Discussion
2-Butanamine on reaction with excess CH3I and Ag2O, H2O followed by heating at 150°C, gives the following major product :
Answer: A
The major product is 1-butene.
This is the Hofmann elimination. Excess CH3I methylates the amine exhaustively to the quaternary salt CH3CH2CH(CH3)N(CH3)3+ I−; moist Ag2O exchanges iodide for hydroxide; and on heating, hydroxide removes a β-hydrogen in an E2 reaction with trimethylamine as the leaving group.
The bulky, positively charged -N(CH3)3+ group makes the base take the most accessible, most acidic β-hydrogen, which is on the terminal CH3, so the less substituted alkene (Hofmann product) dominates. 2-Butene is the more substituted Zaitsev product that a halide elimination would favour. Butane would need a reduction, and 2-butanol a substitution, neither of which occurs.
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