Chap 12: Arenes

Question 1

Why does benzene react with electrophiles?

Answer 1

The delocalised pi e- cloud contributes to high e- density in the benzene ring. This makes benzene highly susceptible to attack by electrophiles (lone pair acceptors).

this is similar to e- in the pi bonds of alkenes contributing to high e- density in the double bond

Question 2

Why does benzene prefer substitution rather than addition reactions?

Answer 2

benzene does not undergo E.A. as the delocalised pi e- cloud results in resonance stability.

the addition reaction will destroy the aromatic ring of delocalised pi e- cloud, resulting in the loss of resonance stability in the addition product formed. the loss of this aromatic hat after is not favourable energetically.

benzene tends to undergo substitution reactions so that its pi e- cloud remains intact to maintain resonance stability

Question 3

Why is benzene less reactive than alkenes?

Answer 3

Aromatic stability renders the pi e- cloud less susceptible to electrophilic attack compared to that of the C=C in alkenes. hence stronger electrophiles are required for benzene to react

Question 4

What are the conditions for Halogenation of benzene?

Answer 4

E.S.
for chlorination-Cl2 (l), anhydrous FeCl3/AlCl3/Fe
for bromination-Br2 (l), anhydrous FeBr3/AlBr3/Fe

• Room temp
• yellowish cl2 decolourised to produce white fumes of hcl
• reddish brown br2 decolourised to produce white fumes of hbr

Question 5

What are the conditions for Friedel-Crafts Alkylation of benzene?

Answer 5

E.S.
CH3Cl, anhydrous FeCl3 or AlCl3 catalyst

• white fumes of hcl produced

Question 6

What are the conditions for nitration of benzene?

Answer 6

E.S.
conc. HNO3, conc. H2SO4

• yellow oil (nitrobenzene formed)

Question 7

What are the conditions to REDUCE nitrobenzene?

Answer 7

[R]
(i) Sn. conc hcl, heat under reflux, followed by NaOH (aq)
(ii) H2 (g), Ni catalyst with heat OR H2 (g), Pt/Pd catalyst, room temperature

Question 8

How does AlCl3 act as a Lewis acid?

Answer 8

AlCl3 acts as a Lewis acid by accepting an electron pair from, chlorine to produce the stronger Cl+ electrophile

Question 9

How does H2SO4 act as a Bronsted-Lowry acid?

Answer 9

conc. H2SO4, being a stronger acid, will react with conc. HNO3 to form NO2+, a stronger electrophile and conc. H2SO4 donates a proton thus acting as a Bronsted-Lowry acid while HNO3 accepts the proton and behave as the base

Question 10

What are the conditions for nitration of methylbenzene?

Answer 10

E.S.
conc. HNO3, conc. H2SO4 at 30°C

a lower temp of 30°C is used for methylbenzene instead of 55°C (because -CH3 is activating and hence milder conditions are needed)

Question 11

What are the conditions for halogenation of methylbenzene?

Answer 11

E.S.
for chlorination -Cl2(g), anhydrous FeCl3/AlCl3/Fe
for bromination: Br2(l), anhydrous FeBr3/AlBr3/Fe

room temp, absence of UV light/in the dark

yellowish green cl2 de colour used to give white fumes of HCl
reddish brown br2 de colour used to give white fumes of HBr

absence of UV light is to prevent FRS from occuring

Question 12

What are the conditions for FRS for side-chain of methylbenzene?

Answer 12

F.R.S
for chlorination - limited Cl2(g)
for bromination - limited Br2(g)

UV light/heat

note: more than 1 Cl or Br atom can be substituted if excess Cl2 or Br2 is used

Question 13

What are the conditions for side-chain [O] of methylbenzene?

Answer 13

[O] (ACIDIC)
KMnO4 (aq), H2SO4 (aq)
Heat under reflux

obv:
- purple KMnO4 decolourised
- white ppt of benzoic acid formed

[O] (ALKALINE)
KMnO4 (aq), NaOH (aq)
Heat under reflux

obv:
- purple KMnO4 decolourised
- brown ppt of MnO2 formed

note: K2Cr2O7 cannot be used as it is a weaker O.A.