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When a tertiary alkylhalide reacts with aqueous koh then why #E2#reaction occurs .As tertiary alkylhalide undergo #E1#reaction What makes the difference?

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  • $\begingroup$ May be ,but I can't understand from it that what happens on reaction of alkylhalide with aqueous koh? because its reaction with alcoholic koh is only what I get $\endgroup$
    – Rabik John
    Commented Feb 12, 2018 at 12:21

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There is a significant difference between two mediums.
For Substitution reactions we need a $Good Nucleophile$, not a $Good Base$.Because we want the attacking species to preferably attack the electrophilic atom(In most of the cases it's Carbon).If it's a good base it will take the proton attached to the Carbon which will not help the Substitution reaction. The medium is aqueous KOH(amount of water is large compared to OH-), in which $H_2O$ is the dominant Nucleophile,as it is not so strong base,so it attacks the Carbon and in the deprotonation step, OH- acts as a base to take H+(as, water is a weak acid, so, conjugate OH- is a strong base).
Now, if you consider alcoholic KOH medium(generally $EtOH$ medium),There is predominant OH- and little amount of $EtO-$ which are both Strong bases. For,the eliminantion reaction we need specifically a proton acceptor. Thus, highly basic medium supports elimination.
Now, if we consider Substitution and Elimination in case of Tertiary alkylhallide, there is always competition between $S_N1$ and $E_1$ .So, the reaction medium and maintaining $Temperature$ is very important. If your solvent medium acts as a good Nucleophile(bad base) then $S_N1$ will occur.But if your medium is highly basic,$E_1$ occurs.
Sometimes, there are solvent mediums such as $MeO-$ or $EtO-$( in some cases) which can act as both Nucleophile and Base.Then,
lowering the temperature follows $S_N1$and at higher temperature $E_1$ occurs for tertiary alkyl hallides.

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  • $\begingroup$ Ok ,I understood what are you saying but there is one doubt in my mind ;Can tertiary alkylhalide undergo $E2$ reactions ,by reacting with aqueous koh? As my book says so? $\endgroup$
    – Rabik John
    Commented Feb 13, 2018 at 4:21
  • $\begingroup$ Aqueos KOH is a solvent which is good at nucleophilic substitution, you need a basic medium. So, I think it's difficult to carry Elimination. More over, In polar solvent Tertiary carbocations are highly stable.So it will definitely choose $E_1$ reaction pathway. Aqueous KOH is also not suitable for $E_2$. You need a highly non-polar basic solvent to do that ,but it is still difficult. $\endgroup$
    – Soumik Das
    Commented Feb 13, 2018 at 4:49

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