Haloalkanes undergo elimination of hydrogen halide (HX) when boiled with alcoholic solution of potassium hydroxide, resulting in the formation of alkenes. As the hydrogen atom present at b-position of haloalkane (i.e., on the carbon atom next to the one carrying the halogen) is removed, these are called b-elimination reactions or 'dehydro halogenation reactions' (removal of hydrogen halide).
Similarly, 1-Chlorobutane on elimination gives 1-butene.
When a haloalkane eliminates hydrogen halide in two different ways, then the preferred alkene is the one in which carbon atoms joined by the double bond are alkylated maximum i.e., contains largest number of alkyl groups. This rule is called 'Saytzeff's rule'. For example,
The alkene with greater number of alkyl groups is the preferred product during dehydrohalogenation of alkyl halide as it is formed faster than the other alkene.
Stability of alkenes
R2C=CR2 > R2C=CHR > R2C=CH2 > RCH=CH2
According to Saytzeff's rule, any alkyl halide which gives a more stable (i.e. more highly substituted) alkene must undergo dehydrohalogenation reaction faster than the one which gives a less stable (i.e. less highly substituted) alkene. The reactivity of haloalkanes towards elimination reaction follows the order: tertiary > secondary > primaryNote: Alcoholic KOH causes 'elimination' in the molecule of haloalkane, while an aqueous solution of the base leads to 'substitution'. In aqueous medium, the base ionizes to give OH- ions which act as nucleophiles and bring about nucleophilic substitution. Also, these ions also take up water molecules and get hydrated. The hydrated ion is not in a position to extract a proton from the b-carbon atom and therefore elimination cannot occur.
Reactions with active metals
Alkyl halides readily combine with active metals like sodium, magnesium, cadmium, lithium etc.
Action with magnesium (formation of Grignard reagent)
When a solution of an alkyl halide in dry ether is treated with magnesium, an alkyl magnesium halide is formed.
Alkyl magnesium halides, generally, represented as RMgX, are known as Grignard reagents. These have great synthetic applications and are useful in the preparation of a large number of organic compounds.
Action with sodium
Two molecules of alkyl halides (same or different) react with metallic
sodium in the presence of ether to form alkanes. This reaction is called Wurtz reaction and is used to prepare symmetrical alkanes.
Reaction with other active metals
(formation of organometallic compounds)
Haloalkanes react with other active metals such as lithium, zinc, mercury, lead, etc. in the presence of dry ether to form the corresponding organometallic compounds. For example,
Tetraethyl lead (TEL) is used as an antiknocking agent in gasoline used for running automobiles.
Reduction
Haloalkanes are reduced to alkanes by reagents like:
- With hydrogen in the presence of a metal catalyst such as nickel, palladium or platinum
- With zinc copper couple and ethyl alcohol (95%)
- Hydroiodic acid in the presence of red phosphorus.
Rearrangement
When a haloalkane is heated at 573K or at a lower temperature in the presence of anhydrous aluminium chloride as catalyst, it undergoes rearrangement to form isomeric haloalkane.
If there is no hydrogen atom on the carbon atom adjacent to the C-X group, rearrangemet occurs in which methyl group migrates. For example,
Problems
1. Complete the following reactions (giving major products) :
Solution
Solution
The structural formula of 2,2,5,5-tetramethylhexane,
suggests that the compound (A) which gives the above compound during Wurtz reaction is:
The complete sequence of reactions is:
Solution
The reaction is:
Molecular mass of iodopropane (CH3CH2CH2I)= 3 x 12 + 7 x 1 + 127 = 170
Molecular mass of propene (CH2-CH=CH2) = 3 x 12 + 6 x 1= 42170 g of 1-iodopropane will give propene = 42 g
But the actual yield is 36% so that
