium on carbon catalyst ( cat. Pd/C, H2)

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35
Q

What do we use in sulfonation reactions?

A

concentrated sulfuric acid is capable of sulfonating benzene itself.
Disproportionation and dehydration of sulfuric acid gives protonated sulphur trioxide (HSO3^+) which can be attacked by benzene.

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36
Q

Is sulfonation reversible? Where does the equilibrium lie?

A

yes - the side that the equilibrium favours depends on the concentration of the sulfuric acid used.

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37
Q

reagents in chlorination reactions:

A

Cl2 and AlCl3

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38
Q

reagents in bromination reactions:

A

Br2 and FeBr3

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39
Q

reagents in nitration reactions:

A

conc HNO3 and conc H2SO4

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40
Q

reagents in sulfonation reactions:

A

conc H2SO4

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41
Q

What do you have to react together to produce an alkylated aromatic? Name of this reaction?

A

treating benzene with alkyl halides, in the presence of a Lewis acid catalyst, e.g. AlCl3 and tert-butyl chloride together.
Reaction: Friedel-Crafts Alkylation

42
Q

What happens when the AlCl3 and tert-butyl chloride react together as the catalyst in Friedel-Crafts alkylation reaction?

A

a chlorine lone pair donates electron density to the empty aluminium p-orbital. Loss of AlCl4^- gives a tertiary carbocation intermediate which is then attacked by benzene.

43
Q

is a primary alkyl halide more reactive than a tertiary alkyl halide?

A

the reactivity series of the alkyl halides goes from tertiary to secondary to primary.

44
Q

What are the 2 major limitations of Friedel-Crafts alkylation?

A
  1. Where possible, the intermediate carbocation will rearrange to a more stable isomer.
  2. the product is more reactive than the starting material which can lead to over-alkylation.
45
Q

What is the difference between Friedel-Crafts alkylation and acylation?

A

Friedel-Crafts acylation uses a acid chloride as the electrophile instead of an alkyl halide like Friedel-Crafts alkylation.

46
Q

What occurs in Friedel-Crafts acylation?

A

Aluminium trichloride (AlCl3) is used as a halophilic Lewis acid to help remove chloride to leave behind a cation: called a acylium ion. This highly reactive electrophile reacts with benzene to give an acylated product.

47
Q

What is Wolff-Kishner reduction?

A 
the resulting ketone from the Friedel-Crafts acylation can be reduced using a Wolff-Kishner reduction.
Uses hydrazine (H2N-NH2), KOH, and EtOH under basic conditions
48
Q

What is Clemmensen reduction?

A

the resulting ketone from the Friedel-Crafts acylation can be reduced using a Clemmensen reduction.
Under acidic conditions, the reduction uses Zn(Hg), and HCl.

49
Q

What is Mozingo reduction?

A

the resulting ketone from the Friedel-Crafts acylation can be reduced using a Mozingo reduction.
Under neutral conditions, it uses HS-CH2-CH2-CH2-S, raney Ni and H2

50
Q

What does the nature of the substituent have on?

A

the rate of reaction and the regioselectivity (where substitution takes place in relation to the substituent)

51
Q

What is the rate of reaction determined by in aromatic chemistry?

A

the nucleophilicity of the aromatic.

52
Q

How does electron-withdrawing groups affect the rate of reaction? Type of effects?

A

EWGs reduce the rate of reaction either through:

  1. inductive effects (i.e. electronegativity) (-CF3)
  2. mesomeric effects (i.e. conjugation) (-NO2)
53
Q

How does electron-donating groups affect the rate of reaction? Type of effects?

A

EDGs increase the rate of reaction either through:

  1. hyperconjugation (-CH3)
  2. conjugation (-OMe)
54
Q

What is the Hammett constant used to determine?

A

how electron-donating or electron-withdrawing a certain substituent is.

55
Q

What does the Hammett constant show a correlation between?

A

Hammett found a linear correlation between the pKa of various benzoic acids and the rate of hydrolysis of their corresponding esters.

56
Q

What is the regioselectivity of the reaction determined by?

A

determined by effective stabilisation of the Wheland intermediate.

57
Q

Where do EWGs and EDGs substitute on the aromatic ring? Any exceptions?

A

EWGs direct substitute to the meta position.
EDGs direct substitute to the ortho and para positions.
Halogens are the exception as they are electronegative, they direct ortho and para.

58
Q

For EWGs why is the meta substitution favoured?

