Aromatic Compounds

Question 1

What is the formula of benzene?

Answer 1

C6H6

Question 2

Describe and explain the structure of a benzene molecule.

Answer 2

6x sigma bonds and a continuous ring shaped pi-bonding system above and below the plane of the ring. This is formed due to the overlap of the p-orbitals of each of the six carbon atoms.

Question 3

Why is benzene considered to be stable?

Answer 3

The electrons in the pi-bond are delocalised across the six carbon atoms, thus the -ve charge is stabilised.

Question 4

What evidence is there for benzene’s structure?

Answer 4

X-ray crystallography: all bond-lengths are the same (intermediate between C-c and C=C)
Bromine water: NOT decolourised by benzene so must have insufficient e- density
Enthalpy of hydrogenation: NOT -360kJmol-1 as Kekule’s model would have implied, but -208kJmol-1 where the difference in energy is the delocalisation energy. This quantifies the extra stability of benzene

Question 5

What is the systematic name for the Kekule model?

Answer 5

Cyclohexa-1,3,5-triene

Question 6

Describe the naming conventions for benzene.

Answer 6

Shorter groups are given as prefixes to benzene. Where the group has chain length >=7 or it includes a functional group, we instead use the phenyl- prefix.

Question 7

What type of reaction does benzene undergo typically?

Answer 7

Electrophilic Substitution

Question 8

Why does benzene undergo substitution rather than addition?

Answer 8

The delocalised pi bonding system has insufficient e- density to induce a dipole in an attacking molecule. Such a reaction would also disrupt the stable pi-bonding system.

Question 9

Describe a typical Electrophilic substitution.

Answer 9

1. Formation of Electrophile E+
2. Electrophile gains 2x electrons from pi-bond to form dative bond to carbon
3. Existing hydrogen atom donates 2x electrons to re-complete pi bond, releasing H+ which bonds with -ve ion formed originally

Question 10

How many electrons in contained within the intermediate of the Electrophilic substitution of benzene?

Answer 10

4x pi-electrons

Question 11

How do we form the Electrophile for the nitration of benzene?

Answer 11

Formed in situ by reacting conc. H2SO4 and HNO3, to form NO2+. This is formed in situ due to its’ instability

Question 12

What are the conditions + reagents for the nitration of benzene?

Answer 12

Reagents: HNO3
Conditions: H2SO4 catalyst (reformed due to reaction of H+, HSO4- at end) and 50 degrees Celsius

Question 13

Why must we use a temperature of 50 degrees Celsius when nitrating benzene?

Answer 13

This temperature ensures the mono=-substitution of NO2+

Question 14

How can we halogenate benzene molecules?

Answer 14

Since Benzene itself doesn’t have sufficient e- density to induce an electrophile from Br2/Cl2, we use AlBr3/AlCl3/etc to form a +ve species. This mono-substitutes.

Question 15

Describe Friedel-Krafts acylation and alkylation.

Answer 15

Alkylation: CH3Cl + ALCl3 catalyst to add CH3 group
Acylation: ClCOCH3 + AlCl3 catalyst to add COCH3 group

Question 16

Why do benzene and Alkenes react differently?

Answer 16

Alkenes have higher electorn density in their pi-bonding system since the electrons are localised over 2x carbons rather than being delocalised over six. They can more readily induce a dipole in an incoming a molecule.

Question 17

What reactant and conditions are used to reduce nitrobenzene into phenylamine?

Answer 17

[H+ such as NaBH4] + Sn/HCl (conc)

Question 18

What typically defines a group as 2/4 directing?

Answer 18

Presence of lone pairs. This makes the group in question electron-donating, so it activates the pi bonding system and directs attacking Electrophile to 2/4 positions.

Question 19

What typically defines a group as 3-directing?

Answer 19

Presence of double or triple bonds. These group are typically electron withdrawing so deactivate the benzene ring. This directs electrophiles to the 3 position.

Question 20

What is a phenol?

Answer 20

An organic compound containing a benzene ring, with a hydroxyl group directly attached to it.

Question 21

Why are phenols more reactive than aromatic alcohols and benzenes alike?

Answer 21

The -OH group donates a pair of electrons to the pi-bonding system, increasing e- density and making the molecule more susceptible to Electrophilic attack.

Question 22

What are some typical comparisons we make between the reactions of phenols and benzene?

Answer 22

• Lower temperatures
• Less concentrated reactants
• No need for a catalyst/carrier/etc

Question 23

How does phenol’s increased e- density increase reactivity?

Answer 23

Incoming molecules are polarised more strongly.

Question 24

Describe the reaction between phenol and bromine water.

Answer 24

3x bromine atoms bond onto phenol in the 2, 4, and 6 positions. This can happen because of the increased e- density and the strong 2,4,6 directing effect of the -OH group promoting multiple substitution.

Question 25

Why are phenols partially soluble?

Answer 25

Their -OH group forms hydrogen bonds so is soluble, but the benzene ring is hydrophobic.

Question 26

How can phenol act a a weak acid?

Answer 26

It partially dissociates to form the phenotype ion in solution.

Question 27

Why does phenol form phenoxide solutions in solution?

Answer 27

When H+ ions are lost, the negative charge remaining on the molecule is stabilised by the pi-bonding ring system. This makes the phenoxide ion particularly stable.

Question 28

Why does phenol dissolve in NaOH?

Answer 28

The phenoxide ion acts as a weka acid, whilst sodium hydroxide is a base. We form sodium phenoxide and water.

Question 29

Why does phenol only act as a weka acid, not a strong one?

Answer 29

Though the phenoxide ion is stable an readily forms, the H+ ion lost in doing so is readily attracted by the O- group to reform the molecule.

Question 30

How can we differentiate between panel and other weak acids like Carboxylic acids?

Answer 30

We use the carbonate test - phenol is too weka and acid to react readily, thus only the Carboxylic acid reacts to release CO2 gas to cause effervescence.