Which alkyl halide has maximum reactivity

Options

A. `CH_3CH_2Br`
B. `CH_3CH_2Cl`
C. `CH_3CH_2I`
D. `CH_3CH_2CH_2Br`

Explanation

To determine which alkyl halide has maximum reactivity, we need to analyze the mechanisms involved in nucleophilic substitution reactions, specifically SN1 and SN2 mechanisms. Here’s a step-by-step solution:

### Step 1: Understanding Alkyl Halides
Alkyl halides are compounds that contain a carbon atom bonded to a halogen atom (F, Cl, Br, or I). The reactivity of these compounds in nucleophilic substitution reactions is influenced by the nature of the halogen.

**Hint:** Remember that the halogen acts as a leaving group during the reaction.

### Step 2: Mechanisms of Nucleophilic Substitution
There are two primary mechanisms for nucleophilic substitution:
- **SN1 Mechanism:** This is a two-step process where the first step involves the formation of a carbocation intermediate after the leaving group departs.
- **SN2 Mechanism:** This is a one-step process where the nucleophile attacks the carbon atom from the opposite side of the leaving group, resulting in a simultaneous bond formation and breaking.

**Hint:** Identify whether the reaction will proceed via SN1 or SN2 based on the structure of the alkyl halide.

### Step 3: Stability of Carbocations
In the SN1 mechanism, the stability of the carbocation is crucial. More stable carbocations (tertiary > secondary > primary) will lead to faster reactions.

**Hint:** Consider the structure of the alkyl halide to determine the stability of the carbocation that would form.

### Step 4: Leaving Group Ability
The reactivity of alkyl halides also depends on the ability of the halogen to act as a leaving group. The order of leaving group ability is:
I⁻ > Br⁻ > Cl⁻ > F⁻

This means that iodine (I) is the best leaving group, followed by bromine (Br), chlorine (Cl), and fluorine (F).

**Hint:** Compare the halogens in the alkyl halides to see which one is the best leaving group.

### Step 5: Comparing Alkyl Halides
Let's consider a few examples of alkyl halides:
- CH3CH2I (ethyl iodide)
- CH3CH2Br (ethyl bromide)
- CH3CH2Cl (ethyl chloride)
- CH3CH2F (ethyl fluoride)

Among these, ethyl iodide (CH3CH2I) has iodine as the leaving group, which is the best leaving group.

**Hint:** Identify which alkyl halide has the best leaving group according to the order established.

### Step 6: Conclusion
Since ethyl iodide (CH3CH2I) has iodine as the leaving group, it will react more readily in both SN1 and SN2 mechanisms compared to the other alkyl halides listed. Therefore, the alkyl halide with maximum reactivity is **ethyl iodide (CH3CH2I)**.

**Final Answer:** The alkyl halide with maximum reactivity is **CH3CH2I (ethyl iodide)**.