A

meta substitution is the lower energy pathway as this avoids a high energy resonance structure, where the positive charge is found on the carbon adjacent to the EWG.

59
Q

For halogens, why are ortho and para the major regioisomers?

A

as orbital overlap of a lone pair on the halogen with the Wheland intermediate can help stabilise it in a similar manner as an electron donating group.

60
Q

Why does selectivity for para substitution over the ortho substitution increase going up the group of halogens?

A

due to the high electronegativity of the fluorine atom reducing electron density at the ortho position.

61
Q

What do activating groups do to a reaction?

A

increase the rate of reaction.

62
Q

What do deactivating groups do to a reaction?

A

decrease the rate of reaction.

63
Q

Common activating groups?

A
  • hydroxy
  • amine
  • amide
  • alkoxide
  • ester
64
Q

Common deactivating groups?

A
  • cyano
  • nitro
  • haloalkyl
  • ammonium
  • carbonyl
65
Q

What do blocking groups do?

A

To improve the ortho/para selectivity of electron-rich aromatics, blocking groups can be used. The reversible nature of the sulfonation reaction makes it perfect for sulfonic acids to be used as blocking groups. Concentrated sulfuric acid can be used to install the blocking group(s), and after the electrophilic aromatic substitution with the desired electrophile, dilute sulfuric acid can be used to remove them

66
Q

What can we use as the most common blocking group and why?

A

sulfonic acids - as the sulfonation reaction has a reversible nature.
Concentrated sulfuric acid can be used to install the blocking group(s)

67
Q

What do we use to remove the blocking group from the aromatic ring?

A

dilute sulfuric acid

68
Q

What are functional group transformations?

A

when they improve the ortho/para selectivity of electron-rich aromatics.
we can install a group that can be reversibly added to an aromatic ring. This blocks the para position ( as para is first preference), which means that any subsequent reaction must go onto the ortho position. Certain functional groups attached to the aromatic ring can be easily converted into other functional groups with contrasting directing effects

69
Q

When do cooperative effects occur?

A

when multiple substituents all direct substitute to the same position.

70
Q

When do competitive effects occur?

A

when multiple substituents direct substitute to different positions, when this occurs electronic and steric factors should be considered to determine where substitution is likely to take place.

71
Q

What takes precedence over the other: electronic factors or steric hinderance?

A

electronic factors typically override steric hinderance.

72
Q

What takes precedence over the other: activating or deactivating groups?

A

activating groups take precedent over deactivating groups

73
Q

What happens when competitive substituents are electronically similar?

A

substitution occurs at the least hindered position.

74
Q

When does iodination occur?

A

in the presence of very electron-rich aromatics, such as phenol and anilines

75
Q

What happens in a Gatterman reaction?

A

either gaseous HCN or ZCN with HCl gas combine to give an imine.
in the presence if a Lewis acid catalyst, the imine is activated to an iminium cation which is electrophilic enough to be attacked by an electron-rich aromatic.

76
Q

What happens in a Vilsmeier-Haack reaction?

A

dimethylformamide (DMF) and phosphorus oxychloride (POCl3) combine to give an iminium directly, which is ten attacked by the aromatic.

77
Q

What is azo-couping?

A

the reaction of electron-rich aromatics with aryldiazonium salts.

78
Q

What are the products from azo-coupling called?

A

called azo dyes.

79
Q

What happens in a Sandmeyer reaction?

A

in the presence of various copper (I) salts, aryl diazonium salts undergo a displacement reaction where the nitrogen cation group is displaced with the counterion of the copper salt.

80
Q

How are diazonium salts made?

A

made from the diazotisation of anilines using sodium nitrite and aqueous acid at low temperatures.

81
Q

What is a Birch reduction?

A

the reduction of benzene and its derivatives using dissolved metal in liquid ammonia conditions to produce an unconjugated diene.

82
Q

What occurs in a Birch reduction?

A

When sodium or lithium is added to ammonia, the metal becomes oxidised and electrons are released into the solution.
These solvated electrons then add into the 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.
The dianion is then protonated by an alcohol present in solution to get the unconjugated diene.

83
Q

In a Birch reduction, when a EWG is present what positions are reduced?

A

the ipso and para carbons are reduced.

84
Q

In a Birch reduction, when a EDG is present what positions are reduced?

A

one ortho and one meta carbon is reduced.

