Chapter 25, Problem 25.57P

Interpretation Introduction

(a)

Interpretation:

The major product of the given reaction is to be predicted and a complete mechanism is to be provided.

Concept introduction:

Reactive metals like Na and Li dissolve in liquid ammonia. This is not a simple physical process. The atoms of the metal ionize and the electron lost is stabilized by extensive solvation by ammonia molecules. The solvated electron behaves like a radical even though it does not belong to a particular atom. It adds to one of the carbons of the triple bond of an alkyne. The bond order decreases to two, the carbon to which the electron adds becomes a highly reactive carbanion and the other carbon is converted to a radical. The configuration of the double bond is trans as a result of the steric repulsion between the two groups attached to the carbons of the initial triple bond.

The carbanion is a strong nucleophile and a base. It extracts a proton from an ammonia molecule to form a neutral alkenyl radical and an amide (${\text{NH}}_{\text{2}}{}^{\text{-}}$) anion. The alkenyl radical adds another electron and is converted to a carbanion. This carbanion also extracts a proton from another ammonia molecule to form a trans alkene.

Overall, the reaction is the addition of two hydrogen atoms to the triple bond to produce a trans alkene.

Interpretation Introduction

(b)

Interpretation:

The major product of the given reaction is to be predicted and a complete mechanism is to be provided.

Concept introduction:

Reactive metals like Na and Li dissolve in liquid ammonia. This is not a simple physical process. The atoms of the metal ionize and the electron lost is stabilized by extensive solvation by ammonia molecules. The solvated electron behaves like a radical even though it does not belong to a particular atom. It adds to one of the carbons of the triple bond of an alkyne. The bond order decreases to two, the carbon to which the electron adds becomes a highly reactive carbanion and the other carbon is converted to a radical. The configuration of the double bond is trans as a result of the steric repulsion between the two groups attached to the carbons of the initial triple bond.

The carbanion is a strong nucleophile and a base. It extracts a proton from an ammonia molecule to form a neutral alkenyl radical and an amide (${\text{NH}}_{\text{2}}{}^{\text{-}}$) anion. The alkenyl radical adds another electron and is converted to a carbanion. This carbanion also extracts a proton from another ammonia molecule to form a trans alkene.

Overall, the reaction is the addition of two hydrogen atoms to the triple bond to produce a trans alkene.