Aromatic Compounds

Question 1

Arene:-

Answer 1

Aromatic hydrocarbon

Question 2

Kekule formula:-

Answer 2

6- carbon ring, each with a carbon-carbon single bond, a carbon-carbon double bond and a bond to a hydrogen

Question 3

Evidence that suggests Kekule’s structure isn’t entirely correct:- (3)

Answer 3

• benzene doesn’t easily undergo the typical reactions of alkenes.
• all the bonds between C atoms should be the same.
• hydrogenation (addition of 3 H2 to form cyclohexane) of benzene is less exothermic than expected by comparison with alkenes.

Question 4

Benzene doesn’t easily undergo the typical addition reactions of alkenes:-

Answer 4

• doesn’t decolourise bromine in the dark.

* reacts with bromine in presence of a catalyst, but it reacts by substitution ratger than addition.

Question 5

Bond length problem with Kekule’s triene structure:-

Answer 5

C-c bonds are longer than c=c, so if Kekule’s was correct, it would have 3 short and 3 long, but all bond lengths in benzene are actually the same.

Question 6

Benzene hydrogenation less exothermic than expected:-

Answer 6

Less heat is released when hydrogen is added to benzene than would be expected for addition of H2 across 3 double bonds. This suggests that the pi-bonds in benzene are more stable than expected

Question 7

Benzen’s structure:- (4)

Answer 7

• planar (all 12 atoms are in the same plane) and all six c-c lengths are the same.
• each c forms 3 sigma bonds.
• each also has 1 electron in a p-orbital, which extends above and below the plane of the molecule.
• these orbitals overlap to form a delocalised pi-electron system.

Question 8

Delocalised pi-electron system:-

Answer 8

Vontinuous pi cloud of electrons above and below plane of molecule (drawn in 3d as a ring above and below the hexagonal shape).

Question 9

Explanation for benzene’s low reactivity:-

Answer 9

Delocalisation spreads out the pi-electrons more than they would be in a localised pi-bond (ie in an alkene). Hence the electron density is lower and they are consequently less attractive to electrophiles.

Question 10

Explanation for benzene’s bond lengths:-

Answer 10

The pi-electrons are evenly distributed around the ring, hence all the c-c bond lengths are the same.

Question 11

Explanation for the DeltaH of hydrogenation:-

Answer 11

Delocalisation lowers the pi-electrons’ energy (more stable state than localised). This means that reactions that break up the delocalised pi-electron system are less favourable than might have been expected (hense DeltaH of hydrogenation is less negative).

Question 12

Differences of aromatic electrophile reaction due to the delocalisation of the pi-electrons:- (2)

Answer 12

• less reactive than alkenes- lower electron density in aromatic ring so electrophiles less strongly attracted- tend to occur much more slowly and require catalysts.
• tend reaction by electrophilic substitution rather than addition.

Question 13

Electrophilic substitution:-

Answer 13

Favourable delocalised pi-system is still present in reaction product, whereas it would be lost in an addition reaction. In contrast, it is generally more favourable in alkenes’ addition reactions as less stable localised pi-bonds are lost and replaced by new sigma-bonds.

Question 14

Overall nitration equation:-

Answer 14

Benzene + HNO3 -> Nitrobenzene + H2O

Question 15

Nitro group:-

Answer 15

NO2

Question 16

Nitration conditions:-

Answer 16

Heat at 50-60oC with a mix of conc nitric acid and conc sulphuric acid.

Question 17

Nitration at higher temp/with long reaction time:-

Answer 17

Further substitutions can occur e.g:-

Benzene + 2HNO3 -> Dinitrobenzene + 2H2O

Question 18

Nitration:- formation of the electrophile:-

Answer 18

HNO3 + H2SO4 ↔️ NO2+

+ H2O + HSO4-

Question 19

Nitration:- reaction with the electrophile:-

Answer 19

• Curly arrow from circle in benzene to NO2+ (electrophile).
• arrow toBenzene w/o full circle inside, + in centre just above bottom of circle, NO2 and H attached at top to same carbon, curly arrow from H’s bond to space inside benzene where circle is incomplete. *carbocation intermediate has 4 pi-electrons, delocalised over 5 electrons, H+ is eliminated to complete reaction.
• arrow to nitrobenzene + H+.

