Experiment 1) 2-bromo-2-methylbutane is reacted with KOH in 1-propanol. (100% el

Question

Experiment 1) 2-bromo-2-methylbutane is reacted with KOH in 1-propanol. (100% elimination)

Experiment 2) 2-bromo-2-methylbutane is reacted with Sodium Methoxide in Methanol (44% substitution, 56% elimination)

A) compare the reaction conditions used in the elimination experiment (experiment 1) versus experiment 2?

B) In experiment 1, elimination was more favored. What parameter in the reaction conditions could have caused a shift in the product in favor of elimination?

C) If kinetic experiments were run on the reactions, they would give a higher rate (faster). True/false and explain why

Explanation / Answer

A)

In both experiments Reactants are one and the same. Reagents are different. In first experiment reagent KOH which contains OH- ions which strong nucleophiles and strong base. In second experiment reagent NaOCH3 which contains OCH3- ions which are also strong nucleophiles and strong bases.inboth experiments solvents are alcohols.

B) In both experiments Reactant alkyl halide is 30 alkyl halide. In both experiments reagents are containing strong nucleophile and strong basic nature, in the case of first experiment OH- is less bulky than OCH3- ions in the second experiment. So in first experiment OH- effectively eliminate the proton from adjacent carbon and gives elimination product.

C) True

Reason: If Hydrogen on Beta carbon is substituted by its isotope H2 then Beta C-H bond can cleaved easily due to heavy mass of H isotope.Hence the reaction is faster

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.