{"id":3286,"date":"2018-07-03T21:18:29","date_gmt":"2018-07-03T21:18:29","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/?post_type=chapter&p=3286"},"modified":"2020-06-23T22:26:51","modified_gmt":"2020-06-23T22:26:51","slug":"8-5-elimination-reactions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/8-5-elimination-reactions\/","title":{"raw":"8.5. Elimination reactions","rendered":"8.5. Elimination reactions"},"content":{"raw":"Alkyl halides undergo elimination via two common mechanisms, known as E2 and E1, which show some similarities to SN<\/sub>2 and SN<\/sub>1, respectively.\u00a0 In E2, elimination shows a second order rate law, and occurs in a single concerted step (proton abstraction at C\u03b1 <\/sub>occurring at the same time as C<\/span>\u03b2<\/sub>-X bond cleavage).\u00a0 In E1, elimination goes via a first order rate law, in two steps (C<\/span>\u03b2<\/sub>-X bond cleavage occurring first to form a carbocation intermediate, which is then 'quenched' by proton abstraction at the alpha-carbon).\u00a0<\/span>\r\n\r\n
\"Fileimage034.png\"These mechanisms are important in laboratory organic chemistry.\u00a0 As explained below, which mechanism actually occurs in a laboratory reaction will depend on the\u00a0 identity of the R groups (ie., whether the alkyl halide is primary, secondary, tertiary, etc.) as well as on the characteristics of the base.\r\n
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E1 and E2 reactions in the laboratory<\/h3>\r\nE2 elimination reactions in the laboratory are carried out with relatively strong bases, such as alkoxides (deprotonated alcohols, -<\/sup>OR).\u00a0 2-Bromopropane will react with ethoxide, for example, to give propene.\r\n\r\n\"Fileimage035.png\"\r\n\r\nPropene is not the only product of this reaction, however - the ethoxide will also to some extent act as a nucleophile in an SN<\/sub>2 reaction.\r\n\r\n\"Fileimage036.png\"\r\n\r\nChemists carrying out laboratory nucleophilic substitution or elimination reactions always have to be aware of the competition between the two mechanisms, because bases can also be nucleophiles, and vice-versa.\u00a0 However, a chemist can tip the scales in one direction or another by carefully choosing reagents.\u00a0 Primary carbon electrophiles like 1-bromopropane, for example, are much more likely to undergo substitution (by the SN<\/sub>2 mechanism) than elimination (by the E2 mechanism) \u2013 this is because the electrophilic carbon is unhindered and a good target for a nucleophile.\r\n\r\n\"Fileimage037.png\"\r\n\r\nSN<\/sub>1 and E1 mechanisms are unlikely with such compounds because of the relative instability of primary carbocations.\r\n\r\nThe nature of the electron-rich species is also critical.\u00a0 Acetate, for example, is a weak base but a reasonably good nucleophile, and will react with 2-bromopropane mainly as a nucleophile.\r\n\r\n\"Fileimage038.png\"\r\n\r\nIn order to direct the reaction towards elimination rather than substitution, heat is often used. This is because elimination leads to an increase in the number of molecules (from two to three in the above example), and thus an increase in entropy.\u00a0 High temperatures favor reactions of this sort, where there is a large increase in entropy.\u00a0 Substitution does not usually involve a large entropy change, so if SN<\/sub>2 is desired, the reaction should be done at the lowest temperature that allows substitution to occur at a reasonable rate.\r\n\r\nAlso, a strong hindered<\/em> base such as tert<\/em>-butoxide can be used.\u00a0 The bulkiness of tert<\/em>-butoxide makes it difficult for the oxygen to reach the carbon (in other words, to act as a nucleophile).\u00a0 It is more likely to pluck off a proton, which is much more accessible than the electrophilic carbon).\r\n\r\n\"Fileimage039.png\"\r\n\r\nE1 reactions occur by the same kinds of carbocation-favoring conditions that have already been described for SN<\/sub>1 reactions (section 8.3.<\/a>):\u00a0 a secondary or tertiary substrate, a protic solvent, and a relatively weak base\/nucleophile.\u00a0 In fact, E1 and SN<\/sub>1 reactions generally occur simultaneously, giving a mixture of substitution and elimination products after formation of a common carbocation intermediate.\u00a0 When tert-butyl chloride is stirred in a mixture of ethanol and water, for example, a mixture of\u00a0 SN<\/sub>1 products (2-methylpropan-2-ol and tert-butyl ethyl ether) and E1 product (2-methylpropene) results.\u00a0 Heat is used if elimination is desired, but mixtures are still likely.\r\n\r\n\"Fileimage040.png\"\r\n
