What is the correct order of reactivity of the following alkyl halides towards ammonolysis reaction?

CH₃I > CH₃Br > CH₃Cl

The correct answer is option 2. CH₃I > CH₃Br > CH₃Cl.

Let us delve into why the order of reactivity of alkyl halides towards ammonolysis is: CH₃I > CH₃Br > CH₃Cl

Leaving Group Ability:

In nucleophilic substitution reactions, the reactivity of alkyl halides largely depends on the ability of the halogen to leave as a halide ion \((X^-)\). The better the leaving group, the more reactive the alkyl halide is.

Size and Bond Strength:

Iodine (I): Iodine is the largest of the halogens and has the weakest bond with carbon (\(C-I\) bond). This weak bond makes it easier for the iodine to leave as \(I^-\), making methyl iodide \((CH_3I)\) the most reactive.

Bromine (Br): Bromine is smaller than iodine but still has a weaker bond with carbon compared to chlorine. The \(C-Br\) bond is stronger than the \(C-I\) bond but weaker than the \(C-Cl\) bond, so \(CH_3Br\) is less reactive than \(CH_3I\) but more reactive than \(CH_3Cl\).

Chlorine \((Cl)\): Chlorine is the smallest and has the strongest bond with carbon (\(C-Cl\) bond). This makes it the least effective leaving group among the halogens, so CH₃Cl is the least reactive in ammonolysis reactions.

Ammonolysis Reaction: In an ammonolysis reaction, an alkyl halide reacts with ammonia \((NH_3)\) to form an amine. The reaction typically involves the nucleophilic attack by ammonia on the carbon atom bonded to the halogen:

\(\text{R-X} + \text{NH}_3 \longrightarrow \text{R-NH}_2 + \text{HX}\)

Here, the halide ion \((X^-)\) leaves, and the reactivity of the alkyl halide depends on how easily the halide ion can dissociate from the carbon.

Reactivity Order Explained:

\(CH_3I\) (Methyl iodide): Iodine is a very good leaving group, making methyl iodide highly reactive in ammonolysis.

\(CH_3Br\) (Methyl bromide): Bromine is a good leaving group, but not as good as iodine. Therefore, methyl bromide is less reactive than methyl iodide but more reactive than methyl chloride.

\(CH_3Cl\) (Methyl chloride): Chlorine is the least effective leaving group among the halogens, so methyl chloride is the least reactive in ammonolysis.

Summary: The reactivity of alkyl halides in ammonolysis is influenced by the leaving group ability. Iodine, being the best leaving group, makes methyl iodide the most reactive. Bromine is next, and chlorine is the least effective leaving group, making methyl chloride the least reactive in ammonolysis reactions.