Animated Solution for Chemistry - Organic Chemistry: The major product formed in the following reaction is

Welcome to a classic organic chemistry puzzle! At first glance, this molecule might look like a simple hydrocarbon chain, but it holds a beautiful secret.

Let's break down our reactant. We are dealing with a di-alkyne, meaning it has two triple bonds.

However, these two triple bonds are not created equal. On the far left, we have a terminal alkyne. This means the triple bond is at the very end of the carbon chain, leaving one carbon attached to a hydrogen atom.

Further down the chain, we find an internal alkyne. This triple bond is sandwiched between other carbon atoms, with no hydrogens directly attached to the alkyne carbons.

Recognizing this structural difference is the absolute key to unlocking this problem. The reagents we are about to add will treat these two functional groups very differently!

Our reaction arrow throws two distinct chemical conditions at us in sequence.

First, we have sodium amide, written as $NaN{H}_2$.

Second, we have sodium metal dissolved in liquid ammonia, written as $Na$ in liq. $N{H}_3$.

You might be tempted to look at the second reagent and immediately think, "Ah, Birch reduction!" But in organic chemistry, sequence is everything. We must process the reagents in the order they are presented.

Let's introduce our first player: $NaN{H}_2$. Sodium amide is an exceptionally strong base.

When a strong base enters the flask, it goes on a hunt for the most acidic proton available.

This is where the difference between our two alkynes becomes crucial. The hydrogen attached to a terminal alkyne is unusually acidic for a hydrocarbon (with a pKa of around 25). This is because the carbon atom is sp-hybridized, holding the electrons tightly and leaving the proton vulnerable.

The strong amide base ($N{H}_2^{-}$) effortlessly plucks this proton away.

What we are left with is an acetylide ion. The terminal alkyne now bears a full negative charge. This might seem like a simple acid-base step, but it is actually a brilliant strategic move. We have just placed a protective shield over our terminal alkyne!

Now, the second set of reagents enters the scene: $Na$ in liquid $N{H}_3$.

When you dissolve sodium metal in liquid ammonia, something magical happens. The sodium atoms give up their valence electrons, creating a deep blue solution of solvated electrons.

These free-floating electrons are powerful reducing agents. This specific combination is the hallmark of the Birch reduction.

The Birch reduction is famous for taking internal alkynes and reducing them to trans-alkenes. The mechanism involves the sequential addition of electrons and protons, and due to electronic and steric repulsions in the radical anion intermediate, the trans geometry is exclusively favored.

So, our internal alkyne is destined to become a trans double bond.

Protection by Deprotonation

But wait! What about our terminal alkyne? Will the solvated electrons reduce it too?

Here is where the magic of our first step shines. Remember, our terminal alkyne is currently sitting as a negatively charged acetylide ion.

The solvated electrons from the Birch reduction are, of course, also negatively charged.

Physics dictates that like charges repel. The electron-rich acetylide ion violently repels the incoming solvated electrons. Because the electrons cannot approach the triple bond, the terminal alkyne is completely immune to the Birch reduction!

This is a classic organic chemistry trick: using deprotonation as a temporary protecting group.

Let's put the pieces together and visualize the final molecule.

During the final workup of the reaction, the acetylide ion will grab a proton from the environment (or from the ammonia itself) and revert back to a neutral terminal alkyne.

Meanwhile, our internal alkyne has been successfully transformed into a trans-alkene.

Let's scan our options. We need a structure with a terminal triple bond on the left and a trans double bond on the right.

Option (A) has two double bonds. Incorrect. Option (C) has a terminal double bond. Incorrect. Option (D) has a cis double bond. Incorrect.

Option (B) perfectly matches our derived structure. It proudly features the untouched terminal alkyne and the newly formed trans double bond.

The chemistry is elegant, the logic is sound, and Option (B) is our undisputed winner!