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Exercise<\/h3>\r\n
A straightforward functional group conversion that is often carried out in the undergraduate organic lab is the phosphoric acid-catalyzed dehydration of cyclohexanol to form cyclohexene.\u00a0 No solvent is necessary in this reaction - pure liquid cyclohexanol is simply stirred together with a few drops of concentrated phosphoric acid. In order to drive the equilibrium of this reversible reaction towards the desired product, cyclohexene is distilled out of the reaction mixture as it forms (the boiling point of cyclohexene is 83 o<\/sup>C, significantly lower than that of anything else in the reaction solution). Any cyclohexyl phosphate that might form from the competing SN<\/sub>1 reaction remains in the flask, and is eventually converted to cyclohexene over time. Draw a mechanism for the cyclohexene synthesis reaction described above.\u00a0 Also, draw a mechanism showing how the undesired cyclohexyl phosphate could form.<\/div>\r\n[reveal-answer q=\"673235\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"673235\"]\r\n

\"image750.png\"<\/p>\r\n \r\n\r\nThe cyclohexyl phosphate could form if the phosphate attacked the carbocation intermediate as a nucleophile rather than as a base:\r\n\r\n\"image752.png\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNext, let\u2019s put aside the issue of competition between nucleophilic substitution and elimination, and focus on the regioselectivity of elimination reactions.\u00a0 In many cases an elimination reaction can result in more than one constitutional isomer or stereoisomer.\u00a0 The elimination products of 2-chloropentane provide a good example:\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\"Fileimage041.png\"\r\n\r\nThis reaction is both regiospecific and stereospecific. In general, <\/em><\/strong>more substituted alkenes are more stable<\/em>, and as a result, the product mixture will contain less 1-butene than 2-butene (this is the regiochemical aspect of the outcome, and is often referred to as Zaitsev's rule<\/strong>). In addition, trans<\/em>-<\/em>alkenes are generally more stable than cis<\/em>-alkenes, so we can predict that more of the trans<\/em> product will form compared to the cis <\/em>product.\r\n\r\nHowever, certain other eliminations (which we will not be studying) favor the least<\/em> substituted alkene as the predominant product, due to steric factors. \u00a0 Such a product is known as the Hoffmann product, and it is usually the opposite of the product predicted by Zaitsev's Rule.\r\n\r\n<\/div>\r\n<\/section>\r\n

E2 eliminations<\/h1>\r\nE2, bimolecular elimination, was proposed in the 1920s by British chemist Christopher Kelk Ingold. Unlike E1<\/a> reactions, E2 reactions remove two substituents with the addition of a strong base, resulting in an alkene.\r\n
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Introduction<\/h3>\r\nE2 reactions are typically seen with secondary and tertiary alkyl halides, but a hindered base is necessary with a primary halide. The mechanism by which it occurs is a single step concerted <\/strong>reaction with one transition state. The rate at which this mechanism occurs is second order kinetics, and depends on both the base and alkyl halide. A good leaving group is required because it is involved in the rate determining step. The leaving groups must be coplanar in order to form a pi bond; carbons go from sp3<\/sup> to sp2 <\/sup>hybridization states.\r\n\r\n<\/div>\r\n
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General Reaction<\/h3>\r\n\"A\r\n\r\nIn this reaction B\u00af represents the base and X represents a leaving group, typically a halogen. There is one transition state that shows the single step (concerted) reaction.\u00a0 The base is forming a bond to the hydrogen, the pi bond is forming, and the C-X bond is beginning to break. In most reactions this requires everything to be in the same plane, and the leaving group 180o<\/sup> to the H that leaves; the H and the X are said to be \"antiperiplanar\".\r\n\r\nIn summary, An E2 reaction has certain requirements to proceed:\r\n