Home Revision A-Level Chemistry Organic Mechanisms Elimination Mechanisms

EXPLAINING THE ELIMINATION REACTIONS PRODUCING ALKENES FROM SIMPLE HALOGENOALKANES

 

This page guides you through the elimination mechanism for the reaction between simple halogenoalkanes like 2-bromopropane and hydroxide ions from, for example, sodium hydroxide.

Elimination involving more complicated halogenoalkanes and the competition between elimination and substitution in these reactions are dealt with on separate pages. These are essential parts of this topic, and you should follow the links at the bottom of this page if you haven't already done so.


          
The elimination reaction involving 2-bromopropane and hydroxide ions

          
The role of the OH- ion in an elimination reaction

          
          
Hydroxide ions have a very strong tendency to combine with hydrogen ions to make water - in other words, the OH- ion is a very strong base.

          
In an elimination reaction, the hydroxide ion hits one of the hydrogen atoms in the CH3 group and pulls it off.  This leads to a cascade of electron pair movements resulting in the formation of a carbon-carbon double bond, and the loss of the bromine as Br-.

      

      

        
        
          

          

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The complete elimination mechanism

          

          
    
            
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    The OH- ion takes one of the hydrogens from the CH3 group, but it only needs the hydrogen nucleus (a hydrogen ion).  That means that the two electrons which originally joined the hydrogen to the carbon aren't being used any more.
    
            
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    Those two electrons move to form a double bond between the two carbon atoms.
    
            
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    The approach of those electrons repels the electrons in the carbon-bromine bond right out onto the bromine, throwing the bromine off as a negative ion.
    
            
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The attack could equally well have been on any of the other hydrogens on the left-hand carbon, or on any on the right-hand one - it simply depends on what the OH- ion hit.

      

      

        
        
          

          

          Taking chemistry further:   The mechanism being described here is known as an E2 mechanism.  The E stands for elimination, and the 2 is because the initial slow part of the reaction involves 2 species.  This mechanism is used by primary and secondary halogenoalkanes.
          

          
There is also an E1 mechanism which would apply to tertiary halogenoalkanes.  (Secondary ones will also do it to some extent.)  No current UK A level syllabus (or its equivalent) is likely to ask you about that.

          
          

          

          

      

      

        

            

              
Where would you like to go now?

            

            
How to deal with more complicated cases  . . . 

            
The competition between elimination and substitution  . . . 

            
To menu of elimination reactions. . .

            
To menu of other types of mechanism. . .

            
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