Let us consider two guys are talking to each other in the kitchen about their favourite golden wrapper chocolate. A candy jar is on the kitchen countertop. So , they both start thinking about their favourite chocolate, which they want to get from the jar (golden wrapper chocolate). Golden wrapper chocolate is the "H" that is attached to the more substituted “C” and there are also some candies which are “H” that are attached to the less substituted “C” . So, the lean guy manages to take his favourite golden wrapper chocolate from the bottom of the jar. But the plump guy cannot reach the bottom of the jar and hence, is unable to get his favourite chocolate. At the end, the plump guy is sad as he did not get his favourite one, and the lean guy is happy for getting it. So, in this analogy, the plump guy which we have considered as a bulky base and the lean guy is a non-bulky base. The golden wrapper chocolate represents the hydrogen that will give the more substituted product (Saytzeff product), while the white candy represents the hydrogen that will give the least substituted product (Hoffman product). This is an analogy for Hofmann elimination. Let us study about this in detail.
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Hofmann elimination is a chemical reaction where alkenes and tertiary amines are obtained from quaternary ammonium salts. In Hofmann’s elimination, the quaternary ammonium salt is initially subjected to methyl iodide. The product formed is further subjected to heat, silver oxide, and water. The reaction of a quaternary ammonium salt with methyl iodide results in the formation of a compound called quaternary ammonium iodide salt. Later on, the hydroxide anion replaces the iodide ion present in the ammonium iodide salt. Once the iodide ion is replaced, an elimination reaction takes place resulting in the formation of an alkene.
As already said earlier, the reactants involved in Hofmann elimination are the quaternary ammonium salts whereas the end products are the alkene and tertiary amines.
Let us understand how Hofmann elimination takes place in quaternary ammonium salts.
Step 1: Exhaustive methylation
Amine is made to react with excess of methyl iodide to form quaternary ammonium salt, the process is known as exhaustive methylation. An unstable intermediate compound consisting of a positive charge is formed as a result of exhaustive methylation, which is known as quaternary ammonium salt.
Step 2: Replacement of the iodide ion
The next step of Hofmann elimination is the replacement of the iodide ion with the hydroxyl ion. Initially, the iodide ion comes into contact with the silver ion resulting in the formation of silver iodide. The proton released from the water molecule comes into contact with the silver oxide ion and the hydroxide ion is present in the quaternary ammonium salt. Silver hydroxide is eliminated from the molecule, resulting in the replacement with hydroxyl ion.
Step 3: Formation of alkene and tertiary amine
The last step involved in Hofmann’s reaction is the formation of an alkene and tertiary amine. The end products are formed when the hydroxyl-containing intermediate is subjected to heat in the presence of water. The amount of heat to be applied to the hydroxide salt should be between 1000C and 2000C. In the presence of water molecules, a tertiary amine separates itself from the compound leaving behind an alkene.
Consider the example given below to understand the mechanism of the Hofmann Elimination reaction better.
Propylamine is made to react with an excess of methyl iodide to form quaternary amine iodide salt which is further reacted with Silver oxide in presence of water. The replacement of the iodide ion with the hydroxyl ion takes place. In the presence of water and excess heat, propene is formed along with trimethylamine. The other byproducts of the Hofmann Elimination reaction include silver iodide and silver hydroxide.
According to Hofmann’s rule, an alkene that is less substituted is the major end product in all types of elimination reaction’s products including Hofmann’s reaction. Hofmann’s rule is applicable to all chemical compounds undergoing an elimination reaction and cyclic transition phase. Hofmann’s rule is mainly used to predict the selectivity of various elimination chemical reactions.
A. Alkene and tertiary amine
B. Secondary amine
D. Alkene and primary amine
Solution: Hofmann elimination is a chemical reaction where alkenes and tertiary amines are obtained from quaternary ammonium salts in the presence of heat and silver oxide.
Q2. Under what conditions, end products alkene and tertiary amine are formed in hofmann Exhaustive methylation?
A. water and strong heat,
B. heating strongly in acidic medium
C. heating strongly in basic medium
D. None of the above.
Solution: In the presence of water and strong heat, end products (alkene and tertiary amine) are formed.
Q3. Predict the major product of the following reaction.
D. None of the above
Q4. Which of the following amines can be used to prepare alkene from Hofmann Exhaustive methylation?
D. All of the above
Solution: Ammonia, primary, secondary or tertiary amines can be used as substrates in Hofmann exhaustive methylation.
Question 1. What happens to silver iodide formed after replacing the iodine ion with the hydroxyl ion?
Answer: Silver iodide is generally insoluble and thus it precipitates out of the solution.
Question 2. What is the role of excess methyl iodide in Hofmann Exhaustive methylation?
Answer: Excess methyl iodide replaces all the -H of amines by -CH3 to form quaternary amines.
Question 3. What is the role of Ag2O in hofmann Exhaustive methylation?
Answer: Ag2O abstracts the proton from the least substituted site to give the Hofmann product, i.e. least substituted alkene. Also, it enables the easy removal of I- by forming its precipitate which is not possible in case of any other bulky bases like t-BuO- .
Question 4. Which mechanism is followed in Hofmann Exhaustive methylation and why?
Answer: Since the reaction occurs in basic medium E2 mechanism is followed.
Substitution vs Elimination
E1 vs E2
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