# Preparation of Haloalkanes

Haloalkanes are known as the replacement of hydrogen atom(s) in aliphatic hydrocarbon by the halogen atom (s) result in the formation of alkyl halide.Methods of preparation of haloalkanes are given below.

## Preparation of haloalkanes from alkanes

Halogen reacts with alkanes in the presence of sunlight or high temperature to give a mixture of mono or polyhaloalkanes. This reaction shows the Free radical reaction mechanism.

For a given alkane, reactivity order of halogen,

$F_{2} > Cl_{2} > Br_{3} > I_{2}$

For a given halogen, reactivity order of H-atom in alkane,

$3^{\circ} > 2^{\circ} > 1^{\circ}>CH_{4}$

Example:-

$CH_{3}CH_{2}CH_{3}\xrightarrow{Cl_{2}/h \nu}\underbrace{CH_{3}CH_{2}CH_{2}Cl}_{\text{ 45 %}}+\underbrace{CH_{3}CHClCH_{3}}_{\text{ 55 %}}$

Chlorination of n-butane give d- and l- optical isomer of sec-butyl chloride hence the mixture is optically inactive(racemic).

Fluorination is not possible as it occurs with breakage of C-C bonds in higher alkanes and this reaction is highly exothermic too.

Iodination of alkanes is a reversible process, therefore it carried out in the presence of oxidizing agents ($HIO_{3}, conc. HNO_{3}$).

The oxidizing agent oxidizes $HI$ produced to $I_{2}$.

## Preparation of haloalkanes from alkenes

### Addition of hydrogen halide to alkene

Hydrogen halide adds to hydrogen halide through MARKOWNIKOFF RULE.

Generalized reaction,

$RCH=CHR{ }’\xrightarrow { }RCH_{2}-CHX-R{ }’$

This reaction follows electrophilic addition mechanism.

### Addition of Halogens to alkene

This addition of halogen to alkene produce vicinal dihalides.

Addition of bromine in $CCl_{4}$ to an alkene result in the discharge of reddish brown color of bromine. This a method to detect the presence of double bond in a molecule.

Example:-

$CH_{2}=CH_{2} + Br_{2} \xrightarrow{CCl_{4}}BrCH_{2}-CH_{2}Br_{2}$

### Preparation of allylic halide by allylic substitution of alkenes

When alkenes are heated with $Cl_{2}$ or $Br_{2}$ at 773 K, they give allylic halide via. Free radical mechanism. In this reaction, the H-atom of allylic carbon is replaced by a halogen atom.

Example:-

$CH_{3}CH_{2}CH=CH_{2} + X_{2} \xrightarrow{773K} CH_{3}CHX-CH=CH_{2} + HX$

Reactivity of different types of hydrogen,

$\text{Benzylic}\cong \text{allylic} > 3^{\circ} > 2^{\circ} > 1^{\circ}>CH_{4} > \text{vinylic} \approx \text{aryl}$

## Preparation of haloalkanes from alkynes

Alkynes react with $Cl-{2}$ or $Br_{2}$ dissolve in $CCl_{4}$ to give vinical tetrahalide.

Example:-

$\underbrace{CH_{3}-C \equiv CH}_{\text{Alkyne}}\xrightarrow{CCl_{4}}\underbrace{CH_{3} – CBr_{2}-CBr_{2}-H}_{\text{ Vicinal tetrahalide}}$

## Preparation of haloalkanes from alcohol

These reactions are nucleophilic substitution reactions and follow $S_{N}1$ or $S_{N}2$ accordingly depending upon the alcohol used.

### Reaction of halogen acids with alcohols

$RCH_{2}OH + HX \xrightarrow{ZnCl_{2}} RCH_{2}X + H_{2}O$

The order of reactivity of $HX$ with a given alcohol is,

$HI > HBr > HCl > HF$

$S_{N}1$ shown by $3^{\circ}$ and $2^{\circ}$ alcohol and $S_{N}2$ shown by $1^{\circ}$ alcohol.

