Answer to Question #31988 in Organic Chemistry for naeim
The reaction most often occurs at an aliphatic sp3 carbon center. The breaking of the C-X bond and the formation of the new C-Nu bond occur simultaneously to form a transition state in which the carbon under nucleophilic attack is pentavalent, and approximately sp2 hybridised. The nucleophile attacks the carbon at 180° to the leaving group, since this provides the best overlap between the nucleophile's lone pair and the C-X σ* antibonding orbital. The leaving group is then pushed off the opposite side and the product is formed.
* It is a one-step process of elimination with a single transition state.
* Typical of secondary or tertiary substituted alkyl halides. It is also observable with primary alkyl halides if a hindered base is used.
* The reaction rate both influenced by the alkyl halide and the base is second order.
* Because E2 mechanism results in the formation of a Pi bond, the two leaving groups (often a hydrogen and a halogen) need to be coplanar. An antiperiplanar transition state has staggered conformation with lower energy and a synperiplanar transition state is in eclipsed conformation with higher energy. The reaction mechanism involving staggered conformation is more favorable for E2 reactions.
* Reaction often present with strong base.
* In order for the pi bond to be created, the hybridization of carbons need to be lowered from sp3 to sp2.
If the carbon you're attacking is tertiary then the reaction is E2.
If the carbon is secondary, it's a mixture of both.
If the carbon is primary, it's MOST LIKELY to undergo SN2 reaction.
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