advanced higher 1 organic organic chemistry · advanced higher 1 organic organic chemistry bonding...

ADVANCED HIGHER 1 Organic

ORGANIC CHEMISTRY

Bonding

In order to form the maximum number of bonds, all atoms maximisetheir number of unpaired electrons by entering an excited state,known as the valence state, just before they bond with another atom.

Carbon does this by promoting a 2S electron into a 2P orbital:

The valency of Carbon is therefore FOUR.

The excited state depicted above, however, is unstable. The electronsare too close to each other. They can move further apart by aprocedure known as hybridisation. This involves mixing atomicorbitals to generate a set of new atomic orbitals called hybridorbitals.

Carbon can form hybrid orbitals in three distinct ways:

SP3 Hybridisation

The mixing of the 2S and all three 2P orbitals:

C-C bonds are formed by end-on overlap of two SP3 hybrid orbitalslying along the axis of the bond:

This is known as sigma σ bonding.

SP2 Hybridisation

The mixing of the 2S and two of the 2P orbitals:

2Px 2Py 2Pz

2S

2Px 2Py 2Pz

2S

σ

2Px 2Py 2Pz

2S SP hybrid orbitals3

2Px 2Py 2Pz

2SSP hybrid orbitals2

2Pz

PZ

SP2


C=C bonds are formed by σ bonding between two SP2 hybrid orbitals andby sideways overlap (pi π bonding) of two of the Pz orbitals:

SP Hybridisation

The mixing of the 2S and one of the 2P orbitals:

C C bonds are formed by σ bonding between two SP hybrid orbitals andby π bonding between four of the Pz orbitals:

Alkanes

Alkanes consist entirely of non-polar bonds. When a non-polar bondbreaks, since both atoms have equal electronegativities, the twoelectrons are shared, one to each atom, forming free radicals(particles containing unpaired electrons) i.e.

The free radicals will then combine with colliding free radicals.

e.g. the chain reaction of Methane with Chlorine.

CH4 + Cl2 -> CH3Cl + HCl

π

σ

PzPy

SP

2Px 2Py 2Pz

2SSP hybrid orbitals

2Py 2Pz

π

σ

A A.. A. A.

Denotes the movement of ONE electron.Bond breaking of this type is known as homolytic fission.


Mechanism :

Initiation Cl2 Cl• Cl•

Propagation CH4 Cl• -> CH3• HCl

CH3• Cl2 -> CH3Cl Cl•

Termination Cl• Cl• -> Cl2CH3• CH3• -> CH3-CH3CH3• Cl• -> CH3Cl

Once the initiation step has produced a few Cl• free radicals byhomolysis of Cl2 the reaction will continue in the absence of heator UV ; although the free radicals are consumed by one of thepropagation steps they are regenerated by the other.

The propagation cycle can continue about 5000 times before Cl• arewiped out in termination steps.

Alkanes also undergo substitution reactions with Bromine by asimilar mechanism.

Alkenes

Alkenes contain at least one C=C bond.

Alkenes can be prepared by dehydration of alcohols:

● pass alcohol vapour over heated Aluminium oxide catalyst

● heat a mixture of alcohol and a dehydrating agent e.g. concentrated Sulphuric acid or concentrated Phosphoric

acid

e.g. dehydration of Butan-1-ol

Alkenes can also be prepared by elimination of hydrogen halidesfrom haloalkanes. The haloalkane is heated with a solution ofPotassium hydroxide in Ethanol solvent e.g.

2C C C

H H

H H

H

H

H

C C C

H H

H

HH

H

H O

Butan-1-ol But-1-ene

C

H

H

OH C

H

H

UV or heat

C C C

H H

H H

HH

H

Br

C C C

H H

H

HH

H

H OK Br+ -

2-Bromopropane Propene

2K OH+ -


Alkenes undergo addition reactions:

Addition of Hydrogen using Nickel catalyst e.g.

Addition of Hydrogen halides e.g.

