Question

The correct decreasing order of reactivity for a given alkyl (R) group in both S\textsubscript{N1 and S\textsubscript{N}2 reaction mechanisms is}

• A. R--I $>$ R--Br $>$ R--Cl $>$ R--F
• B. R--I $>$ R--Cl $>$ R--Br $>$ R--F
• C. R--F $>$ R--Cl $>$ R--Br $>$ R--I
• D. R--F $>$ R--I $>$ R--Cl $>$ R--Br
• E. R--Br $>$ R--I $>$ R--F $>$ R--Cl

Hint:

While alkyl groups ($1^\circ, 2^\circ, 3^\circ$) change their preference between $S_N1$ and $S_N2$, the Halogen preference never changes. Iodine is always the fastest and Fluorine is always the slowest!

Answer 1

The Correct Option is A

Concept: Reactivity in nucleophilic substitution ($S_N1$ and $S_N2$) depends heavily on the "leaving group ability" of the halogen.
• Bond Strength: Reactivity is inversely proportional to the Carbon-Halogen ($C-X$) bond strength. C-F is the strongest bond, making it very difficult to break.
• Size and Polarizability: As we move down the halogen group (F to I), the size of the halide ion increases. Larger ions are better leaving groups because they can stabilize the negative charge over a larger volume.
• Stability: $I^-$ is the weakest base and therefore the most stable leaving group.

Step 1: Evaluate the bond dissociation energies. The $C-I$ bond is the longest and weakest in the series due to the poor orbital overlap between the $p$-orbital of Carbon and the large $5p$ orbital of Iodine. The $C-F$ bond is the shortest and strongest.

Step 2: Compare leaving group ability. In both $S_N1$ (where the bond breaks first) and $S_N2$ (where the bond breaks as the new one forms), the ease of departure of the halogen determines the rate. The order of leaving group ability is: $I^- > Br^- > Cl^- > F^-$. Consequently, the reactivity order follows the same sequence.