1. Halogenoalkanes L.O. To be able to name and draw halogenoalkanes.
To know how halogenoalkanes react and the factors affecting their reactivity.
Pesticides no longer used
Refrigerants
Halon fire extinguishers
Solvents

2. How could you assign primary, secondary and tertiary notation?
HALOGENOALKANES
Alkanes with H atom(s) substituted by HALOGEN atom(s)
A. CH3CH2F
C. CH3CHClCHClCH2CH3
B. CH3CHBrCH3
D. CH3CH(CH3)CH2CH2Cl
Fluoroethane
2,3-dichloropentane
2-bromopropane
1-chloro-3-methylbutane
Name them
How could you assign primary, secondary and tertiary notation?

3. MAKING HALOGENOALKANES
[ X = usually Cl or Br ]
From ALKANES
by Free Radical Substitution
C-H + X2
C-X + HX
uv energy to initiate
e.g. CH3CH3 + Cl2
CH3CH2Cl + HCl
(chloroethane)
Inefficient because multiple substitutions can occur
From ALKENES
by Electrophilic Addition
>C=C< + HX
>CH-CX<
e.g. CH2=CH2 + HBr
CH3CH2Br
(bromoethane)
Mixtures produced if the alkene is unsymmetrical

4. REACTIONS OF HALOGENOALKANES (1)
(a) Carbon-halogen bond is POLAR:
+ C-X -
So, C attacked by ELECTRON-RICH PARTICLES
Such as NUCLEOPHILES.
(b) Carbon-halogen bond is SINGLE
So, C attack results in SUBSTITUTION of X
(a) and (b)  NUCLEOPHILIC SUBSTITUTION reactions
(c) Polarity of C-X bond :
Strength of C-X bond :
Rate of nuc. sub. of C-X :
C-F > C-Cl > C-Br
C-F > C-Cl > C-Br
C-Br > C-Cl > C-F
N.B: rate controlled by strength of C-X bond, NOT polarity

5. REACTIONS OF HALOGENOALKANES (2)Br
If 2 or more carbons in halogenoalkane
HBr can be removed creating a C=C bond
i.e. alkene(s) C
ELIMINATION REACTIONS
as well as NUCLEOPHILIC SUBSTITUTION reactions

6. R-OH R-OH R-X R-NH2 R-CN + HX + K+X- + NH4+X- + K+X-
Nucleophilic substitution: Appropriate nucleophiles :
H2O:
:OH-
:NH3
:CN-
+ H2O
R-OH
+ HX
Heat
(X = any halogen, but usually Cl or Br)
ALCOHOL
+ K+OH-(aq)
R-OH
+ K+X-
Heat
R-X
ALCOHOL
+ 2NH3(ethanol)
R-NH2
+ NH4+X-
(R = any alkyl group, CnH2n+1)
Heat
, excess NH3
AMINE
+ K+CN-(ethanol)
R-CN
+ K+X-
Heat
NITRILE
Notes :
(1) Nitrile, R-CN, contains one more C than original C-X
(2) Ethanol solvent used with NH3 and KCN to avoid alcohol production by reaction with aqueous solvent.

7. Mechanisms for Nucleophilic Substitution Reactions
1. Polar C-Halogen bond breaks heterolytically
 carbocation + bromide H Br C R H R C+
SN1 and Sn2 reactions depending of halogenoalkane
+ :Br-
δ+ δ-
or NH3 molecule bonds to carbocation
 amine after H+ loss
2. OH- ion bonds to carbocation
or CN- ion bonds to carbocation
 alcohol
 nitrile H R C+ H R C+ H R C+
:NH3
:OH-
:CN-
R-CH2-NH3 +
+NH3
R-CH2-OH
R-CH2-CN
R-CH2-NH2
+NH4+

8. Aminoethane (or ethylamine)
Write an equation and name the organic product when:
1. Bromoethane reacts with concentrated ammonia
         
