3.3.3 Halogenoalkanes

Chloroflurocarbons are halogenoalkane molecules with not hydrogen atoms but chlorine and fluorine atoms

Ozone (O3) is in the upper atmosphere and absorbs UV radiation from the sun, UV radiation can cause sunburn and skin cancer

Ozone is formed naturally when an oxygen molecule is broken down into two free radicals by UV radiation. The free radicals attack other oxygen molecules forming ozone. O2 -UV-> O. + O. then O2 + O. -> O3

• Chlorine free radicals (Cl.) are formed in the upper atmosphere when Cl-Cl bonds in CFC’s are broken down by ultraviolet radiation - e.g. CCl3F -UV-> CCl2F. + Cl. • These free radicals are catalysts. They react with the ozone to form an intermediate (ClO.) and an oxygen molecule - e.g. Cl. + O3 -> O2 + ClO. or ClO. + O3 -> 2O2 + Cl. • Since the chlorine is regenerated it can continue to attack other ozone molecules, so many will be destroyed • The overall reaction is… 2O3 -> 3O2 … and Cl. is the catalyst

• CFC’s are unreactive, non-flammable, and non-toxic. They used to be used as coolant gas in fridges, as solvents, and propellants in aerosols • In the 1970’s research by several different scientific groups showed that CFC’s cause damge to the ozone layer • The advantages of CFC’s weren’t worth the envriomental problems caused so they were banned • Chemists have developed safer alternatives to CFC’s which contain no chlorine e.g. HFC’s (hydrofluorocarbons) and hydrocarbons

A halogenoalkane is an alkane with at least one halogen atom inplace of a hydrogen - e.g trichloromethane

• Halogens are significantly more electronegative than carbon, therefore the bond is polar • The δ+ charge on the carbon makes it prone to attack by nucleophiles Halogens are prone to attack from nucleophiles such as OH-, CN- and NH3

A nucleophile is an electron pair donor It donates a pair electons to electron deficient places

• The lone pair of electrons on the nucleophile attack the δ+ carbon in a polar bond, this causes the bond to break • The atom or functional group attached to the δ+ carbon leaves and a new bond between the δ+ carbon and the nucleophile is formed The product of nucleophillic substition reactions depend of what the nucleophile is - e.g. OH-, CN-, or NH3

One of the halogens - e.g. F, Cl, Br, I

Nucleophile

• The δ+ carbon attracts a lone pair of electrons from the OH- ion. • The C-Br bond breaks • The bromine leaves, taking both electrons to become :Br- • A new bond is formed between the carbon and the OH- ion, making an alcohol This is sometimes called hydrolysis beacause the same reaction can occur with water

Splitting a molecule apart by reacting it with water

Warm aqueous sodium or potassium hydroxide

• Warm under reflux • Ethanolic potassium cyanide

• Warm • excess ethanolic ammoinia

• The lone pair of electrons on the ammonia is attracted to the δ+ carbon in the C-X bond • This causes the bond to break, Br leaves taking a lone pair with it to form :Br- • The ammonia molecule forms a bond with the δ+ carbon, the ammonia on the halogenoalkane has a positve charge • Next an ammonia removes a hydrogen atom from the ammonia bonded to the alkane, forming an ammonium ion (NH4+) and an amine group on the alkane • The ammonium ion can react with the bromine ion to form ammonium bromide Overall reaction: CH3CH2Br + 2NH3 -ethanol-> CH3CH2NH2 + NH4Br The amine group on the product still has a ion pair so can act as a nucleophile, so can react with the halogenoalkane itself

• The carbon-halogen bond strength (or enthalpy) decides reactivity, for any reaction to occur this bond must be broken • The C-F bond is the strongest as it has the highest bond enthalpy, therefore fluoroalkanes undergo nucleophillic substitution reactions more slowly than other halogenoalkanes • The C-I bond has the lowest bond enthalpy so iodoalkanes are substiuted more quickly

-Fastest (lowest bond enthalpy)- • Iodoalkanes • Bromoalkanes • Chloroalkanes • Fluoroalkanes -Slowest (highest bond enthalpy)-

• Warm under reflux • Ethanolic/ anhydrous (no water)

• OH- acts as a base and takes a proton, H+ from the carbon adjacent to the δ+ C, this makes water • The adjacent carbon now has a spare electron, so it forms a double bond with the δ+ carbon • To form the double bond, the bond between the δ+ C and the Br atom must be broken, the bromine atom leave as a :Br- ion

CH3CHBrCH3 + KOH -ethanol/reflux-> CH2CHCH3 + H2O + KBr

Nucleophillic substitution: In aqueous conditions, OH- acts as a nucleophile Elimination: Anhydrous conditions, OH- acts as a base If you use both, water and ethanol as solvents, both reactions happen and you get a mixture of products