MECHANISM (TWO steps):

STEP 1:

HBr is polar. The Hδ+ attracts a pair of electrons from thedouble bond.

HBr is an electrophile - a negative seeking reagent.

The alkene is a nucleophile - a postive seeking reagent.

C=CCH

3H

HHH C

H

H

C

H

CH3

H H2Ni

Propene Propane

C=C

H

H H

HBr C

H

H

C

Br

HH

Propene 2-Bromopropane

CH3

CH3

Denotes the movement of TWO electrons.Bond breaking of this type is known as heterolytic fission.

H

C

C

H H

H Br

H

C

C

H H

H

:Br

+-

carbocationintermediate

δ+ δ-

CH3

CH3


Why does the H atom prefer to bond to the bottom Carbon atom?This is because of the inductive effect of the CH3 group.

The CH3 group repels electrons in the

C-C and C=C bonds. The bottom Carbon

thus becomes δ- and thus more likelyto be the nucleophilic source.

Markownikoff's rule can be used to predict which of the twoCarbon atoms is most nucleophilic:

The Hydrogen atom bonds to the Carbon atom that alreadyhas the most Hydrogen atoms bonded to it.

STEP 2:

The Br- ion does a nucleophilic attack at the carbocationintermediate:

Acid catalysed addition of Water

Alkenes react with Water in the presence of H+ catalyst. Hydration of Ethene, with Phosphoric acid catalyst, is used in theindustrial preparation of Ethanol:

C2H4 + H2O -> C2H5OH

Hydration of alkenes also follows Markownikoff's Rule e.g.

H

C

C

H H

CH3

δ-

δ+

C=C

H

H

H O C

H

H

CH

Methylpropene Methylpropan-2-ol

CH3

CH3

CH3

2

H+

CH3

OH

H

C

C

H H

H

+

CH3

Br:-

H

C

C

H H

Br

H

CH3


MECHANISM (THREE steps):

STEP 1:

Nucleophilic attack by the alkene at the H+ catalyst:

STEP 2:

The H2O does a nucleophilic attack at the carbocationintermediate:

STEP 3:

Regeneration of the H+ catalyst:

C

C

H

H

C

C

H H

H

+

carbocationintermediate

CH3

CH3

+

3H C

H

3H C

C

C

H H

H

+

CH33

H C

H

H

O••

C

C

H H

H

+ CH3

H

H

O

CH3

C

C

H H

H

+ CH3

H

H

O

CH3

C

C

H H

H

+CH3

H

HO

CH3


Addition of Halogens e.g.

MECHANISM (TWO steps):

STEP 1:

As Br2 approaches the double bond, electrons in the Br-Br bondare repelled by the double bond electrons causing a δ+ charge todevelop on the Br nearest the double bond. Thus the Brominebecomes electrophilic. It draws an electron pair from the doublebond:

Why do we get a cyclic ion intermediate? Why not just thecarbocation:

Br is nucleophilic: it contains an electron pair which can bedonated to C+ thus closing the ring:

C=C

H H

H H

Br2 C

H

Br

C

H

Br

HH

Ethene 1,2-Dibromoethane

H H

H H

C

C

Br

+

H H

H H

C

C

Br

+

H H

C

C

H H

Br+

••

H H

C

C

H H

Br Br

H H

C

C

H H

Br :Br+ -

cyclic ionintermediate

δ+ δ-


STEP 2:

The Br- ion does a nucleophilic attack at the cyclic ionintermediate:

Haloalkanes

Haloalkanes are classified according to the number of alkyl groups(R) bonded to the Carbon atom containing the halogen atom (X):

The Cδ+ - Xδ- bond is polar ; the positively charged Carbon atom iselectrophilic and is therefore attacked by nucleophiles.

Haloalkanes undergo nucleophilic substitution reactions.

Reaction with hydroxide ion -OH to form alcohols

e.g. the reaction of Bromoethane with hydroxide ion.

The reaction occurs in one step: HO- attacks the Cδ+ forcingelectrons in the C-Br bond on to the Br.

