Types of homolytic fission

• Since bond breaking is an endothermic process, energy is required for homolytic fission to occur

• The amount of energy required depends on the strength of the covalent bond being broken

  • Thermolytic fission: For weaker bonds, simply heating the compound could provide sufficient energy

XX X• + X• 

• Photolytic fission: For stronger bonds such as halogen bonds, exposing the compound to high-energy UV light provides the required energy

XX X• + X• 

Stability of alkanes

• Alkanes are relatively stable / unreactive due to the strengths of the C–C and C–H bonds and their non-polar nature

Strength of bonds

• Alkanes consist of carbon and hydrogen atoms which are bonded together by single bonds

• Unless a lot of heat is supplied, it is difficult to break these strong C-C and C-H covalent bonds

• This decreases the alkanes’ reactivities in chemical reactions

Lack of polarity

• The electronegativities of the carbon and hydrogen atoms in alkanes are almost the same

• This means that both atoms share the electrons in the covalent bond almost equally

 Pauling electronegativity values for the elements

The Pauling Scale shows that the difference in electronegativity between carbon and hydrogen is only 0.4

• As a result of this, alkanes are nonpolar molecules and have no partial positive or negative charges (δ⁺ and δ^–  respectively)

Structural formula of ethane showing bond polarities

Ethane is an example of an alkane that lacks polarity due to almost similar electronegativities of the carbon and hydrogen atoms

• Alkanes, therefore, do not react with polar reagents

  • They have no electron-deficient areas to attract nucleophiles

  • And also lack electron-rich areas to attract electrophiles

• Alkanes only react in combustion reactions and undergo substitution by radicals

Free-radical substitution of alkanes

• Alkanes can undergo free-radical substitution in which a hydrogen atom gets substituted by a halogen (chlorine/bromine)

• Since alkanes are very unreactive, ultraviolet light (sunlight) is needed for this substitution reaction to occur

Proving that energy from UV light is required for radical reactions with halogens

The fact that the bromine colour has disappeared only when mixed with an alkane and placed in sunlight suggests that the ultraviolet light is essential for the free radical substitution reaction to take place

• The free-radical substitution reaction consists of three steps:

  • In the initiation step, the halogen bond (Cl-Cl or Br-Br) is broken by UV energy to form two radicals

    • For more information about the initiation step, see our revision note about Homolytic Fission

  • These radicals create further radicals in a chain reaction called the propagation step

  • The reaction is terminated when two radicals collide with each other in a termination step

Propagation step

• The propagation step refers to the progression (growing) of the substitution reaction in a chain reaction

  • Radicals are very reactive and will attack the unreactive alkanes

  • A C-H bond breaks homolytically (each atom gets an electron from the covalent bond)

  • An alkyl free radical is produced

  • This can attack another halogen molecule to form the halogenoalkane and regenerate the halogen radical

  • This radical can then repeat the cycle

• Industrial processes

• Polymer formation

• Environmental chemistry (ozone depletion)

• Biological processes

Important Equations General format: RH+X2→UV lightRX+HXGeneral format: RH+X2​UV light​RX+HX Where:

• R = alkyl group

• X = halogen

• H = hydrogen

Characteristics:

Have lone pairs or negative charge

Examples: OH⁻, CN⁻, NH₃

Look for lone pairs or negative charges to identify nucleophiles

Diagram to show the process of heterolytic fission

Heterolytic fission forms a positive ion and a negative ion

• In heterolytic fission, a double-headed arrow is used to show the movement of a pair of electrons

• The resulting negative ion is an electron-rich species that can donate a pair of electrons

  • This makes the negative ion a nucleophile

• The resulting positive ion is an electron-deficient species that can accept a pair of electrons

  • This makes the positive ion an electrophile

Heterolytic fission mechanism

• The opposite reaction to heterolytic fission occurs when a nucleophile donates a pair of electrons to the electrophile forming a coordination bond

Electrophilic Addition Reactions

• Electrophilic addition is the addition of an electrophile to an alkene double bond, C=C

• The alkene double bond, C=C, is an area of high electron density which makes it susceptible to attack by electrophiles

• The C=C bond breaks forming a single C-C bond and 2 new bonds from each of the two carbon atoms

• Electrophilic addition reactions include the addition of:

  • Steam, H₂O (g) to form alcohols

  • Hydrogen halides, HX , to form halogenoalkanes

  • Halogens, X₂, to form dihalogenoalaknes

Why does the C=C bond react with electrophiles?

