Alkene from halogenoalkanes

An important method for preparing alkenes involves the elimination (an El or E2 mechanism) of HX from halogenoalkanes, RX (Section 5.3.2). Alcohols (ROH) can also be converted into alkenes by activating the OH group (e.g. by protonation or conversion into a tosylate) to make this into a better leaving group (Section 5.2.2). 

In case of the direct reaction of the natural oil or lower alkyl ester of natural fatty acid and the amine the reaction method for producing the amide derivatives is as follows That is, about 1 mol of the said oils and 1 to 100 equivalent mols of the said amines are mixed in the absence or presence of solvents such alcohols as methanol, ethanol or the like, such aromatic hydrocarbons as benzene, toluene, xylene or the like, such halogenoalkanes as methylene chloride, chloroform, carbon tetrachloride or the like, and such alkenes or alkanes as petroleum ether, benzene, gasoline, ligroin or cyclohexane, such ethers as tetrahyrofuran, dioxane and the like, or a mixture thereof, and the mixture is subjected to the reaction in the absence or presence of catalyst amount or equimolar amount to the amine of an auxiliary agent of condensation, such as alkoholate of alkali metal, i.e. lithium, methylate, lithium ethylate, sodium methylate, sodium ethylate, potassium-t-butylate and the like, or acidic auxiliary agents, i.e. p-toluenesulfonic acid and the like, thereby to yield the amide derivatives. In this reaction, a formal alcohol may be removed from the reaction system. 

Helwig and Hanack (1985) were, however, successful with a synthetic method that was useful for the formation of alkenediazonium salts, namely the elimination of hydrogen halide from 1-halogenoalkanal (4-toluene)sulfonylhydrazones and dissociation of the product, the alkene(4-toluene)sulfonyldiazenes, into alkenediazonium salts with the help of Lewis acids (see Sect. 2.10, Scheme 2-109). By analogy, 1,2-dihalogenoalkanal (4-toluene)sulfonylhydrazones should form alkynediazonium salts. This pathway was indeed successful (2-111). 

The two main mechanisms for elimination reactions of halogenoalkanes are El and E2. In both cases, the mechanism involves the loss of HX from RX (e.g. R2CH-CXR2) to form an alkene (e.g. R2C=CR2). 

A different alkene stereoisomer is obtained from each diastereoisomer of the other (see Section 3.3.2.4j halogenoalkane. 

The addition of HX (X = Cl, Br, I) to an alkene, to form halogenoalkanes, occurs in two steps. The first step involves addition of a proton (i.e. the electrophile) to the C=C bond to make the more stable intermediate carbocation. The second step involves nucleophilic attack by the halide anion. This produces a racemic halogenoalkane because the carbocation is planar and so can be attacked equally from either face. (These addition reactions are the reverse of halogenoalkane eUmination reactions.)... 

A close look at the nature of a nucleophile will emphasize that it shares common features with a Lewis hase (see Chapter 18). Indeed, a nucleophilic species can act as such a base if the reaction conditions are appropriate - it can remove a proton (H ion) from a halc enoalkane and thereby initiate an elimination reaction. In this type of reaction HX is eliminated from the halogenoalkane and an alkene is produced. It is essential to realize that, given the similarity of the reagents involved, the two processes of nucleophilic substitution and elimination are generally in competition with each other. If a primary halogenoalkane is reacted with aqueous alkali (OH (aq)) then the substitution reaction we have discussed earlier is favoured. However, if ethanolic alkali (OH (ethanol)) is used, then the elimination reaction is favoured. 

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