If 1,3- dibromopropane reacts with zinc and NaI, the product obtained is :

Options

A. Propene
B. Propane
C. Cyclopropane
D. Hexane

Answer

To solve the problem of what product is formed when 1,3-dibromopropane reacts with zinc and NaI, we can follow these steps:

### Step 1: Identify the Structure of 1,3-Dibromopropane
1,3-Dibromopropane has the following structure:
- It consists of three carbon atoms (propane) with bromine atoms attached to the first and third carbon atoms.
- The structure can be represented as:

CH2Br-CH2-CH2Br

### Step 2: Understand the Reaction Conditions
The reaction involves:
- Zinc (Zn), which is a reducing agent.
- Sodium iodide (NaI), which can provide iodide ions (I⁻).

### Step 3: Mechanism of the Reaction
1. **Reduction by Zinc**: Zinc will reduce the bromine atoms, leading to the formation of radicals. The bromine atoms will leave as bromine radicals (Br•).

2. **Formation of Radicals**: The reaction will generate carbon radicals at the positions where bromine was originally attached:
- CH2• (from the first carbon)
- CH2 (middle carbon remains unchanged)
- CH2• (from the third carbon)

### Step 4: Radical Coupling
The two carbon radicals (CH2•) can couple together to form a new bond:
- The two CH2 radicals combine to form a new carbon-carbon bond, resulting in cyclopropane.

### Step 5: Final Product
The final product of the reaction is cyclopropane (C3H6), which is a three-membered carbon ring.

### Conclusion
Thus, when 1,3-dibromopropane reacts with zinc and NaI, the product obtained is cyclopropane.