{"id":4509,"date":"2018-07-31T18:46:01","date_gmt":"2018-07-31T18:46:01","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/?post_type=chapter&p=4509"},"modified":"2020-06-23T21:56:55","modified_gmt":"2020-06-23T21:56:55","slug":"9-9-applications-of-eliminations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/9-9-applications-of-eliminations\/","title":{"raw":"9.9. Applications of eliminations","rendered":"9.9. Applications of eliminations"},"content":{"raw":"The eliminations we have studied are more limited in synthetic scope than substitutions, since they only produce alkenes or alkynes.\u00a0 However, if the target product is an alkene, then elimination is almost certainly the method to use (at this point in the course).\u00a0 With alkynes, these are often made from other alkynes (see section 9.8.<\/a>), but they are also accessible from dihaloalkanes.\r\n\r\nThere are two main starting materials: Alkyl halides, and alcohols.\u00a0 Alkyl halides should normally be eliminated using strong base, via an E2 mechanism.\u00a0 Although E1 can work for tertiary alkyl halides, in practice E2 will nearly always give better yields of the alkene, even with tertiary reactants.\u00a0 Alcohols must always be eliminated using strong acid, via an E1 mechanism,\r\n

Zaitsev's Rule<\/h2>\r\nZaitsev's Rule states that an elimination will normally lead to the most stable alkene as the major product<\/strong>.\u00a0 This normally translates to it giving the most substituted alkene.\u00a0 In cases where there is a choice of similarly substituted alkenes, the stability will be increased by conjugation with a nearby double bond.\u00a0 The stability will be decreased by crowding, so cis<\/em>-alkenes are normally slightly less stable than trans<\/em>-alkenes.\r\n\r\nAlkynes do not form E\/Z isomers, but Zaitsev's Rule still applies where there is a choice between an internal (more stable) alkyne and a terminal (less stable) alkyne.\u00a0 Alkyne rearrangements with base are known, but we will not cover these in this course.\r\n

E2 elimination of alkyl halides with strong base - dehydrohalogenation<\/h2>\r\nPrimary, secondary and tertiary alkyl halides all react well in this reaction.\u00a0 Tertiary alkyl halides will scarcely need any heat in order to react.\u00a0 The reaction obeys Zaitsev's Rule, unless a hindered base (such as KOt<\/sup>Bu) is used.\r\n

\"E2<\/h2>\r\n

E1 elimination of alcohols with acid - dehydration<\/h2>\r\nWith alcohols, E2 reactions are not possible because strong base will simply do an acid-base reaction with the alcohol to form an alkoxide -<\/sup>OR.\u00a0 Acid is needed in order to turn the OH into a good leaving group (water), and a strong acid such as H2<\/sub>SO4<\/sub> or H3<\/sub>PO4<\/sub> is usually used.\u00a0 Since E1 reactions work with even very weak bases (water serves here as the base), the reaction is effective for secondary and tertiary alcohols.\u00a0 Primary alcohols may work, but higher temperatures are needed, and there are many side reactions such as rearrangement and ether formation (via SN<\/sub>2).\r\n\r\n\"\"\r\n

E2 Elimination of dihalides for synthesis of alkynes<\/h2>\r\nAlkynes can also be prepared via E2 elimination, though to form two pi bonds it means we must do a double elimination starting with a dihalide.\u00a0 Often a stronger base such as NaNH2<\/sub> is used, though NaOCH3<\/sub> or KOt<\/sup>Bu still work.\u00a0 The dihalide can have the halogens on the same carbon, or on neighboring carbons.\u00a0 Zaitsev's Rule applies as long as rearrangement does not occur (which we will not cover here).\r\n\r\n\"\"\r\n\r\nThere is a useful synthetic \"trick\" that can be done if three moles of NaNH2<\/sub> are used to with a terminal dihalide.\u00a0 The first two moles of base form the terminal alkyne, then the third mole will react with that alkyne to form an acetylide salt.\u00a0 That salt can then be treated with an alkyl halide to produce a longer chain alkyne via an SN<\/sub>2 reaction (see section 9.8.<\/a>).\r\n\r\n\"\"","rendered":"

