Alkyl halides elimination with bases

The rate of the alkylation reaction depends on the enolate concentration, since it proceeds by a SN2-mechanism. If the concentration of the enolate is low, various competitive side-reactions may take place. As expected, among those are E2-eliminations by reaction of the alkyl halide 2 with base. A second alkylation may take place with mono-alkylated product already formed, to yield a -alkylated malonic ester however such a reaction is generally slower than the alkylation of unsubstituted starting material by a factor of about 10. The monoalkylation is in most cases easy to control. Dialkylated malonic esters with different alkyl substituents—e.g. ethyl and isopropyl—can be prepared by a step by step reaction sequence ... 

Section 8 13 When nucleophilic substitution is used for synthesis the competition between substitution and elimination must be favorable However the normal reaction of a secondary alkyl halide with a base as strong or stronger than hydroxide is elimination (E2) Substitution by the Sn2 mechanism predominates only when the base is weaker than hydroxide or the alkyl halide is primary Elimination predominates when tertiary alkyl halides react with any anion... 

The El reaction involves the formation of a planar carbocation intermediate. Therefore, both syn and anti elimination can occur. If an elimination reaction removes two substituents from the same side of the C—C bond, the reaction is called a syn elimination. When the substituents are removed from opposite sides of the C—C bond, the reaction is called an anti elimination. Thus, depending on the substrates El reaction forms a mixture of cis (Z) and trans (E) products. For example, tert-hutyl bromide (3° alkyl halide) reacts with water to form 2-methylpropene, following an El mechanism. The reaction requires a good ionizing solvent and a weak base. When the carbocation is formed, SnI and El processes compete with each other, and often mixtures of elimination and substitution products occur. The reaction of t-butyl bromide and ethanol gives major product via El and minor product via SnI-... 

The rate of this elimination is proportional to the concentrations of both the alkyl halide and the base, giving a second-order rate equation. This is a bimolecular process, with both the base and the alkyl halide participating in the transition state, so this mechanism is abbreviated E2 for Elimination, bimolecular. 

In E2 elimination with bases like KOH and CH30Na, most alkyl halides give Saytzeflf orientation. Certain other compounds (quaternary ammonium salts, Sec. 23.5, for example) give Hofmann orientation. Alkyl sulfonates fall in between. With each kind of compound, orientation is affected—sometimes drastically—by the choice of base and solvent, and by stereochemistry. (The percentage of 1-hexene from 2-hexyl chloride, for example, jumps from 33% in CH ONa/ CH3OH to 91% in /-BuOK//-BuOH, evidently for steric reasons.) In all this, we should remember that orientation is a matter of relative stabilities of competing transition states these stabilities are determined by electronic factors—alkene character and carbanion character—with superimposed conformational factors. 

Alkyl halides react with nucleophiles in substitution reactions and with bases in elimination reactions. 

A tertiary alkyl halide is the least reactive of the alkyl halides in an Sn2 reaction and the most reactive in an E2 reaction (Table 11.5). Consequently, only the elimination product is formed when a tertiary alkyl halide reacts with a nucleophile/base under Sn2/E2 conditions. 

Now that you know that this is an 8 2 reaction (the alkyl halide reacts with a high concentration of a good nucleophile), you can understand why it is best to use primary alkyl halides and methyl halides in the reaction. These alkyl halides are the only ones that form primarily the desired substitution product. A tertiary alkyl halide would form only the elimination product, and a secondary alkyl halide would form mainly the elimination product because the acetylide ion is a very strong base. 

As crowding at the carbon that bears the leaving group decreases, the rate of nucleophilic substitution becomes faster than the rate of elimination. A low level of steric hindrance to approach of the nucleophile is one of the special circumstances that permit substitution to predominate, and primary alkyl halides react with alkoxide bases by an 8 2 mechanism in preference to E2 ... 

A second reaction typical of alkyl halides is elimination of the halogen to regenerate the C=C double bond. Halogen elimination occurs to some extent whenever alkyl haUdes react with base ( OH ), but tends to become the main reaction if the alkyl halide is tertiary ... 

