Elimination rules apply to the process of taking a compound and producing a more simplified and simpler form. It can be done to reduce the number of steps, yield, purity, etc.
Mechanism of the Elimination Reaction:
When elimination takes place, one more molecule is removed and replaced by another compound of a similar or the same structure. Thus, it is known as, “reduction process”. It is always the end result that matters! It is not just a reaction that results in elimination! If the reaction will be reduced from a compound to an atom, it is known as, “elimination.”
The reaction does not take place at the bond which is to be reduced (eliminated). It occurs at a bond that is weaker and less strong than the bond which is to be reduced.
The most reactive carbon in an elimination process is the C-C bond or C-C-C bond. So, the molecule will be in a highly excited state when it takes place. A molecule that contains a high concentration of carbon is the source of energy for the elimination reaction. So, the formation of the carbocation is an important factor in the initiation of the elimination.
The position of the C-C bond to be reduced (eliminated) depends on the structure of the starting material and the product. Generally, the most reactive C-C bond is the end of a ring. Generally, the most reactive C-C bond is the end of a ring.
A structure that contains the highest concentration of energy is an advantage and is more reactive in an elimination reaction. When two or more C-C bonds are involved in the formation of a new C-C bond during elimination, one bond will have a stronger and shorter life. While the bond with a weak bond will be broken and it is eliminated.
Examples of Elimination
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A compound of the formula
CH2 = CH – CO – CH = CH = CH CH2Cl becomes CH3Cl
It is an example of an elimination reaction with two carbons involved (in the form of two bonds that are breaking and one bond which is being formed).
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The same formula in the 1st edition becomes COCl2
when the compound has one carbon that is broken (CH2) and one carbon that is being formed (CH = CHCl).
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The same formula in the 2nd edition becomes CHCl2
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In 1st edition C = C – C – Cl becomes CCl4
It is an example of an elimination reaction with three carbons involved (two bonds that are breaking and one bond which is being formed).
Hofmann Elimination and the Underlying Mechanism
The Hofmann elimination reaction requires a leaving group to be generated in one step. From the point of view of the mechanism, the elimination reaction is a rearrangement reaction. We will now consider the different processes possible and examine how we might predict the most likely reaction to take place. This will give an insight into the nature of the reaction.
If we look at a typical peptide, it is likely that we will observe the formation of the following intermediate.
For the purposes of this article, it is easier to consider the Hofmann elimination of a carboxylic acid (such as an alpha-amino acid) using sodium hydride. The reaction can take place using either the sodium alkoxide or hydroxide as a base, thus, we have an example of the hydroxide route or a route involving a hydroxide species as shown in.
In either case, we have a primary alkyl alkoxide. This can be considered a nucleophile and will react with the reactive center to form a reactive species of the formula (a) shown in This species, known as the “activated species” will react with the side chain of the carboxylic acid or any other reactive center to give the intermediate shown in (b). It is important to realize that the reactive center will not remain static and has a high probability of moving around to react with the activated species. In fact, the reactive center will not react at all and the reaction will be aborted or will proceed to the other side of the molecule.
When looking at a peptide or protein, there are two ways that we can activate the side chain. The first is to activate the carboxylic acid, but before it can react with the active species, we have to generate the leaving group. The second is to generate the active species from the carboxylic acid, and in doing so we generate the leaving group and the active species.
In the context of the example, sodium methoxide is an effective base to generate the sodium hydroxide, an effective nucleophile to react with the carboxylic acid to form the reactive species. An important point here is that both methods for activation are effective, but for our purposes of describing activation of side-chain carboxyl groups in peptides and proteins, we will refer to them as the “base” and “nucleophile” route, respectively.
Hofmann elimination reactions are the elimination reactions of quaternary ammonium salts producing tertiary amines and alkenes. It is also known as Hofmann exhaustive methylation and Hofmann degradation. The products of this reaction, tertiary amines and alkenes are known as Hofmann products. Silver hydroxide and heat is used in the reaction to get the product. This reaction was given by August Wilhelm Von Hofmann in 1851.
Hofmann Elimination Reaction –
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Hofmann Elimination of Cyclic Amines –
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Hofmann Rule
According to Hofmann rule the less stable alkene will be the main product in Hofmann Elimination reactions or other such kinds of elimination reactions. This rule is used in the prediction of regioselectivity of elimination reactions.
So, when you have a bulky leaving group in the elimination reaction, the least substituted alkene will be the main product.
Examples of reactions which follow Hofmann rule –
Cope elimination also follows Hofmann rule.
Reaction –
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Reaction –
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Hofmann Elimination Mechanism
Hofmann Elimination Reaction Mechanism involves E2 elimination mechanism. It involves 3 steps. The Hofmann exhaustive methylation mechanism starts with formation of the ammonium iodide salt then ammonium iodide salt reacts with silver oxide and gives silver iodide as precipitate. Deprotonation of water also takes place by silver oxide. It results in the formation of hydroxide ions. Now, this mixture is heated. This facilitates the elimination reaction and gives rise to alkene.
Here we have explained all three steps in detail –
Step 1.
Ammonium iodide salt is formed in this step. Tertiary amine reacts with methyl iodide and forms ammonium iodide salt.
Reaction –
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Step 2.
In this step substitution of iodide anion with hydroxide formed by deprotonation of water molecules.
Reaction –
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Step 3.
In this step the reaction mixture is heated to start the elimination reaction and to get the required product alkenes.
Reaction –
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In the above step the tertiary amine product is also formed.
E2 transition state has been shown in the diagram below –
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Three steps involved in the Hofmann elimination mechanism have been explained above can be explained in the following way as well –
()
Uses of Hofmann Elimination Reaction
The primary use of the Hofmann Elimination reaction is to synthesize alkenes.
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Physiological importance of Hofmann elimination reaction – the design of skeletal muscle relaxant is superior to tubocurarine because of its less cardiac side effects and self-destruction mechanism into blood by Hofmann Elimination reaction.
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It has uses in the pharmaceutical field as well.
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It is used in the synthesis of anthranilic acid.
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It is used in the formation of primary reactants involved in the production of benzene and its derivatives.
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It is used in the production of precursors of tryptophan.
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It is used in the production of sweetening agents.
Hofmann Elimination reaction is an important name reaction for CBSE Class 12 Board examination perspective. Generally, direct questions are asked related to name reactions in class 12 CBSE board exams. Although, this reaction is an important name in competitive exams point of view as well. Generally, questions based on Hofmann reactions are asked in JEE and NEET exams as well. So, you should understand this reaction nicely and should practice its mechanism many times. Still if you have any doubts refer to free PDFs of study material and NCERT Solutions of class 12 Chemistry available on website.