Now on reaction of ethyl chloride with KOH(Dillute) substitution SN2 occurs. CH3CH2CL + KOH-------->Ch3CH2OH It is mainly because the compound is alycylclic,the concentration of the OH negative ions id plenty to attack the carbon atom and there is no steric hindrance for the backword attack.
In case the reagent used here was Alcoholic KOH then beta elimination wud occur and an alkene wud be formed.
Ch3Ch2Cl + KOH(alc.)--------------->ethene
This is because the concentration of Oh negative ions is less.
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