Question 20

Nitration:-Catalyst regeneration:-

Answer 20

H+ released in stage 2 combines with HSO4- from stage 1 to regenerate H2SO4 catalyst:-
HSO4- + H+ -> H2SO4

Question 21

Halogenation:-

Answer 21

Aromatic compounds react with halogens by substitution. Only takes place in presence of catalyst called a halogen carrier e.g:-
Benzene + Br2 ->(w/AlBr3) Bromobenzene

Question 22

Bromine Halogenation observations:-

Answer 22

Bromine is decolourised (orange -> colourless) and there are steamy fumes of HBr.

Question 23

Halogenation conditions:-

Answer 23

Room temp and presence of halogen carrier catalyst:- AlCl3 or FeCl3 for chlorination, AlBr3 or FeBr3 for bromination.
Halogen carrier requirement is a big difference from alkenes which react by addition without needing a catalyst.

Question 24

Why arenes are less reactive towards electrophiles:-

Answer 24

Delocalisation spreads out the pi-electrons so the electron density in the ring is lower than it would be in isolated pi-bonds. Hence, the electrophile is less strongly attracted to an aromatic ring. To compensate, the halogen carrier polarises the halogen, making it a more potent electrophile. I’m

Question 25

Halogenation electrophilic substitution 1:-

Answer 25

The halogen carrier reversibly reacts with the halogen to create a positively charged halogen ion:-
Br2 + AlBr3 ↔️ Br+ + AlBr4-.
Br+ ion formed will be highly attractive to ring’s pi-electrons and hence be able to act as a powerful electrophile.

Question 26

Halogenation electrophilic substitution 2:-

Answer 26

Attack on aromatic ring by electrophile:- drawn same as Nitration, with Br replacing nitro group (NO2).

Question 27

Halogenation electrophilic substitution 3:-

Answer 27

H+ reacts with AlBr4- :-
H+ + AlBr4- -> AlBr3 + HBr.
Second product (HBr) is formed and catalyst AlBr3 is regenerated.

Question 28

Reaction with Haloalkanes (Friedel-Crafts Alkylation):-

Answer 28

Alkyl group attached to benzene ring by an electrophilic substitution reaction. Halogen carrier catalyst is used and the mechanism is very similar to halogenation.
e.g. C6H6 + CH2CH3Cl ->(w/ AlCl3) benzene with alkyl branch plus HCl.

Question 29

Friedel-Crafts Alkylation mechanism:-

Answer 29

• Halogen carrier catalyst used to generate electrophile:- CH3CH2Cl + AlCl3 ↔️ CH3CH2^+ + AlCl4^-.
• electrophile attacks aromatic ring in normal way.
• catalyst regenerated. H+ + AlCl4^- ↔️ HCl +AlCl3.

Question 30

Why are Friedel-Crafts Alkylation reactions important in organic synthesis?

Answer 30

They form C-C bonds and therefore allow the carbon framework of an organic molecule to be extended l. This allows simple (and often cheap) starting materials to be built up into more complex molecules such as pharmaceuticals.

Question 31

Friedel-Crafts Acylation:-

Answer 31

Reaction with acyl chloride. Product is an aromatic ketone.

Question 32

Acyl chloride:-

Answer 32

R–C==O
|
Cl

R is an alkyl or aryl group.

Question 33

Friedel-Crafts Acylation mechanism:-

Answer 33

•Acyl chloride + halogen carrier react to form an electrophile.
CH3COCl + AlCl3 ↔️ CH3CO^+ + AlCl4^-
•acyl cation (R-CO^+) reacts as electrophile, attacks aromatic ring.
•catalyst regen- H+ +AlCl4^- ↔️ HCl + AlCl3

Question 34

Benzene ring in molecule naming:-

Answer 34

Name of ring changes to prefix phenyl. C6H5 = phenyl group. E.g Phenylethene.

Question 35

Recrystallisation:-

Answer 35

• dissolve impure substance in hot solvent.
• slowly cool- pure product crystallises out. Impurities stay in solution.
• filter off crystals and wash with cool solvent.