The increasing order of reduction of alkyl halides with zinc and dilute `HCl` is

Options

A. `R-CL lt R-I lt R-Br`
B. `R-Cl lt R-Br lt R-I`
C. `R-I lt R- Br lt R-Cl`
D. `R-Br lt R-I lt R-Cl`

Explanation

To determine the increasing order of reduction of alkyl halides with zinc and dilute HCl, we need to analyze the reactivity of different alkyl halides based on the halogen present. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Reaction
Alkyl halides react with zinc and dilute HCl to form corresponding alkanes. The general reaction can be represented as:
\[ R-X + Zn + HCl \rightarrow R-H + ZnXCl \]
where \( R \) is the alkyl group and \( X \) is the halogen.

### Step 2: Analyze the Halogens
The halogens in question are fluorine (F), chlorine (Cl), bromine (Br), and iodine (I). As we move down the group in the periodic table from fluorine to iodine, the size of the halogen increases, which affects the bond strength of the carbon-halogen bond.

### Step 3: Bond Strength Consideration
The bond strength of the carbon-halogen bond decreases as the size of the halogen increases. This means:
- C-F bond is the strongest.
- C-Cl bond is weaker than C-F.
- C-Br bond is weaker than C-Cl.
- C-I bond is the weakest.

### Step 4: Reactivity Order
Since weaker bonds break more easily, the order of reactivity of the alkyl halides with zinc and dilute HCl will be:
- Iodoalkanes (R-I) > Bromoalkanes (R-Br) > Chloroalkanes (R-Cl) > Fluoroalkanes (R-F)

This means that iodoalkanes are the most reactive and fluoroalkanes are the least reactive.

### Step 5: Conclusion
Thus, the increasing order of reduction of alkyl halides with zinc and dilute HCl is:
\[ \text{Fluoroalkane} < \text{Chloroalkane} < \text{Bromoalkane} < \text{Iodoalkane} \]

### Final Answer
The correct increasing order of reduction of alkyl halides is:
\[ \text{R-F} < \text{R-Cl} < \text{R-Br} < \text{R-I} \]