Alkyl Halides and Aryl Halides
6.0 Preparation
6.0 Preparation
From Alcohols
- Reaction with halogen acids: When concentrated halogen acids react with alcohol, in the presence of anhydrous $ZnCl_2$, the hydroxyl group is replaced by a halogen producing alkyl halides. The general reaction is:
Anhydrous $ZnCl_2$ acts as a catalyst for dehydration. A mixture of concentrated $HCl$ and anhydrous $ZnCl_2$ is called Lucas reagent.
Note:
- Groove's process: It is the process of passing $HCl$ gas through anhydrous $ZnCl_2$.
- Pyridine, dimethyl amine and concentrated $H_2SO_4$ can also be used as catalysts for this reaction.
- Tertiary alcohols readily react with conc. $HCl$ even in the absence of $ZnCl_2$.
Heating alcohol with $KBr/NaBr$ and concentrated $H_2SO_4$ produces bromoalkanes. Similarly, Iodoalkanes can be prepared, in the presence of phosphoric acid, by refluxing alcohol with $57\%$ hydriodic acid or potassium iodide.
For this reaction;
- Reactivity of halogen acids: $HI>HBr>HCl$
- Reactivity of alcohols: tertiary > secondary > primary
- Reaction with phosphorus halides: Reactions of alcohols with phosphorus pentachloride or phosphorus trichloride give chloroalkanes $$\begin{equation} \begin{aligned} 3R - OH + PC{l_3} \to 3R - Cl + {H_3}P{O_3} \\ 3C{H_3}C{H_2}C{H_2}OH + PC{l_3} \to 3C{H_3}C{H_2}C{H_2}Cl + {H_3}P{O_3} \\ R - OH + PC{l_5} \to R - Cl + POC{l_3} + HCl \\\end{aligned} \end{equation} $$ The function of red phosphorus and bromine/iodine is to get $PBr_3$ or $PI_3$ during the course of the reaction. (These compounds are relatively unstable and have to be prepared in situ)
- Reaction with thionyl chloride: Alkyl chloride can be formed when alcohol is refluxed with thionyl chloride in the presence of a small amount of pyridine. Pyridine neutralises both the byproducts in the reaction. For this reason, thinoyl is a better reagent for preparing alkyl halide from alcohol. $$R - OH + SOC{l_2} \to R - Cl + S{O_2} + HCl$$
From Alkanes
- Treatment of alkanes with chlorine or bromine in the presence of light forms poly substituted products. Though this method can produce mono-halides, this method is not of much use since the mixture (mono-halide, di-halides etc.) is difficult to separate. This process involves free radical substitution. $$R - H + {X_2} \to R - X + H - X$$ Reactivity of $C-H$ bonds: Allylic = benzylic > tertiary > secondary > primary
Note:
- In case of higher alkanes. different isomeric products are formed even when monosubstituion is carried out. We can find which isotope is the major product following the reactivity order.
- The iodination of alkanes is reversible. It is carried out by heating alkenes in the presence of oxidising agents like concentrated $HNO_3$.
- This process is not preferred for the formation of alkyl fluorides since fluorination of alkanes is explosive.
From Alkenes
Haloalkanes can be formed from the reaction of halogen acids with alkenes. This reaction takes place readily and it involves electrophilic addition. It takes place in the presence of anhydrous $AlCl_3$. $${C_2}{H_4} + HCl \to C{H_3}C{H_2}Cl$$
- Markovnikov's rule is followed for the addition of halogen alkanes in asymmetrical alkenes.
- Addition of $HBr$ to asymmetrical alkenes in the presence of peroxide takes place according to anti Markovnikov's rule. This reaction takes place through free radical addition mechanism.
Halide Exchange
Alkyl bromides or chlorides can be treated with a solution of sodium iodide to produce alkyl iodides in acetone. Halide exchange is preferred for the preparation of alkyl iodides. This reaction is also known as Finkelstein reaction and it is based on the fact the $NaI$ is soluble in acetone but $NaBr$ or $NaCl$ are insoluble. Thus, the forward reaction is favored according to Le Chatelier's principle.
$$\begin{equation} \begin{aligned} R - X + NaI \to R - I + Na - X \\ {C_2}{H_5}Cl + NaI \to {C_2}{H_5}I + NaCl \\\end{aligned} \end{equation} $$
Halide exchange can also be used to synthesize alkyl fluorides. This can be done by heating alkyl chloride or bromide in the presence of a metallic fluoride (AgF, Hg2F2). This reaction is also known as Swarts reaction. $$R - X + AgF \to R - F + Ag - X$$
From Silver Salts of Fatty Acids
Refluxing silver salts of fatty acids with bromine produces bromoalkanes. This reaction is carried out in the presence of $CCl_4$. It is also known as Hunsdiecker reaction. This reaction gives a product with one less carbon atom than the reactant (fatty acids). Chloroalkanes can also be synthesized using this reaction, however, their yield is low.
$$RCOOAg + B{r_2} \to R - Br + C{O_2} + AgBr$$
This reaction cannot be used to produce Iodoalkanes. Iodine forms ester instead of alkyl halide and is called Birnbaum-Simonini reaction.
$$RCOOAg + {I_2} \to RCOOR + C{O_2} + 2AgI$$
Example 1. Sulphuric acid not used during the reaction of alcohols with $KI$. Explain.
Solution: In the conversion of alcohol to alkyl iodide, sulphuric acids cannot be used because it converts $KI$ to the corresponding $HI$ and then oxidises it to iodine.
Example 2. Identify which isomer of pentane will yield the following products on chlorination.
a) Four isomeric monochlorides
b) Three isomeric monochlorides
c) A single monochloride
Solution: a) Methylbutane. $CH_3CH(CH_3)CH_2CH_3$ has four different carbons in a chain and hence there are four hydrogen atoms which can be substituted. Thus, it will yield $4$ isomeric monochlorides.
b) Pentane. In $CH_3(CH_2)_3CH_3$ there are three different carbon atoms and hence $3$ hydrogen atoms are available for substitution. Thus, pentane will yield three isomeric monochlorides.
c) Dimethylpropane. In this isomer, all the hydrogens are considered equivalent. Thus, only one isomeric product will be formed.
