The E2 Reactions ModuleThis module covers "bimolecular elimination" reactions. The purpose of this tutorial is to illustrate this important class of reactions, which the reader should compare with the related E1 and SN1 reactions. Related Topics
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What are E2 Reactions?The term E2 stands for "elimination bimolecular." Like any elimination reaction, the product of an E2 elimination reaction has one more degree of unsaturation than the starting materials did. For instance, the base-induced elimination of "HX" (dehydrohalogenation) of an alkyl halide gives rise to an alkene (illustrated below for the conversion of tert- butyl bromide to isobutylene).
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E2 eliminations, in contrast to E1 reactions are promoted
by strong base. The base, vital to the reaction,
is directly involved in the rate-determining step. The
reaction is bimolecular--that is, it involves "second-order
Kinetics--because two molecules must come
together for the reaction to occur.
The
mechanism of an E2 elimination reaction is shown below:
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Notice that the hydrogen that is removed is on the carbon atom adjacent to the one bearing the halogen. Carbon-carbon double bonds, by definition, exist between two adjacent carbon atoms. As such, the "H" and "X" atoms eliminated during the dehydrohalogenation of an alkyl halide to give an alkene must be on adjacent carbon atoms. Self-Test Question #1There are two elimination products that could be formed by the loss of HBr from t-pentyl bromide. Can you draw them?
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Self-Test Question #2Which of the two products formed in "Self-test question #1" would be expected to predominate in the reaction mixture? (See related page on Zaitsev's rule if you need help with this question.) The two elimination products formed are:
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Self-Test Question #3In the Web page on "Nomenclature," you learned about (or reviewed) the trivial names for twelve alkyl groups having five or fewer carbons. Consider the bromides derived from each of those groups. Which two of them could not possibly undergo an E2 elimination reaction, and why? Choose two and click submit.
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Stereochemical requirements (consequences) of E2 reactionStereochemical evidence indicates that E2 reactions always occur via " periplanar" geometry, that is, the atoms of the H-C-C-X group involved in the reaction must all lie in the same plane. This gives rise to two possible orientations:
Of these two, the "anti periplanar" geometry is obviously the more favorable. |
Self-Test Question #4(This is a two-part question. Follow the directions carefully!)
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Incorrect. |
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Correct! |
Self-Test Question #4(This is a two-part question. Follow the directions carefully!)
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Incorrect. |
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Correct! |
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Incorrect. |
E2 vs. E1 vs. SN2 vs. SN1 reactionsWhen assessing the tendency of an alkyl halide to react by way of any of these four mechanisms, the following generalizations are useful:
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These results are spelled out in more detail in the table below (adapted from McMurry, 4/E, with permission):
Correlation of Structure and Reactivity |
| Halide Type | SN1 | SN2 | E1 | E2 |
|---|---|---|---|---|
| RCH2X (primary) |
Does not occur |
Highly favored |
Does not occur |
Occurs when strong bases are used |
| R2CHX (secondary) |
Can occur with benzylic and allylic halides |
Occurs in competition with E2 reaction |
Can occur with benzylic and allylic halides |
Favored when strong bases are used |
| R3CX (tertiary) |
Favored in hydroxylic solvents |
Does not occur |
Occurs in competition with SN2 reaction |
Favored when bases are used |
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For more information on this topic, please consult the following:
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