Silver oxide, Hofmann elimination

Many of the reactions of amines are familiar from past chapters. Thus, amines react with alkyl halides in S 2 reactions and with acid chlorides in nucleophilic acyl substitution reactions. Amines also undergo E2 elimination to yield alkenes if they are first qualernized by treatment with iodomethane and then heated with silver oxide, a process called the Hofmann elimination. 

The Hofmann elimination is useful synthetically for preparing alkenes since it gives the least substituted alkene. The reaction involves thermal elimination of a tertiary amine from a quaternary ammonium hydroxide these are often formed by alkylation of a primary amine with methyl iodide followed by reaction with silver oxide. The mechanism of the elimination is shown in Scheme 1.13 in this synthesis of 1-methyl-1-... 

This and other evidence (16) indicated that eseretholemethine (VI R = C2H5) was a pseudo base or carbinol amine. Thus, on oxidation with silver nitrate an oxindole was obtained (dehydroeseretholemethine, VII R = C2H5), which on subsequent Hofmann degradation and reduction of the Hofmann elimination product gave 5-ethoxy-l,3-dimethyl-3-ethyloxindole (13). The methyl homolog (VII R = CHa),... 

Hofmann elimination (Section 24.7) a method for effecting the elimination reaction of an amine to yield an alkene. The amine is first treated with excess iodomethane, and the resultant quaternary ammonium salt is heated with silver oxide. 

Apart from the reactions of diazonium salts, a number of other reactions are known in which the C-N bond is broken. The best known of these is the Hofmann elimination of quaternary ammonium hydroxides (Scheme 2.37). An amine is converted by methylation with methyl iodide to the quaternary ammonium salt ( exhaustive methylation ). The iodide, on treatment with moist silver oxide, forms the quaternary ammonium hydroxide which undergoes a bimolecular elimination to form an alkene. The bimolecular elimination of onium salts yields the least alkylated alkene. This substitution pattern is determined by the ease with which a hydrogen atom can be attacked by the base. 

Amines are converted into alkenes by a two-step process called the Hofmann elimination. Reaction of the amine with excess CHjI in the first step yields an intermediate that undergoes E2 reaction when treated with basic silver oxide. Pentyl-amine, for example, yields 1-pentene. Propose a structure for the intermediate, and explain why it undergoes ready elimination. (See Section 24.7.)... 

An example of a reaction that shows this latter regiochemistry is the Hofmann exhaustive methylation, which is sometimes called, rather ambiguously, the Hofmann degradation (this is a term best avoided, because it is sometimes also used to describe the Hofmann rearrangement). In this reaction, an amine is methylated with methyl iodide until the quaternary ammonium iodide is formed. This is then treated with moist silver oxide to convert the iodide to the hydroxide. Write out the final elimination step that occurs on heating. 

For a quaternary ammonium ion to undergo an elimination reaction, the counterion must be hydroxide ion because a strong base is needed to start the reaction by removing a proton from a j8-carbon. Since halide ions are weak bases, quaternary ammonium halides cannot undergo a Hofmann elimination reaction. However, a quaternary ammonium halide can be converted into a quaternary ammonium hydroxide by treating it with silver oxide and water. The silver halide precipitates, and the halide ion is replaced by hydroxide ion. The compound can now undergo an elimination reaction. 

The Hofmann elimination reaction was used by early organic chemists as the last step of a process known as a Hofmaim degradation—a method used to identify amines. In a Hofmann degradation, an amine is exhaustively methylated with methyl iodide, treated with silver oxide to convert the quaternary ammonium iodide to a quaternary ammonium hydroxide, and then heated to allow it to undergo a Hofmann elimination. Once the alkene is identified, working backwards gives the stmcture of the amine. 

This process is called a Hofmann elimination. The function of silver oxide is to convert one ammonium salt into another ammonium salt by exchanging the iodide ion for a hydroxide ion. 

The Hofmann elimination reaction, on the other hand, is a process where a quaternary ammonium salt is decomposed into an olefin and a tertiary amine via exposure to basic conditions e.g., silver oxide and water) (Scheme 2.2). 

In the procedure of Hofmann elimination, the amine is first completely methylated with excess iodomethane (exhaustive methylation) and then treated with wet silver oxide (a source of HO ) to produce the ammonium hydroxide. Heating degrades this salt to the alkene. When more than one regioisomer is possible, Hofmann elimination, in contrast to most E2 processes, tends to give less substituted alkenes as the major products. Recall that this result adheres to Hofmann s rule (Section 11-6) and appears to be caused by the bulk of the ammonium group, which directs the base to the less hindered protons in the molecule. 

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