Arenes

Question 1

Physical properties: Benzene

Answer 1

• Colourless liquid with characteristic odour
• non-polar
• insoluble in water
• less dense than water
• soluble in all organic solvents
• good solvent for organic compounds
• burn with smoky luminous flame due to high C content

Question 2

Benzene undergoes substitution instead of addition reactions

Answer 2

If benzene undergoes addition
=> overall aromatic character destroyed
=> extra stability associated with delocalisation of 6 π electrons is lost
=> hence substitution in ring - retain aromatic character & stability

Question 3

Benzene: characteristic reactions + reason

Answer 3

1. Electrophilic substitution
   => Availability of 6π electrons
   => benzene is electron rich
   => attract electrophiles
2. Reduction

Question 4

RnC: Nitration of benzene

Answer 4

conc HNO₃, H₂SO₄, 55deg C

Question 5

RnC: Halogenation of benzene (CL₂)

Answer 5

CL₂, FeCL₃ (catalyst)

Question 6

RnC: Friedal-craft Alkylation

Answer 6

RCL, ALCL₃ ANHYDROUS, rtp

Question 7

RnC: Friedal-Craft Acylation

Answer 7

RCOX + ALCL₃ (to generate ROC+ electrophile)

Question 8

Activating groups vs Deactivating groups

Answer 8

Activating groups
=> Increases electron density in benzene ring
=> ring more susceptible to electrophilic attack
.
=> helps to disperse +ve charge in carbocation
=> stabilise
=> increase reactivity

Deactivating groups
=> Decreases electron density in benzene ring
=> ring less susceptible to electrophilic attack
.
=> intensifies +ve charge on intermediate carbocation
=> decrease reactivity

Question 9

Inductive effect

Answer 9

Donation/ withdrawal of electrons through σ bonds due to EN difference between atoms

Question 10

Resonance effect

Answer 10

Donation/ Withdrawal of electrons through π bonds due to side-on p-orbital overlap of substituent and benzene ring
=> delocalisation of electrons towards/ away from ring

Question 11

Why alkyl groups electron donating?

Answer 11

• alkyl groups have sp³ hybridised carbon atom
  => less s-character than sp² hybridised carbon atom in ring
  => alkyl groups hold on to electrons less tightly
  => electron more easily donated into ring

Question 12

How are EN atoms electron withdrawing?

Answer 12

X groups more EN than C atom in ring
=> pull electron density away from C through σ bond

Question 13

Why atom with available L.P activating?

Answer 13

Substituent with available L.P on atom directly attached to ring
=> Provided L.P in p-orbital (usually the case)
=> can be delocalised into the ring (resonance)
=> increased electron density in ring
=> activated

Question 14

Deactivating through resonance scenario

Answer 14

Substituent directly attached to ring by atom
=> that is doubly/ triply bonded to
=> a more EN atom
=> π electron of benzene
=> delocalised onto substituent
=> Electron density in ring decreased
=> Ring deactivated

Question 15

Substituent directing effect

Answer 15

For Multisub products
=> have directing effects point towards the same position

Question 16

Benzene: Combustion

Answer 16

Sooty, Smoky flame, due to high % of Carbon in benzene

Question 17

RnC: Reduction of benzene

Answer 17

H₂, Ni catalyst @considerably higher T&P than alkanes
=> since extra E must be supplied to break delocalised π electron system

Question 18

RnC: Side chain Oxidation

Answer 18

Benzene DOES NOT OXIDISE
.
RnC: KMnO₄ (aq), H₂SO₄ (aq), heat or with NaOH (aq)
Observations:
=> Decolorisation of purple KMnO₄ (aq)
=> Brown-black ppt of MnO₂
=> White ppt of benzoic acid upon cooling and & acidification (after recrystallisation)

dicromate is TOO WEAK to oxidise side chain

Question 19

Criteria for side chain oxidation of benzene

Answer 19

H or O atom bonded to benzylic C atom
=> can undergo oxidation when heated with acidified KMnO₄ (aq)

Question 20

Balancing oxidation of alkyl side chain reaction

Answer 20

Organic Balancing method (provided dont need full eqn - which is usually the case)

1. Identify Oxidation or Reduction
   .
   .
   If OX, balance H first using H₂O
   => then balance O with [O]
   .
   If RED, balance O first with H₂O
   => then balance H with [H]