Animated Solution for Chemistry - Organic Chemistry: For the following reactions : $C{H}_{3}C{H}_{2}C{H}_{2}Br+Z^{\ominus }\stackrel{k_s\text{ Substitution}}{\to }C{H}_{3}C{H}_{2}C{H}_{2}Z+B{r}^{\ominus }$ $C{H}_{3}C{H}_{2}C{H}_{2}Br+Z^{\ominus }\stackrel{k_e\text{ Elimination}}{\to }C{H}_{3}CH=C{H}_2+HZ+B{r}^{\ominus }$ where, $Z^{\ominus }=C{H}_{3}C{H}_2{O}^{\ominus }\text{ (A) or }\begin{array}{c}{\text{CH}}_3\\ |\\ {\text{H}}_3\text{C}-\text{C}-{\text{O}}^{\ominus }\\ |\\ {\text{CH}}_3\end{array}\text{ (B)}$ $k_s$ and $k_e$, are respectively the rate constants for substitution and elimination, and $\mu =\frac{k_s}{k_e}$, the correct option is .........

Substitution vs Elimination

Imagine you are standing in front of a locked door. If you are small and agile, you might be able to slip the key in and open it smoothly. But if you are carrying three massive suitcases, you won't even be able to reach the keyhole; instead, you might just knock the door down from the outside. This is exactly the drama that unfolds when different bases attack an alkyl halide.

In this problem, our substrate is 1-bromopropane ($C{H}_{3}C{H}_{2}C{H}_{2}Br$), a primary alkyl halide. It is relatively unhindered, making its $\alpha$-carbon an inviting target for nucleophiles. We are given two different reagents to react with it:

1. Reagent A: Ethoxide ion ($C{H}_{3}C{H}_2{O}^{\ominus }$) 2. Reagent B: tert-Butoxide ion ($(C{H}_3{)}_{3}C{O}^{\ominus }$)

Let's look at Reagent A (Ethoxide). It is a strong base, but more importantly, it is a small, unhindered nucleophile. Because it is small, it can easily bypass the surrounding hydrogen atoms and perform a backside attack on the $\alpha$-carbon, kicking out the bromide ion. This is the classic $S_{N}2$ mechanism. For a primary halide reacting with an unhindered strong base, substitution is the dominant pathway.

Now, let's look at Reagent B (tert-Butoxide). This ion is a behemoth. It has three bulky methyl groups attached to the carbon bearing the oxygen. When it tries to approach the $\alpha$-carbon to perform an $S_{N}2$ attack, it experiences severe steric hindrance. It simply cannot fit. However, it is still a very strong base (even stronger than ethoxide due to the +I effect of the methyl groups). Since it cannot reach the $\alpha$-carbon, it settles for the next best thing: it abstracts a proton from the exposed $\beta$-carbon. This triggers the $E2$ elimination mechanism, forming an alkene.

Understanding $\mu$

The problem defines a ratio $\mu =\frac{k_s}{k_e}$, where $k_s$ is the rate constant for substitution and $k_e$ is the rate constant for elimination.

For Reagent A, substitution is highly favored. Therefore, $k_s(A)$ is very large compared to $k_e(A)$. This makes the fraction ${\mu }_A$ a large number.

For Reagent B, elimination is highly favored because substitution is sterically blocked. Therefore, $k_s(B)$ is extremely small, while $k_e(B)$ is large. This makes the fraction ${\mu }_B$ a very small number.

Comparing the two, it is crystal clear that:

Comparing Elimination Rates

What about the absolute rates of elimination, $k_e(A)$ versus $k_e(B)$?

Even though ethoxide can perform elimination, it prefers substitution. tert-Butoxide, on the other hand, is forced to perform elimination exclusively. Furthermore, tert-butoxide is a stronger base than ethoxide. The strength of the base directly accelerates the rate of the $E2$ reaction. Because tert-butoxide is a stronger base and its energy is entirely channeled into abstracting the $\beta$-proton, the rate of elimination with tert-butoxide is faster than with ethoxide.

Therefore, we conclude that:

Combining both insights, we arrive at the final correct relationship: ${\mu }_A>{\mu }_B$ and $k_e(B)>k_e(A)$, which perfectly matches option (d).