Animated Solution for Chemistry - Organic Chemistry: The reactivity of compound Z with different halogens under appropriate conditions is given below- The observed pattern of electrophilic substitution can be explained by -

Welcome to a fascinating puzzle of organic chemistry! When you look at a molecule like 3-tert-butylphenol, you are not just looking at a static drawing; you are looking at a dynamic battlefield where electronic forces and physical bulk clash to determine the outcome of a chemical reaction.

In this problem, we observe a very peculiar pattern. When we react this molecule with different halogens, we get completely different extents of substitution. Iodine gives a mono-halo derivative, Bromine gives a di-halo derivative, and Chlorine goes all out to give a tri-halo derivative. Why does this happen? Let's break down the forces at play.

The Power of the Phenolic OH

First, let's analyze the electronic environment of our benzene ring. The hydroxyl group ($-OH$) is a powerful activating group. The oxygen atom possesses lone pairs of electrons, which it generously donates into the aromatic ring through resonance (the $+R$ effect).

This influx of electron density doesn't just spread evenly; it specifically concentrates at the ortho and para positions relative to the $-OH$ group. If we number the carbon attached to $-OH$ as position $1$, the activated hotspots for an incoming electrophile are positions $2$, $4$, and $6$.

So, electronically speaking, the ring is screaming, "Come attack me at positions $2$, $4$, and $6$!"

The Bulky tert-Butyl Group

However, chemistry is not just about electronics; it's also about physical space. Enter the tert-butyl group ($-C(C{H}_3{)}_3$) at position $3$. This group is massive. It's like a giant bouncer standing outside a club, physically blocking the entrance to the adjacent doors.

The positions immediately adjacent to the tert-butyl group are positions $2$ and $4$. Because of the sheer bulk of the tert-butyl group, these positions experience severe steric hindrance. Position $2$ is particularly unfortunate—it is sandwiched right between the $-OH$ group and the tert-butyl group, making it the most congested spot on the entire molecule. Position $6$, on the other hand, is far away from the bulky tert-butyl group and remains relatively open and accessible.

A Tale of Three Sizes

Now, let's introduce our attackers: the halogens. The key to solving this puzzle lies in recognizing that halogens come in different sizes. As we go down the periodic table, the atomic radius increases significantly: $Cl<Br<I$.

1. The Giant: Iodine ($I_2$) Iodine is the largest of the three. Imagine trying to park a massive truck in a crowded parking lot. You can only fit into the widest, most open space available. For iodine, positions $2$ and $4$ are simply too crowded due to the tert-butyl group. It can only manage to attack the least hindered spot, which is position $6$. This perfectly explains why we only get a mono-halo substituted derivative with iodine.

2. The Mid-Size: Bromine ($B{r}_2$) Bromine is smaller than iodine. It's like a mid-size SUV. It easily parks in position $6$, and with a little bit of squeezing, it can also fit into position $4$. However, position $2$ is still a no-go zone; it's just too tight. Consequently, bromine manages to substitute at two positions, yielding a di-halo substituted derivative.

3. The Compact: Chlorine ($C{l}_2$) Finally, we have chlorine, the smallest of the bunch. Think of it as a compact motorcycle. It doesn't care about the bouncer. It zips into position $6$, slides into position $4$, and even manages to squeeze into the highly congested position $2$. Because it can access all three electronically activated sites, chlorine produces a tri-halo substituted derivative.

By piecing this all together, we can see that the observed pattern is a beautiful interplay of three distinct factors:

1. The electronic effect of the phenolic group, which sets the stage by activating positions $2$, $4$, and $6$. 2. The steric effect of the tert-butyl group, which creates physical barricades at positions $2$ and $4$. 3. The steric effect of the halogen, where the physical size of the attacking atom determines which barricades it can overcome.

Therefore, options (A), (B), and (C) are the correct explanations for this fascinating chemical behavior. It's a perfect reminder that in organic chemistry, you must always consider both the electronic desires and the physical realities of the molecules involved!