In the reaction `CH_(3) -CH_(2) -Br +2Na+Br -CH_(2)-CH_(3) underset(Delta)overset("dry ether") to A+B`
A and B are __________ respectively

Options

A. n-butane and sodium bromide
B. n-pentane and sodium bromide
C. n-propane and sodium bromide
D. ethane and sodium bromide

Answer

To solve the question regarding the reaction of `CH3-CH2-Br` with `2Na` in the presence of dry ether, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants in the given reaction are `CH3-CH2-Br` (ethyl bromide) and `2Na` (sodium metal) in the presence of dry ether.

2. **Understand the Reaction Type**: This reaction is an example of the Wood's reaction, which is a coupling reaction where two alkyl halides are reacted with sodium to form a higher alkane.

3. **Mechanism of the Reaction**:
- Sodium (Na) donates an electron to the alkyl halide (CH3-CH2-Br), resulting in the formation of a radical (R•) and a bromide ion (Br-).
- The reaction can be represented as:
\[ CH3-CH2-Br + Na \rightarrow CH3-CH2• + NaBr \]
- The radical (CH3-CH2•) can then react with another sodium atom in a similar manner, producing another radical.

4. **Formation of the Alkane**: The two radicals (CH3-CH2•) combine to form a higher alkane:
\[ CH3-CH2• + CH3-CH2• \rightarrow CH3-CH2-CH2-CH3 \]
which is n-butane (C4H10).

5. **Formation of Sodium Bromide**: The bromide ions produced from the reaction will combine with sodium to form sodium bromide (NaBr).

6. **Final Products**: Therefore, the products A and B from the reaction are:
- A = n-butane (C4H10)
- B = sodium bromide (NaBr)

### Final Answer:
A and B are **n-butane** and **sodium bromide**, respectively.

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