2. 2-Chloropropane reacts with hot water   
       
   
3. 3-Iodopentane reacts with warm potassium hydroxide soln
4. 1-Chloropropane reacts with warm potassium cyanide soln.
CH3CH2Br + 2NH3
 CH3CH2NH2 + NH4Br
Aminoethane (or ethylamine)
CH3CHClCH3 + H2O
 CH3CH(OH)CH3 + HCl
Propan-2-ol
CH3CH2CHICH2CH3 + KOH
 CH3CH2CH(OH)CH2CH3 + KI
Pentan-3-ol
CH3CH2CH2Cl + KCN
 CH3CH2CH2CN + KCl
Butanenitrile

9. This is called SN2 process and is illustrated in the next slide
The C-Halogen bond may break FIRST. This is then FOLLOWED by the formation of the C-Nucleophile bond.
This is called an SN1 process and was illustrated in the previous slide
Alternatively, the C-Halogen bond may break SIMULTANEOUSLY with the formation of the C-Nucleophile bond.
This is called SN2 process and is illustrated in the next slide
Whether SN1 or SN2 type applies depends mostly on the type of halogenoalkane being used.
Primary halogenoalkane (e.g. 1-bromobutane)  SN2 route
Tertiary halogenoalkane (e.g. 2-bromo-2-methylpropane)  SN1 route

10. Mechanism for the SN2 Hydrolysis of a Halogenoalkane
Aqueous NaOH or KOH provide the :OH- nucleophile C H X R C H R HO
δ δ-
-HO:
+ X-
Halogenoalkane; X = a halogen
Alcohol
1. C-X bond is polar because halogen atoms are electronegative
2. Lone pair of OH- nucleophile attracted to Cδ+
3. Lone pair of OH- forms C-O bond (i.e. forming an alcohol)
simultaneously causing the C-X bond to break heterolytically,
releasing a halide ion, X-

11. KOH in ethanol equally suitable
ELIMINATION REACTIONS OF HALOGENOALKANES
Removal of H and halogen (e.g. Br) from adjacent C atoms in a halogenoalkane
 alkene(s)
NB H atom may be removed from C on either side of halogen!  mixture of alkenes possible
Eliminated HBr is ACIDIC
 achieved using STRONG BASE
 hot, conc. NaOH(ethanol) suitable.
 converts HX to salt (NaBr) and water H C Br
N.B. OH- is a NUCLEOPHILE as well as being a BASE  elimination will compete with nucleophilic substitution to form an alcohol.
+ Na+OH-(ethanolic)
KOH in ethanol equally suitable C
+ Na+Br HOH

12. Mechanism of Elimination ReactionBr H C
1. C-Halogen bond breaks heterolytically
 carbocation
2. Basic hydroxide ion takes adjacent proton (H+) from carbocation
 water H C+ C
+ :Br-
3. 2e- of C-H bond move towards C+
 alkene
:OH-
REM :OH- is also a nucleophile
- may bond direct to carbocation C
+ HOH
+ :Br-
Giving alcohol too

13. Mixture of but-1-ene and cis and trans but-2-ene
(d) 2-chloro-2-methylpropane : (CH3)2CClCH3
Predict the products when (a)-(d) are reacted with a hot, concentrated, ethanolic solution of the base shown :
(a) Bromoethane : CH3CH2Br
(b) 2-chloropropane : CH3CHClCH3  
(c) 2-bromobutane : CH3CHBrCH2CH3
ethene
CH3CH2Br + NaOH 
CH2=CH2
+ NaBr + H2O
CH3CHClCH3 + KOH 
CH3CH=CH2
+ KCl + H2O
propene
CH3CHBrCH2CH3 + NaOH 
CH3CH2CH=CH2
+ NaBr + H2O
+ CH3CH=CHCH3
Mixture of but-1-ene and cis and trans but-2-ene
(CH3)2CClCH3 + KOH 
(CH3)2C=CH2
+ KCl + H2O
methylpropene