H H

C

C

H H

BrBr:+-

H H

C

C

H H

Br

Br

C

H

H

X

R

C

H

H

Cl

e.g.

Chloroethane

PRIMARY

CH3

C

R

H

X

R

CH

I

SECONDARY

C

R

R

X

R

C

Br

TERTIARY

CH3

2-Iodobutane 3-Bromo 3-methylhexane

C H2 5 3

H C

C H2 5

C H3 7

C

H

Br3

H C

H

HO:-

C

H

:Br

H

HO-

CH3

Bromoethane Ethanol


The above mechanism is described as SN2:

Substitution of Br by theNucleophile HO- with2 particles (HO- and C2H5Br) taking part in the rate determining step

An alternative mechanism, favoured by tertiary and, to someextent, secondary haloalkanes in polar solvents is described asSN1:

The SN1 reaction occurs in two steps:

STEP 1 - the slow, rate determining step.

Polar solvents like Water bond to these ions and prevent reversal:

The inductive effect of the alkyl groups in the carbocation,forcing electrons on to C+, reduces the positive charge and makesreversal less likely. The more alkyl groups, the more stable thecarbocation and the more likely it is to survive and be able totake part in step 2.

STEP 2 - a fast step.

We will assume the SN2 mechanism in all further examples sincethey involve primary haloalkanes.

C :Br

3H C

CH3

CH3

+ -O

H

H

δ+δ−

O

H

H

CHOCH

3

CHO:

3H C

CH3

CH3

+-

CH3

CH3

C :Br

3H C

CH3

CH3

+ -C Br

3H C

CH3

CH3


Reaction with alkoxide ions -OR to form ethers e.g.

The alkoxide ion is produced by reaction of the appropriatealcohol with Sodium e.g. the methoxide ion above is produced byreaction of Methanol with Sodium:

CH3OH + Na -> CH3O-Na+ + H2

Reaction with cyanide ion -CN (in Ethanol solvent) to formnitriles e.g.

Nitriles can be hydrolysed to form amides and, subsequently,carboxylic acids e.g. hydrolysis of Propane nitrile:

C I3

H C

H

NC:-

C :I

H

NC-

CH3

Iodoethane Propane nitrile

H H

3H C C I

H

O:-

C :I

H

-

1-Iodopropane Methoxypropane

OH H3

H C

C H2 5

C H2 5

Propane nitrile

C H2 5

C N

Propanamide

C H2 5

Propanoic acid

C H2 5

H O2

H O2

CO

NH2

CO

OH


Reaction with Ammonia NH3 to form alkyl ammonium saltse.g.

Reaction of the alkyl ammonium salt with alkali produces aminese.g.

Ethers

Ethers have the general formula R1-O-R2 where R1 and R2 are alkylgroups e.g.

Properties:

● They are fairly polar because of their non-linear shape e.g. Ethoxyethane (Diethylether):

C Cl3

H C

H

H N: C :Cl

H

-

CH3

Chloroethane Ethyl ammonium chloride

H H

3H N3

+

C

H

HO:-

CH3

H

N+

H

H

H

C

H

CH3

H

N

H

H

••

Aminoethane

H O2

O

5H C C H

2 52

δ−

δ+δ+

OCH

33H C

O

3H C

O

7H CC H

2 5C H4 93

(Dimethylether)

Methoxymethane

(Ethylmethylether)

Methoxyethane

(Butylpropylether)

Propoxybutane


● Ethers of very low formula mass dissolve in water by hydrogenbonding e.g. Methoxymethane (Dimethylether) dissolves in water:

Higher members are less soluble in water. The increasing length of the alkyl group increases the strength of the van der Waals attraction between the ether molecules themselves. Water molecules are unable to break this attractive force.e.g. Butoxybutane (Dibutylether)

● They have low boiling points due to the lack of hydrogen bonding between the molecules.

