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A solution of KOH hydrolysis CH_(3)CHClCH_(2)CH_(3) and CH_(3)CH_(2)CH_(2)CH_(2)Cl. Which one of these is more easily hydrolysed ? |
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Answer» Solution :Under aqueous conditions, substitution OCCURS by `S_(N)1` mechanism since carbocation intermediates are stabilized by solvation. Further, the reactivity in `S_(N)1` reactions depends upon the STABILITY of CARBOCATIONS. `CH_(3)-CHCl-CH_(2)CH_(3) overset("Ionization")to underset(2^(@)" carbocation (more stable)")(CH_(3)-overset(+)(C)H-CH_(2)CH_(3)+Cl^(-))` `CH_(3)CH_(2)CH_(2)CH_(2)-Cl overset("Ionization")to underset(1^(@)" carbocation (less stable)")(CH_(3)CH_(2)CH_(2)overset(+)(C)H_(2)+Cl^(-))` Now since `CH_(3)CHClCH_(2)CH_(3)` upon ionization gives the more stable `2^(@)` carbocation while `CH_(3)CH_(2)CH_(2)CH_(2)Cl` on ionization gives the less stable `1^(@)` carbocation, therefore, `CH_(3)CHClCH_(2)CH_(3)` hydrolyses FASTER than `CH_(3)CH_(2)CH_(2)CH_(2)Cl`. |
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