Whenever, there is the reagent "alcoholic KOH", we prefer E2 Elimination reaction. However in case of 3 degree alkyl halides, like tert-butyl chloride, why do we not consider SN1 pathway? There is a polar protic solvent(alcohol) and the carbocation is stable, yet E2 dominates. What's the exact reason for this?
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1$\begingroup$ Tertiary alkyl halides exhibit a significant preference for elimination reactions compared to substitution reactions. This preference arises primarily from the steric hindrance caused by the bulky substituents surrounding the central carbon atom. These bulky groups create a congested environment that hinders nucleophilic attack in both SN1 and SN2 mechanisms. $\endgroup$– RonithCommented May 15 at 13:13
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$\begingroup$ Right, Understood it. Just another question, if there was a 3 degree carbon with three tert-butyl groups such that there is no Hydrogen for Elimination, will SN1 happen, or will be there be no reaction? $\endgroup$– EagerToLearnCommented May 16 at 12:46
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$\begingroup$ Try checking this example out: chem.libretexts.org/Bookshelves/Organic_Chemistry/… $\endgroup$– RonithCommented May 16 at 13:21
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$\begingroup$ Thanks, Gained some new insights. Suppose we take Br-C(CR3)3 instead of Br-C(CH3)3. In the first compound, Elimination isn't possible. But is SN1 possible, or is too crowded? Can we say the reaction undergoes 100% SN1? $\endgroup$– EagerToLearnCommented May 16 at 13:40
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