Q21E - Organic Chemistry (Mcmurry)

Question

Predict the product(s) and show the complete electron-pushing mechanism for each of the following dissolving metal reductions.

Step-by-Step Solution

Answer

a). 

b).

c).

General mechanism for Li, NH 3 with alkyne

The Lithium atom acts as an electron donor.   acts as a proton donor in the reaction. The general mechanism for the reduction of alkynes is given below. Reduction with   gives products with trans stereochemistry. 

Reaction of given alkyne with lithium metal and ammonia

In the first step of the reaction, the Li atom donates an electron to form ${\mathrm{Li}}^{+}$ ion and one pi bond breaks. This results in the generation of a radical on one carbon atom along with the formation of carbanion on the adjacent carbon. The carbanion formed extracts a proton from ${\mathrm{NH}}_3$. In the following steps the Lithium atom donates one electron. Therefore the carbon radical forms a carboanion. The carboanion form abstracts a proton from ${\mathrm{NH}}_3$  to give the required alkene.

Reaction of given alkyne with lithium metal and ammonia

In the first step of the reaction, the Li atom donates an electron to form ${\mathrm{Li}}^{3+}$ ion and one pi bond breaks. This results in the generation of a radical on one carbon atom along with the formation of carbanion on the adjacent carbon. The carbanion formed extracts a proton from ${\mathrm{ND}}_3$. In the following steps the Lithium atom donates one electron. Therefore the carbon radical forms a carbanion. The carbonaton form abstracts a proton from ${\mathrm{NiD}}_3$ to give the required alkene.

Reaction of given alkyne with lithium metal and ammonia

In the first step of the reaction, the Li atom donates an electron to form ${\mathrm{Li}}^{+}$ ion and one pi bond breaks. This results in the generation of a radical on one carbon atom along with the formation of carbanion on the adjacent carbon. The carbanion formed extracts a proton from ${\mathrm{NH}}_3$. In the following steps the Lithium atom donates one electron. Therefore the carbon radical forms a carbanion. The carboanion form abstracts a proton from ${\mathrm{NH}}_3$ to give the required alkene.