I think in sn1 reaction the order of reactivity is 3degree>2>1. Am i correct?
Yes it is for tertiary . The first one reacts faster according to Saidsef rule
Saytzeff Rule implies that base-induced eliminations (E2) will lead predominantly to the olefin in which the double bond is more highly substituted, i.e. that the product distribution will be controlled by thermodynamics.
The use of sterically hindered bases raises the activation energy barrier for the pathway to the product predicted by Saytzeff's Rule. Thus, a sterically hindered base will preferentially react with the least hindered protons, and the product distribution will be controlled by kinetics.
Im sorry, it was SN2 i was talking about. In SN2 order of reactivity is 1 degree>2>3.
A good nucleophile favours SN2 while a weak nucleophile favours SN1. In the given question we presume KOH is good nucleophile, so the reaction should be SN2. The decreasing relative reactivity of alkyl halides in SN2 follows the sequence
Methyl halide > Primary > Secondary > Tertiary > Neopentyl halide.
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