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Alkyl halides Zaitsev’s rule

Elimination Reactions of Alkyl Halides Zaitsev s Rule 383... [Pg.383]

Alkenes are prepared by P elimination of alcohols and alkyl halides These reactions are summarized with examples m Table 5 2 In both cases p elimination proceeds m the direction that yields the more highly substituted double bond (Zaitsev s rule)... [Pg.221]

Here s an example how might we prove that E2 elimination of an alkyl halide gives the more highly substituted alkene (Zaitsev s rule, Section 11.7) Does reaction of 1-chloro-l-methylcyclohexane with strong base lead predominantly to 1-methyl cyclohexene or to methylenecyclohexane ... [Pg.453]

If an alkyl halide contains more than two carbons in its chain, and the carbon atoms adjacent to the carbon atom bonded to the halogen each have hydrogen atoms bonded to them, two products will form. The major product is predicted by Zaitsev s Rule, which states that the more highly branched alkene will be the major product. For example, in the dehydrohalogenation reaction between 2-chlorobutane and sodium methoxide, the major product is 2-butene. [Pg.49]

In Chapter 7, we saw that eliminations of alkyl halides usually follow Zaitsev s rule that is, the most substituted product predominates. This rule applies because the most-substituted alkene is usually the most stable. In the Hofmann elimination, however, the product is commonly the /east-substituted alkene. We often classify an elimination as giving mostly the Zaitsev product (the most-substituted alkene) or the Hofmann product (the least-substituted alkene). [Pg.905]

Zaitsev s rule In the elimination of HX from an alkyl halide, the more highly substituted alkene product predominates,... [Pg.434]

Zaitsev s rule cannot be used to predict the major products of the foregoing reactions because it does not take into account the fact that conjugated double bonds are more stable than isolated double bonds (Section 8.3). Therefore, if there is a double bond or a benzene ring in the alkyl halide, do not use Zaitsev s rule to predict the major (most stable) product of an elimination reaction. [Pg.405]

Why do alkyl halides follow Zaitsev s rule, while quaternary amines violate the rule When hydroxide ion starts to remove a proton from the alkyl bromide, the bromide ion immediately begins to depart and a transition state with an alkene-like stmc-ture results. The proton is removed from the 8-carbon bonded to the fewest hydrogens in order to achieve the most stable alkene-like transition state. [Pg.890]

List the alkenes that would be formed when each of the following alkyl halides is subjected to dehydrohalogenation with potassium ethoxide in ethanol and use Zaitsev s rule to predict the major product of each reaction (a) 2 bromo-3-methylbutane and (b)2 -bromo-2,3-dimethylbutane. [Pg.300]


See other pages where Alkyl halides Zaitsev’s rule is mentioned: [Pg.937]    [Pg.937]    [Pg.413]    [Pg.937]    [Pg.413]    [Pg.1029]    [Pg.937]    [Pg.937]    [Pg.413]    [Pg.937]    [Pg.413]    [Pg.1029]    [Pg.397]    [Pg.511]    [Pg.397]    [Pg.414]    [Pg.414]    [Pg.589]   
See also in sourсe #XX -- [ Pg.263 , Pg.264 ]

See also in sourсe #XX -- [ Pg.257 ]




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