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NCERT Class XII Chemistry
Chapter - Haloalkanes and Haloarenes
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Question : 20 of 31
Marks: +1, -0
The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
Solution:  
Formation of alcohols from the reaction between alkyl chlorides and aqueous KOH is an example of simple nucleophilic substitution.
CH3−CH2Cl+KOH→H2OCH3−CH2−OH+KCl\mathrm{CH_3-CH_2Cl} + \mathrm{KOH} \xrightarrow{\mathrm{H_2O}} \mathrm{CH_3-CH_2-OH} + \mathrm{KCl}
But when aqueous KOH is replaced by alcoholic KOH, alkenes are formed instead of alcohols due to elimination of HCl from an alkyl halide.
CH3−CH2Cl+KOH→CH2=CH2\mathrm{CH_3-CH_2Cl} + \mathrm{KOH} \rightarrow \mathrm{CH_2=CH_2}
This can be exaplained if we consider the size of the nucleophile in the two reactions. In the aqueous medium the Nu⊖\overset{\ominus}{\mathrm{Nu}} is O⊖H\overset{\ominus}{\mathrm{O}}\mathrm{H} which is relatively small while in the alcoholic medium the Nu⊖\overset{\ominus}{\mathrm{Nu}} is C2H⊖5\mathrm{C}_2\overset{\ominus}{\mathrm{H}}_5 which is bulky in nature.
Thebulky Nu⊖\overset{\ominus}{\mathrm{Nu}} will always find it easier to abstract a proton than to attack a tetravalent carbon to produce a substitution product.
If C2H⊖5\mathrm{C}_2\overset{\ominus}{\mathrm{H}}_5 was to attack the carbon carrying halogen, steric repulsions would hinder the attack and prevent substitution product.
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