A secondary alkyl halide, compared with a primary alkyl halide, reacts slower in an Sn2 reaction and faster in an E2 reaction (page 467). Thus, a secondary alkyl halide forms both substitution and elimination products under Sn2/E2 conditions. The relative amounts of the two products depend on the strength and bulk of the nucleophile/base. The stronger and bulkier the nucleophile/base, the greater the percentage of the elimination product. 

Readily available A A -dimethyldithiocarbamoylacetonitrile (15) is alkylated by alkyl halide and aqueous base longer reaction times lead to high-yield dialkylation. Elimination of the cyano and dithiocarbamoyl groups with consequent aldehyde or ketone formation is achieved simply on reaction with sodium hydroxide in hot ethanol or on treatment with A -bromosuccinimide (Scheme 20). ... 

In some respects, the alkylation of enolate anions resembles nucleophilic substitution. We recall that many nucleophiles displace leaving groups from primary alkyl halides by an Sj 2 mechanism (Section 9.3). A similar reaction occurs with secondary alkyl halides, but competing elimination reactions also occur. Primary alkyl halides react with carbanions, such as the alkynide ion, by an Sj 2 mechanism. (Secondary alkyl halides react not only in displacement reactions but also in elimination reactions because the alkynide ion is a strong base.)... 

When we discussed elimination reactions in Chapter 5 we learned that a Lewis base can react with an alkyl halide to form an alkene In the present chapter you will find that the same kinds of reactants can also undergo a different reaction one m which the Lewis base acts as a nucleophile to substitute for the halo gen substituent on carbon... 

Substitution can take place by the S l or the 8 2 mechanism elimination by El or E2 How can we predict whether substitution or elimination will be the principal reac tion observed with a particular combination of reactants The two most important fac tors are the structure of the alkyl halide and the basicity of the anion It is useful to approach the question from the premise that the characteristic reaction of alkyl halides with Lewis bases is elimination and that substitution predominates only under certain special circumstances In a typical reaction a typical secondary alkyl halide such as iso propyl bromide reacts with a typical Lewis base such as sodium ethoxide mainly by elimination... 

FIGURE 8 11 When a Lewis base reacts with an alkyl halide either substitution or elimination can occur Sub stitution (Sn2) occurs when the Lewis base acts as a nu cleophile and attacks carbon to displace bromide Elimi nation (E2) occurs when the Lewis base abstracts a pro ton from the p carbon The alkyl halide shown is iso propyl bromide and elimi nation (E2) predominates over substitution with alkox ide bases... 

As a practical matter elimination can always be made to occur quantitatively Strong bases especially bulky ones such as tert butoxide ion react even with primary alkyl halides by an E2 process at elevated temperatures The more difficult task is to find condifions fhaf promofe subsfifufion In general fhe besf approach is fo choose condi lions lhal favor fhe 8 2 mechanism—an unhindered subslrale a good nucleophile lhal IS nol slrongly basic and fhe lowesl praclical lemperalure consislenl wilh reasonable reaclion rales... 

The acidity of acetylene and terminal alkynes permits them to be converted to their conjugate bases on treatment with sodium amide These anions are good nucleophiles and react with methyl and primary alkyl halides to form carbon-carbon bonds Secondary and tertiary alkyl halides cannot be used because they yield only elimination products under these conditions... 

Secondary and tertiary alkyl halides are not suitable because they react with alkox ide bases by E2 elimination rather than by 8 2 substitution Whether the alkoxide base IS primary secondary or tertiary is much less important than the nature of the alkyl halide Thus benzyl isopropyl ether is prepared m high yield from benzyl chloride a pri mary chloride that is incapable of undergoing elimination and sodium isopropoxide... 

We see that a secondary alkyl halide is needed as the alkylating agent The anion of diethyl malonate is a weaker base than ethoxide ion and reacts with secondary alkyl halides by substitution rather than elimination Thus the synthesis of 3 methylpentanoic acid begins with the alkylation of the anion of diethyl mal onate by 2 bromobutane... 

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