The reaction of $1^{\circ}$ and $2^{\circ}$ alcohol with $HX$ requires the catalyst, $ZnCl{2}$ which is known as Groove’s Process.

The importance of $ZnCl_{2}$is, the $Zn^{2+}$ ion can coordinate to the oxygen atom of alcohol. This makes alcohol as a good leaving group and hence leads to carbocation formation. This formed carbocation reacts with chloride ion to form alkyl chlorides.

$3^{\circ}$ alcohols, the reaction simply occur by shaking with concentrated HCl at room temperature.

The order of reactivity of alcohols with a given haloacid is,

$3^{\circ} > 2^{\circ} > 1^{\circ}$

### Reaction of phosphorus halides with alcohols

This follows $S_{N}1$ reaction mechanism. The advantage of phosphorus halide is that they don’t undergo rearrangement.

$ROH \xrightarrow[P_{4}/X_{2},X= \text{Br,I}]{PX{3}}RCl + POCl_{3} + HCl$

$ROH \xrightarrow{PCl_{5}}RCl + POCl_{3} + HCl$

### Darzen’s reaction for preparation of Haloalkanes

In Darzen’s reaction haloalkanes are prepared by using $SOCl_{2}$ in the presence of either base or pyridine. For the preparation haloalkanes, it is the excellent method because the other two products are gases. Thus, it gives pure allyl halides.

$R-OH + SOCl_{2} \xrightarrow{\Delta} R-Cl +\underbrace{ SO_{2}\uparrow + HCl \uparrow }_{\text{Gases}}$

## Preparation of haloalkanes from silver salts of carboxylic acid

This reaction is known as Borodine-Hunsdiecker reaction. This reaction is mainly used for the preparation of alkyl bromide. It follows Free radical mechanism.

$RCO_{2}Ag +Br_{2} \xrightarrow{CCl_{4}} R-Br + CO_{2} \uparrow+ AgBr$

The yield of alkyl bromide is better than that of alkyl chloride. As the $O-Cl$ bond is weaker than the $O-Br$ bond.

Birnbaum-Simonini reaction

The silver salt of carboxylic acid gives ester as the main product instead of alkyl iodide on reaction with $I_{2}$.

$2RCO_{2}Ag + I_{2} \xrightarrow{} RCO_{2} + CO_{2} + 2AgI$

## Preparation of haloalkanes from dihalides, Grignard reagent, and ether

These reactions follow Nucleophilic substitution reactions.

### Haloalkanes preparation from dihalides

$Dihalides \xrightarrow[HCl]{Zn-Co} \underbrace{R-X}_{Haloalkane}$

### Haloalkanes preparation  from Grignard reagent

$\underbrace{RMgX}_{\text{Grignard ragent}}\xrightarrow{X_{2}} \underbrace{R-X}_{Haloalkane}$

### Haloalkanes preparation from Ether

$\underbrace{ROR}_{\text{Ether}} \xrightarrow[HCl]{Zn-Co} \underbrace{R-X}_{Haloalkane}$

## Preparation of haloalkanes by halide exchange method

### Finkelstein reaction

Alkyl chlorides/ bromides reacts with $NaI$ in dry acetone to give alkyl iodide.

$R-X + NaI \rightarrow{\text{Dry acetone}}R-I + NaX$

(X=Cl, Br)

### Swarts reaction

Alkyl florides are prepared by heating an alky chloride/bromide in the presence of a metallic fluoride ($AgF,Hg_{2}F_{2},CoF_{2}, or SbF_{3}$).

$R-Br + AgF \xrightarrow[acetone]{Dry} R-F + AgBr$

$F^{-}$ – act as strong halogen.

$Ag^{+}$ – act as weak metal.

Thus, $F^{-}$ Replace $Br^{-}$ from $R-Br$.