In elimination reaction, some molecules leave the compound leading to the formation of double or triple bond.
α and β
- α−carbon : The carbon atom which carries the halogen is called an alpha-carbon (α−carbon).
- β−carbon : The carbon atom next to the α−carbon is called a β−carbon.
- β−hydrogen : The hydrogen atoms attached to the β−carbon are called β−hydrogens.
Dehydrohalogenation or β−Elimination
When alkyl halides having β−hydrogen are heated with some base such as alcoholic KOH or alkoxide etc., the halogen is eliminated from α−position and a hydrogen from β−position leading to the formation of alkenes.
This reactions is called β−elimination because the relative position of eliminated substituents is 1,2 (see attacking species and types of organic reactions ).
Let us take the example of 2-chlorobutane. 2-Chlorobutane has two β-carbon atoms, and gives two different products on elimination.
In such reactions, the major product is that alkene which has lesser number of hydrogen atoms on the doubly bonded carbon atoms. This rule is known as Saytzeff's rule. Let us apply Saytzeff's rule to the above example:
Applying Saytzeff's rule to the previous example
What will be the major product in the following reaction :
Ease of dehydrohalogenation
The alkyl halide that gives alkene with lesser number of hydrogen atoms on the doubly bonded carbon atoms undergoes elimination faster. Study the following example :
Question : Out of 2-chloropropane and 2-chloro-2-methylpropane, which one undergoes dehydrohalogenation faster?
Answer : 2-Chloropropane is a secondary alkyl halide and yields propene on elimination.
2-Chloro-2-methylpropane is a tertiary alkyl halide and yields 2-methylpropene on elimination.
Since 2-methylpropene has fewer number of hydrogen atoms on doubly bonded carbon atoms, 2-chloro-2-methylpropane will undergo dehydrohalogenation faster than 2-chloropropane.
In general, the ease of dehydrohalogenation follows the order :
tertiary > secondary > primary
For a given halogen, the order is :
R−I > R−Br > R−Cl
Question : The treatment of alkyl chloride with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are the major products. Explain why?
Answer : In aqueous solution, KOH is ionised into K+ and OH−. OH− being a nucleophile leads to nucleophilic substitution reaction with alkyl chlorides to form alcohols. Moreover, the OH− ions in aqueous solution are highly hydrated. This reduces the basic character of OH− ions which, therefore, fail to remove hydrogen from β-carbon to form an alkene.
An alcoholic solution of KOH, on the other hand, contains alkoxide (RO−) ions that are much stronger base than OH− ions; hence, alcoholic solution of KOH preferentially eliminates a molecule of HCl from an alkyl chloride to form an alkene.
Question : What mass of propene is obtained from 34.0 g of 1-iodopropane on treating with ethanolic KOH, if yield is 36%?
Answer : The molecular mass of 1-iodopropane is 170 (3×12 + 7×1 + 127) and that of propene is 42 (3×12 + 6×1).
170 g of 1-iodopropene gives 42 g of propene.
34.0 g of 1-iodopropane gives 34×(42/170) = 8.4 g of propene.
If yield is 36%, amount of propene formed = 8.4×(36/100) = 3.024 g.
Hence, 3.024 g of propene is obtained from 34.0 g of 1-iodopropane on treating with ethanolic KOH, if yield is 36%.
Revision (Includes Earlier Reactions)
You should be able to answer the following questions provided you have studied previous tutorials.
Take a close look at the products :
Convert ethane to dibromoethane.
Convert 2-chloropropane to 1-propanol.
Convert tert-butyl bromide to isobutyl bromide