The trans-alkenes are formed by the reduction of alkynes with
(a) H2-Pd/C, BaSO4
(b) NaBH4
(c) Na/li

Q: Why doesn't the reduction go all the way to an alkane?

The Birch reduction stops at the alkene stage because the resulting alkene is much harder to reduce than the initial alkyne. The radical anion of an alkene is significantly less stable, and liquid ammonia is not a strong enough acid to protonate it effectively.

Q: Why is the trans isomer formed exclusively?

The stereochemistry is determined during the formation of the vinylic radical intermediate. To minimize steric repulsion (crowding) between the two bulky alkyl groups attached to the double bond, they position themselves as far apart as possible, which is the trans geometry.

Q: What does Lindlar's catalyst do?

Lindlar's catalyst ($H_2$ with $Pd/BaSO_4$ poisoned with quinoline) is used to reduce alkynes to cis-alkenes. Because the reaction happens on the solid surface of the palladium metal, both hydrogen atoms are delivered to the same side of the alkyne, resulting in syn-addition.

Q: Can we use Lithium (Li) or Potassium (K) instead of Sodium (Na)?

Yes! Any Group 1 alkali metal (Li, Na, or K) dissolved in liquid ammonia will provide the solvated electrons necessary to drive the Birch reduction and yield a trans-alkene.

Q: What happens if we use this reagent on a terminal alkyne?

Terminal alkynes have a highly acidic hydrogen atom attached to the triple bond. Instead of reducing the triple bond, the strongly basic amide ions ($NH_2^-$) or the sodium metal itself will simply deprotonate the alkyne, forming a stable sodium acetylide salt. No reduction occurs!

Question

The trans-alkenes are formed by the reduction of alkynes with

Options:

(A) $H_2$-Pd/C, $BaSO_4$

(B) $NaBH_4$

(C) Na/liq. $NH_3$

(D) Sn-HCl

Sigma Percentile · Accepted Answer

### The Art of Alkyne Reduction

Alkynes, with their carbon-carbon triple bonds, are highly unsaturated and energy-rich molecules. When we want to reduce them, we have a choice: do we go all the way down to an alkane, or do we stop at the alkene stage? And if we stop at the alkene, can we control whether we get a *cis* or a *trans* geometry?

The answer is a resounding yes! The key lies in choosing the right chemical reagent. In this problem, we are specifically looking for the reagent that yields a **trans-alkene**.

### The Magic of Birch Reduction

The correct reagent for this transformation is **Sodium metal dissolved in liquid ammonia** ($Na/	ext{liq. } NH_3$). This classic reaction is known as the **Birch Reduction**.

Unlike standard catalytic hydrogenation (which uses $H_2$ gas and a metal surface to deliver two hydrogen atoms from the same side, yielding a *cis*-alkene), the Birch reduction operates through a completely different pathway: a **single-electron transfer mechanism**.

### Step-by-Step Mechanism

Let's break down the beautiful choreography of this reaction:

1. **The First Electron Transfer:** Sodium is an alkali metal, meaning it is highly electropositive and eager to give up its single valence electron. It donates this electron into the lowest unoccupied molecular orbital (LUMO)—the $\pi^*$ antibonding orbital—of the alkyne. This breaks one of the $\pi$ bonds, creating a **radical anion** intermediate ($R-\dot{C}=C^--R'$). This species is highly reactive, possessing both an unpaired electron and a negative charge.

2. **The First Protonation:** The radical anion is a very strong base. It abstracts a proton ($H^+$) from the solvent, ammonia ($NH_3$). This neutralizes the negative charge, leaving behind a **vinylic radical** ($R-\dot{C}=CH-R'$).

**The Crucial Stereochemical Step:** Here is where the magic happens! The vinylic radical can exist in either a *cis* or *trans* geometry. However, the bulky alkyl ($R$) groups repel each other sterically. To minimize this steric hindrance, the molecule adopts the much more stable **trans** configuration, placing the $R$ groups on opposite sides of the double bond.

3. **The Second Electron Transfer:** Another sodium atom approaches and donates a second electron to the unpaired radical. This converts the vinylic radical into a **vinylic anion** ($R-C^-=CH-R'$). Because the *trans* geometry was already established and is sterically favored, the anion maintains this configuration.

4. **The Final Protonation:** Finally, this strongly basic vinylic anion abstracts one last proton from another ammonia molecule. This completes the reaction, yielding the final **trans-alkene** product.

### The Verdict

Because the Birch reduction proceeds via a radical intermediate that naturally relaxes into the least sterically hindered conformation, it exclusively produces *trans*-alkenes via an **anti-addition** of hydrogen.

Therefore, the correct option is **(c) Na/liq. $NH_3$**.

*(Bonus Tip: If you ever need to synthesize a cis-alkene from an alkyne, you would use Lindlar's catalyst—$H_2$ with $Pd/BaSO_4$ poisoned with quinoline—which forces a syn-addition!)*

Animated Solution for Chemistry - Organic Chemistry: The trans-alkenes are formed by the reduction of alkynes with

Select Answer:

Visualized Solution

\text{The Objective}

\text{The Reagent: Birch Reduction}

\text{Step 1: Electron Transfer}

\text{Step 2: First Protonation}

\text{Step 3: Second Electron Transfer}

\text{Step 4: Final Protonation}

\text{Conclusion}

\text{The Way Forward}

The Sigma Insight: Hydrocarbons

The Art of Alkyne Reduction

The Magic of Birch Reduction

Step-by-Step Mechanism

The Verdict

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