15.1 Making halogenoalkanes

The free-radical substitution of alkanes by $Cl_2$ or $Br_2$ in UV light.
The electrophilic addition of an alkene with a halogen, $X_2$ , or a hydrogen halide, $HX$ , at room temperature.
Substitution of an alcohol:
- With $HX$ $H X$ (or with $KX/NaX$ and $H_2SO_4$ or $H_3PO_4$ to make $HX$ )
  - $C_2H_5OH + HCl \rightarrow C_2H_5Cl + H_2O$
  - Type: Nucleophilic substitution
  - Condition: Heat under reflux
- With $PCl_3$ $PC l_{3}$ and heat
  - $3C_2H_5OH + PCl_3 \rightarrow 3C_2H_5Cl + H_3PO_3$
- With $PCl_5$ $PC l_{5}$ at room temperature
  - $C_2H_5OH + PCl_5 \rightarrow C_2H_5Cl + HCl + POCl_3$
- With $SOCl_2$ $SOC l_{2}$
  - $C_2H_5OH + SOCl_2 \rightarrow C_2H_5Cl + HCl + SO_2$


Primary Haloalkanes	$X$ -bonded C bonded to one alkyl group	$C_2H_5Cl$
Secondary Haloalkanes	$X$ -bonded C bonded to two alkyl groups	$CH_3CH(CH_3)Cl$
Tertiary Haloalkanes	$X$ -bonded C bonded to three alkyl groups	$C(CH_3)_3Cl$

15.2 Nucleophilic substitution reactions

🔥 Nucleophilic substitution

The mechanism of the organic reaction in which a nucleophile attacks a carbon atom carrying a partial positive charge, resulting in the replacement of an atom carrying a partial negative charge by the nucleophile.

Haloalkanes are reactive due to the polar nature of the $C-X$ $C - X$ bond (between a carbon and a halogen).
- Carbon $\rightarrow$ partial positive charge
- Halogen $\rightarrow$ partial negative charge

Substitution reactions with aqueous alkali, $OH^-(aq)$

Reagent: Aqueous sodium hydroxide (or a different alkali).
- Behaves as a nucleophile $\rightarrow$ donates a pair of electrons to the $\delta^+ C$ .
Conditions: Heating.
The halogen atom in the halogenoalkane is replaced by a hydroxyl ( $-OH$ ) group, forming an alcohol.
This process is known as hydrolysis.

$C_2H_5Br + NaOH \rightarrow C_2H_5OH + NaBr$

Hydrolysis of a bromoalkane with aqueous sodium hydroxide

The $C-F$ bond is the strongest.
The $C-I$ bond is the weakest $\rightarrow$ broken easily during reactions.

Substitution with cyanide ions, $CN^-$ (in ethanol)

Reagent: Ethanolic $NaCN$ , $KCN$ or any such compound.
Conditions: Heating; under reflux.
Adds an extra carbon atom to the compound (original haloalkane carbon chain).

$C_2H_5Br + CN^- \rightarrow C_2H_5CN + Br^-$

Substitution with ammonia, $NH_3$ (in ethanol)

Reagent: Ethanolic $NH_3$ .
Conditions: Heat; pressure.
A primary amine is formed.
If the ammonia is not in excess, a mixture of amines will be formed $\rightarrow$ primary amines will act as nucleophiles and attack haloalkanes to form secondary amines.

$C_2H_5Br + NH_3 \rightarrow C_2H_5NH_2 + HBr$ (excess $NH_3$ )

$2C_2H_5Br + NH_3 \rightarrow (C_2H_5)_2NH + 2HBr$ (not excess $NH_3$ )

15.3 Mechanism of nucleophilic substitution of halogenoalkanes

The nucleophile attacks the carbon atom bonded to the halogen.
- Nucleophiles are electron pair donors which are attracted to electron-deficient atoms.
The carbon-halogen bond is polarised $\rightarrow$ the halogen is more electronegative than carbon.
The halogen atom is replaced by the nucleophile.

Mechanism for primary halogenoalkanes ( $S_N2$ )

The carbon bonded to the halogen is attached to one alkyl group.
The nucleophile donates a pair of electrons to the $\delta^+$ carbon atom $\rightarrow$ forms a new covalent bond.
The $C-X$ $C - X$ bond breaks at the same time.
- The $X$ atom takes both the electrons (heterolytic fission) $\rightarrow$ an $X^-$ ion is formed.
$S$ - Substitution
$N$ - Nucleophilic
$2$ - The reaction rate (determined by the slow step in the mechanism) involves two reacting species.
- Dependent on the concentrations of the haloalkanes and the hydroxide ions.

SN2 mechanism for nucleophilic substitution of a primary halogenoalkane Fig 16.1 Example of $S_N2$ mechanism

Mechanism for tertiary halogenoalkanes ( $S_N1$ )

The carbon bonded to the halogen is attached to three alkyl groups.
The reaction involves a two-step mechanism.
Step 1: Breaking of the carbon-halogen bond.
- Forms a tertiary carbocation.
- $Br^-$ forms by heterolytic fission.
Step 2: The tertiary carbocation is attacked by the nucleophile.
The ‘1’ in $S_N1$ means the reaction rate depends on one reagent only (the concentration of the haloalkane in the slow step).
Tertiary carbocations are the most stable due to the positive inductive effect of the alkyl groups attached to the carbon atom bonded to the halogen.
- They release electrons towards the C atom $\rightarrow$ reducing its charge density.
Both the $S_N1$ and $S_N2$ mechanisms are likely to be a part of the nucleophilic substitution of secondary halogenoalkanes.

SN1 mechanism for nucleophilic substitution of a tertiary halogenoalkane Fig 16.2 Example of $S_N1$ mechanism

15.4 Elimination reactions

Involves the loss of a small molecule from the original organic molecule.
- A hydrogen halide (e.g. $HCl$ ; $HBr$ ) in this case.
Reagent: Ethanolic $NaOH$ .
Conditions: Heat.
A hydrogen halide is eliminated.
The ethanolic $OH^-$ ion acts as a base $\rightarrow$ it accepts an $H^+$ ion from the haloalkane to form water and a salt.
Note: The $C-X$ bond breaks heterolytically $\rightarrow$ forms a $Br^-$ ion and produces an alkene.

$C_2H_5Br + NaOH(\text{ethanol}) \rightarrow CH_2=CH_2 + NaBr + H_2O$