Animated Solution for Chemistry - Organic Chemistry: Treatment of buta-1,3-diyne with ${\text{NaNH}}_2$ (2 equivalents), followed by reaction with excess of trans-${\text{CH}}_3{\text{–CH=CH–CH}}_2\text{-Br}$ gives $\mathbf{X}$ as the major product. The maximum number of carbon atoms that are collinear (in a straight line) in $\mathbf{X}$ is ____.

Imagine you are looking at a molecule of buta-1,3-diyne. It is a simple yet fascinating structure, essentially two acetylene molecules fused together. The terminal hydrogens on this molecule are like low-hanging fruit for a strong base. Because they are attached to $sp$ hybridized carbons, the electrons in the $\text{C-H}$ bond are held very close to the carbon nucleus, making these protons surprisingly acidic for a hydrocarbon.

Acidic Alkynes and The Base

When we introduce sodium amide (${\text{NaNH}}_2$) into the mix, we are bringing in a chemical sledgehammer. Sodium amide is an exceptionally strong base. Since the problem specifies that we are using 2 equivalents of ${\text{NaNH}}_2$, it will swoop in and abstract both of those terminal acidic protons.

This double deprotonation leaves us with a dianion: ${}^{\ominus }\text{C}\equiv \text{C}-\text{C}\equiv {\text{C}}^{\ominus }$. This intermediate is a powerful, double-ended nucleophile, absolutely hungry for positive charge and ready to attack.

Enter our electrophile: excess trans-1-bromo-2-butene. This molecule is a primary allylic halide. The primary nature means there is very little steric hindrance blocking the approach of our nucleophile. Furthermore, the adjacent double bond stabilizes the transition state of the reaction. This makes it a perfect, textbook candidate for an $S_{N}2$ reaction.

Our dianion acts like a perfectly choreographed dancer, attacking two molecules of the allylic halide simultaneously from both ends. The bromide leaving groups are kicked out, and we form our major product, Product X.

Collinearity and Hybridization

Now, let's look at the masterpiece we've created. Product X is a long, symmetrical carbon chain with a central diyne unit. The question asks a very specific geometric question: What is the maximum number of carbon atoms that are collinear (in a straight line)?

This is where VSEPR theory shines. We must analyze the hybridization of the carbon atoms. The central diyne unit consists of four carbons involved in triple bonds. These are $sp$ hybridized carbons. An $sp$ hybridized carbon is the epitome of linearity; it demands a bond angle of exactly $180^{\circ }$.

Because all four of these central carbons are $sp$ hybridized and bonded to each other, they form a perfectly straight, rigid rod: $-\text{C}\equiv \text{C}-\text{C}\equiv \text{C}-$.

But don't stop at just the $sp$ carbons! The atoms they hold hands with must also lie on that exact same line to satisfy the $180^{\circ }$ geometry. The carbons immediately adjacent to the diyne unit are $s{p}^3$ hybridized (specifically, the $-{\text{CH}}_2-$ groups from the allylic halide). Because they are directly attached to the terminal $sp$ carbons of the diyne, they are forced to sit on the same linear axis.

However, the $s{p}^3$ carbons themselves have a tetrahedral geometry with bond angles of approximately $109.5^{\circ }$. This means the next carbons in the chain will bend away from our straight line.

Therefore, our collinear chain consists of the 4 central $sp$ carbons plus the 2 adjacent $s{p}^3$ carbons.