Hofmann Elimination

Exhaustive methylation converts an amine into a quaternary ammonium iodide, which is then exchanged to the hydroxide and heated, producing an alkene by an E2-type elimination known as the Hofmann elimination. Unlike typical E2 reactions that follow the Zaitsev rule, this reaction follows the Hofmann rule and gives predominantly the least substituted alkene. The reason is the very bulky trimethylammonium leaving group: the base must abstract a beta hydrogen, and it does so most easily at the least hindered, most accessible beta carbon, which usually bears more hydrogens and fewer alkyl groups. This steric preference, together with the poor leaving-group character of the neutral amine, channels the elimination toward the terminal, less substituted alkene. The Zaitsev option is wrong because it predicts the opposite, more substituted product. Markovnikov addition describes an electrophilic addition, not an elimination, so it is irrelevant. The claim that an anti-periplanar arrangement is impossible is false; E2 still requires it, and that geometry is attainable. This is a classic NCERT and JEE Advanced contrast between Hofmann and Zaitsev orientation. As a plausibility check, the dominance of the terminal alkene confirms that steric accessibility of the beta hydrogen, not alkene stability, dictates the outcome.