● They are generally unreactive but:-

(1) they are highly flammable

(2) they slowly react with the Oxygen in the air forming explosive peroxides (-O-O-)

Ethers make good solvents because of their polarity, low boilingpoints and low reactivities.

Alcohols

The presence of the OH group enables alcohols to form hydrogenbonds.

Hydrogen bonding occurs between molecules in liquid and solidalcohols e.g. in Ethanol

OCH

33H C

H

H

O

O H

H

O

O

van der Waals forces

CH2CH

2CH

2CH

3

CH2CH

2CH

2CH

3

CH3CH

2CH

2CH

2

CH3CH

2CH

2CH

2

CH3

CH2

O

H

CH3

CH2

O

H

hydrogen bond


Alcohols thus have higher melting points and boiling points thanother organic compounds of comparable formula mass and shape e.g. BP of Ethanol is 78 0C; BP of isomeric Dimethylether is -24 0C.

Hydrogen bonding is also responsible for the solubility of thelower alcohols in water e.g. Ethanol

Higher members are less soluble in water. The increasing length ofthe alkyl group increases the strength of the van der Waalsattraction between the alcohol molecules themselves. Watermolecules are unable to break this attractive force e.g. Pentan-1-ol is only sparingly soluble in water

Aldehydes and Ketones

Both aldehydes and ketones contain the highly polar

carbonyl group Cδ+ =Oδ−.

Lower members dissolve in water by

hydrogen bonding e.g. Propanone

CH3

CH2

O

H

O

H

H

CH2

CH2

CH2

CH2

CH3

O

H

CH2

CH2

CH2

van der Waals forces

CO

CH3

3H C

δ−

HH

Oδ+

C H4 9

CO

H

Pentanal(Aldehyde)

C H3 7

CO

Pentan-2-one(Ketone)

CH3

Examples:-

CH3

CH2

O

H O

H

hydrogen bond

H


Inter-molecular bonding in solids and

liquids is by attraction between the

polar bonds e.g. Propanone

Aldehydes and ketones are more polar and therefore have higherboiling points than the parent alkanes:

Aldehydes and ketones cannot form inter-

molecular hydrogen bonds and therefore

have lower boiling points than the

corresponding alcohol.

Reactions of Aldehydes and Ketones

1. Reduction

Aldehydes are prepared by oxidation of primary alcohols e.g.

Ketones are prepared by oxidation of secondary alcohols e.g.

CH2

BP -42 C0

CO

3H C

BP 49 C0

CO

CH3

3H C

BP 56 C0

H

CH3

CH2

CH3

Propane Propanal Propanone

BP 97 C0

CH3

CH2

CH2

OH

Propan-1-ol

CO

CH3

3H C

CO

CH3

3H C δ+

δ−

δ−

δ+

CC H2 5

H

H

OH C H2 5

CO

H

Oxidising agent

Propan-1-ol Propanal

CC H2 5

H

OH C H2 5

COOxidising agent

Butan-2-ol Butanone

CH3

CH3


Since oxidation involves LOSS OF HYDROGEN, both aldehydes and ketonescan be converted back to their parent alcohols by reduction (GAIN OFHYDROGEN). This is done using a solution of Lithium aluminium hydridein Ethoxyethane solvent e.g.

2. Oxidation

Aldehydes can be oxidised to carboxylic acids by a variety ofoxidising agents.

e.g. the oxidation of Propanal to Propanoic acid

(1) With acidified Potassium dichromate.

(2) With Benedict's solution - a solution containing OH- and Cu2+.Cu2+ is complexed with the citrate ion to prevent the reactionbetween Cu2+ and OH-.

(3) With ammoniacal Silver(I) nitrate (Tollen's reagent) - a solution containing OH- and Ag+. Ag+ is complexed with Ammonia toprevent the reaction between Ag+ and OH-.

Ketones CANNOT BE OXIDISED and therefore do not undergo the abovethree reactions. Since these reactions all involve colour changesthey can be used to easily distinguish between aldehydes and ketones.

