can you summarise why trans alkene is formed in birch reduction?

Why Does Birch Reduction of Alkynes Give Trans-Alkenes?

Birch reduction employs alkali metal (Na, K, or Li) in liquid $NH_3$ plus an alcohol (ROH). Key mechanistic ideas:

1. Electron–Proton Sequence (E–P–E–P)
   • Step 1: $M \;\rightarrow\; M^+ + e^-$ (electron injection).
   • Step 2: Proton donation from ROH.
   • Steps 3 & 4: Another electron, then another proton.
   This staggered E–P–E–P order means the two protons almost never hit the same face.
2. Radical Anion Geometry
   • After the first electron, the alkyne forms a linear radical anion.
   • Protonation bends the carbon now bearing $H$, forcing the remaining anion center to the opposite side.
3. Electrostatic Repulsion
   • The anion produced after the second electron strongly repels the proton source on the near side, encouraging approach from the far side.
4. Anti Addition Overall
   • Because the two $H$ atoms add from opposite faces, stereochemistry is anti. For alkynes, anti addition converts the triple bond into a trans-alkene (E-alkene).
   • Contrast: Lindlar catalyst performs syn addition (both $H$ atoms on the same face), giving cis-alkene.
   Chemically, the sequence can be written:

$\text{RC}\equiv\text{CR'} \xrightarrow[]{e^-} \boxed{\text{RC}\equiv\text{CR'}^{\,\bullet-}} \xrightarrow[]{H^+} \text{RCH}=\text{C}^{\bullet}\text{R'} \xrightarrow[]{e^-} \text{RCH}=\text{C}^-\text{R'} \xrightarrow[]{H^+} \text{RCH}=\text{CHR'}$

The boxed radical anion grabs proton from the same face as electron entry; the second proton arrives from the opposite face. Result: trans product.

🌟 Birch Reduction — Story Time for a 10-Year-Old

Imagine an alkyne (a molecule with a very tight double knot of two rubber bands) sitting on a bench. Two friends, Sodium metal (Na) and Liquid Ammonia ($NH_3$), come along with a soft drink (an alcohol like $CH_3OH$). These friends do not hug at the same time.

1. First, Sodium tosses one tiny electron onto the alkyne from one side, making it a bit shaky (it becomes a radical anion).
2. Second, the soft drink (alcohol) gives one proton (an $H^+$) to calm the shaky spot—but only on the side the electron landed.
3. Another Sodium throws a second electron—again from the outside—turning that spot into an anion.
4. Finally, another proton jumps in—but this time it lands on the opposite side (because the first side is already crowded and negatively charged).
   Because the two protons landed on opposite sides of the original triple bond, the rubber bands relax into a neat trans (opposite-side) double knot—a trans-alkene.
   So, the trans product appears simply because each proton arrives at a different time and from a different face of the molecule.

Step-by-Step Mechanistic Solution

1. Start with alkyne $RC\equiv CR'$.
2. First electron from Na produces radical anion:

$RC\equiv CR' + e^- \longrightarrow RC\equiv CR'^{\,\bullet-}$

3. First proton from ROH adds to the carbon bearing more electron density (same face as electron came):

$RC\equiv CR'^{\,\bullet-} + H^+ \longrightarrow RCH= C^{\bullet}R'$

4. Second electron converts radical into vinyl anion:

$RCH= C^{\bullet}R' + e^- \longrightarrow RCH= C^-R'$

5. Second proton approaches the opposite face due to electrostatic repulsion and steric hindrance, giving trans product:

$RCH= C^-R' + H^+ \longrightarrow RCH=CHR'$

6. Stereochemical outcome: Because steps 3 and 5 occur from opposite faces, the final alkene is E (trans).

Final Answer: Birch reduction of alkynes forms trans-alkenes due to stepwise anti addition of two protons following two separate single-electron transfers.