Animated Solution for Chemistry - Hydrocarbons: For the given reaction

Welcome to a fascinating journey through organic synthesis! In this problem, we are presented with a sequential two-step reaction. Our starting material is 1-bromo-propene ($C{H}_3-CH=CHBr$). This molecule is a vinylic halide, meaning the halogen (bromine) is directly attached to one of the carbon atoms forming the double bond.

Our objective is to determine the final major product, labeled as $(A)$, after treating this starting material with two specific sets of reagents in sequence. The first reagent is sodium amide ($NaN{H}_2$), and the second condition involves passing the intermediate through a red hot iron tube at $873\text{ K}$. Let's break down this transformation step by step.

Dehydrohalogenation

The first reagent we encounter is $NaN{H}_2$. Sodium amide is an exceptionally strong base. In organic chemistry, when a strong base interacts with a molecule containing a halogen and a neighboring hydrogen atom, the stage is set for an elimination reaction. Specifically, this is a $\beta$-elimination.

Here is how it works: The strong amide ion ($N{H}_2^{-}$) acts as a base and abstracts a proton ($H^{+}$) from the $\beta$-carbon (the carbon adjacent to the one holding the bromine). Simultaneously, the bromide ion ($B{r}^{-}$) leaves from the $\alpha$-carbon. This concerted removal of $HBr$ forces the electrons to collapse and form a new $\pi$ bond.

Since our starting material already had a double bond, the addition of another $\pi$ bond converts it into a triple bond. The resulting intermediate molecule is propyne ($C{H}_3-C\equiv CH$).

This completes the first phase of our synthesis. We have successfully transformed a vinylic halide into a terminal alkyne.

Cyclic Trimerization

Now we move to the second set of conditions. We take our newly synthesized propyne and pass it through a red hot iron tube at a high temperature of $873\text{ K}$. This is a classic, textbook reaction condition that every chemistry student must recognize.

Whenever you expose alkynes to a red hot iron tube, the molecules undergo cyclic trimerization. This is a remarkable process where three separate alkyne molecules join hands to form a single aromatic ring.

Let's visualize the mechanism. Three molecules of propyne align themselves. The high thermal energy causes the $\pi$ electrons in their triple bonds to shift. These electrons reorganize to form new $\sigma$ bonds between the adjacent molecules, effectively stitching them together into a six-membered benzene ring.

Where do the Methyl Groups go?

The formation of the benzene ring is only half the story. Each of our three propyne molecules carries a methyl group ($C{H}_3$). As the ring forms, these methyl groups must find their places on the newly created aromatic structure.

This is where steric hindrance plays a crucial role. If the methyl groups were to attach to adjacent carbons (like positions 1, 2, and 3), they would bump into each other, creating severe steric strain and making the molecule highly unstable.

To minimize this crowding and achieve the most thermodynamically stable configuration, the methyl groups position themselves as far apart from each other as possible. They arrange themselves symmetrically around the ring, landing on positions 1, 3, and 5.

The resulting molecule is 1,3,5-trimethylbenzene, which is commonly known by its trivial name, Mesitylene.

This beautifully symmetric aromatic compound is our final major product $(A)$. When we look at the given options, we can see that option (d) perfectly depicts the structure of Mesitylene.

As a final thought, consider what would happen if our starting material was simply ethyne ($HC\equiv CH$) instead of propyne. Passing ethyne through the red hot iron tube would result in the trimerization yielding pure benzene, without any substituents. Understanding these subtle variations is the key to mastering organic chemistry!