CC H2 5

H

OHC H2 5

CO

Butan-2-olButanone

CH3

CH3

Li AlH4

+ -

CO

3H C

H

Cr O2 7

2-CO

3H C

OH

Cr3+

H+

H O2

orange green

CO

3H C

H

Cu2+

CO

3H C

OH

(Cu ) O+OH-

H O2

blue brick-red

2

CO

3H C

H

Ag+

CO

3H C

OH

AgOH-

H O2

colourless silver mirror


3. Addition

Both aldehydes and ketones undergo nucleophilic addition reactions.e.g. reaction of Ethanal with acidified Sodium cyanide.

Mechanism (two steps)

STEP 1:

Nucleophilic attack by the cyanide ion at the Cδ+ of the C=O group:

STEP 2:

Ketones undergo nucleophilic additionmore slowly.

The presence of TWO alkyl groups:

● hinders attack by the nucleophile ● reduces the δ+ charge on the

carbonyl Carbon atom due to the alkyl group's inductive effect

C

O

CH3

H

:CN

CH3

C

:O

CN

H

-

δ+

δ−-

CO

CH3

H

CN H CH3

C

OH

CN

H+-

Ethanal Ethanal cyanohydrin

CH3

C

:O

CN

H

-

H+

CH3

C

OH

CN

H

C

O

CH3

:CN-

δ+

δ−

CH3


4. Condensation

Both aldehydes and ketones undergo condensation reactions. e.g.reaction of Propanone with 2,4-Dinitrophenylhydrazine.

Mechanism (three steps)

STEP 1:

Nucleophilic attack by 2,4-Dinitrophenylhydrazine at the Cδ+ of theC=O group:

STEP 2:

Proton transfer:

STEP 3:

Proton transfer and subsequent elimination of water:

The melting points of the resulting 2,4-Dinitrophenylhydrazones areused to identify carbonyl compounds.

O

NO2

O N2

N N

H

HH

NO2

O N2

N N

H

C

CH3

CH3

H O2

C

CH3

CH3

Propanone 2,4-dinitrophenylhydrazone

O

NO2

O N2

N N:

H

HH

C

CH3

CH3

δ+ δ−O:

NO2

O N2

N N

H

HH

C

CH3

CH3

+ -

O:

NO2

O N2

N N

H

HH

C

CH3

CH3

+ -

H+

OH

NO2

O N2

N N

H

H

C

CH3

CH3

OH

NO2

O N2

N N

H

H

C

CH3

CH3

NO2

O N2

N N

H

C

CH3

CH3

H O2

H+


Carboxylic Acids

One method of preparing Carboxylic acids is by the oxidation ofaldehydes e.g.

In pure carboxylic acids, hydrogen bonding produces dimers. Thelarger surface area of these dimers results in stronger van derWaals bonding between the dimers and high boiling points e.g.Ethanoic acid BP 118 0C.

Dimerisation does not occur in aqueous solution. Instead, hydrogenbonds form between the acid and water molecules thus explainingthe appreciable solubility of the lower acids in water:

As the chain length increases water solubility decreases.

Reactions of Carboxylic Acids

1. Weak Acids

e.g. Ethanoic acid, Ka = 1.7 x 10-5 mol l-1

OCO

H3

H Cδ+

δ−H O

CH3

CO

δ−

δ+

OCO

H3

H C δ+

H

Oδ−

H

C H2 5

CO

H

Oxidising agent

Propanal

C H2 5

CO

OH

Propanoic acid

OCO

H3

H CO:

CO

H3

H C +-

Ethanoate ion


Ionisation in carboxylic acids occurs more readily thanionisation in alcohols e.g. Ethanol Ka = ~ 1.0 x 10-16 mol l-1

Resonance in the ethanoate ion reduces the negative charge on O-:

This renders O less attractive to attack by H+ and thusreduces reversal. There is therefore a fairly highconcentration of H+ ions free in a solution of Ethanoic acidin water.

