Hofmann Elimination and E2 Elimination are both types of bimolecular elimination reactions, but they differ in their mechanisms, products, and reaction conditions. Let's discuss each reaction in detail.Hofmann Elimination:Hofmann Elimination, also known as Hofmann Degradation or Exhaustive Methylation, is a reaction that involves the conversion of a primary amine to an alkene. The mechanism of Hofmann Elimination involves three main steps:1. Formation of a quaternary ammonium salt: The primary amine reacts with an excess of a methylating agent, such as methyl iodide CH3I , to form a quaternary ammonium salt. This step involves the successive alkylation of the amine nitrogen until all available lone pairs are consumed.2. Formation of a quaternary ammonium hydroxide: The quaternary ammonium salt is then treated with a strong base, such as silver oxide Ag2O or potassium hydroxide KOH , to form a quaternary ammonium hydroxide.3. Elimination: The quaternary ammonium hydroxide undergoes an elimination reaction to form an alkene and a tertiary amine. The least substituted alkene Hofmann product is the major product in this reaction, as the reaction proceeds through an anti-periplanar transition state with the least steric hindrance.E2 Elimination:E2 Elimination, also known as bimolecular elimination, is a reaction that involves the conversion of an alkyl halide to an alkene. The mechanism of E2 Elimination involves a single concerted step:1. Elimination: A strong base abstracts a -hydrogen from the alkyl halide, while the halogen leaves as a leaving group, forming a double bond between the - and -carbons. The reaction proceeds through an anti-periplanar transition state, and the most substituted alkene Zaitsev product is usually the major product due to the stability provided by hyperconjugation.Comparison:1. Substrate: Hofmann Elimination involves primary amines, while E2 Elimination involves alkyl halides.2. Reaction conditions: Hofmann Elimination requires a methylating agent and a strong base, while E2 Elimination requires only a strong base.3. Major product: Hofmann Elimination forms the least substituted alkene Hofmann product , while E2 Elimination forms the most substituted alkene Zaitsev product .Examples:Hofmann Elimination:R-NH2 + 3 CH3I R-N+ CH3 3I- quaternary ammonium salt R-N+ CH3 3I- + Ag2O R-N+ CH3 3OH- quaternary ammonium hydroxide R-N+ CH3 3OH- R-CH=CH2 + N CH3 3 least substituted alkene and tertiary amine E2 Elimination:R-CH2-CH2-Br + KOH R-CH=CH2 + KBr + H2O most substituted alkene and potassium bromide