• Alkenes are unsaturated molecules that contain a C=C bond

• The atoms around the carbon-carbon double bond adopt a planar arrangement and the bond angle is 120^o

Diagram to show the planar arrangement of the C=C bond

The bond angles are 120^o

• The presence of the C=C bond gives alkenes a number of chemical properties that are not seen in alkanes

• Since the alkene contains π-bonds, it is possible to break the weaker π-bond and form stronger  σ-bonds with other species without forcing any atoms on the molecule to break off

• As a result alkenes (unlike alkanes) are capable of undergoing addition reactions

• The ability of alkenes to undergo addition means that they are much more reactive than alkanes

Diagram to show the general equation for addition reactions across the C=C

Addition reactions in alkenes

Addition of water

• When alkenes are treated with steam at 300 ^oC, a pressure of 60 atmospheres and sulfuric acid (H₂SO₄) or phosphoric acid (H₃PO₄) catalyst, the water is added across the double bond in a reaction known as hydration

• An alkene is converted into an alcohol

• The reaction processes via an intermediate in which H⁺ and HSO₄^-  ions are added across the double bond

• The intermediate is quickly hydrolysed by water, reforming the sulfuric acid

• The following equation shows the conversion of ethene to ethanol

CH₂CH₂  CH₃CH₂OH

ethene               ethanol

• This is a very important industrial reaction for producing large quantities of ethanol, a widely used solvent and fuel

• The process is much faster and higher yielding that producing ethanol by fermentation

Addition of halogens

• The reaction between alkenes and halogens is known as halogenation

• It is an example of an electrophilic addition where an electrophile ('electron seeker') joins onto to a double bond

• The C=C double bond is broken, and a new single bond is formed from each of the two carbon atoms

• The result of this reaction is a dihalogenoalkane

This reaction occurs readily at room temperature and is the basis for the test for unsaturation in molecules

Halogenation in alkenes

• Halogens can be used to test if a molecule is unsaturated (i.e. contain a double bond)

• Br₂ is an orange or yellow solution, called bromine water

• The unknown compound is shaken with the bromine water

• If the compound is unsaturated, an addition reaction will take place and the coloured solution will decolourise

Diagram to show the colour change that occurs when testing for unsaturation

The bromine water test is the standard test for unsaturation in alkenes

Addition of hydrogen halides

• Alkenes will react readily with hydrogen halides such as HCl and HBr to produce halogenoalkanes

• This reaction is known as hydrohalogenation 

• It is also an electrophilic addition reaction that occurs quickly at room temperature

• Haloalkanes will undergo nucleophilic substitution reactions due to the polar C-X bond (where X is a halogen)

Partial positive C atom and partial negative X atom

Due to large differences in electronegativity between the carbon and halogen atom, the C-X bond is polar

Diagram to show nucleophilic substitution where :Nu^– represents the nucleophile

General Mechanism for Nucleophilic Substitution

Hydrolysis of Haloalkanes 

• The nucleophile in this reaction is the hydroxide, OH^– ion

• An aqueous solution of sodium hydroxide (NaOH) or potassium hydroxide (KOH) with ethanol is used

• This reaction is very slow at room temperature, so the reaction mixture is warmed

• This is an example of a hydrolysis reaction and the product is an alcohol

CH₃CH₂Br + OH^– → CH₃CH₂OH + :Br^–

bromoethane      →       ethanol

• :Br^– is the leaving group 

  • Halogens make good leaving groups as they form relatively weak bonds with carbon

  • Their higher electronegativity also means the bonded electrons are drawn towards the halogen atom making the carbon partially positive, δ+, and susceptible to nucleophilic attack

• The rate of this reaction depends on the type of halogen in the haloalkane 

• The stronger the C–X bond, the slower the rate of the reaction

• In terms of bond enthalpy, C–F > C–Cl > C–Br > C–I

• Fluoroalkanes do not react at all, but iodoalkanes have a very fast rate of reaction

The nucleophilic substitution mechanisms for the above reactions are as follows:

Nucleophilic substitution mechanism of bromoethane with a hydroxide ion 

Nucleophilic Substitution with OH^–, the bond that forms and the bond that breaks must both involve the carbon atom that is bonded to the leaving group

Neutral nucleophiles

• When the nucleophile is neutral, e.g. H₂O, the initial product is positive

• The positive product then deprotonates, losing H⁺, and forms a neutral product

  • CH₃CH₂Cl + H₂O → CH₃CH₂OH + :H⁺

Mechanism Steps:

1. Initial Approach

   • Electrophile approaches electron-rich site

   • π bond electrons are the target

   • Formation of induced dipole

2. Bond Formation

   • π electrons attack electrophile

   • Formation of σ bond

   • Generation of intermediate species

3. Product Formation

   • Addition of nucleophile

   • Formation of final product

Bromine Addition to Ethene: C2H4+Br2→C2H4Br2C2​H4​+Br2​→C2​H4​Br2​

• Electron pair acceptors

• Examples: BF₃, AlCl₃, H⁺

• Properties:

  • Empty orbital available

  • Electron deficient

  • Form coordinate bonds

Lewis Bases:

• Electron pair donors

• Examples: NH₃, H₂O, OH⁻

• Properties:

  • Lone pair of electrons

  • Can form coordinate bonds

Lewis Acid-Base Reactions: BF3+NH3→BF3:NH3BF3​+NH3​→BF3​:NH3​

Common Ligands

• Water molecules frequently act as ligands

• Each water molecule makes a single bond with the metal ion using one of the lone pairs on the oxygen atom