Resonance cannot occur in the ethoxide ion. Reversal is muchmore extensive and there are thus fewer H+ ions in solution ofEthanol in water.

Carboxylic acids undergo the normal reactions of acids e.g.

(a) Reaction with metals

e.g. CH3COOH + Na -> CH3COO-Na+ + H2

(b) Reaction with bases

e.g. CH3COOH + Na+OH- -> CH3COO-Na+ + H2O

e.g. CH3COOH + Mg2+CO32- -> (CH3COO-)2Mg2+ + CO2 + H2O

2. Reaction with Alcohols

Carboxylic acids undergo condensation reactions with alcohols, in the presence of concentrated Sulphuric acidcatalyst, to form esters e.g.

Ethoxide ion

C

H

H

C

H

H

O

H

H C

H

H

C

H

H

O:H H+-

O:

C

O

3H C

-O

C

O:

3H C

-

OCO

HC H2 5

CH3

HO OCO

C H2 5

H O

Propanoic acid Methanol Methyl propanoate

2

H+

CH3


The reaction is slow. Esters are therefore often prepared bythe much faster reaction between an alcohol and an acidchloride e.g.

Since esterification is reversible carboxlic acids can beprepared by hydrolysis of the appropriate ester e.g.

3. Reaction with Ammonia

Carboxylic acids react with Ammonia forming the ammonium salte.g.

Subsequent heating of the ammonium salt produces an amide:

4. Reduction

Carboxylic acids can be reduced to the corresponding primaryalcohol with a solution of Lithium aluminium hydride inEthoxyethane solvent e.g.

O

C

O

H HOO

C

O

H O

Methanoic acidEthyl methanoate

2

H+

C H2 5

C H2 5

H H

ClCO

C H2 5

CH3

HO OCO

C H2 5

HCl

Propanoyl chloride Methanol Methyl propanoate

CH3

OCO

HC H3 7

Butanoic acid

NH3

OCO

C H3 7

Ammonium butanoate

NH4

+-

OCO

C H3 7

NH4

+-

NHCO

C H3 7

2

H O2

Butanamide

C

H

OHCO

EthanolEthanoic acid

OH

Li AlH4

+ -

3H C

3H C

H


Amines

Amines are compounds of general formula NR3 where R = alkyl or H.

Amines are classified according to the number of alkyl groups (R)bonded to the Nitrogen atom:

Primary and secondary amines, but not tertiary amines, associateby hydrogen bonding e.g. 1-Aminopropane

As a result, primary and secondary amines have higher boilingpoints than isomeric tertiary amines and alkanes with comaparableformula mass e.g.

C H3 7

N

H

HC H3 7

N

H

H

C H3 7

N

H

H

N CH3

3H C

3H C

C H4 10

1-Aminopropane N,N-Dimethylaminomethane Butane

BP 49 C0 BP 7.5 C0 BP 0 C0

N

R

H

H

e.g.

Aminomethane

PRIMARY SECONDARY TERTIARY

N

R

H

R N

R

R

R

N

H

H N

H

N

3H C

3H C

3H C

C H2 5

3H C

3H C

(Methylamine)

N-Methyl aminomethane

(Dimethylamine)

N,N-Dimethyl aminoethane

(Ethyldimethylamine)


Amine molecules can hydrogen bond with water molecules thusexplaining the appreciable solubility of the lower amines in watere.g. Aminoethane

Reactions of Amines

The Nitrogen atom has a lone pair of electrons which can acceptprotons e.g.

Amines are thus weak bases.

They react with Water to form alkaline solutions e.g.

They react with other acids to form salts.

Example 1.

Reaction of N-Ethylaminoethane with dilute Hydrochloric acid:

Example 2.