The eliminations we have studied are more limited in synthetic scope than substitutions, since they only produce alkenes or alkynes.\u00a0 However, if the target product is an alkene, then elimination is almost certainly the method to use (at this point in the course).\u00a0 With alkynes, these are often made from other alkynes (see section 9.8.<\/a>), but they are also accessible from dihaloalkanes.<\/p>\n

There are two main starting materials: Alkyl halides, and alcohols.\u00a0 Alkyl halides should normally be eliminated using strong base, via an E2 mechanism.\u00a0 Although E1 can work for tertiary alkyl halides, in practice E2 will nearly always give better yields of the alkene, even with tertiary reactants.\u00a0 Alcohols must always be eliminated using strong acid, via an E1 mechanism,<\/p>\n

Zaitsev’s Rule<\/h2>\n

Zaitsev’s Rule states that an elimination will normally lead to the most stable alkene as the major product<\/strong>.\u00a0 This normally translates to it giving the most substituted alkene.\u00a0 In cases where there is a choice of similarly substituted alkenes, the stability will be increased by conjugation with a nearby double bond.\u00a0 The stability will be decreased by crowding, so cis<\/em>-alkenes are normally slightly less stable than trans<\/em>-alkenes.<\/p>\n

Alkynes do not form E\/Z isomers, but Zaitsev’s Rule still applies where there is a choice between an internal (more stable) alkyne and a terminal (less stable) alkyne.\u00a0 Alkyne rearrangements with base are known, but we will not cover these in this course.<\/p>\n

E2 elimination of alkyl halides with strong base – dehydrohalogenation<\/h2>\n

Primary, secondary and tertiary alkyl halides all react well in this reaction.\u00a0 Tertiary alkyl halides will scarcely need any heat in order to react.\u00a0 The reaction obeys Zaitsev’s Rule, unless a hindered base (such as KOt<\/sup>Bu) is used.<\/p>\n

\"E2<\/h2>\n

E1 elimination of alcohols with acid – dehydration<\/h2>\n

With alcohols, E2 reactions are not possible because strong base will simply do an acid-base reaction with the alcohol to form an alkoxide –<\/sup>OR.\u00a0 Acid is needed in order to turn the OH into a good leaving group (water), and a strong acid such as H2<\/sub>SO4<\/sub> or H3<\/sub>PO4<\/sub> is usually used.\u00a0 Since E1 reactions work with even very weak bases (water serves here as the base), the reaction is effective for secondary and tertiary alcohols.\u00a0 Primary alcohols may work, but higher temperatures are needed, and there are many side reactions such as rearrangement and ether formation (via SN<\/sub>2).<\/p>\n

\"\"<\/p>\n

E2 Elimination of dihalides for synthesis of alkynes<\/h2>\n

Alkynes can also be prepared via E2 elimination, though to form two pi bonds it means we must do a double elimination starting with a dihalide.\u00a0 Often a stronger base such as NaNH2<\/sub> is used, though NaOCH3<\/sub> or KOt<\/sup>Bu still work.\u00a0 The dihalide can have the halogens on the same carbon, or on neighboring carbons.\u00a0 Zaitsev’s Rule applies as long as rearrangement does not occur (which we will not cover here).<\/p>\n

\"\"<\/p>\n

There is a useful synthetic “trick” that can be done if three moles of NaNH2<\/sub> are used to with a terminal dihalide.\u00a0 The first two moles of base form the terminal alkyne, then the third mole will react with that alkyne to form an acetylide salt.\u00a0 That salt can then be treated with an alkyl halide to produce a longer chain alkyne via an SN<\/sub>2 reaction (see section 9.8.<\/a>).<\/p>\n

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