• The lone pair is donated to the partially filled d-subshell of the transition metal ion

Representing complex ions

• Square brackets are used to group together the ligands and metal ion in a representation of the geometrical arrangement

• The overall charge on the complex ion is the sum of the oxidation states of all the species present

• If the ligands are neutral then the overall charge will be the same as the oxidation state of the metal ion

Coordination number

• The coordination number is the number of coordinate bonds to the metal ions

• This number can be the same as the number of ligands if they are monodentate

  • It can be different if bidentate or multidentate ligands are present

• Complexes with a coordination number of 4 commonly have a tetrahedral shape, while complexes with a coordination number of 6 have an octahedral shape

Charges of complex ions

• The charge of a complex ion depends upon three factors:

  • The charge of the central metal ion

  • The charge of the ligands

  • The coordination number

• If these factors are known then we can deduce the charge on the complex ion

• Electrophilic addition is the addition of an electrophile (or Lewis acid) to an alkene double bond, C=C

• The alkene double bond, C=C, is an area of high electron density which makes it susceptible to attack by electrophiles

• The C=C bond breaks forming a single C-C bond and 2 new bonds from each of the two carbon atoms

• Electrophilic addition reactions include the addition of:

  • Hydrogen, H₂ (g)

  • Steam, H₂O (g)

  • Hydrogen halides, HX

  • Halogens, X₂

Different reactions of the alkenes including reagents and conditions

Alkene electrophilic addition reaction overview

Addition of hydrogen halides

• A hydrogen halide molecule is polar as the hydrogen and halogen atoms have different electronegativities

  • For example, in a molecule of hydrogen bromide, HBr, the bromine atom has a stronger pull on the electrons in the H-Br bond

  • As a result of this, the Br atom has a partial negative charge and the H atom a partial positive charge

  Partial charges in the H–Br bond

Due to differences in electronegativities of the hydrogen and bromine atom, HBr is a polar molecule

• In electrophilic addition reactions with hydrogen halides, the H atom acts as an electrophile and Lewis acid by accepting a pair of electrons from the C=C bond in the alkene

  • The H-Br bond breaks heterolytically, forming a Br^- ion

• This results in the formation of a highly reactive carbocation intermediate which reacts with the bromide ion, Br^-

• For example, the mechanism for the electrophilic addition of hydrogen bromide and ethene is:

Electrophilic addition mechanism

• Benzene undergoes a wide range of reactions including combustion - (complete and incomplete) and nitration

• Nitration involves the substitution of a hydrogen atom from the benzene ring with an electrophilic atom or group of atoms

Nitration of benzene

In nitration reactions, a nitro (-NO₂) group replaces a hydrogen atom on the arene

• Nitration is an electrophilic substitution reaction

  • This is different to the reactions of unsaturated alkenes, which involve the double bond breaking and the electrophile atoms 'adding across' the carbon atoms

Electrophilic substitution reaction

• The electrophilic substitution reaction in arenes consists of three steps:

  1. Generation of an electrophile

  2. Electrophilic attack

  3. Regenerating aromaticity

Generation of an electrophile

• The delocalised π system is extremely stable and is a region of high electron density

• Consequently, the first step of an electrophilic substitution reaction involves the generation of an electrophile

  • An electrophile can be a positive ion or the positive end of a polar molecule

• The electrophile for nitration is the nitronium ion, NO₂⁺ 

  • This is produced in situ, by adding a mixture of concentrated nitric acid (HNO₃) and concentrated sulfuric acid (H₂SO₄), at a temperature between 25 and 60 ^oC, to the reaction mixture

Electrophilic attack

• The second step in nitration is when a pair of electrons from the benzene ring is donated to the electrophile to form a covalent bond

• This disrupts the aromaticity in the ring as there are now only four π electrons and there is a positive charge spread over the five carbon atoms

The electrophilic attack of the nitronium ion by benzene

Electrons from the benzene π bonding system attack the electrophile

Regenerating aromaticity

• In the third step of electrophilic substitution, the aromaticity of the benzene ring system is restored

• This happens by heterolytic cleavage of the C-H bond

  • This means that the electrons in this bond go into the benzene π bonding system

Breaking a C-H bond to restore aromaticity

The C-H bond breaks heterolytically to restore the aromaticity of the benzene π bonding system

• Alkanes, therefore, do not react with polar reagents

  • They have no electron-deficient areas to attract nucleophiles

  • And also lack electron-rich areas to attract electrophiles

• Alkanes only react in combustion reactions and undergo substitution by radicals

Free-radical substitution of alkanes

• Alkanes can undergo free-radical substitution in which a hydrogen atom gets substituted by a halogen (chlorine/bromine)

• Since alkanes are very unreactive, ultraviolet light (sunlight) is needed for this substitution reaction to occur