Reaction of Aminomethane with Ethanoic acid:

C H2 5

N

H

H

H

H

O

H OHN:

H

H

3H C

:OHN

H

H

CH3

+H

-

N:

H

H

3H C

H+ N

H

H

CH3

+H

N:

H

H+ N

H

+H

C H2 5

C H2 5

Cl- Cl-

C H2 5

C H2 5

N:

H

3H C

H

O

H CH3

CO

N

H

+H

CH3

H :O

CH3

CO

-


Aromatics

Compounds containing a benzene ring are described as aromatic;compounds which do not, are described as aliphatic.

The Carbon atoms in Benzene are SP2 hybridised.Bonding within the ring is a combination of σ bonding between the SP2 hybrid orbitals and π bonding between the Pz orbitals:

C-H bonds are formed by σ overlap:

The 6 electrons in the π orbital move around the ring - theydelocalise. Not confined to a single bond where they would beattracted by only two nuclei, these delocalised electrons areattracted by six nuclei and are therefore stabilised and stronglyattracted to the ring. It is therefore paradoxical that it it theseelectrons which take part in the reactions of Benzene!

π

σC C

C

CC

C

H H

H

HH

H

σ

σ bonding π bonding


Reactions of Benzene

In most of the reactions of Benzene, the ring remains intact, heldtogether by the delocalised electrons. Reactions usually involvesubstitution of the Hydrogen atoms.

Example

Reaction with concentrated Nitric acid using concentrated Sulphuricacid catalyst:

The reaction is known as nitration.

Mechanism (three steps)

STEP 1:

Nitric acid and Sulphuric acid react:

HNO3 + H2SO4 -> HSO4- + H2O + NO2+

STEP 2:

The nitronium ion NO2+ is an electrophile. It attracts two of thedelocalised electrons in the Benzene ring and bonds to one of theCarbon atoms:

STEP 3:

The delocalisation has been disrupted! Two electrons must re-enterthe ring immediately! The HSO4- ion attacks:

This is substitution of Hydrogen by the electrophile NO2+ and istherefore known as electrophilic substitution.

H OHNO 3

H SO42

NO2

2

Nitrobenzene

NO2

+

NO2

+ H

NO2

+ H

NO2

4:O HS-

H SO2 4


Other electrophilic substitution reactions of Benzene

With Chlorine or Bromine e.g.

With conc. Sulphuric acid, a reaction known as sulphonation:

With Chloroalkanes to form alkyl benzenes e.g.

Other Aromatic Compounds

Hydroxybenzene (Phenol)

Phenol (Ka = 1.28 x 10-10 mol l-1) is a stronger acid than aliphaticalcohols.

Ionisation in Phenol occurs more readily than ionisation in alcohols

HCl

Chlorobenzene

Cl

Cl2

H O

Benzene sulphonic acid

H SO2 4

SO H3

2

HCl

Ethyl benzene

C H2 5Cl

C H2 5

O H+

Phenoxide ion

O:-

Ethoxide ion

C

H

H

C

H

H

O

H

H C

H

H

C

H

H

O:H H+-

K = 1.28 x 10-10

mol l-1a

K ~ 1.0 x 10 mol l-1a-16

Phenol

Ethanol

H


Delocalisation of electrons into the Benzene ring in the phenoxideion reduces the negative charge on O-:

This renders O less attractive to attack by H+ and thus reducesreversal. There is therefore a fairly high concentration of H+ ionsfree in a solution of Phenol in water.

Delocalisation cannot occur in the ethoxide ion. Reversal is muchmore extensive and there are thus fewer H+ ions in a solution ofEthanol in water.

Aminobenzene (Aniline)

Aniline is a weaker base than aliphatic amines.

Donation of the lone pair is more difficult in Aniline:

Delocalisation of the lone pair into the Benzene ring in Anilinemakes donation of the lone pair to H+ more difficult.

Delocalisation cannot occur in aliphatic amines. Donation of thelone pair to H+ is therefore easier.

O:-

N:

H

H

N:+

Aniline

H

H

N

H

H

HH+

N:+

Aminomethane

H

H

N

H

H

HH+CH3

CH3

difficult

easy

advanced higher 1 organic organic chemistry · advanced higher 1 organic organic chemistry bonding...

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