Halogenalkanes (Haloalkanes)
Halogen alkanes are named by adding the prefix chloro/bromo instead of
methyl. E.g.
Similar to alcohols halogenalkanes can also be primary, secondary or tertiary, depended on the position of the C-Cl bond.
C-Cl Bond Polarity
The C-Cl bond is polar because there is a large difference in electronegativity, which results in the Cl atom having a higher ‘concentration’ of electrons and results in a permanent dipole.
Because of this polarity, halogenalkanes can undergo neucleophilic substitution
reactions. A reaction mechanism that is similar to electrophilic addition
for alkanes.
Neucleophilic Substitution
Heterolytic Fission
Unequal breaking of a covalent bond resulting in a positive and negative
ion.
Nucleophile
An electron pair donor
1. Hydrolysis with alkali
Conditions: Heat, water solvent
Overall Reaction
CH3CH2Br + NaOH 
CH3CH2OH + NaBr
Again this reaction actually occurs in a series of steps
1. H–O- Na+
Na+ + H–O-
2.
3. Na+ + Br-
NaBr
Step 1
The NaOH is actually ionic Na+OH- and there are
also OH- ions in the water (the NaOH is aqueous in water for
this reaction) the OH- is referred to as a nucleophile –
An electron pair donor
Step 2
The OH ion is attracted to the Cd+ and forms a dative covalent
bond with it. The C-Br bond undergoes heterolytic fission to result in
CH3CH2OH and Br-.
Step 3
The Na+ and Br- react to form NaBr
2. Hydrolysis with water
Conditions: mixed with ethanol to force the hydrated and hydrophobic layers
to mix.
CH3CH2Cl + H2O
CH3CH2OH + HCl
CH3CH2Br + H2O
CH3CH2OH + HBr
CH3CH2I + H2O
CH3CH2OH + HI
There are different rates of hydrolysis dependent on which halogen alkane reacts. Iodine has the quickest rate of hydrolysis because it has a lower bond enthalpy, due to its larger atom and lower electronegativity. Therefore this acts as proof that the C-I bond is a weaker covalent bond.
You can test the rate of hydrolysis by using silver nitrate:
Ag+(aq) + Cl- AgCl(s)
White precipitate
Ag+(aq) + Br- AgBr(s)
Cream precipitate
Ag+(aq) + I- AgI(s)
Yellow precipitate
The yellow precipitate will form first, then the cream one and then finally the white precipitate.
3. Reaction with excess ammonia
Conditions: ethanol solvent
CH3CH2Br + 2NH3
CH3CH2NH2 + NH4Br
Forms an amine, in this case ethylamine.
4. Elimination Reaction
Conditions: ethanol solvent, heat
CH3CH2Br + NaOH
CH2=CH2 + NaBr + H2O
Forms an alkene.
Recap - Reactions that produce halogenalkanes
Alkanes
Free radical substitution – Cl2 in the presence of UV
light
Alkene
Electrophilic addition – HCl or Cl2 via Refulx
Alcohol
Substitution – HBr made in situ.