In Part 1, draw a mechanism for the reaction of butan-2-ol with sodium amide. In the box to the left, draw any necessary curved arrows. Show the products of the reaction in the box to the right. Include any nonzero formal charges and all lone pairs of electrons. In Part 2, check the box to indicate which side of the reaction is favored at equilibrium.

Question

Ana Bright

Chemistry

29 December, 11:57

In Part 1, draw a mechanism for the reaction of butan-2-ol with sodium amide. In the box to the left, draw any necessary curved arrows. Show the products of the reaction in the box to the right. Include any nonzero formal charges and all lone pairs of electrons. In Part 2, check the box to indicate which side of the reaction is favored at equilibrium.

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Pooh Bear · Answer 1

Formation of sodium butan-2-olate and ammonia is favoured

Explanation:

Butan-2-ol gives acid-base reaction with sodium amide to produce sodium butan-2-olate and ammonia as product. Amide anion is a strong base. Hence it deprotonates butan-2-ol. At equilibrium, product side is favoured (i. e. formation of sodium butan-2-olate and ammonia) because amide anion is stronger base than butan-2-olate anion due to lesser electronegativity of N atom and butan-2-ol is more acidic than ammonia due to higher electronegativity of O atom. Reaction has been shown below.