Alkyl Halides and Aryl Halides
10.0 Reaction with Metals
10.0 Reaction with Metals
Reaction with magnesium:
The reaction of haloalkanes with magnesium powder, in the presence of dry ether forms Grignard reagents. Grignard reagents are highly reactive and form hydrocarbons on reaction with any aqueous solutions. Even water, alcohol, and amines are acidic enough to react with grignard reagents to form the corresponding hydrocarbons.
$$\begin{equation} \begin{aligned} R - X + Mg \to R - Mg - X \\ RMgX + {H_2}O \to RH + Mg(OH)X \\\end{aligned} \end{equation} $$
Reaction with sodium:
The reaction of haloalkanes with sodium in presence of dry ether forms hydrocarbons with twice the number of carbon atoms present in the substrate. This reaction is also known as Wurtz reaction.
$$2R - X + 2Na \to R - R + 2NaX$$
Reaction with lead-sodium alloy:
The reaction of lead-sodium alloy with ethyl chloride, in the presence of dry ether, forms tetra ethyl lead. This is used as an antiknock compound in petrol.
$$4{C_2}{H_5}Cl + 4Pb(Na) \to {({C_2}{H_5})_4}Pb + 4NaCl + 3Pb$$
Reduction
Reduction of haloalkanes (with nascent hydrogen) in the presence of $Ni/Pd$ form alkanes.
$$R - X + 2[H] \to R - H + HX$$
Friedel-Craft Reaction
Haloalkanes react with benzene in the presence of anhydrous aluminum halides to form alkyl benzene.
$$R - X + {C_6}{H_6} \to {C_6}{H_5}R + HX$$
Uses of Ethyl Chloride
Ethyl chloride is used as a refrigerant and a local anesthetic. It is also used in the preparation of tetra ethyl lead, and the preparation of Grignard reagent.
Example 1. What will be the product when the optically active form of $C_4H_9Br$ undergoes dehydrohalogenation?
Solution: $CH_3CH_2CHBrCH_3$ is the optically active form of $C_4H_9Br$ since the carbon attached to bromine is a chiral carbon.
On dehydrohalogentaion, two products will be formed, $CH_3CH=CHCH_3$ and $CH_3CH_2CHCH_2$.
The major product, according to Sayzeff rule would be $CH_3CH=CHCH_3$ as it has a greater number of alkyl groups attached to the double bonded carbon.
Example 2. Free radical halogenation of alkenes is usually not utilized to prepare haloalkanes in a laboratory. Explain.
Solution: Free radical halogenation of alkenes is not a preferred method to prepare haloalkanes. This is because the reaction results in the formation of my isomeric monosubstituted products because of the substitution of different hydrogen atoms. In addition, poly halogenation may also take place.
Example 3. What are the three possible outcomes of a substitution reaction at the chiral carbon of a haloalkane?
Solution:
1) Retention of configuration: This occurs when substitution is at the same side of the leaving group.
2) Inversion of configuration: This occurs when the nucleophile attacks at the opposite side pf the leaving group.
3) Racemisation: This occurs when both the above configurations are obtained in a $50:50$ ratio.
