Animated Solution for Chemistry - Organic Chemistry: LIST-I contains reactions and LIST-II contains major products. Match each reaction in LIST-I with one or more product in LIST-II and choose the correct option.

Welcome to an epic journey through the battlegrounds of organic chemistry! In this matrix match problem, we are tasked with predicting the major products of four distinct reactions. To conquer this, we must master the delicate balance between substitution ($S_{N}1$, $S_{N}2$) and elimination ($E_1$, $E_2$) pathways. The key to victory lies in analyzing the steric hindrance of both our nucleophiles (or bases) and our electrophiles (the alkyl halides).

Let's dive into the molecular arena and break down each reaction step-by-step.

The Battle of Bulky Groups

In reaction P, we are mixing sodium tert-butoxide with tert-butyl bromide.

Imagine the scene: tert-butoxide is a massive, bulky molecule acting as a strong base. On the other side, tert-butyl bromide is a tertiary alkyl halide, meaning its reactive carbon is heavily guarded by three bulky methyl groups.

When these two titans meet, a direct backside attack (substitution) is geometrically impossible due to severe steric hindrance. The bulky base simply cannot reach the central carbon. Instead, it takes the path of least resistance: it acts as a base and plucks off an exposed $\beta$-hydrogen from one of the methyl groups.

This triggers an $E_2$ elimination, collapsing the molecule into a double bond and kicking out the bromide ion. The resulting major organic product is isobutylene (Product 4). Meanwhile, the tert-butoxide ion, having gained a proton, transforms into tert-butanol (Product 1).

Thus, reaction P perfectly matches with products 1 and 4.

Acidic Cleavage of Ethers

Moving to reaction Q, we encounter tert-butyl methyl ether reacting with strong hydrobromic acid (HBr). This is a classic acidic ether cleavage.

The first step is always the protonation of the ether oxygen, turning it into a good leaving group. Now, the molecule faces a choice: how should it break apart?

Because one of the attached groups is a tert-butyl group, the molecule can undergo heterolytic cleavage to form a highly stable tertiary carbocation. This incredible stability strongly drives the reaction down the $S_{N}1$ pathway.

The ether splits, releasing methanol as a leaving group and leaving behind the tert-butyl carbocation. The waiting bromide ion swiftly attacks this carbocation, forming tert-butyl bromide (Product 2).

Therefore, reaction Q matches exclusively with product 2.

Strong Base meets Tertiary Halide

In reaction R, we react tert-butyl bromide with sodium methoxide (NaOMe).

Methoxide is a small, powerful nucleophile and a strong base. However, our electrophile is once again a tertiary alkyl halide. Even though the nucleophile is small, the tertiary carbon is simply too crowded for an $S_{N}2$ backside attack to occur.

Whenever you have a strong base reacting with a tertiary halide, $E_2$ elimination will always dominate. The methoxide ion acts as a base, abstracting a $\beta$-hydrogen to form a double bond.

The major product is once again isobutylene (Product 4), with methanol and sodium bromide forming as byproducts.

So, reaction R matches perfectly with product 4.

The Williamson Ether Synthesis

Finally, we arrive at reaction S: sodium tert-butoxide reacting with methyl bromide (MeBr).

This is the legendary Williamson Ether Synthesis. Here, we have a very bulky base (tert-butoxide) but a completely unhindered electrophile (methyl bromide).

The critical detail here is that methyl bromide has zero $\beta$-hydrogens. Because there are no $\beta$-hydrogens to abstract, elimination ($E_2$) is structurally impossible! The bulky tert-butoxide has absolutely no choice but to act as a nucleophile.

It performs a clean $S_{N}2$ attack on the exposed methyl carbon, displacing the bromide ion. The result is the formation of an ether linkage, specifically tert-butyl methyl ether (Product 3).

Thus, reaction S matches with product 3.

By carefully analyzing the steric environments and the nature of the reagents, we have successfully mapped the entire matrix:

P $\to$ 1, 4 Q $\to$ 2 R $\to$ 4 S $\to$ 3

This problem beautifully illustrates that in organic chemistry, the structure of the molecules dictates their destiny. Always watch out for steric hindrance!