The Hofmann elimination reaction and the E2 reaction are both elimination reactions that involve the removal of a proton and a leaving group from adjacent carbon atoms, resulting in the formation of a double bond. However, there are some key differences between the two reactions in terms of their mechanisms, selectivity, and the products formed.Hofmann Elimination Reaction:The Hofmann elimination reaction involves the conversion of an amine to an alkene through the formation of a quaternary ammonium salt, followed by the elimination of a leaving group. The mechanism proceeds as follows:1. Formation of a quaternary ammonium salt: A tertiary amine reacts with an alkyl halide to form a quaternary ammonium salt.2. Treatment with a strong base: The quaternary ammonium salt is treated with a strong base, such as silver oxide Ag2O or potassium hydroxide KOH , which abstracts the least sterically hindered proton from the -carbon.3. Elimination of the leaving group: The electrons from the C-H bond form a double bond between the - and -carbons, and the leaving group the amine is eliminated, resulting in the formation of an alkene.The Hofmann elimination reaction typically favors the formation of the less substituted alkene Hofmann product due to the steric hindrance around the leaving group.E2 Reaction:The E2 reaction is a bimolecular elimination reaction that involves the abstraction of a proton from the -carbon and the simultaneous elimination of a leaving group from the -carbon, resulting in the formation of a double bond. The mechanism proceeds as follows:1. Treatment with a strong base: A strong base, such as ethoxide ion EtO- or hydroxide ion OH- , abstracts a proton from the -carbon.2. Elimination of the leaving group: The electrons from the C-H bond form a double bond between the - and -carbons, and the leaving group usually a halide is eliminated, resulting in the formation of an alkene.The E2 reaction typically favors the formation of the more substituted alkene Zaitsev product due to the stability provided by hyperconjugation.Example of Hofmann Elimination Reaction:Tertiary amine: CH3 3NAlkyl halide: CH3IQuaternary ammonium salt: [ CH3 4N]+I-Base: Ag2OProduct: CH2=C CH3 2 less substituted alkene Example of E2 Reaction:Substrate: 2-bromo-2-methylpentaneBase: EtO-Product: 2-methyl-2-pentene more substituted alkene Factors affecting selectivity:1. Steric hindrance: Hofmann elimination favors the less substituted alkene due to steric hindrance around the leaving group, while E2 reaction favors the more substituted alkene due to hyperconjugation.2. Base size and strength: Bulky bases tend to favor the formation of the less substituted alkene in E2 reactions, while smaller, strong bases favor the more substituted alkene.3. Leaving group: A good leaving group is essential for both reactions, with halides being common leaving groups in E2 reactions and amines in Hofmann elimination reactions.