Hofmann’s elimination is the process of creating tertiary amines and alkenes from the treatment of quaternary ammonium with excess methyl iodide, and the treatment of the resulting compound with silver oxide, water, and heat. The product formed from the treatment of quaternary ammonium with excess methyl iodide is called quaternary ammonium iodide salt. The replacement of the iodine by a hydroxyl anion is followed by an elimination reaction to form the alkene as well.
What is Hofmann Elimination?
As discussed earlier, the process of producing tertiary amines and alkenes from quaternary ammonium is called Hofmann elimination. This process can also be referred to as exhaustive methylation. The Hofmann rule states that the major alkene product is the least substituted and least stable product when it comes to asymmetrical amines. The Hofmann elimination process is named after its discoverer, the German chemist August Wilhelm Von Hofmann. The Hofmann elimination can be illustrated as follows.
It can be noted that methyl iodide is used in excess because it has no beta hydrogens and therefore cannot compete in the elimination reaction. If the given alkyl group contains two different sets of beta hydrogens which are available for the elimination process, the alkene isomer having less substituted double bond is formed as the major product. An important example of the Hofmann elimination process or the exhaustive methylation process is the synthesis of trans-cyclooctene which is shown below.
It is important to note that the neutral NR3 group acts as a better leaving group than the anionic NH2– or the NR2– groups. The outcome of the reaction is also affected by steric effects caused by large leaving groups.
Hofmann Elimination Mechanism
The Hofmann elimination mechanism or the exhaustive methylation mechanism begins with the attack of the amine with a beta-hydrogen on the methyl iodide to form the ammonium iodide salt. Now, the iodide reacts with the silver oxide to form silver iodide. This silver iodide is insoluble and hence is precipitated out of the solution. A silver oxide ion deprotonates the water to form a hydroxide ion. This mixture is now heated. The heat facilitates the elimination reaction, the hydroxide now picks up the beta-hydrogen from the ammonium ion. It also releases an amine to give the final alkene product. The mechanism can be illustrated via the following steps:
The ammonium iodide salt is formed from the treatment of the amine with a beta-hydrogen or quaternary ammonium with excess methyl iodide, as shown below.
The substitution of the iodide ion with the hydroxide ion is achieved by the reaction between the iodide and silver oxide and the successive deprotonation of water by the silver oxide ion. This step can be illustrated as follows.
The mixture is heated to facilitate an elimination reaction and form the required product.
Thus, the required olefin product is achieved. The tertiary amine product is also formed. The Hofmann elimination process can be categorized as an elimination reaction as well as an olefination reaction.
Frequently Asked Questions
What is the Hofmann Rule?
The Hofmann rule states that the major product in Hofmann eliminations and other similar elimination reactions is the less stable alkene (or the alkene featuring a lesser substituted double bond).
This rule is used to predict the regio selectivity of some elimination reactions. The pyrolysis of esters also obeys this rule. In general, the elimination reactions that take place via a cyclic transition phase obey the Hofmann rule.
What Products are Formed in the Hofmann Elimination Process?
The products formed from the exhaustive methylation process are an alkene and a tertiary amine. Initially, a quaternary ammonium iodide salt is obtained when quaternary ammonium is treated with methyl iodide.
The addition of silver oxide and water to the quaternary ammonium iodide salt and the subsequent heating of the mixture results in an elimination reaction that affords the required products.
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