Alkynes on reduction with sodium in liquid ammonia forms

Options

A. cis alkene
B. cis alkane
C. trans alkene
D. trans alkane

Answer

To solve the question regarding the reduction of alkynes with sodium in liquid ammonia, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactant in this case is an alkyne, which has the general formula R-C≡C-R'.

2. **Understand the Reaction Conditions**: The reaction involves sodium (Na) and liquid ammonia (NH₃). This combination is known as the Barge reduction.

3. **Reaction Mechanism**: During the Barge reduction, sodium donates electrons to the alkyne, resulting in the formation of a radical anion. This radical anion then undergoes protonation by ammonia.

4. **Formation of the Product**: The product formed from this reaction is a cis-alkene. However, due to the specific conditions of the Barge reduction, the product is actually a trans-alkene. The general structure of the product can be represented as R-C=C-H.

5. **Conclusion**: Therefore, the reduction of an alkyne with sodium in liquid ammonia results in the formation of a trans-alkene.

### Final Answer:
The product formed from the reduction of alkynes with sodium in liquid ammonia is a **trans-alkene**.

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