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Dehydrohalogenation of ethyl chloride with alcoholic KOH

Why elimination occurs instead of nucleophilic substitution when ethyl chloride reacts with alcoholic potassium hydroxide?


Ethyl chloride can undergo both nucleophilic substitution as well as elimination reactions with strong alkali like KOH. 

When it undergoes nucleophilic substitution, ethyl alcohol is formed as major product.

 nucleophilic substitution

Mechanism of nucleophilic substitution: 

SN2 mechanism

Above mechanism depicts SN2 path. Usually primary alkyl halides show Nucleophilic bimolecular substitutions since the formation of carbocation is less easy. Hence bond breaking and making occurs in one step.

However, due to presence of β-hydrogen, ethyl chloride can also undergo elimination reaction to give ethylene in presence of strong base like KOH.


Since KOH is a strong base, it can also abstract β-hydrogen and thus by favoring elimination of HCl molecule (dehydrohalogenation).

Mechanism of dehydrohalogenation: 

The elimination of hydrogen halide may occur by E1 or E2 mechanism. 

E1 path

E1 path

E2 path

E2 path

Since ethyl chloride is a primary alcohol, the preferred path is E2


Now we can safely assume that:

i) When water is used as solvent, the likeliness of nucleophilic attack by H2O as well as OH- is more. That leads to formation of nucleophilic substituted product. 

ii) However in water, the elimination is not favored due to formation of water as one of the product. It is according to le Chatelier’s principle. Excess of water will shift the equilibrium to the left side.

iii) When the reaction is carried out in alcohol, less number of water molecules are available and thus by increasing the chances of attack of OH- ion on β-hydrogen.

Related questions 

1) Why tertiary alkyl halides undergo elimination reactions in presence of strong bases rather than nucleophilic substitution?

2) Why primary alkyl halides take SN2 or E2 paths?

3) Write rate expressions for SN1 and SN2 mechanisms.




Author: Aditya vardhan Vutturi