85
Q

What are the 2 types of reaction manifolds that a nucleophilic aromatic substitution undergo? what intermediates do they go through?

A
  • elimination - addition (goes through a benzyne intermediate)
  • addition - elimination (goes through a Meisenheimer intermediate)
86
Q

What conditions and reagents does the elimination - addition (benzyne intermediate) use?

A

under harsh, basic conditions, aromatics with a good leaving group such as a halogen, undergo nucleophilic substitution.
strong bases such as sodium amide (NaNH2) or potassium tert-butoxide are required - and in this instance the base also acts as a nucleophile.

87
Q

What is the 2 step mechanism for the elimination - addition (benzyne intermediate)?

A
  1. the strong base deprotonates the proton ortho to the leaving group to give an aryl anion.
  2. Syn-elimination occurs to give an aromatic intermediate with a triple bond = benzyne.
  3. Addition of the nucleophile to the electrophilic triple bond and subsequent deprotonation gives the substituted product.
88
Q

Why is the triple bond formed in the elimination - addition reaction not a true alkyne?

A

because all the carbon atoms within the bond are not sp hybridised and the bond is not linear.
The carbon atoms are still sp2 hybridised and a pi-bond is still incorporated into the aromatic ring.

89
Q

Why can benzyne be attacked at either end of the triple bond?

A

benzyne is symmetrical

90
Q

What does the fact that benzyne can be attacked at either end of the triple bond mean for mono-substituted halobenzene’s and disubstituted halobenzene’s?

A

because benzyne is symmetrical, mono-substituted halobenzenes is inconsequential but for disubstituted halobenzenes, product mixtures and issues of regioselectivity can occur.

91
Q

How many intermediates can be formed with para-disubstituted halobenzenes? Any effects caused?

A

1 benzyne intermediate can be formed.
As the triple bond is too far away from the substituent there is minimal electronic and steric factors that can affect the regioselectivity.
A mixture of meta and para products is formed.

92
Q

How many intermediates can be formed with meta-disubstituted halobenzenes? Any effects caused?

A

2 benzyne intermediates can be formed.
If the substituent is electronegative (or is stabilising through inductive effects) the triple bond will be formed closest to the substituent.
If the substituent is electropositive (or is electron-donating though inductive effects) then the triple bond is formed away from the substituent preferentially.

93
Q

How many intermediates can be formed with ortho-disubstituted halobenzenes? Any effects caused?

A

1 benzyne intermediate can be formed.
With respect to the regioselectivity, electronic effects are more important than steric effects.
So, if the substituent is electronegative, substitution will be favoured at the meta position, as the resulting negative charge at the ortho position will be stabilised.
If the substituent is electropositive, substitution will be favoured at the ortho position as the negative charge at the meta position will be less destabilised.

94
Q

What are disadvantages of the elimination - addition (benzyne intermediate) reaction?

A
  • this method uses harsh basic conditions.
  • there is minimal control of regioselectivity for meta-disubstituted halobenzenes and the nucleophile also being the base.
95
Q

What are the other 2 common methods of the elimination - addition (benzyne intermediate) reaction?

A
  • a 2-diazonium benzoate zwitterion which composes to benzyne upon heating.
  • a 2-(trimethylsilyl)phenyl triflate which collapses to benzyne upon addition of a fluoride source (e.g. tetrabutylammonium fluoride)
96
Q

What does the addition-elimination (Meisenheimer intermediate) have in addition to the halogen?

A

EWGs which are ortho or para to the halogen.

97
Q

What happens in the addition-elimination (Meisenheimer intermediate) reaction?

A
  1. the nucleophile attacks the carbon which is directly bonded to the halogen to give a negatively charged intermediate (Meisenheimer intermediate).
  2. elimination of the halogen then occurs with restoration of the aromaticity.
98
Q

How can EWGs stabilise negative charge?

A

through conjugation.

99
Q

In addition-elimination (Meisenheimer intermediate) reactions what is the order of reactivity of the halogens?

A

F - most activating
Cl ≈ Br
I

100
Q

What are the 2 reasons why the halogen used in the addition-elimination (Meisenheimer intermediate) reaction has an effect on the rate?

A
  1. Electronegativity - the Meisenheimer intermediate is negatively charged, the more electronegative F atom can better stabilise this intermediate.
  2. Sterics - F is also smaller, allowing the nucleophile unhindered access to attack the pi* antibonding orbital.

CH10135 (7 decks)

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