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Organic Chemistry > Aromatics > Flashcards

Aromatics Flashcards

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1
Q

What is the molecular formula of benzene

A

C6H6

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2
Q

Describe the geometry of benzene

A
  • Benzene consists of a flat, regular hexagon of carbon atoms, each of which is bonded to a single hydrogen atom.
  • The bond angles between all bonds is 120 degrees.
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3
Q

Describe the two reasons why the structure of benzene puzzled organic chemists

A

1) Despite being unsaturated, it does not regularly undergo addition reactions.
2) All of the carbon atoms are equivalent, which implies that all of the carbon-carbon bonds are the same

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4
Q

Describe the structure of the C-C bonds in benzene

A
  • The C—C bonds in benzene are intermediate between a single and double bond (the bond length is also between the expected length for a single and double bond).
  • This is electron delocalisation and makes benzene unusually stable.
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5
Q

Explain the electron delocalisation that occurs in benzene

A
  • Each carbon atom has 3 covalent bonds: one to a hydrogen atom and the other two to carbon atoms
  • The fourth electron of each carbon atom is in a p-orbital.
  • There are 6 p-orbitals in total (one for each carbon) and each contains a single electron.
  • These p- orbitals overlap and the electrons in them are delocalised.
  • This forms a region of electron density above and below the carbon ring.
  • This means that overall, each C-C bond in benzene is in between a single and double bond.
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6
Q

What is electron delocalisation and what does this mean in the case of benzene

A
  • Deocalisation means that electrons are spread over more than two atoms.
  • In the case of benzene this means over the 6 carbon atoms that form the ring.
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7
Q

Why is the delocalised electron system in benzene important

A

It makes benzene (and it’s derivative aromatics) an unusually stable molecule

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8
Q

What thermochemical evidence do we use to prove the stability of benzene

A

The enthalpy change of hydrogenation reactions

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9
Q

Describe and explain the thermochemical evidence for the stability of benzene

A
  • The enthalpy change for the hydrogenation of cyclohexene (1 double bond) is -120 KjMol-1
  • So the enthalpy change for the hydrogenation of of a ring with 3 alternate double bonds (hypothetical non-delocalised benzene) should be -360 KJmol-1
  • However the actual enthalpy change for the hydrogenation of benzene is -208 KJmol^-1.
  • Thus proves the delocalised ring theory and proves that benzene is more stable than the initially proposed ring structure.
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10
Q

Describe the two factors that determine the reactivity of aromatic compounds

A

1) The ring is an area of high electron density, because of the delocalised bonding, and is therefore attacked by electrophiles.
2) The aromatic ring is very stable. It needs energy to be put in to break the ring before the system can be destroyed which is called the delocalisation energy. This means that the ring almost always stays intact in the reactions of arenes.

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11
Q

What is the most common type of reaction that aromatic systems undergo

A

Electrophilic substitution reactions

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12
Q

Describe the solubility of arenes and the state of benzene at room temperature

A
  • Arenes are non-polar so they do not mix with water but do mix with other hydrocarbons and non-polar solvents.
  • benzene is a colourlesss liquid at room temperature
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13
Q

Describe the melting and boiling point of benzene and this compares to those of hexane

A
  • Benzene boils at 353K and freezes at 297K.
  • It’s boiling point is comparable with that of hexane but it’s melting point is much higher than hexanes.
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14
Q

Why does benzene have a comparable boiling point to hexane but a much higher melting point than hexanes

A
  • This is because benzenes flat,hexagonal molecules pack very closely together in the solid state, making them hard to separate.
  • The separation of molecules is what must happen in order for melting to occur.
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15
Q

Describe the combustion of arenes

A
  • Arenes burn in air with flames that are noticeably smoky.
  • This is because they have a high carbon:hydrogen ratio compared with alkanes.
  • There is usually unburnt carbon that remains when they burn in air- this produces soot.
  • A smoky foam suggests an aromatic compound.
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16
Q

Explain why aromatics often undergo electrophilic substitution reactions

A
  • The delocalised system of the aromatic ring has a high electron density that attracts electrophiles.
  • The electrons in the delocalised ring are also attracted to the electrophile.
17
Q

Give two examples of important electrophilic substitution reactions of aromatics

A

1) Nitration
2) Friedel- Crafts acylation

18
Q

Describe generally what happens (the mechanism) for the electrophilic substitution of an aromatic compound

A
  • The delocalised system of the aromatic ring has a high electron density and attracts electrophiles and is itself attracted to the electrophile.
  • A bond forms between one of the carbon atoms and the electrophile.
  • To do this, the carbon uses electrons from the delocalised system which destroys the system.
  • To get back to the stability of the aromatic system, the carbon loses a H+ ion with the electron in the C—H bond returning to the delocalised system.
  • The sum of these reactions is the substitution of H+ by the electrophile.
19
Q

Summarise what the nitration of an aromatic is

A

The substitution of a NO2 group for one of the hydrogen atoms on an arene ring.

20
Q

What is the electrophile in the nitration of an aromatic

A

NO2+

21
Q

What is the overall equation for the generation of the NO2+ electrophile for nitration

A

H2SO4 + NHO3 —> NO2+ + HSO4- + H2O

22
Q

Explain (in words) the steps in the generation of the NO2+ electrophile needed for the nitration of an aromatic

A
  • A reaction mixture of concentrated sulphuric acid and concentrated nitric acid is needed.
  • Because sulfuric acid is a stronger acid than nitric acid and donates a proton H+ to HNO3 to form H2NO3 +
  • The H2NO3 + then loses a molecule of water to give NO2+ which is called the nitronium ion or nitryl cation
23
Q

Write the equations for the two stages in the generation of the NO2+ electrophile needed for the nitration of aromatics

A

1) H2SO4 + HNO3 —> H2NO3+ + HSO4-
2) HNO3+ —> NO2+ + H2O

24
Q

Why is the sulfuric acid a catalyst in the nitration of benzene despite being used in the formation of the electrophile NO2 +

A

Because the HSO4- produced when the electrophile is formed then reacts with the H+ from the aromatic, therefore regenerating the catalyst.

25
Q

What are two key uses of nitrated arenes

A
  • Nitration is an important step in the production of explosives like TNT
  • Nitration is the first step in making aromatic arenes, which in turn are used to make industrial dyes.
26
Q

What is the reagent and catalyst for the nitration of arenes

A
  • Reagent: concentrated nitric acid
  • Catalyst: concentrated sulfuric acid
27
Q

What is the catalyst for Friedel Crafts acylation reactions

A

Aluminium chloride

28
Q

What product does Friedel Crafts acylation form

A

Acyl-substituted arenes

29
Q

Describe how the electrophile RCO+ is formed in Friedel Crafts acylation

A
  • Acyl chlorides provide the RCO group.
  • These Acyl chlorides react with AlCl3 to form AlCl4- and RCO+
  • This reaction takes place because the aluminium atom in the aluminium chloride has only six electrons in its outer main level and readily accepts a lone pair from the chlorine atom of ROCl
  • RCO+ is a good electrophile that is attacked by the benzene ring to form substitution products.
  • The AlCl3 os a catalyst because it is reformed by the reaction of the AlCl4- with the H+ from the benzene ring to form AlCl3 and HCl
30
Q

Why is Friedel Crafts acylation useful

A

It is a useful step in the synthesis of new substituted aromatic compounds

Organic Chemistry (14 decks)

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